JEE Advanced Physics-Mechanics-II [3 ed.] 9789390577149

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3

THIRD EDITION

JEE ADVANCED

PHYSICS

2 kg A

1m B

C

Mechanics – II Rahul Sardana

About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide. Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world. We believe learning opens up opportunities, creates fulfilling careers and hence better lives. We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum. Superior learning experience and improved outcomes are at the heart of everything we do. This product is the result of one such effort. Your feedback plays a critical role in the evolution of our products and you can contact us - [email protected]. We look forward to it.

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Copyright © 2021 Pearson India Education Services Pvt. Ltd Although the author and publisher have made every effort to ensure that the information in this book was correct at the time of editing and printing, the author and publisher do not assume and hereby disclaim any liability to any party for any loss or damage arising out of the use of this book caused by errors or omissions, whether such errors or omissions result from negligence, accident or any other cause. Further, names, pictures, images, characters, businesses, places, events and incidents are either the products of the author’s imagination or used in a fictitious manner. Any resemblance to actual persons, living or dead or actual events is purely coincidental and do not intend to hurt sentiments of any individual, community, sect or religion. In case of binding mistake, misprints or missing pages, etc., the publisher’s entire liability and your exclusive remedy is replacement of this book within reasonable time of purchase by similar edition/reprint of the book. ISBN 978-93-905-7714-9 First Impression Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128. Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India. Registered office: The HIVE, 3rd Floor, No. 44, Pillaiyar Koil Street, Jawaharlal Nehru Road, Anna Nagar, Chennai 600 040, Tamil Nadu, India. Phone: 044-66540100 Website: in.pearson.com, Email: [email protected] Compositor: SRS Global, Puducherry Printed in India

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CONTENTS Chapter Insight Preface

xiii

xix

About the Author

CHAPTER

1

F01_Mechanics II__Prelims.indd 5

xx

WORK, ENERGY, POWER AND CONSERVATION OF ENERGY 1.1 Work, Energy, Power and Law of Conservation of Energy .

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Introduction .

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1.1

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. 1.1

Work Done by a Constant Force .

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. 1.1

Positive and Negative Work .

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Zero Work Done .

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. 1.3

Work Done by Friction .

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Work Done by Static Friction .

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. 1.4

Work Done by Kinetic Friction

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. 1.5

Dependence of Work on Frame of Reference.

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Work Done by Gravity .

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. 1.6

Work Done by Pseudo Force .

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. 1.6

Work Done by a Variable Force .

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. 1.8

Work Done by Spring Force

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. 1.9

Work Done as Area Under f-x Graph .

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1.10

Energy.

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1.13

Concept of Kinetic Energy and Work Energy Theorem (or the Classical Work Energy (CWE) Theorem) . .

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1.13

Importance of the Work Energy Theorem .

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1.14

Conservative and Non-conservative Forces .

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1.17

Potential Energy .

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1.17

The Potential Energy is Defined Only for Conservative Forces

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1.18

Conservative System and Concept of Potential Energy (U) .

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1.18

Gravitational Potential Energy (Near the Earth’s Surface) .

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1.19

Spring Potential Energy or Elastic Potential Energy .

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1.19

Power .

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1.20

Law of Conservation of Mechanical Energy .

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1.23

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vi Contents

Modified Work-energy Theorem (MWET)

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1.34

Law of Conservation of Mechanical Energy .

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1.35

Work-energy Theorem for Non-conservative System . . . . . . . . . . . . r Relation Between FC and U . . . . . . . . . . . . . . . . . . . .

1.35 1.42

Potential Energy Curve .

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1.43

Nature of Force .

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1.43

Types of Equilibrium .

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1.43

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Motion in a Vertical Circle

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1.46

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1.46

Velocity at the Point P

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1.46

Tension in the String at Any Point P

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1.46

Tension at the Lowest Point L .

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1.47

Tension at the Highest Point H .

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1.47

Condition for Looping the Loop .

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1.47

Tension and Velocity at the Point M (Midway between L and H) .

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1.47

For 0 < u
vA , so ω should be counterclockwise and vcm should be leftwards. Hence, we have vB = vcm + Rω = 6

…(1)

and vA = Rω − vcm = 2

…(2)

Mechanics II_Chapter 3_Part 1.indd 18

The negative sign with vcm indicates that it is directed leftwards. Rω − v

ω v

⇒ v + Rω

2/9/2021 6:38:02 PM

Chapter 3: Rotational Dynamics ILLUSTRATION 8

Consider a hoop of radius R having centre of mass velocity v0. Let the hoop roll on a surface without sliding. Assuming a point P on the hoop to be at the origin initially. Find the velocity of this point as a function of time. Also find the acceleration of this point P. Prove that the trajectory followed by this point is a cycloid and the total distance travelled by this point is 8R (in one complete cycle). SOLUTION

Since the hoop is rolling without slipping, therefore the arc length AP is equal to the displacement PA from the origin to point A i.e., v0 t

v0t

θ=

 v = vP = vx2 + vy2

⇒

⎛ v t⎞ ⎛ v t⎞ ⎛ v t⎞ vP = v0 1 + cos 2 ⎜ 0 ⎟ − 2 cos ⎜ 0 ⎟ + sin 2 ⎜ 0 ⎟ ⎝ R ⎠ ⎝ R ⎠ ⎝ R ⎠

⇒

⎛ v t⎞ vP = v0 2 − 2 cos ⎜ 0 ⎟ ⎝ R ⎠

⇒

⎛ v t⎞ vP = 2v0 1 − cos ⎜ 0 ⎟ ⎝ R ⎠

⇒

⎛ v t⎞ vP = 2v0 2 sin 2 ⎜ 0 ⎟ ⎝ 2R ⎠

v0 t R

…(6)

Now let us calculate the acceleration of the point P. Since

=v t AP 0

⇒

and

⎛ v t⎞ vP = 2v0 sin ⎜ 0 ⎟ ⎝ 2R ⎠ vt Since θ = 0 R ⎛ v t⎞ ⎛θ⎞ vP = 2v0 sin ⎜ 0 ⎟ = 2v0 sin ⎜ ⎟ ⎝ 2R ⎠ ⎝ 2⎠

P (x, y)

Rθ = v0 t

…(5)

⇒

v0

⇒

dy ⎛ v t⎞ = v0 sin ⎜ 0 ⎟ ⎝ R ⎠ dt   Since v = vp = vx iˆ + vy ˆj

and vy =

3.19

ax =

dvx d⎛ ⎛ v t⎞⎞ = ⎜ v0 − v0 cos ⎜ 0 ⎟ ⎟ ⎝ R ⎠⎠ dt dt ⎝

…(1)

Co-ordinates of P at t are ( x , y ) where x = v0 t − R sin θ ⇒

⎛ v t⎞ x = v0 t − R sin ⎜ 0 ⎟ ⎝ R ⎠

…(2) = 0° vP = 0

y = R − R cos θ ⇒

⎛ v t⎞ y = R − R cos ⎜ 0 ⎟ ⎝ R ⎠

…(3)

The (2) and (3) combination of x and y will give rise to the trajectory called Cycloid. Since, vx = ⇒

Mechanics II_Chapter 3_Part 1.indd 19

⇒

ax =

dvx v02 ⎛ v t⎞ = sin ⎜ 0 ⎟ ⎝ R ⎠ dt R

…(7)

Similarly

v ⎡ dx ⎛ v t⎞ ⎤ = v0 − 0 ⎢ R cos ⎜ 0 ⎟ ⎥ ⎝ R ⎠⎦ dt R⎣

⎛ v t⎞ vx = v0 − v0 cos ⎜ 0 ⎟ ⎝ R ⎠

vP =

ay = …(4)

dvy dt

=

2 d⎛ ⎛ v0 t ⎞ ⎞ v0 ⎛ v t⎞ v sin cos ⎜ 0 ⎟ …(8) = ⎜⎝ ⎟⎟ ⎜ 0 ⎝ R ⎠ dt ⎝ R ⎠⎠ R

Since, a = ax2 + ay2

2/9/2021 6:38:15 PM

3.20

JEE Advanced Physics: Mechanics – II

⇒

a=

v02 R

⎛ω t⎞ ⇒ vP = 2v0 sin ⎜ 0 ⎟ ⎝ 2 ⎠

…(9)

where, θ is the angle between the point P at time t and the vertical line passing through the centre of the hoop. (c) If at t = 0 , the point P lies at the top of the hoop, then In time dt,

⎛ v t⎞ ⎛θ⎞ vP = 2v0 cos ⎜ 0 ⎟ = 2v0 cos ⎜ ⎟ ⎝ 2R ⎠ ⎝ 2⎠

dr = vP dt ⇒

2π R v0

⇒

⎛ω t⎞ ⇒ vP = 2v0 cos ⎜ 0 ⎟ ⎝ 2 ⎠

⎛ v t⎞ dr = 2v0 sin ⎜ 0 ⎟ dt ⎝ 2R ⎠

r = 2v0

∫ 0

⎡ ⎢ − cos ⎛ v0 t ⎞ ⎜⎝ ⎟ ⎢ ⎛ v0 t ⎞ 2R ⎠ sin ⎜ dt = 2v0 ⎢ ⎟ ⎝ 2R ⎠ ⎢ ⎛⎜ v0 ⎞⎟ ⎢⎣ ⎝ 2R ⎠

⇒

⎤ ⎡ 2π R ⎞ ⎛ v r = −4 R ⎢ cos ⎜ 0 × − cos 0 ⎥ ⎟ R v 2 ⎝ 0 ⎠ ⎣ ⎦

⇒

r = −4 R ( −2 ) = 8 R

⇒

r = 8R

2π R v0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦

where θ is the angle between the point P at time t and the vertical line passing through the centre of the hoop.

UNIFORM PURE ROLLING Uniform Pure Rolling or simply “pure rolling” means that no relative motion exists at the point of contact between the body and the surface. Let a disc of radius R rolls without slipping on a horizontal stationary surface/ground. For the disc to roll without slipping, we must have

(a) Curve Rectification 2 2 2 ⇒ ( dr ) = ( dx ) + ( dy ) 2

⎛ dy ⎞ ⇒ dr = 1+ ⎜ ⎟ dx ⎝ dx ⎠

( vnet )P = vsurface

⇒ dr = 1+ m2 dx

∫ ∫

⇒ r = dr =

f

1+ m2 dx =

2

⎛ dy ⎞ 1+ ⎜ ⎟ dx ⎝ dx ⎠

∫ i

dr

f dy

dx

(b) If at t = 0 , the point P lies at the bottom of the hoop, then ⎛ v t⎞ ⎛θ⎞ vP = 2v0 sin ⎜ 0 ⎟ = 2v0 sin ⎜ ⎟ ⎝ 2R ⎠ ⎝ 2⎠

Mechanics II_Chapter 3_Part 1.indd 20

⇒

vcm − Rω = 0

⇒

vcm = Rω

So, vcm = Rω is the condition for a body to be in pure rolling on a stationary horizontal surface/ground. It is sometimes simply called as ROLLING. CASE OF FORWARD SLIPPING If vP > vQ ⇒

vcm − Rω > 0

⇒

vcm > Rω

2/9/2021 6:38:25 PM

Chapter 3: Rotational Dynamics

CASE OF BACKWARD SLIPPING (also called FORWARD ENGLISH) If vP < vQ ⇒

vcm − Rω < 0

⇒

vcm < Rω

Substituting the value of aB from equation (3) in equation (2), we get ⎤ ⎛ VA2 ⎞ 1⎡ α = ⎢ aA sin θ + ⎜ + aA cos θ ⎟ cot θ ⎥ 2 ⎝ ⎠ l⎣ l sin θ ⎦ lω 2 =

ILLUSTRATION 9

A rod AB of length l is lying as shown in the Figure. At the instant shown, the point A has a speed VA and acceleration aA. Calculate the acceleration of point B and angular acceleration of the rod.

VA

3.21

VA2

l sin 2 θ

In ground frame, net acceleration of point B will be the vector sum of all the three accelerations shown in Figure.

aA

ILLUSTRATION 10

SOLUTION

A bobbin having inner radius r and outer radius R is placed on a rough horizontal surface. A light string wrapped on its inner core connects a block with the bobbin as shown in Figure.

By definition of relative velocity, we have    aBA = aBG − aAG    ⇒ aBG = aBA + aAG The angular velocity ω of the rod is

ω=

VA l sin θ

The components of the accelerations along the rod and perpendicular to the rod are shown in Figure.

Now system is released from rest and bobbin moves on the horizontal surface without sliding. If the string does not slide from bobbin, calculate a A. SOLUTION

So, we have aB sin θ − aA cos θ = lω 2 and aA sin θ + aB cos θ = lα ⇒

aB =

…(1) …(2)

⎞ 1 aA cos θ + lω 2 ⎛ VA2 …(3) =⎜ + aA cos θ ⎟ 2 ⎝ l sin θ ⎠ sin θ sin θ

Mechanics II_Chapter 3_Part 1.indd 21

Since the bobbin is in pure rolling, point P should be instantaneous centre (IC) of zero velocity. So, for pure rolling acceleration of the point of contact is zero, hence we have   aP = 0     ⇒ aP = aPC + aC = 0

2/9/2021 6:38:31 PM

3.22

JEE Advanced Physics: Mechanics – II

⇒

Rα − A = 0

⇒

A = Rα

…(1)

Acceleration of the point Q with respect to the centre of mass C is shown in Figure.

As the particle moves along x-axis, its position vector turns clockwise, thus decreasing the angle θ . The dθ time rate of change of this angle, i.e. equals the dt angular velocity. So,  ⎛ dθ ⎞ ω = ⎜ ⎟ kˆ ⎝ dt ⎠ Since y = x tan θ

|ac| = A = Rα , |aQ| = a = α (R − r)

⇒

x = y cot θ

⇒

dx dθ = v = − ycosec 2θ dt dt

(

)

{

d ( cot θ ) = −cosec2θ dθ

}

⇒

vy y dθ v sin 2 θ ⎧⎪ =− =− 2 ∵ sin θ = 2 dt y x + y 2 ⎨⎪ x + y2 ⎩

⎫⎪ ⎬ ⎭⎪

⇒

 ⎛ vy ⎞ ˆ ω = −⎜ 2 k 2 ⎝ x + y ⎟⎠

   aQC = aQ − aC ⇒

∵

   aQ = aQC + aC

where, aQ = a , aQC = − rα and aC = A = Rα aQ = a = A − r = ( R − r ) α Since point Q is directly connected with block, it means the magnitude of acceleration of the point Q should be equal to the acceleration of the block aQ = a = α ( R − r )

…(2)

From Equation (1) and (2), we get a R−r r = = 1− A R R ILLUSTRATION 11

A particle of mass m is moving in x -y plane with a speed v, along x-axis. At a given instant of time, its position co-ordinates are ( x , y ). Calculate the angular velocity of the position vector of the particle and the angular acceleration of the particle with respect to origin.

  dω Also, angular acceleration α = dt  ⎡ ⇒ α = −vykˆ ⎢ ( −1 ) x 2 + y 2 ⎣

(

⇒

 α=

2xv 2 y

(x

2

+y

)

2 2

)

−2

⎛ dx ⎞ ⎤ ⎜⎝ 2x ⎟⎠ ⎥ dt ⎦

kˆ

RIGID BODY CONSTRAINT Consider a rigid body on which two points A and   B having velocities vA and vB are taken as shown in Figure.

SOLUTION

The position of x and velocity of the particle are shown in Figure. Since the body is rigid, so the distance between points A and B is constant and hence the component of velocity of A and B along the line joining A to B must be equal, i.e. vA cos θ A = vB cos θB

Mechanics II_Chapter 3_Part 1.indd 22

2/9/2021 6:38:42 PM

Chapter 3: Rotational Dynamics

However, the perpendicular components may or not be equal. If vA sin θ A ≠ vB sin θB and if we observe the motion of the rigid body from the frame of point A, then the only possible way in which point B and all other points on the body can move is in a circle with A as the centre.

If the point A is moving with a speed VA, calculate the speed of point B and angular velocity of the rod at this instant. SOLUTION

Let us consider the point B to move downwards with speed VB . According to the rigid body constraint, the component of velocity of A and B along the AB must be equal. So, we have

So, angular velocity of the line AB (or angular) velocity of the body is given by v sin θB − vA sin θ A ω= B 

Conceptual Note(s) Angular velocity of the body means the rate of change of orientation of any line drawn on the body. Now let us try to find the angular velocity of the body from the frame of point B.

3.23

VB sin θ = VA cos θ cos θ sin θ

⇒

VB = VA

⇒

VB = VA cot θ

…(1)

The components of velocity of A and B perpendicular to the rod will contribute to the rotation of the body, so we have

ω=

Thus, it can be concluded that angular velocity of the body remains same for all the points on the body.

( vrel )⊥ l

=

VB cos θ + VA sin θ l

Using equation (1), we get ILLUSTRATION 12

A rod AB of length l has its one end lying on the horizontal floor and the other end leaning against the vertical wall as shown in Figure.

ω= ⇒

ω=

( VA cos2 θ + VA sin 2 θ ) / l sin θ

VA l sin θ

Also, we see that, tan θ =

Mechanics II_Chapter 3_Part 1.indd 23

⇒

x = l cos θ

⇒

dx ⎛ dθ ⎞ = −l sin θ ⎜ ⎝ dt ⎟⎠ dt

y x

2/9/2021 6:38:50 PM

3.24

JEE Advanced Physics: Mechanics – II

⇒

ω =

vA l sin θ

VB = VC2 + ( l − h 2 ) ω 2 2

Both the methods fetch us same results. ILLUSTRATION 13

A rod AB of length l is lying as shown in the Figure. The end A of the rod has speed VA as shown in Figure.

⇒

VB = VA

1 ⎛ l−h 2⎞ +⎜ ⎟ 2 ⎝ 2h ⎠

2

RIGID BODY CONSTRAINT FOR ACCELERATION Just like we saw the constraint relation between the velocities of any two points on a rigid body, we can also find some relations between the acceleration, angular acceleration, velocity and angular velocity. Consider two points A and B, separated by a distance l, having accelerations aA and aB as shown in Figure.

Calculate the speed of point B and the angular velocity of the rod SOLUTION

Since we observe that the velocity of the point C is along the rod, so by using the rigid body constraint, we get VA cos 45° = VC ⇒

VC =

VA 2

Since AC = h 2, so the angular velocity ω is

ω=

VA sin 45° h 2

V = A 2h

Once we know the angular velocity, then we can calculate the speed of point B by using the concept of relative motion and the rigid body constraint.

Since, B is moving in a circle with angular velocity ω when observed from reference frame attached to A. So, we have aA cos θ A − aB cos θB ≠ 0 As a matter of fact, aA cos θ A − aB cos θB directed radially inverse provides centripetal acceleration or normal acceleration and hence aA cos θ A − aB cos θB = lω 2 The components of accelerations perpendicular to the line joining A to B will provide the tangential acceleration α to the body, so tangential acceleration of B in frame of A is aA sin θ A − aB sin θB = lα Just like angular velocity, it can also be proved that angular acceleration of the body remains same in the frame of all the points on the rigid body.

So, the linear tangential velocity of the point B is rω = ( l − h 2 ) ω and hence speed of point B is

Mechanics II_Chapter 3_Part 1.indd 24

ILLUSTRATION 14

The angular velocity and angular acceleration of the pivoted rod are ω and α respectively. Calculate the x and y components of acceleration of the end B of the rod.

2/9/2021 6:39:00 PM

Chapter 3: Rotational Dynamics

3.25

Since, v = Rω dv dω =R dt dt

⇒

a = Rα

So, acceleration of the ends of rope is related to angular acceleration of the pulley as a = Rα .

SOLUTION

The acceleration of the point A is zero. The tangential and radial accelerations of the point B w.r.t. the point A are

( aBA )T

⇒

= lα and ( aBA )R = lω 2

Since, by definition, we have    aBA = aB − aA    ⇒ aB = aBA + aA

COAXIAL PULLEYS Consider an arrangement of two pulleys fixed on the same shaft (co-axial). When the shaft turns with angular speed ω , the two pulleys will also turn with angular speed ω . Consider the following Figure.

Then resolving the vectors in x and y directions, we get

As the shaft turns, the two pulleys also turn with same angular speed ω .

( aB )x = lα cos θ − lω 2 sin θ and

( aB )y = lα sin θ + lω 2 cos θ

ATWOOD’S MACHINE (SIMPLY PULLEY) It is a simple pulley, hinged at the centre so as to turn freely without friction. The pulley has a mass M, radius R and it can be in the shape of a disc or a ring or a cylinder. Sufficient friction is present between the rope and the pulley surface so that the rope passing over its surface does not slip on the pulley surface. Due to this, the speed of the rope will be same as the speed of a point on the surface of the pulley as shown in Figure.

DISCS CONNECTED BY A ROPE OR CHAIN Consider two discs/wheels of radii r1 and r2 mounted on two separate fixed shafts turning with angular speeds ω1 and ω 2. Let the two discs be connected to each other through a rope or a belt or a chain as shown in Figure.

If the rope or belt or the chain does not slip over the two discs, the linear speed of any section of the rope is same as that of a point on the periphery of the two discs. So, we have v = ω1 r1 = ω 2 r2 and a = α 1 r1 = α 2 r2 ILLUSTRATION 15

Two wheels 1 and 2 of radii r1 and r2 are connected by a belt as shown in Figure.

Mechanics II_Chapter 3_Part 1.indd 25

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JEE Advanced Physics: Mechanics – II

If the wheel 1 is rotating with uniform angular velocity ω and the belt does not slide over the wheels, calculate the speed of a point on the belt, the angular velocity of the wheel 2 and acceleration of the point P with respect to the point Q.

⇒

    vA − vcm = ω × rAO

SOLUTION

⇒

    vA = vcm + ω × rAO

Since the belt is not sliding, so we have

Differentiating, we get

   vA cm = ω × rAO

⇒

 dvcm d     d   + ( ω × rAO ) aA = ( vcm + ω × rAO ) = dt dt dt    dr dω    aA = acm + × rAO + ω × AO dt dt       aA = acm + α × rAO + ω × vA cm

⇒

       aA = acm + α × rAO + ω × ( ω × rAO )

vA = vB

⇒

⇒

r1ω1 = r2ω 2

⇒

ω2 =

r1ω r2

Since ω is constant, so tangential acceleration of any point on the wheels is zero and hence we have   aP = r1ω 2 ( iˆ ) and aQ = r2ω 22 ( −iˆ ) So, acceleration of point P w.r.t. point Q is r ⎞    ⎛ aPQ = aP − aQ = r1ω 2 ⎜ 1 + 1 ⎟ iˆ r2 ⎠ ⎝

ACCELERATION AND VELOCITY OF ANY POINT ON A RIGID BODY Consider a rigid body moving in space such that at  any instant, the velocity of its centre of mass is vcm  and angular velocity is ω . To describe the motion of any point A with respect to a stationary point (or ground) we must understand that in a rigid body, the relative position of the points cannot change. This simply implies that, the relative velocity of point A (or any other point that lies on the body) w.r.t. centre of mass is tangential and is given by

Mechanics II_Chapter 3_Part 1.indd 26

Also, we note that the velocity of any point A w.r.t. B is given by      vAB = vA − vB = ω × rAB    where, rAB = rA − rB is the position vector of A w.r.t. B. In general, we can say that the general motion of a body can be called as a combination of rotational motion and translational motion. ILLUSTRATION 16

A thin uniform rod AB of mass m and length L is placed on a smooth horizontal table. A constant horizontal force of magnitude F starts acting on the rod at one of the ends AB. Initially, the force is perpendicular to the length of the rod. Taking the moment at which force starts acting as t = 0, calculate the distance moved by centre of mass of the rod in time t0, magnitude of initial acceleration of the end A of the rod. SOLUTION

Consider the rod to be placed in x -y plane (horizontal) with initial orientation along y-axis. Further, assuming that force acting on the rod is along the x-axis as shown in Figure.

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Chapter 3: Rotational Dynamics

3.27

The displacement is given by ⇒

xcm =

1 1F 2 acm t02 = t0 2 2m

{∵ acm = constant }

The initial acceleration of end A of the rod is     {∵ ω = 0 } aA = acm + α × rAO

The acceleration of centre of mass is given by  F Fˆ  acm = = i = constant m m

⇒

F  ⎛ 6F ˆ ⎞ ⎛ L ˆ ⎞ k⎟ ×⎜ j⎟ aA = iˆ + ⎜ − m ⎝ mL ⎠ ⎝ 2 ⎠

⇒

4F ˆ  i aA = m

Test Your Concepts-II

Based on Rotational Kinematics, Combined Effect of Rotation and Translation Motion (Solutions on page H.150) 1.

2.

3.

The angular position of a point on the rim of a rotating wheel is given by θ = 4t − 3t 2 + t 3, where θ is in radian and t is in second (a) Calculate the angular velocity at t = 2 s and t = 4 s. (b) Calculate the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4 s (c) Calculate the instantaneous angular acceleration at the beginning and the end of this time interval. A uniform disc of radius r spins with angular velocity ω and angular acceleration α . If the centre of mass of the disc has linear acceleration a, calculate the magnitude and direction of acceleration of the points 1, 2, 3 and 4.

4.

5.

A disc starts rotating with constant angular acceleration of π rads −2 about a fixed axis perpendicular to its plane and through its centre. (a) Find the angular velocity of the disc after 4 s. (b) Find the angular displacement of the disc after 4 s and (c) Find the number of turns accomplished by the disc in 4 s. A rod AB of length l is supported against smooth horizontal surface. The rod is inclined at angle θ with horizontal. Now it released so that end A starts sliding on horizontal surface. Find the velocity of centre of mass of the rod in terms of its angular velocity ω .

Calculate the linear acceleration of the particle P at t = 1 s if a = 2 ms −2, ω = ( 2t ) rads −1 and CP = 1 m as shown in Figure. Express your result in terms of iˆ and ˆj. 6.

Mechanics II_Chapter 3_Part 1.indd 27

A flywheel of radius 20 cm starts from rest, and has a constant angular acceleration of 60 rads −2. Find

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3.28

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JEE Advanced Physics: Mechanics – II

(a) the magnitude of the net linear acceleration of a point on the rim after 0.15 s (b) the number of revolutions completed in 0.25 s A particle A moves along a circle of radius  R = 10 cm so that its radius vector r relative to O rotates with constant angular velocity ω 0 = 0.2 rads −1 . Find the modulus of the velocity of the particle and modulus and direction of its total acceleration. r

9.

a and A respectively. Calculate the ratio of the acceleration of the bobbin to the acceleration of the block. A disc of radius R start at time t = 0 moving along the positive x-axis with linear speed v0 and angular speed ω 0 . Find the x and y co-ordinates of the bottommost point at any time t. y

A

ω0

O C

8.

A bobbin with inner radius r and outer radius R is arranged with light strings and a block as shown in Figure.

v0

C

x

O

10. A uniform rod of length l is spinning with an 2v angular velocity ω = while its centre of mass l moves with a velocity v. Calculate the velocity of the end A of the rod.

11. Calculate the linear velocity of particle P shown in Figure, if v = 2 ms −1 , ω = 5 rads −1 and CP = 1 m . Express your result in terms of iˆ and ˆj .

The string does not slide over the cylinder and when the system is released from rest, then the block and bobbin move down with accelerations

ROTATIONAL KINETIC ENERGY (R.K.E.) Consider a system made of particles of masses m1, m2, m3, m4, ......, mn placed at distances r1, r2, …, rn from the axis of rotation. Let the particles rotate about the axis of rotation with uniform angular velocity ω such that the tangential velocities of the particles are v1, v2, …, vn respectively. From above consideration we conclude that the system possesses kinetic energy due to rotational motion of the particles and hence this kinetic energy is called rotational kinetic energy denoted by K R.

Mechanics II_Chapter 3_Part 1.indd 28

Axis of rotation

m2 m4

r2

r1 r3

r4

m1 m3

A discrete system of particles

⇒

K rotational = K R =

1 1 1 m1v12 + m2 v22 + .... + mn vn2 2 2 2

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Chapter 3: Rotational Dynamics

But

v1 = r1ω , v2 = r2ω , …, vn = rnω

⇒

KR =

1 m1 r12 + m2 r22 + ... + mn rn2 ω 2 2

⇒

KR =

1 2 Iω 2

(

3.29

The translational kinetic energy of rod is

)

K translational = KT =

2

1 ⎛ 3ω l ⎞ 9 2 2 m⎜ ⎟ = mω l 2 ⎝ 2 ⎠ 8

The rotational kinetic energy of rod is

For a system rotating about a point, we must keep in mind that the translational kinetic energy of a rigid body having mass M and centre of mass velocity vcm is given by K translational = KT =

1 2 Mvcm 2

and the rotational kinetic energy of the body about its centre of mass is 1 K R = I cm ω 2 2 So, the total energy of a body in rotation about an axis can be written as K = K R + KT ILLUSTRATION 17

A rod of mass m and length l is connected with a light rod of length l. The composite rod is made to rotate with angular velocity ω as shown in Figure.

Krotational =

1 1 ⎛ ml 2 ⎞ 2 1 I cm ω 2 = ⎜ mω 2 l 2 ⎟⎠ ω = ⎝ 2 2 12 24

Total kinetic energy of the rod is K total = K translational + K rotational ⇒

K total =

1 7 9 mω 2 l 2 + mω 2 l 2 = mω 2 l 2 8 24 6

The total kinetic energy of the rod can also be thought to be the rotational kinetic energy of the rod about the fixed axis. Hence, we have 1 IPω 2 2 where I P is the moment of inertial of the rod about the axis passing through the point P. According to parallel axis theorem, we have K total =

⎛3 ⎞ I P = I cm + md 2 = I cm + m ⎜ l ⎟ ⎝2 ⎠ ⇒

IP =

2

ml 2 9 2 7 2 + ml = ml 12 4 3

Hence total kinetic energy of the rod is Calculate the translational kinetic energy, rotational kinetic energy and total kinetic energy of rod. SOLUTION

Translational kinetic energy of rod, K translational =

1 2 mvcm 2

Velocity of centre of mass is ⎛3 ⎞ vcm = ω ⎜ l ⎟ ⎝2 ⎠

K total =

1⎛ 7 2⎞ 2 7 2 2 ⎜ ml ⎟⎠ ω = mω l 2⎝ 3 6

MODIFIED WORK ENERGY THEOREM (MWET) AND CONSERVATION OF MECHANICAL ENERGY If Wext is the work done by external forces, Wnc is the work done by non-conservative forces, Wps is the work done by pseudo forces and Wint is the work done by internal forces, then according to Modified Work Energy Theorem (MWET) studied earlier, we have Wext + Wnc + Wps + Wint = ΔU + ΔK However, in the absence of dissipative (non-conservative), external, pseudo forces, work done by thems is zero and

Mechanics II_Chapter 3_Part 1.indd 29

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JEE Advanced Physics: Mechanics – II

if work done by internal forces is zero, then the total mechanical energy of a system is conserved, i.e. ΔK + ΔU = 0 ⇒

K f + U f = Ki + Ui

⇒

( U + K )initial = ( U + K )final

(a) What is the angular velocity of the disk when the particle is at its lowest point? (b) At this point, what force must be exerted on the particle by the disk to keep it on the disk? SOLUTION

(a) By Law of Conservation of Mechanical Energy, we have

ILLUSTRATION 18

Loss in Gain in ⎛ ⎞ ⎛ ⎞ ⎜ Gravitational ⎟ ⎜ Rotational ⎟ ⎜ Potential Energy ⎟ = ⎜ Kinetic Energy ⎟ ⎜ ⎟ ⎜ ⎟ of Particle ⎝ ⎠ ⎝ of disc + Particle ⎠

A uniform circular disc of mass m, radius R and centre O is free to rotate about a smooth, horizontal axis which is tangential to the disc at a point A. The disc is held in a vertical plane with A below O and is then slightly displaced from this position. Find the angular velocity of the disc when its plane is next vertical.

⇒ mg ( 2R ) =

SOLUTION

⇒ ω=

By Law of Conservation of Mechanical Energy, we have ⎛ Loss in Gravitational ⎞ ⎛ Gain in Rotational ⎞ ⎜ Potential Energy of ⎟ ⎜ Kinetic Energy of ⎟ = ⎜ Centre of Mass ⎟⎟ ⎜⎜ Disc about the said ⎟⎟ ⎜ of the Disc ⎝ ⎠ ⎝ Axis of Rotation ⎠

O Axis

⇒ ⇒ ⇒

A Initially

m R O M

⇒

Initial

⇒

Finally A

⇒ F = mg +

O

g 5R

ILLUSTRATION 19

A uniform disk of mass M and radius R is pivoted so that it can rotate freely about a horizontal axis through its centre and normal to the plane of the disk. A small particle of mass m is attached to the rim of the disk at the top directly above the pivot. The system is given a gentle start and the disk begins to rotate.

O ω m Finally

F − mg = mRω 2

ω

1 2 Iω 2 1⎛ 1 ⎞ mg ( 2R ) = ⎜ mR2 + mR2 ⎟ ω 2 ⎠ 2⎝ 4

Mechanics II_Chapter 3_Part 1.indd 30

8 mg ( 2m + M ) R

(b) If F be the force exerted by disk on the particle (upwards), then

⇒ F=

mgh =

ω=4

1⎛ 1 2 2⎞ 2 ⎜⎝ MR + mR ⎟⎠ ω 2 2

8m2 g ( 2m + M )

mg ( 10 m + M ) ( 2m + M )

ILLUSTRATION 20

ABC is a triangular framework of three uniform rods each of mass m and length 2l. It is free to rotate in its own plane about a smooth horizontal axis through A normal to the plane ABC. If it is released from rest when AB is horizontal and C is above AB, find the maximum velocity of C in the subsequent motion. SOLUTION

I A = I AB + I AC + I BC ⇒

IA =

1 1 ⎞ 2 2 ⎛ 1 2 m ( 2 ) + m ( 2 ) + ⎜ m ( 2 ) + md 2 ⎟ ⎝ 12 ⎠ 3 3

where d = 3

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Chapter 3: Rotational Dynamics

3.31

2⎞ 4 2 4 2 ⎛1 2 m + m + ⎜ m + m (  3 ) ⎟ ⎝3 ⎠ 3 3

⇒

IA =

⇒

I A = 6 m 2

Assume the horizontal line passing through A to be the zero Potential Energy Level (ZPEL) as shown in Figure. Calculate the angular velocity of the rod as the function of angle of rotation θ and the maximum angular displacement of the rod.

C 2

3

A

ω A

B

SOLUTION 2 3

B

O

3

According to modified work energy theorem, we have Wext = ΔU + ΔK , where Fext = F Δr = Fl sin θ

C

1 (  3 ) above ZPEL and finally 3 i.e., when C attains maximum velocity, the CG is 2 (  3 ) below ZPEL. So, in the process the CG has 3 fallen through Then initially CG is

h=

2 ( 3 )+ 1 ( 3 ) =  3 3 3

Applying Law of Conservation of Mechanical Energy, we get Gain in Loss in ⎛ ⎞ ⎛ ⎞ ⎜ Gravitational ⎟ ⎜ Rotational ⎟ ⎜ Potential Energy ⎟ = ⎜ Kinetic Energy ⎟ ⎜ ⎟ ⎜ ⎟ of CG ⎝ ⎠ ⎝ of system ⎠ ⇒ ⇒

( 3m ) g (  3 ) = ω=

1( 6 m 2 ) ω 2 2

g 3 

So, velocity of C at this instant is vC = ( 2 ) ω = 2 g 3 ILLUSTRATION 21

A uniform rod of mass m and length l is pivoted smoothly at O. A horizontal force acts at the bottom of the rod.

Mechanics II_Chapter 3_Part 2.indd 31

ΔU is the rise in potential energy of the centre of mass of the rod, so ⎛ ΔU = mg ⎜ ⎝

l⎞ ⎟ ( 1 − cos θ ) 2⎠

ΔK is the rotational kinetic energy of the rod about rotational axes passing through hinge and can also be called as the total kinetic energy of the rod, so ΔK =

1 IO ω 2 2

⇒

Fl sin θ =

⇒

ω=

mgl 1 ⎛ ml 2 ⎞ 2 ( 1 − cos θ ) + ⎜ ⎟ω 2 2⎝ 3 ⎠

6 F sin θ 3 g ( 1 − cos θ ) − ml l

…(1)

At maximum angular displacement, the angular velocity of the rod is zero. Substituting ω = 0 in equation (1), we get 0=

3g 6F ( 1 − cos θ ) sin θ − ml l

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JEE Advanced Physics: Mechanics – II

⇒

3g 6F ( 1 − cos θ ) sin θ = ml l

⇒

2F ⎡ ⎛θ⎞ ⎛θ⎞⎤ 2 sin ⎜ ⎟ cos ⎜ ⎟ ⎥ = ⎝ 2⎠ ⎝ 2⎠ ⎦ m ⎢⎣

⇒

⎛ θ ⎞ 2F tan ⎜ ⎟ = ⎝ 2 ⎠ mg

⇒

⎛ 2F ⎞ θ = 2 tan −1 ⎜ ⎝ mg ⎟⎠

⎛ ⎛θ⎞⎞ g ⎜ 2 sin 2 ⎜ ⎟ ⎟ ⎝ 2⎠⎠ ⎝

ILLUSTRATION 22

A block of mass m = 4 kg is attached to a spring of spring constant ( k = 32 Nm −1 ) by a rope that hangs over a pulley of mass M = 8 kg. If the system starts from rest with the spring unstretched, find the speed of the block after it falls 1 m. Treat the pulley as a 1 disc, so I = MR2. 2

Please note that in this problem, the radius R was not required. Substituting m = 4 kg, M = 8 kg, k = 32 Nm −1 and x = 1 m, we get 1⎛ 8⎞ 2 1 2 ⎜⎝ 4 + ⎟⎠ v + ( 32 )( 1 ) − ( 4 )( 10 )( 1 ) = 0 2 2 2 ⇒

4v 2 + 16 − 40 = 0

⇒

v = 2.4 ms −1

TOTAL ENERGY OF A BODY IN PURE ROLLING When a body rolls without slipping, then it possesses simultaneous translational motion (of the CM) and rotational motion of the entire body. So, Total K.E. = Rotational K.E. + Translational K.E. 1 1 2 I CM ω 2 + mvCM 2 2

⇒

Total K.E. = E =

⇒

E=

1 2 1 Iω + mv 2 2 2

⇒

E=

1( 2 ) 2 1 mk ω + mv 2 2 2

⇒

E=

1( 2 ) 2 1 2 mk ω + m ( Rω ) 2 2

⇒

E=

1( 2 mk + mR2 ) ω 2 2

k

m

SOLUTION

Since the rim of the pulley moves at the same speed as the block, the speed of the block and the angular velocity of the pulley related by v = ω R. When the block falls by a distance x, its potential energy decreases, so ΔU g = − mgx, the potential 1 energy of the spring increases, so ΔU s = + kx 2, 2 both the block and the pulley gain kinetic energy, so 1 1 ΔK = mv 2 + Iω 2. Applying Law of Conservation 2 2 of Mechanical Energy, we get

Version I

ΔK + ΔU = 0 , 2

⇒

1 1 ⎛ v⎞ 1 mv 2 + I ⎜ ⎟ + kx 2 − mgx = 0 2 2 ⎝ R⎠ 2

⇒

1⎛ M⎞ 2 1 2 ⎜ m + ⎟⎠ v + kx − mgx = 0 2⎝ 2 2

Mechanics II_Chapter 3_Part 2.indd 32

…(1)

⇒

E=

R2 ⎞ 1 ( 2 )⎛ mk ⎜ 1 + 2 ⎟ ω 2 ⎝ 2 k ⎠

E=

R2 ⎞ 1 2⎛ Iω ⎜ 1 + 2 ⎟ ⎝ 2 k ⎠

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Chapter 3: Rotational Dynamics

Version II

⇒

E=

⎛ 1 R2 ⎞ ⎛ v 2 ⎞ mk 2 ⎜ 1 + 2 ⎟ ⎜ 2 ⎟ ⎝ 2 k ⎠⎝ R ⎠

E=

⎛ 1 k2 ⎞ mv 2 ⎜ 1 + 2 ⎟ ⎝ 2 R ⎠

Conceptual Note(s) If KR stands for rotational kinetic energy and K T for translational kinetic energy, then we have 1 1 2 KR = Icmω 2 and K T = mvcm , then 2 2 (i) for a ring,

KR =1 KT

K 1 (ii) for a disc, R = KT 2 K 1 (iii) for a cylinder, R = KT 2 (iv) for a shell,

KR 2 = KT 3

K 2 (v) for a solid sphere, R = KT 5

⇒

I 0 = mR2 + mR2 = 2mR2

⇒

K=

⇒

K = mR2ω 2 = mv02

3.33

1( 2mR2 ) ω 2 2

ILLUSTRATION 24

A carpet of mass M of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet R when its radius reduces to . 2 SOLUTION

Let M be the initial mass of the carpet of radius R and length l (say), then M = ( π R2 l ) ρ When the gentle push is given to the carpet and it unrolls such that its radius reduces to R 2, then the new mass of the rolling portion is 2

M ⎛ R⎞ M ′ = π ⎜ ⎟ lρ = ⎝ 2⎠ 4 By law of conservation of energy, we have ⎛ Loss in ⎞ = ⎛ Gain in ⎞ + ⎛ Gain in ⎞ ⎝ GPE of CM ⎠ ⎝ RKE of CM ⎠ ⎝ TKE of CM ⎠

ILLUSTRATION 23

A hoop of mass m and radius R rolls without slipping with velocity v0. Find its kinetic energy. SOLUTION

METHOD I: 1 1 mvc2 + I cω 2 2 2 1 1 K = mvo2 + ( mR2 ) ω 2 2 2 Since v0 = ω R K=

⇒

K=

1 1 mvo2 + mvo2 2 2

⇒

⎛1 ⎞ 1 2 MgR − M ′gR ′ = ⎜ I ′ω 2 ⎟ + M ′vcm ⎝2 ⎠ 2

⇒

7 1⎛ 1 ⎛ 1⎛ M⎞ 2 ⎞ ⎞⎛v ⎞ MgR = ⎜ M ′R ′ 2 ⎟ ⎜ cm ⎟ + ⎜ ⎜ ⎟v ⎠ ⎝ R′ ⎠ ⎝ 2 ⎝ 4 ⎠ cm ⎟⎠ 8 2⎝ 2

⇒

7 1⎛ M⎞ 2 ⎛ M⎞ 2 MgR = ⎜ ⎟⎠ vcm + ⎜⎝ ⎟ vcm ⎝ 8 2 8 8 ⎠

⇒

7 ⎛ M⎞ ⎛ 3⎞ 2 MgR = ⎜ v ⎝ 8 ⎟⎠ ⎜⎝ 2 ⎟⎠ cm 8

⇒

vcm =

2

⇒ K = mvo2 METHOD II: K=

1 I 0ω 2 where I 0 = I c + mR2 2

Mechanics II_Chapter 3_Part 2.indd 33

14 gR 3

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JEE Advanced Physics: Mechanics – II

ILLUSTRATION 25

A small steel sphere of mass m and radius r rolls without slipping on the surface of a large hemisphere of radius R (  r ) whose axis of symmetry is vertical. It starts at the top from the rest. (a) What is the kinetic energy at the bottom? (b) What fraction is the rotational kinetic energy of the total kinetic energy? (c) What fraction is the translational kinetic energy of the total kinetic energy? (d) Calculate the normal force that the small sphere will exert on the hemisphere at its bottom. How the results will be affected if r is not very small as compared to R. SOLUTION

Dotted line is the axis of symmetry, which is vertical.

17 mg even if r 7 is not very small as compared to R. The normal reaction will still be

ILLUSTRATION 26

A uniform solid cylinder of mass M and radius 2R rests on a horizontal table top. A string is attached by a yoke to a frictionless axle through the centre of the cylinder so that the cylinder can rotate about the axle. The string runs over a pulley in the shape of a disk of mass M and radius R that is mounted on a frictionless axle through its centre. A block of mass M is suspended from the free end of the string. The string does not slip over the pulley surface and the cylinder rolls without slipping on the table top. After the system is released from rest, what is the magnitude of the downward acceleration of the block? M R 2R

(a) In pure rolling (on stationary ground) work done by friction is zero. Hence, mechanical energy remains conserved. So K.E. at bottom = decrease in P.E. = mg ( R − r ) (b) In pure rolling ⇒ (c)

KR 2 = (for a sphere) KT 5

KR 2 = K R + KT 7

KT 5 = K R + KT 7

(d) KT = ⇒ ⇒

5 (Total energy) 7

1 5 mv 2 = mg ( R − r ) 2 7 2

mv 10 = mg R−r 7

⇒ N − mg =

10 mv = mg R−r 7

17 ⇒ N= mg 7

Mechanics II_Chapter 3_Part 2.indd 34

2

M

SOLUTION

When the block moves down by h (say), in time t, then it loses potential energy. This potential energy lost by block is gained as 1 (a) translational kinetic energy, Mv 2 by the block 2 1 (b) rotational kinetic energy, I1ω12 by the pulley 2 disk 1 1 Mv 2 + I 2ω 22 by the (c) rolling kinetic energy, 2 2 cylinder So, by Law of Conservation of Energy, we get Mgh = where h = I1 =

1 1 1 ⎛1 ⎞ Mv 2 + I1ω12 + ⎜ Mv 2 + I 2ω 22 ⎟ ⎝ ⎠ 2 2 2 2

v at 1 2 v at = at , v = at , ω1 = = , ω 2 = 2 R R 2R 2R 1 1 2 MR2 and I 2 = M ( 2R ) 2 2

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Chapter 3: Rotational Dynamics

So, we get

3.35

ILLUSTRATION 28 2

2

( 2R ) ⎛ at ⎞ R2 ⎛ at ⎞ ⎛1 ⎞ 2 g ⎜ at 2 ⎟ = ( at ) + ⎜ ⎟ ⎜⎝ ⎟⎠ + ⎝2 ⎠ 4 R 4 ⎝ 2R ⎠ ⇒

g = 2a +

⇒

a=

2

a a + 2 2

g 3

A semi-circular track of radius R = 62.5 cm is cut in a block. Mass of block, having track, is M = 1 kg and rests over a smooth horizontal floor. A cylinder of radius r = 10 cm and mass m = 0.5 kg is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in Figure.

ILLUSTRATION 27

A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearing. A  light cord passes around the equator of the shell, over a pulley of rotational inertia I and radius r and is attached to a small object of mass m that is otherwise free to fall under the influence of gravity. There is no friction of pulley’s axle and the cord does not slip on the pulley. Calculate speed of the object after it has fallen a distance h from rest.

When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip. Calculate velocity of axis of cylinder and velocity of the block when it reaches bottom of the track. Also find force applied by block on the floor at that

(

)

moment g = 10 ms −2 . SOLUTION

Using Law of Conservation of Linear Momentum, we get mv1 = Mv2 SOLUTION

By law of conservation of energy, we observe that the loss in gravitational potential energy of the block equals the sum of gain in kinetic energy of the block, gain in rotational kinetic energy of the pulley and the gain in rotational kinetic energy of the shell. ⇒

mgh =

where, ω =

1 1 1 ⎛ 2 MR2 ⎞ 2 mv 2 + Iω 2 + ⎜ ⎟ (ω ′ ) 2 2 2⎝ 3 ⎠

v=

Applying Law of Conservation of Mechanical Energy, we get mg ( R − r ) =

mgh m I M + + 2 2r 2 3

Mechanics II_Chapter 3_Part 2.indd 35

1 1 1 mv12 + Iω 2 + Mv22 2 2 2

…(2)

v 1 2 mr and ω = r , where vr is the veloc2 r ity of cylinder axis relative to block

where, I =

vr = v1 + v2

v v and ω ′ = r R

After substituting these values in the above equation and solving, we get

…(1)

…(3) ω

v1

v2

Solving equations (1), (2) and (3) with given data, we get v1 = 2 ms −1

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JEE Advanced Physics: Mechanics – II

If IIAOR is the Moment of Inertia of body about Instantaneous Axis of Rotation (IAOR), then

and v2 = 1.5 ms −1 Further, we have

IIAOR = ICG + md 2 = mk 2 + mR2

mvr2 N − mg = R−r ⇒

N = mg +

mvr2

R−r

= ( 0.5 )( 10 ) +

( 0.5 )( 3.5 )

KE =

2

0.525

The Instantaneous Axis of Rotation (IAOR) passes through a point called the Instantaneous Centre of Zero Velocity (IC). So, from above we conclude that pure rolling is equivalent to the case of pure rotation about a new axis (called Instantaneous Axis of Rotation, IAOR) that passes through the point of contact (because it was the only point which is at rest in pure rolling).

⇒ N = 16.67 N

Conceptual Note(s) AN OBSERVATION For a body in pure rolling motion, we have E=

1( 2 mk + mR2 )ω 2 2

Now ( mk 2 + mR2 ) is actually moment of inertia of body (in combined effect of rotation and translation) about axis that passes through the point of contact P, which happens to be at rest (in case of pure rolling), so E=

1 1 ( Iabout P )ω 2 = ( mk 2 + mR2 )ω 2 2 2 Rω

vcm

ω vcm

INSTANTANEOUS AXIS OF ROTATION (IAOR) The combined effect of translation of the centre of mass and rotation of a rigid body about an axis through the centre of mass is equivalent to a pure rotation with the same angular speed about an axis passing through a point of zero velocity. Such an axis is called the Instantaneous Axis of Rotation (IAOR).

CM

R Rω

IC ω

ω

(2R)ω = 2v

Rω P IC

rω

IC

ω

⎛ Pure rotation ⎞ ⎜ about IAOR ⎟ ⎟ So, ( Rotation ) + ( Translation ) ≡ ⎜ ⎜ passing ⎟ ⎜⎝ through IC ⎟⎠ ⇒

E=

1 1 2 mvcm + I cm ω 2 2 2

⇒

E=

1 1( 2 mk + mR2 ) ω 2 = I IAORω 2 2 2

r

Mechanics II_Chapter 3_Part 2.indd 36

v

vcm

This axis passing through P is always normal to the plane used to represent motion and the intersection of this axis with the plane is the location of instantaneous centre of zero velocity ( IC ). 2R

1 ( IIAOR )ω 2 2

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Chapter 3: Rotational Dynamics

Problem Solving Technique(s) WORD OF ADVICE (a) Although the concept of IC is conveniently used to determine the velocity of any point in a body, however generally, IC does not have zero acceleration and therefore, it should not be used for finding the acceleration of any point in the body. (b) When a body is subjected to general plane motion, the point determined as the instantaneous centre of zero velocity for the body can only be used for an instant of time. Since the body changes its position from one instant to the next, then for each position of the body a unique instantaneous centre must be determined. The locus of points which defines the ΙC during the body’s motion is called a centrode. Thus, each point on the centrode acts as the ΙC for the body only for an instant of time.

LOCATION OF THE IC The location of the IC is determined by using the fact that the relative position vector extending from the IC to a point is always perpendicular to the velocity of the point. Then the following three possibilities exist. CASE-1 When the velocity of a point (generally the centre of mass) on the body and the angular velocity of the body are known.

ω CM

v

v

CM v r =ω

ω IC Combined Effect of Rotation & Tranlation

Mechanics II_Chapter 3_Part 2.indd 37

Single Effect of Rotation about IC

3.37

When v and ω are known, then IC is located along  the line drawn perpendicular to v at CM, such that v the distance from CM to IC is, r = . ω Note that IC lie on that side of CM which causes rotation about the IC, which is consistent with the   direction of motion caused by ω and v. CASE-2 When the lines of action of two non-parallel velocities are known. Consider the body shown in Figure where the line of   action of the velocities vA and vB are known.

STEP-1: Draw perpendiculars at A and B to these lines of action. STEP-2: The point of intersection of these perpendiculars, locates the IC at the instant considered. If vA and vB originate at perpendicular distances rA and rB from IC, then vA = rAω and vB = rBω CASE-3 When the magnitude and direction of two parallel velocities are known. When the velocities of points A and B are parallel separated by perpendicular distance d and have known magnitudes vA and vB, then the location of the IC is determined by the following steps. STEP-1: Always draw the velocities to the scale. STEP-2: Join the points of origin of the two velocities i.e., tails of velocity vectors by a straight line. STEP-3: Now, join the heads of the two velocity vectors by straight line. STEP-4: IC is located at the intersection of the lines in STEP-2 and STEP-3.

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JEE Advanced Physics: Mechanics – II

To locate position of IC in (a), we have

⎛ Combined Effect ⎞ ⎛ Single Effect of ⎞ So, ⎜ of Rotation ⎟ ≡ ⎜ Rotation about ⎟ ⎜ and Translation ⎟ ⎜ IC or IAOR ⎟ ⎠ ⎝ ⎠ ⎝ Hence, if IC is at a distance x from centre of mass, then ( vΙC )net = 0 i.e., vcm − xω = 0 ⇒

x=

vcm ω ω

vA = xω and vB = ( d − x ) ω

ω

vcm

CM

vcm

CM x

To locate position of IC in (b), we have v x = ωcm

xω

vcm

IC

IC lies in the portion or IC lies towards the portion that rotates opposite (i.e., has tangential velocity opposite) to vcm . ILLUSTRATION 29

vA = xω and vB = ( d + x ) ω

A hoop of radius R rolls over a horizontal plane with a constant velocity v without slipping. Find the velocity of point A of the loop as shown in Figure.

As a special case, if the body is translating, vA = vB and the IC would be located at infinity, in which case ω = 0.

Conceptual Note(s) Q. Why do we use the concept of IAOR for bodies in combined rotation and translation? A. When a body is under the combined influence of rotational and translational effect, then there exists a point (on the body or off the body) which happens to have zero net velocity, see diagram. This makes us think about a point called Instantaneous Centre (IC) of zero velocity and the axis passing through IC, normal to plane of motion, is called Instantaneous Axis of Rotation (IAOR). So, the combined effect of rotation and translation is just equivalent to a single effect of rotation about IAOR (as if the body is pinned to IC and is just rotating about IC.

Mechanics II_Chapter 3_Part 2.indd 38

SOLUTION

The velocity of the point A can be obtained by two methods. METHOD I:  Velocity of point A w.r.t. centre of mass C is vAC given by    vAC = vA − vC ⇒

   vA = vAC + vC

   Now vC = vCM = v and vAC = Rω , perpendicular  to the line CA. So, vA is actually the magnitude of   the resultant of vAC and vC as shown in Figure.

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Chapter 3: Rotational Dynamics

  If θ is the angle between vPC and vC, then θ = 2α . ⇒

 vA = v 2 + R2ω 2 + 2v ( Rω ) cos ( 2α )

For pure rolling condition, v = Rω ⇒ ⇒ ⇒

 vA = v 2 + v 2 + 2v 2 cos ( 2α )  ⎛ 2α ⎞ vA = v 2 ( 1 + cos 2α ) = v 2 × 2 cos 2 ⎜ ⎝ 2 ⎟⎠  vA = 2v cos α

METHOD II: A body in pure rolling can be considered to be in pure rotation about a point having zero velocity (in this case the point of contact), i.e. instantaneous centre (IC) of zero velocity. So, we have

3.39

Since velocity of IC is zero, so we have vIC = 0 ⇒

vcm − xω = 0

⇒

⎛v ⎞ vcm = x ⎜ cm ⎟ ⎝ 4R ⎠

⇒

x = 4R

Problem Solving Technique(s) If the body is placed on a surface which is in motion, then for no slipping, the point of contact must move with the velocity of the surface.

( vP )net = velocity of surface ILLUSTRATION 31

The disc of mass m, radius r is confined to roll without slipping at A and B. If the plates have the velocities shown, determine the angular velocity of the disc. Also find the velocity of centre of mass of disc, the location of instantaneous centre of zero velocity and total energy of the disc. v A

vA = rω = ( 2R cos α ) ω = 2 ( Rω ) cos α ⇒

vA = 2v cos α

r

{∵ vcm = v = Rω }

B

2v

ILLUSTRATION 30

A wheel of radius R is moving on the ground. If vcm be the velocity of centre of mass of the wheel and v ω = cm be the angular velocity of the wheel, then 4R locate its instantaneous centre of zero velocity.

SOLUTION

Let v0 be the velocity of centre of mass and ω 0 the angular velocity of the disc as shown in Figure then, for no slipping at A and B, we have

( vB )net = 2v

SOLUTION

Let the instantaneous centre of zero velocity be located at a distance x from the centre of the wheel as shown in Figure.

Mechanics II_Chapter 3_Part 2.indd 39

⇒

v0 − rω 0 = 2v

…(1)

( vA )net = −v

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JEE Advanced Physics: Mechanics – II

⇒

v0 + rω 0 = −v

…(2)

where IG =

Solving (1) and (2), we get 3v v ω0 = − and v0 = 2r 2 A

v

v0

ω0 B

2v

⇒

E=

1⎛ 1 2 1 2⎞ 2 ⎜ mr + mr ⎟⎠ ω 2⎝ 2 9

⇒

E=

1 ⎛ 11 2 ⎞ ⎛ 3v ⎞ ⎟ ⎜ mr ⎟⎠ ⎜⎝ 2 ⎝ 18 2r ⎠

⇒

E=

1 ⎛ 11 2 ⎞ ⎛ 9v 2 ⎞ ⎜ mr ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 ⎝ 18 4r

⇒

E=

11 mv 2 16

Also, we ca directly find the value of ω by using

ω= ⇒

ω=

( v r )⊥

⊥ separation between the velocities 3v 2r

v = xω x=

⇒ ⇒

( 2r − x ) =

SOLUTION

Rolling can be considered as pure rotation about point of contact, so

E=

1 1 mv02 + Iω 2 2 2

E=

1 ⎛ v 2 ⎞ 1 ⎛ 1 2 ⎞ ⎛ 9v 2 ⎞ m ⎜ ⎟ + ⎜ mr ⎟ ⎜ 2 ⎟ ⎠ ⎝ 4r ⎠ 2 ⎝ 4 ⎠ 2⎝ 2

E=

2

⇒

9mv 11 mv + = mv 2 8 16 16

where I IAOR = IG + md

⇒

Krolling =

⎞ 3 1 ⎛ mR2 + mR2 ⎟ ω 2 = mR2ω 2 ⎜⎝ ⎠ 2 2 4

v = ωR

…(1)

…(2)

From (1) and (2), we get Krolling 2

3 = mv 2 4

Kinetic energy of the rod is A

v

IC

Krod =

2r 3

G

⇒ B

Mechanics II_Chapter 3_Part 2.indd 40

1 IPω 2 2

The rod translates with the velocity v, hence velocity of centre of disc will also be v.

1 ( IIAOR ) ω 2 2

4r 3

Krolling =

2

Alternatively, we have E=

A uniform disc of mass m is fitted (pivoted smoothly) m with a rod of mass . If the bottom of the rod is 2 pulled with a velocity v , it moves without changing its angle of orientation and the disc rolls without sliding. Calculate the kinetic energy of the rod disc system.

2r v = , from A 3v 2r 3

4r , from B 3 The total energy of the disc is OR

2

ILLUSTRATION 32

Now let IC be located at a distance x from A or ( 2r − x ) from B, then

⇒

1 2 2r r mr and d = r − = 2 3 3

1 ⎛ m ⎞ 2 mv 2 ⎜ ⎟v = 2⎝ 2 ⎠ 4

…(3)

K total = K rolling + Krod = mv 2

2v

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Chapter 3: Rotational Dynamics ILLUSTRATION 33

A rotating disc moves in the positive direction of the

⇒

ω = αt =

x-axis. Find the equation y ( x ) describing the position of the instantaneous axis of rotation, if at the initial moment, the centre c of the disc was located at the point O after which it moved with constant velocity v while the disc started rotating counterclockwise with a constant angular acceleration α . Assume the initial angular velocity to be zero. y

O

αx v

3.41

{∵ ω 0 = 0 }

y IC

yω O

v

y c

v

x

ω

Now, if IC is located at a distance y from centre of disc, then we have c v

vIC = 0

x

⇒

yω − v = 0

⇒

y=

v ω

⇒

y=

v v2 = αx v αx

⇒

xy =

SOLUTION

The position of disc at time t is x = vt x ⇒ t= v Since the disc is rotation with constant angular acceleration α , so

ω = ω0 + αt

v2 = constant α

This is the desired x -y equation. This equation represents a rectangular hyperbola.

Test Your Concepts-III

Based on Instantaneous Axis of Rotation, Pure Rolling and Conservation of Energy 1.

A uniform ball of radius r rolls without slipping along the loop-the-loop track in Figure. It starts from rest at height h above the bottom of the loop. If the ball is not to leave the track at the top of the loop, what is the least value h can have (in terms of the radius R  r of the loop)? What would h have been if the ball were to slide along a frictionless track instead of rolling?

2.

A

3.

Mechanics II_Chapter 3_Part 2.indd 41

(Solutions on page H.152) A solid ball rolls down a parabolic path ABC from a height h as shown in Figure. The portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ?

A rod of mass m is kept on a cylinder and sphere each of radius R. The masses of the sphere and cylinder are m1 = 4m and m2 = 5m respectively. If the speed of the rod is v, find the KE of the system. Assume that the surfaces do not slide relative to each other.

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3.42

4.

5.

6.

7.

JEE Advanced Physics: Mechanics – II

A uniform sphere of radius r starts rolling down without slipping from the top of another sphere of radius R. Find the angular velocity of the sphere after it leaves the surface of the larger sphere. A bead of mass m is welded at the periphery of the smoothly pivoted disc mass m and radius R. Find the speed of the bead at its lowest position.

A uniform rod of length 2l and mass m is free to rotate in a vertical plane about a smooth fixed horizontal axis perpendicular to the rod through one end of the rod. Initially, the rod is held in horizontal position and is then released. Find the maximum angular velocity of the rod in the subsequent motion. A uniform disc of mass M and radius R is pivoted about the horizontal axis normal to its plane and passing through its centre C. A point of mass m glued to the disc at its rim, as shown in Figure, is released from rest. Find the angular velocity of the disc when m reaches the point B.

Calculate the angular velocity of the rod, linear velocity of centre of mass of the rod and the kinetic energy of the rod. 9. A uniform ring of radius a and mass m is free to rotate in a vertical plane about a fixed smooth axis which is perpendicular to the plane of the ring and passes through a point A on the ring. A particle of mass m is attached to the ring at B, where AB is a diameter. When the ring is hanging in a position of stable equilibrium the particle is struck a blow which gives it a velocity 3 ga. Find the vertical height above A to which the particle rises. 10. A solid disk is rolling without slipping on a level surface at a constant speed of 2 ms −1. How far can it roll up a 30° ramp before it stops? Take g = 10 ms −2. 11. A disc of mass m and radius R has a spring of constant k attached to its centre, the other end of the spring being fixed to a vertical wall (shown in Figure). If the disc rolls without slipping on a level floor, how far to the right does the centre of mass move, if initially the spring was unstretched and the angular speed of the disc was ω 0?

M m

C

B

8.

A uniform rod of mass m, length l is kept on a smooth horizontal surface. The ends A and B of the rod move with speeds v and 2v respectively perpendicular to the rod as shown in Figure.

12. The 9 kg cradle is supported as shown by two uniform disks that roll without sliding at all surfaces of contact. The mass of each disk is m = 6 kg and the radius of each disk is r = 80 mm . Assuming that the system is initially at rest, find the velocity of the cradle after it has moved 250 mm. 30 N

A

Mechanics II_Chapter 3_Part 2.indd 42

r

B

2/9/2021 6:38:35 PM

Chapter 3: Rotational Dynamics

TORQUE

⊕

The rotational effect of force is called Torque. It is also called as the rotational analogue of force. Force acting on a body provides linear acceleration to it and correspondingly torque acting on a body provides angular acceleration to it. It is an axial vector whose direction can be found by using the Right-Hand Thumb Rule. To find the direction of torque simply curl the fingers of the right hand in the sense of rotation in which the force tends to rotate the system about a given axis, then the direction of the thumb gives the direction of the torque. Mathematically, the moment of force is called torque i.e. torque is equal to the product of force and the perpendicular distance of the force from the axis of rotation (AOR). Mathematically, if θ is angle   between r and F, then

τ = Fr⊥ = F ( r sin θ ) where r⊥ is the perpendicular distance from the origin to the line of action of the force as shown in Figure. It is also called the lever arm.

iˆ    τ = r×F = x Fx

○-

ˆj y Fy

3.43

⊕

kˆ z Fz

In the most simplest manner, if we know the sense of rotation (Clockwise or Anticlockwise) then, to find  the direction of τ we just curl the fingers of Right Hand along the sense of rotation and then the thumb  gives direction of τ . For a system with moment of inertia I having an  angular acceleration α , we have   τ = Iα In Magnitude τ = Iα (Direction can be found by using Right Hand Thumb Rule). This is also called Newton’s Second Law of Rotational Motion. ILLUSTRATION 34

The position vector of point of application of force   F = iˆ + 2 ˆj − 3 kˆ N about O is r = 2iˆ + 3 ˆj − kˆ m . Calculate the torque of a force about a point O .

(

)

(

)

SOLUTION

iˆ ˆj kˆ    Torque τ = r × F = 2 3 −1 1 2 −3  ˆ ⇒ τ = i ( −9 + 2 ) + ˆj ( −1 + 6 ) + kˆ ( 4 − 3 ) The above definition of torque i.e. τ = rF sin θ , may be also be interpreted as

τ = r ( F sin θ ) = rF⊥ The turning effect of a force about the origin is produced only by the perpendicular component ( F⊥ = F sin θ ) as shown in Figure. F⊥ = Fsinθ

(AOR)

r

θ

F|| = Fcosθ

θ

O

Vectorially,    τ = r×F   So, if r = xiˆ + yjˆ + zkˆ and F = Fx iˆ + Fy ˆj + Fz kˆ , then

Mechanics II_Chapter 3_Part 2.indd 43

⇒

(

)

 τ = −7 iˆ + 5 ˆj + kˆ Nm

Conceptual Note(s) (a) Cases for zero torque  Since, τ = rF sinθ  So, for τ = 0 , we have the following three cases: CASE-1: F = 0 i.e., no force acts on system CASE-2: r = 0 i.e., force acts at AOR CASE-3:   F ≠ 0, r ≠ 0 but angle between F and r is either   0° or 180° i.e., F acts along r or opposite to  r . Such forces are called Radial Forces. So, torque due to Radial Forces is always zero.

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JEE Advanced Physics: Mechanics – II

 (b) SI unit of τ is Nm (not joule)       (c) τ = r × F, so τ is ⊥ to r as well as F i.e.,     τ ⋅ r = 0 and τ ⋅ F = 0

ILLUSTRATION 35

Calculate the torque about point O and A due to the force as shown in Figure.

FORCE COUPLE A pair of forces each of same magnitude and acting in opposite direction is called a force couple. The torque due to couple is the product of magnitude of either force and the perpendicular separation between the forces. SOLUTION

Torque about point O      τ = r0 × F, r0 = iˆ + ˆj , F = 5 3iˆ + 5 ˆj

⇒

⇒

(

) (

)

 τ = iˆ + ˆj × 5 3iˆ + 5 ˆj = 5 ( 1 − 3 ) kˆ

⎛ Torque ⎞ ⎛ Magnitude ⎞ ⎛ ⊥ Distance ⎞ ⎜ due to ⎟ = ⎜ of one ⎟ × ⎜ between their ⎟ ⎜⎝ couple ⎟⎠ ⎜⎝ force ⎟⎠ ⎜⎝ lines of action ⎟⎠

Torque about point A      τ = ra × F, ra = ˆj , F = 5 3iˆ + 5 ˆj

τ = F ( 2d )

⇒

Conceptual Note(s) (a) A couple does not exert a net force on an object even though it exerts a torque. (b) Net torque due to a force couple is same about any point.

x1

(

)

 τ = ˆj × 5 3iˆ + 5 ˆj = 5 ( − 3 ) kˆ

ILLUSTRATION 36

Calculate the torque about point A, O and B due to the force as shown in Figure.

x2 y2 y1

SOLUTION

⇒ τ A = F ( x1 + x 2 ) = Fd

Torque about point A      τ A = rA × F, rA = 3iˆ, F = 10iˆ

Torque about B is τ B = y1F − y2F

⇒

Torque about A is τ A = x1F + x 2F

⇒ τ B = F ( y1 − y2 ) = Fd (c) If net force acting on a system is zero, torque is same about any point. (d) A consequence is that, if Fnet = 0 and τ net = 0 about one point, then τ net = 0 about any point.

Mechanics II_Chapter 3_Part 2.indd 44

 τ A = 3iˆ × 10iˆ = 0

Torque about point B      τ B = rB × F, rB = 5 ˆj , F = 10iˆ ⇒

 τ B = 5 ˆj × 10iˆ = −50 kˆ

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Chapter 3: Rotational Dynamics

Torque about point O      τ O = rO × F, rO = 3iˆ + 5 ˆj , F = 10iˆ  ⇒ τ O = 3iˆ + 5 ˆj × 10iˆ = −50 kˆ

(

)

ILLUSTRATION 37

A particle having mass m is projected with a velocity v0 from a point P on a horizontal ground making an angle θ with horizontal. Calculate the torque about the point of projection acting on the particle when it is at its maximum height.

SOLUTION

When the particle reaches the maximum height as shown in Figure, then

3.45

having two components, the horizontal component Px and the vertical component Py. Then P = Px2 + Py2 Please note that the force due to the hinge can be assumed to be in any direction. The forces acting on the rod are drawn in the FBD shown in Figure.

Please note that, Px and Py have been assigned arbitrary directions. If the answers for Px, Py or both come out to be negative, then the direction will be opposite to the direction assumed. For translatory equilibrium, we have ∑ Fx = 0 and ∑ Fy = 0

V0

When ∑ Fx = 0, we get Px + F sin θ = 0 r⊥ = R/2

⎛ v 2 sin 2θ ⎞ ⎛ R⎞ τ = Fr⊥ = mg ⎜ ⎟ = mg ⎜ 0 ⎝ 2⎠ 2 g ⎟⎠ ⎝ ⇒

τ=

mv02 sin 2θ 2

ILLUSTRATION 38

A uniform bar of mass m and length l is hinged at one end and held in the position shown by applying a constant force F at the other end, perpendicular to the rod.

⇒

Px = − F sin θ

…(1)

Negative sign simply shows that Px is directed to the right. When ∑ Fy = 0, we get Py + F cos θ − mg = 0 ⇒

Py = mg − F cos θ

…(2)

To calculate Px, we must find the angle θ . For that, let us apply the condition for rotational equilibrium. For rotational equilibrium, we have Στ = 0

(about any point)

So, lets calculate the net torque acting on the rod about the hinge, then we get

Calculate the magnitude and direction of the force exerted by the hinge on the rod. SOLUTION

The two known forces that act on the rod are its weight mg acting vertically downward and the applied force F. Let the force applied by the hinge be

Mechanics II_Chapter 3_Part 2.indd 45

⎛ l ⎞ − ( mg ) ⎜ cos θ ⎟ + Fl = 0 ⎝2 ⎠ mg ⇒ F= cos θ 2 From equations (1) and (3), we get Px = − F 1 −

4F 2 m2 g 2

=−

…(3)

F m2 g 2 − 4 F 2 mg

2/9/2021 6:39:07 PM

3.46

JEE Advanced Physics: Mechanics – II

ILLUSTRATION 39

Calculate minimum coefficient of friction between a thin homogeneous rod and the floor for which a person can slowly lift the rod upwards without slipping on the floor by applying a force perpendicular to the rod.

⇒

d ( 2 tan α + cot α ) = 0 dα

⇒

2 sec 2 α − cosec 2α = 0

⇒

tan α =

1 2

Substituting this value in equation (1), we get

SOLUTION

Consider the equilibrium of the rod when it is inclined at an angle α with the horizontal. The forces acting on the rod are shown in Figure.

μmin =

1 2 2

ILLUSTRATION 40

Consider two heavy right circular rollers of radii R and r respectively at rest on a rough horizontal plane as shown in Figure.

Let us consider torque about the point of intersection O ′ of the force of gravity mg and the force F applied by the person perpendicular to the rod. Since the torque due to these forces about this point O ′ is zero, so doing this eliminates the unknown force F. At equilibrium, we have

( Στ )about O ′ = 0 Since, moment arm for N about O ′ is l cos α , while ⎛ l ⎞ + l sin α ⎟ that for friction about O ′ is ⎜ ⎝ sin α ⎠ ⇒

⎛ 1 ⎞ Nl cos α − fl ⎜ + sin α ⎟ = 0 ⎝ sin α ⎠

⇒

cos α sin α ⎞ ⎛ cos α sin α ⎞ ⎛ f = N⎜ = N⎜ ⎟ 2 ⎝ 1 + sin α ⎠ ⎝ 2 sin 2 α + cos 2 α ⎟⎠

⇒

1 ⎛ ⎞ f = N⎜ ⎝ 2 tan α cot α ⎟⎠ 1 ⎛ ⎞ ≤ μN N⎜ ⎝ 2 tan α cot α ⎟⎠

⇒

μ≥

1 2 tan α + cot α

For μ to be minimum, we must have 2 tan α + cot α = MAXIMUM

Mechanics II_Chapter 3_Part 2.indd 46

SOLUTION

Let N1 be the reaction between the two cylinders and N 2 be the normal reaction between the smaller cylinder and the horizontal surface. The forces acting on the smaller cylinder have been shown in Figure. N1

Since, f ≤ μ N ⇒

The larger roller has a string wound around it to which a horizontal force P can be applied. Assume the coefficient of friction μ has the same value for all surfaces in contact, calculate the necessary condition under which the larger roller can be pulled over the smaller one. Assume the smaller cylinder should neither roll nor slide.

N2

O2

O1 O2

…(1) At equilibrium, we have ∑ Fx = N1 cos α − μ N 2 − μ N1 cos ( 90 − α ) = 0 …(1)

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Chapter 3: Rotational Dynamics

Also, torque about O2 (centre of smaller sphere) is zero, so we get ∑ τ = μ N1 r − μ N 2 r = 0 ⇒

N1 = N 2

…(2)

Substituting equation (2) in equation (1), we get

μ N1 + μ N1 sin α = N1 cos α ⇒

μ ( 1 + sin α ) = cos α

…(3)

Also, we see that sin α = ⇒

R−r R+r

cos α = 1 − sin 2 α =

2 Rr R+r

Substituting these values in equation (3), we get R − r ⎞ 2 Rr ⎛ μ ⎜ 1+ ⎟= ⎝ R+r⎠ R+r ⇒

μ=

Net torque about the point O is ⎛L ⎞ τ O = Mg ⎜ sin θ ⎟ ⎝2 ⎠ Using the Second Law of Motion τ O = IO α MgL ML2 sin θ = α 2 3 3 g sin θ ⇒ α= 2L (b) From above, we have 3 g sin θ α= 2L dω 3 g sin θ ⇒ ω = dθ 2L 3g ⇒ ω dω = sin θ dθ 2L Integrating within appropriate limits, we get ω

∫

r R

0

Hence the required condition is μ ≥

r R

ILLUSTRATION 41

A uniform rod of length L and mass M is pivoted freely at one end as shown in Figure. (a) Find the angular acceleration of the rod when it is at angle θ to the vertical. (b) Assuming the rod to start from the vertical position, find the angular velocity as the function of θ . (c) Find the tangential linear acceleration of the free end when the rod is horizontal. SOLUTION

(a) Figure shows the rod at an angle θ to the vertical.

3.47

⇒

θ

3g ω dω = sin θ dθ 2L

∫ 0

2

3g ω =− cos θ 2 2L

⇒ ω=

θ

= 0

3g ( 1 − cos θ ) 2L

3g ( 1 − cos θ ) L

The above result can also be obtained by using the Law of Conservation of Mechanical Energy, where we use, ⎛ Loss in GPE ⎞ ⎛ Gain in RKE ⎞ ⎜⎝ of CM of Rod ⎟⎠ = ⎜⎝ of Rod ⎟⎠ ⇒ Mg ⇒ ω=

L 1 1 1 ( 1 − cos θ ) = Iω 2 = ⎛⎜⎝ ML2 ⎞⎟⎠ ω 2 2 2 2 3 3g ( 1 − cos θ ) L

3g π , so α = . So, 2 2L the tangential linear acceleration is

(c) When the rod is horizontal θ =

θ O

Mechanics II_Chapter 3_Part 2.indd 47

at = α L =

α Mg L sinθ 2

3g 2

This is greater than the acceleration of an object falling freely.

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3.48

JEE Advanced Physics: Mechanics – II

ILLUSTRATION 42

A thin rod of mass M and length l is pivoted at one end. Initially the rod makes an angle of 60° with the vertical and is then released. Calculate the magnitude and direction of the force on the pivot when the rod is horizontal.

⇒

Fy = Mg −

Mg 3 Mg = 4 4 Fy

Fx

α Mg



So,

F = Fx2 + Fy2 =

10 Mg 4

⎛ Mg ⎞ ⎜⎝ ⎟ 4 ⎠ =1 tan α = = Fx ⎛ 3 Mg ⎞ 3 ⎜⎝ ⎟ 4 ⎠ Fy

SOLUTION

Let ω be the angular velocity of rod in horizontal position, then

τ α= = I

l

( Mg ) 2 Ml 2 3

…(1)

⇒

⎛ l ⎞ 1 Mg ⎜ sin ( 30° ) ⎟ = Iω 2 ⎝2 ⎠ 2

⇒

⎛ l ⎞ 1⎛ 1 ⎞ Mg ⎜ ⎟ = ⎜ Ml 2 ⎟ ω 2 ⎝ 4⎠ 2⎝ 3 ⎠

60°

ω=

α

ILLUSTRATION 43

A thin uniform rod of mass m and length l is placed on smooth horizontal surface. It is acted upon by a force F as shown in Figure.

C

Calculate linear and angular acceleration of the rod. Find the location of a point on the rod which has zero acceleration.

 sin(30°) 2

SOLUTION

3g 2l

…(2)

⎛ l⎞ ⎛ l ⎞ ⎛ 3g ⎞ 3 = Mg Fx = M ⎜ ⎟ ω 2 = M ⎜ ⎟ ⎜ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ 2l ⎟⎠ 4 ⎛ Mg − Fy = Mat = M ( α ) ⎜ ⎝

Mechanics II_Chapter 3_Part 2.indd 48

Fy

Fx

⎛ Loss in GPE of ⎞ ⎛ Gain in RKE ⎞ ⎜⎝ CM of Rod ⎟⎠ = ⎜⎝ of the Rod ⎟⎠

⇒

⎛ 1⎞ α = tan −1 ⎜ ⎟ ⎝ 3⎠ F

3g = 2 l

By Law of Conservation of Energy

O

⇒

l⎞ ⎟ 2⎠

According to the problem, the external force is acting on rod along the x-direction, hence the centre of mass of the rod will accelerate along the x-direction. Also, we observe that due to this force, a net torque will act on the rod in clockwise sense. Hence the motion of the rod will be combined rotational and translational.

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Chapter 3: Rotational Dynamics

3.49

SOLUTION

Torque about hinged point is ⎛ l ⎞ mgl τ net = τ mg = mg ⎜ sin 30° ⎟ = ⎝2 ⎠ 4

According to Newton’s Second Law for translational motion, we have F = ma F …(1) m According to Newton’s Second Law for rotational motion, we have ⇒

a=

⎛ ml 2 ⎞ Since τ net = IO α = ⎜ α ⎝ 3 ⎟⎠

τ = I cm α ⇒

2 ⎛ l ⎞ ml F⎜ ⎟ = α ⎝ 4⎠ 12

⇒

α=

3F ml

⇒ …(2)

Point of zero acceleration on the rod should be between C and B, as the vectors due to acceleration due to translation and acceleration due to rotation act in opposite directions as shown in Figure. At point P, aP = 0 ⇒

acm − α x =

⇒

l x= 3

F ⎛ 3F ⎞ −⎜ ⎟x=0 m ⎝ ml ⎠

ILLUSTRATION 44

A rod of mass m and length l (hinged at O) is released from the rest from horizontal position as shown in Figure.

When the rod makes an angle of 30° with the vertical, calculate the angular acceleration, velocity of point P, acceleration of centre of mass and the horizontal component of force applied by hinge.

Mechanics II_Chapter 3_Part 2.indd 49

α=

mgl 4 3 g τ = = IO ml 2 3 4l

…(1)

By conservation of mechanical energy, the loss in gravitational potential energy of the centre of mass of the rod equals the gain in rotational kinetic energy of the rod. Loss of potential energy of the centre of mass of the rod is ⎛ − ΔU = mg ⎜ ⎝

l⎞ 3 ⎟⎠ cos 30° = mgl 4 2

Gain in rotational kinetic energy of the rod is KR =

1 1 ⎛ ml 2 ⎞ 2 IO ω 2 = ⎜ ⎟ω 2 2⎝ 3 ⎠

⇒

mgl 3 ⎛ ml 2 ⎞ 2 =⎜ ω ⎝ 6 ⎟⎠ 4

⇒

ω2 =

6g 3 4l

⇒

ω=

6 3g 4l

⇒

vp = lω =

6 3 gl 4

The tangential acceleration of the centre of mass of the rod is l ⎛ l ⎞ ⎛ 3g ⎞ 3g at = α = ⎜ ⎟ ⎜ = ⎝ 2 ⎠ ⎝ 4l ⎟⎠ 2 8

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3.50

JEE Advanced Physics: Mechanics – II

The centripetal acceleration of the centre of mass of the rod is ⎛ ac = rω 2 = ⎜ ⎝

l ⎞ ⎛ 6 3g ⎞ 3 3g ⎟⎜ ⎟= 2 ⎠ ⎝ 4l ⎠ 4

⇒

acm = ac2 + at2 =

− mg 3 ˆ 15mg ˆ i+ j 16 8

Comparing, we get V − mg =

So, total acceleration of centre of mass of rod is ⇒

− Hiˆ + ( V − mg ) ˆj =

g g 117 9 + 108 = 8 8

⇒

V=

15mg mg 3 , H= 8 16

mg 3 23 mg , H= 16 8

ILLUSTRATION 45

A uniform rod of mass m and length l which can rotate freely in vertical plane without friction, is hinged at its lower end on a table. A sphere of mass m and radius one third the length of the rod is placed in contact with the vertical rod and a horizontal force F is applied at the upper end of the rod as shown in Figure.

ac

 To find the x and y components of acm, we know that    acm = ac + at , where

(

 iˆ ac = ac − cos 60°iˆ + sin 60° ˆj

(

)

3 3g ˆ  ac = −i + 3 ˆj 8  andiˆ at = − at cos 30°iˆ + sin 30° ˆj ⇒

(

⇒

3g  at = − 16

(

)

3iˆ + ˆj

)

)

(

)

3g 3 3g ˆ  acm = −i + 3 ˆj − 8 16

(

)

(

3iˆ + ˆj

)

(

)

⇒

 1 1 acm = − ac + at 3 iˆ + ac 3 − at ˆj 2 2

⇒

9 3 g ˆ 15 g ˆ  acm = − i+ j 16 16

⇒

 243 + 225 g 117 acm = g = 256 8

If horizontal component of force applied by the hinge is H and the vertical component force applied by the hinge is V, then from Newton’s Second Law, we have   Fnet = macm ⇒

⎛ −9 3 ˆ 15 g ˆ ⎞ − Hiˆ + Vjˆ − mgj = m ⎜ i+ j⎟ ⎝ 16 8 ⎠

Mechanics II_Chapter 3_Part 2.indd 50

Calculate the acceleration of the sphere just after the force starts acting. Also, calculate the horizontal component of hinge reaction acting on the rod just after force F starts acting. SOLUTION

The forces acting on the rod and the sphere are shown in Figure.

aP

The torque equation for rod about O is ⎛ τ = Fl − N ⎜ ⎝ ⇒

Fl −

l⎞ ⎟ = IO α 3⎠

Nl ml 2 = α 3 3

2/9/2021 6:39:50 PM

Chapter 3: Rotational Dynamics

mlα N = F− …(1) 3 3 The motion of the sphere is translational only, so the acceleration ( a ) of the sphere equals the acceleration of point P, i.e. ⇒

lα 3 From equations (1) and (2), we get a=

ma = F −

N 3

For sphere, N = ma

…(2)

⇒

3.51

2 ⎛ L ⎞ 1 ⎛ mL ⎞ 2 mg ⎜ ⎟ = ⎜ ω ⎝ 2 ⎠ 2 ⎝ 3 ⎟⎠

3g L Since the centre of mass of rod is moving in a circular L path of radius , so we have 2 ⇒

ω=

…(3) …(4)

Substituting equation (3) in (4), we get a=

3F 3F and N = 4m 4

Acceleration of centre of mass of rod is 3 a 9F ⎛ l⎞ acm = ⎜ ⎟ α = = ⎝ 2⎠ 2 8m If the horizontal component of hinge reaction is N x , then for the rod we have 3F F− + N x = macm 4 3F ⎛ 9F ⎞ + Nx = m ⎜ ⎝ 8 m ⎟⎠ 4

⇒

F−

⇒

Nx =

9F F 7 F − = 8 4 8

T − mg = mω 2

L 2

⇒

⎛ L ⎞ ⎛ 3g ⎞ T = mg + m ⎜ ⎟ ⎜ ⎝ 2 ⎠ ⎝ L ⎟⎠

⇒

T = 4 mg

ILLUSTRATION 47

A light rod is connected rigidly with two identical particles each of mass m. The free end of the rod is smoothly pivoted at O. The rod is released from rest from its horizontal position at t = 0 as shown in Figure.

ILLUSTRATION 46

A rod of mass m and length l is hinged about its one of the ends. The rod is released from horizontal position. When the rod becomes vertical, find the angular speed of the rod and the hinge reaction SOLUTION

Applying conservation of mechanical energy, we see that the loss in gravitational potential energy equals gain in rotational kinetic energy.

Calculate the initial angular acceleration of the rod and the initial reaction offered by the pivot. SOLUTION

The net torque about O is

τ O = mgl + mg ( 2l ) = 3 mgl

This torque produces an angular acceleration α given by the relation τ α= IO 2

where, IO = ml 2 + m ( 2l ) = 5ml 2

Mechanics II_Chapter 3_Part 2.indd 51

2/9/2021 6:39:59 PM

3.52

JEE Advanced Physics: Mechanics – II

⇒

α=

3 mgl 5ml

2

=

3g 5l

Initially just when the rod is released, then ω = 0 but 3g α = , so 5l Rx = mlω 2 + m ( 2l ) ω 2 = 0 The various forces acting on the rod are shown in Figure.

The mass of the cord wound on the perimeter of the cylinder will be m ( l − x ) R2 l Let tension at the upper end of the cord in contact with the cylinder be T and its downward acceleration be a, then according to Newton’s Second Law, we have m ′′ =

⎛m ⎞ ⎛m ⎞ ⎜⎝ x ⎟⎠ g − T = ⎜⎝ x ⎟⎠ a l l

mg + mg − Ry = ma1 + ma2 where a1 = lα and a2 = 2lα ⇒

Ry = m ( 2 g − 3lα )

Substituting α =

3g , we get 5l

mg Ry = 5

When the cord accelerates downwards, the cylinder will be rotating with an angular acceleration a α= R The tension in the string provides necessary torque to the cylinder to rotate, so we have

τ = TR = Iα

I=

1 1 m MR2 + m ′′ R2 = MR2 + ( l − x ) R2 2 2 l

…(3)

Substituting equation (3) in (2), we get m ⎛1 ⎞ TR = ⎜ MR2 + ( l − x ) R2 ⎟ α ⎝2 ⎠ l ⎛ MR m ( ⎞ T=⎜ + l − x )R⎟ α ⎝ 2 ⎠ l Substituting equation (4) in (1), we get ⇒

…(4)

⎞ ⎛m ⎞ ⎛ m ⎞ ⎛ MR m ( + l − x ) R ⎟ α = ⎜ x ⎟ Rα ⎜⎝ xg ⎟⎠ − ⎜⎝ ⎠ ⎝ l ⎠ 2 l l

SOLUTION

At any instant, let m ′ be the mass of cord hanging from the cylinder as shown in Figure.

…(2)

where, I is the moment of inertia of the cylinder plus the part of cord wound on it, so

ILLUSTRATION 48

A uniform cylinder of radius R and mass M can rotate freely about a stationary horizontal axis O. A  thin cord of length l and mass m is wound on the cylinder in a single layer. Calculate the angular acceleration of the cylinder as a function of the length x of the hanging part of the cord. Assume that the wound part of cord has its centre of mass at the axis of the cylinder.

…(1)

⇒

α=

2mgx ( 2m + M ) Rl

ILLUSTRATION 49

A uniform solid cylinder A of mass m1 can freely rotate about a horizontal axis fixed to a mount B of mass m2. A constant horizontal force F is applied to end P of a light thread tightly wound on the cylinder as shown in Figure.

⇒

m′ =

m x l

Mechanics II_Chapter 3_Part 2.indd 52

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Chapter 3: Rotational Dynamics

The friction between the mount and the supporting horizontal surface is supposed to be absent. Calculate the acceleration of this point P and the kinetic energy of this system t seconds after beginning of motion.

3.53

ω0 R

SOLUTION

Applying Newton’s Second Law to the cylinder mount system, we get F = ( m1 + m2 ) a ⇒

a=

F m1 + m2

For cylinder, we have ⎛1 ⎞ τ = FR = Iα = ⎜ m1 R2 ⎟ α ⎝2 ⎠ ⇒

α=

The coefficient of kinetic friction between the corner walls and the cylinder is μ . Calculate the normal reaction imparted by the wall to the cylinder, by the ground to the cylinder. Also calculate the angular acceleration of the cylinder and the number of turns accomplish by the cylinder before stopping. SOLUTION

Let us draw the free body diagram for the situation shown.

2F m1 R

Acceleration of point P is aP = a + Rα ⇒

aP =

⇒

aP =

2F F + m1 + m2 m1 F ( 3 m1 + 2m2 ) m1 ( m1 + m2 )

In t seconds, displacement of point P is 1 s = aP t 2 2 The kinetic energy of system equals the work done by external force, so we get

Since the centre of mass of cylinder does not accelerate, so N 2 − μ N1 = 0 Also, N1 + μ N 2 − mg = 0 ⇒

⎡ 1 ⎛ F ( 3 m1 + 2m2 ) ⎞ 2 ⎤ K = F⎢ ⎜ ⎟t ⎥ ⎢⎣ 2 ⎝ m1 ( m1 + m2 ) ⎠ ⎥⎦

⇒

K=

F 2 t 2 ( 3 m1 + 2m2 ) 2m1 ( m1 + m2 )

ILLUSTRATION 50

A uniform cylinder of mass m , radius R is spun about its axis to the angular velocity ω 0 and then placed into a corner, as shown in Figure.

Mechanics II_Chapter 3_Part 2.indd 53

…(1)

N1 + μ N 2 = mg

…(2)

From (1) and (2), we get N1 =

ΔK = K − 0 = Fs ⇒

N 2 = μ N1

mg 1+ μ

2

and N 2 =

μ mg 1 + μ2

So, τ cm = ( μ N1 ) R + ( μ N 2 ) R ⇒

⎛ 1+ μ ⎞ τ cm = μ mgR ⎜ ⎝ 1 + μ 2 ⎟⎠

⇒

⎛ 1+ μ ⎞ ⎛1 2⎞ ⎜⎝ mR ⎟⎠ α = μ mgR ⎜ 2 ⎝ 1 + μ 2 ⎟⎠

⇒

⎛ 2μ g ⎞ ⎛ 1 + μ ⎞ α=⎜ ⎝ R ⎟⎠ ⎜⎝ 1 + μ 2 ⎟⎠

⇒

α=

2μ ( 1 + μ ) g

(1 + μ2 ) R

, clockwise sense

2/9/2021 6:40:13 PM

3.54

JEE Advanced Physics: Mechanics – II

Now n0 = Since, ω

2

θ 2π

− ω 02

SOLUTION

Initially 2Ti sin α = mg = 2αθ

⇒

⎛ −2 μ ( 1 + μ ) g ⎞ 0 2 − ω 02 = 2 ⎜ ⎟θ 2 ⎝ (1 + μ )R ⎠

⇒

θ=

⇒

n0

⇒

mg 2 sin α

Ti =

…(1)

When the string OB is cut, let the tension in string OA be T f , then equations of motion for centre of mass of rod are

( 1 + μ 2 ) Rω 02 4μ ( 1 + μ ) g

T f cos α

ax =

( 1 + μ 2 ) Rω 02 =

…(2)

m

8πμ ( 1 + μ ) g

O Tf sinα

GENERAL MOTION OF A RIGID BODY Till now, we have been learning ways to analyse the rotatory motion of a rigid body about a fixed axis of rotation. Now, let us describe the motion of a rigid body free to move in a plane. The basic laws of dynamics are applicable in this case also. You can use the following two results.   1. ∑ Fnet = macm   2. ∑ τ net = Iα Please note that the second result must be applied about an axis taken through the centre of mass. This is so as the points on the body are accelerating and thus non-inertial. Although centre of mass also accelerates but the torque due to pseudo force about the centre of mass will be zero

Tf

α

A

x

C α

B

Tf cosα

y

mg

mg − T f sin α

ay =

…(3)

m

If α be the angular acceleration of the rod about its centre of mass C, then

α=

( Tf sin α ) l = 3Tf sin α

…(4) 2 ml m ( 2l ) 12 At the instant OB is cut, point A moves perpendicular to OA, so acceleration of point A along AO should be zero, so ax cos α + lα sin α = ay sin α

ILLUSTRATION 51

A uniform rod AB of mass m and length 2l is suspended by two strings OA and OB of equal length attached to a fixed point O. The rod is at rest in a horizontal position and each string makes an angle α with the horizontal. If the string OB is cut, in what ratio the tension in OA will be instantaneously reduced?

Substituting the values, we get T f cos 2 α m

+

3T f sin 2 α m

=

( mg − Tf sin α ) sin α m

α

α

O

ax ay

A

Mechanics II_Chapter 3_Part 2.indd 54

α

α

Solving this equation, we get B

Tf =

mg sin α 2

cos α + 4 sin 2 α

…(5)

2/9/2021 6:40:20 PM

Chapter 3: Rotational Dynamics

From equations (1) and (5), we get Tf Ti

=

2 sin 2 α cos 2 α + 4 sin 2 α

=

3.55

ILLUSTRATION 53

2 sin 2 α 1 + 3 sin 2 α

ILLUSTRATION 52

A uniform rod of mass m and length l is acted upon by the forces F and 2F as shown in Figure.

The uniform 50 kg pole ABC is balanced in the vertical position. A 500 N horizontal force is suddenly applied at B. If the coefficient of kinetic friction between the pole and the ground is 0.3, determine the initial acceleration of point A. Take g = 10 ms −2. A 4m

500 N

B 2m

C

SOLUTION

N = mg = 500

Calculate the linear and angular acceleration of the rod. Also find the value of x for which the point P does not accelerate. ⇒

SOLUTION

Applying Newton’s Second Law for translational motion, we get 2F + F = ma ⇒

a=

500 − μ N = max

3F m

…(1)

Applying Newton’s Second Law for rotational motion to get the torque equation about centre of mass, we have

ax =

500 − μ N m

( 500 ) − ( 0.3 )( 500 )

= 7 ms −2 50 Now, let us calculate torque due to forces about the centre of mass or centre of gravity G of the rod, then ⇒

ax =

τ = ( 500 ) ( 1 ) − ( 0.3 )( 500 )( 3 ) ⇒

τ = 500 − 450 = 50 Nm A

⎛ ml 2 ⎞ ⎛ l⎞ 2 F ⎜ ⎟ − Fx = ⎜ α ⎝ 2⎠ ⎝ 12 ⎟⎠ ⇒

α=

12 F ( l − x ) ml

⇒

aP = 0 = −α x + a

⇒

αx = a

⇒

12 F ( l − x ) x ml x=

2

l 2

Mechanics II_Chapter 3_Part 2.indd 55

=

G B

…(2)

2

If point P does not accelerate, then we have     aP = aP , C + aC = 0

⇒

α

3F m

mg

500 N

N

μN

Since α =

C

τ I

50 1 ( 50 )( 6 )2 12 1 ⇒ α = rads −2 3 So, acceleration of point A is ⇒

α=

⎛ 1⎞ aA = ax − 3α = 7 − ( 3 ) ⎜ ⎟ = 6 ms −2 (to the right) ⎝ 3⎠

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3.56

JEE Advanced Physics: Mechanics – II

ILLUSTRATION 54

A hemisphere of radius r and mass M is pulled by means of a string (Figure) so that it moves with a uniform velocity. If μ is the coefficient of friction with the surface, find the angle of inclination of the hemisphere. Given that the centre of gravity of the 3r from centre. hemisphere lies at a distance 8

angle θ. Assuming that the rod has not started to slip, deduce an expression, in terms of θ , for the angular acceleration, and hence determine the reaction normal to the rod. Show that the rod begins to slip when 4μ tan θ = , where μ is the coefficient of friction. 13 SOLUTION

Let C be the centre of mass of rod then by Law of Conservation of Mechanical Energy, we have 1 I 0ω 2 = mg ( a sin θ ) 2

θ

SOLUTION

The hemisphere is moving with uniform velocity. Hence, it is in equilibrium, both rotational as well as translational. So, N = mg

…(1)

F = μN

…(2)

⇒

2 ⎞ 1 ⎛ m ( 4a ) + ma 2 ⎟ ω 2 = mga sin θ ⎜⎝ ⎠ 2 12

⇒

7 2 aω = g sin θ 6

⇒

ω=

6 g sin θ 7a a O

C

ω

3a

θ

a

O

C mg

O

mg

μN

N

For rotational equilibrium about the bottom-most point,

∑τ ⇒

bottom-most point

=0

⎛ 3r ⎞ F ( r − r sin θ ) = mg ⎜ sin θ ⎟ ⎝ 8 ⎠

τ mga cos θ 3 g cos θ = = 7 I 7a ma 2 3 Applying Newton’s Second Law for translation motion along y-axis, we get Since, α =

θ

G

ΣFy = may ⇒

…(3)

mg cos θ − N = mat

θ

Solving equations (1), (2) and (3), we get sin θ =

N

μN O

8μ 3 + 8μ

ILLUSTRATION 55

A rough uniform rod, of mass m and length 4a, is held on a horizontal table perpendicularly to an edge of the table, with a length 3a projecting horizontally over the edge. If the rod is released from rest and allowed to turn about the edge without slipping then find its angular speed after turning through an

Mechanics II_Chapter 3_Part 2.indd 56

an at

θ

C

y

mg x

Since at = rα = aα ⇒

N = mg cos θ − mat = mg cos θ − m ( aα )

⇒

N = mg cos θ −

3 mg cos θ 4 = mg cos θ 7 7

Rod begins to slip when

μ N − mg sin θ = man

2/9/2021 6:40:32 PM

Chapter 3: Rotational Dynamics

The acceleration of point A on the rod will be same as the acceleration of the cart, so we have    aA = aAC + aC

Since an = rω 2 = aω 2 ⇒ ⇒ ⇒ ⇒

3.57

4 μ mg cos θ − mg sin θ = m ( aω 2 ) 7 6 mg sin θ 4 μ mg cos θ − mg sin θ = 7 7 4 13 μ mg cos θ = mg sin θ 7 7 4μ tan θ = 13

⇒ ⇒

αL −a 2 Lα a′ = a − 2

− a′ =

…(4)

Solving equations (1), (2), (3), and (4), we get a=

ILLUSTRATION 56

A uniform slender bar AB of mass m is suspended as shown from a small cart of the small cart of the same mass m. Neglecting the effect of friction, determine the accelerations of points A and B immediately after a horizontal force F has been applied at B.

7F 18 F ,α= 5m 5mL

⇒

a ′ = aA = a −

⇒

aB = a +

Lα 2F =− 2 5m

Lα 16 F = 2 5m

CONCEPT OF TOPPLING In our practical life, we have observed that when a force F is applied on a block A of smaller width and greater height, it is more likely to topple down before sliding in comparison to another block B of broader base/bigger width and lower height where the chances of sliding are more compared to toppling.

SOLUTION

Let the acceleration of bar be a, cart be a ′ and α be the angular acceleration of the bar, then

F A

F

B

So, let us analyse these situations with the help of examples using basic concepts of torque. For this let us consider a block PQRS having centre of mass at C. N S

for the bar, we have F − F ′ = ma

…(1)

for the cart, we have F ′ = ma ′

Applying the torque equation for rod about the centre of mass, we get

⇒

L ML2 = α 2 12

F + F′ =

MLα 6

Mechanics II_Chapter 3_Part 2.indd 57

C

…(3)

F h

f P

…(2)

( F + F′ )

R

Q mg

Let a force F be applied normally to face QR at height h above ground. Let the frictional force f be sufficient enough to prevent sliding. Now, if we say that the normal reaction also passes through C, then we have to face a problem that in spite of the fact that translational equilibrium exists i.e., F = f and N = mg, we have an unbalanced CW torque due to

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JEE Advanced Physics: Mechanics – II

F about the point Q. This unbalanced CW torque has a tendency to topple the block about point Q. So, to cancel the effect of unbalanced torque due to F about the point Q, the normal reaction N shifts towards the right, say by l, such that

N F C

( mg ) l = Fh

f mg

N S

R C

F



h

f 2

P

Q

mg

Now, if F or h (or both) are increased then correspondingly l also increases, however cannot go beyond the right side of the block ( QR ). So, in the extreme case the normal reaction passes through Q and beyond this the block will topple down. N S C P

A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If μ is the coefficient of friction, then under what conditions the cylinder will (a) slide before toppling (b) topple before sliding. SOLUTION

(a) The cylinder will slide if

R

F h

f

ILLUSTRATION 57

mg sin θ > μ mg cos θ ⇒ tan θ > μ

…(1)

Q mg

Now, if F or h (or both) are further increased, then the block will topple down. It is due to this reason, that the block with broader base has less chances of toppling in comparison to a block of smaller base because the larger base has more margin to accommodate the shifting of the normal reaction.

Conceptual Note(s) Why rolling is easy in comparison to sliding? From arguments and reasons discussed, we observe that in case of a rolling body the normal reaction has a zero margin to shift. So, even if the body is in translational equilibrium ( F = f and N = mg ), still an unbalanced torque (due to F and f ) is left behind that rotates the body in clockwise sense. The instant the body starts rolling, the force of friction gets adjusted in magnitude and direction such that pure rolling starts (if it is sufficient enough) or the body starts sliding if it is not sufficient enough).

Mechanics II_Chapter 3_Part 2.indd 58

The cylinder will topple if h

( mg sin θ ) 2 > ( mg cos θ ) r 2r …(2) h Thus, the condition of sliding is tan θ > μ and 2r condition of toppling is tan θ > . Hence, the h cylinder will slide before toppling if 2r μ< h (b) The cylinder will topple before sliding if ⇒ tan θ >

μ>

2r h

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Chapter 3: Rotational Dynamics ILLUSTRATION 58

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly 3a above the above the centre of the face, at a height 4 base. What is the minimum value of F for which the cube begins to topple about an edge? SOLUTION

3.59

To balance the torque and force of gravitational pull, normal reaction should have magnitude equal to the weight of the block and effectively normal reaction should pass through the centre of the block as shown in figure (b). ILLUSTRATION 59

A uniform square block of side a and mass m is lying on a horizontal surface. A small block of mass m is placed on the top corner of the block. Find the distance by which normal reaction is shifted.

In the limiting case normal reaction will pass through O. The cube will topple about O if torque of F exceeds the torque of mg. N a 2 mg

⇒

⎛ 3a ⎞ ⎛ a⎞ F ⎜ ⎟ > mg ⎜ ⎟ ⎝ 4 ⎠ ⎝ 2⎠

⇒

F>

3a 4

F

SOLUTION

For translational equilibrium of the system, we have N1 = 2mg

O

2 mg 3

So, the minimum value of F is

2 mg 3

SHIFTING OF NORMAL REACTION AND TOPPLING When a surface comes into contact with other surface then one surface applies normal reaction on other surface. The normal reaction does not act only at one point but it is distributed over the whole surface in contact. Consider a uniform block lying on a horizontal surface. Then the normal reaction is distributed uniformly over the surface as shown in figure  (a). Now as the body is in translatory and rotatory equilibrium thus the net force and torque on the body should be zero.

For rotational equilibrium of the system, we have a⎞ ⎛ τ 0 = ⎜ N1x − mg ⎟ = 0 ⎝ 2⎠ ⇒

a⎞ ⎛ ⎜⎝ mg ⎟⎠ a 2 = x= 2mg 4

Here the normal reaction is still distributed but now its distribution is non uniform as shown in Figure.

ILLUSTRATION 60

A cubical block of side a is held at rest, against a rough vertical wall by applying a force F acting along the centre. The mass of the block is m. Taking acceleration due to gravity as g, calculate the

Mechanics II_Chapter 3_Part 2.indd 59

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(a) minimum coefficient of friction between block and the wall. (b) torque by normal reaction about centre of mass. SOLUTION

of normal has a limit. It cannot shift beyond point A. Now if F is further increased then normal cannot shift further to balance the torque. Hence now the block will start rotating about point A. This is called toppling.

The situation is shown in the Figure. fs

ILLUSTRATION 61

Let C be the centre of mass of the block. For equilibrium of the block, we have ΣFx = 0

⇒

F−N =0

…(1)

ΣFy = 0

⇒

f s = mg

…(2)

Στ = 0

⇒

⎛ a⎞ − fs ⎜ ⎟ + τ N = 0 ⎝ 2⎠

…(3)

A square plate of side a and mass m is lying on a horizontal floor. A force F is applied at the top. Find the maximum force that can be applied on the square plate so that the plate does not topple about A. SOLUTION

For translational equilibrium of the system, we have N = mg f =F

While writing equation (3), we have taken clockwise torque to be negative. Please note that the normal reaction N due to the wall on the block does not pass through the centre C. ⇒

N = F and f s = mg

(a) Since f s ≤ μ N ⇒ mg ≤ μ F mg ⇒ μ≥ F ⎛ (b) − f s ⎜ ⎝

a⎞ ⎟ + τN = 0 2⎠

τ N = fs ×

a a = mg (in counter clockwise sense) 2 2

TOPPLING Consider a square plate shown in figure. A force is applied at the top as shown in figure. Now normal reaction will shift to balance the torque produced by F and friction about centre of mass. Now if force F increases then normal will shift further but shifting

Mechanics II_Chapter 3_Part 2.indd 60

For rotational equilibrium of the system, we have

⇒ ⇒

a a a τ0 = 0 = F + f − N 2 2 2 a Fa = mg 2 mg F= 2

We can also think that when the body is in complete equilibrium, then net torque will be zero about any point taken. So net torque about the point A is zero. ⇒

Fa = mg

⇒

F=

a 2

mg 2

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Chapter 3: Rotational Dynamics ILLUSTRATION 62

To avoid sliding, F < μ mg

A force F is applied at the top most point of a cylinder of radius r and height h as shown in Figure.

⇒

3.61

mgr < F < μ mg h r ⇒ τ mg ⇒ Fh > mgr ⇒ F>

mgr h

(about centre)

{∵

fr = F }

If a string is given an angular displacement θ (measured in radian) then the torsion produced tries to restore it to its original configuration and this torque (torsion produced) is directly proportional to θ. Mathematically,

τ ∝θ ⇒

τ = −Cθ

where, C is called the Torsional Constant or the String Constant of the string. The negative sign shows that the torque is restoring in nature. If dW is the infinitesimal work done to give the string an angular displacement θ , then dW = τ dθ cos ( 180° ) = −τ dθ ⇒

W=

∫ dW = 2 C (θ 1

2 2

− θ12

)

If θ1 = 0 and θ 2 = θ , then W=

Mechanics II_Chapter 3_Part 2.indd 61

1 2 Cθ 2

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JEE Advanced Physics: Mechanics – II

ILLUSTRATION 63

ILLUSTRATION 64

A uniform cylinder of mass M and radius R, initially at rest is mounted so as to rotate freely about a horizontal axis that passes through the central axis of the cylinder. A constant force F acts on the cylinder as shown in Figure.

A motor rotates a pulley of radius 25 cm at 20 rpm. A rope around the pulley lifts a 50 kg block, as shown in Figure. What is the power output of the motor? SOLUTION

The tension in the rope is equal to the weight since there is no acceleration. Thus, T = 500 N. R

T

Calculate the angular velocity of the cylinder as it rotates by an angle of π . SOLUTION

∫

Work done by a constant force F is W = τ dθ , where

τ = Fr⊥ = FR ⇒

W=

∫

π

∫

⇒ ⇒

0

⇒

ωf =

2 FRπ = I

{cylinder was initially at rest} 2 FRπ MR2 2

=

4 FRπ MR2

4 Fπ MR

POWER The rotational power or simply the Power is defined as the rate at which work is done by a torque. The SI unit of power is watt and here too 1 W = 1 Js −1 W ⎛θ⎞ = τ ⎜ ⎟ = τω av ⎝ t⎠ t  dW  dθ   Instantaneous Power Pins = =τ⋅ = τ ⋅ω dt dt Average Power Pav =

Mechanics II_Chapter 3_Part 2.indd 62

2π N 60

2π ( 20 ) 2π = rads −1 60 3 The power required is ⎛ 2π ⎞ P = τω = 125 Nm −1 ⎜ rads −1 ⎟ = 260 W ⎝ 3 ⎠

(

1 2 1 2 Iω f − Iω i 2 2

1 2 Iω f = FRπ 2

ωf =

Angular velocity, ω =

ω=

FRdθ = FR dθ = FRπ

According to Work-Energy theorem, we have W=

Therefore, τ = TR = ( 500 )( 0.25 ) = 125 Nm .

)

WORK-ENERGY PRINCIPLE According to this theorem, work done by a torque equals the change in rotational kinetic energy. 1 2 1 2 Iω − Iω 0 2 2 So, in complete analogue to the Work Energy Theorem studied in Translational Motion, we can say that the net rotational work done by the forces is equal to the change in rotational kinetic energy of the body. i.e.

W = τθ =

Wrot = ΔKrot For a rolling body, we have, work done equals the change in total energy possessed by the rolling body. So, 1 ⎛1 Wtotal = ΔK = ⎜ mv 2 − mu2 2 ⎝2

k2 ⎞⎛ ⎟ ⎜⎜ 1 + 2 R ⎠⎝

⎞ ⎟⎟ ⎠

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Chapter 3: Rotational Dynamics

Further,

MODIFIED NEWTON’S SECOND LAW FOR FIXED AXIS ROTATION

mg − T = ma

Since torque is a rotational analogue of force, therefore, Newton’s Second Law for rotational motion is given by

Ia

⇒

mg −

⇒

⎛ I ⎞ mg = ⎜ 2 + m ⎟ a ⎝R ⎠

It is valid in two situations.

⇒

a=

(a) The axis is fixed in position and direction. (b) The axis passes through the centre of mass and is fixed in direction only.

Put this value of a in (1), we get

τ net

3.63

⎛a ⎞ = I cm α = I cm ⎜ cm ⎟ ⎝ R ⎠

T=

The equation τ cm = I cmα cm is valid even if the centre of mass is accelerating.

= ma

R2

g I ⎞ ⎛ ⎜⎝ 1 + ⎟ mR2 ⎠ mg ⎛ mR ⎞ ⎜⎝ 1 + ⎟ I ⎠ 2

=

=

g M⎞ ⎛ ⎜⎝ 1 + ⎟ 2m ⎠ mg 2m ⎞ ⎛ ⎜⎝ 1 + ⎟ M⎠

Special Case If both the cylinder and the point mass possess equal mass, then m = M and hence,

MOTION OF A POINT MASS ATTACHED TO A CYLINDER WITH A THREAD Consider a point mass m attached to a thread wound over a cylinder of radius R, mass M, moment of ⎛ 1 ⎞ inertia I ⎜ = MR2 ⎟ and radius of gyration k. The ⎝ 2 ⎠

ATWOOD’S MACHINE

point mass ascends down with acceleration a and the cylinder rotates while the thread unwinds. The tension in the thread provides necessary torque to the cylinder to rotate with angular acceleration α . So,

The free body diagrams of the pulley and the blocks are shown in Figure. Note that tensions on two sides of the pulley are different. Why? Applying Newton’s Second Law on the pulley, we get

T=

α= a R R

M M, R

a

T

mg Cross Sectional View

T2

⇒

⎛ a⎞ TR = I ⎜ ⎟ ⎝ R⎠

⇒

T=

Ia R2

Mechanics II_Chapter 3_Part 2.indd 63

T1

m2g

a1 m1g

Therefore, ( T1 − T2 ) R = Iα

Further τ = Iα TR = Iα

T1 a2

τ = TR sin 90 = TR ⇒

T2

α

m

m Front view

τ = T1 R − T2 R = ( T1 − T2 ) R Since, τ = Iα

T

AOR (⊥ to plane of page)

mg 2g and a = 3 3

{

∵α =

a R

}

…(1)

⇒

T1 − T2 =

Ia

R2 Applying Newton’s Law on the blocks, we get

…(1)

T2 − m2 g = m2 a

…(2)

m1 g − T1 = m1 a

…(3)

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JEE Advanced Physics: Mechanics – II

Solving equations (1), (2), (3), we get a=

⎛ m1 − m2 ⎜ I ⎜⎝ m1 + m2 + 2 R

Mg − T = Ma

⎞ g ⎟ ⎟⎠

1 MR2 ⎯⎯ →0 2

a=

Iα 1 where I = MR2 R 2

T=

MRα 2 Condition of no slipping, a = α r

For pulley (generally assumed to be a disc or a cylinder)

⇒

⇒ ⇒

⎛ m − m2 ⎞ g a≅⎜ 1 ⎝ m1 + m2 ⎟⎠

I=

1 MR2 2

T=

…(2) …(3)

Solving equation (1), (2) and (3) we obtain a=

2g Mg and T = 3 3

ILLUSTRATION 66

⎛ m1 − m2 ⎞ g ⎜ M⎟ ⎜⎝ m1 + m2 + ⎟ 2 ⎠

APPLICATION OF NEWTON’S SECOND LAW IN ROLLING MOTION (a) Ignoring Rotation, write Fnet = Macm for the object as if it were a point−mass. (b) Ignoring Translation, write τ = Icmα as if the object were only rotating about the centre of mass. (c) Use of no slip condition i.e. vcm = Rω or acm = Rα (d) Solve the resulting equations simultaneously for any unknown.

A thread is wound around two discs on either side. The pulley and the two discs have the same mass and radius. There is no slipping at the pulley and no friction at the hinge. Find out the accelerations of the two discs and the angular acceleration of the pulley.

1

2

SOLUTION

Let R be the radius of the discs and T1 and T2 be the tensions in the left and right segments of the rope. T1

ILLUSTRATION 65

T2

α1

A solid cylinder of mass M has a string wrapped several times around its circumference. The free end of string is attached to the ceiling and the cylinder is released from rest. Find the acceleration of the cylinder and the tension in the string.

a1

R

Mg

mg

α1

mg

a2

Acceleration of disc 1, a1 =

Mechanics II_Chapter 3_Part 2.indd 64

…(1)

Also, τ = TR = Iα

If the pulley had no mass, then I=

SOLUTION

mg − T1 m

…(1)

Acceleration of disc 2, a2 =

mg − T2 m

…(2)

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Chapter 3: Rotational Dynamics

Angular acceleration of disc 1, 2T TR τ = 1 = 1 1 I mR2 mR 2 Similarly, angular acceleration of disc 2,

α1 =

2T α2 = 2 mR

3.65

SOLUTION

…(3)

The forces acting on the system are shown in Figure.

…(4)

T

T

T

T R

M

where, both α 1 and α 2 are clockwise.

Mg

α

Applying Newton’s Second Law for rotational motion, we get 2 ( TR ) = Iα T1

T1

⇒

Angular acceleration of pulley,

α=

( T2 − T1 ) R = 2 ( T2 − T1 )

1 mR mR2 2 For no slipping, Rα 1 − a1 = a2 − Rα 2 = Rα

…(5) …(6)

Solving these equations, we get

α = 0 and a1 = a2 =

2g 3

2T =

…(1) R2 Applying Newton’s Second Law for translational motion, we get

⇒ ⇒

Mg − 2T = Ma Ia Mg − 2 = Ma R Mg a= I ⎞ ⎛ ⎜⎝ M + 2 ⎟⎠ R

1 MR2 2 Mg 2g ⇒ a= = M 3 M+ 2 Put this value of a in (1), we get

Since, I =

OBSERVATION Since, both the discs are in identical situation, T1 = T2 and α = 0, i.e., each of the discs falls independently and identically.

2T = ⇒ ILLUSTRATION 67

Find the acceleration and tension in the system shown. Assume that the cylinder remains straight and the strings are same.

R M

Mechanics II_Chapter 3_Part 3.indd 65

Ia

⇒

I ⎛ 2g ⎞ ⎜ ⎟ R2 ⎝ 3 ⎠

M ⎛ 2g ⎞ ⎜ ⎟ 2 ⎝ 3 ⎠ Mg T= 6 2T =

{

∵I=

1 MR2 2

}

ILLUSTRATION 68

The arrangement shown in Figure consists of two identical uniform solid cylinders, each of mass M, radius R on which two light threads are wound symmetrically. Find the tension of each thread in the process of motion. The friction in the axle of the upper cylinder is assumed to be absent.

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3.66 JEE Advanced Physics: Mechanics – II

⇒

Applying Newton’s Second Law for rotational motion, we get

R

M

A = Rα + Rα = 2 a = 2Rα

τ = ( 2T ) R = Iα ⇒

M

SOLUTION

If a be the acceleration at which the thread from the upper cylinder is released and A be the total acceleration of the lower cylinder, then n at ⎞ ⎛ Acceleration ⎞ ⎛ Acceleration of CM ⎟ ⎜ which thread ⎟ ⎜ A=⎜ + of lower ⎟ ⎜ is being released ⎟ ⎜ cylinder ⎟ ⎜ by upper cylinder ⎟ ⎝ ⎠ ⎝ ⎠

α=

a R

R

M

2T =

Ia

…(1) R2 Applying Newton’s Second Law for translational motion, we get

R

T

T

T

T

Mg − 2T = M ( A ) = M ( 2 a ) Ia

⇒

Mg −

⇒

Mg −

⇒

M⎞ ⎛ Mg = ⎜ 2 M + ⎟ a ⎝ 2 ⎠

⇒

a=

⇒

A = 2a =

⇒

T=

R2

= 2 Ma

M a = 2 Ma 2

2g 5 4g M ⎛ 2g ⎞ and 2T = ⎜ ⎟ 5 2 ⎝ 5 ⎠

Mg 10

R M Mg

Test Your Concepts-IV

Based on Torque and Applications 1.

2.

A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the 4 vertical. Given, cos ( 37° ) = and g = 10 ms −2 . 5 A thin horizontal uniform rod AB of mass m and length  can rotate freely about a vertical fixed axis passing through its end A . At a certain moment the end B starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed

Mechanics II_Chapter 3_Part 3.indd 66

3.

(Solutions on page H.155) in the horizontal plane. Find the angular velocity of the rod as a function of its rotation angle ϕ counted relative to the initial position. The system in Figure is released from rest. The 30 kg body is 2 m above the floor and is connected through an ideal string passing over the pulley (a uniform disk with a radius of 10 cm and mass 5 kg ) to another body of mass 20 kg. Find the speed of the 30 kg body just before it hits the floor and the angular speed of the pulley at that time, the tensions in the strings and the time it takes for the 30 kg body to reach the floor.

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Chapter 3: Rotational Dynamics

7. 30 kg 2m

20 kg

4.

A slender rod of mass m and length  is released from rest at θ = 0°. Calculate the normal and frictional forces, which are exerted on the rod by the ground as it falls downward as a function of θ . Is it possible for the rod to slip as it falls? Give reason in support of your answer.

8.

3.67

A uniform solid cylinder of mass m and radius a is free to rotate about its axis which is smooth and vertical. A light inextensible string is wrapped around the cylinder and its free end is pulled horizontally with a constant force 2mg. Calculate the angular velocity of the cylinder when the free end of the string has moved through a distance 4a. A uniform cylinder of mass m1 and radius R is pivoted on frictionless bearings. A string wrapped around the cylinder connects to a mass m2, which is on a frictionless incline of angle θ , as shown in Figure.

m2

θ θ

5.

In the arrangement shown in Figure the mass of the uniform solid cylinder of radius R is equal to m and the masses of two bodies are equal to m1 and m2. The thread slipping and the friction in the axle of the cylinder are supposed to be absent. Find the angular acceleration of the cylinder and T the ratio of tensions 1 of the vertical sections of T2 the thread in the process of motion.

9.

This system is released from rest with m2 at a height h above the bottom of the incline. Find the acceleration of m2, tension in the string, total energy of the system when m2 is at height h, total energy when m2 reaches bottom of the incline with speed v and the speed v. A solid hemisphere rests on a truck as shown in Figure. If the coefficient of friction μ is just sufficient to prevent slipping, calculate the acceleration of the truck.

1 2

m1 m2

6.

A horizontal force F is applied to the sphere shown in Figure. Calculate the (a) value of F required to hold the sphere in equilibrium. (b) frictional force of the incline on the sphere. F

R

Truck

10. A uniform rod of mass m and length  is held horizontally by two vertical strings of negligible mass, as shown in Figure. Immediately after the right string is cut, find the

M

θ

Mechanics II_Chapter 3_Part 3.indd 67

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JEE Advanced Physics: Mechanics – II

(a) linear acceleration of the free end of the rod. (b) linear acceleration of the centre of mass of rod. (c) tension in the left string. 11. A uniform cylinder of mass M and radius R has a string wrapped around it. The string is held fixed and the cylinder falls vertically, as shown in Figure. Calculate the acceleration of the cylinder and tension in the string.

around the discs in opposite directions. Disc A is attached to the ceiling in such a way that it can rotate freely about its axis. The disc, B, initially held at same height as A, is then released to fall so that string unwinds from both the discs. Find the angular and linear accelerations of falling disc and tension in the string. Assume that string does not slip and motion is confined in the same vertical plane.

A

B

12. A uniform solid cylinder of mass m and radius a starts rotating from rest freely about its axis of symmetry under the action of a constant torque 4mga. Find the angular velocity of the cylinder at t = 4 s from the start. 13. A cord is wrapped around the rim of a solid cylinder of radius 0.25 m and a constant force of 40 N is exerted on the cord as shown in Figure.

15. A ring of mass m and radius r has a particle of mass m attached to it at a point A. The ring can rotate about a smooth horizontal axis which is tangential to the ring at a point B diametrically opposite to A. The ring is released from rest when AB is horizontal. Find the angular velocity and the angular acceleration of the body when AB π has turned through an angle . 3

B 40 N

40 N

If the cylinder is mounted on frictionless bearings and moment of inertia of cylinder is 4 kgm2 , then (a) use the work-energy principle to calculate the angular velocity of the cylinder after 5 m of cord have been unwound. (b) if 40 N force is replaced by a 40 N weight, calculate the angular velocity of cylinder after 5 m of cord has unwound. Take g = 10 ms −2. 14. Two identical uniform discs A and B each of mass m and radius R are held, as shown in Figure with the help of a long massless string which is wrapped

Mechanics II_Chapter 3_Part 3.indd 68

A

16. A uniform slender bar is released from rest in the horizontal position as shown in Figure. For what value of x the angular acceleration is maximum? Determine the corresponding angular acceleration α . x

 G

O

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Chapter 3: Rotational Dynamics

3.69

ACCELERATED PURE ROLLING

Now three cases arise

Till now we had been discussing uniform pure rolling where v and ω are constants. However, if an external force is applied to a rigid body, then the motion no longer is uniform pure rolling. Now for pure rolling on a stationary rough ground we must have

(i) If a = Rα , then no need of friction and f = 0. (ii) If a < Rα , then to support the linear motion and to oppose the angular motion, friction f acts forwards (along a). (iii) If a > Rα , then to support the angular motion and to oppose the linear motion, friction f acts backwards (opposite to a).

v = Rω

…(1)

Taking derivative on both sides of (1) w.r.t. time t, we get dv ⎛ dω ⎞ = R⎜ ⎝ dt ⎟⎠ dt ⇒

F + f = Ma

a = Rα

…(2)

So, in addition to v = Rω to be obeyed at every instant of time, the relation a = Rα is also obeyed for pure rolling to take place and this is called Accelerated Pure Rolling.

Problem Solving Technique(s) Here friction plays an important role in maintaining the pure rolling. The friction sometimes may act backward and sometimes may act forward or under certain conditions may also be zero, because the basic nature of friction is to self-adjust (up to a certain maximum limit) and it has a tendency to stop the relative motion between the two surfaces / bodies in contact. Let us now have a more clear understanding of this concept by applying a force F to the top most point of a rigid body of mass M, radius R and moment of inertia I about an axis passing through its centre of mass, as shown in Figure. F C

α C

a f

Now, this applied force F produces in the body a/an (a) linear acceleration ( a ) and (b) angular acceleration ( α )

Mechanics II_Chapter 3_Part 3.indd 69

So, we observe that the friction f can be zero, forward or backward depending upon value of I, M and R. Applying Newton’s Second Law for linear motion, we get …(3)

Applying Newton’s Second Law for rotational motion, we get

τ = Iα ⇒

FR ( CW ) + fR ( CCW ) = Iα

⇒

FR − fR = Iα

…(4)

For pure rolling to take place, we have a = Rα ⇒

⎛ a⎞ FR − fR = I ⎜ ⎟ ⎝ R⎠

⇒

⎛ I ⎞ F− f =⎜ 2 ⎟a ⎝R ⎠

…(5)

From (3) and (5), we get ⎛ MR2 − I ⎞ f =⎜ F ⎝ MR2 + I ⎟⎠

…(6)

From (6), we observe that if we have (i) I = MR2, i.e., a force F is applied at top of a ring, then f = 0 and the ring will roll without slipping. (ii) I < MR2 i.e., a force F is applied at top of a solid sphere, a shell, a cylinder or a disc, then f is positive i.e., f acts forwards. (iii) If I > MR2 i.e., f acts backwards. Although for the axis passing through CM, we cannot have I > MR2. So, we conclude that f is either zero or forwards i.e., along F and so it supports linear motion.

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ILLUSTRATION 69

ILLUSTRATION 70

A force F is applied at centre of a uniform round object of mass m radius R and moment of inertia I about its centre of mass I cm. Find (if cm2 = k). mR

A force F is applied on a disc at its centre. Find acceleration of centre of mass in the case of pure rolling and also find minimum coefficient of friction required for pure rolling.

(a) Acceleration of centre of the round object if it rolls without slipping. (b) Minimum coefficient of friction required so that the round object rolls without slipping.

SOLUTION

Let the friction force acting on disc be f r, acceleration of centre of mass is a and angular acceleration is α . We have three unknowns

SOLUTION

Equation for translatory motion F − f = ma

…(1)

Equation for rotatory motion fR = I cm α

…(2)

Condition for no slipping a = Rα

…(3)

(2) α

(1) a

(3) f r

We require 3 equations to solve them Translatory Motion F − f r = ma

…(1)

Rotatory Motion rf r = By solving the equations F− a=

f =

I cm a R2

mR2 ( 1 + k )

Thus μmin mg = ⇒

a = rα

F F F = = I cm I cm ⎤ m ( 1 + k ) ⎡ m+ 2 m ⎢1+ R mR2 ⎥⎦ ⎣ =

kF 1+ k

⎛ k ⎞ F μmin = ⎜ ⎝ 1 + k ⎟⎠ mg

Mechanics II_Chapter 3_Part 3.indd 70

kF (1 + k )

…(2)

Condition of no slipping

= ma

I cm F

mr 2 α 2

…(3)

By solving (2) and (3) fr =

ma 3 ma , F= 2 2

m 2F 2F F , fr = × , fr = 3m 2 3m 3 Now, f r ≤ μ N ⇒

a=

⇒

F ≤ μ mg 3

⇒

μ≥

F 3 mg

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Chapter 3: Rotational Dynamics

Problem Solving Technique(s) (a) In accelerated pure rolling the velocity of the bottommost point is zero but despite the relation a = Rα , the acceleration of the bottommost point is not zero. Because acceleration of any point P can be given as    aP = aC + aPC  Here, aPC has two components: Tangential acceleration at = rα (which is perpendicular to CP ) and radial or normal acceleration an = rω 2 (which is along PC )

C

r P R

r

R

A

In case of pure rolling, problems can also be solved by using the energy conservation principle (provided no other dissipative forces are present). So, in this case we can also use the Law of Conservation of Energy to get ⎛ Decrease ⎞ ⎛ Increase ⎞ ⎛ Increase in ⎞ ⎜ in GPE ⎟ = ⎜ in KE ⎟ + ⎜ KE of Rolling ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ of A ⎠ ⎝ of A ⎠ ⎝ of Spool ⎠ (d) If aCM is acceleration of CM of spool, then acceleration of B is ⎛ Acceleration at ⎞ ⎛ Acceleration ⎞ ⎜ which thread is ⎟ of CM ⎟ + ⎜ aB = ⎜ ⎟ ⎜ of spool ⎟ ⎜ being released ⎟ ⎝ ⎠ ⎝ from spool ⎠ aB = aCM + rα A r

B m

If

A

P

v, a

v, a

   Since, aPC = aP − aC    ⇒ aP = aPC + aC     ⇒ aP = aC + ( aPC )t + ( aPC )n   For the bottommost point, aC + ( aPC )t = 0  Because aC = a {in forward direction}  and ( aPC )t = Rα {in backward direction}   and since, a = Rα therefore, aC + ( aPC )t = 0  But ( aPC )n ≠ 0 , because it is Rω 2 towards centre. Thus, acceleration of bottommost point is Rω 2 towards centre. (b) If acm is acceleration of CM of A then acceleration of B is also aCM = a.

3.71

aCM = a aB = a + rα

a

In this particular case, aB = −a + rα

B

(c) Similarly, in the problems like shown in Figure, it is wrong to say that acceleration of point P is equal to acceleration of block A . Although we can write,

{CM of spool moves backwards} (e) In cases where pulley is having some mass and friction is sufficient enough to prevent slipping, the tension on two sides of the pulley will be different and rotational motion of the pulley is also to be considered.

a A = a + rα

Mechanics II_Chapter 3_Part 3.indd 71

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(f) At a given instant the value of ω for a rigid body will be same for every point. (g) The torque equation ( τ = Iα ) can be applied only about two points. These are, (i) centre of mass (ii) point about which body is in pure rotation.

For no slipping condition, we have a1 = 2Rα Since 2R = 0.2 m ⇒

2R

SOLUTION

Let, a1 = acceleration of centre of mass of reel a2 = acceleration of 1 kg block α = angular acceleration of reel (clockwise) T = tension in the string and f = force of friction Free body diagram of reel is as shown where, only horizontal forces are taken. Equations of motion for reel are T − f = 3 a1

a1 = 0.27 ms −2 and a2 = 0.135 ms -2 ILLUSTRATION 72

A uniform solid cylinder of mass M and radius R rolls a rough inclined plane with its axis perpendicular to the line of greatest slope as shown in Figure.

As the cylinder rolls it winds up a light string which passes over a light and smooth pulley and attached to a mass m, the part of the string between pulley and cylinder being parallel to the line of greatest slope. Prove that the tension in the string is ⎛ ( 3 + 4 sin θ ) Mmg ⎞ T=⎜ ⎟⎠ ⎝ 3 M + 8m SOLUTION

If cylinder rolls down at acceleration a, mass m goes up at acceleration 2a so we use

…(1)

τ f ( 2R ) − TR 0.2 f − 0.1T f T = = − …(2) = I 0.6 3 6 I Free body diagram of mass is also shown here. α=

a1 T

T − mg = m ( 2 a )

…(1)

Mg sin θ − T − f = Ma

…(2)

1 ⎛ a⎞ MR2 ⎜ ⎟ ⎝ R⎠ 2

…(3)

fR − TR =

Adding equations (2) and (3), we get

T

α

Mg sin θ − 2T =

a2

3 Ma 2

2 4 T g sin θ − 3 3M From equation (1), we use ⇒

10 N

Equation of motion for mass is

Mechanics II_Chapter 3_Part 3.indd 72

…(5)

R

A

10 − T = a2

a2 = a1 − 0.1α

Solving equations (1), (2), (3), (4) and (5), we get

A thin light thread is wound on a reel of mass 3 kg and moment of inertia 0.6 kgm 2. The inner radius is R = 10 cm and peripheral radius is 2R = 20 cm. The reel is placed on a rough table and the friction is enough to prevent slipping. Find the acceleration of the centre of reel and of hanging mass of 1 kg.

f

…(4)

and a2 = a1 − Rα ⇒

ILLUSTRATION 71

a1 = 0.2α

…(3)

a=

4 T ⎞ ⎛2 T − mg = 2m ⎜ g sin θ − ⎟ ⎝3 3 M⎠

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Chapter 3: Rotational Dynamics

8m 4 mg sin θ − T 3 3M

⇒

T − mg =

⇒

8m ⎞ ⎛4 ⎞ ⎛ T ⎜ 1+ ⎟ = mg ⎜⎝ sin θ + 1 ⎟⎠ ⎝ 3M ⎠ 3

⇒

T=

⇒

I = Mk 2

3 M + 8m

BODY ROLLING WITHOUT SLIPPING ON AN INCLINED PLANE METHOD I: Using the concept of Newton’s Second Law (for translation and rotation motion) Consider a body of M radius R and moment of inertia I rolling without slipping on an inclined plane (making an angle θ with the horizontal). For the body to roll without slipping necessary friction ( f ) must be present. From the Figure we observe that all forces (other than friction) are acting along the radius and hence will not produce torque in the body.

⇒

a

f =

θ

Mg

However, friction being a tangential force will produce a torque τ given by

τ = fR

⇒

Iα = fR

a = fR R Ia ⇒ f = 2 R Further ⇒

I

Mg sin θ

…(3)

⎛ R2 ⎞ + 1 ⎜⎝ ⎟ k2 ⎠

⇒

( M g sin θ ) R = ( M R2 + M k 2 ) Ra

⇒

a=

⇒

a=

g sin θ R2 R2 + k 2 g sin θ 1+

⇒

f =

⇒

f =

Further

τ = Iα

…(2)

⎛ k2 ⎞ 1 + ⎜⎝ ⎟ R2 ⎠

τ = ( Mg sin θ ) R = I IAORα

τ = fR sin 90 ⇒

g sin θ

METHOD II: Using the concept of IAOR At the point of contact velocity is zero, so the net torque due to the various forces about the IAOR is

sθ

co

Mg

a=

Put (2) in (1), we get

N

θ

Mg sin θ I ⎞ ⎛ ⎜⎝ M + 2 ⎟⎠ R

If k is the radius of gyration of the rolling body, then

Mmg ( 4 sin θ + 3 )

sin Mg

a=

3.73

Ia R

2

k2 R2 =

( MR2 ) R

2

g sin θ ⎛ k2 ⎞ 1 + ⎟ ⎜⎝ R2 ⎠

Mg sin θ ⎛ R2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ k

If μ be the coefficient of friction, then …(1)

Mg sin θ − f = Ma Ia

⇒

Mg sin θ −

⇒

⎛ I ⎞ Mg sin θ = ⎜ 2 + M ⎟ a ⎝R ⎠

R2

Mechanics II_Chapter 3_Part 3.indd 73

= Ma

μ=

f , where N = Mg cos θ N

Using (3), we get

μ=

tan θ ⎛ R2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ k

…(4)

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JEE Advanced Physics: Mechanics – II

CONDITION FOR A BODY TO ROLL WITHOUT SLIPPING

For a body rolling on a surface,

For a body to roll without slipping, the force of friction f calculated above must be less than or equal to the maximum value of friction i.e. μ Mg cos θ . ⇒

⇒

Mg sin θ ⎛ R2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ k

μ≥

≤ μ Mg cos θ

(a) if the surface is smooth, then, a = g sinθ if surface is smooth (b) if pure rolling takes place, then friction must be sufficient enough to prevent slipping, i.e. μ > μmin then a=

tan θ ⎛ R2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ k

and f =

For various rolling bodies a, f and the condition to roll without slipping are shown in the table. From table, we conclude aSPHERE > aDISC = aCYLINDER > aSHELL > aRING Hence tSPHERE < tDISC = tCYLINDER < tSHELL < tRING i.e. the sphere reaches the first and the ring reaches the last down an incline (while rolling).

OBSERVATION

g sinθ g sinθ = Ι ⎞ ⎛ ⎛ k2 ⎞ ⎜⎝ 1+ ⎟ + 1 MR2 ⎠ ⎜⎝ R2 ⎟⎠ mg sinθ ⎛ R2 ⎞ ⎜ 1+ k 2 ⎟ ⎝ ⎠

(c) if the surface is rough, but friction is insufficient to prevent slipping i.e. μ < μmin then forward slipping will take place and maximum friction will act in this case and then the acceleration of the body is a = g sinθ − μ g cosθ The various parameters for bodies of different shapes are given in the Table.

M, R, I a

θ

Condition to Roll Without Slipping

a=

f =

g sin θ ⎛ k2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ R Mg sin θ ⎛ R2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ k

Condition to Roll without Slipping is ⎡ ⎛ R2 ⎞ ⎤ θ = tan ⎢ μ ⎜ 1 + 2 ⎟ ⎥ k ⎠⎦ ⎣ ⎝ −1

Mechanics II_Chapter 3_Part 3.indd 74

Ring  k2   2  1 R 

a=

f =

1 g sin θ 2

1 Mg sin θ 2

θ = tan −1 ( 2μ )

Disc  k2 1   2   2 R

a=

f =

2 g sin θ 3

1 Mg sin θ 3

θ = tan −1 ( 3 μ )

Cylinder  k2 1   2   2 R

a=

f =

2 g sin θ 3

1 Mg sin θ 3

θ = tan −1 ( 3 μ )

Shell  k2 2   2   3 R

a=

f =

3 g sin θ 5

2 Mg sin θ 5

⎛ 5μ ⎞ θ = tan −1 ⎜ ⎝ 2 ⎟⎠

Sphere  k2 2   2   5 R

a=

f =

5 g sin θ 7

2 Mg sin θ 7

⎛ 7μ ⎞ θ = tan −1 ⎜ ⎝ 2 ⎟⎠

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Chapter 3: Rotational Dynamics ILLUSTRATION 73

Angular acceleration of shell is

A shell is released from the top of an inclined plane of inclination θ . Calculate the minimum coefficient of friction between the shell and the plane to prevent sliding. Assuming the friction coefficient between the shell to be half the calculated minimum value, then calculate kinetic energy of the shell as it moves a length l down the incline.

1 fR 5 mgR sin θ 3 g sin θ α= = = 2 10 R I mR2 3 Time of sliding is t=

SOLUTION

For pure rolling of shell, we have mg sin θ − f = macm where, acm =

g sin θ

(

1 + k 2 R2

For hollow sphere

k2 R

2

=

2l = a

3.75

5l 2 g sin θ

Speed attained by shell as it travels a distance l of inclined is v = at

)

⇒ 2 , so we get 3

v=

8 gl sin θ 5

Angular speed attained by shall as it travels a distance l

ω = αt

⎛ g sin θ ⎞ mg sin θ − f = m ⎜ ⎝ 1 + ( 2 3 ) ⎟⎠ ⇒

2 f = mg sin θ 5

⇒

ω=

⇒

ω=

3 g sin θ 10 R

5l 2 g sin θ

6 gl sin θ 40 R2

Final kinetic energy of balls is K=

1 1 mv 2 + Iω 2 2 2 1 ⎛ 8 gl sin θ ⎞ 1 ⎛ 2 2 ⎞ ⎛ 9 gl sin θ ⎞ m⎜ ⎟⎠ + ⎜⎝ mR ⎟⎠ ⎜ ⎝ 40 R2 ⎟⎠ 2 ⎝ 5 2 3

For no sliding to take place, we must have f < fl ( = μ mg cos θ )

⇒

K=

⇒

2 mg sin θ < μ mg cos θ 5

⇒

⎛4 3 ⎞ K = ⎜ + ⎟ gl sin θ ⎝ 5 40 ⎠

⇒

μ>

⇒

K=

⇒

μmin =

2 tan θ 5 2 tan θ 5

ROLLING WHEEL ON A MOVABLE PLANK

μmin 1 = tan θ , then friction on shell is 2 5 1 f = μ mg cos θ = mg sin θ 5 Acceleration of shell is If μ =

a = g sin θ − μ g cos θ =

Mechanics II_Chapter 3_Part 3.indd 75

7 mgl sin θ 8

Till now we have seen round objects rolling on a fixed surface. Surfaces being either horizontal or inclined but fixed. But what if the surfaces themself move.

4 g sin θ 5

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JEE Advanced Physics: Mechanics – II

Consider a round body of radius R moving on a plank. Let the centre of the round object has velocity v0 and the round object has angular velocity ω . The plank is moving with a speed vp as shown. Now for pure rolling, the two points in contact should have same velocity. Thus

For plank, we have F − f = Ma1

…(2)

Applying torque equation on disc about its centre of mass, we get

τ = lα

v0 − Rω = vp and a0 − Rα = ap

mR2 α 2 For pure rolling, we have ⇒

where a0 is acceleration of centre, α is angular acceleration and ap is acceleration of plank. For forward slipping, we have

fR =

…(3)

a1 = a + Rα

…(4)

Solving the four equations, we get

v0 − Rω > vp a0 − Rα > ap

f =

if v0, ω and vp satisfy the relation v0 − Rω = vp.

2F Fm and α = ( m + 3M ) R m + 3M

ILLUSTRATION 75

For backward slipping, we have

In the arrangement shown a spool of mass m, moment of inertia I is placed on a plank of mass M. There is no friction between the plank and the surface inclined at an angle θ . If friction between spool and plank is sufficient enough to prevent slipping, find the angular acceleration of the spool and the value of

v0 − Rω < vp a0 − Rα < ap if v0, ω and vp satisfy the relation v0 − Rω = vp. ILLUSTRATION 74

Consider a disc of mass m and radius R placed on a rough plank of mass M which in turn is placed on a smooth horizontal surface.

Now plank is pulled by a force F and disc starts to roll on the plank. If there is no friction anywhere then calculate the force of friction acting on the disc. Also calculate the angular acceleration of the disc.

⎛ m⎞ for which plank will ratio ⎜ ⎝ M ⎟⎠

(i) ascend (ii) descend (iii) will remain stationary SOLUTION

SOLUTION

Let us apply Newton’s laws on disc in horizontal direction

The free body diagrams for spool and plank are shown here. According to Newton’s Second Law, we have a1

α C

m

B A

f

mgsinθ f

M

f = ma

Mechanics II_Chapter 3_Part 3.indd 76

…(1)

a2 mgsinθ

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Chapter 3: Rotational Dynamics

for m, mg sin θ + f − T = ma1

…(1)

for M, Mg sin θ − f = Ma2

…(2)

3.77

without slipping. Neglect friction between ground and wedge. Take g = 10 ms −2.

Taking moment of forces about centre of spool, we get T ( R ) − f ( 2R ) = Iα

…(3)

30°

For no slipping at point B, we have a1 − Rα = 0

…(4)

For no slipping at point A, we have a1 − 2Rα = a2

…(5)

From equations (4) and (5), we get a2 = − Rα

…(6)

Solving above equations, we get ⎛ ( m − M ) g sin θ ⎞ α=⎜ R ⎝ mR2 + MR2 + 1 ⎟⎠ CASE-1: ⎛ m⎞ 1 For m > M i.e., ⎜ ⎝ M ⎟⎠ α is positive, a2 is negative So, the plank will ascend. CASE-3: For wedge, the equations of motion are N ′ = N cos ( 30° ) + mg + f sin ( 30° )

RT (2R)T

N sin 30° − f cos 30° = ma

a1

m = 1, For m = M i.e., M α = 0, a2 = 0 So, the plank will remain stationary.

⇒

3f N − = 10 a 2 2

⇒

N − 3 f = 20 a

For sphere, the equations of motion are, ( m = 10 kg ) ΣFx = mar

ILLUSTRATION 76

A sphere of mass 10 kg is placed on the inclined surface of a rough wedge of inclination 30°. If the mass of the wedge is also 10 kg, calculate the acceleration of the wedge when sphere is allowed to roll down

Mechanics II_Chapter 3_Part 3.indd 77

…(1)

⇒

mg sin 30° + ma cos 30° − f = mar

⇒

50 + 5 3 a − f = 10 ar

…(2)

Also, ΣFy = 0

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JEE Advanced Physics: Mechanics – II

⇒

N + ma sin 30° = mg cos 30°

⇒

N + 5 a = 50 3

…(3)

fR f τ = = …(4) I 2 mR2 4 R 5 For no slipping between the wedge and the sphere, we have Further, α =

ar = Rα

…(5)

Since the string attached the mass m to the highest point of the cylinder, so vm = vCM + Rω Differentiating we get a2 = a1 + Rα

…(4)

For M to roll without slipping, we have a1 = Rα

…(5)

Solving equations (1), (2), (3), (4) and (5), we get

Solving equations (1), (2), (3), (4) and (5), we get a = 2.11 ms −2

a2 =

ILLUSTRATION 77

In the arrangement shown in Figure, the string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a light, frictionless pulley and is connected to a falling weight. Find the acceleration of the falling mass m in terms of only the mass of the cylinder M, the mass m and g. M

8 mg 3 M + 8m

ILLUSTRATION 78

A cylinder of mass M, radius R is sandwiched between two planks having masses M and 2M. Two constant horizontal forces F and 2F are applied on the planks as shown. Determine the acceleration of the centre of mass of cylinder, top plank and bottom plank if there is no slipping at the top and bottom of cylinder. Also find the friction between the planks and cylinder and the angular acceleration of the cylinder. M

F

m1

M R 2M

2F

SOLUTION

Let T be the tension in the string and f the force of (static) friction, between the cylinder and the surface. If a1 be the acceleration of centre of mass of cylinder towards right a2 be the downward acceleration of block m and α be the angular acceleration of cylinder (clockwise) α

T a1

smooth

SOLUTION M

α

a2

a2

f2 2F

f

A

f1

2M

f1

F

M, R

B

f2

a3

mg

For block, mg − T = ma2

…(1)

For cylinder, T + f = Ma1

…(2)

(T − f )R α=

…(3)

1 MR2 2

Mechanics II_Chapter 3_Part 3.indd 78

Equations of motion for plank M is, F + f1 = Ma1

…(1)

for cylinder M is, f1 + f 2 = Ma2

…(2)

for plank 2M is, 2 F − f 2 = 2 Ma3

…(3)

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Chapter 3: Rotational Dynamics

We have six unknowns, f1, f 2, a1, a2, a3 and α . Solving the above six equations, we get

Further for the cylinder, we have

α=

⇒

α=

( f1 − f 2 ) R 1 MR2 2

a1 =

2 ( f1 − f 2 )

For no slipping condition, at the points of contact A and B, we have a2 + Rα = − a1

…(5)

and a2 − Rα = a3

21F 23 F F , a2 = , a3 = 26 M 26 M 26 M

f1 = −

…(4)

MR

3.79

5F 3F 11F , f2 = and α = − 26 13 13 MR

Negative sign with f1 indicates its direction is opposite to the direction shown in Figure and negative sign with α shows it is in CW sense.

…(6)

Test Your Concepts-V

Based on Uniform and Accelerated Pure Rolling 1.

A heavy roll of wrapping paper of mass M, radius R, in the form of a solid cylinder is resting on a horizontal tabletop. If a horizontal force F is applied uniformly to the paper, as shown in Figure, find the linear acceleration of the centre of the roll and the angular acceleration around the centre of the roll. The coefficient of kinetic friction between the paper and the table is μK . Assume the roll of paper does slip on the surface.

3.

(Solutions on page H.159) In Figure the cylinder of mass 10 kg and radius 10 cm has a tape wrapped round it. The pulley weighs 100 N and has a radius 5 cm. When the system is released, the 5 kg mass comes down and the cylinder rolls without slipping. Find the acceleration and velocity of the mass as a function of time. 20 cm

5 kg

4. 2.

A spool of thread of mass m is placed on an inclined smooth plane set at an angle θ to the horizontal. The free end of the thread is attached to the wall as shown in Figure. Find the acceleration of the centre of mass of the spool, if its moment of inertia about its axis is I and the radius of the wound thread layer is r.

6.

r

θ

Mechanics II_Chapter 3_Part 3.indd 79

5.

A lawn roller in the form of a thin-walled hollow cylinder of mass M is pulled horizontally with a constant horizontal force F applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force. A solid cylindrical wheel of mass M and radius R is pulled by a force F applied to the centre of the wheel at 37° to the horizontal. If the wheel is to roll without slipping, what is the maximum value of F ? The coefficients of static and kinetic fric3 tion are μ S = 0.40 and μK = 0.30. sin( 37° ) = . 5 A sphere, a disc, and a hoop made of homogeneous materials have the same radius ( 10 cm ) and mass ( 3 kg ). They are released from rest at the top of a 30° incline and roll down without slipping through a vertical distance of 2 m .

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JEE Advanced Physics: Mechanics – II

(a) What are their speeds at the bottom? (b) Find the frictional force f in each case. (c) If they start together at t = 0, at what time does each reach the bottom? A Yo-Yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be MR2 taken to be . The Yo-Yo is placed upright on 2 a table and the string is pulled with a horizontal force F as shown. The coefficient of friction between the Yo-Yo and the table is μ . What is the maximum value of F for which the Yo-Yo will roll without slipping?

b R

8.

A cylinder of mass M and radius R is lying on a rough horizontal plane. It has a plank lying at its top as shown in Figure. A force F is applied on the plank such that the plank moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact. Find the acceleration of the cylinder and the frictional forces at the two contacts. F m

M

θ

is tied to a string which is wrapped around a disk capable of rotating about a horizontal axis. The disk has a mass M = 5 kg and a radius R = 0.2 m. Initially the string is taut. If the mass is released, calculate its acceleration. Take g = 9.8 ms −2.

m

30°

10. Find the acceleration of the cylinder of mass m and radius R and that of plank of mass M placed on smooth surface if pulled with a force F shown in Figure. Given that sufficient friction is present between cylinder and the plank surface to prevent sliding of cylinder. m M

F

11. A ball of mass m and radius r rolls along a circular path of radius R. Its speed at the bottom ( θ = 0 ) of the path is v0. Find the tangential and the normal force exerted by the path on the ball as a function of θ .

R

θ R

9.

A block of mass m = 1 kg slides down the surface of a smooth incline as shown in Figure. The block

ANGULAR MOMENTUM (L) The rotational effect of linear momentum is called Angular Momentum. When a body rotates about some point/axis, then the momentum associated with the body due to its rotation is called Angular Momentum. An external torque is required to change angular momentum just like an external force is required to change the linear momentum.

Mechanics II_Chapter 3_Part 3.indd 80

CASE-1: Angular Momentum of a Particle About Some Point  Angular Momentum, L of a particle about an arbitrary point O is the moment of linear momentum taken about that point. of ⎞ ⎛ Linear ⎞ ⎛ Distance L=⎜ ⊥⎜  ⎟ ⎟⎠ p O Momentum from ⎝ ⎠ ⎝ ⇒

L = pr⊥

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Chapter 3: Rotational Dynamics

⇒

L = pr sin θ

{in magnitude}

where r⊥ = r sin θ is called the moment arm. Vectorially    L=r×p    ⇒ L = m( r × v )   If r = xiˆ + yjˆ + zkˆ and v = vx iˆ + vy ˆj + vz kˆ , then ⊕

iˆ  L=m x vx

|

ˆj y vy

⊕

kˆ z vz

 The direction of L is also found by Right Hand Thumb Rule. p = mv

r O

θ

r⊥ = r sinθ

The SI unit of angular momentum is kgms −2. Please note that angular momentum is defined always with respect to a point. CASE-2: Angular Momentum of a Rigid Body Rotating About a Fixed Axis  The total angular momentum L of a system of particles relative to a given origin is the sum of the angular momentum of the particles.

3.81

CASE-3: Angular Momentum of System in Combined Rotation and Translation If a body is in combined rotation and translation like rolling and we are asked to find the angular momentum of the body about any fixed point or a reference point, then     Ltotal = LCM + M ( rCM × vCM ) i.e., total angular momentum of system/body is equal to the sum of the angular momentum of the CM about that point and the angular momentum of the system about the CM. So, for a rigid body undergoing linear and rotational motion, the total angular momentum may be split into two parts  (a) the orbital angular momentum, LO and  (b) the spin angular momentum Lcm . The orbital angular momentum is the angular momentum of the centre of mass motion about an origin O in an inertial frame. The spin angular momentum is the angular momentum relative to the centre of mass. The orbital term treats the system as a point particle at the centre of mass, whereas the spin term is the sum of the angular momenta of the particles relative to the centre of mass. The total angular momentum relative to the origin O in an inertial frame is the sum of both the angular momenta i.e.,      L = LOrbital + LSpin = LO + Lcm

ω

ILLUSTRATION 79 ri

mi

A rigid body rotating about a fixed axis

A particle of mass m is projected from origin O with speed u at an angle θ with positive x-axis. Positive y-axis is in vertically upward direction. Find the angular momentum of particle at any time t about O before the particle strikes the ground again.

   L = Σ ( ri × pi )   Since ri and pi are perpendicular, so L = Σri pi , where pi = mi vi = mi ri ω

(

)

⇒

L = Σmi ri 2 ω

⇒

L = Iω , where I = Σmi ri 2

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SOLUTION

   L = m( r × v ) 1  ⎛ ⎞ Here, r ( t ) = xiˆ + yjˆ = ( u cos θ ) tiˆ + ⎜ ut sin θ − gt 2 ⎟ ˆj ⎝ ⎠ 2  and v ( t ) = vx iˆ + vy ˆj = ( u cos θ ) iˆ + ( u sin θ − gt ) ˆj iˆ ⇒

ˆj

 1   L = m ( r × v ) = m ( u cos θ ) t ( u sin θ ) t − gt 2 2 u cos θ u sin θ − gt

kˆ 0 0

 1 ⇒ iˆ L = − m ( u cos θ ) gt 2 kˆ 2 A solid sphere of mass M and radius R rolls without slipping on a horizontal surface as shown in Figure. Find the total angular momentum of the sphere with respect to the origin O fixed on the ground. ω v

O

SOLUTION

Let us assume the clockwise sense of rotation positive. Orbital angular momentum about O is LO = MvR Spin angular momentum about centre of mass is 2 MR2ω 5 The total angular momentum is 2 L = LO + Lcm = MvR + MvR 5 For pure rolling, we have v = ω R Lcm = Iω =

L=

2 MvR 5

ILLUSTRATION 81

A disc of mass M and radius R rotating at an angular velocity ω about an axis perpendicular to its R plane at a distance from the centre, as shown in 2 Figure. Calculate the angular momentum of the disc about the axis shown.

Mechanics II_Chapter 3_Part 3.indd 82

R

SOLUTION

The moment of inertia of the disc about the given axis may be found from the parallel axes theorem, equation I = I cm + Md 2, where h is the distance between the given axis and a parallel axis through the centre of mass. Here h =

ILLUSTRATION 80

⇒

R/2

2

R 1 3 ⎛ R⎞ , therefore, I = MR2 + M ⎜ ⎟ = MR2 ⎝ 2⎠ 2 2 4

The angular momentum is L = Iω =

RELATION BETWEEN L AND

3 MR2ω 4

τ

As force changes the linear momentum of a particle, torque changes the angular momentum of a particle.    Since, L = r × p …(1) The rate of change of angular with time is found by taking the derivative on both sides of (1) with respect to time. So  dL d   = (r × p) dt dt   d (   )  dB dA  Using, A×B = A× + × B , we get dt dt dt    dL  dp dr  + ×p =r× dt dt dt    dp  dL     dr  ⇒ = r ×F+v×p ∵ = F and =v dt dt dt  dL       ⇒ = r × F + m(v × v ) {∵ p = mv } dt    Since, v × v = 0  dL    ⇒ = r×F =τ dt   dL ⇒ τ= dt

{

}

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Chapter 3: Rotational Dynamics ILLUSTRATION 82

  dL Show that the equation τ = , can be applied to the dt motion of a projectile. SOLUTION

The change in angular momentum of the projectile is produced by the torque exerted by the force of gravity. In Figure, we take the initial point as the origin. At a  later time, r = xiˆ + yjˆ . y v mg x

 Since the force on the particle is F = − mgjˆ , the gravitational torque on it is  τ=

If vi = 0 and v f = v, then we have J = mv ⇒

v=

J m

Just like translational the impulse momentum theorem in translational motion, we have angular impulse angular momentum theorem in rotation according to which angular impulse equals the change in angular momentum of the body and hence Angular Impulse (AI) = τΔt = ΔL ⇒

AI = τΔt = ΔL = L f − Li = I ( ω f − ω i )

Since, τ = Fr⊥ r

O

3.83

( xiˆ + yjˆ ) × (−mgjˆ) = −mgxkˆ

The rate of change of the angular momentum    L = r × p is    dL  dp  dv =r× = mr × dt dt dt  dv But the acceleration is = − gjˆ . So, dt   dL  dv = mr × = m xiˆ + yjˆ × − gjˆ = − ( mgx ) kˆ dt dt   dL Hence the equation τ = is applicable here. dt

(

) ( )

ANGULAR IMPULSE In complete analogy with the linear momentum, angular impulse is defined as   ΔL = τ ext dt  Since we know that linear impulse J equals the change in linear momentum, so we have      J = F Δt = Δp = m ( v f − vi )

⇒

AI = ( Fr⊥ ) Δt = Jr⊥

⇒

AI = Jr⊥ = L f − Li = I ( ω f − ω i )

{∵

F Δt = J }

If Li = 0, then L f = L = Iω ⇒

AI = Jr⊥ = Iω

Jr⊥ I In Figure (a): A linear impulse J is applied at centre of mass C of the rigid body. ⇒

ω=

Just after hitting, it will have only translational motion and its linear velocity will be given by J v= m In Figure (b): A linear impulse J is applied at point P, at a perpendicular distance r⊥ = CP.

∫

In one dimension, we can simply write this as J = Δp = p f − pi = m ( v f − vi )

Mechanics II_Chapter 3_Part 3.indd 83

Just after hitting it will have both translational and rotational motion. Its linear velocity v and angular velocity ω will be given by v=

Jr J and ω = ⊥ m I

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JEE Advanced Physics: Mechanics – II

If r⊥ is increased (keeping J to be constant) then v will remain same but ω will increase. So, the translational kinetic energy will have the same value but rotational kinetic energy will be more.

Problem Solving Technique(s) Since, Angular Impulse (AI) = τΔt = ΔL Now, following three cases can be considered. (a) If torque is constant, then angular impulse (AI) is obtained by directly multiplying this constant torque with the given time interval. (b) If torque is a function of time, then angular impulse (AI) is given by tf

∫

Angular Impulse (AI) = τ dt ti

(c) If torque versus time graph is given, then angular impulse (AI) can be obtained by the area under that graph.

LAW OF CONSERVATION OF ANGULAR MOMENTUM Since, by definition we know that   dL τ= dt

For no external torque acting on system, we have   τ =0 So, (1) gives   dL 0= dt  L = constant (both in magnitude and direction) ⇒  So, for no external torque acting on the system L is conserved both in magnitude and direction. This law is the rotational analogue of the Law of Conservation of Linear Momentum. If no external torque acts on system, then

In all three cases, angular impulse is equal to the change in angular momentum.

L is conserved both in magnitude and direction

ANGULAR IMPULSE-ANGULAR MOMENTUM THEOREM This theorem is the rotational analogue of the Impulse Momentum theorem studied already in Newton’s Laws of Motion. Using Newton’s Second Law for Rotation Motion, we have   dL τ ext = dt   dL = τ dt       ⇒ ΔL = L f − Li = I ( ω − ω 0 ) = τ dt

∫

The net angular impulse acting on a rigid body is equal to the change in angular momentum of the body. This is called the Angular Impulse Angular Momentum theorem for rotational dynamics. Please be careful to understand and see the  hidden fundamental of the axis of rotation i.e. all L f ,   Li and τ must be about identical AOR to use the results as they are. Else suitable modifications have to be made in the results.

Mechanics II_Chapter 3_Part 3.indd 84

…(1)

⇒

= constant = 2 2 1 1

mv1r1 = mv2 r2 i.e., vr sinθ = constant

= ∑ mi vi ri

system about specified AOR

about specified AOR

Analogy between Rotational and Linear Dynamics. Quantity

1. Inertia

Linear

Rotational

m

∑ m r or ∫ r dm 2 i i

2

2. Newton’s Second Law

Fext = ma   dp Fext = dt

τ ext = Iα   dL τ ext = dt

(Continued)

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Chapter 3: Rotational Dynamics

Quantity

3. Work

4. Kinetic Energy

Rotational

  Wlin = F ⋅ dl

∫

Wrot = τ dθ

1 2 mv 2

K rot =

K lin =

5. Work Energy Theorem 6. Impulse

Linear

∫

Wlin = ΔK lin

∫

I = Fext dt = Δp

1 2 Iω 2

Wrot = ΔK rot

∫

J = τ ext dt = ΔL

7. Momentum

p = mv

L = Iω

8. Impulse Momentum Theorem

  I = Δp

  J = ΔL

  P = F⋅v

  P = τ ⋅ω

9. Power

3.85

(b) The kinetic energies before and after the collision are 1 Ki = I i ω i2 = 18 J 2 1 K f = I f ω 2f = 12 J 2 The change is ΔK = K f − Ki = −6 J . In order for the two discs to spin together at the same rate, there had to be friction between them. The lost kinetic energy is converted with thermal energy. ILLUSTRATION 84

A man of mass m = 80 kg runs at a speed u = 4 ms −1 along the tangent to a disc-shaped platform of mass M = 160 kg and radius R = 2 m. The platform is initially at rest but can rotate freely about an axis 1 through its center. Take I = MR2. 2

ILLUSTRATION 83

A disc of moment of inertia 4 kgm 2 is spinning freely at 3 rads −1. A second disc of moment of inertia 2 kgm 2 slides down the spindle and they rotate together. (a) What is the angular velocity of the combination? (b) What is the change in kinetic energy of the system? SOLUTION

(a) Since there are no external torques acting, we may apply the conservation of angular momentum. I f ω f = Iiω i

( 6 kgm2 ) ω f = ( 4 kgm2 ) ( 3 rads−1 ) Thus, ω f = 2 rads −2

A

B

Mechanics II_Chapter 3_Part 3.indd 85

Calculate the angular velocity of the platform after the man jumps on. If the man now walks to the centre, then calculate the new angular velocity. Treat the man as a point particle. SOLUTION

Before trying to attempt this problem let us have a selfanalysis done and answer the following questions. Q. Can we apply the conservation of linear momentum? A. No, it cannot be applied because the axle exerts an external force on the system i.e. man + platform . Q. Can we apply the conservation of angular momentum? A. Yes, since the axle does not exert any torque, we may use the conservation of angular momentum. Q. Can we apply conservation of kinetic energy for the collision between the man and the platform? A. No. Let us consider the origin to be at the centre of platform. When the man runs in a straight line along

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JEE Advanced Physics: Mechanics – II

the tangent, then his initial angular momentum about this origin is L = r⊥ p, where r⊥ = R so

S V

Li = muR After he jumps on, one must take into account his contribution mR2 to the moment of inertia. The final angular momentum, L = Iω , is, ⎛1 ⎞ L f = ⎜ MR2 + mR2 ⎟ ω ⎝2 ⎠ By Law of Conservation of Angular Momentum, we have L f = Li , wo

ω=

mu ⎛M ⎞ + m⎟ R ⎜⎝ ⎠ 2

Substituting, m = 80 kg, R = 2 m, we get

ω=

M = 160 kg,

u = 4 ms −1,

( 80 ) ( 4 ) = 1 rads −1 ⎛ 160 ⎞ + 80 ⎟ 2 ⎜⎝ ⎠ 2

When the man reaches the center, his contribution to the moment of inertia is zero. The final angular momentum and the initial momentum in this case are given by

B

A O

C

L/2

(a) Determine the angular velocity ω in terms of V and L . (b) If the insect reaches the end B when the rod has turned through an angle of 90°, determine V. SOLUTION

In this problem we will denote angular momentum by its standard symbol H because L has been used for length of the rod. M

ω V

A

O L/2

C L/4

B

A

O

Just after collision

Angular momentum of the system (rod + insect) about the centre of the rod O will remain conserved just before collision and after collision i.e., Hi = H f . ⇒

MV

⎛ MR2 ⎞ Lf = ⎜ ω = 320ω 2 ⎝ 2 ⎟⎠ 2

2 ⎡ ML2 L ⎛ L⎞ ⎤ = Ιω = ⎢ + M⎜ ⎟ ⎥ω ⎝ 4⎠ ⎥ 4 ⎢⎣ 12 ⎦

⇒

MV

L 7 = ML2ω 4 48

⇒

ω=

12 V 7 L

ω 2 = 2 rads −2 ILLUSTRATION 85

A homogeneous rod AB of length L = 1.8 m and mass M is pivoted at the centre O in such a way that it can rotate freely in the vertical plane (shown in Figure). The rod is initially in the horizontal position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the points O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity ω .

Mechanics II_Chapter 3_Part 3.indd 86

B

L/4

Just before collision

⎛1 ⎞ Li = ⎜ MR2 + mR2 ⎟ ω1 = 640 kgm 2 s −1 ⎝2 ⎠

Applying angular momentum conservation L f = Li we get

C

…(1)

Due to the torque of weight of insect about O, angular momentum of the system will not remain conserved (although angular velocity ω is constant). As the insect moves towards B, moment of inertia of the system increases, hence, the angular momentum of the system will increase. Let at time t the insect be at a distance x from O and by then the rod has rotated through an angle θ . Then, angular momentum at that moment, ⎤ ⎡ ML2 H=⎢ + Mx 2 ⎥ ω ⎣ 12 ⎦

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3.87

Chapter 3: Rotational Dynamics

⇒

dH dx = 2 Mω x dt dt

⇒

τ = 2Mω x

{

dx dt

∵

⇒

dx Mgx cos θ = 2 Mω x dt

⇒

⎛ g ⎞ dx = ⎜ cos θ dt ⎝ 2ω ⎟⎠

⇒

g dx dθ = cos θ dθ dt 2ω

⇒

ω

⇒

⇒

} m

{∵ θ = ω t }

g

ω=

=

g 2ω 2

sin θ

⇒

B During impact

F

m

vCM

v B

Just after impact

…(2)

0

⇒ ⇒

V = 3.5 ms −1

…(3)

The total linear momentum of Ball + Rod system is conserved. Using Angular Impulse − Angular Momentum Theorem, we get For Ball

2 g 12 V = L 7 L 7 7 V= 2 gL = 2 × 10 × 1.8 = 3.5 ms −1 12 12

mu = mvcm + mv ⎛ Total Initial ⎞ ⎛ Total Final ⎞ ⎜ Momentum of ⎟ = ⎜ Momentum of ⎟ ⎜⎝ Ball + Rod ⎟⎠ ⎜⎝ Ball + Rod ⎟⎠

π 2

Substituting in equation (1), we get

( mv ) l − ( mu ) l = −τ t ⇒

( mv ) l − ( mu ) l = − ( Fl ) t

…(4)

For Rod Iω − 0 = ( Fl ) t

…(5)

From (4) and (5), we get

( mv ) l − ( mu ) l = − Iω

BALL ROD COLLISIONS Consider a uniform rod AB of mass M, length L. Let a ball of mass m moving with initial velocity v, hit the rod at a point P (other than centre of mass of the rod). Now two cases arise. CASE-1: When the Rod is not Hinged For Ball

Mechanics II_Chapter 3_Part 3.indd 87

F

ω

mv − mu = − mvcm

2g L

mv − mu = − Ft

P

C

From (1) and (2), we get

0

L 4

u

C

mvcm − 0 = Ft

2

x

C

A

For Rod

∫ dx = 2ω ∫ cosθ dθ L 2

A

B Just before impact

π 2

L 4

⇒

dH =τ dt

g dx cos θ = dθ 2ω

L 2

A

{ ω = constant }

…(1)

⇒

( mu ) l = Iω + ( mv ) l

…(6)

⎛ Total Initial ⎞ ⎛ Total Final ⎞ ⎜ Angular ⎟ ⎜ Angular ⎟ ⎜ Momentum ⎟ = ⎜ Momentum ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ of Ball + Rod ⎠ ⎝ of Ball + Rod ⎠ So, the total Angular Momentum of Ball + Rod is conserved.

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JEE Advanced Physics: Mechanics – II

So, if the rod is not hinged, then we observe both linear and angular momentum of Rod + Ball to be conserved.

A

 C m

⎛ Total Initial Energy ⎞ ⎛ Total Final Energy ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ of Ball + Rod of Ball + Rod ⎟⎠ 1 1 1 1 2 mu2 + 0 = mv 2 + mvcm + Iω 2 2 2 2 2

⇒

⎤ ⎥ ⎥⎦ at the point of impact

v B

Just after impact

⎤ ⎥ ⎥⎦ at the point of impact

⎛ lω − v ⎞ e = −⎜ ⎝ 0 − u ⎟⎠

e=0

CASE-2: When the Rod is Hinged For Ball Using Impulse − Momentum Theorem, we get

⇒

v = lω

ILLUSTRATION 86

…(1)

Using Angular Impulse − Angular Momentum Theorem, about hinge, we get …(2)

For Rod Using Angular Impulse − Angular Momentum Theorem, about hinge, we get …(3)

From (2) and (3), we get {about Hinge}

So, if the rod is hinged, then only angular momentum is conserved only about the hinge because τ due to the forces acting on the hinge about the hinge is zero.

Mechanics II_Chapter 3_Part 3.indd 88

m

Subcase-2(c) If the collision is perfectly inelastic, then

v = vcm + lω

( mu ) l = ( mv ) l + Iω

B During impact

⎡ ( v2 )n − ( v1 )n e = −⎢ ⎢⎣ ( u2 )n − ( u1 )n ⇒

e=0

Iω − 0 = ( Fl ) t

F

Subcase-2(b) If the collision is inelastic, then we have

Subcase-1(c) For perfectly inelastic collision, we have

( mv ) l − ( mu ) l = −τ t = − ( Fl ) t

F

1 1 1 mv 2 = mv 2 + Iω 2 2 2 2

⎡ ( v + lω ) − v ⎤ e = − ⎢ cm ⎥ 0−u ⎣ ⎦

mv − mu = − Ft

P

C

Subcase-2(a) If the collision is elastic, then we have

(Net velocity of rod at P will be vcm + lω , as it is under combined influence of translation and rotation.

⇒

u

B Just before impact

Subcase-1(b) If collision is inelastic ⎡ ( v2 )n − ( v1 )n e = −⎢ ⎢⎣ ( u2 )n − ( u1 )n

A

ω

Subcase-1(a) If the collision is elastic, then we have

⇒

A

A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end ’ A ’ of the rod with a velocity v0 in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest. m (a) Find the ratio M (b) A point P on the rod is at rest immediately after collision. Find the distance AP. (c) Find the linear speed of the point P after a time πL after the collision. 3V0 SOLUTION

(a) Let just after collision, velocity of centre of mass of rod is v and angular velocity about centre of mass is ω . Applying following three laws:

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Chapter 3: Rotational Dynamics m

v0 L 2

ω

CM

CM

v

x

L 2 Before collision

After collision

(i) External force on the system (rod + mass) in horizontal plane along x-axis is zero. ∴ Applying Conservation of Linear Momentum in x-direction. mvo = mv …(1) (ii) Net torque on the system about CM of rod is zero ∴ Applying Conservation of Angular Momentum about CM of rod, we get ⎛ L⎞ mvo ⎜ ⎟ = Ιω ⎝ 2⎠

MLω …(2) 6 (iii) Since, the collision is elastic, kinetic energy is also conserved. So, ⇒ mvo =

1 1 1 mvo2 = Mv 2 + Ιω 2 2 2 2 ⇒ mvo = Mv 2 +

mvo v ⇒ x= = M ω 6 mvo ML L ⇒ x= 6 L L ⇒ AP = + 2 6 2 ⇒ AP = L 3 πL (c) After time t = 3 vo A

v

ω

ML2 2 ω 12

P

mvo 6mvo m 1 = , v= and ω = M 4 M ML (b) Point P will be at rest if xω = v A L 2

π 2 Therefore, situation will be as shown below: So, resultant velocity of point P will be ⇒ θ=

⇒

v 2  ⎛ m⎞ vp = 2v = 2 ⎜ v = vo = o ⎝ M ⎟⎠ o 4 2 2 vo  vp = 2 2

ILLUSTRATION 87

A rigid massless rod of length L joins two particles each of mass m. The rod lies on a frictionless table, and is struck by a particle of mass m and velocity v0, moving as shown. After the collision, the projectile moves straight back. m

ω CM P

6 mvo π L . ML 3vo

⎛ 1⎞ m 1 ⎛ m⎞ = 2π ⎜ ⎟ = ⇒ θ = 2π ⎜ ⎝ M ⎟⎠ ⎝ 4⎠ M 4

…(3)

From equations (1), (2) and (3), we get the following results

xω

A

P

angle rotated by rod, θ = ω t =

L ML2 ω ⇒ mvo = 2 12

3.89

v v

x

L m

v0

45° m

Mechanics II_Chapter 3_Part 3.indd 89

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JEE Advanced Physics: Mechanics – II

Find the angular velocity of the rod about its centre of mass after the collision, assuming that mechanical energy is conserved. SOLUTION

Applying Law of Conservation of Linear Momentum, we get mv0 = 2mv1 − mv2 ⇒

2v1 − v2 = v0

…(1)

(a) Describe quantitatively the motion of the system after the boy is on the plank. Neglect friction with the ice. (b) One point on the plank is at rest immediately after the collision. Where is it? SOLUTION

Let C be the centre of mass of boy plus plank. Let C be at a distance x from the end where the boy steps on the plank, then ⎛ l⎞ m(0 ) + M ⎜ ⎟ ⎝ 2⎠ x= m+ M

Applying Law of Conservation of Angular Momentum about centre of mass C of light rod and the two identical particles, we get 2

⎛ L⎞ ⎛ 1 ⎞ ⎛ L⎞ ⎛ L⎞ ⎛ 1 ⎞ = 2m ⎜ ⎟ ω − mv2 ⎜ ⎟ ⎜ mv0 ⎜ ⎟ ⎜ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎝ 2⎠ ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⇒

v0 = 2Lω − v2

⇒

⎛ M ⎞ l BC = x = ⎜ ⎝ M + m ⎟⎠ 2 A

…(2)

Since the mechanical energy is conserved so, the collision is elastic, hence e = 1 at point of impact along common normal direction ⇒

⎛ Relative Speed ⎞ ⎛ Relative Speed ⎞ ⎜⎝ of Approach ⎟⎠ = ⎜⎝ of separation ⎟⎠

⇒

⎛ L ⎞⎛ 1 ⎞ v0 = v2 + v1 + ⎜ ω ⎟ ⎜ ⎝ 2 ⎠ ⎝ 2 ⎟⎠

⇒

ωL v0 = v1 + v2 + 2 2

v

B

So, distance of centre of mass C from the middle of the rod ( O ) is ⎛ l ⎞ ⎛ m ⎞ l OC = ⎜ − x ⎟ = ⎜ ⎝2 ⎠ ⎝ M + m ⎟⎠ 2

…(3)

Applying the Law of Conservation of Linear Momentum, we get

( M + m ) v = mv0

4 2 v0 7 L

⇒

ILLUSTRATION 88

A boy of mass m runs on ice with velocity v0 and steps on the end of a plank of length l and mass M which is perpendicular to his path. M

⎛ m ⎞ v=⎜ v ⎝ M + m ⎟⎠ 0

mv0 ( BC ) = I system ω 2 2 ⎡ ⎛ M ⎞ 2 ⎛ l ⎞ 2 Ml 2 mMv0 l ⎛ m ⎞ ⎛l ⎞⎤ = ⎢m⎜ + + M⎜ ω ⎜ ⎟ ⎜ ⎟ ⎟ ⎝ M + m ⎠ ⎝ 4 ⎟⎠ ⎥⎦ 2( M + m ) 12 ⎣ ⎝ M+m⎠ ⎝ 4⎠ 2 2 ⎡ ⎛ M ⎞ 2 ⎛ l ⎞ 2 Ml 2 ⎛ m ⎞ ⎛l ⎞⎤ + + M m ⎢ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎥ ω 4 12 M+m 4 ⎦ M+m ⎣

Substituting,

m v0

…(1)

Applying the Law of Conservation of Angular Momentum about point C , we get

⇒ 

Mechanics II_Chapter 3_Part 3.indd 90

ω

C m

Solving equations (1), (2) and (3), we get

ω=

O

mv0 = v from equation (1), we get M+m

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Chapter 3: Rotational Dynamics

3.91

v l ⎛ 4m + M ⎞ = ⎜ ⎟ ω 6⎝ M+m ⎠ Now, we have the plank divided in two portions (a) White portion, below C till the end B. (b) Grey shaded portion, above C till the end A. mv0 ⎞ ⎛ Since v ⎜ = is actually the velocity of centre of ⎝ M + m ⎟⎠ mass of boy plus plank and so every point of the boy + plank system has a forward velocity v . However, the lower white portion has a tangential velocity rω where r is measured from C to B. This rω is forward and so we cannot expect the resultant of v and rω both forwards to be zero.

SOLUTION

System is free to rotate but not free to translate. During collision, net torque on the system ( rod A + rod B + mass m ) about point P i.e., hinge is zero, so angular momentum is conserved about P. If ω be the angular velocity of system just after collision, then Li = L f ⇒

However, in the upper grey portion, all points move forward with velocity v but simultaneously the upper grey portion has a tangential velocity backwards. Now wherever (say the point P) the forward velocity v equals the backwards tangential velocity xω (where x is the distance of point P from C), then at that point net velocity is zero, so

⇒

v = xω v x= ω

⇒

BP = BC + x

⇒

⎛ M ⎞ ⎛ l ⎞ l ⎛ 4 m + M ⎞ 2l + = BP = ⎜ ⎝ M + m ⎟⎠ ⎜⎝ 2 ⎟⎠ 6 ⎜⎝ M + m ⎟⎠ 3

mv ( 2l ) = Iω

where, I is the moment of inertia of system about P, so ⎛ l2 ⎛ l ⎞ 2 ⎞ ⎛ l2 ⎞ 2 I = m ( 2l ) + mA ⎜ ⎟ + mB ⎜ + ⎜ + l⎟ ⎝ 12 ⎝ 2 ⎠ ⎟⎠ ⎝ 3⎠ Given l = 0.6 m, m = 0.05 kg, mA = 0.01 kg mB = 0.02 kg. Substituting the values, we get I = 0.09 kgm 2 Therefore, from equation (1), we get

ILLUSTRATION 89

Two uniform rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly joined end to end. The combination is pivoted at the lighter end, P as shown in Figure, such that it can freely rotate about point P in a vertical plane. A small object of mass 0.05 kg , moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity of the object so that the system could just be raised to the horizontal position?

Mechanics II_Chapter 3_Part 3.indd 91

and

ω= ⇒

2mvl ( 2 ) ( 0.05 )( v )( 0.6 ) = I 0.09

ω = 0.67 v

…(1)

Now, after collision, mechanical energy will be conserved. ⎛ Increase in ⎞ ⎛ Decrease in ⎞ ⎜ Therefore, ⎜ ⎟⎠ = GPE of CM ⎟ RKE ⎝ ⎜ of both Rods ⎟ ⎝ ⎠

2/9/2021 6:40:53 PM

3.92

JEE Advanced Physics: Mechanics – II

⇒

1 2 l⎞ ⎛ ⎛ l⎞ Iω = mg ( 2l ) + mA g ⎜ ⎟ + mB g ⎜ l + ⎟ ⎝ ⎝ 2⎠ 2 2⎠

⇒

ω2 =

⇒

ω2 =

⇒

gl ( 4 m + mA + 3 mB ) I

( 9.8 )( 0.6 ) ( 4 × 0.05 + 0.01 + 3 × 0.02 ) 0.09

ω = 4.2 rads

−1

= 17.64

Further by Law of Conservation of Energy, we have

…(2)

⎛ Loss in RKE ⎞ ⎛ Gain in GPE ⎞ ⎜⎝ of Block about O ⎟⎠ = ⎜⎝ of CM of Block ⎟⎠

Equating equations (1) and (2), we get v= ⇒

4.2 ms −1 0.67

v = 6.3 ms −1

⇒

where Δh =

ILLUSTRATION 90

An ice cube of mass M and with sides of length a is sliding without friction across a countertop with a speed v0 when it hits a ridge E at the edge of the counter (see the Figure). This collision causes the cube to tilt as shown. Show that the minimum value of v0 needed for the cube to fall off the table is given by v0 = 1.1ag Given that the moment of inertia of the cube about an axis passing through its centre of gravity and parallel 1 to horizontal surface is Ma 2 6

1 2 Iω = Mg ( Δh ) 2 a 2

−

a 2 2

⇒

1⎛ 2 a⎞ ⎛ a 2 ⎞ ⎛ 3 v0 ⎞ − ⎜ Ma ⎟⎠ ⎜⎝ ⎟ = Mg ⎜ ⎝ 2 2 ⎟⎠ 2⎝ 3 4 a ⎠

⇒

v0 = 1.1ag

ILLUSTRATION 91

A rod of mass M and length L is falling vertically with speed v. Suddenly its one end gets stuck in a frictionless hook. Find the angular velocity of the rod just after its end gets stuck.

SOLUTION

SOLUTION

By Law of Conservation of Angular Momentum around the axis through E , we get Li = L f ⇒

⎛ Mv0 ⎜ ⎝

2 ⎛ Ma 2 a⎞ ⎛ a ⎞ ⎞ + M⎜ ω ⎟⎠ = I 0ω = ⎜ ⎝ 2 ⎟⎠ ⎟⎠ 2 ⎝ 6

⇒

⎛ Mv0 ⎜ ⎝

a⎞ 2 2 ⎟ = Ma ω 2⎠ 3

⇒

ω=

3 v0 4 a

Mechanics II_Chapter 3_Part 3.indd 92

Since all the impulse forces during the impact are applied by the hook so they don’t produce any torque about the hook and hence angular momentum of the rod about the hook is conserved. Initial angular momentum about hook is ⎛ L⎞ Hi = mvr⊥ = Mv ⎜ ⎟ ⎝ 2⎠ Final angular momentum about the hook is ⎛ ML2 ⎞ H f = Iω = ⎜ ω ⎝ 3 ⎟⎠ ⇒

MvL ML2 = ω 2 3

⇒

ω=

3v 2L

2/9/2021 6:41:10 PM

Chapter 3: Rotational Dynamics

3.93

For disc 2, we get

ILLUSTRATION 92

Two discs of radii R and 2R are pressed against each other. Initially disc with radius R is rotating with angular velocity ω and another disc was stationary. Both discs are hinged at their respective centres and free to rotate about them. Moment of inertia of smaller is I and bigger disc is 2I about their respective axis of rotation. Find the angular velocity of the bigger disc after long time.

∫ f ( 2R ) dt = 2Iω

2

⇒

2I ( ω − ω 1 ) = 2I ω 2

⇒

ω1 + ω 2 = ω

⇒

2ω 2 + ω 2 = ω

⇒

ω2 =

ω 3

ILLUSTRATION 93

SOLUTION

Let R1, R2 be the horizontal and vertical components of the reaction offered by the first hinge to the pulley and R3, R4 be the horizontal and vertical components of the reaction offered by the second hinge to the pulley. The only force which is producing any torque about the centre of first disk is friction as shown in Figure. R2

f R1

m1g

R3

A uniform rod AB of length 2l and mass 2m is suspended freely at A and hangs vertically at rest when a particle of mass m is fired horizontally with speed v to strike the rod at its mid-point. If the particle is brought to rest by the impact, calculate the impulsive reaction at A, the initial angular speed of the rod and the maximum angle the rod makes with the vertical in the subsequent motion. SOLUTION

At the instant of collision, if impulsive reaction at A is J and impulse between particle and rod be J1, then we have

R4 m2g

Let the disc rotate with angular velocity ω1 and ω 2 as shown in Figure. by impulse-momentum theorem J1 = mv and by angular impulse-angular momentum theorem Since v is same at the point of contact, so we have

J1l =

ω 1 R = ω 2 ( 2R ) ⇒

ω2 =

ω1 2

Now, total angular impulse provided by the friction is equal to change in angular momentum of the disc. So, for disc 1, we get

∫

fRdt = I ( ω − ω1 )

Mechanics II_Chapter 3_Part 3.indd 93

⇒

ω=

2

2 m ( 2l ) ω 3 3 J1 3v = 8 ml 8l

For rod, we have J1 − J = 2m ( lω ) ⇒

⎛ 3v ⎞ mv J = J1 − 2ml ⎜ = ⎝ 8l ⎟⎠ 4

2/9/2021 6:41:25 PM

3.94 JEE Advanced Physics: Mechanics – II

If rod gets displaced by an angle before coming to rest, then by law of conservation of energy, we have 1 2 Iω = mgh 2

mv = mv ′ +

2 Mlω 3

…(1)

Since collision is elastic, so we have 2

⇒

⇒

v=

2

1 ⎛ ( 2m )( 2l ) ⎞ ⎛ 3v ⎞ ⎜ ⎟⎠ ⎜⎝ ⎟ = 2mgl ( 1 − cos ϕ ) 2⎝ 3 8l ⎠

ωl − v′ 2

From equations (1) and (2), we get

2

3v 32 gl

⇒

cos ϕ = 1 −

⇒

⎛ 3v 2 ⎞ ϕ = cos −1 ⎜ 1 − 32 gl ⎟⎠ ⎝

mv = mv ′ +

ILLUSTRATION 94

A thin uniform square plate with side l and mass M can rotate freely about a stationary vertical axis coinciding with one of its sides. A small ball of mass m flying with velocity v at right angles to the plate strikes elastically at centre of the square plate. Calculate the velocity of the ball v ′ after the impact and the horizontal component of the force which the axis will exert on the plate after the impact.

…(2)

2 M ( v + v′ ) ( 2 ) 3

4 ⎞ 4 ⎛ Mv = ⎜ m + M ⎟ v ′ ⎝ 3 3 ⎠

⇒

mv −

⇒

⎛ 3m − 4 M ⎞ v′ = ⎜ v ⎝ 3 m + 4 M ⎟⎠

From, equation (2), we get

⇒

ω=

2 2v 3m − 4 M ⎞ ( v + v ′ ) = ⎛⎜ 1 + ⎟ ⎝ l l 3m + 4 M ⎠

ω=

12mv l ( 3m + 4 M )

Force due to axis on the plate is 72 Mm2 v 2 ⎛ l⎞ F = M ⎜ ⎟ ω2 = 2 ⎝ 2⎠ l ( 3m + 4 M )

SOLUTION

Applying conservation of angular momentum, we get 2 ⎛ l⎞ ⎛ l ⎞ Ml mv ⎜ ⎟ = mv ′ ⎜ ⎟ + ω ⎝ 2⎠ ⎝ 2⎠ 3

Test Your Concepts-VI

Based on Angular Momentum and Its Conservation 1.

2.

A small ball is suspended from a point O by a light thread of length  . Then the ball is drawn aside so that the thread deviates through an angle θ from the vertical and set in motion in a horizontal direction at right angles to the vertical plane in which the thread is located. What is the initial velocity that has to be imparted to the ball so that π it could deviate through the maximum angle of 2 with the vertical in the process of motion? A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal

Mechanics II_Chapter 3_Part 3.indd 94

3.

(Solutions on page H.162) surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. (a) What is the final angular velocity of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision? A wheel of moment of inertia I and radius R is rotating about its axis at an angular speed ω 0. It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel.

2/9/2021 6:41:37 PM

Chapter 3: Rotational Dynamics

4.

5.

A uniform circular disc of mass m and radius a is rotating with constant angular velocity ω in a horizontal plane about a vertical axis through its centre A. A particle P of mass 2m is placed gently a on the disc at a point distant from A. If the par2 ticle does not slip on the disc, find the new angular velocity of the rotating system. A pulley in the form of a uniform disc of mass 2m and radius r, is free to rotate in a vertical plane about a fixed horizontal axis through its centre. A light inextensible string has one end fastened to a point on the rim of the pulley and is wrapped several times round the rim. The portion of string not wrapped round the pulley length 8r and carries a particle of mass m at its free end. The particle is held close to the rim of the pulley and level with its centre. If the particle is released from this position find the initial angular velocity of the pulley and the impulse of the sudden tension in the string when it becomes taut.

4r

6.

7.

A horizontally oriented uniform disc of mass M and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which a small body of mass m can slide without friction. A light thread running down through the hollow axle of the disc is tied to the body. Initially the body is located at the edge of the disc and the whole system is rotated with an angular velocity ω 0 . Then, by means of a force F applied to the lower end of the thread the body is slowly pulled towards the axis of rotation. Find the (a) angular velocity of the system in its final state. (b) work performed by the force F. A particle is projected horizontally along the interior of a smooth hemispherical bowl of radius r which is kept at rest. Find the minimum initial speed v0 required for the particle to just reach the top of the bowl. The initial angular position of the particle is θ0.

Mechanics II_Chapter 3_Part 3.indd 95

r

8.

9.

3.95

θ0

A uniform rod AB of length 2 and mass m is rotating in a horizontal plane about a vertical axis through A, with angular velocity ω , when the midpoint of the rod strikes a fixed nail and is brought immediately to rest. Find the angular impulse exerted by the nail. The assembly of two discs as shown in Figure is placed on a rough horizontal surface and the front disc is given an initial angular velocity ω 0. Determine the final linear and angular velocity when both the discs start rolling. It is given that friction is sufficient to sustain rolling in the rear wheel from the beginning of motion.

10. A man of mass m1 stands on the edge of a horizontal uniform disc of mass m2 and radius R which is capable of rotating freely about a stationary vertical axis passing through its centre. The man walks along the edge of the disc through an angle θ relative to the disc and then stops. Find the angle ϕ through which the disc turned by the time the man stopped. 11. The axis of a cylinder of radius R and moment of inertia about its axis I is fixed at centre O as shown in Figure. Its highest point A is in level with two plane horizontal surfaces. A block of mass M is initially moving to the right without friction with speed v1. It passes over the cylinder to the dotted position. Calculate the speed v2 in the dotted position and the angular velocity acquired by the cylinder if at the time of detaching from cylinder block stops slipping on it.

2/9/2021 6:41:46 PM

3.96

JEE Advanced Physics: Mechanics – II

12. A uniform rod of mass m and length  rests on a smooth horizontal surface. One of the ends of the rod is struck in a horizontal direction at right angles to the rod. As a result, the rod obtains velocity v0. Find the force with which one half of the rod will act on the other in the process of motion. 13. A rod of mass M, length  lies on horizontal table and is free to move on the table. A ball of mass m, moving perpendicularly to the rod at a distance d from its centre with speed v collides elastically with it as shown in Figure. What quantities are conserved in the collision? What must be the mass of the ball so that it remains at rest immediately after collision?

14. A uniform rod AB, of mass m and length 4a, is smoothly pivoted at a point O of its length, where AO = a and hangs at rest in equilibrium position with A above pivot O. A horizontal impulse of magnitude J is imparted to the rod at its centre of mass. Find the initial angular velocity of the rod. If the rod describes complete revolutions in the subsequent motion, find an inequality for J in terms of a, m and g. 15. A smooth rod rotates freely in a horizontal plane with the angular velocity ω 0 about a stationary vertical axis O, relative to which the rod’s moment of inertia is I . A small ring the mass m is located on the rod close to the rotation axis and is tied to it by a thread. When the thread is burned, the ring starts sliding along the rod. Find the velocity vr of the ring relative to the rod as a function of its distance r from the rotation axis.

d m

v

ROLLING WITH SLIPPING (v0 > Rω 0) CASE-1: When, v0 > Rω0 Since v0 > Rω 0, so the point P has a tendency to slip forward and hence sliding friction (later on called as friction) acts backwards as shown.

So, to attain v = Rω at some later time t, f must be directed such that the role of f is (a) to decrease v0 to v (say) (b) to increase ω 0 to ω (say) such that at t , we get v = Rω (condition for pure rolling) From impulse momentum theorem and angular impulse angular momentum theorem, we get

ROLE OF FRICTION Since f is acting opposite to v0, therefore it will have a tendency to decrease v0. Also, when f acts backwards then it will provide a torque that will be acting in the clockwise sense and hence will have a tendency to increase ω 0, i.e., at some later time, t (say), we can have v = rω Since, v0 > Rω 0 {at t = 0}

Mechanics II_Chapter 3_Part 3.indd 96

mv − mv0 = − ft

…(1)

Iω − Iω 0 = +τ t = ( fR ) t

…(2)

But at t, we have v = Rω

…(3)

So, we use (1), (2), (3) to get the desired results. Also, work done by friction equals change in KE, so 1 1 ⎞ ⎛1 ⎞ ⎛1 W f = ⎜ mv 2 + Iω 2 ⎟ − ⎜ mv02 + Iω 02 ⎟ ⎝ ⎠ ⎝2 ⎠ 2 2 2       ω =v R

…(4)

ω0 ≠ v R

2/9/2021 6:41:58 PM

Chapter 3: Rotational Dynamics

μN = μ g, so Since, f = μ N = μ mg, retardation a = m distance travelled in time t is 1 1 ⎛ μ mg ⎞ 2 s = v0 t + ( − a ) t 2 = v0 t + ⎜ − ⎟t 2 2⎝ m ⎠

…(1)

Iω − Iω 0 = −τ t = − ( fR ) t

…(2)

v = Rω

{∵ τ = Iα }

…(3)

So, we use (1), (2), (3) to get the desired results. Also work done by friction equals change in KE, so ⇒

where τ = fR = Iα

1 1 ⎞ ⎛1 ⎞ ⎛1 W f = ⎜ mv 2 + Iω 2 ⎟ − ⎜ mv02 + Iω 02 ⎟ ⎝ ⎝2 ⎠ 2  2 2 ⎠    ω =v R

fR ( μ mg ) α= = R I I

1 ⎛ μ mgR ⎞ 2 So, θ = ω 0 t + ⎜ ⎟t 2⎝ I ⎠

mv − mv0 = ft But at t, we have

1 θ = ω0t + αt2 2

⇒

From impulse momentum theorem and angular impulse angular momentum theorem, we get

…(5)

If θ is the total angle traversed, then

1⎛ τ ⎞ θ = ω0t + ⎜ ⎟ t2 2⎝ I ⎠

3.97

…(6)

CASE-2: When, v0 < Rω0 Since v0 < Rω 0, so the point P has a tendency to slip backward and hence sliding friction acts forward, as shown in Figure.

…(4)

ω0 ≠ v R

μN Since, f = μ N = μ mg and a = = μ g, so distance m travelled in time t is 1 s = v0 t + ( a ) t 2 2 ⇒

{∵

f accelerates v0 }

1 ⎛ μ mg ⎞ 2 s = v0 t + ⎜ ⎟t 2⎝ m ⎠

…(5)

If θ is the angle traversed, then

⇒

1 θ = ω0t + αt2 2 1⎛ τ ⎞ θ = ω0t + ⎜ − ⎟ t2 2⎝ I ⎠

{∵ τ = Iα }

where, τ = fR = Iα

ROLE OF FRICTION Since f is acting along v0, so it will have a tendency to increase v0. Also, when f acts forwards, along v0, then it will provide a torque that will be acting in the counter-clockwise sense and hence will have a tendency to decrease ω 0, i.e., at some later time, t (say), we can have v = rω Since, v0 < Rω 0 {at t = 0 } So, to attain v = Rω at some later time t, f must be directed such that the role of f is (a) to increase v0 to v (say) (b) to decrease ω 0 to ω (say) Such that at t, we get v = Rω (condition for pure rolling)

Mechanics II_Chapter 3_Part 4.indd 97

fR ( μ mg ) R = I I

⇒

α=

⇒

1 ⎛ μ mgR ⎞ 2 θ = ω0t − ⎜ ⎟t 2⎝ I ⎠

…(6)

ILLUSTRATION 95

A sphere of mass m and radius R is placed at rest on a plank of mass M which is placed on a smooth horizontal surface as shown in Figure. The coefficient of friction between the sphere and the plank is μ . At t = 0, a horizontal velocity v0 is given to the plank. Find the time after which the sphere starts rolling. m R M

v0

2/9/2021 6:36:07 PM

3.98 JEE Advanced Physics: Mechanics – II SOLUTION

For the Sphere: ac =

α=

f = μg m fR τc 5 μg = = I c 2 mR2 2 R 5

stationary, spinning in place. After a short time it begins to move forward and eventually reaches the point where it rolls without slipping. Find the final velocity of the wheel in terms of the initial angular MR2 . velocity ω 0. Take I = 2 SOLUTION

We assume that wheel is initially rotating clockwise. Let the wheel starts rolling after a time t. Then, Using Impulse Momentum Theorem For translation: Impulse = Δp = p f − pi ⇒

Angular Impulse = ΔL = L f − Li

vc = ac t = μ gt

⇒ − fRt = Iω − Iω 0

5 μg t 2 R

The velocity of the point of contact is 5 7 v = vc + ω R = μ gt + μ gt = μ gt 2 2

Note that clockwise angular momentum is considered as positive. Since I =

MR2 2

− ( fR ) t =

For the Plank:

μ mg f Retardation a = = M M

t=

v0 ⎛7 m⎞ ⎜⎝ + ⎟⎠ μ g 2 M

ILLUSTRATION 96

A wheel is held by a handle on its axle and given an initial angular velocity ω 0. The wheel is then placed in contact with the ground. At first the wheel remains

Mechanics II_Chapter 3_Part 4.indd 98

v =ωR

ω0

ω f

f

Initially at t = 0

Finally at time t

f

μ mg 7 v = μ gt = v0 − t 2 M

…(2)

y

N

Condition of pure rolling

MR2 ( ω − ω0 ) 2

v=0

μ mg Instantaneous velocity v = v0 − t M

⇒

…(1)

For rotation:

After time t

ω = αt =

ft = Mv − 0

x

For the condition of pure rolling, we have v = ωR

…(3)

Using equations (1), (2) and (3), we get − Mω R2 = ⇒

ω=

MR2 ( ω − ω0 ) 2

ω0 3

The linear velocity of the wheel is v = Rω =

Rω 0 3

2/9/2021 6:36:17 PM

Chapter 3: Rotational Dynamics

Solving equations (3), (4) and (5) we get,

ILLUSTRATION 97

A billiard ball of radius R, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance h above the centre line as shown in Figure. The ball leaves the cue with a speed v0 and because of its forward English (backward slipping) eventu9 ally acquires a final speed v0. Find h. 7 h

4 R 5

h=

ILLUSTRATION 98

A solid sphere of radius r is gently placed on a rough horizontal ground with an initial angular speed ω 0 and no linear velocity. If the coefficient of friction is μ, find the time t when the slipping stops. In addition, state the linear velocity v and angular velocity ω at the end of slipping.

SOLUTION

Let ω 0 be the angular speed of the ball just after it leaves the cue. The maximum friction acts in forward direction till the slipping continues. Let v be the linear speed and ω the angular speed when slipping ceases, so F h

R

3.99

v0

ω

v = 9 v0 7

ω0

SOLUTION

METHOD I: Let m be the mass of the sphere. According to the problem v0 < rω 0, so it is a case of backward slipping and hence force of friction is in forward direction. Limiting friction will act in this case.

Fmax

v = Rω v ⇒ ω= R 9 Given, v = v0 …(1) 7 9 v0 ⇒ ω= …(2) 7 R Applying impulse momentum theorem, we get Fdt = mv0

…(3)

Applying angular impulse angular momentum theorem, we get

τ dt = Iω 0 2 mR2ω 0 …(4) 5 Angular momentum about bottommost point will remain conserved, so ⇒

Fhdt =

Li = L f ⇒

Iω 0 + mRv0 = Iω + mRv

⇒

2 2 ⎛ 9 v0 ⎞ 9 mR2ω 0 + mRv0 = mR2 ⎜ + mRv0 …(5) ⎝ 7 R ⎟⎠ 7 5 5

Mechanics II_Chapter 3_Part 4.indd 99

v fmax

Linear acceleration a =

f μ mg = = μg m m

fr τ 5 μg = = I 2 mr 2 2 r 5 Slipping ceases when v = rω Angular retardation α =

⇒

( at ) = r ( ω 0 − α t )

⇒

5 μ gt ⎞ ⎛ μ gt = r ⎜ ω 0 − ⎟ ⎝ 2 r ⎠

⇒

7 μ gt = rω 0 2

⇒

t=

⇒

v = at = μ gt =

2 rω 0 7 μg

and ω =

2 rω 0 7

v 2 = ω0 r 7

2/9/2021 6:36:30 PM

3.100 JEE Advanced Physics: Mechanics – II

METHOD II: Net torque on the sphere about the bottommost point i.e., the point of contact is zero. Therefore, angular momentum of the sphere will remain conserved about the bottommost point.

Now, τ due to friction is

τ = R ( μ mg cos θ ) ⇒

α=

⇒

α=

Li = L f ⇒

Iω 0 = Iω + mrv

⇒

2 2 2 mr ω 0 = mr 2ω + mr ( ω r ) 5 5

⇒

ω=

⇒

2 2 ω 0 and v = rω = rω 0 7 7

α=

( r μ mg cos θ ) = 5 g μ cos θ 2 mR2 5

2R

(

5 g ( 0.5 ) 1

2

)

2R 5g

…(2)

4R 2

For translatory motion v ′ = v − at = Rω − at

ILLUSTRATION 99

A solid sphere of radius R is projected along an inclined plane as shown in Figure.

When v ′ = 0, then t=

2 ⎛ Rω ⎞ Rω Rω 2 = = 1.5 g 1.5 ⎜⎝ g ⎟⎠ a

For rotatory motion, we have

ω ′ = ω − αt ⇒ If coefficient of friction is μ = 0.5 then calculate the time when linear velocity of the sphere becomes zero. Also calculate the angular velocity of the sphere at that instant.

t=

ω = α

ω 5g

=

4 2Rω 5g

4R 2 Linear velocity of sphere becomes zero before angular velocity become zero Angular velocity at time t when linear velocity is zero.

SOLUTION

Velocity of lower most point is not zero, hence it is not the case of pure rolling and friction between the sphere and the incline is kinetic in nature. Since velocity of lower most point in forward direction therefore friction will act in backward direction. Let retardation is a and angular retardation is α

⎛ 5 g ⎞ ⎛ 2Rω ⎞ ω′ = ω − ⎜ ⎝ 4 R 2 ⎟⎠ ⎜⎝ 1.5 g ⎟⎠ ⇒

fr

⎛ ω′ = ω − ⎜ ⎝

5⎞ ⎛ ω ⎞ ω ⎟⎜ ⎟= 4 ⎠ ⎝ 1.5 ⎠ 6

ILLUSTRATION 100

⇒

mg sin θ + μ mg cos θ = ma

⇒

a = g ( sin θ + μ cos θ ) =

Mechanics II_Chapter 3_Part 4.indd 100

g 2

( 1 + 0.5 ) =

1.5 g 2

…(1)

A uniform round object of mass m, radius R and moment of inertia about its centre of mass I cm is thrown with speed v0 without any rotation on a rough horizontal surface of coefficient of friction μ . I Assuming cm2 = k, calculate the time after which mR

2/9/2021 6:36:44 PM

Chapter 3: Rotational Dynamics

slipping stops. Also calculate the speed and angular speed of the object when the slipping stops. SOLUTION

3.101

ILLUSTRATION 101

A solid sphere starts to roll without slipping on a rough inclined plane as shown in Figure.

The direction of friction is such that it opposes the translational motion of object and provides torque that supports the rotational motion of the object as shown in Figure.

Applying Newton’s Second Law for translational motion, we get f = μ N = μ mg

Calculate the friction force acting on the sphere, minimum coefficient of friction μ0 for pure rolling of the μ sphere and the work done by friction if μ = 0 and 4 the displacement of centre of mass is x. SOLUTION

Acceleration of centre of mass of the sphere is

f = μg m

⇒

a=

⇒

v = v0 − at = v0 − μ gt

a= …(1)

g sin θ 5 g sin θ = 2 7 1+ 5

Applying Newton’s Second Law for rotational motion, we get fR = I cm α

μ mgR I cm

⇒

α=

⇒

⎛ μ mgR ⎞ ω = αt = ⎜ t ⎝ I cm ⎟⎠

Applying Newton’s Second Law for translational motion, we get

For pure rolling, we have

Mg sin θ − f r = Ma

v = Rω ⇒

v0 − μ gt =

⇒

Rμ mgRt I cm

⇒

1⎞ ⎛ v0 = μ gt ⎜ 1 + ⎟ ⎝ k⎠

⇒

t=

I cm ⎫ ⎧ = k⎬ ⎨∵ 2 ⎩ mR ⎭

v0 k μg (1 + k )

v = v0 − μ g ⇒

v=

v0 k

(1 + k ) μg

⎛ 1+ k − k ⎞ = v0 ⎜ ⎝ 1 + k ⎟⎠

v0 v0 v and ω = = 1+ k R R(1 + k )

Mechanics II_Chapter 3_Part 4.indd 101

5 Mg sin θ 2 Mg sin θ = 7 7

2 Mg sin θ ≤ μN 7 2 Mg sin θ ≤ μ Mg cos θ 7

Since, f r = ⇒ ⇒

Substituting this value in equation (1), we get

f r = Mg sin θ −

μ≥

2 tan θ 7

2 tan θ 7 If μ < μmin, then there will be slipping and hence friction will be kinetic in nature, so we have ⇒

μ0 =

⎛ 2 tan θ ⎞ f k = μ Mg cos θ = ⎜ Mg cos θ ⎝ 7 4 ⎟⎠

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3.102 JEE Advanced Physics: Mechanics – II

Mg sin θ 14

⇒

fk =

⇒

Mg sin θ − f k Mg sin θ − a= = M M

⇒

13 g sin θ a= 14

⇒

α=

⇒

⎛ 5 g cos θ ⎞ ⎛ 2 sin θ ⎞ ⎛ α=⎜ ⎝ 2R ⎟⎠ ⎜⎝ 7 cos θ ⎟⎠ ⎜⎝

⇒

13 g sin θ a 26 R ⎛ 13 ⎞ ⎛ 28 R ⎞ 14 R= = =⎜ ⎟⎜ ⎝ 14 ⎠ ⎝ 5 ⎟⎠ 5 g sin θ α 5 28 R

Mg sin θ 14

μ MgR cos θ 5 μ g cos θ = 2 2R 2 MR 5

SOLUTION

We must know that maximum torque will act on cylinder when it is ta the extreme positions of oscillatory motion. The cylinder moves under the influence of pseudo force with respect to platform as shown in Figure.

1 ⎞ 5 g sin θ ⎟= 4⎠ 28 R

x a Since, = = 5.2R θ α ⇒

The platform is given a motion in the x-direction given by x = A cos ( ω t ). There is no slipping between the cylinder and platform. Calculate the maximum torque acting on the cylinder during its motion.

Thus, equation of motion of cylinder is mAω 2 − f = mamax and

fR =

1 ⎛a ⎞ mR2 ⎜ max ⎟ ⎝ R ⎠ 2

Aω 2 =

3 amax 2

amax =

2 Aω 2 3

Work done by friction is Rotation

…(2)

Adding (1) and (2), we get

x = 5.2R θ W fr = ( W fr )

…(1)

+ ( W fr )

Trans

⇒

W fr = ( μ mg cos θ ) Rθ + ( μ mg cos θ ) x cos ( 180° )

⇒

W fr = μ mg ( Rθ − x ) cos θ

⇒

4.2 ⎛ x ⎞ W fr = μ mg cos θ ⎜ − x⎟ = − ( μ mg cos θ ) x ⎝ 5.2 ⎠ 5.2

⇒

W fr = −

21 μ mgx cos θ 26

ILLUSTRATION 102

A cylinder of mass m and radius R is resting on a horizontal platform (which is parallel to xy plane) with its axis fixed along the y-axis and free to rotate about its axis as shown in Figure.

⇒

Maximum angular acceleration is

α max =

amax 2 Aω 2 = R 3R

Maximum torque on cylinder is

τ max = Iα max ⇒

τ max =

⎛ 2 Aω 2 ⎞ 1 MR2 ⎜ ⎝ 3 R ⎟⎠ 2

⇒

τ max =

1 mARω 2 3

ILLUSTRATION 103

A plank of mass m1 with a uniform sphere of mass m2 placed on it rests on a smooth horizontal plane as shown in Figure. A constant horizontal force F is applied to the plank. Calculate the acceleration of the plank and the centre of the sphere if there is no sliding between the plank and the sphere.

Mechanics II_Chapter 3_Part 4.indd 102

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Chapter 3: Rotational Dynamics m2

3.103

a2 Fpseudo = m2 a1

m2

m1

SOLUTION

FROM FRAME ATTACHED TO GROUND For the two bodies, equations of motion from the ground reference frame for accelerations a1 and a2 shown in Figure are m2 asphere = a2

aplank = a1 m1

F − f = m1 a1

…(1)

f = m2 a2

…(2)

2 ⎛ a −a ⎞ fR = − m2 R2 ⎜ 1 2 ⎟ ⎝ R ⎠ 5

…(3)

From (1) and (2), we get F = m1 a1 + m2 a2

…(4)

From (2) and (3), we get

Also note that since friction on sphere is acting towards right, so it must act on the plank towards left. Let plank moves toward right with an acceleration a1, due to which the sphere will experience a pseudo force m2 a1 towards left, and hence it will roll towards left with respect to plank with an acceleration a2. Since we have used the concept of pseudo force, so we must say that the acceleration a2 must be with respect to the plank. Let the angular acceleration of the sphere during rolling be α , so we have a2 = Rα For translational motion of plank, we have F − f = m1 a1

For translational motion of sphere with respect to plank we have m2 a1 − f = m2 a2

7 a2 = 2 a1

fR = Iα …(5)

⇒

⎛2 ⎞⎛ a ⎞ fR = ⎜ m2 R2 ⎟ ⎜ 2 ⎟ ⎝5 ⎠⎝ R⎠

⇒

f =

Using equation (4), we get ⎛ 2a ⎞ F = m1 a1 + m2 ⎜ 1 ⎟ ⎝ 7 ⎠ ⇒

a1 =

7F 7 m1 + 2m2

From equation (5), we get 2F a2 = 7 m1 + 2m2 FROM FRAME ATTACHED TO PLANK Since the sphere does not slip over the plank surface, so it is the case of pure rolling and hence we can take friction on sphere in any direction, say towards right as shown in Figure.

Mechanics II_Chapter 3_Part 4.indd 103

…(2)

For rotational motion of sphere with respect to plank, we have

5m2 a2 = 2m2 a1 − 2m2 a2 ⇒

…(1)

2 m2 a2 5

…(3)

From equation (2) and (3), we get 2 m2 a1 − m2 a2 = m2 a2 5 7 a2 5 Using equations (1), (2) and (3), we get ⇒

a1 =

…(4)

2 F − m2 a2 = m1 a1 5 ⇒

F=

2 2 ⎛5 ⎞ m2 a2 + m1 a1 = m2 ⎜ a1 ⎟ + m1 a1 ⎝7 ⎠ 5 5

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3.104 JEE Advanced Physics: Mechanics – II

⇒

a1 =

SOLUTION

7F 7 m1 + 2m2

Let the impulse imparted by the cue to the ball be J and initial angular speed of ball be ω 0, then applying the impulse momentum theorem, we get

From equation (4), we get

⇒

a2 =

5 5⎛ 7F ⎞ a1 = ⎜ 7 7 ⎝ 7 m1 + 2m2 ⎟⎠

a2 =

5F 7 m1 + 2m2

J = mv0

Applying the angular impulse angular momentum theorem, we get

Since, a2 is acceleration of sphere relative to the plank, so net acceleration of the sphere is

( anet )sphere = a1 − a2 =

…(1)

2F 7 m1 + 2m2

2 mR2ω 0 5 From equations (1) and (2), we get Jh =

ω0 =

Please note that in the equation a2 = Rα , a2 is the acceleration of cylinder with respect to plank. In problems where rolling of a body takes place in an inertial (ground) frame or a non-inertial (accelerated) frame, then in the equation a = Rα , the acceleration a must be relative, i.e. with respect to inertial (ground) frame or with respect to the non-inertial (accelerated) frame.

…(2)

5v0 h

2R 2 Applying conservation of angular momentum about point of contact P on the ground, we get Linitial = Lfinal ⎛ 2mR2 ⎞ ⎛ 5v0 h ⎞ where, Linitial = mv0 R + ⎜ and ⎝ 5 ⎟⎠ ⎜⎝ 2R2 ⎟⎠ 2 ⎛ 9v ⎞ ⎛ 9v ⎞ Lfinal = m ⎜ 0 ⎟ R + mR2 ⎜ 0 ⎟ ⎝ 7 ⎠ ⎝ 7R ⎠ 5

ILLUSTRATION 104

A billiard ball of mass m, radius R initially at rest is given a sharp horizontal impulse by a cue. The cue is held horizontally a distance h above the centre line as shown in Figure. v =

9 v 7 0

The ball leaves the cue with a speed v0 and because of its forward English eventually acquires a final 9 speed of v0, calculate the value of h. 7

⇒

2 2 ⎛ 5v h ⎞ ⎛ 9v ⎞ ⎛ 9v ⎞ mv0 R + mR2 ⎜ 02 ⎟ = m ⎜ 0 ⎟ R + mR2 ⎜ 0 ⎟ ⎝ 7 ⎠ ⎝ 7R ⎠ ⎝ 2R ⎠ 5 5

⇒

h ⎞ ⎛ 9 18 ⎞ ⎛ ⎜⎝ 1 + ⎟⎠ = ⎜⎝ + ⎟⎠ R 7 35

⇒

4 ⎛ 63 ⎞ h=⎜ −1 R = R ⎝ 35 ⎟⎠ 5

Test Your Concepts-VII

Based on Rolling with Slipping 1.

A horizontal plank having mass m lies on a smooth horizontal surface. A sphere of same mass and radius r is spun to an angular frequency ω 0 and gently placed on the plank as shown in Figure. If coefficient of friction between the plank and the sphere is μ , find the distance moved by the plank

Mechanics II_Chapter 3_Part 4.indd 104

(Solutions on page H.165) till the sphere starts pure rolling on the plank. The plank is long enough.

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Chapter 3: Rotational Dynamics

2.

3.

4.

5.

A disc of radius R and mass m is projected on to a horizontal floor with a backward spin such that its centre of mass speed is v0 and angular velocity is ω 0. What must be the minimum value of ω 0 so that the disc eventually returns? A solid cylinder of mass m and radius R is set in rotation about its axis with an angular velocity ω 0 , then lowered with its lateral surface onto a horizontal plane and released. The coefficient of friction between the cylinder and the plane is μ . Calculate the time for which the cylinder will move with sliding and the total work done by friction. A bowling ball is thrown straight down the alley. When it starts, its centre of mass has a speed v0 and it is sliding without rotating. Determine how far the ball moves down the alley before it starts rolling without slipping. Express your answer in terms of v0, g and μK . A cylinder of mass m and radius R is rotated about its axis with angular velocity ω 0 (as shown in Figure). It is lowered on a rough inclined plane at an angle 30° with horizontal and having the 1 . The point of initial coefficient of friction μ = 3 contact of cylinder and incline is at a height of 3R from horizontal. Calculate the total time taken by the cylinder to reach the bottom of incline.

3R 30°

6.

A sphere of radius R, mass M is projected along a rough horizontal surface (having coefficient of friction μ), with initial velocities shown in Figure. v0

If the final velocity of the sphere is zero, then calculate (in terms of v0, R, ω 0, μ and M ) the

Mechanics II_Chapter 3_Part 4.indd 105

7.

8.

9.

3.105

required magnitude of ω0, the time t required by sphere to come to rest and the distance travelled by sphere before it stops. A bowling ball is thrown straight down the alley. When it starts, its centre of mass has a speed v0 and it is sliding without rotating. Determine how far the ball moves down the alley before it starts rolling without slipping. Express your answer in terms of v0, g and μK . A plank of mass M is placed on a smooth surface over which a cylinder of mass m and radius R is placed as shown in Figure.

The plank is now pulled towards right with an external force F. If cylinder does not slip over the surface of plank, calculate the linear acceleration of plank and cylinder. Also calculate the angular acceleration of the cylinder. A bowler projects a ball of mass M, radius R along an alley with forward velocity v0 and a backspin ω 0 . The coefficient of friction between the alley and the ball is μ . Calculate the time t at which the ball will start rolling with slipping and the speed of the ball at time t. ω0

v0

10. A uniform circular cylinder of mass m and radius r is given an initial angular velocity ω 0 and no initial translational velocity. It is placed in contact with a plane inclined at an angle α to the horizontal. If there is a coefficient of friction μ for sliding between the cylinder and plane, find the distance the cylinder moves up before sliding stops. Also calculate the maximum distance it travels up the plane. Assume μ > tanα .

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3.106 JEE Advanced Physics: Mechanics – II

SOLVED PROBLEMS Solving equations (1) and (2), we get

PROBLEM 1

A uniform rod AB of length 2l and mass m is turning freely in a horizontal plane about a vertical axis 3v through A, with angular velocity . A particle P, l also having mass m, is moving with constant velocity v in the same horizontal plane. The particle and the rod are moving towards each other and when AB is perpendicular to the path of the particle, P collides with point B of the rod. If the coefficient of restitu1 tion between the rod and the particle is . Find the 2 speed of the particle after impact. SOLUTION

Angular momentum of the system about A is conserved, so Iω − mv ( 2l ) = Iω ′ + mv ′ ( 2l )

v′ =

13 v 8

PROBLEM 2

Two heavy metallic plates are joined together at 90° to each other. A laminar sheet of mass 30 kg is hinged at the line AB joining the two heavy metallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its centre of mass is 1.2 kgm 2. Two rubber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5 m from the line AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rads −1 and turns back. If the impulse on the sheet due to each obstacle is 6 Ns. A

′

v′ P

2

2

⇒

m ( 2l ) ⎛ 3v ⎞ m ( 2l ) ω ′ + 2mv ′l ⎜⎝ ⎟⎠ − 2mvl = 3 l 3

⇒

2mvl =

⇒

2v =

4 lω ′ + 2v ′ 3

⇒

2v =

4 lω ′ + 2v ′ 3

4 2 ml ω ′ + 2mv ′l 3 …(1)

v ′ − 2ω ′l v ′ − 2ω ′l = 3 v 7v ⎛ ⎞( ) v+⎜ 2l ⎝ l ⎟⎠

1 = 2

⇒

7 v = 2v ′ − 4ω ′l

⇒

4ω ′l = 2v ′ − 7 v

Mechanics II_Chapter 3_Part 4.indd 106

(a) find the location of the centre of mass of the laminar sheet from AB. (b) at what angular velocity does the laminar sheet come back after the first impact? (c) after how many impacts, does the laminar sheet come to rest?

Let r be the perpendicular distance of centre of mass from the line AB and ω be the angular velocity of the sheet just after colliding with rubber obstacle for the first time. The linear velocity of centre of mass before and after collision are

Relative Velocity of Separation oach Relative Velocity of Appro

⇒

Q

SOLUTION

At point B, we have e=

B

vi = ( r ) ( 1 rads −1 ) = r and v f = rω   where, vi and v f will be in opposite directions. Now, by Impulse − Momentum Theorem, we have …(2)

⎛ Linear Impulse ⎞ ⎛ Change in Linear ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ Momentum of CM ⎟⎠ on CM

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3.107

Chapter 3: Rotational Dynamics

(

)

⇒

6 = m v f + vi = 30 ( r + rω )

⇒

r (1 + ω ) =

1 5

SOLUTION

…(1)

Similarly, by Angular Impulse − Angular Momentum Theorem, we have ⎛ Angular Impulse ⎞ ⎛ Change in Angular ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ Momentum about AB ⎟⎠ about AB ⇒

⎛ Perpendicular ⎞ ⎛ Angular ⎞ ⎛ Linear ⎞ ⎜ ⎟ distance of = × ⎜⎝ impulse ⎟⎠ ⎜⎝ impulse ⎟⎠ ⎜ impulse from AB ⎟ ⎝ ⎠

⇒

6 ( 0.5 m ) = I AB ( ω + 1 )

(

)(1 + ω )

3 = ICM + Mr 2

⇒

3 = ( 1.2 + 30 r 2 ) ( 1 + ω ) r = 0.4 m OR r = 0.1 m

But at r = 0.4 m, ω comes out to be negative

( −0.5 rads −1 ) which is not acceptable. Therefore,

(a) r = distance of CM from AB = 0.1 m (b) Substituting r = 0.1 m in equation (1), we get ω = 1 rads −1 i.e., the angular velocity with which sheet comes back after the first impact is 1 rads −1. (c) Since, the sheet returns with same angular velocity of 1 rads −1, the sheet will never come to rest. PROBLEM 3

A uniform circular disc of mass M and radius a is pivoted at a point O on its circumference so that it can rotate about the tangent at O, which is horizontal, the centre of the disc describing a vertical circle of centre O in a plane perpendicular to the tangent. The point diametrically opposite to O is A and the disc is just displaced from rest when A is vertically above O. Find the angular velocity of the disc when A is vertically below O. At this instant a particle of mass M travelling with velocity u in the opposite direction of motion of the centre of the disc hits the disc at its centre and adheres to it. Find the angular velocity of the system immediately before and after the impact. If the disc just reaches its initial position show that

Mechanics II_Chapter 3_Part 4.indd 107

1 2 Iω 2

…(1)

5 1 Ma 2 + Ma 2 = Ma 2 4 4

where, I =

Substituting in equation (1), we get 2 Mga =

1⎛ 5 2⎞ 2 ⎜ Ma ⎟⎠ ω 2⎝ 4

g 5a

(ii) By Law of Conservation of Angular Momentum about an axis through O, we get

…(2)

Solving equations (1) and (2) for r, we get

u = ( 3 2 + 5 ) ag

Mg ( 2 a ) =

⇒ ω=4

(Initial angular velocity = 1 rads −1 )

⇒

(i) By Law of Conservation of Mechanical Energy, we get

C O u

M

Initially

aω

O

ω′

M

Bullet Embeds at Centre

Just Before Impact

Just After Impact

Outside the paper, towards the reader Inside the paper, away from the reader

O

ω

⎛5 ⎞ ⎛5 ⎞ Mu ( a ) − ⎜ Ma 2 ⎟ ω = ⎜ Ma 2 + Ma 2 ⎟ ω ′ ⎝4 ⎠ ⎝4 ⎠ 9 5 ⇒ u − ω a = ω ′a 4 4 ⇒ ω′ =

4u − 5ω a 4 ⎛ 5 ⎞ = ⎜ u − aω ⎟⎠ 9a 9a ⎝ 4

Substituting the value of ω we get

ω′ =

g ⎞ 4 ⎛ 4 u − 5 ag ⎜ u − 5a ⎟= 9a ⎝ 5a ⎠ 9a

(

)

(iii) Again, applying Law of Conservation of Mechanical Energy, we get 2 Mg ( 2 a ) = ⇒ 4 Mga =

1⎛ 5 2 2 2⎞ ⎜ Ma + Ma ⎟⎠ ( ω ′ ) 2⎝ 4

9 2 Ma 2 ( ω ′ ) 8

⇒ 32 g = 9 a ( ω ′ )

2

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3.108 JEE Advanced Physics: Mechanics – II

⇒ ω′ = ⇒ 4

This will last till pure rolling begins, so

32 g 2g =4 9 a 9a

(

v = rω

4 2g u − 5 ag = 9 a 9a

⇒

)

Substituting the values, we get

2 ag 9

⇒ u − 5 ag = 9

t1 =

⇒ u = ( 3 2 + 5 ) ag

⇒

A sphere of radius r is projected up an inclined plane ⎛ 1⎞ for which μ = ⎜ ⎟ tan θ with a velocity v0 and initial ⎝ 7⎠ angular velocity ω 0 such that, v0 > rω 0. Prove that friction firstly acts down the incline and afterwards acts down the incline. Further prove that the total 17 v0 + 4 rω 0 . time of rise is T = 18 g sin θ

SOLUTION

Initially i.e., at t = 0, we have v0 > rω 0, therefore, there is forward slipping. Hence, friction will be down the incline. Once, v = rω , force of friction becomes up the incline, so

μ mg cos θ + mg sin θ 8 = g sin θ m 7

{

1 ∵ μ = tan θ 7

τ μ rmg cos θ = 2 2 I mr 5 5 g sin θ α= 14 r

}

Also, α =

{

1 ∵ μ = tan θ 7

α ω0

Initially

Mechanics II_Chapter 3_Part 4.indd 108

16 5 v0 + rω 0 21 21 Once v = rω i.e., the condition of pure rolling is obtained, then minimum value of μ required to maintain pure rolling is ⇒

v=

μmin =

tan θ 1+

mr I

2

=

tan θ tan θ = 5 3.5 1+ 2

1 tan θ and since μgiven < μmin 7 so slipping will occur even after that and so the force of friction is upwards and maximum. So, linear retardation is 6 a ′ = g sin θ − μ g cos θ = g sin θ 7 If t2 be the further time taken by sphere to rise, then However, μgiven = μ =

θ

a

⎛8 ⎞ ⎛ 2 ⎞ ⎛ v − rω 0 ⎞ v = v0 − ⎜ g sin θ ⎟ ⎜ ⎟ ⎜ 0 ⎝7 ⎠ ⎝ 3 ⎠ ⎝ g sin θ ⎟⎠

v0

ω0

⇒

2 ( v0 − rω 0 ) 3 g sin θ

Since v = v0 − at1

PROBLEM 4

a=

( v0 − at1 ) = r ( ω 0 + α t1 )

v0

}

0 = v − a ′ t2 ⇒

5 16 v + rω v 21 0 21 0 t2 = = 6 a′ g sin θ 7

⇒

t2 =

5v0 + 16 rω 0 18 g sin θ

So, we get, T = t1 + t2 =

17 v0 + 4 rω 0 18 g sin θ

PROBLEM 5

A vertically oriented uniform rod of mass M and length l can rotate about its upper end. A horizontally flying bullet of mass m strikes the lower end of the rod and gets embedded in it as a result of which the rod swings through an angle α . Assuming that m  M, find the

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Chapter 3: Rotational Dynamics

(a) velocity of the flying bullet. (b) momentum increment in the system “bullet-rod” during the impact and the cause that changes the momentum. (c) distance x from the upper end of the rod the bullet must strike for the momentum of the system “bullet-rod” to remain constant during the impact. SOLUTION

(a) Applying Law of Conservation of Angular Momentum about its upper end, we get

∑ mvr

⊥

= ( I system ) ω

O

O

O

ω

⇒

⇒

α

⎛ l ⎞ (b) ΔP = Pf − Pi = M ⎜ ω ⎟ − mv ⎝2 ⎠ ⇒ ΔP =

1 Ml ⎛ 3 m ⎞ ⎛ v ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ − mv = mv 2 M l 2

⇒ ΔP =

1 ⎛ M⎞ m⎜ ⎟ 2 ⎝ m⎠

⇒ ΔP = M

1 ⎛α⎞ gl sin ⎜ ⎟ ⎝ 2⎠ 6

The force exerted by the support is responsible for the change in momentum. (c) For Angular Momentum to be conserved, we have

⎛ Ml 2 ⎞ ⇒ mvx = Iω = ⎜ ω ⎝ 3 ⎟⎠

v

O

…(1)

O x

⇒

m

3m v M l

v

After the bullet gets embedded in the rod, the rod swings through an angle α , so by Law of Conservation of Energy we have ⎛ Loss in ⎞ ⎛ Gain in ⎞ ⎛ Gain in ⎞ ⎜ RKE of ⎟ = ⎜ GPE of ⎟ + ⎜ GPE of ⎟ ⎜ Rod + Bullet ⎟ ⎜ CM of Rod ⎟ ⎜ Bullet ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 ⎛ Ml 2 ⎞ 2 l 2 2 ⎜ ⎟ ω + ml ω = Mg ( 1 − cos α ) + 2⎝ 3 ⎠ 2 mgl ( 1 − cos α ) Since m  M , so the underlined terms are neglected ⇒

…(3)

ω

⎛ Ml 2 ⎞ ω ⇒ ( mv ) l = ⎜ ⎝ 3 ⎟⎠

⇒

2 ⎛α⎞ gl sin ⎜ ⎟ ⎝ 2⎠ 3

Li = L f

ω

m

⇒ ω=

3.109

1 ⎛ Ml 2 ⎞ 2 l ⎜⎝ ⎟⎠ ω = Mg ( 1 − cos α ) 2 3 2

…(2)

For Linear Momentum to be conserved, we have Pi = Pf ⎛ l ⎞ ⇒ mv = Mvc = M ⎜ ω ⎟ ⎝2 ⎠

…(4)

From equations (3) and (4), we have x=

2 l 3

PROBLEM 6

A uniform rod of mass m and length l is suspended by the two wires at P and Q. The wire at Q suddenly breaks. Find the tension T in the wire at P immediately after the wire at Q is broken. R

S

Solving equations (1) and (2), we get M 2 ⎛α⎞ v= gl sin ⎜ ⎟ ⎝ 2⎠ m 3

Mechanics II_Chapter 3_Part 4.indd 109

30°

30° P

Q

2/9/2021 6:38:47 PM

3.110 JEE Advanced Physics: Mechanics – II SOLUTION

PROBLEM 7

Equation of motion for the centre of mass of rod are

A uniform rod of length 4l and mass m is free to rotate about a horizontal axis passing through a point distant l from its one end. When the rod is horizontal, its angular velocity is ω as shown in Figure. Calculate

R

Tsin30° T

α

30° Tcos 30° P

C

Q

x y

mg

3T T cos 30° = m 2m mg − T sin 30° T ay = = g− m 2m ax =

ω

…(1) …(2)

Angular acceleration of rod about its centre of mass C is ( T sin 30° ) l τ 2 = 3T α= = …(3) 2 I ml ml 12 Acceleration of point P immediately after the wire Q breaks will be perpendicular to RP. Thus, along RP its component should be zero. Further    aPC = aP − aC    ⇒ aP = aC + aPC  where, aC has two components ax and ay.  Similarly aPC has also two components at and an, l where an = 0, because an = ω 2 and ω is zero at the 2 l instant wire Q breaks. Now, at = α in the direction 2 shown in Figure. R

ax

P

at sin ( 30° ) + ax cos ( 30° ) = ay cos ( 60° )

T=

2 mg 7

Mechanics II_Chapter 3_Part 4.indd 110

m ( 4l ) + ml 2 12 If at be the tangential acceleration of the centre of mass of rod, then at = lα =

3 g 7

{downwards}

H

C

O

2

B

mg

3 mg 7

⇒ V=

Since, acceleration of P along RP is zero so

Solving above equations we get,

3g 7 l

=

2

4 mg …(1) 7 If H be the horizontal reaction (towards C) at axis, then

ay

ay l 3 ax = α+ 4 2 2

mgl

(a) α =

mg − V = mat =

60°

⇒

SOLUTION

Let V be the vertical reaction (upwards) at axis O, then

at 30°

(a) reaction of axis at this instant, (b) acceleration of centre of mass of the rod at this instant, (c) reaction of axis and acceleration of centre mass of the rod when rod becomes vertical for the first time. (d) minimum value of ω so that centre of rod can complete circular motion.

…(4)

H = man = mlω 2

…(2)

So, total reaction at axis is ⎛ 7 lω 2 ⎞ 4 N = H + V = mg 1 + ⎜ 7 ⎝ 4 g ⎟⎠ 2

2

2

2/9/2021 6:39:05 PM

Chapter 3: Rotational Dynamics

(b) The acceleration of centre of mass of rod is aC = at2 + an2 2

2 ⎛ 3g ⎞ ⇒ aC = ⎜ + ( lω 2 ) ⎝ 7 ⎟⎠

(c) Let ω ′ be the angular speed of the rod when it becomes vertical for the first time. Then by Law of Conservation of Mechanical Energy, we have 1 ( 2 I ω ′ − ω 2 ) = mgl 2 2mgl ⇒ ω ′2 = ω 2 + I 2 mgl 6g ⇒ ω ′2 = ω 2 + = ω2 + 7 2 7l ml 3 Acceleration of centre of mass at this instance is 6g 7 Let V be the reaction (upwards) at axis at this instant, then, aC = lω ′ 2 = lω 2 +

V − mg = maC = mlω 2 + ⇒ V=

6 mg 7

The drum is given an initial angular velocity such that the block X starts moving up the plane. (i) Find the tension in the string during the motion. Take g = 9.8 ms −2. (ii) At a certain instant of time the magnitude of the angular velocity of Y is 10 rads −1. Calculate the distance travelled by X from the instant of time until it comes to rest. SOLUTION

Mass of block X is m = 0.5 kg Mass of drum Y, M = 2 kg Radius of drum, R = 0.2 m Angle of inclined plane, θ = 30 o Let ’ a ’ be the linear retardation of block X and α the angular retardation of drum Y. Then, a = Rα Since, mg sin 30 o − T = ma ⇒

(d) By Law of Conservation of Mechanical Energy, we have

⇒

2mgl = I

T= 2mgl = 7 2 ml 3

6g 7l

A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination 30° to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2 kg and of radius 0.2 m as shown in Figure. Y

30°

Mechanics II_Chapter 3_Part 4.indd 111

…(3)

1 ⎡ Mmg ⎤ 2 ⎢⎣ M + 2m ⎥⎦

Substituting the value, we get ⎛ 1 ⎞ ⎧ ( 2 )( 0.5 )( 9.8 ) ⎫ T=⎜ ⎟⎨ = 1.63 N ⎝ 2 ⎠ ⎩ 2 + ( 0.5 )( 2 ) ⎬⎭

PROBLEM 8

X

τ TR = I 1 MR2 2 2T α= MR

…(2)

Solving equations (1), (2) and (3) for T, we get

1 2 Iω min 2

⇒ ω min =

mg − T = ma 2

…(1)

Also α =

13 mg + mlω 2 7

mgl =

3.111

⇒

T = 1.63 N

From equation (3), angular retardation of drum

α=

( 2 )( 1.63 ) 2T = = 8.15 rads −2 MR ( 2 )( 0.2 )

Linear retardation of block is a = Rα = ( 0.2 )( 8.15 ) = 1.63 ms −2 At the moment when angular velocity of drum is

ω o = 10 rads −1

2/9/2021 6:39:25 PM

3.112 JEE Advanced Physics: Mechanics – II

The linear velocity of block will be vo = ω o R = ( 10 )( 0.2 ) = 2 ms

−1

Now, the distance(s) travelled by the block until it comes to rest will be given by v2 s= o 2a ⇒

{ using

2

v =

vo2

− 2 as with v = 0

}

( 2 )2 s= m = 1.22 m 2 ( 1.63 )

PROBLEM 9

A thin uniform rod of mass m and length l rotates with the constant angular velocity ω about the vertical axis passing through the rod’s suspension point O. In doing so, the rod describes a conical surface with a half aperture angle θ . Find the angle θ as well as the magnitude and direction of the reaction force N at the point O.

The gravity and the centrifugal forces are the only torque producing forces relative to the point O. So, if τ C is the total torque produced due to centrifugal forces acting on the rod, then

τ mg = τ C mgl sin θ = τ c …(1) 2 Let us now calculate τ c by integration The elementary torque due to the centrifugal force that acts on the infinitesimal rod element of length dx located at the distance x from the point O and ⎛ m ⎞ having mass dm ⎜ = dx ⎟ is equal to, ⎝ l ⎠ ⇒

dτ C = ( dFc ) x cos θ ⎛m ⎞ where dFC = ( dm )( x sin θ ) ω 2 = ⎜ dx ⎟ ( x sin θ ) ω 2 ⎝ l ⎠ ⇒

O

θ

dτ c =

Integrating this expression over the whole length of the rod, we get

τc =

mω 2 l 2 sin θ cos θ 3

…(2)

From equation (1) and (2), we get

SOLUTION

Let us consider the frame rotating about the vertical axis along with the rod. In this reference frame, the rod experiences not only the gravity mg and the reaction force N but also the centrifugal force Fc , directed radially outwards as shown in Figure. N

N|| = mg

cos θ = ⇒

3g

…(3)

2ω 2 l

⎛ 3g ⎞ θ = cos −1 ⎜ ⎝ 2ω 2 l ⎟⎠

Further, writing equations for centre of mass of the rod, we get N = mg

θ′ O

N⊥

and N ⊥ = man =

θ C mg

⇒

N=

N2

+

mω 2 l sin θ 2

N ⊥2

…(4)

( mg )

=

Fc

Since the rod is observed to be in equilibrium in the given reference frame, the resultant torques of all forces relative to any point and the resultant of all the forces are equal to zero.

Mechanics II_Chapter 3_Part 4.indd 112

mω 2 sin θ cos θ x 2 dx l

2

⎛ mω 2 l ⎞ +⎜ sin θ ⎟ ⎝ 2 ⎠

2

θ dm

dFc

2/9/2021 6:39:38 PM

Chapter 3: Rotational Dynamics

Substituting the value of θ from equation (3), we get

For no slipping, we have a2 = a1 + Rα

7 g2 mω 2 l 1+ 2 4ω 4 l 2

N=

3.113

…(4)

Solving the above four equations, we get

For its direction, we observe that mg cos θ ′ = N Substituting the values, we get

7 ⎞ ⎛ F = μs g ⎜ M + m ⎟ ⎝ 2 ⎠

7 ⎞ ⎛ Thus, maximum value of F can be μ s g ⎜ M + m ⎟ ⎝ 2 ⎠ PROBLEM 11

cos θ ′ = 4 cos θ 9 + 7 cos 2 θ The centrifugal force Fc does not pass through point C but below it. (Think, why?) PROBLEM 10

Determine the maximum horizontal force F that may be applied to the plank of mass m for which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are μ s and μ k respectively. M

Two thin circular discs of mass 2 kg and radius 10 cm each are joined by a rigid massless rod of length 20 cm. The axis of the rod is along the perpendicular to the planes of the disc through their centres. This object is kept on a truck in such a way that the axis of the object is horizontal and perpendicular to the direction of motion of the truck. Its friction with the floor of the truck is large enough so that the object can roll on the truck without slipping. Take X-axis as the direction of motion of the truck and Z-axis as the vertically upwards direction. If the truck has an acceleration 9 ms −2, calculate

R m

F

SOLUTION

The free body diagrams for the sphere and the plank are shown in Figure. a2

α a1

F

(i) the force of friction on each disc. (ii) the magnitude and direction of the friction torque acting on each disc about the centre of mass O of the object. Express the torque in the ˆ ˆj and kˆ vector form in terms of unit vectors i, in X, Y and Z directions. SOLUTION

For sphere: Linear acceleration

μ Mg a1 = s = μs g M Angular acceleration α=

( μs Mg ) R = 5 μs g

2 2 R MR2 5 For plank: Linear acceleration a2 =

F − μ s Mg m

Mechanics II_Chapter 3_Part 4.indd 113

Given mass of disc m = 2 kg and radius R = 0.1 m …(1)

…(2)

(i) FBD of any one disc is Frictional force on the disc should be in forward direction. Let a0 be the linear acceleration of centre of mass of disc and α the angular acceleration about its centre of mass. Then, f f = m 2 fR 2f 2f τ α= = = = = 10 f 2 I mR 2 mR 2 ( 0.1 ) a0 =

…(3)

…(1) …(2)

2/9/2021 6:39:50 PM

3.114 JEE Advanced Physics: Mechanics – II

Since, there is no slipping between disc and truck.

f

a0

a = 9 ms−2

 and τ 1 =

( 0.6 )2 + ( 0.6 )2

= 0.85 Nm    iˆ τ 2 = r2 × f = 0.6 − ˆj − kˆ Similarly,

  and τ 1 = τ 2 = 0.85 Nm

z

(

)

Truck y

x

Therefore, ⎛ Acceleration of the ⎞ ⎛ Acceleration of ⎞ = ⎜⎝ ⎟⎠ contact point ⎟⎠ ⎜⎝ the truck ⇒ a0 + Rα = a

Show that if a rod held at angle θ to the horizontal and released, its lower end will not slip if the friction coefficient between rod and ground is greater than 3 ⎛ sin 2θ ⎞ ⎜ ⎟. 2 ⎝ 1 + 3 sin 2 θ ⎠ SOLUTION

⎛ f⎞ ⇒ ⎜ ⎟ + ( 0.1 ) ( 10 f ) = a ⎝ 2⎠ ⇒

PROBLEM 12

Point A is momentarily at rest, so

3 f =a 2

α=

2a 2 × 9 = N 3 3 ⇒ f =6 N ⇒ f =

τ mg ( l 2 ) cos θ 3 g cos θ = = I 2l ml 2 3

If at be the tangential acceleration of the centre of mass of rod C, then

Since, this force is acting in positive x-direction, so in vector form,  f = 6iˆ N

at =

( )

3 l α = g cos θ 2 4

20 cm = 0.2 m z

N

2

at

y x

P f

A

f

   (ii) τ = r × f  where, f = 6iˆ N (for both the discs)

( )

mg

θ

Q

atsinθ

C an

atcosθ

Now, μ N = max or μ N = maC sin θ ⇒

μN =

3 mg sin θ cos θ 4

Further, mg − N = may

Therefore, frictional torque on disk 1 about point O (centre of mass).    iˆ τ 1 = r1 × f = −0.1 ˆj − 0.1kˆ × 6iˆ Nm

3 mg cos 2 θ 4 Dividing equation (1) by (2), we get

(

 iˆ τ 1 = 0.6 kˆ − 0.6 ˆj

(

)

)

 ⇒ iˆ τ 1 = 0.6 kˆ − ˆj Nm

Mechanics II_Chapter 3_Part 4.indd 114

⇒

) ( )

y

f

  iˆ rP = r1 = −0.1 ˆj − 0.1 kˆ   andiˆ rQ = r2 = 0.1 ˆj − 0.1 kˆ

(

x

⇒

…(1)

N = mg − mat cos θ N = mg −

…(2)

3 sin θ cos θ 3 sin θ cos θ μ= 4 = 3 2 4 − 3 cos 2 θ 1 − cos θ 4

2/9/2021 6:40:04 PM

Chapter 3: Rotational Dynamics

⇒

μ=

⇒

μ=

3 sin θ cos θ 2

1 + 3 sin θ

=

Now, Ft + Mg sin θ = Mat

3 ⎛ 2 sin θ cos θ ⎞ ⎜ ⎟ 2 ⎝ 1 + 3 sin 2 θ ⎠

3 ⎛ sin ( 2θ ) ⎞ ⎜ ⎟ 2 ⎝ 1 + 3 sin 2 θ ⎠

1 Ft = − Mg sin θ 4 Here negative sign implies that direction of Ft is opposite to the component Mg sin θ .

A uniform stick of length L , mass M hinged at one end is released from rest at an angle θ0 with the vertical. When the angle with the vertical is θ , the hinge exerts a force Fr along the stick and Ft perpendicular to the stick. Calculate Fr and Ft. SOLUTION

PROBLEM 14

A man pushes a cylinder of mass m1 with the help of a plank of mass m2 as shown. There is no slipping at any contact. The horizontal component of the force applied by the man is F, find F

Let C be the centre of mass of the rod and ω be the angular speed of rod about point O at angle θ . By Law of Conservation of Mechanical Energy, we have

⇒

L 1 ⎛ ML2 ⎞ 2 ( cos θ − cos θ0 ) = ⎜⎝ ⎟ω 2 2 3 ⎠

3g ω = ( cos θ − cos θ0 ) L 2

…(1)

⎛ L⎞ Since, Fr − Mg cos θ = M ⎜ ⎟ ω 2 ⎝ 2⎠

…(2)

O

θ

c

θ0

m1

(a) the acceleration of the plank and the centre of mass of the cylinder, and (b) the magnitudes and directions of frictional forces at contact points.

We can choose my arbitrary directions of frictional forces at different contacts. In the final answer the negative values will show the opposite directions. Let the friction between plank and cylinder, i.e. at contact point B be f1, the friction between cylinder and ground, i.e. at contact point A be f 2, acceleration of the plank be ap = a1, acceleration of centre of mass of cylinder be ac = a2 and angular acceleration of cylinder about its CM be α . Directions of f1 and f 2 are as shown in Figure.

1 Fr = Mg ( 5 cos θ − 3 cos θ0 ) 2

θ

m2

SOLUTION

From equations (1) and (2), we get

O

…(4)

From equation (3) and (4), we get

PROBLEM 13

Mg

3.115

Fr

c

c′ A A′

m2

Angular acceleration of rod at this instant, ⎛L ⎞ Mg ⎜ sin θ ⎟ ⎝2 ⎠ 3 g sin θ = 2 2 L ML 3 Tangential acceleration of centre of mass C is

τ α= = I

⎛ L⎞ 3 at = rα = ( α ) ⎜ ⎟ = g sin θ ⎝ 2⎠ 4

Mechanics II_Chapter 3_Part 4.indd 115

m1

…(3)

Since, there is no slipping anywhere, so acceleration of plank equals the acceleration of top point of cylinder, so we have

2/9/2021 6:40:17 PM

3.116 JEE Advanced Physics: Mechanics – II SOLUTION

a1 = 2a2

When the rod first becomes horizontal, this situation is shown in Figure.

a2

A

a1 = 2 a2 Also, a1 =

F − f1 m2

…(2)

a2 =

f1 + f 2 m1

…(3)

If I be moment of inertia of cylinder about CM, then

α= ⇒ ⇒

α=

( f1 − f 2 ) R = ( f1 − f 2 ) R I

2 ( f1 − f 2 )

a2 = Rα =

2 ( f1 − f 2 )

(a) Solving equations (1), (2), (3) and (5), we get 8F 4F and a2 = 3 m1 + 8 m2 3 m1 + 8 m2

A (2m)g

mg

Finally

By Law of Conservation of Mechanical Energy, we have Loss in ⎛ ⎞ ⎛ Gain in ⎞ ⎛ Gain in ⎞ ⎜ RKE of ⎟ = ⎜ GPE of ⎟ + ⎜ GPE of ⎟ ⎜ Particle + Rod ⎟ ⎜ CM of Rod ⎟ ⎜ Particle ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(

)

⇒

1 I A ω 02 − ω 2 = ( 2m ) gl + mg ( 2l ) 2

⇒

2 1 ⎛ 2 m ( 2l ) ⎞ 2 ⎞ ⎛ 49 g − ω 2 ⎟ = 4 mgl + m ( 2l ) ⎟ ⎜ ⎜⎝ ⎠ ⎝ ⎠ 2 3 4l

⇒

ω2 =

…(5)

m1

P

Fx

m Initially

…(4)

Acceleration of bottom most point of cylinder is zero.

a1 =

2m

m1 R2 2

m1 R

r

Fy

…(1)

221g 20l

In the horizontal position of the rod, we have

τ = ( 2m ) gl + mg ( 2l ) = 4 mgl Since τ = Iα

Rα

(b) fB = f1 =

a2

m1 F 3 m1 F and f A = f 2 = 3 m1 + 8 m2 3 m1 + 8 m2

Since, all quantities are positive, they are correctly shown in the above Figures. PROBLEM 15

A uniform rod AB of length 2l and mass 2m has a particle of mass m attached at B. The rod is free to rotate in a vertical plane about a horizontal axis through A. When the rod is hanging at rest, with B 7 g . Find below A, it is given an angular velocity 2 l the reaction at the axis when the rod first becomes horizontal. Take g = 10 ms −2.

Mechanics II_Chapter 3_Part 4.indd 116

⇒

α=

2mgl + mg ( 2l ) 3g τ = = 2 5 l I ⎛ 2m ( 2l ) 2⎞ + m ( 2l ) ⎟ ⎜⎝ ⎠ 3

Let the CM of Rod plus Particle be at point P, a distance r from A. Then r= ⇒

( 2 m ) l + m ( 2l ) 3m

=

4l 3

⎛ 4l ⎞ ⎛ 3 g ⎞ ay = at = rα = ⎜ ⎟ ⎜ = 0.8 g ⎝ 3 ⎠ ⎝ 5l ⎟⎠

Since 3 mg − Fy = ( 3 m ) at ⇒

Fy = 3 mg − 3 m ( 0.8 g )

⇒

Fy = 0.6 mg , upwards

⎛ 4l ⎞ ⎛ 221g ⎞ Also, ax = ac = rω 2 = ⎜ ⎟ ⎜ = 14.73 g ⎝ 3 ⎠ ⎝ 20l ⎟⎠

2/9/2021 6:40:32 PM

Chapter 3: Rotational Dynamics

Force exerted by hinge in x-direction is

⇒

Fx = Fc = ( 3 m ) ac = 44.2mg

v = vA − μ gt = 31.3 −

F = Fx2 + Fy2 = 442 mN PROBLEM 16

A solid sphere is released from a height of 49 m from horizontal on an inclined plane as shown in Figure. All the surfaces are smooth except AB which has 2 μ = and length of AB is 49 m. After passing over 3 the rough surface, the sphere climbs to the right portion BD of incline. C 49 m

α

The remaining distance of 49 − 36 = 13 m the velocity is maintained at 22.36 ms −1, because of pure rolling the friction is zero. So, the height raised above the horizontal is H=

2

v 2 ( 22.36 ) = 25 m = 2g 2 × 10

From B to A, let pure rolling start at time t′, then v ′ = Rω ′

α

⇒

v − μ gt ′ = −ω R + α Rt ′

⇒

t′ =

2v v + Rω = 3.5 μ g 3.5 μ g

B

ω

(a) Find the height from horizontal to which it climbs on BD. (b) Now the sphere returns and goes up the incline AC. Find the linear velocity of sphere at A during the left-out motion. SOLUTION −1

From A to B, let pure rolling starts after time t. v = Rω

v

v

A

⇒

t′ =

ω

B

μmg

2 × 22.36 = 1.91 s 3.5 × ( 2 3 ) × 10

So, v ′ = v − μ gt

From C to A we have, velocity at A is vA = 2 × g × h = 2 × 10 × 49 = 31.3 ms

2 × 10 × 1.34 = 22.36 ms −1 3

where ω ′ = −ω + α t ′ and v ′ = v − μ gt ′

D

49 m

s = 36 m

Since velocity v = vA − at, so we have

So, total hinge force is given by

A

3.117

⇒ ⇒

2 × 10 × 1.91 = 9.62 ms −1 3 1 2 2 s ′ = 22.36 × 1.91 − × × 10 × ( 1.91 ) 2 3

v ′ = 22.36 −

⇒

vA − at = R ( α t )

⇒

⇒

⎛ μ mgR ⎞ vA − μ gt = R ⎜ t = 2.5 μ gt 2 ⎝ 2mr 5 ⎟⎠

Since s ′ is less than 49 m, velocity of sphere at A

⇒

t=

vA 31.3 = 1.34 s = 3.5 μ g 3.5 × ( 2 3 ) × 10

Now, s = vA t − ⇒

1 ( μ g ) t2 2

1⎛ 2⎞ 2 s = ( 31.3 )( 1.34 ) − ⎜ ⎟ ( 10 ) × ( 1.34 ) 2⎝ 3⎠

Mechanics II_Chapter 3_Part 4.indd 117

s ′ = 30.53 m

during the remaining journey is 9.62 ms −1. PROBLEM 17

A uniform disc of mass m and radius R is projected horizontally with velocity V0 on a rough horizontal floor, so that it starts off with a purely sliding motion at t = 0. After t0 second, it acquires a purely rolling motion as shown in Figure.

2/9/2021 6:40:47 PM

3.118 JEE Advanced Physics: Mechanics – II

Substituting in equation (1), we get V0

⎛ V ⎞ V = V0 − μ g ⎜ 0 ⎟ ⎝ 3μ g ⎠

t=0

t = t0

(a) Calculate the velocity of the centre of mass of the disc at t0. (b) Assuming the coefficient of friction to be μ , calculate t0. Also calculate the work done by the frictional force as a function of time and the total work done by it over a time t much longer than t0. SOLUTION

Between the time t = 0 to t = t0, there is forward sliding, so friction f is leftwards and maximum, i.e. μ mg. For time t > t0, friction f will become zero, because now pure rolling has started i.e., there is no sliding (no relative motion) between the points of contact. V0

V

ω

a

α

t = t0

t=0

So, for time t < t0, the linear retardation and the angular acceleration respectively are f a = = μg m

{∵ f

2 μ gt0 R For pure rolling to take place V = Rω i.e.,

V0 − μ gt0 = 2 μt0

⇒

t0 =

V0 3μ g

Mechanics II_Chapter 3_Part 4.indd 118

1 1 1 mV 2 + Iω 2 − mV02 2 2 2 2

1 1 2 1⎛ 1 ⎞ ⎛ 2 μ gt ⎞ ⇒ W = m ( V0 − μ gt ) + ⎜ mR2 ⎟ ⎜ − mV02 ⎠ ⎝ 2 ⎟⎠ 2 2⎝ 2 2

(

m 2 V0 + μ 2 g 2 t 2 − 2V0 μ gt + 2 μ 2 g 2 t 2 − V02 2 μ mgt ⇒ W= ( 3 μ gt − 2V0 ) 2 ⇒ W=

…(1) …(2)

)

For t > t0, friction force is zero i.e., work done in friction is zero. Hence the energy will be conserved. Therefore, total work done by friction over a time t much longer then t0 is total work done up to time t0 (because beyond the work done by friction is zero) which is equal to W=

mμ gt0 ( 3 μ gt0 − 2V0 ) 2

Substituting t0 =

Now let V be the linear velocity and ω , the angular velocity of the disc at time t = t0 then

and ω = α t0 =

V=

⇒ W=

= μ mg }

fR 2μ g τ α= = = 2 R I mR 2

V = V0 − at0 = V0 − μ gt0

2 V0 3 For t ≤ t0, the linear velocity of disc at any time t is 2 μ gt V = V0 − μ gt and angular velocity is ω = α t = . R From work-energy theorem, work done by friction up to time t equals the change in kinetic energy, so we have ⇒

W= ⇒

V0 , we get 3μ g

mV0 ( V0 − 2V0 ) 6

W=−

mVo2 6

PROBLEM 18

A solid cylinder of mass m is rolling on a rough horizontal surface with velocity v0. It collides elastically with a cubical block of same mass at rest. The height of centre of mass of both the bodies is same. Assume that there is no friction between the cylinder and the block and coefficient of friction between all other surfaces is μ, find the

2/9/2021 6:41:08 PM

Chapter 3: Rotational Dynamics

m v0

Since, s = vr t1 +

m

⇒ (a) velocity of block when cylinder starts pure rolling again. (b) time after first collision at which the second collision will take place. SOLUTION

In head on elastic collision between two equal masses velocities are exchanged. v0

⇒

v

…(1)

where v = at1 = ( μ g ) t1 and ⎛ fR ⎞ ω = ω0 − αt = ω0 − ⎜ t ⎝ I ⎟⎠ 1 ⇒

⎛ 2μ g ⎞ ω = ω0 − ⎜ t ⎝ R ⎟⎠ 1

So, from (1) i.e., v = Rω , we get 2 μ gt1 ⎞ ⎛ μ gt1 = R ⎜ ω 0 − ⎟ ⎝ R ⎠ ⇒

2 μ gt1 ⎞ ⎛v μ gt1 = R ⎜ 0 − ⎟ ⎝ R R ⎠

⇒

v t1 = 0 3μ g

Since, v = μ gt1 ⇒

⎛ v ⎞ v v = ( μg )⎜ 0 ⎟ = 0 3 ⎝ 3μ g ⎠

v02 v2 2 v02 − 0 = 3 μ g 9μ g 9 μ g

1 2 ar t 2 Substituting the values, we get − s = vr t +

−

s

v = Rω

1 2 ar t1 2

After the cylinder starts pure rolling friction on cylinv der becomes zero and vr = v ′ − v = 0 , ar = − μ g 3 Let the second collision takes place after time t2, then

v′

Let the cylinder have radius R and let it start rolling after time t1, then

s=

2 v02 v0 1 = t2 − μ gt22 9 μg 3 2

⇒

⎛ 5 + 1⎞ t2 = ⎜ v0 ⎝ 3 μg ⎟⎠

⇒

t = t1 + t2 =

(

5 + 2 ) v0 3 μg

PROBLEM 19

A uniform thin rod of mass m and length l is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Describe the trajectory of the centre of mass. Find the velocity of centre of mass of the rod at the instant when it makes an angle θ with horizontal. Also find the velocity of centre of mass when it touches the surface. SOLUTION

Since the floor is smooth so the mechanical energy of the rod will remain conserved. Further, no horizontal force acts on the rod, hence the centre of mass moves vertically downwards in a straight line. Thus, velocities of centre of mass and the lower end B are in the directions shown in Figure.

During this time t1, the block moves forward under a retardation μ g, so if v ′ be the velocity of block when cylinder starts pure rolling at t1, then v ′ = v0 − μ gt1 = v0 −

v0 2v0 = 3 3

After the first collision, we have vr = v0 , ar = −2 μg

Mechanics II_Chapter 3_Part 4.indd 119

3.119

CM

h =  (1− sinθ ) 2

IC VC

ω

 sin θ 2

θ B

VB

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3.120 JEE Advanced Physics: Mechanics – II

The location of IC at this instant is found by draw  ing perpendiculars to vC and vB at respective points and IC is located at the point of intersection of these two perpendiculars. So, the rod may now be assumed to be in pure rotational motion about IAOR passing through IC with angular speed ω . Applying Law of Conservation of Mechanical Energy. we get ⎛ Decrease in GPE ⎞ ⎛ Increase in RKE ⎞ ⎜⎝ of CM of Rod ⎟⎠ = ⎜⎝ of Rod about IAOR ⎟⎠ 1 I IAORω 2 2

⇒

mgh =

⇒

⎞ l 1 ⎛ ml 2 ml 2 mg ( 1 − sin θ ) = ⎜ + cos 2 θ ⎟ ω 2 ⎠ 2 2 ⎝ 12 4

⇒

8 2 2 3 mga ma ω = 3 2

⇒

ω2 =

Now, α =

⇒

τ mg ( a sin 60° ) + mg ( 2 a sin 60° ) = 16 2 I ma 3 9 3 g 32 a

12 g ( 1 − sin θ ) l ( 1 + 3 cos 2 θ )

 vc =

B

⇒

3 gl ( 1 − sin θ ) cos θ

60°

( 1 + 3 cos θ ) 2

3 gl 4

PROBLEM 20

A uniform rod AB of mass m and length 2a has a particle of mass m attached to the end B . The rod can rotate in a vertical plane about a smooth axis through A. If the body is slightly displaced from the position in which B is vertically above A, find the magnitude of the reaction at the axis, which is horizontal, when π the rod has rotated through radians. 3

A

A

Force applied by hinge in x-direction Fx = 2max = 2m ( an sin 60° − at cos 60° ) ⇒

⎛ 27 3 27 3 ⎞ Fx = 2m ⎜ − ⎟g ⎝ 64 128 ⎠

⇒

Fx =

54 3 mg 128

3a 2 Fx

SOLUTION

By Law of Conservation of Energy, we get

⇒

B

2

When the rod touches the surface, then θ = 0°, so  vc =

…(2)

27 3 27 ⎛ 3a ⎞ ⎛ 3a ⎞ at = ⎜ ⎟ α = g and an = ⎜ ⎟ ω 2 = g ⎝ 2 ⎠ ⎝ ⎠ 64 2 32

 ⎛ l ⎞ Now, vc = ⎜ cos θ ⎟ ω ⎝2 ⎠ ⇒

…(1)

Now centre of mass of the system (point C) is at a 3a from point A. Acceleration of C has distance of 2 two components at and an, where

Solving this equation, we get

ω=

α=

9 g 16 a

A

C 60° an

a 2 60°

x

at

y

Fy

⎛ Gain in ⎞ ⎛ Loss in ⎞ ⎛ Loss in ⎞ ⎜ RKE of ⎟ = ⎜ GPE of ⎟ + ⎜ GPE of ⎟ ⎜ Rod ⎟ ⎜ CM of Rod ⎟ ⎜ Particle ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Similarly, Fy + ( 2m ) g = ( 2m ) ay ⇒

1 ⎡ m ( 2a ) 2⎤ + m ( 2a ) ⎥ ω 2 = ⎢ 2⎣ 3 ⎦ mg ( 2 a ) ( 1 − cos 60° ) + mg ( a ) ( 1 − cos 60° )

Fy = 2m ( at sin 60° + an cos 60° ) − 2mg

⇒

270 ⎛ 81 27 ⎞ Fy = 2m ⎜ + g − 2mg = mg − 2mg ⎝ 128 64 ⎟⎠ 128

2

Mechanics II_Chapter 3_Part 4.indd 120

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Chapter 3: Rotational Dynamics

14 mg 128

⇒

Fy =

So,

F = Fx2 + Fy2 =

⇒ mg 559 32

l v0 = v + ω cos θ 2 Solving equations (1), (2) and (3), we get

A rod of length l forming an angle θ with the horizontal strikes a frictionless floor at A with its centre of mass velocity v0 and no angular velocity. Assuming that the impact at A is perfectly elastic, find the angular velocity of the rod immediately after the impact.

A

θ

⎛ Relative speed ⎞ ⎛ Relative speed ⎞ ⎜⎝ of approach ⎟⎠ = ⎜⎝ of separation ⎟⎠

⇒

PROBLEM 21

v0

3.121

ω=

…(3)

6v0 cos θ

l ( 1 + 3 cos 2 θ )

PROBLEM 22

A cylinder of mass M and radius R is rotated in a uniform V shaped groove with constant angular velocity ω . The coefficient of friction between the cylinder and each surface is μ . What torque must be applied to the cylinder to keep it rotating? ω

SOLUTION

45°

If v be the linear velocity of rod after impact (upwards), ω be the angular velocity of rod and J be the linear impulse at A during impact, then by Impulse − Momentum Theorem, we have v

45°

SOLUTION

The forces acting on the cylinder are shown in Figure.

ω J

θ

v0

θ

A During impact

A

1

After impact

J = mv − ( − mv0 )

⇒

J = m ( v + v0 )

…(1)

For vertical equilibrium of cylinder, we have N1 cos 45° + N 2 cos 45° + μ N1 cos 45° =

Further by angular impulse angular momentum theorem, we have 2

ml ⎛ l ⎞ J ⎜ cos θ ⎟ = Iω = ω ⎝2 ⎠ 12

…(2)

Since the collision is elastic, so at the point of impact, e = 1.

μ N 2 cos 45° + Mg ⇒

N1 ( 1 + μ ) + N 2 ( 1 − μ ) = 2 Mg

…(1)

For horizontal equilibrium of cylinder, N1 sin 45° = N 2 sin 45° + μ N1 sin 45° + μ N 2 sin 45° ⇒

v ω 2

N1 ( 1 − μ ) − N 2 ( 1 + μ ) = 0

…(2)

Solving these two equations, we get θ

Mechanics II_Chapter 3_Part 4.indd 121

N2 mg

J = Δp = p f − pi ⇒

N1

θ

2 N1 ( 1 + μ 2 ) = 2 Mg ( 1 + μ )

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3.122 JEE Advanced Physics: Mechanics – II

⇒

2 Mg ( 1 + μ )

N1 =

2 (1 + μ2 )

Hence, the mechanical energy of the cylinder will be conserved i.e.,

( PE + KE ) = ( PE + KE )2

From (2), we get 2 Mg ( 1 − μ )

N2 =

2 (1 + μ2 )

The required torque is,

τ = μ N1 R + μ N 2 R = μ R ( N1 + N 2 ) ⇒

τ=

2 μ MgR

(1 + μ2 )

PROBLEM 23

A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in Figure.

1 1 mgR + 0 = mgR cos θ + Iω 2 + mv 2 2 2 1 2 where, I = mR and for no slipping at point of 2 v contact, we have ω = R ⇒

⇒

2 1⎛ 1 ⎞⎛ v ⎞ 1 mgR = mgR cos θ + ⎜ mR2 ⎟ ⎜ 2 ⎟ + mv 2 ⎠⎝ R ⎠ 2 2⎝ 2

⇒

3 2 v = gR ( 1 − cos θ ) 4

⇒

v2 =

⇒

v2 4 = g ( 1 − cos θ ) R 3

4 gR ( 1 − cos θ ) 3 …(1)

R

θ N=0 mg

There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine: (a) the angle θc through which the cylinder rotates before it leaves contact with the edge. (b) the speed of the centre of mass of the cylinder before leaving contact with the edge, and (c) the ratio of the translational to rotational kinetic energies of the cylinder when its centre of mass is in horizontal line with the edge.

At the time of leaving contact, normal reaction N = 0 and θ = θc hence, mg cos θ =

mv 2 R

v2 = g cos θ R From equations (1) and (2), ⇒

…(2)

4 g ( 1 − cos θc ) = g cos θc 3

SOLUTION

⇒

The cylinder rotates about the point of contact as shown in Figure.

7 cos θc = 1 4

⇒

cos θc =

4 7

PROBLEM 24

A spherical ball of radius r and mass m collides with a plank of mass M kept on a smooth horizontal surface. Before impact, the centre of the ball has a velocity v0 and angular velocity ω 0 as shown in Figure.

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Chapter 3: Rotational Dynamics

3.123

Solving equations (1), (2), (3) and (4), we get v1 =

v0

The normal velocity is reversed with same magnitude and the ball stops rotating after the impact. Find the distance on the plank between first two impacts of the ball. The coefficient of friction between the ball and the plank is μ . Assume that plank is large enough.

Now, actual path of the ball is a projectile whose time of flight will be

Therefore, the required distance is

y v1

s = vr T x

⇒

v2

Let the horizontal velocities of the ball and the plank be v1 and v2 in opposite directions as shown in Figure, then by Law of Conservation of Linear Momentum in horizontal direction, we get mv1 = Mv2

…(1)

According to Impulse Momentum Theorem, Change in ⎛ Linear Impulse ⎞ ⎛ ⎞ ⎜ of the ball in ⎟ = ⎜ Linear Momentum ⎟ ⎜ vertical direction ⎟ ⎜ in vertical direction ⎟ ⎝ ⎠ ⎝ ⎠ ⇒

J = Ndt

⇒

J = 2mv0

s=

4 ⎛ M + m ⎞ v0 rω 0 ⎜ ⎟ 5⎝ M ⎠ g

PROBLEM 25

A uniform rod of length L and weight W stands vertically touching a vertical wall (y-axis). When slightly displaced, its lower end begins to slide along the floor (x-axis). Obtain an expression for the angular velocity ( ω ) of the rod as a function of θ . Determine the distance moved by the lower end at which the rod no longer touches the vertical wall. Neglect friction everywhere. y

…(2)

Change in ⎛ Linear Impulse ⎞ ⎛ ⎞ ⎜ on the ball in ⎟ ⎜ Linear Momentum ⎟ ⎜ horizontal ⎟ = ⎜ in horizontal ⎟ ⎜ ⎟ ⎜ ⎟ direction direction ⎝ ⎠ ⎝ ⎠ ⇒

2v0 g

2⎛ M+m⎞ ⎛ M+m⎞ vr = v1 + v2 = ⎜ v = rω ⎝ M ⎟⎠ 1 5 ⎜⎝ M ⎟⎠ 0

v0

μN

g

=

Relative velocity of ball with respect to plank in horizontal direction is

The forces during impact are shown in Figure.

μN

2vy

T=

SOLUTION

N

2 m⎛2 ⎞ rω 0 and v2 = ⎜ rω 0 ⎟⎠ 5 M⎝ 5

μ J = mv1

θ

x

SOLUTION

…(3)

According to Angular Impulse Angular Momentum Theorem, we have

The position of instantaneous axis of rotation (IAOR) is shown in Figure.

Change in ⎛ Angular Impulse ⎞ ⎛ ⎞ ⎜ on the ball about ⎟ = ⎜ Angular Momentum ⎟ ⎜ Centre of Mass ⎟ ⎜ about Centre of Mass ⎟ ⎝ ⎠ ⎝ ⎠ ⇒

( μ J ) r = Iω 0 = ⎛⎜⎝

Mechanics II_Chapter 3_Part 4.indd 123

2 2⎞ mr ⎟ ω 0 ⎠ 5

…(4)

2/9/2021 6:42:04 PM

3.124 JEE Advanced Physics: Mechanics – II

Fx = 0

l ⎛ l ⎞ C ≡ ⎜ cos θ , sin θ ⎟ ⎝2 ⎠ 2

⇒

ax = 0

l r = = half of the diagonal 2 All surfaces are smooth. Therefore, mechanical energy will remain conserved. So, ⎛ Decrease in GPE ⎞ ⎛ Increase in RKE ⎞ ⎜⎝ of CM of Rod ⎟⎠ = ⎜⎝ of Rod about IAOR ⎟⎠ . l 1 mg ( 1 − sin θ ) = Iω 2 2 2

⇒

ml 2 where, I = + mr 2 12

{about IAOR}

5

Substitute, ax = 0 in equation (4), we get −ω 2 cos θ − α sin θ = 0 ⇒

−

3g 3g cos θ ( 1 − sin θ ) + sin θ cos θ = 0 l 2l

Solving this, we get 2 3 At this moment the lower end has moved a distance d given by, sin θ =

l 1 ⎛ ml 2 ⎞ 2 mg ( 1 − sin θ ) = ⎜ ⎟ω 2 2⎝ 3 ⎠

d = l cos θ =

3 g ( 1 − sin θ ) l ( 3 g 1 − sin θ ) ⇒ ω2 = l Differentiating w.r.t. time, we get

ω=

3g ⎛ dω ⎞ ⎛ dθ ⎞ =− cos θ ⎜ 2ω ⎜ ⎝ dt ⎟⎠ ⎝ dt ⎟⎠ l Since,

dω =α dt

…(2)

{the angular acceleration of rod}

dθ =ω {the angular velocity of rod} dt Substituting in equation (2), we have 3g cos θ 2l

5 l 3

PROBLEM 26

and

α=−

2

θ

…(1)

ml 2 ml 2 ml 2 ⇒ I= + = 12 4 3 Substituting in equation (1), we get

⇒

3

A cylinder of mass M is placed on top of a block of mass 2M. The friction coefficient between the block and the horizontal plane is zero and the friction coefficient between the cylinder and the top surface of block is μ. A small body of mass M travelling with a velocity v0 hits the block horizontally such that the collision is completely elastic. Find the maximum value of v0 for which the cylinder does start pure rolling before it topples from the top of the block. M d

…(3)

M

v0

2M

The x-coordinate of centre of mass at angle θ is x=

l cos θ 2

Differentiating twice w.r.t. time, we get

ω 2l αl cos θ − sin θ …(4) 2 2 The rod will lose contact with the vertical wall at the instant when normal reaction by the wall (along x-axis) becomes zero. So, ax = −

Mechanics II_Chapter 3_Part 4.indd 124

SOLUTION

Let v be the velocity of block after collision. Applying the equation of head on elastic collision between two particles, we get v=

2 2M v0 = v0 M + 2M 3

The forces acting on the cylinder and the block till pure rolling starts are shown in Figure.

2/9/2021 6:42:12 PM

Chapter 3: Rotational Dynamics a1

3.125

ω

ω0 2

v

a2

v

μ MgR

α=

2

MR 2

=

Initially (at t = 0)

Then by law of conservation of angular momentum applied to the initial and the final situations, we get

2μ g R

Let the pure rolling starts after time t, then v − a2 t = a1t + ( α t ) R ⇒

t=

v = a1 + a2 + α R

⇒

t=

4v0 2v = 7 μ g 21μ g

Finally (at t0) when the discs attain common angular velocity

at rest

μ Mg a1 = = μg M μ Mg μ g = a2 = 2M 2

v μg μg + + 2μ g 2

I ω 0 = 2I ω

ω0 2 The angular velocity of each disc varies due to the torque τ of the friction forces. To calculate τ , let us take an elementary ring with radii r and r + dr as shown in Figure. ⇒

ω=

1 ⎛ ⎞ 1 Now, d = ⎜ vt − a2 t 2 ⎟ − a1t 2 ⎝ ⎠ 2 2 1 ( a1 + a2 ) t 2 2

⇒

d = vt −

⇒

8v02 12v02 d= − 63 μ g 441μ g

⇒

441μ gd = 56v02 − 12v02

⇒

44v02 = 441μ gd

⇒

v0 =

⇒

21 μ gd 2 11

( v0 )min

The torque of the friction forces acting on the given ring is equal to dτ = fr⊥ = μ ( dN ) r = ⎡⎣ μ ( dm ) g ⎤⎦ r ⇒

where, m is the mass of each disc. Integrating this with respect to r between 0 and r0, we get

21 μ gd = 2 11

PROBLEM 27

A uniform disc of radius r0 lies on a smooth horizontal plane. A similar disc spinning with the angular velocity ω 0 is carefully lowered onto the first disc. How soon do both discs spin with the same angular-velocity if the friction coefficient between them is equal to μ . SOLUTION

If I be the moment of inertia of each disc relative to common rotation axis as shown in Figure.

Mechanics II_Chapter 3_Part 4.indd 125

⎛ mg ⎞ ⎛ 2 μ mg ⎞ 2 dτ = μ ⎜ 2 ⎟ ( 2π rdr ) r = ⎜ r dr ⎝ π r0 ⎠ ⎝ r02 ⎟⎠

⇒

τ=

2 μ mgr0 = constant 3

α=

τ ( 2 μ mgr0 3 ) 4 μ g = = = constant I 3 r0 mr02 3

(

)

The angular speed of lower disc increases from zero ω to 0 with constant angular acceleration α , so 2

⇒

ω0 = 0 + αt 2 3r ω ω t= 0 = 0 0 2α 8μ g

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3.126 JEE Advanced Physics: Mechanics – II

PRACTICE EXERCISES SINGLE CORRECT CHOICE TYPE QUESTIONS This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

Three identical rods, each of length l, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is 3 l (B) l (A) 2 2 l l (C) (D) 2 3

(A) (B) (C) (D) 5.

One quarter sector is cut from a uniform disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is

A spherical body of radius R is allowed to roll without slipping down an incline to reach the bottom with a speed v0 . The incline is then made smooth by waxing and the body is allowed to slide without rolling now 5 to reach the bottom with a speed v0 . The radius of 4 gyration of the body about an axis passing through its centre is 2 3 R (B) (A) R 5 4 (C)

6.

(A)

MR 2

2

MR2 (C) 8 3.

4.

(B) (D)

MR 4

2 MR

7.

Two particles of equal mass m at A and B are connected by a rigid light rod AB, lying on a smooth horizontal table. An impulse J is applied at A in the plane of the table and perpendicular at AB. Then the velocity of particle at A is J (A) zero (B) 2m J 2J (D) (C) m m

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 126

(D)

4R 3

(B)

5 R 2

R 17 2

R 15 2 An impulse J is applied on a ring of mass m along a line passing through its centre O. The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is (C) R 13

2

Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

4 R 3

A uniform disc of radius R lies in x -y plane with its centre at origin. Its moment of inertia about the axis x = 2R and y = 0 is equal to the moment of inertia about the axis y = d and z = 0, then d equals (A)

2

0.42 m from mass of 0.3 kg 0.70 m from mass of 0.7 kg 0.98 m from mass of 0.3 kg 0.98 m from mass of 0.7 kg

(D)

J

(A)

J m

O

(B)

J 2m

J J (D) 4m 3m A wheel is rolling without sliding on a horizontal surface. The centre of the wheel moves with a constant speed v0 . Consider a point P on the rim which is at (C)

8.

2/9/2021 6:37:13 PM

3.127

Chapter 3: Rotational Dynamics the top at time t = 0. The square of speed of point P is plotted against time t. The correct plot is (R is radius of the wheel) (A) 2 v

4v 20

πR 2v0

ladder is μ . The angle θ at which the ladder will stay in equilibrium is (A) θ = tan −1 ( μ )

(B) θ = tan −1 ( 2 μ )

⎛ μ⎞ (C) θ = tan −1 ⎜ ⎟ ⎝ 2⎠

⎛ 1 ⎞ (D) θ = tan −1 ⎜ ⎝ 2 μ ⎟⎠

11. A horizontal turn table in the form of a disc of radius r carries a gun at G and rotates with angular velocity ω 0 about a vertical axis passing through the centre O. The increase in angular velocity of the system if the gun fires a bullet of mass m with a tangential velocity v with respect to the gun is (moment of inertia of gun + table about O is I 0 )

3πR 2v 0

(B) 4v 02

G O

πR v0

(C)

v2

(A)

v 2r

(B)

mvr 2I 0

(C)

2mvr I0

(D)

mvr I 0 + mr 2

4v 20

(D)

πR

2πR

v0

v0

12. A rod of mass m and length l is hinged at one of its end A as shown in figure. A force F is applied at a distance x from A. The acceleration of centre of mass ( a ) varies with x as

v2

A

2v 20

x πR

F

v0

πR

2 v0

9.

2π R

v0

A thin uniform rod of mass m moves translationally with acceleration a due to two antiparallel forces of lever arm l. One force is of magnitude F and acts at one extreme end. The length of the rod is F ⎞ ⎛ (A) l ⎜ 1 + ⎟ ⎝ ma ⎠

F ⎞ ⎛ (B) 2l ⎜ 1 + ⎟ ⎝ ma ⎠

l⎛ F ⎞ (C) ⎜ 1+ ⎟ 2⎝ ma ⎠

mal (D) ma + F

10. A ladder of length l and mass m is placed against a smooth vertical wall, but the ground is not smooth. Coefficient of friction between the ground and the

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 127

(A)

(B)

a

a

3F Slope = 2m

x

(C)

a Slope =

x

3F m

x

(D)

3F m

Slope =

a

x

2/9/2021 6:37:23 PM

3.128 JEE Advanced Physics: Mechanics – II 13. A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about the origin O is y

ω M O

R

2 (B) I sin θ

(C) I cos 2 θ

⎛θ⎞ (D) I cos 2 ⎜ ⎟ ⎝ 2⎠

17. Two uniform rods of equal length but different masses are rigidly joined to form an L-shaped body, which is then pivoted as shown. If in equilibrium the body is in M the shown configuration, ratio will be m O

x

(B) MR2ω

⎛ 3⎞ (C) ⎜ ⎟ MR2ω ⎝ 2⎠

(D) 2 MR2ω

m

90° 30

°

⎛ 1⎞ (A) ⎜ ⎟ MR2ω ⎝ 2⎠

14. A table fan, rotating at a speed of 2400 rpm is switched off and the resulting variation of the rpm with time is shown in the figure. The total number of revolutions of the fan before it comes to rest is

(A) 420 (C) 190

(A) I

(B) 280 (D) 380

15. A stick of length L and mass M lies on a frictionless horizontal surface on which it is free to move in any way. A ball of mass m moving with speed v collides elastically with the stick as shown in Figure. If after the collision ball comes to rest, then what should be the mass of the ball?

M

(A)

2

(B) 2

(C)

3

(D) 3

18. The minimum value of l so that truck can avoid hitting the dead end, without toppling the block kept on it is

(A)

2 hv 2 bg

(B)

bv 2 2 hg

(C)

2bv 2 hg

(D)

hv 2 2bg

19. A disc of radius R, having angular velocity ω rolls without slipping along positive x-axis. The velocity of point P at the instant shown in figure is y

(A) m = 2 M

(B) m = M

M M (D) m = 2 4 16. Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to (C) m =

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 128

P ω r θ R x

 (A)iˆ vP = ω ⎡⎣ ( r sin θ ) iˆ + ( r cos θ ) ˆj ⎤⎦  (B)iˆ vP = ω ⎡⎣ ( r sin θ ) iˆ − ( r cos θ ) ˆj ⎤⎦  (C)iˆ vP = ω ⎡⎣ ( R + r sin θ ) iˆ − ( r cos θ ) ˆj ⎤⎦  (D)iˆ vP = ω ⎡⎣ ( R + r sin θ ) iˆ + ( r cos θ ) ˆj ⎤⎦

2/9/2021 6:37:33 PM

Chapter 3: Rotational Dynamics

3.129

20. AB and CD are two identical rods each of length l and mass m joined to form a cross. The moment of inertia of these two rods about a bisector of the angle between the rods is C

B

O

(A)

ml 2 6

(B)

ml 2 3

(C)

ml 2 12

(D)

2ml 2 3

v A

v sin α sin β

v cos α (C) cos β

(B)

v sin β sin α

(C)

Mv02 2

(D) Mv02

(B) v1 > v2 (C) v1 < v2 (D) data is insufficient 23. A uniform ring of mass M and radius R is in uniform pure rolling motion on a horizontal surface. The velocity of the centre of ring is v0 . The kinetic energy of the segment ACB is

v

O

At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains  ω throughout. The magnitude of the torque ( τ ) on the system about O, as a function of time is best represented by which plot? (B)

(A)

|τ |

|τ |

v cos β (D) cos α

22. A solid sphere and a hollow sphere of equal mass and radius are placed over a rough horizontal surface after rotating them about their respective centre of mass with same angular velocity ω 0 . Once the pure rolling starts let v1 and v2 be the linear speeds of their centres of mass. Then (A) v1 = v2

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 129

Mv02 Mv02 + 2 π

ω



β

(A)

(B)

z

21. A rod of length l slides down along the inclined wall as shown in figure. At the instant shown in figure the speed of end A is v, then the speed of B will be α

Mv02 Mv02 − 2 π

24. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω , as shown in the figure.

D

B

(A)

O

T

t

(C)

O

T

t

(D) |τ | O

|τ |

T

t

O

T

t

25. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as beads slide down are

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3.130 JEE Advanced Physics: Mechanics – II y

x

(A) angular velocity and total energy (kinetic and potential). (B) total angular momentum and total energy. (C) angular velocity and moment of inertia about the axis of rotation. (D) total angular momentum and moment of inertia about the axis of rotation. 26. A uniform rod of length 2a and mass m lies at rest on a smooth horizontal table. A perfectly elastic particle having same mass as the rod moving with speed u on the table in a direction perpendicular to the rod, strikes one end of the rod. The final kinetic energy of the rod is 4 mu2 13 4 (C) mu2 25

(A)

1 mu2 4 8 (D) mu2 25

(B)

27. A uniform horizontal circular platform of mass 200 kg is rotating at 10 rpm about a vertical axis passing through its centre. A boy of mass 50 kg is standing on its edge. If the boy moves to the centre of the platform, the frequency of rotation would be (A) 7.5 rpm

(B) 12.5 rpm

(C) 15 rpm

(D) 20 rpm

28. A plank P is placed on a hollow cylinder C, which rolls on a horizontal surface as shown. No slippage is there at any of the surfaces in contact. Both have equal mass say M (each) and if v is the velocity of centre of mass of the cylinder C, then the ratio of the kinetic energy of plank P to the cylinder C is

(A) 1 : 1 (C) 3 : 8

(B) 2 : 1 (D) 8 : 11

(A)

π MR2 12

⎛ 8 10π ⎞ (C) ⎜ − MR2 ⎝ 3 16 ⎟⎠

⎛4 π⎞ (B) ⎜ − ⎟ MR2 ⎝ 3 4⎠ ⎛4 π⎞ (D) ⎜ − ⎟ MR2 ⎝ 3 6⎠

30. A flat rail road car is accelerating along the positive x-axis with an acceleration ap . A sphere is placed over the car. The friction between car and sphere is not sufficient to support pure rolling of sphere. The correct statement is (A) The sphere will slip and force of friction on sphere is along −x direction (B) The sphere will slip and force of friction on sphere is along +x direction (C) Acceleration of sphere is along −x direction (D) None of the above 31. A hollow straight tube of length 2l and mass m can turn freely about its centre on a smooth horizontal table. Another smooth uniform rod of same length and mass is fitted into the tube so that their centres coincide. The system is set in motion with an initial angular velocity ω 0 . The angular velocity of the tube at the instant when the rod slips out of the tube ω0 ω (B) (A) 0 4 5 ω0 ω0 (C) (D) 7 2 32. A cubical block of side a is moving with velocity v on a horizontal smooth plane as shown in figure. It hits a ridge at point O. If the moment of inertia of the block about an axis passing through centre of gravity 1 is ma 2, then the angular speed of the block after it hits 6 O is

29. Four holes of radius R are cut from a thin square plate of side 4R and mass M. The moment of inertia of the remaining portion about z-axis is

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 130

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Chapter 3: Rotational Dynamics

(A) (C)

3v 4a 3v 2a

(B)

3v 2a

(D) zero

33. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle θ with the vertical ⎛ 1⎞ (A) θ = cos −1 ⎜ ⎟ ⎝ 3⎠ ⎛ 2⎞ (B) θ = cos −1 ⎜ ⎟ ⎝ 3⎠ (C) θ = 60° (D) θ = 0 34. Two men each of mass m stand on the rim of a horizontal circular disc, diametrically opposite to each other. The disc has a mass M and is free to rotate about a vertical axis passing through its centre of mass. Each man starts simultaneously along the rim clockwise and reaches its original starting position on the disc. The angle turned through by the disc with respect to the ground (in radian) is

πm M+m 4π m (C) 2M + m (A)

2π m 4m + M 8π m (D) 4m + M (B)

35. A cubical block of side L rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

(A) infinitesimal

(B)

mg 4

mg (D) mg ( 1 − μ ) 2 36. Three identical cylinders of radius R are in contact. Each cylinder is rotating with angular velocity ω . A thin belt is moving without sliding on the cylinders. If P and Q are two points of belt which are in contact with the cylinder, then the magnitude of velocity of point P with respect to Q. (C)

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 131

3.131

(A) 2Rω

(B) Rω

Rω (D) 3Rω 2 37. A balance is made of a rigid rod free to rotate about a point not at the centre of the rod. When an unknown mass m is placed in the left hand pan, it is balanced by a mass m1 placed in the right hand pan, and similarly when the mass m is placed in the right pan, it is balanced by a mass m2 in the left hand pan. Neglecting the masses of the pans, m is (C)

1 ( m1 + m2 ) 2 1 (C) m12 + m22 2

(A)

(B)

m1m2

(D)

m12 + m22 2

38. A uniform sphere of radius R is placed on a rough horizontal surface and given a linear velocity v0 and angular velocity ω 0 as shown. For the sphere to come to rest after moving some distance to the right, we have

v0

(A) v0 = Rω 0

(B) 5v0 = 2Rω 0

(C) 2v0 = 5Rω 0

(D) 2v0 = Rω 0

39. A stone of mass 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can stand is 16 N. The maximum velocity of revolution that can be given to the stone without breaking the string is (A) 20 ms −1 (C) 14 ms −1

(B) 16 ms −1 (D) 12 ms −1

40. At some instant, a particle is moving along a straight line 2x − 3 y = 2 and its co-ordinates on that line are ( 4, 2 ) . Now at another instant the same particle is moving along a straight line 3 x + 4 y = 7 and its coordinate are ( 1, 1 ) . The co-ordinates of the axis about which it is in pure rotation are

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3.132 JEE Advanced Physics: Mechanics – II ⎛ 8 16 ⎞ (A) ⎜ , ⎝ 9 9 ⎟⎠

⎛ 5 6 ⎞ , (B) ⎜ ⎝ 17 17 ⎟⎠

(A) μ > 1

(B) μ


1 2

41. A disc is rotating with an angular velocity ω 0 . A  constant retarding torque is applied on it to stop ω the disc. The angular velocity becomes 0 after n 2 rotations. How many more rotations will it make before coming to rest? n n (B) (A) 3 2 (C) n (D) 2n 42. A wire of mass m and length l is bent in the form of a quarter circle. The moment of inertia of this wire about an axis passing through the centre of the quarter circle and perpendicular to the plane of the quarter circle is

( Take π

2

= 10

)

(A)

3 ml 2 5

(B) ml 2

(C)

ml 2 5

(D)

2ml 2 5

R is removed from a circular 3 disc of radius R and mass 9M . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

43. A small disc of radius

45. A disc of mass m0 rotates freely about a fixed horizontal axis through its centre. A thin cotton pad is fixed to its rim, which can absorb water. The mass of water dripping onto the pad is μ per second. The time after which the angular velocity of the disc gets reduced to half of its initial value is

(A)

2m0 μ

(B)

m0 2μ

(C)

m0 μ

(D)

3 m0 μ

46. Three point masses each of mass m, are placed at the corners of an equilateral triangle of side l. The moment of inertia of this system about an axis along one side of the triangle is 3 2 (B) ml (A) ml 2 4 (C) 3 ml 2

(D)

R/3

47. A uniform rod PQ of length L is hinged at one end P. The rod is kept in the horizontal position by a massless string tied to point Q as shown in the figure. If the string is cut, the initial angular acceleration of the rod will be

O R

(A) 4 MR2

(B)

40 MR2 9

(C) 10 MR2

(D)

37 MR2 9

44. A force F is applied on the top of a cube as shown in figure. The coefficient of friction between the cube and the ground is μ . If F is gradually increased, the cube will topple before sliding if

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 132

3 2 ml 2

g L 2g (C) 3L (A)

2g L 3g (D) 2L (B)

48. A ball of mass m = 0.5 kg is attached to the end of a string having length L = 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N.

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Chapter 3: Rotational Dynamics The  maximum possible value of angular velocity of ball, in rads −1 is

L

m

(A) 9 (C) 27

(B) 18 (D) 36

49. A uniform ladder of length 5 m is placed against the wall as shown in the Figure.

3.133

(A) 3 v , anticlockwise 4l (B)

4v , anticlockwise 3l

(C)

3v , clockwise 4l

(D)

4v , clockwise 3l

52. A uniform circular disc of radius r placed on a rough horizontal plane has initial velocity v0 and an angular velocity ω 0 as shown. The disc comes to rest after moving some distance in the direction of motion. Then

v0

If coefficient of friction μ is the same for both the walls, the minimum value of μ for the ladder not to slip is (A) μ =

1 2

(B) μ =

1 3

(C) μ =

1 4

(D) μ =

1 5

50. A thin circular ring of mass M is rotating about its axis with a constant angular velocity ω . The two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity

ωM (A) M+m (C)

ωM M + 2m

(A) the friction force acts in the forward direction (B) the point of contact of disc with ground has zero velocity rω (C) v0 must be equal to 0 in magnitude 2 (D) v0 must be equal to 2rω 0 in magnitude 53. A solid sphere of radius r rolls without slipping from rest from a height h of an inclined track at the bottom of which there is a loop of radius R much larger than the radius of sphere, as shown in figure. The minimum value of h for the sphere to complete the loop is r

ω ( M − 2 m) (B) M + 2m (D)

ω ( M + 2 m) M

51. A bar of mass m, length l is in pure translatory motion with its centre of mass velocity v. It collides with and sticks to another identical bar at rest as shown in figure. Assuming that after collision it becomes one composite bar of length 2l, the angular velocity of the composite bar will be

h R

(A) 2.1R

(B) 2.3 R

(C) 2.5R

(D) 2.7 R

54. A solid sphere of radius R is resting on a smooth horizontal surface. A constant force F is applied at a height h from the bottom. Choose the correct alternative F h

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 133

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3.134 JEE Advanced Physics: Mechanics – II (A) (B) (C) (D)

Sphere will always slide whatever be the value of h Sphere will roll without slipping when h ≥ 1.4 R Sphere will roll without slipping if h = 1.4 R None of the above

55. A wire of length l and mass m is first bent in a circle, then in a square and then in an equilateral triangle. The moment of inertia in these three cases about an axis perpendicular to their planes and passing through their centres of mass are I1, I 2 and I 3 respectively. Then maximum of them is (B) I 2 (A) I1 (C) I 3

(D) data insufficient

56. A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX′ is X

X′

rolling. The least time will be taken in reaching the bottom by (A) the solid sphere (B) the ring (C) the disc (D) All will take the same time 60. In PROBLEM 59, the least value of kinetic energy at the bottom of the incline will be achieved by (A) the ring (B) the disc (C) the solid sphere (D) All will achieve the same kinetic energy 61. The string of a pendulum, having bob of mass m, is displaced through 90° from the vertical and then released. The minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is (B) 3mg (A) mg (C) 5mg

(A)

ρL3 8π 2

(B)

ρL3 16π 2

(C)

5ρL3 16π 2

(D)

3 ρL3 8π 2

57. A playground merry-go-round is at rest, pivoted about a frictionless axis. A child of mass m runs along the path tangential to the rim with speed v and jumps on to merry-go-round. If R be the radius of merry-goround and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is (A)

mvR mR2 + I

(B)

mvR I

(C)

mR2 + I mvR

(D)

I mvR

58. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω . The force exerted by the liquid at the other end is (A)

Mω 2 L 2

(B) Mω 2 L

Mω 2 L Mω 2 L2 (D) 4 2 59. A solid sphere, a ring and a disc all having same mass and radius are placed at the top of an incline and released. The coefficient of friction between the objects and the incline is same but not sufficient to allow pure (C)

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 134

(D) 6mg

62. Two men A and B are carrying a uniform bar of length L on their shoulders. The bar is held horizontally such ⎛ 1⎞ that A gets ⎜ ⎟ th load. If A is at one end of the bar, the ⎝ 4⎠ distance of B from that end is (A)

L 3

(B)

L 2

(C)

2L 3

(D)

3L 4

63. A body having moment of inertia about its axis of rotation equal to 3 kgm 2 is rotating with angular velocity equal to 3 rads −1 . The kinetic energy of this body is the same as that of a body of mass 27 kg moving with a speed of (B) 0.5 ms −1 (A) 1 ms −1 −1 (C) 1.5 ms (D) 2.0 ms −1 64. A particle is moved in a circle with a constant angular velocity. Its angular momentum is L. If the radius of the circle is halved keeping the angular velocity same, the angular momentum of the particle will become L L (B) (A) 4 2 (C) L (D) 2L 65. In the figure shown, the spherical body and block each have a mass m. The moment of inertia of the spherical body about centre of mass is 2mR2. The spherical body rolls without slipping on the horizontal surface. The ratio of kinetic energy of the spherical body to that of block is

2/9/2021 6:38:40 PM

Chapter 3: Rotational Dynamics

69. Two discs have same mass and thickness. Their materials are of densities d1 and d2 . The ratio of their moments of inertia about and axis passing through the centre and perpendicular to the plane

2R

R

1 3 2 (C) 3

1 2 3 (D) 4

(A)

66. A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle (Neglect the force due to gravity). The length of the string is reduced gradually, keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Ar n , where A is a constant, r is the instantaneous radius of the circle, then n is (B) −1 (A) 1 (C) −2 (D) −3 67. Average torque on a projectile of mass m, initial speed u and angle of projection θ is launched between initial and final positions P and Q as shown in figure. The average torque on the projectile about the point of projection is y

u

θ

Q

(A) mu2 cos θ (C)

(A) d1 : d2

(B) d2 : d1

(C) 1 : d1d2

(D) d1d2 : 1

70. A cube of mass m and side a is moving along a plane with constant speed v0 as shown in figure. The magnitude of angular momentum of the cube about z-axis would be

(B)

P

3.135

y v0

30°

O

(A)

x

b

mv0b 2

a⎞ ⎛ (C) mv0 ⎜ b − ⎟ ⎝ 2⎠

3 mv0b 2

(B)

(D) None of these

71. Two masses m and M are connected by a light string that passes through a smooth hole O at the centre of a table. Mass m lies on the table and M hangs vertically. m is moved round in a horizontal circle with O as the centre. If l is the length of the string from O to m then the frequency with which m should revolve so that M remains stationary is

x

(B) mu2 sin θ

mu2 sin ( 2θ ) 2

(D)

mu2 cos θ 2

68. A spool of mass M and radius 2R lies on an inclined plane as shown in figure. A light thread is wound around the connecting tube of the spool and its free end carries a weight of mass m. The value of m so that system is in equilibrium is R 2R

M

m

α

(A) 2M sin α (C) 2M tan α

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 135

(B) M sin α (D) M cos α

(A)

1 2π

Mg ml

(B)

1 π

Mg ml

(C)

1 2π

Ml mg

(D)

1 Ml π mg

72. A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about its axis is I . A weight mg is attached to the end of the cord and is allowed to fall from rest. The angular velocity of the wheel, when the weight has fallen through a distance h, is

2/9/2021 6:39:01 PM

3.136 JEE Advanced Physics: Mechanics – II

(A)

2gh I + mr

(B)

2mgh I + mr 2

(C)

2mgh I + 2mr 2

(D)

2gh

(A) (B) (C) (D)

73. A hoop rolls without slipping on a horizontal ground having centre of mass speed v0 . The speed of a particle P on the circumference of the hoop at angle θ is

always down the incline. always up the incline. initially up the incline and then becomes zero. initially up the incline and then becomes down the incline.

77. In the figure shown, the plank is being pulled to the right with a constant speed v. If the cylinder does not slip then R

v0

v

θ P

(A) v0 sin θ

(B) v0 cos θ

⎛θ⎞ (C) 2v0 sin ⎜ ⎟ ⎝ 2⎠

⎛θ⎞ (D) 2v0 cos ⎜ ⎟ ⎝ 2⎠

(A) the speed of the centre of mass of the cylinder is 2v (B) the speed of the centre of mass of the cylinder is v v (C) the angular velocity of the cylinder is R (D) the angular velocity of the cylinder is zero

74. Two rings each of mass m and radius r are placed such that their centres are at a common point and their planes are normal to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to plane of one of the ring is (A) 2mr 2 (C)

(B) mr 2

1 mr 2 2

(D)

3 2 mr 2

(

)

75. A particle of mass 2 kg located at the position iˆ + ˆj m has a velocity 2 iˆ − ˆj + kˆ ms −1. Its angular momentum

(

)

about z-axis in kgm 2s −1 is (A) zero

(B) +8

(C) −4

(D) −8

76. A uniform cylinder of mass M and radius R, rolls without slipping down an incline making an angle θ with horizontal. The cylinder is connected to a spring of force constant k at the centre, the other side of which is connected to a fixed support at A. The cylinder is released when the spring is unstretched. The force of friction ( f ) is A

θ

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 136

78. A cylinder is released from rest from the top of an incline plane of inclination 60° where friction coef2 − 3x . Find the ficient varies with distance x as μ = 3 distance travelled by the cylinder on incline before it starts slipping 1 1 m m (B) (A) 3 3 (C) 3 m

(D)

3m

79. A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ω A and ω B. Then (B) ω A = ω B (A) ω A < ω B (C) ω A = ω (D) ω = ω B 80. Masses of 1 g, 2 g, 3 g, …… 100 g are suspended from the 1 cm, 2 cm, 3 cm, …… 100 cm marks of a light metre scale. The system will be supported in equilibrium at (A) 60 cm (B) 67 cm (C) 55 cm (D) 72 cm 81. Two cylinders having radii 2R and R and moment of inertias 4I and I about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity ω 0 . The small cylinder is moved to the right until it touches the large cylinder and is made to rotate

2/9/2021 6:39:24 PM

Chapter 3: Rotational Dynamics by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. During this process 4

the surface and the ring is μ . The time after which its angular speed is reduced to half is ω0 g ω μR (B) (A) 0 2g 2μR (C)

R

2R

82. A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod in horizontal direcl tion. The speed of particle P at a distance from the 6 π ml is centre towards A of the rod after time t = 12 J J m

(B)

(C)

J 2 m

(D)

2J m J 2m

83. A circular platform of radius 2 m and moment of inertia 200 kgm 2 is mounted on a vertical frictionless axle. It is initially at rest. A 70 kg man stands on the edge of the platform and begins to walk along the edge at speed O relative to the ground. The angular velocity of the platform is (B) CD (A) 0.4 rads −1 (C) 2ma 2

2ω 0 R μg

ω0R 2μ g

ω0 3v and length is bent 8 4a to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and normal to the plane of hexagon is 12 2 (A) mL (B) 6 mL2 5 (D) 30 mL2 (C) 20 mL2

(D) O

87. A uniform disc of radius R lies in x -y plane with its centre at the origin. Its moment of inertia about z-axis is equal to its moment of inertia about line y = x + c . Then c equals R R (B) − (A) 2 2 R (C) + (D) −R 4 88. Four spheres each of mass M and diameter 2r , are placed with their centres on four corners of a square of side a(> 2r ). The moment of inertia of the system about one side of square 2 ( 2 2 ( 2 (A) M 5r + 4 a 2 ) M 5r + 2 a 2 ) (B) 5 5 2 ( 2 2 ( 2 M 2r + 5 a 2 ) M 4r + 5a2 ) (C) (D) 5 5 89. A sphere is rotating between two rough inclined plane as shown in figure. The coefficient of friction between 1 each plane and the sphere is . If f1 and f 2 be the fric3 f tion forces at P and Q. Then 1 is f2

84. In PROBLEM 83, when the man has walked once around the platform, so that he is at his original position on it, then his angular displacement relative to ground is 6π 5π (B) (A) 5 6 4π 5π (C) (D) 4 5 85. A ring of radius R is first rotated with an angular velocity ω 0 and then carefully placed on a rough horizontal surface. The coefficient of friction between

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 137

(D)

86. A uniform thin bar of mass

(A) angular momentum of system is conserved (B) kinetic energy is conserved (C) neither the angular momentum nor the kinetic energy is conserved (D) both the angular momentum and kinetic energy are conserved

(A)

3.137

Q

P 30° 30°

(A)

1 + 3 2

(C) 1 +

4 3

(B) 2 +

1 3

(D) 1 + 2 3

2/9/2021 6:39:48 PM

3.138 JEE Advanced Physics: Mechanics – II 90. A plank of mass M is placed on a smooth inclined, plane and a sphere of mass m is then placed on the plank. Assuming sufficient friction to be present between the sphere and plank the frictional force on sphere when the plank and sphere are released from rest is m

M

θ

(A) zero (C) up the incline

(B) horizontal (D) down the incline

94. A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is K . The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is K (B) (A) 2K 2 K (C) (D) 4K 4 95. A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in Figure. The ratio of moment of inertia of the loop about x-axis to that about y-axis is

91. A thin rod of mass m and length 2l is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to ω in time t, the torque acting on it is (A)

ml 2ω 12t

(B)

ml 2ω 3t

(C)

ml 2ω t

(D)

4 ml 2ω 3t

92. A ring of mass m and radius R has three particles attached to the ring as shown in the figure. The centre of the ring has a speed v0 . Assume slipping to be absent, the kinetic energy of the system is m

2m

m

(A) 4 mv02

(B) 6 mv02

(C) 8 mv02

(D) 12mv02

93. Two cones A and B are made of two different materials, the density of A being greater than that of B. The height of B is greater than that of A but their base areas and masses are the same. The correct statement about the moment of inertia of the two cones about their axis is (A) A will have larger moment of inertia than B (B) B will have larger moment of inertia than A (C) in such a situation, it is dependent up on the height of the cone, the mass of the cone and radius of the base (D) the moment of inertia of the two will be the same as it is not dependent on height of the cone but depends only upon the mass and the base area

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 138

(A) > 1

(B) < 1

(C) = 1

(D) =

1 2

96. A homogeneous rod AB of length L and mass M is pivoted at the centre O in such a way that it can rotate freely in a vertical plane. The rod AB is initially in horizontal position. An insect S of same mass falls vertically with a speed V at the point C, midway between points O and B. Immediately after the insect falls, the angular velocity of the system is

7 V 12 L 3V (C) 2L (A)

12 V 7 L 2V (D) 3L

(B)

97. A circular loop of wire of mass m and radius r is making n revolutions per second about a point on its rim. Its rotational kinetic energy is (A) π 2 mr 2n2 (B) 2π 2 mr 2n2 (C) 4π 2 mr 2n2 (D) 8π 2 mr 2n2

2/9/2021 6:40:05 PM

Chapter 3: Rotational Dynamics 98. One end of a uniform rod of length l and mass m is hinged at A. It is released from rest from horizontal position AB as shown in figure. The force exerted by the rod on the hinge when it becomes vertical is A B

3.139

101. A solid sphere of mass 2 kg rolls up a 30° incline with an initial speed of 10 ms −1 . The maximum height reached by the sphere is ( g = 10 ms −2 ) (A) 3.5 m (C) 10.5 m

(B) 7.0 m (D) 14.0 m

102. A small ball of mass m, radius r is released from rest from a point A to roll inside a hemispherical shell of radius R as shown in figure. The angular velocity of centre of the ball in position B about the centre of the shell is A

3 mg 2 (C) 3 mg

5 mg 2 (D) 5 mg

(B)

(A)

99. The acceleration a of the plank P required to keep the centre C of a cylinder which rolls without slipping on the plank in a fixed position during the motion is C

a

g sin θ 2 (C) 2g sin θ

P

(B) g sin θ (D)

2g sin θ

100. A uniform circular disc of mass 2m and radius R is placed freely on a horizontal smooth surface. A  particle of mass m is connected to the circumference of the disc with a massless string. Now an impulse J is applied on the particle in the direction as shown in Figure.

g 5( R − r )

(A) 2 (C)

θ

(A)

B

2g 5( R − r )

(B)

10 g 7(R − r )

(D)

5g 2( R − r )

103. In PROBLEM 102, the normal force between the ball  dL is and the shell in position dt 10 (A) mg mg (B) 7 12 17 (C) mg (D) mg 7 7 104. An inclined plane makes an angle of 60 o with horizontal. A disc rolling down this inclined plane without slipping has a linear acceleration equal to g 3g (B) (A) 3 4 (C)

g 3

(D)

g 2

105. Locus of all the points in a plane on which the moment of inertia of a rigid body is same throughout is (A) a straight line (B) a circle (C) a parabola (D) an ellipse If J = 10 Ns, m = 10 kg and R = 25 cm, then acceleration of centre of mass of the disc just after application of impulse is (A) 1 ms −2

(B) 2 ms −2

(C) 3 ms −2

(D) 4 ms −2

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 139

106. A rigid spherical body is spinning around an axis without any external torque. Due to temperature its volume increases by 3%. Then percentage change in its angular speed is (B) −1% (A) −2% (C) −3% (D) 1%

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3.140 JEE Advanced Physics: Mechanics – II 107. A uniform rod AB of mass m and length 2a is falling with velocity v without rotation in gravity free space with AB horizontal. Suddenly the end A gets hit by an obstacle when the speed of the rod is v. Assume the end A to hinge about the obstacle, the angular speed with which the rod begins to rotate is v 4v (A) (B) 2a 3a v 3v (D) (C) 3a 4a 108. A ring of mass m is rolling without slipping with linear velocity v as shown in figure. A rod of identical mass is fixed along one of its diameter. The total kinetic energy of the system is

(A) I 2 > I1 > I 3 (B) I 3 > I1 > I 2 (C) I 3 < I 2 < I1 (D) I1 = I 2 = I 3 111. A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights h1 and h2 respectively. At the bottom of the plane the two objects have same linear velocities, then h1 : h2 is (A) 2 : 3 (B) 7 : 5 (C) 14 : 15

(D) 15 : 14

112. A time varying force F = 2t is applied on a spool as shown in figure. The angular momentum of the spool at time t about bottommost point is F = 2t r R

7 mv 2 5 4 mv 2 (C) 3

2 mv 2 3 5 (D) mv 2 3 (B)

(A)

109. A uniform rod of mass m and length l is suspended by means of two light inextensible strings as shown in figure. Tension in one string immediately after the other string is cut is

A

mg 4 (C) mg 110.

ABC is a right angled triangular plate of uniform thickness. I1, I 2 and I 3 are moments of inertia about 3 AB, BC and AC respectively. Given that sin ( 37° ) = , 5 then which of the following relation is correct? A

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 140

(C) ( R + r ) t 2

(D)

37°

C

( R + r )2 r

t2

R 2t 2 r

113. The moment of inertia of a uniform rod of length 2l and mass m about an axis passing through its centre and inclined at an angle α is

B

(B)

B

(B)

A

mg 2 (D) 2mg

(A)

r 2t 2 R

(A)

ml 2 sin 2 α 3 K (C) 4 (A)

α

B

C

(B)

ml 2 sin 2 α 12

(D) vc

114. A rod of length l is given two velocities v1 and v2 in opposite directions at its two ends at right angles to the length. The distance of the instantaneous axis of rotation from v1 is l (A) zero (B) 2 ⎛ v1 ⎞ (C) ⎜ l ⎝ v1 + v2 ⎟⎠

⎛ v2 ⎞ l (D) ⎜ ⎝ v1 + v2 ⎟⎠

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Chapter 3: Rotational Dynamics 1 of nth its present radius without any change in its mass, the duration of the day will approximately become 24 24 (A) hour hour (B) n n2 (C) 24n hour (D) 24n2 hour

115. If the radius of the earth suddenly contracts to

116. A solid sphere of mass 5 kg and radius 1 m after rotating with angular speed ω 0 = 40 rads −1 is placed between two smooth walls on a rough ground. Distance between the walls is slightly greater than the diameter of the sphere. If the coefficient of friction between the sphere and the ground is μ = 0.1, then the sphere will stop rotating after time

120. A homogeneous cylinder of mass M and radius R is pulled on a horizontal plane by a horizontal force F acting through its centre of mass. Assuming the cylinder to roll without slipping, the angular acceleration of the cylinder is (A)

3F 2 MR

(B)

2F 3 MR

(C)

F 2 MR

(D)

3F 4 MR

121. Moment of inertia I of a solid sphere about an axis parallel to a diameter and at a distance x from it varies as (B)

(A)

t

t

(C) (A) 8 s (C) 16 s

(D)

(B) 12 s (D) 20 s

117. A wheel rotates with constant acceleration of 2.0 rads −2 . If the wheel starts from rest, the number of revolutions it makes in the first ten second will be approximately (A) 8 (B) 16 (C) 24 (D) 32 118. Two loops P and Q are made from a uniform wire. The radii of P and Q are R1 and R2 respectively, and their moments of inertia are I1 and I 2 respectively. If I2 R = 4 , then 2 is I1 R1 (A) 4 2 3

(B) 41 3

(C) 4 −2 3

(D) 4 −1 3

119. In the arrangement shown, a smooth inclined plane of inclination θ is fixed in a car and a sphere is set in pure rolling on the incline. The value of a (the acceleration of car in horizontal direction) for which the sphere will continue pure rolling is

t

t

122. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity ω 0 . When the tortoise move along a chord of the platform with a constant velocity (with respect to the platform). The angular velocity of the platform ω ( t ) will vary with time t as (A)

ω (t)

(B)

ω0

ω (t) ω0

t

(C)

ω (t)

t

(D)

ω0

ω (t) ω0

t a

α

(A) g cos θ (C) g cot θ

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 141

3.141

(B) g sin θ (D) g tan θ

t

123. A particle of mass 5 g is moving with a uniform speed of 3 2 cms −1 in the xy-plane along the line y = x + 4. Magnitude of its angular momentum about origin in gcms −2 is (A) ZERO (C) 30 2

(B) 30 (D) 60

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3.142 JEE Advanced Physics: Mechanics – II 124. When a mass is rotated in a plane about a fixed point, its angular momentum is directed along the (A) radius (B) tangent to orbit (C) axis of rotation (D) line at an angle of 45° to plane of rotation v and mass m 5a is bent to form of a rectangular frame ABCD with AB = 2. The moment of inertia of this wire frame BC about the smaller side is 6 6 ml 2 (B) (A) ml 2 81 217 7 7 (C) ml 2 (D) ml 2 81 162

125. A uniform thin wire of length ω =

126. Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = Mv is imparted to the body at one of its end, the its angular velocity is M

v L v (C) 3L

2v L v (D) 4L (B)

127. A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are (A) up the incline while ascending and down the incline while descending (B) up the incline while ascending as well as descending (C) down the incline while ascending and up the incline while descending (D) down the incline while ascending as well as descending 128. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω . Another disc of same dimen1 sions but of mass M is placed gently on the first 4 disc co-axially. The angular velocity of the system is 4 (A) 2ω ω (B) 5 1 3 (C) ω (D) ω 4 3

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 142

(A) mgRt sin θ − μmgRt cos θ (B) mgRt sin θ (C) mgRt sin θ + μmgRt cos θ (D) mgRt ( 1 − μ 2 ) sin θ 130. A disc of radius 0.1 m rolls without sliding on a horizontal surface with a velocity of 6 ms −1. It then ascends a smooth continuous track as shown in figure. Taking g = 10 ms −2 , the height upto which it will ascend is

M J = Mv

(A)

129. A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle θ with horizontal. The coefficient of friction between the ring and the plane is μ . Initially, the point of contact of ring and plane is P. Angular momentum of the ring about point P as a function of time t is

6 ms−1

(A) 0.9 m (C) 2.4 m

(B) 1.8 m (D) 2.7 m

131. Two equal and opposite forces act on a rigid body at a certain distance. Then (A) the body is in equilibrium (B) the body will rotate about its centre of mass (C) the body may rotate about any point other than its centre of mass (D) the body cannot rotate about its centre of mass 132. A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP, vQ and vC be the magnitude of velocities of points P , Q and C respectively, then Q C P

(A) vQ > vC > vP (B) vQ < vC < vP (C) vQ = vP , vC = (D) vQ < vC > vP

1 vP 2

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Chapter 3: Rotational Dynamics 133. Two particles A and B are moving with constant   velocities v1 = 1 ms −1 ˆj and v2 = 2 ms −1 iˆ respectively in XY plane. At time t = 0, the particle A is at coordinates ( 0 , 0 ) m and B is at ( 0 , − 4 ) m. The angular velocity of B with respect to A at t = 2 s is 1 rads −1 (B) 1 rads −1 (A) 2 (C) 2 rads −1 (D) 4 rads −1 134. A force F is applied at the top of a ring of mass M and radius R placed on a rough horizontal surface as shown in figure. Friction is sufficient to prevent slipping. The friction force acting on the ring is F

F towards right 2 2F (C) towards right 3 (A)

(B)

F towards left 3

(D) zero

135. Two rods OA and OB of equal length and mass are lying in xy plane as shown in figure. Let I x , I y and I z be the moment of inertias of both the rods about x, y and z-axis respectively. Then y A

B

45°

45° O

x

(A) I x = I y > I z

(B) I x = I y < I z

(C) I x > I y > I z

(D) I z > I y > I x

136. A carpet of mass M made of inextensible material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. The horizontal velocity of the axis of cylindrical part of the carpet R when its radius reduces to is 2 7 (B) gR (A) gR 3 (C)

8 gR 3

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 143

(D)

14 gR 3

3.143

137. A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with a constant angular acceleration α. If the coefficient of friction between the rod and the bead is μ, and gravity is neglected, then the time after which the bead starts slipping is (A) (C)

μ α 1 μα

(B)

μ α

(D) infinitesimal

138. A solid sphere rolls down two different inclined planes of same height but of different inclinations. In both cases (A) the speed and time of descend will be same (B) the speed will be same but time of descend will be different (C) the speed will be different but time of descend will be same (D) speed and time of descend both are different 139. The linear velocity of a particle moving with angular   velocity ω = 2kˆ at position vector r = 2iˆ + 2 ˆj is

(

(A) 4 iˆ − ˆj (C) 4iˆ

)

(

(B) 4 ˆj − iˆ

)

(D) −4iˆ

140. A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (A) Centre of circle (B) On the circumference of the circle (C) Inside the circle (D) Outside the circle  141. A force F = ai + 3 j + 6 k is acting at a point  r = 2i − 6 j − 12k . The value of a for which angular momentum is conserved is (B) 1 (A) 0 (D) 2 (C) −1 142. A solid cylinder of mass 50 kg and radius 0.5 mis free to rotate about its axis which is horizontal. A string is wound around the cylinder with one end attached to it and other hanging freely. The tension in the string required to produce an angular acceleration of 2 revs −2 in the cylinder is (B) 157 N (A) 78.5 N (C) 314 N (D) 628 N 143. A uniform cylindrical disc of radius R and mass M is pulled over a horizontal frictionless surface by a constant force. The force is applied by means of a  string wound around the disc as shown in the

2/9/2021 6:41:50 PM

3.144 JEE Advanced Physics: Mechanics – II figure. If it starts from rest at t = 0, the linear and angular displacements respectively at time t are

⎛ F ⎞ ⎛ F ⎞ (A) ⎜ ⎟ t 2 , ⎜ ⎟ t 2 ⎝ M⎠ ⎝ M⎠

⎛ F ⎞ ⎛ F ⎞ 2 (B) ⎜ t, t ⎝ 2 M ⎟⎠ ⎜⎝ 2 MR ⎟⎠

⎛ F ⎞ ⎛ F ⎞ 2 (C) ⎜ ⎟ t 2 , ⎜ t ⎝ M⎠ ⎝ MR ⎟⎠

⎛ 2F ⎞ 2 ⎛ 2F ⎞ 2 (D) ⎜ t ,⎜ t ⎝ M ⎟⎠ ⎝ MR ⎟⎠

144. A cube is placed on a rough inclined plane of inclination θ as shown in figure. The coefficient of friction between the cube and the plane is μ . If the angle θ is gradually increased, the cube slides before toppling when

1 2 1 (D) μ > 2

(A) μ < 1

(B) μ
1

145. Assuming the centre of mass of a hemisphere to lie 3R at a height from the base, the radius of gyration 8 of a solid hemisphere of mass M and radius R about 3 an axis parallel to the diameter at a distance R from 4 this plane is given by P

147. A sphere of mass M and radius R moves on a horizontal surface with a velocity V and then climbs up an inclined plane up to a height h where it stops. The height upto which it rises will be

(A) directly proportional to the square of the velocity and inversely proportional to the angle of the inclination (B) directly proportional to the velocity and inversely proportional to its mass (C) directly proportional to the square of and the angle of the velocity and independent of mass the inclination (D) directly proportional to its velocity and inversely proportional to the angle of the inclination 148. The moment of inertia of the body about an axis is 1.2 kgm 2 . Initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rads −2 must be applied about the axis for the duration of (B) 4 s (A) 2 s (C) 8 s (D) 10 s 149. A uniform rod of length l is pivoted at point A. It is struck by a horizontal force which delivers an impulse J at a distance x from point A as shown in figure. The impulse delivered by pivot is zero when x equals A

x

Q J

3R 4

3R (A) 10 5R (C) 8

5R (B) 4 2 R (D) 5

146. A particle moves in a circular path with decreasing speed. Choose the correct statement (A) Angular momentum remains constant  (B) Acceleration ( a ) is towards the centre (C) Particle moves in a spiral path with decreasing radius (D) The direction of angular momentum remains constant

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 144

l 2 2l (C) 3

l 3 3l (D) 4

(A)

(B)

150. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity ω, in gravity free space. The increase in length of the spring will be ω k

m

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Chapter 3: Rotational Dynamics

(A)

mlω 2 k

(B)

mlω 2 k − mω 2

(A) d =

(C)

mlω 2 k + mω 2

(D) None of these

(C) d = l

151. A uniform cube of side a and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to one face at a point that lies directly above a the centre of the face at a height above the centre. 4 The minimum value of F for which the cube begins to topple above an edge without sliding is 1 1 mg (B) mg (A) 4 2 2 (C) mg (D) 2mg 3 152. The ratio of moment of inertia of the ring about an axis passing through its rim and perpendicular to its plane and that about its diameter is (B) 4 : 1 (A) 1 : 4 (C) 1 : 2 (D) 2 : 1 153. A plank with a uniform sphere placed on it resting on a smooth horizontal plane. Plank is pulled to right by a constant force F. If sphere does not slip over the plank. Which of the following is incorrect?

3.145

l 3 2l (D) d = 3

l 2

(B) d =

155. A railway track is banked for a speed v, by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. The radius of curvature of the track is r. (A)

h v2 = d rg

2 ⎧ ⎛ h⎞⎫ v (B) tan ⎨ sin −1 ⎜ ⎟ ⎬ = ⎝ ⎠ d ⎭ rg ⎩ 2 ⎛ h⎞ v (C) tan −1 ⎜ ⎟ = ⎝ d ⎠ rg

(D)

h v2 = r dg

156. Of a wedge shown in figure the portion AB is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB = BC , then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches the point C is A B

D

(A) Acceleration of the centre of sphere is less than that of the plank (B) Work done by friction acting on the sphere is equal to its total kinetic energy (C) Total kinetic energy of the system is equal to work done by the force F (D) None of the above 154. The uniform bar of length l and mass m is suspended from a very thin axle that passes through a hole near the top end A of the bar as shown. An impulsive blow having magnitude P is applied at right angles to the bar in order to start the bar rotating about A without breaking the axle. For this

C

3 5 8 (C) 3 (A)

(B)

7 5

(D) 5

157. A disc of radius R rolls on a horizontal ground with linear acceleration a and angular acceleration α as shown in figure. The magnitude of acceleration of point P shown in figure at an instant when its linear velocity is v and angular velocity is ω will be P r

v, a

O

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 145

(A)

( a + rα )2 + ( rω 2 )

(C)

r 2α 2 + r 2ω 4

2

ar R (D) rα

(B)

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3.146 JEE Advanced Physics: Mechanics – II 158. In both the cases, B is mid-point of A and C, while all other factors are same, except that in Case (i) AB is rough and BC is smooth while in Case (ii) AB is smooth and BC is rough. The kinetic energy of the ball on reaching the bottom

B

h

B

h C

C

Case (i)

(A) (B) (C) (D)

Case (ii)

(A) μ g

is same in both the cases is greater in Case (i) is greater in Case (ii) None of these

(C)

159. A wheel of radius R rolls on the ground with a uniform velocity v. The relative acceleration of topmost point of the wheel with respect to the bottommost point is v2 2v 2 (B) (A) R R (C)

v2 2R

(D)

4v 2 R

160. Two cylinders having radii 2R and R and moment of inertias 4I and I about their central axes are supported by axles perpendicular to their planes. The large cylinder is initially rotating clockwise with angular velocity ω 0 . The small cylinder is moved to the right until it touches the large cylinder and is made to rotate by the frictional force between the two. Eventually slipping ceases and the two cylinders rotate at constant rates in opposite directions. The final angular velocity of the small cylinder is ω0

4I I

R

ω0 4 ω0 (C) 2 (A)

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 146

161. A sphere S of mass M is given a finite angular velocity about a horizontal axis through its centre. Now it is gently placed on a plank P of same mass. The coefficient of friction between the two is μ and the plank rests on a smooth horizontal surface as shown. The initial acceleration of the sphere relative to the plank will be

(B) ω 0 (D)

ω0 8

(D) ZERO

162. A solid cylinder R is free to rotate about its axis which is horizontal. A string is wound around it and a mass m is attached to its free end. When m falls through a distance h, its speed at that instant is proportional to 1 (B) (A) R R 1 (C) (D) None of these R2 163. A wheel of mass 40 kg and radius of gyration 0.5 m comes to rest from a speed of 1800 rpm in 30 s. Assuming that the retardation is uniform, the value of the retarding torque, in Nm, is (B) 20π (A) 10π (C) 30π (D) 40π 164. A solid cylinder of mass M and radius R rolls down the inclined plane from height h without slipping. The speed of its centre of mass when it reaches the bottom is 4 (B) gh (A) 2gh 3 (C)

2R

2 μg 5

(B) 2 μ g

3 gh 4

(D)

4g h

165. A uniform rod AB of length L having mass m are lying on a smooth table. A small particle of mass m strikes the rod with a velocity v0 at point C, a distance x from the centre O. The particle comes to rest after collision. The value of x, so that point A of the rod remains stationary just after collision is

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Chapter 3: Rotational Dynamics B m

169. A disc of radius r rolls without slipping on a rough horizontal floor. If velocity of its centre of mass is r v0 , then velocity of point P at distance such that 2 ∠POQ = 60° as shown in the figure is

v0 C x O

Q

O

A

(A)

L 3

(B)

L 6

(C)

L 4

(D)

L 12

(A) v0

166. A wheel of radius r rolls without slipping with a speed v on a horizontal road. When it is at a point A on the road, a small blob of mud separates from the wheel at its highest point and lands at point B on the road (A) AB = v (C) AB = 4v

r g r g

3.147

(B) AB = 2v

r g

(D) AB = 8v

r g

167. A mass M is moving with a constant velocity parallel to the x-axis. Its angular momentum with respect to the origin (A) is zero (B) remains constant (C) increases (D) decreases 168. A billiard ball of mass m and radius r, hit by a cue at a height h above the centre, acquires a linear velocity v0 . The angular velocity ω 0 acquired by the ball is (A)

2v0 h 5r 2

(B)

5v0 h 2r 2

(C)

2v0 r 2 5h

(D)

5v0 r 2 2h

(C)

v0 7 2

60° P

v0

v0 2 v0 3 (D) 2 (B)

170. A particle is projected with velocity v at an angle of ϕ with horizontal. The average angular velocity of the particle from the point of projection to impact equals g cos ϕ g (B) (A) ϕv v sin ϕ g gϕ (C) (D) vϕ v sin ϕ 171. A uniform stick of length F3 and mass m lies on a smooth table. It rotates with angular velocity ω about an axis perpendicular to the table and passes through one end of the stick. The angular momentum of the stick about the end is (B) F2 (A) ml 2ω (C) I1 + I 2

(D)

ml 2ω 6

MULTIPLE CORRECT CHOICE TYPE QUESTIONS This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

The moment of inertia of a solid cylinder of massM, length 2R and radius R about an axis passing through the centre of mass and perpendicular to the axis of the cylinder is I1 and about an axis passing through one end of the cylinder and perpendicular to the axis of cylinder is I 2

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 147

(A) I 2 − I1 = MR2

2.

(B) I 2 > I1

I 19 (C) 2 = (D) I1 − I 2 = MR2 I1 12  Thetorque  τ ona body about a given point is found to be A × L where A is a constant vector and L is angular momentum of the body about that point. From this it follows that

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3.148 JEE Advanced Physics: Mechanics – II   dL is perpendicular to L at all instants of time. dt   (B) the component of L in the direction of A does not change with time.  (C) the  magnitude of L does not change with time. (D) L does not change with time.

(A)

3.

(A) The hinge reaction at A on the rod AB is downward (B) The hinge reaction at B on the rod AB is upward (C) The hinge reaction at B on the rod AB is downward (D) The angular momentum of the system about point O is not along the rod AB.

A triangular wedge ABC of mass m and sides 2a lies on a smooth horizontal plane as shown. Three point masses of mass m each strike the wedge at A, B and C with speeds v as shown. After the collision, the particles come to rest. Select the correct alternative(s). m

A

y

v

5.

60°

x

G

B

60°

v

m v

60°

A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v, respectively, strike the bar as shown in the figure and stick to it after collision. Denoting angular velocity (about the centre of mass), total energy and the centre of mass velocity by ω , E and Vc respectively, we have after collision

C

m

(A) The centre of mass of ABC remains stationary after collision (B) The centre of mass of ABC moves with a velocity v along x-axis after collision (C) The triangular wedge rotates with an angular 2 3mva about its centre of mass velocity ω = I (here, I is the moment of inertia of triangular wedge about its centroid axis perpendicular to its plane) I ⎛ ⎞ from G on (D) A point lying at a distance of ⎜ ⎝ 2 3 ma ⎟⎠

(A) Vc = 0 (C) ω = 6.

perpendicular bisector of BC (below G) is at rest just after collision 4.

Figure shows a horizontal rod AB which is free to rotate about two smooth bearing system. Two identical uniform rods each of mass m are attached to rod AB. All the dimensions are given in the Figure. The system is rotating with constant angular velocity ω in such a way that the upper rod is coming outward from the plane of the paper in the position shown. Gravity can be assumed to be absent in the experiment, then choose the correct option(s).

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 148

v 5a

(B) ω =

3v 5a

(D) E =

3 mv 2 5

A particle of mass m is travelling with a constant  velocity v = v0 i along the line y = b , z = 0. If dA be the area swept out by the position vector from origin to the particle in time dt and L be the magnitude of angular momentum of particle about origin at any time t. Then (A) L ≠ constant (B) L = constant dA 2L = dt m dA L = (D) dt 2m (C)

7.

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then,

2/9/2021 6:43:25 PM

Chapter 3: Rotational Dynamics C

B

A

    (A) VC − VA = 2 ( VB − VC )     (C) VC − VA = 2 VB − VC 8.

    (B) VC − VB = VB − VA    (D) VC − VA = 4 VB

A rod of length l is balanced in a vertical plane by applying a horizontal force F to its end lying on the surface as shown in Figure.

3.149

  1F i . (A) Acceleration of point O is a = 3m   F (B) Acceleration of point O is a = i . m (C) The magnitude of angular acceleration of the 1 Fb frame is α = . 3 mr 2 (D) The magnitude of angular acceleration of the Fb frame is α = . mr 2 10. A body of mass m, radius R rolls on a horizontal surface with linear velocity v as shown in Figure. ω v

P

The linear mass density of the rod varies with distance ⎛λ ⎞ r from the end lying on the surface as λ ( r ) = ⎜ 20 ⎟ r , ⎝ l ⎠ where λ 0 is a positive constant. If the applied force equals 4 3 times the weight of the rod and the rod only translates, then select the correct option(s). (A) θ = 37° 4g (B) Acceleration of rod is 3 λ g (C) normal force is 0 2 (D) normal force is λ 0 g 9.

Each of the three balls has a mass m and is welded to the rigid equiangular from of negligible mass. The assembly rests on a smooth horizontal surface. A force F is suddenly applied to one bar as shown.

If L1 and L2 be the magnitudes of the respective angular momenta of the body about centre of mass and the point of contact P. Then (A) L2 = 2L1, for all cases (B) L2 = 2L1, if radius of gyration K = R (C) L2 > 2L1, if radius of gyration K < R (D) L2 > 2L1, if radius of gyration K > R 11. A and B are two solid spheres of equal masses. A rolls down an incline plane without slipping from a height of 3 m. B falls vertically from the same height. Then on reaching the ground (A) A can do more work than B (B) B can do more work than A (C) both perform equal work (D) both will have different linear speeds 12. A solid body rotates about a stationary axis so that its angular velocity depends upon the rotation angle ϕ as ω = ω 0 − aϕ , where a and ω 0 are positive constants. Initially that is at t = 0, ϕ = 0. (A) ϕ =

ω0 ( 1 − e − at ) a

(B) ϕ =

ω0 ( 1 + e at ) a

(C) ω = ω 0 e − at (D) ω = ω 0 e at

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3.150 JEE Advanced Physics: Mechanics – II 13. A string is wrapped over a uniform cylinder, as shown in Figure (side view). When cylinder is released, string unwraps without any slipping and cylinder comes down. Select the correct statements(s).

P

θ O

(A) Work done by tension force on the cylinder is zero. (B) Work done by tension is negative (C) Ratio of rotational kinetic energy and translational 1 kinetic energy is 2 (D) Ratio of rotational kinetic energy to translational kinetic energy is 2 14. The density of a rod AB increases linearly from A to B. Its midpoint is O and its centre of mass is at C. Four axes pass through A , B, O and C, all perpendicular to the length of the rod. The moments of inertia of the rod about these axes are I A, I B, IO and IC respectively (A) I A > I B

(B) I A < I B

(C) IO >IC

(D) IO < IC

15. A thin uniform rod of mass m and length l is free to rotate about its upper end. When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact (A) the angular momentum of the rod is Jl 3J (B) the angular velocity of the rod is ml 3J2 2m (D) the linear velocity of the midpoint of the rod is 3J 2m

(C) the kinetic energy of the rod is

16. A disc of radius R rolls on a horizontal surface with linear velocity v and angular velocity ω . A point P on the circumference of the disc at angle θ shown has a vertical velocity. Then,

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 150

v

ω

π ⎛ v ⎞ − sin −1 ⎜ ⎝ Rω ⎟⎠ 2

⎛ v ⎞ (A) θ = π + sin −1 ⎜ ⎝ Rω ⎟⎠

(B) θ =

⎛ v ⎞ (C) θ = π − cos −1 ⎜ ⎝ Rω ⎟⎠

⎛ v ⎞ (D) θ = π + cos −1 ⎜ ⎝ Rω ⎟⎠

17. A cylinder C of mass 8m is rolling without sliding over two horizontal planks A and B having masses 2m and m, moving with uniform velocities −viˆ and 2viˆ respectively. Select the correct option(s).

(A) The instantaneous axis of rotation is located at a 4R distance of below the topmost contact point. 3 3v (B) Angular velocity of the cylinder is 2R (C) Translational kinetic energy of the system is 4 mv 2 (D) Kinetic energy of the system is

9 mv 2 2

18. A uniform disc is rotating at a constant speed in a vertical plane about a fixed horizontal axis passing through the centre of the disc. A piece of the disc from its rim detaches itself from the disc at the instant when it is at horizontal level with the centre of the disc and moving upward. Then about the fixed axis, the angular speed of the (A) remaining disc decreases (B) remaining disc increases (C) remaining disc stays constant (D) broken away piece decreases initially and later increases

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Chapter 3: Rotational Dynamics 19. A constant force F is applied at the top of a ring of mass M and radius is R as shown. The angular momentum of particle about point of contact at time t F

(A) (B) (C) (D)

(C) The friction force on the cylinder acts backward F (D) The magnitude of the friction force is 3 24. A spool of wire rests on a horizontal surface as shown in figure. As the wire is pulled, the spool does not slip at contact point P. On separate trials, each one of the forces F1 , F2, F3 and F4 is applied to the spool. For each one of these forces the spool

is constant increases linearly with time is 2FRt decreases linearly with time

F3 F2

F4

20. A particle moves in a circle of radius r with angular   velocity ω . At some instant its velocity is v and radius  vector with respect to centre of the circle is r. At this  particular instant centripetal acceleration ac of the particle is     (B) ω × v (A) v × ω       (C) v × ( r × ω ) (D) ω × ( ω × r ) 21. A disc rolls without slipping in the absence any external force on a (A) rough inclined plane (B) smooth inclined plane (C) rough horizontal surface (D) smooth horizontal surface 22. A pan containing a layer of uniform thickness of ice is placed on a circular turntable with its centre coinciding with the centre of the turntable. The turntable is now rotated at a constant angular velocity about a vertical axis passing through its centre and the driving torque is withdrawn. There is no friction between the table and the pivot. The pan rotates with the table. As the ice melts (A) the angular velocity of system increases (B) the angular velocity of system decreases (C) the angular velocity remains unchanged (D) the moment of inertia of system increases 23. A solid cylinder of mass M and radius R is rolled horizontally on a rough surface as shown in the figure. Choose the correct alternative(s) cm

3.151

F1 P

(A) (B) (C) (D)

will rotate anticlockwise if F1 is applied will rotate anticlockwise if F3 is applied will rotate clockwise if F4 is applied will not rotate if F2 is applied

25. A rod of length l is standing vertically on a frictionless surface. It is disturbed slightly from this position. Let ω and α be the angular speed and angular acceleration of the rod when the rod turns through an angle θ with the vertical. If the speed of the centre of mass is v and the value of acceleration of centre of mass of the rod is a, then (A) a =

lα lω 2 sin θ + cos θ 2 2

lω 2 lα sin θ + cos θ 2 2 lω (C) v = cos θ 2 lω (D) v = sin θ 2 (B) a =

26. In the figure the blocks have unequal masses m1 and m2 ( m1 > m2 ). m1 has a downward acceleration a. The pulley P has a radius R, and some mass. The string does not slip on the pulley

F

Rough

F M 2 F (B) The acceleration of the centre of mass is 3M

(A) The acceleration of the centre of mass is

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 151

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3.152 JEE Advanced Physics: Mechanics – II (A) the two sections of the string have unequal tensions (B) the two blocks have accelerations of equal magnitude a (C) the angular acceleration of P is R ⎛ m − m2 ⎞ g (D) a < ⎜ 1 ⎝ m1 + m2 ⎟⎠

30. A disc of mass M and radius R moves in the x -y plane as shown in the figure. The angular momentum of the disc at the instant shown is y A

27. A solid sphere of radius R is rolled by a force F acting at the top of the sphere as shown in the figure. There is no slipping and initially sphere is in the rest position then

ω

3R O

F

Rough

(A) work done by force F when the centre of mass moves a distance S is 2FS

x

B

4R

(A)

5 MR2ω about O 2

(B)

7 MR2ω about O 2

(C)

MR2ω about A 2

(D) 4 MR2ω about A

31. A horizontal force F acts on one side of a hexagonal body that lies on a horizontal rough surface as shown in Figure. Select the correct statement(s).

(B) speed of centre of mass (CM) when CM moves a 20 FS distance S is 7M (C) work done by the force F when CM moves a distance S is FS (D) speed of the CM when CM moves a distance S is 4FS M 28. If a circular concentric hole is made in a disc, then about an axis passing through the centre of the disc and perpendicular to its plane the (A) radius of gyration increases (B) radius of gyration decreases (C) moment of inertia increases (D) moment of inertia decreases 29. The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plate is

(A) For toppling, the minimum value of coefficient of 1 friction is . 2 3 (B) For toppling, the minimum value of coefficient of 2 friction is . 3 mg , then (C) If μ = 2 μmin and applied force F = 3 6g . angular acceleration of the body is 17 a mg , then (D) If μ = 2 μmin and applied force F = 3 6g . angular acceleration of the body is 5a 32. The end B of the rod AB which makes angle θ with the floor is being pulled with a constant velocity v0 as shown. The length of the rod is l. At the instant when θ = 37° , the y A

(A) I1 + I 2

(B) I 3 + I 4

(C) I1 + I 3

(D) I1 + I 2 + I 3 + I 4

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 152

O

θ

v0 B

x

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Chapter 3: Rotational Dynamics

(A) velocity of end A is constant 4 (B) velocity of end A is v0 downwards 3 (C) angular velocity of rod is constant 5 v0 (D) angular velocity of rod is 3 l 33. Four particles each of mass m are placed at four corners of a square ABCD of side a. If the point O happens to be the centre of the square, then the moment of inertia of all four particles about an axis passing through (A) A and B is 2ma 2 (B) A and C is ma 2 (C) O and perpendicular to plane of square is 2ma 2 (D) O and parallel to CD is ma 2 34. In pure rolling, the fraction of total energy associated with rotation is α for a ring and β for a solid sphere. Then 1 1 (B) α = (A) α = 2 4 2 2 (C) β = (D) β = 5 7

3.153

 L0 pointing upward. The wheel is inverted about its centre by180°. (A) The angular momentum of the system is conserved. (B) The angular momentum of the system is not conserved. (C) The final angular momentum of student plus  stool will be2L0 . (D) The final angular momentum of student plus stool will be zero. 37. Which of the following statement(s) is/are correct for a spherical body rolling without slipping on a rough horizontal surface at rest? (A) The speed of some of the point(s) is (are) zero (B) The acceleration of a point in contact with ground is zero (C) Work done by friction may or may not be zero (D) Friction force may or may not be zero 38. A ring rolls without slipping on a horizontal surface. At any instant, its position is as shown in the figure

35. A spool of wire rests on a horizontal surface as shown in Figure.

As the wire is pulled, the spool does not slip at contact point P. On separate trials, each one of the forces F1 , F2, F3 and F4 is applied to the spool. The direction of friction force is (A) towards left if F1 is applied (B) towards left if F2 is applied (C) towards right if F3 is applied

(A) section ABC has greater kinetic energy than section ADC (B) section BC has greater kinetic energy than section CD (C) section BC has the same kinetic energy as section DA (D) the section AB, BC, CD and DA have the same kinetic energy 39. A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. The magnitude of angular momentum of the projectile about point of projection when the particle is at maximum height h is (A) zero

(D) may be right or left or friction may be zero if F4 is applied

(B)

mv 3 4 2g

36. A student holds the axle of a spinning bicycle wheel while seated on a pivoted stool. The student and the stool are initially at rest while the wheel is spinning in a horizontal plane with an angular momentum

(C)

mv 3 2g

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 153

(D) m 2 gh 3

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3.154 JEE Advanced Physics: Mechanics – II 40. Two forces F1 and F2 are acting on a rod abc as shown in figure. Select the correct statement(s). F1

43. A force F is applied on the plank such that the hollow hemispherical shell of mass m = 5 kg is in equilibrium as shown in Figure.

c a

b F2

(A) If F1 ≠ F2, then τ a ≠ τ b ≠ τ c (B) If F1 ≠ F2, then τ a = τ c ≠ τ b (C) If F1 = F2, then τ a = τ b = τ c (for both the forces) (D) If F1 = F2, then τ a = τ c ≠ τ b 41. A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same mass and radius as the disc, is now gently placed on the disc. After some time, the two rotate with a common angular velocity (A) some friction exists between the disc and the ring. (B) the angular momentum of the ‘disc plus ring’ is conserved. 2 times (C) the final common angular velocity is 3 initial angular velocity of the disc. 2 (D) times the initial kinetic energy changes to heat. 3 42. A uniform rod of mass M and length L is held vertically on a smooth horizontal surface. When the rod is released, choose the correct alternative(s).

The coefficient of friction μ between hemispherical shell and plank is same as between plank and ground. Friction is just sufficient to prevent the slipping. Taking g = 10 ms −2 , select the correct statement(s). (A) Minimum coefficient of friction to prevent 1 slipping is μmin = 4 (B) Minimum coefficient of friction to prevent 1 slipping is μmin = 2 (C) Acceleration of plank is 10 ms −2 (D) Magnitude of applied force is 100 N 44. A particle of mass 2 kg is attached to a string of length 1 m, moves in a horizontal circle just like a conical pendulum. The string makes an angle θ = 30 o with the

(

)

vertical. Select the correct alternative(s) g = 10 ms −2 . P

θ

L

(A) The centre of mass of the rod accelerates in the vertical direction (B) Initially, the magnitude of the normal reaction is Mg (C) When the rod becomes just horizontal, the magniMg tude of the normal reaction becomes 2 (D) When the rod becomes just horizontal, the magniMg tude of the normal reaction becomes 4

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 154

(A) The vertical component of angular momentum of mass about the point of support P is approximately 1.7 kgm 2s −1. (B) The horizontal component of angular momentum of mass about the point of support P is approximately 2.9 kgm 2s −1 .   dL (C) Magnitude of , where L is the angular momendt tum of mass about point of support P, is approximately 10 kgm 2s −2 .   dL (D) τ = will not hold good in this case. dt

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Chapter 3: Rotational Dynamics

3.155

REASONING BASED QUESTIONS This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) Bubble (B) Bubble (C) Bubble (D)

If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

1.

Statement-1: If earth shrink (without change in mass) to half it’s present size. Length of the day would become 6 hours. Statement-2: As size of earth changes its moment of inertia changes.

9.

2.

Statement-1: A ring moving down on a smooth inclined plane will be in slipping motion. Statement-2: Work done by friction in pure rolling motion is zero.

10. Statement-1: A solid sphere rolling on a rough horizontal surface. Acceleration of contact point is zero. Statement-2: A solid sphere can roll on the smooth surface.

3.

Statement-1: A non-uniform sphere is placed such that its centre is at the origin of coordinate system. If I x and I y be moment of inertia about x-axis and y-axis respectively then moment of inertia about z-axis is I x + I y.

11. Statement-1: A disc is rolling on an inclined plane without slipping. The velocity of centre of mass is V . These other points on the disc lie on a circular arc having same speed as centre of mass. Statement-2: When a disc is rolling on an inclined plane. The magnitude of velocities of all the point from the contact point is same, having distance equal to radius r.

Statement-2: According to perpendicular axis theory I z = I x + I y when object is lying in x -y plane. 4.

Statement-1: A sphere is performing pure rolling on a rough horizontal surface with constant angular velocity. Frictional force acting on the sphere is zero. Statement-2: Velocity of contact point is zero.

5.

Statement-1: If momentum of system is zero than kinetic energy must be zero. Statement-2: If kinetic energy of system is zero than momentum must be zero.

6.

Statement-1: A ladders is more likely to slip when a person is near the top than when he is near the bottom. Statement-2: The friction between the ladder and floor decreases as he climbs up.

7.

Statement-1: If rod is thrown upward with initial angular velocity and velocity of centre of mass then its momentum changes but angular velocity remains same. Statement-2: Torque on rod about centre of mass due to gravitational force is zero.

8.

Statement-1: The mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing its moment of inertia. Statement-2: For then the moment of inertia of everybody about an axis passing through its centre of mass would be zero.

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 155

Statement-1: A disc is rolling on a rough horizontal surface. The instantaneous speed of the point of contact during perfect rolling is zero with respect to ground. Statement-2: The force of friction can help in achieving pure rolling condition.

12. Statement-1: The velocity of a body at the bottom of an inclined place of given height, is more when it slides down the plane, compared to, when it rolling down the same plane. Statement-2: In rolling down, a body acquires both, kinetic energy of translation and rotation. 13. Statement-1: The force of friction in the case of a disc rolling without slipping down on inclined plane is 1 g sin α . 3 Statement-2: When the disc rolls without slipping, friction is required because for rolling condition velocity of point of contact is zero. 14. Statement-1: A horizontal force F is applied such that the block remains stationary because N will produce torque. Statement-2: The torque produced by friction force is equal and opposite the torque produce due to normal reaction ( N ). a

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3.156 JEE Advanced Physics: Mechanics – II 15. Statement-1: When a diver dives, the rotational kinetic energy of diver increases, during several somersaults. Statement-2: When diver pulls his limbs, the moment

of inertia decreases and on account of conservation of angular momentum his angular speed increases.

LINKED COMPREHENSION TYPE QUESTIONS This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options) F

Comprehension 1 A uniform rod AB of mass 3m and length 4l, which is free to turn in a vertical plane about a smooth horizontal axis through A, is released from rest from the horizontal position. When the rod first becomes vertical, a point C of the rod, where AC = 3l , strikes a fixed peg. In Situation-I, for an impulse J1 exerted by the peg on the rod, the rod is brought to rest by the peg. However, in Situation- II, for an impulse J 2 exerted by the peg on the rod, the rod rebounds and next comes to instantaneous rest when inclined to the π downward vertical at an angle of radian. Based on above 3 information, answer the following questions. 1.

A

3.

Tangential acceleration of centre of mass is 3F F (A) (B) 4m m 2F 4F (C) (D) 3m 3m

4.

The component of reaction at hinge in vertical direction is 4 (A) mg (B) mg 3 mg 2 (C) (D) mg 2 3

5.

Component of reaction at hinge in horizontal direction is F (B) F (A) 4 F F (D) (C) 3 2

The value of J1 in Situation-I is (B)

(A) m 3 gl (C) 8 m 2.

R

gl 3

m 3 gl 8

(D) 3 m

gl 8

The value of J 2 in Situation-II is (A) m

gl 3

(C) 2m ( 2 + 1 )

(B) 4 m ( 2 + 1 ) gl 3

(D) 4 m

2 gl 3

gl 3

Comprehension 2 When a body is hinged at a point and a force is acting on the body, in such a way that the line of action of force is at some distance from the hinged point, the body will start rotating about the hinged point. The angular acceleration of the body can be calculated by finding the torque of that force about the hinged point. A disc of mass m and radius R is hinged at point A at its bottom and is free to rotate in the vertical plane. A force of magnitude F is acting on the ring at the top most point. Based on above information, answer the following questions.

Mechanics II_Chapter 3_Part 5 Exercises_1.indd 156

Comprehension 3 A uniform rod AB of mass 3m and length 2l is lying at rest on a smooth horizontal table with a smooth vertical axis through the end A. A particle of mass 2m moves with speed 2u across the table and strikes the rod at its midpoint C. In Situation-I, the particle strikes the rod normally, the impact is perfectly elastic such that the speed of the particle after impact is v1. In Situation-II, the particle strikes the rod such that its path before impact is inclined at 60° to AC, the impact is still perfectly elastic such that the speed of the particle after impact is v2 . Based on above information, answer the following questions.

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Chapter 3: Rotational Dynamics 6.

The value of v1 in Situation-I is u 2 u (D) 4 (B)

(A) u (C) 7.

2u 3

The value of v2 in Situation-II is u 2u (A) (B) 3 3 u 3u (C) (D) 2 2

0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar one with a velocity of 10 ms −1 , and the other with 6 ms −1 , as shown in figure. The first particle strikes the bar at points A and the other point B. Points A and B are at a distance of 0.5 m from the centre of the bar. The particles strike the bar at the same instant of time and stick to the bar on collision. Based on above information, answer the following questions. 10 ms−1

Comprehension 4

m

μ

m Smooth

8.

The time t at which the pure rolling begins is rω 0 2rω 0 (B) (A) 9μ g 9μ g rω 0 (C) 3μ g

9.

A

6 ms−1

A long horizontal plank of mass m is lying on a smooth horizontal surface. A sphere of same mass m and radius r is spinned about its own axis with angular velocity ω 0 and gently placed on the plank. The coefficient of friction between the plank and the sphere is μ . After some time, the pure rolling of the sphere on the plank starts. Based on above information, answer the following questions.

2rω 0 (D) μg

The velocity of the sphere, after the pure rolling begins is 2 rω 0 (A) rω 0 (B) 9 9 rω 0 2 (D) rω 0 (C) 3 3

10. The displacement of the plank, till the sphere starts pure rolling is 2 ⎛ r 2ω 02 ⎞ r 2ω 02 (B) (A) 81μ g 27 ⎜⎝ μ g ⎟⎠ 4 ⎛ r 2ω 02 ⎞ (C) 81 ⎜⎝ μ g ⎟⎠

2 ⎛ r 2ω 02 ⎞ (D) 81 ⎜⎝ μ g ⎟⎠

Comprehension 5 A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16 kg and length is

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 157

3 m. Two particles, each of mass

3.157

B

11. The velocity of the centre of mass of system, just after impact is (B) 2 ms −1 (A) 1 ms −1 −1 (C) 3 ms (D) 4 ms −1 12. The angular velocity of the system, just after impact is (B) 4 rads −1 (A) 8 rads −1 (C) 2 rads −1

(D) 1 rads −1

13. The loss of kinetic energy of the system in the above collision process is (A) 1.72 J (B) 2.72 J (C) 3.36 J (D) 4.36 J

Comprehension 6 A uniform disk of mass M and radius R is pivoted so that it can rotate freely about a horizontal axis through its centre and perpendicular to the plane of the disk. A small particle of mass m, the half of the mass of the disc, is attached to the rim of the disk at the top directly above the pivot. The system is given a gentle start and the disk begins to rotate about O. Based on above information, answer the following questions. 14. The angular velocity of the disk when the particle is at its lowest point is g 3g (B) (A) R R (C)

2g R

(D)

g 2R

15. At the lowest point, the force that must be exerted by the disc on the particle to keep it on the disk is Mg 2 Mg (A) (B) 3 2 Mg 3 Mg (D) (C) 3 2

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3.158 JEE Advanced Physics: Mechanics – II

Comprehension 7 A uniform rod of mass 2 kg and length l is suspended by two smooth hinges A and B as shown in figure. A force F = 4 N l is applied downwards at a distance from hinge B. Due to 4 the application of this force F, the hinge B breaks. At this instant, applied force F is also removed. The rod starts to rotate downward about hinge A. Take g = 10 ms −2 . Based on above information, answer the following questions.

A

B F=4N

16. The reaction at hinge A, before hinge B breaks, is (A) 10 N (B) 11 N (C) 12 N (D) 24 N 17. The reaction at hinge A, just after hinge B breaks is (A) 35 N (B) 25 N (C) 15 N (D) 5 N 18. The acceleration of the end point of rod having small mass dm when the rod becomes vertical is (B) 20 ms −2 (A) 100 ms −2 −2 (C) 30 ms (D) 40 ms −2

Comprehension 8 A vertically oriented uniform heavy rod of mass M and length L can rotate about its upper end hinged to a rigid support. A fast moving light ball of mass m strikes the lower end of the rod with velocity u and sticks to it, as a result of which the rod swings through an angle θ . Based on above information, answer the following questions. 19. The angular velocity of the system just after impact is mu Mu (A) (B) 3 ML 3 mL 3mu 3Mu (C) (D) ML mL 20. The velocity, u of the ball is gL ⎛θ⎞ sin ⎜ ⎟ ⎝ 2⎠ 3

(A)

M m

(C)

M 2 ⎛θ⎞ gL sin ⎜ ⎟ ⎝ 2⎠ m 3

(B)

m 2 ⎛θ⎞ gL sin ⎜ ⎟ ⎝ 2⎠ M 3

(D)

m ⎛θ⎞ gL sin ⎜ ⎟ ⎝ 2⎠ M

22. The distance x from the upper end of the rod where the ball must strike the rod such that the momentum of the ball + rod system remains constant during the impact is L L (A) (B) 3 6 2L L (D) (C) 3 2

Comprehension 9 Moment of inertia of a straight wire about an axis perpendicular to the wire and passing through one of its end is I . This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and normal to its plane is I1. Now the same wire is bent into a ring of two turns and the moment of inertia about an axis passing through the ring normal to its plane is I 2 . Based on above information, answer the following questions. 23. The value of I1 is ⎛ 3 ⎞ (A) ⎜ 2 ⎟ I ⎝π ⎠

⎛ 3 ⎞ (B) ⎜ 2 ⎟ I ⎝ 4π ⎠

⎛ π2 ⎞ (C) ⎜ I ⎝ 3 ⎟⎠

⎛ 4π 2 ⎞ (D) ⎜ I ⎝ 3 ⎟⎠

24. The value of I 2 is ⎛ π2 ⎞ (A) ⎜ I ⎝ 3 ⎟⎠

⎛ π2 ⎞ (B) ⎜ I ⎝ 12 ⎟⎠

⎛ 3 ⎞ I (C) ⎜ ⎝ 16π 2 ⎟⎠

⎛ 3 ⎞ (D) ⎜ 2 ⎟ I ⎝ 8π ⎠

Comprehension 10 A fixed wedge has two semi-circular tracks of radius R1 and R2 , such that R1 = 2R2. The track of radius R1 is frictionless. A small block of mass m is held on the top of the larger track. A spherical ball of radius r (  R2 ) is held on the top of the other track. Both the block and the ball are released simultaneously. Friction in the track containing the ball is large enough to avoid slipping of the ball. Based on above information, answer the following questions. R2 R1

21. The momentum change in the ball-rod system during the impact (A) m

gL ⎛θ⎞ sin ⎜ ⎟ ⎝ 2⎠ 6

⎛θ⎞ (C) m gL cos ⎜ ⎟ ⎝ 2⎠

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 158

⎛θ⎞ (B) m gL sin ⎜ ⎟ ⎝ 2⎠ gL ⎛θ⎞ sin ⎜ ⎟ (D) M ⎝ 2⎠ 6

25. Out of the block and the ball, (A) the block reaches the bottom of the track first. (B) the ball reaches the bottom of the track first.

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Chapter 3: Rotational Dynamics (C) both reach the bottom of the track simultaneously. (D) it is difficult to arrive at a conclusion, because some vital information is not provided in the paragraph. 26. Assuming the block and the ball to reach the bottom of their respective tracks simultaneously, then the ratio R1 is R2 (A)

1 5

(B)

5 7

(C)

7 5

(D)

9 5

Comprehension 11 The angular momentum of a flywheel having a moment of inertia of 0.4 kgm 2 decreases from 3 kgm 2s −1 to 2 kgm 2s −1 in a period of 2 second. Based on above information, answer the following questions. 27. The average torque acting on the flywheel during this period is (A) 10 Nm (B) 2.5 Nm (C) 0.5 Nm (D) 5 Nm 28. If the angular acceleration is uniform, the number of revolutions the flywheel would have turned in that interval is (A) 2 (B) 4 (C) 12 (D) 6 29. The work done is (A) 3.125 Nm (B) 12.5 Nm (C) 62.5 Nm (D) 6.25 Nm

Comprehension 12 A rod of mass m and length l in placed on a smooth horizontal table. A particle of same mass m strikes the rod with velocity v0 in a direction perpendicular to the rod at distance ⎛ l⎞ x ⎜ < ⎟ from its centre and sticks to the rod. Let ω be the ⎝ 2⎠ angular speed of system just after collision, then based on above information, answer the following questions. l 30. When x is increased from 0 to , the angular speed ω 2 (A) will first decrease and then increase (B) will first increase and then decrease (C) will continuously increase (D) will continuously decrease

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3.159

31. Assume that the particle is still sticking to the rod, the maximum possible value of impulse obtained by varying x, that can be imparted to the particle during collision is 2mv0 mv0 (A) (B) 2 3 4 mv0 3 mv0 (D) (C) 4 5

Comprehension 13 A solid sphere is kept over a smooth surface as shown in figure. It is hit by a cue at height h above the centre C. In R R case 1, h = and in case 2, h = . Suppose in case 1 the 4 2 sphere acquires a total kinetic energy K1 and in case 2 total kinetic energy is K 2 . h C

Assuming that in both the cases the sphere is hit by the same impulse, answer the following questions. 32. The appropriate comparison relation between K1 and K 2 is (A) K1 = K 2

(B) K1 > K 2

(C) K1 < K 2

(D) data insufficient

33. If the surface is rough, then after hitting the sphere, in which case the force of friction is in forward direction. (A) In case 1 (B) In case 2 (C) In both the cases (D) In none of the cases

Comprehension 14 An L shaped frame is free to rotate in a vertical plane about a horizontal axis passing through a smooth hinge O. Each side of the frame has a length L and mass m. The frame is let to fall with one side horizontal and the other vertical. Based on above information, answer the following questions. B

A

O L/2

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3.160 JEE Advanced Physics: Mechanics – II 34. The angular acceleration of the allowed to fall is 4g (B) (A) 3L g (C) (D) 2L

frame just after it is 9g 8L 3g 2L

35. The speed of the end A with which it strikes the ground is (B) 2 gL (A) 1.1 gL (D)

(C) 3.2 gL

gL

Comprehension 15 Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictionless table ( x -y plane ) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity ω . At time T , when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Based on above information, answer the following questions. y A

ω

x

F B

C



36. The magnitude of the horizontal force exerted by the hinge on the body is (A) mlω 2 (C)

3 mlω 2

(B)

mlω 2 3

(D)

mlω 2 3

37. The x-component of force exerted by the hinge on the body immediately after time T is F , along x-axis (A) 2 F (B) , along negative x-axis 2 F (C) , along x-axis 4 F (D) , along negative x-axis 4

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38. The y-component of the force exerted by the hinge on the body immediately after time T is (A) mlω 2 (C)

3 mlω 2

(B)

mlω 2 3

(D)

mlω 2 3

Comprehension 16 In the arrangement shown, F = 10 N, R = 1 m, M = 2 kg and moment of inertia of the body about an axis passing through O and perpendicular to plane of body is I = 4 kgm 2 . O is the centre of mass of the body. If the ground is smooth, the kinetic energy of the body at t = 2 s is K1. However, when the ground is sufficiently rough the kinetic energy of the body at t = 6 s is K 2 . Based on above information, answer the following questions.

39. The value K1 equals (A) 25 J (C) 75 J

(B) 50 J (D) 100 J

40. The value K 2 equals (A) 10.33 J (C) 16.67 J

(B) 250 J (D) 150 J

Comprehension 17 A solid sphere of mass m, radius R is rolling without slipping on rough horizontal surface as shown in figure. It collides elastically with another identical sphere at rest. There is no friction between the two spheres.

v

ω

Based on above information, answer the following questions. 41. Linear velocity of first sphere after it again starts rolling without slipping is 2 2 (B) Rω (A) Rω 5 7 7 7 (C) (D) Rω Rω 10 5

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Chapter 3: Rotational Dynamics 42. The net angular impulse imparted to the second sphere by the external forces is 2 5 (B) mvR (A) mvR 7 7 2 7 (C) mvR mvR (D) 5 10

Comprehension 18 On an inclined planeAB of length 5 m, a body is released from the point A. Assuming friction to be sufficient enough for pure rolling to take place, answer the following questions.

47. The moment of inertia of the rod about an axis passing through A and perpendicular to AB is (A)

7 Ma 2 18

(B)

5 Ma 2 18

(C)

1Ma 2 12

(D)

1Ma 2 18

48. If on the application of impulse, the end B passes through a point vertically above A, then the minimum value of J is M 90 ag 7 M 30 ag (C) 7

(A) A

3.161

M 70 ag 9 M 80 ag (D) 5 (B)

Comprehension 20 B

43. The maximum time which a body (that is capable of rolling) can take to reach the bottom is (A) 8 s (B) 6 s (C) 4 s (D) 2 s 44. If four bodies like ring, disc, solid sphere and hollow sphere (all having identical mass, radius and coefficient of friction) are taken and the angle θ is now gradually increased, then the body that will be the first to start slipping is (A) Ring (B) Disc (C) Solid sphere (D) Hollow sphere

Comprehension 19 A thin rod AB of mass M, length a has a variable mass per ρ (a + x) , where x is the distance measured from unit length 0 a the endA of the rod and ρ0 is a constant. The rod is freely pivoted at A and is hanging in equilibrium when it is struck by a horizontal impulse of magnitude J at the point B. Based on above information, answer the following questions. 45. The total mass M of the rod is (B) aρ0 (A) 2 aρ0 aρ0 3 aρ0 (D) (C) 3 2 46. The centre of mass of the rod is at the point at distance from the end A given by a a (A) (B) 9 3 5a 2a (C) (D) 9 3

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 161

In the diagram shown, a plank of mass m is lying at rest on a smooth horizontal surface. A disc of same mass m and radius r is rotated to an angular speed ω 0 and then gently placed on the plank. It is found that finally slipping ceases and 50% of the total kinetic energy of the system is lost. Assume that the plank is long enough and the coefficient of friction between disc and plank is μ . Based on above information, answer the following questions.

49. The final velocity of the plank is rω 0 rω 0 (A) (B) 4 10 rω 0 rω 0 (C) (D) 2 2 10 50. Time when slipping ceases is rω 0 (B) (A) 2μ g (C)

rω 0 4μ g

(D)

rω 0 10 μ g rω 0 2 10 μ g

51. Magnitude of the change in angular momentum of the disc about center of mass of the disc is (A)

3 2 mr ω 0 4

(C) zero

3 2 mr ω 0 8 1 (D) mr 2ω 0 2 (B)

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3.162 JEE Advanced Physics: Mechanics – II 52. Distance moved by the plank from the placing of disc on the plank till the slipping ceases between disc and plank is r 2ω 0 2 (A) 16 μ g (C)

2m

r 2ω 0 2 (B) 8μ g

r 2ω 0 2 32 μ g

(D)

r 2ω 0 2 200 μ g

Comprehension 21 A rod AB of length 2 m and mass 2 kg is lying on smooth horizontal x -y plane with its centre at origin O as shown figure. An impulse J of magnitude 10 Ns is applied perpendicular to AB at A. Based on above information, answer the following questions. y J

A

x

O

m 45°

55. The acceleration of the block and the cylinder when both move separately is g g g g , (B) (A) , 2 2 3 3 2 3 2g g g 2g , (D) , (C) 3 2 2 2 2 3 56. The acceleration of the block and the cylinder when both move together is 5g 3g (B) (A) 3 2 5 2 5g 3g (D) (C) 2 2

Comprehension 23 B

53. The distance of point P from centre of the rod which is at rest just after the impact is 1 1 m m (B) (A) 4 2 1 2 (C) m (D) m 3 3 π s 54. Co-ordinates of point A of the rod after time t = 45 will be 1 ⎤ ⎡1 (A) ⎢ m, m ⎥ 2 ⎦ ⎣2 3 ⎞⎤ ⎡⎛ 3 (B) ⎢ ⎜ m, m ⎟ ⎥ 2 ⎠⎦ ⎣⎝ 4 1 ⎤ ⎡⎛ π 1 ⎞ (C) ⎢ ⎜ + ⎟ m, m ⎥ 2 ⎦ ⎣⎝ 6 2 ⎠ ⎡⎛ π 3⎞ 1 ⎤ (D) ⎢ ⎜ + m, m ⎥ ⎟ 2 ⎠ 2 ⎦ ⎣⎝ 9

Comprehension 22 On a rough inclined plane of inclination 45°, a block of mass m and a cylinder of mass 2m are released. The coefficient of friction between all the surfaces of contact is 0.5. Based on above information, answer the following questions.

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 162

A uniform disc of mass M = 2.5 kg and radius R = 0.2 m is mounted on an axle supported on fixed frictionless bearings. A light cord wrapped around the rim is pulled with a force 5 N. On the same system of pulley and string instead of pulling it down, a body of weight 5 N is suspended. If the first process is termed A and the second B. M

M R

R

5 N (Pull)

mg = 5 N

[A]

[B]

Based on above information, answer the following questions. 57. The tangential acceleration will be (A) equal in the processes A and B (B) greater in process A than in B (C) greater in process B than in A (D) independent of the two processes 58. The mechanical energy is conserved (A) in both A and B (B) in B only (C) in A only (D) neither in A nor in B

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Chapter 3: Rotational Dynamics

3.163

Comprehension 24

Comprehension 26

A plank of length 40 m and mass 2 kg is kept on a horizontal frictionless surface. A cylinder of mass 2 kg is placed on the plank. The coefficient of friction between the two surfaces is μ = 0.1. Plank is suddenly given a velocity v = 10 ms −1 towards left. Based on above information, answer the following questions.

A solid sphere of radius 1 m and mass 2 kg has linear velocity v0 = 4 ms −1 and angular velocity ω 0 = 9 rads −1 as shown. The ground on which it is moving, is smooth. It collides elastically with a rough wall of coefficient of friction μ . Based on above information, answer the following questions. ω0

2 kg v0 2 kg

 = 40 m

59. Initially the acceleration of point of contact with respect to the plank is (B) 1 ms −2 (A) 2 ms −2 (C) 4 ms −2

(D) 6 ms −2

60. The time after which pure rolling starts is (A) 2 s (B) 2.5 s (C) 3.5 s (D) 4.5 s 61. The time after which the plank and cylinder separates (A) 6.75 s (B) 4 s (C) 3.25 s (D) 5.25 s

64. If the sphere after colliding with the wall rolls without slipping in opposite direction, then 1 1 (A) μ = (B) μ = 4 3 2 1 (D) μ = (C) μ = 2 3 65. The net linear impulse imparted by the wall on the sphere during impact is (B) 4 5 Ns (A) 32 Ns (C) 4 17 Ns

(D) 15 2 Ns

Comprehension 25

Comprehension 27

A disc of mass 10 kg and radius 1 m is set into pure rolling on a rough horizontal surface as shown in the figure. At a particular instant, point P (figure) which is at a distance R from the centre O is found to have a speed of 2 ms −1 2 and acceleration of 10 ms −2 . Based on above information, answer the following questions.

The radius of a wheel is R and its radius of gyration about its axis passing through its center and perpendicular to its plane is K . If the wheel is rolling without slipping. Based on above information, answer the following questions.

ω F

R/2 R/2

R

(C) O P

62. Acceleration of the centre of mass of the disc is (A) 4 ms −2 (B) 8 ms −2 (C) 12 ms −2

(D) 20 ms −2

R from the centre (as 2 shown in the figure) causes this motion, the value of F is (A) zero (B) 60 N (C) 120 N (D) 200 N

63. If a horizontal force F, applied at

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 163

66. The ratio of its rotational kinetic energy to its translational kinetic energy is K2 R2 (B) (A) R2 K2 R2 R + K2 2

(D)

K2 R + K2 2

67. The ratio of the rotational kinetic energy to total kinetic energy is K2 R2 (B) (A) 2 2 2 R +K R + K2 1 (C) (D) None of these R2 + K 2 68. The ratio of the translational kinetic energy to the total kinetic energy is K2 R2 (B) (A) 2 R + K2 R2 + K 2 1 (C) (D) None of these R2 + K 2

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3.164 JEE Advanced Physics: Mechanics – II

Comprehension 28

Comprehension 30

A wedge of mass 4m is placed at rest on a smooth horizontal surface. A uniform solid sphere of mass m and radius r is placed at rest on the flat portion of the wedge at the point Q as shown in the figure. A sharp horizontal impulse P is given to the sphere at a point below h = 0.4 r from the center of the sphere. The radius of curvature of the curved portion of the wedge is R. Coefficient of friction to the left side of point Q is μ and to the right side of point Q is zero. Based on above information, answer the following questions.

A disc of mass m and radius R is placed over a plank of same mass m. There is sufficient friction between disc and plank to prevent slipping. A force F is applied at the centre of the disc. Based on above information, answer the following questions.

73. Acceleration of the plank is F 3F (A) (B) 4m 4m F 3F (C) (D) 2m 2m

r P

h Q

F

4m

69. The maximum height to which the center of mass of the sphere will climb on the curved portion of the wedge is 2P 2 (A) 5m 2 g

P2 (B) 5m 2 g

P2 (C) 2m 2 g

(D) None of these

70. Kinetic energy of the system when sphere is at the highest point is (A)

P2 10 m

(B)

P2 5m

(C)

3P 2 10 m

(D)

3P 2 5m

Comprehension 29 A rod of mass 4 kg , length 2 m is kept vertically inside a smooth spherical shell of radius 2 m. The rod starts slipping inside the shell. Based on above information, answer the following questions. 71. Angular speed of the rod ( in rads −1 ) in the position when it becomes horizontal is

74. Acceleration of the disc is F (A) 4m 3F (C) 4m

(B)

F 2m

(D) F

75. The angular acceleration of the disc is F F (A) (B) mR 2mR 3F 3F (D) (C) 2mR mR 76. Force of friction between the disc and the plank is F F (A) (B) 4 3 2F F (C) (D) 2 3

Comprehension 31 A uniform rod AB of mass 3m and length 2l is free to turn in a vertical plane about a smooth horizontal axis through A. A particle of mass m is attached to the rod at B. When the rod is hanging in equilibrium, it is set moving with kg , where k is a positive real l number. Based on above information, answer the following questions.

an angular velocity ω 0 =

(A) 3.2 (C) 6.8

(B) 4.6 (D) 7.2

72. Velocity of centre of mass of the rod ( in ms −1 ) at this instant is approximately (A) 4.7 (B) 5.5 (C) 6.2 (D) 10.2

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 164

77. For k = 2, the height of B above the level of A when the rod first comes to instantaneous rest is 3l 4l (B) (A) 5 5 6l 8l (D) (C) 5 5

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Chapter 3: Rotational Dynamics 78. The minimum value of k for which the particle describes complete circle about A is 3 5 (B) (A) 2 2 7 9 (C) (D) 2 2

Comprehension 32 A uniform disc of radius 1 m and mass 2 kg is mounted on an axle supported on fixed frictionless bearings. A light cord is wrapped around the rim of the disc and a mass of 1 kg is tied to the free end. If it is released from rest, the tension in the cord is T . If the applied torque acts for 4 second, the total angular displacement in this interval of time is θ . The work done by the applied torque is W and the increase in rotational kinetic energy is ΔK. Based on above information, answer the following questions. 79. The value of T is (A) 10 N (C) 40 N

(B) 5 N (D) 15 N

84. The acceleration of the system will be g g (B) (A) 4 8 7g g (C) (D) 24 2 85. The minimum value of friction coefficient to prevent slipping is 7 5 (B) (A) 7 5 3 5 3 5 (C) (D) 7 12 3

Comprehension 34 A disc of mass M and radius R is given a velocity v0 and an angular velocity ω 0 as shown in the figure. Then it is kept over a rough horizontal surface where coefficient of friction is μ . After some time slipping ceases, based on above information, answer the following questions. R

80. θ equals (A) 40 rad (C) 20 rad

(B) 80 rad (D) 10 rad

81. W has a value given by (A) 400 J (C) 100 J

(B) 50 J (D) 200 J

82. ΔK equals (A) 400 J (C) 200 J

(B) 100 J (D) 80 J

θ

83. The moment of inertia of the system about the centre of ring will be (A) 10 kgm 2 (B) 20 kgm 2

Positive v0 x-direction

86. If v0 = Rω 0, then time after which pure rolling starts is (A)

v0 μg

(B)

3v0 μg

(C)

2 v0 3 μg

Comprehension 33 Four identical rods of mass M = 6 kg each are welded at their ends to form a square and then welded to a massive ring having mass m = 4 kg having radius R = 1 m. If the system is allowed to roll down the incline of inclination θ = 30° . Based on above information, answer the following questions.

3.165

(D) pure rolling will never start 87. Velocity of centre of disc at the moment when angular velocity of disc becomes zero is v (A) v0 (B) 0 2 v0 v0 (C) (D) 3 4 88. Velocity of centre of mass of disc when pure rolling begins is v (B) 0 (A) v0 2 v0 v0 (C) (D) 3 4

(C) 40 kgm 2 (D) 60 kgm 2

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3.166 JEE Advanced Physics: Mechanics – II

Comprehension 35

89. The angular momentum vector is (A) 12 kgm 2s −1, out of the plane of the figure

A particle P with a mass 2 kg has position vector r = 3 m and velocity V = 4 ms −1 shown. It is accelerated by the force F = 2 N. All three vectors lie in the common plane. Based on above information, answer the following questions.

(B) 12 kgm 2s −1, into the plane of the figure (C) zero (D) 24 kgm 2s −1, out of the plane of the figure

y

90. The torque is (A) 6 Nm, out of the page (B) 3 Nm, out of the page (C) 6 Nm, into the page (D) 3 Nm, into the page

°

30 °

30

P

45°

x

MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles will look like the following:

A B C D

1.

p p p p p

q q q q q

r

s

t

r r r r

s s s s

t t t t

A smooth ball of mass m moving with a uniform velocity v0 strikes a smooth uniform rod AB of equal mass m, lying on a frictionless horizontal table. The ball strikes the rod at one end A, perpendicular to the rod, as shown in Figure. The collision is perfectly elastic. If K i, K f , KT , J, pi , Lcm, Li denote the initial kinetic energy, final kinetic energy, final kinetic energy of translation of rod, impulse delivered to rod, initial momentum of ball, angular momentum of centre of mass and initial angular momentum of ball about CM of rod, then match the physical quantities pertaining to this situation given in COLUMN-I with their values given in COLUMN-II. m

v0

A

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 166

(A) (B)

Kf Ki J pi

(C) Lcm Li (D)

2. m

B

COLUMN-I

Kf KT

COLUMN-II (p) 2 5 (q)

3 5

(r) 9 25 (s) 3

A uniform disc of mass 10 kg , radius 1 m is placed on a rough horizontal surface. The coefficient of friction between the disc and the surface is 0.2. A horizontal time varying force is applied at the centre of the disc whose variation with time is shown in graph. Match the contents of COLUMN-I with their respective time values in COLUMN-II.

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Chapter 3: Rotational Dynamics

3.

COLUMN-I

COLUMN-II

(A) Disc rolls without slipping

(p) at t = 7 s

(B) Disc rolls with slipping

(q) at t = 3 s

(C) Disc starts slipping at

(r) at t = 4 s

(D) Friction force is 10 N at

(s) at t = 6 s

Match the following quantities of COLUMN-I to their respective match(es) in COLUMN-II. COLUMN-I

COLUMN-II

(A) Axial vector

(p) Rotational K.E.

(B) Scalar

(q) Translational K.E.

5.

COLUMN-I (Axis of rotation)

COLUMN-II (Moment of Inertia about axis of rotation)

(C)

(r)

7 2 ml 3

(D)

(s)

8 2 ml 3

A disc rolls without slipping on the ground such that velocity of centre of mass of the disc is v0 . A point P on circumference of disc at angle θ with the vertical have a speed vp . Assuming that initially the point P is in contact with the ground, match the following.

C

4.

P

(s) Torque

A square frame is made by using four uniform rods each of mass m, length l. Match the moment of inertia for the frame about axis specified in COLUMN-I to their respective values given in COLUMN-II. COLUMN-I (Axis of rotation)

COLUMN-II (Moment of Inertia about axis of rotation)

(A)

(p)

2 2 ml 3

v0

θ

(C) Turning ability of force (r) Angular momentum (D) A rolling body can have

6.

COLUMN-I

COLUMN-II

(A) If θ = 60°

(p) vP = 2v

(B) If θ = 90°

(q) vP = v

(C) If θ = 120°

(r) vP = 2v

(D) If θ = 180°

(s) vP = 3v

Match the moment of inertia of bodies about specified axis with their respective values. COLUMN-I

(B)

3.167

COLUMN-II

1 (A) Moment of inertia of a circular (p) MR2 2 disc of mass M and radius R about a tangent parallel to plane of disc

5 (q) ml 2 3 (Continued)

(B) Moment of inertia of a solid sphere of mass M and radius R about a tangent.

(q)

7 MR2 5

(Continued)

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 167

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3.168 JEE Advanced Physics: Mechanics – II

COLUMN-I

COLUMN-II

(C) Moment of inertia of a circular disc of mass M and radius R about a tangent perpendicular to plane of disc

(r)

8.

5 MR2 4

COLUMN-I

COLUMN-II

(A)

(p) up the incline

1 (s) MR2 4

(D) Moment of inertia of a cylinder of mass M and radius R about its axis

Rolling upwards 3 2 (t) MR 2

7.

The inclined surfaces shown in COLUMN-I are sufficiently rough and COLUMN-II gives direction of frictional forces for situations in COLUMN-I. Match the two columns.

(B)

The diagram shows a uniform smooth solid cylinder A of radius 4 m rolling without slipping on the 8 kg plank, which in turn is supported by a fixed smooth surface. The blocks B and C both accelerate down with 6 ms −2 and 2 ms −2 respectively.

(q) down the incline

Kept in rotating position (C)

(r) maximum friction will act

Kept in translational position (D)

(s) required value of friction will act Kept in translational position

The angular acceleration of the cylinder A is α , the acceleration of centre of mass of cylinder is a, the ratio m of mass of cylinder A to the mass of block B is A . The mB length of the unwrapped thread between the cylinder and the block at t = 0 is l0 = 20 m and at t = 2 s is l (assume the system to be released from rest), match the following

9.

Four thin uniform rods of equal length l and mass m each form a square as shown in Figure. Moment of inertia about axes 1, 2, 3 and 4 are say I1, I 2 , I 3 and I 4 . Then, match the following 3

4

1 2

COLUMN-I

COLUMN-II

(A) α ( in rads −2 )

(p) 8

COLUMN-I

COLUMN-II

(B) 2 a ( in ms −2 )

(q) 2

(A) I1

(p)

(C) mA mB

4 2 ml 3

(r) 4

(B) I 2

(q)

2 2 ml 3

(D) ( l − l0 )

(s) 1

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 168

(Continued)

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Chapter 3: Rotational Dynamics

COLUMN-I

COLUMN-II

(C) I 3

(r)

(D) I 4

(s)

COLUMN-I

COLUMN-II

10 2 ml 3

(A) I1

(p)

21 MR2 32

5 2 ml 3

(B) I 2

(q)

I1 2

(C) ( I 3 + I 4 )

(r)

15 MR2 32

(D) ( I 2 − I 3 )

(s)

31 MR2 22

10. A uniform disc is acted upon by some forces and it rolls on a horizontal plank without slipping from north to south. The plank, in turn lies on a smooth horizontal surface. Match the following regarding this situation. COLUMN-I

COLUMN-II

(A) Frictional force on the disc by the surface

(p) May be directed towards north

(B) Velocity of the lowermost point of the disc

(q) May be directed towards south

3.169

12. A solid sphere is placed on a rough horizontal ground as shown in Figure.

(C) Acceleration of centre (r) May be zero of mass of the disc (D) Vertical component of the acceleration of centre of mass

(s) Must be zero

11. From a uniform disc of mass M and radius R, a concentric disc of half the radius is cut out. For the remaining annular disc the moment of inertia about axis 1 is I1 about 2 is I 2 about 3 is I 3 and about 4 is I 4 . The axes 1, 2 are perpendicular to the disc and axes 3, 4 are in the plane of the disc. Axes 2, 3 and 4 intersect at a common point. Match the contents of COLUMN-I with those in COLUMN-II.

If E is the centre of sphere such that DE > EF and a linear impulse can be applied either at point A, B or C then match the following two columns. COLUMN-I

COLUMN-II

(A) Sphere will acquire maximum angular speed if impulse is applied at

(p) A

(B) Sphere will acquire maximum linear speed if impulse is applied at

(q) B

(C) Sphere can roll without slipping (r) C if impulse is applied at (D) Sphere can roll with forward slipping if impulse is applied at

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 169

(s) at any point A, B or C

2/9/2021 6:42:08 PM

3.170 JEE Advanced Physics: Mechanics – II 13. A solid sphere is rotating about an axis as shown in Figure. An insect follows the dotted path on the circumference of sphere as shown. Match the following Insect

COLUMN-I

COLUMN-II

(A) Moment of inertia

(p) will remain constant

(B) Angular velocity

(q) will first increase then decrease

(C) Angular momentum

(r) will first decrease then increase

(D) Rotational kinetic energy

(s) will continuously decrease (t) will continuously increase

15. A solid sphere, a hollow sphere, a cylinder all of same mass and radius are released from a rough inclined plane on which all of them roll down without slipping. On reaching the bottom of the plane, match the contents of COLUMN-I with those of COLUMN-II. COLUMN-I

COLUMN-II

(A) Time taken to reach the bottom

(p) Maximum for solid sphere

(B) Total kinetic energy

(q) Maximum for hollow sphere

(C) Rotational kinetic energy

(r) Maximum for cylinder

(D) Translational kinetic energy

(s) Same for all

16. In each case, there is sufficient friction for regular rigid uniform disc to undergo pure rolling on a rigid horizontal surface. Match the situations in COLUMN-I to the quantities in COLUMN-II. COLUMN-I

COLUMN-II

(A) h

14. A thin walled cylindrical shell of mass m and radius R is placed on a rough inclined plane so that it rolls without slipping. Match the following table

(p) The direction of static friction may be forward or may be backward or static friction may be zero

F R

(B)

(q) The direction of static friction is towards backward

F

30°

R

COLUMN-I

COLUMN-II

(A) Linear acceleration of centre of mass

(p) is directly proportional to m

(B) Angular acceleration

(q) is inversely proportional to m

(C) Rotational kinetic (r) is directly energy at any instant proportional to R2 (D) Translational kinetic (s) is inversely energy at any instant proportional to R

(C)

(r) The angular acceleration will be clockwise

R h

F

(D) F

h

(s) Acceleration of the centre mass will be along direction F

R

(t) constant

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 170

12-Feb-21 4:23:20 PM

Chapter 3: Rotational Dynamics 17. A small solid ball rolls down along sufficiently rough surface from point 1 to point 3 as shown in Figure.

From point 3 onwards it moves under the influence of gravity. Match the following two columns. COLUMN-I

COLUMN-II

(A) Rotational kinetic energy of ball at point 2

1 (p) mgh 7

(B) Translational kinetic energy of ball at point 3

(q)

2 mgh 7

5 (C) Rotational kinetic energy of ball (r) mgh 7 at point 4 (D) Translational kinetic energy of ball at point 4

(s)

5 mgh 14

18. A particle moving on the smooth horizontal table strikes the rod AB at an end A perpendicularly and then stops. Match the following. COLUMN-I

COLUMN-II

(A) Angular momentum of system (particle + rod) about any point is

(p) Conserved for elastic collision

(B) Linear momentum of system (particle + rod) is

(q) Not conserved for elastic collision

(C) Angular momentum of rod about its centre of mass is

(r) Conserved for inelastic collision

(D) Kinetic energy of system (s) Not conserved (particle + rod), except for inelastic during collision is collision 19. For the following statements, except the gravity and contact force between the contact surfaces, no other force is acting on the body. Match the quantities in COLUMN-I to their respective match(es) in COLUMN-II.

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 171

3.171

COLUMN-I

COLUMN-II

(A) When a sphere is in pure–rolling on a fixed horizontal surface.

(p) Upward direction

(q) vcm > Rω (B) When a cylinder is in pure rolling on a fixed inclined plane in upward direction then friction force acts in (C) When a cylinder is in pure rolling down a fixed incline plane, friction force acts is

(r) vcm < Rω

(D) When a sphere of radius (s) No frictional force acts. R is rolling with slipping on a fixed horizontal surface, the relation between vcm and ω is 20. A disc of radius R is rolling without slipping with an angular acceleration α on a horizontal plane. Four points are marked at the end of horizontal and vertical diameter of a circle of radius r ( < R ) on the disc. Let horizontal and vertical directions be chosen as x and y axis as shown in Figure. When angular velocity of the disc is ω , then acceleration of points 1, 2, 3 and 4 are     a1 , a2, a3 and a4 respectively. Match the accelerations given in COLUMN-I with their respective vectors given in COLUMN-II.

COLUMN-I

COLUMN-II

 (A) a1

(p) ( R − r ) α iˆ + ( rω 2 ) ˆj

 (B) a2

(q) ( R + r ) α iˆ − ( rω 2 ) ˆj

 (C) a3

(r) ( Rα − rω 2 ) iˆ − ( rα ) ˆj

 (D) a4

(s) ( Rα + rω 2 ) iˆ + ( rα ) ˆj

2/9/2021 6:42:24 PM

3.172 JEE Advanced Physics: Mechanics – II 21. A rectangular slab ABCD has dimensions a × 2 a as shown in Figure.

Match the radius of gyration about the axis specified in COLUMN-I with their respective values in COLUMN-II.

COLUMN-I

COLUMN-II

(A) About axis 1

(p)

(B) About axis 2

(q)

2a 3

(C) About axis 3

(r)

a 2

(D) About axis 4

(s)

a 3

a 12

INTEGER/NUMERICAL ANSWER TYPE QUESTIONS In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1.

A small ball of mass 1 kg is attached to one end of a 1 m long light string and the other end of the string is hung from a point. When the resulting pendulum is 30° from the vertical, calculate the magnitude of torque, in Nm, about the point of suspension. Take g = 10 ms −2 .

2.

A uniform rod is falling without rotation on a smooth horizontal plane. Assuming the collision to be perfectly elastic, the angular velocity of the rod after striking the table is maximum when the rod makes an ⎛ 1 ⎞ angle cos −1 ⎜ with the horizontal just before strik⎝ ∗ ⎟⎠ ing where ∗ is not readable. Find ∗.

3.

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its center of mass (see figure). If the coefficient of friction is μ = 0.4, the maxib mum possible value of 100 × for a box not to topple a before moving is ……

4.

R starts rolling down 16 without slipping from the top of another sphere of radius R = 1 m. Find the angular velocity of the sphere, in rads −1, after it leaves the surface of the larger sphere.

A uniform sphere of radius =

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 172

5.

Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N × 10 −4 kgm 2 , find N .

6.

A uniform disc of mass 40 kg and radius 0.5 m can turn about a smooth axis through its centre and perpendicular to the disc. A constant torque is applied to the disc for 2 s from rest and the angular 240 velocity at the end of that time is revolutions per π minute. Find the magnitude of the torque, in Nm. If the torque is then removed and the disc is brought to rest in t  second by a constant force of 10 N applied tangentially at a point on the rim of the disc, find t, in seconds.

7.

One end of a straight uniform 1 m long bar is pivoted on horizontal surface. It is released from rest when it makes an angle 30° from the horizontal (see figure). Its angular speed when it hits the table is given as n s −1 , where n is an integer. The value of n is ……

2/9/2021 6:42:37 PM

Chapter 3: Rotational Dynamics

8.

A cylinder of mass m is kept on the edge of a plank of mass 2m and length 12 metre, which in turn is kept on smooth ground. Coefficient of friction between the plank and the cylinder is 0.1. The cylinder is given an impulse, which imparts it a velocity 7 ms −1 but no angular velocity. Find the time, in second, after which the cylinder falls off the plank. g = 10 ms −2 .

(

11. A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of mass 0.1 kg moving in a straight line with velocity 80 ms −1 hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rads −1) of the rod immediately after the collision will be ……

)

7 ms−1

m

2m 12 m

9.

3.173

A uniform circular disc has radius R and mass m, A  particle, also of mass m, is fixed at a point A on the edge of the disc as shown in Figure. The disc can rotate freely about a fixed horizontal chord PQ that is R at a distance from the centre C of the disc. The line 4 AC is perpendicular to PQ. Initially the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. The linear speed of the particle as it reaches its lowest position is xgR . Find x.

12. A flywheel whose moment of inertia about its axis of rotation is 16 kgm 2 is rotating freely in its own plane about a smooth axis through its centre. Its angular velocity is 9 rads −1 when a torque is applied to bring it to rest in t0 second. Find t0 , in seconds, if (a) the torque is constant and of magnitude of 4 Nm (b) the magnitude of the torque after t second is given by 8t 13. A block of mass m height 2 h = 2.4 m and width 2b = 1.2 m rests on a flat truck which moves horizontally with constant acceleration a as shown in figure. Determine

A A

R C

R/4

P

Q

10. A thin uniform rod AB of mass m = 1 kg moves translationally with acceleration a = 2 ms −2 due to two anti-parallel forces F1 and F2. The distance between the points at which these forces are applied is equal to l = 20 cm. Besides, it is known that F2 = 5 N. Find the length of the rod, in metre. F1

A  F2

B

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 173

Truck

a fmax

(a) the value of the acceleration, in ms −2 , at which slipping of the block on the truck starts if the coefficient of friction is μ = 0.4. (b) the value of the acceleration a, in ms −2 , at which block topples about A, assuming sufficient friction to prevent slipping. (c) the shortest distance, in metre, in which the truck can be stopped from a speed of 72 kmhr −1 with constant deceleration so that the block is not disturbed. Given that g = 10 ms −2 14. A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis as 5 revolutions per minute (rpm). The person now starts moving towards the center of the platform. The rotational speed (in rpm) of the platform when the person reaches its centre is ……

2/9/2021 6:42:56 PM

3.174 JEE Advanced Physics: Mechanics – II 15. For the system shown in figure a hoop of mass M = 1 kg radius r = 0.2 m is attached to a block of mass m = 0.5 kg. Calculate. M r

30 kg

Hoop

m

(a) the linear acceleration of hoop, in ms −2 . (b) the angular acceleration of the hoop in rads −2. (c) the tension in the rope, in newton. Neglect the mass of small pulley and the friction between the hoop and the horizontal surface. Masses of the pulley and string and friction between the hoop and the horizontal surface are negligible. 16. A ball rolls without sliding over a rough horizontal floor with velocity v0 = 7 ms −1 towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is e = 0.7 , calculate velocity v of the ball, in ms −1, along after the collision. 17. A 100 N circular plate of radius 0.6 m is suspended from a pin at A. If the pin is connected to a truck which is given an acceleration a = 0.9 ms −2, determine the horizontal and vertical components of reaction, in newton, at A and the acceleration of the centre of mass ( C ) of the plate in cms −2 . Assume, the plate to be originally at rest. Also find the angular acceleration in rads −2 of the plate. a

(a) the speed of the 30 kg body, in ms −1, just before it hits the floor and the angular speed of the pulley, in rads −1, at that time (b) the tension, in newton, in the strings and (c) the time, in second, it takes for the 30 kg body to reach the floor. 19. A circular disc of mass M and radius R is rotating about its axis with angular speed ω 1. If another stationR ary disc having radius and same mass M is dropped 2 co-axially on to the rotating disc. Gradually both discs attain constant angular speed ω 2 . The energy lost in the process is p% of the initial energy. Value of p is …… 20. A sphere starts from rest at the upper end of the track shown in the accompanying figure. It rolls until it leaves the track at the right end, where the track is horizontal. Calculate the distance d to the right, in metre, of this point at which the sphere strikes the ground. Take g = 10 ms −2 .

48 m

20 m d

Truck

A

0.6

m

C

18. The system in figure is released from rest. The 30 kg body is 2 m above the floor and is connected through an ideal string passing over the pulley (a uniform disk with a radius of 10 cm and mass 20 kg ) to another body of mass 10 kg . Find

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 174

2m

10 kg

21. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is IO and I P , respectively. Both these axes are perpendicular to the plane of the lamina. Find the ratio IP to the nearest integer. IO

2/9/2021 6:43:12 PM

Chapter 3: Rotational Dynamics 22. ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid points of AV and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I 0 . If part ADE is removed, the moment of inertia of the remaining part about the same NI 0 axis is where N is an integer. Value of N is …… 16

23. One fourth length of a uniform rod of mass m and length l is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread as shown in the figure. The thread is then burnt and the rod starts rotating about the edge. The angle between the rod and the horizontal when it is ∗μ , where ∗ is not about to slide on the edge is tan θ = 13 readable and μ is the coefficient of friction between the rod and the surface. Find ∗. /4

24. A bowling ball is thrown straight down the rough alley. The coefficient of friction between the ball and the alley is 0.75 . When the ball starts, its centre of mass has a speed 12 ms −1 and it is sliding without rotating. Calculate the distance moved by the ball (in metre) down the alley before it starts rolling without slipping. 25. A block of mass m = 1 kg slides down the surface of a smooth incline as shown in figure. The block is tied to a string which is wrapped around a disk capable of rotating about a horizontal axis. The disk has a mass M = 3 kg and a radius R = 0.2 m. Initially the string is taut. If the mass is released, calculate its acceleration, in ms −2 , if g = 10 ms −2 .

3.175

26. A massless equilateral triangle EFG of side a (as shown in Figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG N is ma 2 where N is an integer. The value of N is …… 20

27. A uniform disk of mass M = 40 g and radius R = 0.5 cm is pivoted so that it can rotate freely about a horizontal axis through its centre and normal to the plane of the disk. A small particle of mass m = 5 g is attached to the rim of the disk at the top directly above the pivot. The system is given a gentle start and the disk begins to rotate. (a) What is the angular velocity of the disk, in rads −1, when the particle is at its lowest point? (b) At this point, what force, in newton, must be exerted on the particle by the disk to keep it on the disk? Take g = 10 ms −2 28. Two identical uniform discs A and B each of mass 2 kg and radius 0.5 m are held, as shown in figure with the help of a long massless string which is wrapped around the discs in opposite directions. Disc A is attached to the ceiling in such a way that it can rotate freely about its axis. The disc, B, initially held at same height as A, is then released to fall so that string unwinds from both the discs. Find the angular acceleration, in rads −2 linear acceleration of falling disc in ms −2 and tension in the string, in newton. Assume that string does not slip and motion is confined in the same vertical plane and take g = 10 ms −2 .

A

B m

30°

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 175

29. A grindstone in the form of a solid cylinder has a radius of 0.2 m and a mass of 30 kg . (a) What constant torque, in newton metre, will bring it from rest to an angular velocity of 250 revmin −1 in 10 s ?

2/9/2021 6:43:31 PM

3.176 JEE Advanced Physics: Mechanics – II (b) Through what angle, in radian, has it turned during that time? (c) Calculate the work done, in joule, by the torque. 30. A uniform rod AB of mass m = 2 kg and length  = 1 m is placed on a sharp support O such that a = 0.4 m and b = 0.6 m . A spring of force constant k = 600 Nm −1 is attached to end B as shown in Figure. a

b

A

B O

To keep the rod horizontal, its end A is tied with a thread such that the spring is elongated by 1 cm. Calculate reaction of support O, in newton, at the instant thread is burnt. 31. A solid fly wheel of 20 kg mass and 120 mm radius revolves at 600 rpm . Calculate the force (in newton) with which the brake lining should be pressed against the flywheel to stop the flywheel in 3 s, if the coefficient of friction is 0.1. 32. The polar ice caps contain about 3 × 1019 kg of ice. This contributes essentially nothing to the moment of inertia of the earth because it is located at the poles, close to the axis of rotation. Estimate the expected change in the length of the day (in second) if the polar ice caps melt, distributing the water uniformly over the entire surface of the earth if mass and radius of earth are 6 × 10 24 kg and 6.4 × 106 m.

 33. A force F = iˆ + 2 ˆj + 3 kˆ N acts at a point 4iˆ + 3 ˆj − k m . Then the magnitude of torque about the point iˆ + 2 ˆj + kˆ m will be x Nm. The value of x is ……

(

(

)

(

)

)

34. A man stands at the centre of a circular platform holding his arms extended horizontally with 4 kg block in each hand. He is set rotating about a vertical axis at 0.5 revs −1. The moment of inertia of the man plus platform is 1.6 kgm 2, assumed constant. The blocks are 90 cm from the axis of rotation. He now pulls the blocks inwards towards his body until they are 15 cm from the axis of rotation. Calculate (a) his new angular velocity, in rads −1. (b) the initial and final kinetic energy of the man and platform (in joule). (c) the work done by the man (in joule) to pull in the blocks. 35. A cylinder of mass r = 0.1 m and mass M = 2 kg is placed such that it is in contact simultaneously with a vertical and a horizontal surface as shown in Figure. ⎛ 1⎞ The coefficient of static friction μ = ⎜ ⎟ for both the ⎝ 3⎠ surfaces. Find the distance d, in cm, from the centre of the cylinder at which a force F = 40 N should be applied so that the cylinder just starts rotating in the anticlockwise direction. Take g = 10 ms −2 . F = 40 N

d

ARCHIVE: JEE MAIN 1.

[Online September 2020] Shown in Figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling. The rod is of mass m and has another weight of mass 2 m hung at a distance of 75 cm from A. The tension in the string at A is

(A) 2mg

(B) 0.5mg

(C) 0.75mg

(D) 1mg

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 176

2.

[Online September 2020] A uniform cylinder of mass M and radius R is to be pulled over a step of height a ( a > R ) by applying a force F at its center O perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is

2/9/2021 6:43:49 PM

Chapter 3: Rotational Dynamics

(A) Mg 1 − (C) Mg 3.

4.

a2 R2

a R

5.

2

⎛ R ⎞ (B) Mg ⎜ −1 ⎝ R − a ⎟⎠ ⎛ R− a⎞ (D) Mg 1 ⎜ ⎝ R ⎟⎠

2

[Online September 2020] Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centers. The moment of inertia and angular velocity of the first disc are 0.1 kgm 2 and 10 rads −1 respectively while those for the second one are 0.2 kgm 2 and 5 rads −1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is 10 2 J (B) J (A) 3 3 5 20 J (D) J (C) 3 3 [Online September 2020] A block of mass m = 1 kg slides with velocity v = 6 ms −1 on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. If the rod has mass M = 2 kg, and length l = 1 m, the value of θ is approximately (Take g = 10 ms −2 )

(A) 69° (C) 55°

ml 2 2 ω sin θ cos θ about the center of mass 12 (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of θ is then such that

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 177

g 2lω 2 2g (C) cos θ = 3lω 2 (A) cos θ =

3g 2lω 2 g (D) cos θ = 2 lω (B) cos θ =

6.

[Online September 2020] Consider two uniform discs of the same thickness and different radii R1 = R and R2 = α R made of the same material. If the ratio of their moments of inertia I1 and I 2 , respectively, about their axes is I1 : I 2 = 1 : 16 then the value of α is (B) 2 (A) 2 (C) 4 (D) 2 2

7.

[Online September 2020] For a uniform rectangular sheet shown in Figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the center of mass) and O′ (corner point) is

(B) 63° (D) 49°

[Online September 2020] A uniform rod of mass m, length l is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ it (see figure). To find θ , equate the rate of change of angular momentum (direction going into the paper)

3.177

1 2 1 (C) 8

(A)

8.

2 3 1 (D) 4 (B)

[Online September 2020] A wheel is rotating freely with an angular speed ω on a shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is

2/9/2021 6:44:08 PM

3.178 JEE Advanced Physics: Mechanics – II 1 4 5 (D) 6

(A) 0 (C) 9.

(B)

3 4

[Online September 2020] Shown in Figure is a hollow ice cream cone (it is open at the top). If its mass is M, radius of its top, R and height, H , then its moment of inertia about its axis is

(A)

MR2 2

(B)

MH 2 3

(C)

MR2 3

(D)

M ( R2 + H 2 ) 4

10. [Online September 2020] Four point masses, each of mass m, are fixed at the corners of a square of side l. The square is rotating with angular frequency ω , about an axis passing through one of the corners of the square and parallel to its diagonal, as shown in Figure. The angular momentum of the square about this axis is

12. [Online January 2020] As shown in Figure, a bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be

(A)

1 2 gh r 3

(B) r

3 4 gh

(C)

1 4 gh r 3

(D) r

3 2 gh

13. [Online January 2020] The radius of gyration of a uniform rod of length l, l about an axis passing through a point away from 4 the center of the rod, and perpendicular to it, is (A)

l 8

(C) l

(B) l 3 8

(D)

7 48

l 4

14. [Online January 2020] Mass per unit area of a circular disc of radius a depends on the distance r from its center as σ ( r ) = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is ⎛ A aB ⎞ ⎛ A aB ⎞ (A) 2π a 2 ⎜ + (B) π a 4 ⎜ + ⎟ ⎟ ⎝ 4 ⎝ 4 5 ⎠ 5 ⎠ (A) 2ml 2ω 2

(C) ml ω

(B) 3 ml 2ω 2

(D) 4 ml ω

11. [Online September 2020] The linear mass density of a thin rod AB of length L x⎞ ⎛ varies from A to B as λ ( x ) = λ 0 ⎜ 1 + ⎟ , where x is the ⎝ L⎠ distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is 5 3 ML2 (B) ML2 (A) 12 7 2 7 (C) ML2 (D) ML2 5 18

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 178

⎛ aA B ⎞ + ⎟ (C) 2π a 4 ⎜ ⎝ 4 5⎠

⎛ A B⎞ (D) 2π a 4 ⎜ + ⎟ ⎝ 4 5⎠

15. [Online January 2020] A uniform sphere of mass 500g rolls without slipping on a plane horizontal surface with its center moving at a speed of 5.00 cms −1. Its kinetic energy is (A) 8.75 × 10 −4 J (B) 8.75 × 10 −3 J (C) 6.25 × 10 −4 J

(D) 1.13 × 10 −3 J

16. [Online January 2020] Three solid spheres each of mass m and diameter d are stuck together such that the lines connecting the centers from an equilateral triangle of side of length I d. The ratio 0 of moment of inertia I 0 of the system IA

2/9/2021 6:44:30 PM

Chapter 3: Rotational Dynamics about an axis passing the centroid and about center of any of the spheres I A and perpendicular to the plane of the triangle is

13 23 23 (C) 13

15 13 13 (D) 15

(A)

(B)

17. [Online January 2020] A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its center of mass (see figure). A massless string is wrapped over its rim and two blocks of masses m1 and m2 ( m1 > m2 ) are attached to the ends of the string. The system is released from rest. The angular speed of the wheel m1 descents by a distance h is

3 2 3 (C) 5 (A)

3.179

1 2 8 (D) 5

(B)

19. [Online April 2019] A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time τ = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to

(A) 0.3

(B) 0.02

(C) 0.28

(D) 0.5

20. [Online April 2019] A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the hsph is given by incline. The ratio hcyl

1

m1 + m2 ⎡ ⎤2 (A) ⎢ 2 ⎥ gh + m + m R I ( ) 2 ⎣ 1 ⎦ 1

⎡ 2 ( m1 − m2 ) gh ⎤ 2 (B) ⎢ ⎥ 2 ⎣ ( m1 + m2 ) R + I ⎦

1

⎡ 2 ( m1 + m2 ) gh ⎤ 2 (C) ⎢ ⎥ 2 ⎣ ( m1 + m2 ) R + I ⎦

1 ⎤2

⎡ ( m1 − m2 ) (D) ⎢ ⎥ gh 2 ⎣ ( m1 + m2 ) R + I ⎦ 18. [Online April 2019] A thin circular plate of mass M and radius R has its density varying as ρ ( r ) = ρ0 r with ρ0 as constant and r is the distance from its centre. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = a ( MR2 ). The value of the coefficient a is

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 179

(A)

4 5

(C) 1

2 5 14 (D) 15

(B)

21. [Online April 2019] A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ , where θ is the angle by which it has rotated, is given as kθ 2. If its moment of inertia is I , then the angular acceleration of the disc is (A)

2k θ I

(B)

k θ I

(C)

k θ 2I

(D)

k θ 4I

2/9/2021 6:44:48 PM

3.180 JEE Advanced Physics: Mechanics – II 22. [Online April 2019] The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane (1) a ring of radius R, R (2) a solid cylinder of radius and 2 R (3) a solid sphere of radius . 4 If, in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is (B) 10 : 15 : 7 (A) 14 : 15 : 20 (C) 4 : 3 : 2 (D) 2 : 3 : 4 23. [Online April 2019] A thin smooth rod of length L and mass M is rotating freely with angular speed ω 0 about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be Mω 0 Mω 0 (B) (A) M + 3m M+m Mω 0 Mω 0 (C) (D) M + 6m M + 2m 24. [Online April 2019] Moment of inertia of a body about a given axis is 1.5 kgm 2 . Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rads −2 must be applied about the axis for a duration of (B) 2 s (A) 2.5 s (C) 5 s (D) 3 s 25. [Online April 2019] Two coaxial discs, having moments of inertia I1 and I1 are rotating with respective angular velocities ω 1 2 ω and 1 about their common axis. They are brought in 2 contact with each other and thereafter they rotate with a common angular velocity. If E f and Ei are the final and initial total energies, then ( E f − Ei ) is I1ω 12 12 I1ω 12 (C) 6 (A) −

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 180

3 I1ω 12 8 I ω2 (D) − 1 1 24

(B)

26. [Online April 2019] A particle of mass m is moving along a trajectory given by x = x0 + a cos ω 1t and y = y0 + b sin ω 2t . The torque, acting on the particle about the origin, at t = 0 is (A) − m x bω 2 − y aω 2 kˆ (B) m ( − x b + y a ) ω 2 kˆ

(

(C)

0

2

0

+my0 aω 12 kˆ

1

)

0

0

1

(D) zero

27. [Online April 2019] A thin disc of mass M and radius R has mass per unit area σ ( r ) = kr 2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is MR2 MR2 (B) (A) 3 6 (C)

MR2 2

(D)

2 MR2 3

28. [Online April 2019] The time dependence of the position of a particle of  mass m = 2 is given by r ( t ) = 2tiˆ − 3t 2 ˆj . Its angular momentum, with respect to the origin, at time t = 2 is (A) −34 ( kˆ − iˆ ) (B) 48 iˆ − ˆj

(

(C) 36 kˆ

)

(D) −48 kˆ

29. [Online April 2019] A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in Figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5 s, is closed to

(A) 4.0 × 10 −6 Nm

(B) 7.9 × 10 −6 Nm

(C) 2.0 × 10 −5 Nm

(D) 1.6 × 10 −5 Nm

30. [Online April 2019] A solid sphere of mass M and radius R is divided into 7M two unequal parts. The first part has a mass of 8 and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere.

2/9/2021 6:45:13 PM

Chapter 3: Rotational Dynamics Let I1 be the moment of inertia of the disc about its axis and I 2 be the moment of inertia of the new sphere I about its axis. The ratio 1 is given by I2 (A) 140 (B) 185 (C) 285 (D) 65 31. [Online April 2019] A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc ⎛σ ⎞ varies as ⎜ 0 ⎟ , then the radius of gyration of the disc ⎝ r ⎠ about its axis passing through the centre is

(A)

a+b 2

(B)

a 2 + b 2 + ab 2

(C)

a+b 3

(D)

a 2 + b 2 + ab 3

32. [Online January 2019] A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rads −1) will be g = 10 ms −2

(

(A)

)

20 3

(B)

30

30 30 (D) 2 2 33. [Online January 2019] To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is μ , the torque, applied by the machine on the mop is μ FR μ FR (B) (A) 2 3 μ FR 2 (C) μ FR (D) 6 3 (C)

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 181

3.181

34. [Online January 2019] A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is F 2F (B) (A) 2 MR 3 MR F 3F (C) (D) 3 MR 2 MR 35. [Online January 2019] A rigid massless rod of length 3l has two masses attached at each end as shown in Figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be

g 2l g (C) 13l

(A)

g 3l 7g (D) 3l (B)

36. [Online January 2019] Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

152 MR2 15 17 MR2 (B) 15 209 MR2 (C) 15 137 (D) MR2 15 (A)

37. [Online January 2019] An equilateral triangle ABC is cut from a thin solid sheet of wood (see figure). D, E and F are the midpoints of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I 0 . If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I . Then

2/9/2021 6:45:34 PM

3.182 JEE Advanced Physics: Mechanics – II 41. [Online January 2019] A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly at its opposite ends (see figure). The moment of inertia of the system about the axis OO′, passing through the centre of D1 as shown in Figure, will be 3 I0 4 I (C) I = 0 4

15 I0 16 9 (D) I = I0 16

(A) I =

(B) I =

38. [Online January 2019]   A slab is subjected to two forces F1 and F2 of same  magnitude F as shown in Figure. Force F2 is in xy plane   while force F1 acts along z-axis at the point 2 i + 3 j . The moment of these forces about point O will be

(



F1

(A) (C)

( 3iˆ − 2 ˆj + 3kˆ ) F ( 3iˆ + 2 ˆj + 3kˆ ) F

)



F2

(B) (D)

( 3iˆ + 2 ˆj − 3kˆ ) F ( 3iˆ − 2 ˆj − 3kˆ ) F

39. [Online January 2019] A string is wound around a hollow cylinder of mass 5 kgand radius 0.5 m. If the string is now pulled with a horizontal force of 40 N and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)

(A) 16 rads −2

(B) 20 rads −1

(C) 12 rads −2

(D) 10 rads −2

40. [Online January 2019] The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) π π (B) (A) 8 6 π π (D) (C) 3 4

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 182

(A) 3 MR2 (C) MR2

4 MR2 5 2 MR2 (D) 3

(B)

42. [Online January 2019] Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I . The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I , is (B) 12 cm (A) 14 cm (C) 16 cm (D) 18 cm 43. [Online January 2019] The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is I ( x ). Which one of the graphs represents the variation of I ( x ) with x correctly? (A)

(B)

(C)

(D)

44. [Online January 2019] A particle of mass 20 g is released with an initial velocity 5 ms −1 along the curve from the point A, as shown in Figure. The point A is a height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be (Take g = 10 ms −2 )

2/9/2021 6:45:49 PM

Chapter 3: Rotational Dynamics

3.183

47. [Online 2018] A force of 40 N acts on a point B at the end of an L-shaped object, as shown in Figure. The angle θ that will produce maximum moment of the force about point A is given by

(A) 2 kgm 2s −1

(B) 3 kgm 2s −1

(C) 8 kgm 2s −1

(D) 6 kgm 2s −1

45. [2018] Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is

1 2 (C) tan θ = 2 (A) tan θ =

(B) tan θ = 4 (D) tan θ =

1 4

48. [Online 2018] A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of ( m, x ) values is recorded. The appropriate variables that give a straight line, when plotted are 19 MR2 2 73 (C) MR2 2 (A)

55 MR2 2 181 (D) MR2 2

(B)

46. [2018] From a uniform circular disc of radius R and mass R 9M , a small disc of radius is removed as shown in 3 Figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

(A) m,

1 x2

(C) m, x

(B) m, x 2 (D) m,

1 x

49. [Online 2018] A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is g = 10 ms −2

(

(A) 0.7 (C) 0.3

(A) 4 MR2

(B)

40 MR2 9

(C) 10 MR2

(D)

37 MR2 9

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 183

)

(B) 0.5 (D) 0.6

50. [Online 2018] A thin rod MN, free to rotate in the vertical plane about the fixed end N , is held horizontal. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to (see figure)

2/9/2021 6:45:59 PM

3.184 JEE Advanced Physics: Mechanics – II

(A) cos α (C)

cos α

(B) sin α (D)

sin α

51. [Online 2018] A thin uniform bar of length L and mass 8m lies on a smooth horizontal table. Two point masses m and 2m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance L L and respectively from the centre of the bar. If the 3 6 bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be

6v 5L v (C) 5L (A)

v 6L 3v (D) 5L (B)

54. [2017] A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (shown in Figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is

3g sin θ 2l 3g cos θ (C) 2l (A)

2g sin θ 3l 2g cos θ (D) 3l

(B)

55. [Online 2017] A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in Figure. The body is released from rest. Then the acceleration of the body is

52. [Online 2018] A thin circular disk is in the xy plane as shown in Figure. The ratio of its moment of inertia about z and z′ axes will be

2mg 2m + M 2 Mg (C) 2M + m (A)

(A) 1 : 4 (C) 1 : 2

(B) 1 : 3 (D) 1 : 5

53. [2017] The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I . l What is the ratio such that the moment of inertia is R minimum? (A)

3 2

(C) 1

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 184

(B)

3 2

(D)

3 2

2 Mg 2m + M 2mg (D) 2M + m (B)

56. [Online 2017] In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct? (A) Left arm is shorter than the right arm (B) Left arm is longer than the right arm (C) Every object that is weighed using this balance appears lighter than its actual weight (D) Both the arms are of same length

2/9/2021 6:46:09 PM

Chapter 3: Rotational Dynamics 57. [Online 2017]

R A circular hole of radius is made in a thin uniform 4 disc having mass M and radius R, as shown in Figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is

(A)

219 MR2 256

(B)

197 MR2 256

(C)

19 MR2 512

(D)

237 MR2 512

58. [2016] A particle of mass m is moving along the side of a square of side a, with a uniform speed v in the x -y plane as shown in Figure.

are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to

(A) (B) (C) (D)

turn left and right alternately turn left turn right go straight

60. [Online 2016] A cubical block of side 30 cm is moving with velocity 2 ms −1 on a smooth horizontal surface. The surface has a bump at a point O as shown in Figure. The angular velocity (in rads −1) of the block immediately after it hits the bump, is

(A) 13.3 (C) 9.4

Which of the following statements is false for the  angular momentum L about the origin?  mvR (A) L = kˆ when the particle is moving from D 2 to A  mvR ˆ (B) L = − k when the particle is moving from 2 A to B  ⎛ R ⎞ (C) L = mv ⎜ − a ⎟ kˆ when the particle is moving ⎝ 2 ⎠ from C to D  ⎛ R ⎞ (D) L = mv ⎜ + a ⎟ kˆ when the particle is moving ⎝ 2 ⎠ from B to C 59. [2016] A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 185

3.185

(B) 5.0 (D) 6.7

61. [Online 2016] Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to (Take the radius of the drum to be 1.25 m and its axle to be horizontal) (A) 27.0 (B) 0.4 (C) 1.3 (D) 8.0 62. [2015] From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is (A)

MR2 32 2π

(B)

MR2 16 2π

(C)

4 MR2 9 3π

(D)

4 MR2 3 3π

2/9/2021 6:46:26 PM

3.186 JEE Advanced Physics: Mechanics – II 63. [Online 2015] A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping then the value of F is (B) 2ma (A) ma 5 3 (C) ma (D) ma 2 3 64. [Online 2015] Consider a thin uniform square sheet made of a rigid material. If its side is a, mass m and moment of inertia I about one of its diagonals, then ma 2 (A) I > 12 (C) I =

ma 2 12

ma 2 ma 2 (B) hC ; K B > KC

(B) hA > hC ; KC > K A

(C) hA = hC; K B = KC

(D) hA < hC ; K B > KC

11. [IIT-JEE 2005]

R is removed from a circular 3 disc of radius R and mass 9M . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is A small disc of radius

R/3

R

(C) 10 MR2

40 MR2 9 37 MR2 (D) 9

(B)

12. [IIT-JEE 2005] A particle moves in a circular path with decreasing speed. Choose the correct statement (A) Angular momentum remains constant  (B) Acceleration ( a ) is towards the centre

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 189

(A) vQ > vC > vP (B) vQ < vC < vP (C) vQ = vP , vC = (D) vQ < vC > vP

1 vP 2

15. [IIT-JEE 2003] A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (A) Centre of circle (B) On the circumference of the circle (C) Inside the circle (D) Outside the circle

O

(A) 4 MR2

P

16. [IIT-JEE 2003] Consider a body, shown in Figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = Mv is imparted to the body at one of its end, then its angular velocity is L M

J = Mv

2/9/2021 6:47:04 PM

3.190 JEE Advanced Physics: Mechanics – II

2v L v (D) 4L

v L v (C) 3L (A)

(B)

17. [IIT-JEE 2002] A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity ω 0 . When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform). The angular velocity of the platform ω ( t ) will vary with time t as (A)

ω (t)

(B)

(C)

MR2 8

(D)

MR2 4 2 MR2

20. [IIT-JEE 2000] A thin wire of length L and uniform linear mass density ρ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX′ is X′

t

(D) 0

0

t

ρL3 8π 2 5ρL3 (C) 16π 2 (A)

0

t

t

18. [IIT-JEE 2002] A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are (A) up the incline while ascending and down the incline while descending (B) up the incline while ascending as well as descending (C) down the incline while ascending and up the incline while descending (D) down the incline while ascending as well as descending 19. [IIT-JEE 2001] One quarter sector is cut from a uniform disc of radius R. This sector has massM. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 190

MR2 2

(B)

ω0

(C)

(A)

ρL3 16π 2 3 ρL3 (D) 8π 2 (B)

21. [IIT-JEE 2000] A cubical block of side L rests on a rough horizontal surface with coefficient of friction μ. A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

(A) infinitesimal (C)

mg 2

(B)

mg 4

(D) mg ( 1 − μ )

22. [IIT-JEE 2000] An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities that are conserved as beads slide down are

2/9/2021 6:47:12 PM

Chapter 3: Rotational Dynamics (A) angular velocity and total energy (kinetic and potential). (B) total angular momentum and total energy. (C) angular velocity and moment of inertia about the axis of rotation. (D) total angular momentum and moment of inertia about the axis of rotation. 23. [IIT-JEE 1999] A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane as shown in Figure. The magnitude of angular momentum of the disc about the origin O is y

ω M

R

x

O

⎛ 1⎞ (A) ⎜ ⎟ MR2ω ⎝ 2⎠

(B) MR2ω

⎛ 3⎞ (C) ⎜ ⎟ MR2ω ⎝ 2⎠

(D) 2 MR2ω

24. [IIT-JEE 1999] A smooth sphere A is moving on a frictionless horizontal plane with angular speed ω and centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ω A and ω B. Then (B) ω A = ω B (A) ω A < ω B (C) ω A = ω

(D) ω = ω B

25. [IIT-JEE 1999] A cubical block of side a is moving with velocity v on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. If the moment of inertia of the block about an axis passing through centre of gravity 1 is ma 2, then the angular speed of the block after it hits 6 O is

(A) (C)

3v 4a 3v 2a

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 191

(B)

3v 2a

(D) ZERO

3.191

26. [IIT-JEE 1998] Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to (B) I sin 2 θ (A) I ⎛θ⎞ (C) I cos 2 θ (D) I cos 2 ⎜ ⎟ ⎝ 2⎠ 27. [IIT-JEE 1997] A mass M is moving with a constant velocity parallel to the x-axis. Its angular momentum with respect to the origin (A) is zero (B) remains constant (C) increases (D) decreases 28. [IIT-JEE 1995] Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of (A) 0.42 m from mass of 0.3 kg (B) 0.70 m from mass of 0.7 kg (C) 0.98 m from mass of 0.3 kg (D) 0.98 m from mass of 0.7 kg 29. [IIT-JEE 1992] A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω . The force exerted by the liquid at the other end is Mω 2 L (A) (B) Mω 2 L 2 (C)

Mω 2 L 4

(D)

Mω 2 L2 2

30. [IIT-JEE 1983] A thin circular ring of mass M is rotating about its axis with a constant angular velocity ω . The two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity ωM ω ( M − 2 m) (B) (A) M+m M + 2m ωM ω ( M + 2 m) (C) (D) M + 2m M

2/9/2021 6:47:23 PM

3.192 JEE Advanced Physics: Mechanics – II

Multiple Correct Choice Type Problems In this section each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

[JEE (Advanced) 2020] A rod of mass m and length L, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it. The combined system now rotates with angular speed ω about the pivot. The maximum angular speed ω M is achieved for x = x M. Then

(A) Instantaneous torque about the point in contact with the floor is proportional to sin θ (B) The trajectory of the point A is parabola (C) The mid-point of the bar will fall vertically downward (D) When the bar makes an angle θ with the vertical, the displacement of its mid-point from the initial position is proportional to ( 1 − cos θ ) 4.

(A) ω =

3vx L2 + 3 x 2

(C) x M = 2.

L 3

(B) ω =

12vx L2 + 12x 2

(D) ω M =

v 3 2L

[JEE (Advanced) 2019] A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical? (g is the acceleration due to gravity) 2g L (B) The radial acceleration of the rod’s centre of mass 3g will be 4 3g (C) The angular speed of the rod will be 2L (D) The normal reaction force from the floor on the Mg rod will be 16 (A) The angular acceleration of the rod will be

3.

[JEE (Advanced) 2017] A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in Figure). At some instant of time, the angle made by the bar with the vertical is θ . Which of the following statements about its motion is/are correct?

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 192

[JEE (Advanced) 2016] Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l = 24 a through their centers. This assembly is laid on a firm and flat surface and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω . The angular momentum of the entire assembly about the  point O is L (see the figure). Which of the following statement(s) is(are) true?

(A) The center of mass of the assembly rotates about ω the z-axis with an angular speed of 5 (B) The magnitude of angular momentum of center of mass of the assembly about the point O is 81ma 2ω . (C) The magnitude of angular momentum of the ω assembly about its center of mass is 17 ma 2 2  (D) The magnitude of the z-component of L is 55ma 2ω . 5.

[JEE (Advanced) 2012] The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with

2/9/2021 6:47:31 PM

Chapter 3: Rotational Dynamics

3.193

C

angular speed ω . The ring and disc are separated by 2 frictionless ball bearings. The system is in the x -z plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface,

B

A

    (A) VC − VA = 2 ( VB − VC )     (C) VC − VA = 2 VB − VC 8.

    (B) VC − VB = VB − VA    (D) VC − VA = 4 VB

[IIT-JEE 2006] A solid sphere is in pure rolling motion on an inclined surface having inclination θ

(A) the point O has a linear velocity 3Rω iˆ (B) the point P has a linear velocity

11 ˆ 3 Rω i + Rω kˆ 4 4

(C) the point P has a linear velocity

13 3 Rω iˆ − Rω kˆ 4 4

P has (D) the point ⎛ ⎞ 3 1 ⎜⎝ 3 − ⎟ Rω iˆ + Rω kˆ 4 ⎠ 4 6.

a

linear

velocity

[IIT-JEE 2011] A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 ms −1 . A small ball of mass 0.1 kg , moving with velocity 20 ms −1 in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 ms −1 . Immediately after the collision,

(A) the ring has pure rotation about its stationary CM (B) the ring comes to a complete stop (C) friction between the ring and the ground is to the left (D) there is no friction between the ring and the ground 7.

[IIT-JEE 2009] A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then,

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 193

(A) frictional force acting on sphere is f = μmg cos θ (B) f is dissipative force (C) friction will increase its angular velocity and decrease its linear velocity (D) If θ decreases, friction will decrease 9.

[IIT-JEE 1998]  The torque τ on a body about a given point is found to     be A × L where A is a constant vector and L is angular momentum of the body about that point. From this it follows that   dL is perpendicular to L at all instants of time. (A) dt   (B) the component of L in the direction of A does not change with time.  (C) the  magnitude of L does not change with time. (D) L does not change with time.

10. [IIT-JEE 1992] The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plate is

(A) I1 + I 2

(B) I 3 + I 4

(C) I1 + I 3

(D) I1 + I 2 + I 3 + I 4

2/9/2021 6:47:41 PM

3.194 JEE Advanced Physics: Mechanics – II 11. [IIT-JEE 1991] A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v, respectively, strike the bar as shown in Figure and stick to it after collision. Denoting angular velocity (about the centre of mass), total energy and the centre of mass velocity by ω , E and Vc respectively, we have after collision

(A) Vc = 0 (C) ω =

v 5a

(B) ω = (D) E =

2

12. [IIT-JEE 1990] A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. The magnitude of angular momentum of the projectile about point of projection when the particle is at maximum height h is (A) ZERO (C)

mv 3 2g

(B)

Linked Comprehension Type Questions This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1

3v 5a 3 mv 5

the same height. The hollow cylinder will reach the bottom of the inclined plane first. Statement-2: By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

mv 3 4 2g

One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity ω 0 . The rotating rings rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is μ and the acceleration due to gravity is g.

(D) m 2 gh 3

Reasoning Based Questions This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE. 1.

[IIT-JEE 2008] Statement-1: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 194

1.

[JEE (Advanced) 2017] The total kinetic energy of the ring is 1 2 2 (B) Mω 02 ( R − r ) (A) Mω 02 ( R − r ) 2 3 (C) Mω 02 R2 (D) Mω 02 ( R − r 2 ) 2

2.

[JEE (Advanced) 2017] The minimum value of ω 0 below which the ring will drop down is (A)

g 2μ ( R − r )

(B)

3g 2μ ( R − r )

(C)

g μ(R − r )

(D)

2g μ(R − r )

2/9/2021 6:47:48 PM

Chapter 3: Rotational Dynamics

3.195

Comprehension 2

Comprehension 3

A frame of the reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of a non-inertial frame of reference.  The relationship between the force Frot experienced by a particle  of mass m moving on the rotating disc and the force Fin experienced by the particle in an inertial frame        of reference is, Frot = Fin + 2m ( vrot × ω ) + m ( ω × r ) × ω ,  where, vrot is the velocity of the particle in the rotating frame  of reference and r is the position vector of the particle with respect to the centre of the disc

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a  thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in Figure.

Now, consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its centre. We assign a coordinate system with the origin at the centre of the disc, the X-axis along the slot, the Y-axis perpendicular to the slot and the z-axis along the rotation axis ( ω = ω kˆ ). A  small block of mass m is gently placed in the slot at  ⎛ R⎞ r = ⎜ ⎟ iˆ at t = 0 and is constrained to move only along the ⎝ 2⎠ slot. 3.

4.

[JEE (Advanced) 2016] The distance r of the block at time t is R R cos 2ωt (B) cos ωt (A) 2 2 R ( ωt R ( 2ω t (C) e + e −ω t ) (D) e + e −2ω t ) 4 4 [JEE (Advanced) 2016] The net reaction of the disc on the block is (A)iˆ mω 2 R sin ω tjˆ − mgkˆ 1 mω 2 R ( eω t − e −ω t ) ˆj + mgkˆ (B) 2 1 mω 2 R ( e 2ω t − e −2ω t ) ˆj + mgkˆ (C) 2 iˆ − mω 2 R cos ω tjˆ − mgkˆ (D)

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 195

When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed ω in this case. Now, consider two similar systems as shown in Figure : Case (a) the disc with its face vertical and parallel to x -z plane; Case (b) the disc with its face making an angle of 45° with x -y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P and the systems are rotated with constant angular speed ω about the z-axis.

5.

[IIT-JEE 2012] Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct? (A) It is 2ω for both the cases ω (B) It is ω for case (a) and for case (b) 2 (C) It is ω for case (a) and 2ω for case (b) (D) It is ω for both the cases

2/9/2021 6:47:56 PM

3.196 JEE Advanced Physics: Mechanics – II 6.

[IIT-JEE 2012] Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? (A) It is vertical for both the cases (a) and (b) (B) It is vertical for case (a) and is at 45° to the x -z plane and lies in the plane of the disc for case (b) (C) It is horizontal for case (a) and is at 45° to the x -z plane and is normal to the plane of the disc for case (b) (D) It is vertical for case (a) and is at 45° to the x -z plane and is normal to the plane of the disc for case (b)

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after series of calculations based on the data provided in the question(s). 1.

[JEE (Advanced) 2020] Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the centre, initially only the left finger slips with respect to the scale and the right finger dos not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance xR from the centre ( 50.00 cm ) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of xR (in cm) is ……

2.

[JEE (Advanced) 2015] Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds v1 and v2 , respectively and always remain in contact with the surfaces. If they reach B and D with the same linear speed and v1 = 3 ms −1, then v2 in ms −1 is g = 10 ms −2

Comprehension 4 Two discs A and B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectively about the common axis. Disc A is imparted an initial angular velocity 2ω using the entire potential energy of a spring compressed by a distance x1. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance x2 . Both the discs rotate in the clockwise direction. 7.

[IIT-JEE 2007] x The ratio 1 is x2 (A) 2 (C)

8.

9.

2

[IIT-JEE 2007] When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is 2 Iω 9 Iω (B) (A) 3t 2t 9 Iω 3 Iω (C) (D) 4t 2t [IIT-JEE 2007] The loss of kinetic energy during the above process is (A)

(

1 2 1 (D) 2

(B)

Iω 2 2

Iω 2 (C) 4

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 196

(B)

Iω 2 3

Iω 2 (D) 6

3.

)

[IIT-JEE 2014] A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rads −1 is

2/9/2021 6:48:06 PM

Chapter 3: Rotational Dynamics

3.197

IO and I P , respectively. Both these axes are perpendicI ular to the plane of the lamina. Find the ratio P to the IO nearest integer.

2R

4.

5.

6.

[JEE (Advanced) 2014] A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 ms −1 with respect to the ground. The rotational speed of the platform in rads −1 after the balls leave the platform is

[JEE (Advanced) 2013] A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rads −1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rads −1) of the system is [IIT-JEE 2012] A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in Figure. The moment of inertia of this lamina about axes passing through O and P is

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 197

O

2R

P

7.

[IIT-JEE 2011] Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N × 10 −4 kgm 2 , find N .

8.

[JEE (Advanced) 2011] A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in Figure.

The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 ms −2 . The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is ⎛ P⎞ ⎜⎝ ⎟⎠ . The value of P is 10

2/9/2021 6:48:13 PM

3.198 JEE Advanced Physics: Mechanics – II

ANSWER KEYS—TEST YOUR CONCEPTS AND PRACTICE EXERCISES Test Your Concepts-I (Based on Moment of Inertia and Applications)

4. (a) 4π rads −1 , (b) 8π rad, (c) 4 5.

ωl cos θ 2

1. 23 kgm 2, 45 kgm 2, 27 kgm 2 , 0

6. (a) 20.2 ms −2 , (b) 0.3 rev

2. 34.6 kgm 2, 23 kgm 2

7. 4 cms −1 , 1.6 cms −2, towards centre C r 8. R+r

3.

ml 2 3

4. (a) 2 ML2 , (b) 5.

π2 6

9. v0t − R sin ( ω 0t ) , R − R cos ( ω 0t )

1 ML2 3

10. v

(

6. m ( a 2 − b 2 ) 2

7. (a) 8 md , (b) 4 m

Test Your Concepts-III (Based on Instantaneous Axis of Rotation, Pure Rolling and Conservation of Energy)

2

8. σπ R3 ( 2 h + R )

1. 2.7 R, 2.5R 5 2. h 7 17 3. mv 2 8

9. 14 mR2 10.

2 ⎛ a5 − b 5 ⎞ M⎜ ⎟ 5 ⎝ a3 − b 3 ⎠

11.

1 ⎛ R 4 − a 4 − 2 a 2b 2 ⎞ M⎜ ⎟⎠ 2 ⎝ R2 − a2

⎛ 1 16 ⎞ 12. MR2 ⎜ − 2 ⎟ ⎝ 2 9π ⎠ 2 a 13. ( m1 + m2 ) 4 R 14. 2

4.

10 g ( R + r ) 17 r 2

5.

8 gR 3

6.

3g 2

7.

2

2 MR 3 11 2 16. mL 12 m 2  , 17. 2 6 15.

)

11. 2iˆ + 3iˆ − 4 ˆj = 5iˆ − 4 ˆj ms −1

4 mg

( 2m + M ) R

3v v 1 , , mv 2 l 2 2 9. 2.5 a 10. 0.6 m 8.

3m 2k 12. 0.745 ms −1 11. Rω 0

Test Your Concepts-II (Based on Rotational Kinematics, Combined Effect of Rotation and Translation Motion) −1

−1

−2

−2

1. (a) 4 rads , 28 rads (b) 12 rads (c) 6 rads , 18 rads   2. a1 = ( a + Rω 2 ) iˆ + ( Rα ) ˆj , a2 = ( a + Rα ) iˆ − ( Rω 2 ) ˆj ,   a3 = ( a − Rω 2 ) iˆ − ( Rα ) ˆj , a4 = ( a − Rα ) iˆ + ( Rω 2 ) ˆj iˆ3. ( 1.2iˆ − 4.4 ˆj ) ms −2

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 198

−2

Test Your Concepts-IV (Based on Torque and Applications) 1. 9 2dm 6F sin ϕ m 3. 2.73 ms −1, 27.3 rads −1, 233.5 N, 238.2 N, 1.47 s 2.

2/9/2021 6:48:30 PM

Chapter 3: Rotational Dynamics

4.

mg 3 3 ( 1 − 3 cos θ )2 , mg sin θ ⎛⎜ cos θ − 1 ⎞⎟ , Yes ⎝2 ⎠ 4 2

2 ( m2 − m1 ) g m ( m + 4 m2 ) , 1 5. ( 2m1 + 2m2 + m ) R m2 ( m + 4m1 )

8.

8.

2m2 g sin θ m1m2 g sin θ , , mgh, mgh, 2m2 + m1 2m2 + m1

3 g sin θ 8 − 3 cos θ 3g 3g mg 10. (a) , (b) , (c) 2 4 4 2 g Mg , 11. 3 3 32g 12. a

2 gh m 1+ 1 2m2

9.

13. (a) 8.16 rads −1, (b) 8 rads −1 2 g 4 g mg , , 14. 5R 5 5 15.

6g 3 3g , 11r 11r

g 3 16. ,  2 3

10.

F 3m + M

11.

mg mv02 2 ( 17 cos θ − 10 ) + mg sin θ , (R − r) 7 7

Test Your Concepts-VI (Based on Angular Momentum and its Conservation)

1. 2.

F 2⎛ F ⎞ − μk g , ⎜ − μk g ⎟ ⎠ M R⎝ M g sin θ Ι 1+ mr 2

3.

4 gt 4 g, 11 11

4.

F F , 2M 2

5. 0.79 Mg 6. (a) 5.3 ms −1, 5.1 ms −1, 4.43 ms −1 (b) 4.2 N , 4.9 N , 7.4 N (c) 1.5 s, 1.6 s, 1.8 s 3 μ MgR 7. R + 2b

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 199

2g sec θ

1.

2. (a)

2v 1 , (b) 9L 9

Iω 0 I + mR2 ω 4. 2 3.

5. 2

g , 2m gr r

2m ⎞ 2m ⎞ 1 ⎛ 2 2⎛ 6. (a) ⎜ 1 + ⎟ ω 0 , (b) mω 0 R ⎜⎝ 1 + ⎟ ⎝ M⎠ 2 M⎠ 7.



Test Your Concepts-V (Based on Uniform and Accelerated Pure Rolling)

4 F cos θ , 3 MF cos θ and MF cos θ 3 M + 8m 3 M + 8m 3 M + 8m

9. 1.4 ms −2

6. (a) Mg tan θ , (b) 0 2g 7. 4 a

3.199

2 gr sec θ 0

8.

4 mω 3

9.

ω0R ω0 , 6 6

⎛ 2m1 ⎞ θ 10. ⎜ ⎝ m2 + 2m1 ⎟⎠ 11.

v1 I ⎞ ⎛ R⎜ 1 + ⎟ ⎝ MR2 ⎠

9mv02 2 13. Linear momentum, angular momentum and kinetic M 2 energy, 12d 2 +  2 12.

14. 15.

7 ga 3J , J > 2m 7 ma 3

ω0r 1+

mr 2 Ι

2/9/2021 6:48:47 PM

3.200 JEE Advanced Physics: Mechanics – II

Test Your Concepts-VII (Based on Rolling with Slipping) 1.

2ω 02 r 2 81μ g

2v0 R ω0R 1 , − mω 02 R2 3. 3μ g 6

6.

5v0 v0 v02 , , 2R μ g 2 μ g

7.

12v02 49 μ k g

8.

3F F F , , 3M + m 3M + m R( 3M + m )

9.

2 ⎛ v0 + Rω 0 ⎞ 1 , ( 5v0 + 2Rω 0 ) 7 ⎜⎝ μ g ⎟⎠ 7

2.

12v02 4. 49 μ k g 5.

10.

R ω0R +6 g g

r 2ω 02 ( μ cos α − sin α )

2 g ( 3 μ cos α − sin α )

2

,

r 2ω 02 ( μ cos α − sin α ) 4 g sin α ( 3 μ cos α − sin α )

Single Correct Choice Type Questions 1. C

2. A

3. C

4. C

5. B

6. B

7. B

8. B

9. B

10. D

11. D

12. B

13. C

14. B

15. D

16. A

17. C

18. D

19. C

20. C

21. D

22. C

23. A

24. B

25. B

26. D

27. C

28. B

29. C

30. B

31. C

32. A

33. B

34. D

35. C

36. D

37. B

38. B

39. D

40. D

41. B

42. D

43. A

44. D

45. B

46. B

47. D

48. D

49. B

50. C

51. A

52. C

53. D

54. C

55. A

56. D

57. A

58. A

59. D

60. A

61. B

62. A

63. A

64. A

65. C

66. D

67. C

68. A

69. B

70. D

71. A

72. B

73. C

74. D

75. D

76. D

77. C

78. A

79. C

80. B

81. C

82. C

83. B

84. B

85. D

86. C

87. A

88. D

89. C

90. A

91. B

92. B

93. D

94. B

95. B

96. B

97. C

98. B

99. C

100. D

101. B

102. B

103. D

104. C

105. B

106. A

107. D

108. D

109. A

110. C

111. C

112. C

113. A

114. C

115. B

116. C

117. B

118. B

119. D

120. B

121. B

122. C

123. D

124. C

125. D

126. A

127. D

128. B

129. B

130. B

131. B

132. A

133. B

134. D

135. B

136. D

137. A

138. B

139. B

140. A

141. C

142. B

143. C

144. A

145. D

146. D

147. C

148. A

149. C

150. B

151. C

152. B

153. D

154. D

155. B

156. D

157. A

158. B

159. B

160. B

161. B

162. D

163. B

164. B

165. B

166. C

167. B

168. B

169. C

170. D

171. B

Multiple Correct Choice Type Questions 1. A, B

2. A, B, C

3. B

4. A, B, D

6. B, D

7. B, C

8. A, B, C

9. A, C

5. A, C, D 10. B, D

11. C, D

12. A, C

13. A, C

14. A, C

15. A, B, C, D

16. C, D

17. A, B, C, D

18. C, D

19. B, C

20. B, D

21. A, C, D

22. B, D

23. B, C, D

24. B, D

25. A, D

26. A, B, C, D

27. A, B

28. A, D

29. A, B, C

30. A, C

31. A, C

32. B, D

33. A, B, C, D

34. A, D

35. A, B, D

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 200

2/9/2021 6:48:52 PM

Chapter 3: Rotational Dynamics

36. A, C

37. A, D

38. A, B

39. B, D

41. A, B, D

42. A, B, D

43. B, D

44. A, B, C

3.201

40. A, C

Reasoning Based Questions 1. A

2. B

3. D

4. B

5. D

11. A

12. A

13. B

14. A

15. A

6. C

7. A

8. A

9. B

10. D

Linked Comprehension Type Questions 1. C

2. B

3. D

4. B

5. C

6. C

7. B

8. B

9. A

10. D

11. D

12. C

13. B

14. C

15. D

16. B

17. D

18. C

19. C

20. C

21. D

22. C

23. B

24. D

25. B

26. C

27. C

28. A

29. D

30. B

31. A

32. C

33. B

34. B

35. A

36. C

37. D

38. C

39. B

40. D

41. B

42. A

43. D

44. A

45. D

46. C

47. A

48. B

49. A

50. C

51. B

52. C

53. C

54. D

55. D

56. B

57. B

58. A

59. C

60. B

61. A

62. C

63. C

64. A

65. C

66. A

67. A

68. B

69. A

70. C

71. A

72. B

73. A

74. C

75. B

76. A

77. C

78. B

79. B

80. A

81. D

82. C

83. B

84. C

85. D

86. C

87. B

88. C

89. A

90. B

Matrix Match/Column Match Type Questions 1. A → (r)

B → (p)

C → (p)

D → (s)

2. A → (q, r)

B → (p)

C → (s)

D → (q)

3. A → (r, s)

B → (p, q)

C → (s)

D → (p, q, r, s)

4. A → (s)

B → (q)

C → (p)

D → (r)

5. A → (q)

B → (p)

C → (s)

D → (r)

6. A → (r)

B → (q)

C → (t)

D → (p)

7. A → (s)

B → (r)

C → (q)

D → (p)

8. A → (p, s)

B → (p, r)

C → (q, r)

D → (p, r)

9. A → (q)

B → (s)

C → (q)

D → (r)

10. A → (p, q, r)

B → (p, q, r)

C → (p, q, r)

D → (s)

11. A → (r)

B → (p)

C → (p)

D → (q)

12. A → (p)

B → (s)

C → (p)

D → (r)

13. A → (q)

B → (r)

C → (p)

D → (r)

14. A → (t)

B → (s)

C → (p)

D → (p)

15. A → (q)

B → (s)

C → (q)

D → (p)

16. A → (p, r, s)

B → (q, r, s)

C → (q, r, s)

D → (p, r, s)

17. A → (q)

B → (s)

C → (p)

D → (s)

18. A → (p, r)

B → (p, r)

C → (q, s)

D → (p, s)

19. A → (s)

B → (p)

C → (p)

D → (q, r)

20. A → (q)

B → (r)

C → (p)

D → (s)

21. A → (q)

B → (s)

C → (s)

D → (p)

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 201

2/9/2021 6:48:52 PM

3.202 JEE Advanced Physics: Mechanics – II

Integer/Numerical Answer Type Questions 1. 5 6. 20, 8

2. 3

3. 75

7. 15

8. 2.25

4. 40

5. 9

9. 5

10. 1

11. 20

12. (a) 36, (b) 6

13. (a) 4, (b) 5, (c) 50

14. 9

15. (a) 1, (b) 5, (c) 1

16. 1.5

17. 100, 3, 30, 1

18. (a) 4, 40 (b) 500, 180 (c) 1

19. 20

20. 40

21. 3

22. 11

23. 4

24. 4, 8

25. 2

26. 25

27. (a) 4, (b) 90

28. 8, 8, 4

29. (a) 1.37, (b) 131, (c) 206

30. 20

31. 251.43

32. 0.7

33. 195

34. (a) 14.3, (b) 40, 181 (c) 141

35. 6

ARCHIVE: JEE MAIN 1. D

2. D

3. D

4. B

5. B

6. B

7. B

8. C

9. A

10. B

11. D

12. C

13. B

14. A

15. A

16. A

17. B

18. D

19. D

20. D

21. A

22. Bonus

23. C

24. B

25. D

26. C

27. B

28. D

29. C

30. A

31. D

32. B

33. D

34. B

35. C

36. D

37. B

38. A

39. A

40. B

41. A

42. C

43. C

44. D

45. D

46. A

47. A

48. D

49. D

50. D

51. A

52. B

53. A

54. A

55. A

56. A

57. D

58. A, C

59. B

60. B

61. A

62. C

63. C

64. C

65. C

66. C

67. B

68. C

69. B

70. C

71. D

72. D

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1. A

2. D

3. C

4. D

5. A

6. B

7. B

8. D

9. A

10. D

11. A

12. D

13. B

14. A

15. A

16. A

17. C

18. B

19. A

20. D

21. C

22. B

23. C

24. B

25. A

26. A

27. B

28. C

29. A

30. C

Multiple Correct Choice Type Problems 1. A, C, B

2. B, C, D

3. A, C, D

4. A, C

6. A, C

7. B, C

8. C, D

9. A, B, C

11. A, C, D

5. A, B 10. A, B, C

12. B, D

Reasoning Based Questions 1. D

Linked Comprehension Type Questions 1. A

2. C

3. C

4. B

5. D

6. A

7. C

8. A

6. 3

7. 9

8. 4

9. B

Integer/Numerical Answer Type Questions 1. 25, 6

2. 7

Mechanics II_Chapter 3_Part 6 Exercises_2.indd 202

3. 2

4. 4

5. 8

2/9/2021 6:48:52 PM

CHAPTER

4

Gravitation and Satellites

Learning Objectives After reading this chapter, you will be able to: After reading this chapter, you will be able to understand concepts and problems based on: (a) Newton’s Laws of Gravitation (h) Relation Between Gravitational Field and (b) Principle of Superposition Gravitational Potential (c) Gravitational Field Strength (i) Conservation Laws for Gravitational Systems (d) Acceleration Due to Gravity and its Variation ( j) Escape Speed (e) Gravitational Potential Energy (k) Satellites and Kepler’s Laws (f) Gravitational Self Energy (l) Double Star System or Binary Star System (g) Gravitational Potential All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main and Advanced) are also given.

NEWTON’S LAW OF GRAVITATION The force of gravitation between any two particles having masses m1 and m2 separated by a distance r is attractive and acts along the line joining the particles. The magnitude of the force is given by F=

Gm1 m2

…(1)

r2

where G is a universal constant called Gravitational Constant having the same value for all pairs of particles. Its numerical value is 6.67 × 10 −11 Nm 2 kg −2 . m1

m2 F12

F21

The gravitational forces between two particles always form an action-reaction pair. The first particle exerts

Mechanics II_Chapter 4_Part 1.indd 1

a force on the second particle that is directed toward the first particle along the line joining the two. Similarly, the second particle exerts a force on the first particle that is directed towards the second particle along the line joining the two. These forces are equal in magnitude but oppositely directed. In the vector form, equation (1) is written as r Gm1 m2 F=− rˆ …(2) r2 where we must remember that rˆ has its origin at the source of the force. It is important to realize that Newton’s Law of Gravitation is stated only for point particles. For two arbitrary bodies, as shown in figure, there is no unique value for the separation r . So, the equation Gm1 m2 F= cannot be used for arbitrary bodies. r2

09-Feb-21 6:27:12 PM

4.2

JEE Advanced Physics: Mechanics – II r=?

SOLUTION

Since, m1 = m2 = ( volume ) ( density )

To compute the force between them requires integral calculus. However, for the special case of a uniform spherical mass distribution, r may be taken as the distance to the center. Also, when the separation between two objects is very much larger than their sizes, they may be approximated as point masses and then, equation may be used.

Problem Solving Technique(s) (a) The gravitational force between two particles is independent of the presence of other bodies or the properties of the intervening medium. (b) Gravitational force is conservative force, therefore work done in displacing a body from one place to another is independent of path. It depends only on initial and final positions. (c) The mass of air bubble in material medium is negative. (d) The gravitational force obeys Newton’s Third Law i.e.   F12 = − F21 ⇒

m1a1 = m2 a2

⇒

⎛4 ⎞ m1 = m2 = ⎜ π r 3 ⎟ ρ ⎝3 ⎠

⇒

F=

⇒ ⇒

Gm1 m2

r2 ⎛4 ⎞⎛ 4 ⎞ G ⎜ π r 3 ⎟ ⎜ π r 3 ⎟ ρ2 ⎝3 ⎠⎝ 3 ⎠ F= 2 r

2r

F ∝ r4

ILLUSTRATION 2

The figure shows a uniform rod of length l whose mass per unit length is λ and the total mass is M . What is the gravitational force of the rod on a particle of mass m0 located a distance d from one end of the rod, as shown in the figure? d m0

SOLUTION

Consider an infinitesimal element on the rod having length dx at a distance x from the point mass m0 . If dF is the force between the infinitesimal element of mass dm and m0 , then dF =

(e) Gravitational force is a central force.

Gm0 dm x2

, where dm = λ dx dx

m0

PROPERTIES OF G (a) Its value does not depend upon place, time and masses of the bodies and hence it is called a Universal Constant. (b) The value of G is 6.67 × 10 −11 Nm 2 kg −2 (c) Its dimensional formula is M −1 L3 T −2 . (d) Its value is extremely small. Hence, we do not experience it in our daily life. (e) G is numerically equal to gravitational force of potential between two bodies each of mass 1 kg placed at a distance of 1 m.

x

To obtain the total force on m0 due to the rod, we integrate the above expression for force within appropriate limits. So, d+l

F=

∫ d

⇒

F=

G ( λ dx ) m0 x

2

1 ⎞ Gm0 ( λ l ) ⎛1 = Gλ m0 ⎜ − = ⎝ d d + l ⎟⎠ d(d + l)

Gm0 M d(d + l)

ILLUSTRATION 1

PRINCIPLE OF SUPERPOSITION

Spheres of the same material and same radius r are touching each other. Show that gravitational force between them is directly proportional to r 4 .

Experiments show that when several particles interact, the force between a given pair is independent of the other particles present.

Mechanics II_Chapter 4_Part 1.indd 2

09-Feb-21 6:27:15 PM

Chapter 4: Gravitation and Satellites

The net force F1 on the point mass m1 due to the other particles as shown in the figure is found by first calculating its potential with each of the other particles one at a time. The net force on m1 is the vector sum of the pair-wise potentials. r r r r F1 = F12 + F13 + ...... + F1n …(3)

m1

F23

m1

F23 =

r3

F14

Gm2 m1

L2 Gm2 m3 2

L

m3

= 1.33 × 10 −10 N = 1.01 × 10 −10 N

r r r The net force on m2 is F2 = F21 + F23 Its components are

F13

F12

F2 x = − F21 cos 60° + F23 cos 60° = −1.6 × 10 −11 N F2 y = − F21 sin 60° − F23 sin 60° = −2.03 × 10 −10 N

r2

(

m2

)

iˆ F2 = − 1.6iˆ + 20.3 ˆj × 10 −11 N Thus,

The net force on m1 is r r r r F1 = F12 + F13 + F14

ILLUSTRATION 4

Problem Solving Technique(s) 1. Set up a convenient coordinate system 2. Indicate the directions of the forces acting on the particle under consideration. 3. Calculate the (scalar) magnitudes of the forces. 4. Find the net force by using the component method. ILLUSTRATION 3

Three point particles with masses m1 = 4 kg , m2 = 2 kg and m3 = 3 kg are at the corners of an equilateral triangle of side L = 2 m , as shown in figure. Find the net force on m2 . m2

y

F21 x

m3

r4

m2

y

F21 =

m4

4.3

Three particles each of mass m are placed at the three corners of an equilateral triangle of side a . Find the force exerted by this system on another particle of mass m placed at (a) the centre of the triangle and (b) mid-point of a side. SOLUTION

To solve the above problem, we apply the gravitational potential which follow the principle of superposition. (a) When another mass m is placed at O , it exper r r riences three forces FA , FB and FC . Since AO , r r r BO and CO are equal hence FA = FB = FC . Angle between any two forces is same i.e. 120° . Therefore, the resultant force exerted by the system on particle at O is zero. y

x

A m1

m3

FA

SOLUTION

The first two steps have already been done in the figure. The magnitudes of forces are

Mechanics II_Chapter 4_Part 1.indd 3

O FB B

FC D

C

x

09-Feb-21 6:27:19 PM

4.4

JEE Advanced Physics: Mechanics – II

(b) In this case the particle is placed at point D , which is equidistant from B and C . r r FB = FC But they are opposite in direction. Therefore, the effective force at D will be due to mass m at A. 3a By geometry of the figure AO = a sin 60 = 2 4Gm2 Therefore, FA = along DA 3 a2

GRAVITATIONAL FIELD STRENGTH (Eg) Every mass particle is surrounded by a space within which its influence can be felt. This region or space is said to be occupied with gravitational field. Each point in the field is associated with a (vector) force which is experienced by a unit mass placed at that point and is called the gravitational field strength. The region of space around a body in which its gravitational influence is experienced by other bodies is the gravitational field of that body. OR

( )

OR The gravitational field is also defined as the force ⎛ F ⎞ per unit test mass ⎜ placed in the gravitational ⎝ m0 ⎟⎠ influence of source mass. Mathematically,

Problem Solving Technique(s) If a system has got a number of masses, then the total gravitational field at a point P due to all the masses is found in accordance with the Principle of Superposition. So,      E g = E g1 + E g2 + E g3 + … + E gn i.e. net field at a point P is the vector sum of the fields due to individual masses (each calculated as if others were absent) at the point of consideration. ILLUSTRATION 5

Find the field strength at a point along the axis of a thin rod of length L and mass M , at a distance d from one end.

First we need to find the field due to an element of length dx . The rod must be thin if we are to assume that all points of the element are at the same distance from the field point. The mass of the element is ⎛ M⎞ dm = ⎜ dx , so contribution to the field is ⎝ L ⎟⎠ dg =

G ( λ dx ) x2

=

GM dx L x2 d

L

F Gm = 2 m0 r

dx

where m = Source Mass, m0 = Test Mass.

Intensity of gravitational field

GM g= L

L+ d

dx

∫x

2

d

Notice that when d for a point particle. (Distance)

dg

x

The total field strength is

Eg

Mechanics II_Chapter 4_Part 1.indd 4

)

SOLUTION

The gravitational force of attraction acting on a body of unit mass at any point in the gravitational field is r defined as the intensity of gravitational field Eg at that point.

Eg =

(

Its SI unit is newton/kg Nkg −1 and its dimensional formula is LT −2 . It is a vector quantity and its direction is always towards the centre of the body producing the field. The value of Eg is zero as r → ∞ . The graph between Eg and r .

=

1 ⎞ GM ⎛ 1 GM ⎜ − ⎟= L ⎝ d L + d ⎠ d(L + d) L , we find g →

GM d2

, the result

r

09-Feb-21 6:27:22 PM

Chapter 4: Gravitation and Satellites

Problem Solving Technique(s) (a) For calculating the gravitational field at a point P at distance r due to a source mass m we proceed as follows. STEP-1: Place a test mass of 1 kg at point P. STEP-2: Calculate the force between m and 1 kg. STEP-3: This force is the value of Eg due to m at the point P. STEP-4: The direction of force experienced by the 1 kg mass is the direction of Eg due to m at the point P. (b) If we are to find the gravitational field due to assembly of masses m1, m2, …, mn at a point P, then we apply Superposition Principle and then     E p = E1 + E 2 + … + E n    where we find E 1 , E 2 , … E n at point P by the technique mentioned above. (c) If we are to find the gravitational field at a point P due to a uniform mass distribution, then we calculate the field due to an infinitesimal element of the distribution at the point P and then integrate it within appropriate limits. r

dm

∫r

dm 2

For mass distributed on a wire with linear mass density λ , Eg = G

λd

∫r

ILLUSTRATION 6

Two identical uniform rods of length l and mass M are placed along the same line so that their closer ends are a distance d apart, as shown in the figure.

2

For mass distributed on a surface with surface mass density σ ,

∫ A

Mechanics II_Chapter 4_Part 1.indd 5

σ dA r2

d

ROD 1

ROD 2

(a) Find the gravitational force of attraction between the rods. M2 (b) Show that lim F = G 2 . d →∞ d SOLUTION

E=

GM x(x + l)

A

B

C

ROD 1

P

Q

D

ROD 2 x

dx

Now consider an infinitesimal element of mass dm at a distance x from end B of rod 1. Then force on this element is ⎛M ⎞ dx ⎟ GM ⎜ ⎠ dF = ( dm ) E = ⎝ l x( x + l )



Eg = G

ρdV r2

If the mass distributions are uniform, then mass densities can be taken out of the integral to get desired results.

(d) Volume mass density, ρ

(c) Surface mass density, σ

∫ V

P

P dv

∫

Eg = G

d (b) Linear mass density, λ

(a) Continuous distribution r dA

E = dE = G

For mass distributed on a volume with volume mass density ρ ,

(a) The gravitational field due to thin rod at a distance x from one of its end is

P

P

r

4.5

⇒ dF =

GM 2 dx GM 2 = 2 l x( x + l ) l

d+l

⇒ F=

∫ d

1 ⎤ ⎡1 ⎢⎣ x − x + l ⎥⎦ dx d+l

GM 2 ⎡ ⎛ x ⎞⎤ dF = 2 ⎢ log e ⎜ ⎝ x + l ⎟⎠ ⎥⎦ d l ⎣

09-Feb-21 6:27:25 PM

4.6

JEE Advanced Physics: Mechanics – II

⇒ F=

GM 2 l2

FIELD DUE TO SPHERE AND SHELL

⎡ ( d + l )2 ⎤ log e ⎢ ⎥ ⎣ d ( d + 2l ) ⎦

For an external point ( r > R ) , a sphere (solid or hollow) behaves as if whole of its mass is concentrated at its centre, so the gravitational field outside the sphere/shell is given by

(b) The above expression can be written as, 2 ⎡⎛ l⎞ ⎢ ⎜⎝ 1 + ⎟⎠ GM d F = 2 log e ⎢ 2l ⎞ ⎢⎛ l ⎢⎣ ⎜⎝ 1 + d ⎟⎠ 2

⇒ F=

GM 2 l2

⎤ ⎥ ⎥ ⎥ ⎥⎦

Eoutside =

{for r > R } r2 In case of a spherical shell, for an internal point, ( r < R ) the gravitational field inside is zero. So,

⎡⎛ l 2 2l ⎞ ⎛ 2l ⎞ ⎤ log e ⎢ ⎜ 1 + 2 + ⎟ ⎜ 1 − ⎟ ⎥ d ⎠⎝ d⎠⎦ d ⎣⎝

Einside = 0

⎛ GM l2 ⎞ ⇒ F = 2 log e ⎜ 1 + 2 ⎟ ⎝ l d ⎠ 2

{

⇒ F≈ ⇒ F≈

neglecting the higher terms of

GM 2 ⎛ l 2 ⎞ ⎜ ⎟ l2 ⎝ d2 ⎠

GM

l d

}

⎧ ⎛ l2 ⎞ l2 ⎫ ⎨ log e ⎜ 1 + 2 ⎟ ≈ 2 ⎬ ⎝ d ⎠ d ⎭ ⎩

In case of a spherical volume distribution of mass (i.e., a solid sphere) for an internal point ( r < R ) , the portion of the sphere that lies outside the radius r will not contribute to the field (because the field inside a spherical shell is zero), so Einside =

GM 2

GM ′ r2

d2

GRAVITATIONAL FIELD LINES A gravitational field line (or a field line) is a line straight or curved such that a unit mass placed in the field of another mass would always move along this line. Field lines for an isolated mass m are radially inwards. If two masses are placed close to each other then field lines are shown in figure. Two field lines never cross each other. The direction of the field can just be found by drawing a tangent to the field line at the point of consideration.

Field inside a solid sphere

where M ′ = mass of sphere of radius r . Now if M is the mass of solid sphere, then

ρ=

M M = V ⎛ 4⎞ 3 ⎜⎝ ⎟⎠ π R 3

and M ′ =

m

The Gravitational field lines (for two equal masses).

Mechanics II_Chapter 4_Part 1.indd 6

⇒

4 3 M π r ρ = 3 r3 3 R

Einside =

G ⎛ M 3 ⎞ GM r ⎟= 3 r ⎜ r 2 ⎝ R3 ⎠ R

i.e., intensity inside a solid sphere varies linearly with distance from the centre. So, it is minimum ( Ecentre = 0 ) at the centre and maximum at the GM ⎞ ⎛ surface ⎜ Esurface = 2 ⎟ . This is shown graphically ⎝ R ⎠ in the figure.

09-Feb-21 6:27:28 PM

4.7

Chapter 4: Gravitation and Satellites Eg

M2

Eg Spherical shell

Solid sphere GM R2 Eg∝r

GM R2 Eg ∝ 1 r2

Eg = 0

M1 P3

r (a)

(b)

Problem Solving Technique(s) Gravitational Field For a Shell ⎡ GM ⎢ 2 r Eg = ⎢ ⎢ ⎢0 ⎢⎣

r ≥R

( outside and at surface ) r R ρ0 is a constant. A test mass m0 can undergo circular motion under the influence of the gravitational field of particles. Calculate its speed v as a function of distance r from the centre of the system. Also draw a plot of speed versus distance r . SOLUTION

For r ≤ R ,

m0 v 2 Gm0 M = r r2

…(1)

⎛4 ⎞ where, M = ⎜ π r 3 ⎟ ρ0 ⎝3 ⎠ Substituting in Equation (1), we get v2 = r

⎛4 ⎞ G ⎜ π r 3 ⎟ ρ0 ⎠ ⎝3 2 r

09-Feb-21 6:27:31 PM

4.8

JEE Advanced Physics: Mechanics – II

⇒

v=

⇒

v∝r

If, E1 be the field due to hole and E2 be the field due to the remaining portion, then we have

4 4 π G ρ0 r 2 = r π G ρ0 3 3

E = E1 + E2

i.e. v -r graph is a straight line passing through origin. For r > R , we have M = Mtotal

⇒

m0 v 2 = r v=

⇒

v∝

4 = π R3 ρ0 3

⎛4 ⎞ Gm0 ⎜ π R3 ⎟ ρ0 ⎝3 ⎠ 2 r

⇒

E2 = E − E1

⇒

E2 =

GM x

where, M =

4π Gρ0 R3 3r

m=

2

−

Gm R⎞ ⎛ ⎜⎝ x − ⎟⎠ 2

2

…(1)

4 π R3 ρ0 and 3 3

4 ⎛ R⎞ π ⎜ ⎟ ρ0 3 ⎝ 2⎠

Substituting the values in equation (1), we get

1 r

The corresponding v -r graph is shown in Figure.

⎛ π Gρ0 R3 E2 = − ⎜ ⎝ 6

⎞⎡ 8 ⎤ 1 − 2⎥ ⎟⎠ ⎢ 2 x ⎥ ⎢ ⎛⎜ x − R ⎞⎟ ⎝ ⎠ ⎢⎣ ⎥⎦ 2

Problem Solving Technique(s)

ILLUSTRATION 9

A sphere of radius R has its centre at the origin. It has uniform mass density ρ0 except that there is a spheri1 1 cal hole of radius r = R whose centre is at x = R , 2 2 as shown in figure. Find the gravitational field at points on the x-axis for x > R . y

r R

x

SOLUTION

Let, E be the gravitational field at x due to the complete sphere.

Mechanics II_Chapter 4_Part 1.indd 8

For symmetrical mass distributions, if we are asked to calculate the gravitational field at a point which lies symmetrically with respect to the mass distribution, then we proceed as follows STEP-1: Consider an infinitesimal element having a mass dm (say) that lies at a distance r from the point P. If dE is the gravitational field due to this element then Gdm dE = 2 r STEP-2: Now consider another identical mirror infinitesimal element having same mass and located at the same distance from P. Then field due to this element will also be dE. STEP-3: On resolving dE due to both the elements along suitable axis we observe that one set of components cancels and the net gravitational field will then be calculated by taking the integral of the component of the gravitational field due to the contribution of the single element. Enet =

∫ ( contribution due to a single element )

09-Feb-21 6:27:35 PM

Chapter 4: Gravitation and Satellites

However, in the case of unsymmetrical mass distributions, if we have to calculate the gravitational field at any point P, then we proceed as follows: STEP-1: Consider an infinitesimal element having a mass dm (say) that lies at a distance r from the point P. If dE is the gravitational field due to this element then Gdm dE = 2 r STEP-2: Resolve dE along the selected axes (say x and y) so as to get the components dEx and dEy. STEP-3: Integrate dEx and dEy separately to get Ex and Ey.

GRAVITATIONAL FIELD AT THE AXIS OF A CIRCULAR UNIFORM RING Consider a ring of mass M , radius a and uniform linear mass density λ . Then M 2π a

Ex =

(a

…(1)

Q

a dE cosθ x

dE sin θ dE θ θ dE

y

⇒

x

Mechanics II_Chapter 4_Part 1.indd 9

)

0

2

+ x2 )

( 2π a )

(a

2

+ x2 )

32

(along +x axis)

Remark(s)

( x 2 + a2 )3 2 ≅ x 3 . GM

⇒ Ex 

x2

This result under the specified condition just matches with the field due to a point mass at a distance x from it. dE =0 (b) For this field to be a MAXIMUM dx ⇒

d ⎡ ( 2 2 −3 2 ⎤ ⎣x x +a ) ⎦=0 dx

⇒

( x 2 + a2 )− 3 2 + x ⎛⎜ − 3 ⎞⎟ ( x 2 + a2 )−5 2 ( 2 x ) = 0 ⎝ 2⎠

3x 2 x 2 + a2

⇒ x 2 + a2 = 3x 2 ⇒ x=±

Let us calculate the gravitational field due to this ring at a point lying on the axis of the ring at a distance x from the centre of the ring. Since this point is located symmetrically with respect to the ring so let us consider two elements of length dl placed symmetrically on the two diametrically opposite ends. If dE is the field due to each such element, then dE sin θ components will cancel out such that the net field is just due to dE cos θ components summed up over the entire ring. So, Gdm Enet = Ex = dE cos θ = cos θ 2 2 2 a +x

∫(

Gλ cos θ

dl = )∫ (a

GMx

Ex = Eaxis =

⇒ 1=

dE sin θ

∫

+x

2

But from equation (1), we have λ ( 2π a ) = M and also, x we have from the figure, cos θ = , so we get 2 a + x2

dm a2 + x2

2

2π a

(a) For the point P to lie at far off distance from the centre of the ring i.e., for x >> a we have

STEP-4:   Then E = E x iˆ + E y ˆj such that E = E = E 2x + E 2y and  Ey if E makes an angle β with x-axis, then tan β = . Ex

λ=

⇒

Gλ cos θ

4.9

a 2

i.e., the field will attain a maximum value at a x=± along the axis of the ring and the maxi2 mum value equals Emax given by Emax =

⇒ Emax =

GM ( a / 2 ) ⎛ 2 a2 ⎞ ⎜a + ⎟ ⎝ 2 ⎠

32

2GM 3 3a2

09-Feb-21 6:27:39 PM

4.10

JEE Advanced Physics: Mechanics – II

ILLUSTRATION 10

Calculate the gravitational field due to a uniform rod AB at a point P at perpendicular distance a from the rod as shown in figure. Assume that the rod has a linear mass density λ .

Also, we observe that x tan θ = a ⇒ x = a tan θ ⇒

dx = a sec 2 θ dθ

⇒

Ex = Gλ

⇒

Gλ Ex = a

A

a

N

α β

P

⇒ B

Here it is very important to note the unsymmetrical placement of the point P and hence we must calculate the components of the gravitational fields Ex and Ey separately. For this, let us consider an infinitesimal element of length dx at a distance x as shown. A

y

D C

r

dE a

dE sin θ dθ

θ dE cosθ

The net gravitational field at point P due to this infinitesimal element is dE . This dE is resolved into components. (a) dEx = dE cos θ (b) dEy = dE sin θ The net field E can be calculated by finding Ex and Ey from the above expressions. So G ( λ dx )

∫ dE cosθ , where dE = r dx cos θ dx cos θ = Gλ ∫ r = Gλ ∫ ( a + x )

Ex =

Mechanics II_Chapter 4_Part 1.indd 10

∫ cosθ dθ

−β

Gλ ( sin β + sin α ) a

2

2

∫ dE = ∫ dE sin θ dx sin θ = Gλ ∫ (a + x )

⇒

Ey

2

2

y

2

2

Since x = a tan θ ⇒

dx = a sec 2 θ dθ

⇒

Ey = Gλ

⇒

Gλ Ey = a

P

B

Ex

α

Ey =

x

r

⇒

a 2 sec 2 θ

Similarly, let us calculate the value of Ey , given by

SOLUTION

dx

Ex =

∫

a sec 2 θ cos θ dθ

∫

a sec 2 θ sin θ dθ a 2 sec 2 θ

α

∫ sin θ dθ

−β

(

)

⇒

Ey =

Gλ − cos θ a

⇒

Ey =

Gλ ( cos β − cos α ) a

α −β

So, the gravitational field due to a rod of length having uniform mass density λ at a point P , that subtends an angle α at one end and β at the other is given by Ex =

Gλ ( sin α + sin β ) and a

Ey =

Gλ ( cos β − cos α ) a

ILLUSTRATION 11

Mass M is distributed uniformly along a line of length 2L . A particle of mass m0 is at a point that is a distance a above the centre of the line on its perpendicular bisector (point P in figure).

09-Feb-21 6:27:46 PM

Chapter 4: Gravitation and Satellites ILLUSTRATION 12

P

Calculate the gravitational field at the centre of a uniform wire in the form of an arc of radius r . The wire subtends an angle ϕ at the centre.

a L

L

SOLUTION

M

For the gravitational force that the line exerts on the particle, calculate the components perpendicular and parallel to the line. What happens when a L ?

Consider an element dm ( = λ dx ) of the wire that makes an angle θ with y-axis and subtends an angle dθ at the centre. Then dm = λ ( rdθ )

SOLUTION

Since, dE =

Consider an infinitesimal element of mass dm , as shown in figure.

⇒

dE =

m

Gdm r

2

λ ( rdθ ) r2

=

Gλ dθ r y

θ F

= dq

a

Consider another identical mirror element of the already taken infinitesimal element, then we observe that the components dF cos θ (i.e., parallel to line) due to these components cancel each other and the net force will be perpendicular to the rod.

⇒

Fnet =

⎛ contribution due to ⎞ ⎜⎝ a single element ⎟⎠ = Gm0 Mdx

∫ 2L ( a

2

−L

Fnet

GMm0 a = 2L

Fnet =

a

+ x 2 ) a2 + x 2

L

∫ (a

−L

GMm0 a L2 + a 2

Mechanics II_Chapter 4_Part 1.indd 11

=λ

dx

r dE dθ

ϕ dE sin θ

x

Again, take another element which is the mirror image of the element already taken. On resolution, we observe that the x components cancel, while the net field is equal to integral of contribution due to a single element. Hence

( a2 + x 2 )

L

θ θ

dE sin θ

Gm0 dM

∫

dE cos θ

dx

⎛ M⎞ dx Then dm = ⎜ ⎝ 2L ⎟⎠

Fnet =

dq

x

λd

dE dθ

θ x

dF =

4.11

dx 2

+ x2 )

32

x=L

∫

dF sin θ

E = Ey = ⇒

E=

∫ dE cosθ +

⇒

E=

x= −L

∫ dE cosθ

Gλ r

ϕ 2

∫ cosθ dθ

−

ϕ 2

ϕ 2

⇒

Gλ E= sin θ r

⇒

E=

Gλ ⎡ ⎛ ϕ⎞⎤ ⎛ϕ⎞ sin ⎜ ⎟ − sin ⎜ − ⎟ ⎥ ⎢ ⎝ 2⎠ ⎦ ⎝ ⎠ 2 r ⎣

⇒

E=

Gλ ⎛ϕ⎞ 2 sin ⎜ ⎟ ⎝ 2⎠ r

⇒

E=

2Gλ ⎛ϕ⎞ sin ⎜ ⎟ ⎝ 2⎠ r

−

ϕ 2

09-Feb-21 6:27:49 PM

4.12

JEE Advanced Physics: Mechanics – II

Problem Solving Technique(s) (a) For semi-circle, ϕ = π 2Gλ So, E = r (b) For Quarter-circle ϕ = ⇒ E=

2Gλ

⇒

dE =

Gσ ( 2π ydy ) x

( x2 + y2 )

32

Please note that, here once point P is taken, then value of x remains fixed. So,

π 2

R

E = Eaxis =

∫ dE 0

r 2 (c) For circle, ϕ = 2π

R

⇒

⇒ E=0

E=

∫ dE = 2Gσπ x∫ ( x 0

R

ILLUSTRATION 13

⇒

Calculate the gravitational field at point P that lies on the axis of a uniformly disc at a distance x from the centre. Assume the disc to have a radius R and uniform surface mass density σ . SOLUTION

To find gravitational field at point P due to this disc let us consider an elemental ring of radius y and width dy . If dm and dA be the mass and area of this infinitesimal element, then dm = σ dA = σ ( 2π ydy ) {Q dA = 2π y dy}

E = Gσπ x

∫ (x 0

⇒

⇒

2 ydy 2

+ y2

)

ydy 2

+ y2

)

32

32

⎡ 2 E = Gσπ x ⎢ − 2 ⎢ x + y2 ⎣

y=R

y =0

⎤ ⎥ ⎥ ⎦

⎞ ⎛ x E = 2Gσπ ⎜ 1 − ⎟ ⎝ x 2 + R2 ⎠

Since, σ = E=

M

π R2

, so the above expression becomes

⎞ 2GM ⎛ x 1− ⎟ 2 ⎜ 2 2 R ⎝ x +R ⎠

If we assume that the complete disc subtends an apex angle α at P , as shown in the figure, then R

σ x

Now we know that gravitational field strength due to a ring of radius R , mass M , at a distance x from its centre on its axis can be given as Eaxis =

GMx

( x2 + R )

2 32

(Derived earlier)

So, due to the infinitesimal elemental ring the gravitational field strength dE at point P is dE =

G ( dm ) x

( x2 + y2 )

Mechanics II_Chapter 4_Part 1.indd 12

32

cos α =

E

α

P

x 2

R + x2

So, the gravitational field can also be expressed as E=

⎞ 2GM 2GM ⎛ x ( 1 − cos α ) 1− ⎟= 2 ⎜ R ⎝ R2 R2 + x 2 ⎠

ILLUSTRATION 14

A spherical body of radius R is made of material having constant density ρ . The body is in equilibrium under its own gravity. If P ( r ) is the pressure

09-Feb-21 6:27:54 PM

Chapter 4: Gravitation and Satellites

at a distance r from the centre of the body inside it, then calculate P ( r ) . SOLUTION

Gravitational field at a distance r due to mass ⎛ 4 ⎞ m ⎜ = π r 3 ρ ⎟ is ⎝ 3 ⎠ ⎛4 ⎞ Gρ ⎜ π r 3 ⎟ ⎝3 ⎠ 4Gρπ r E= = 2 3 r Consider a small element of width dr and area ΔA at a distance r from the centre. Pressure force on this element is due to the gravitational force on dm from m inwards towards the centre.

4.13

This acceleration is called acceleration due to gravity. Its magnitude g is independent of the mass, size, shape and composition of the body. It is directed radially inward to the centre of the earth. If in Newton’s law of gravitation one body (say earth) is taken as ‘reference body’, the force with which the reference body attracts any other body towards its centre is called force due to gravity (of reference body) and the phenomenon ‘gravity’. Now if the body is free to move, force of gravity (of reference body) in accordance with Newton’s Second law will produce an acceleration in it. The acceleration produced in a body by the force of gravity of reference body (usually earth) is called acceleration due to gravity and is represented by g . If the reference body has mass M and radius R, the force on a body of mass m at the surface of reference body by Newton’s law of gravitation will be ⎛ GMm ⎞ F=⎜ ⎝ R2 ⎟⎠ and if g is acceleration due to gravity of reference body on its surface, by Newton’s Second law,

⇒

( dP ) ΔA = E ( dm )

⎛4 ⎞ where dm = ( ΔA ) ( dr ) ρ and m = ⎜ π r 3 ⎟ ρ ⎝3 ⎠ ⇒

⎛4 ⎞ − dP ΔA = ⎜ Gπρr ⎟ ( ρΔAdr ) ⎝3 ⎠ P

⇒

r

∫

− dP = 0

∫ R

⇒

−P =

⇒

P=

⎛ 4Gρ 2π ⎞ ⎜⎝ ⎟ rdr 3 ⎠

4Gρ 2π ( 2 r − R2 ) 3×2

2Gρ 2π ( 2 R − r2 ) 3

ACCELERATION DUE TO GRAVITY (g) Earth attracts all bodies towards its centre. This property of the earth is called ‘gravity’ and the force with which it attracts a body is called the ‘force of gravity’ acting on that body. Thus, when a body falls freely r towards the earth’s surface, the force of gravity F r produces an acceleration g in it given by r r F g= m

Mechanics II_Chapter 4_Part 1.indd 13

F = mg ⇒

g=

F GM = m R2

This is the relation between g and G showing that g depends on reference body. Also, from this relation we observe that actually the acceleration due to gravity is also the measure of the gravitational field of the earth So, F GM = m R2 It is independent of mass, shape, size, etc., of ‘falling body’, i.e., a given reference body produces same acceleration in a light and a heavy falling body. g = Eg =

VARIATION IN g With Altitude As for an external point a spherical distribution of mass behaves as if the whole of its mass were concen⎛ GM ⎞ trated at the centre, i.e., g = Eg = ⎜ 2 ⎟ ⎝ r ⎠ So, at the surface of earth

09-Feb-21 6:27:59 PM

4.14

JEE Advanced Physics: Mechanics – II

g=

Z

GM R

2

N

and for a height h above the surface of earth ⎡ GM ⎤ gh = ⎢ 2 ⎥ ⎣ (R+ h) ⎦ ⇒

C

{ as r = R + h }

−2

⇒

h⎞ ⎛ gh = g ⎜ 1 + ⎟ ⎝ R⎠

⇒

2h ⎞ ⎛ gh = g ⎜ 1 − ⎟ ⎝ R⎠

O

⎛ GM ⎞ g = Eg = ⎜ 3 ⎟ r ⎝ R ⎠

Z′

Now rω 2

{Q r = R − d }

R gd ⎛ R − d ⎞ =⎜ ⎟ g ⎝ R ⎠

⇒

g ′ ≈ g − Rω 2 cos 2 ϕ

So, with increase in depth below the surface of earth g decreases and at the centre of earth it becomes zero.

r = R cos ϕ }

Note that the vector g ′ is not exactly towards the center of earth. Therefore, we note that the decrease in g is maximum at equator and zero at poles. As we go from equator to the pole, value of g increases by Δg = ( g p − g e ) = Rω 2 Δg Rω 2 = g g

1 291

ILLUSTRATION 15

Find the height at which the gravitational field of the earth becomes one fourth the field at the surface.

The acceleration due to gravity is the measure of the gravitational field. Let the gravitational field be one fourth the field at the surface at a height h above the earth’s surface. So, we have gh =

Due to Rotation of Earth The earth rotates from west to east on its axis, due to which every object on the surface of earth experiences a centrifugal force in the reference frame of the earth. The effective value of acceleration due to gravity at a place of latitude ϕ is given by

Mechanics II_Chapter 4_Part 1.indd 14

{Q

SOLUTION

d⎞ ⎛ gd = g ⎜ 1 − ⎟ ⎝ R⎠

(

g 2 − 2 gRω 2 cos 2 ϕ

g′ ≅

The fractional change in the value of g is

(R − d)

g2 + ω 2r

g

⇒

⎛ GM ⎞ So, at the surface of earth g = ⎜ 2 ⎟ and for a point ⎝ R ⎠ at a depth d below the surface,

g′ =

E

At the poles ϕ = 90° i.e. g P = g

As in case of spherical distribution of mass, for an internal point

3

F

At equator ϕ = 0° i.e. gE ≈ g − Rω 2

With Depth

GM

m

S

So, with increase in height, g decreases. However, if h R , then g gh = 2 ⎡ ⎛ h⎞⎤ 1 + ⎜ ⎟ ⎢ ⎝ R⎠ ⎥ ⎣ ⎦

⇒

ϕ

W

gh R2 = g ( R + h )2

gd =

P Fc

)

2

+ 2 grω 2 cos ( π − ϕ )

g gR2 = 4 ( R + h )2

⇒

1 ⎛ R ⎞ =⎜ ⎟ 4 ⎝ R+ h⎠

⇒

R 1 =± R+ h 2

⇒

h=R

2

09-Feb-21 6:28:05 PM

Chapter 4: Gravitation and Satellites

4.15

ILLUSTRATION 16

ILLUSTRATION 18

Assuming earth to be a sphere of uniform mass density, how much would a body weigh half way down the centre of the earth if it weighed 100 N on the surface?

A planet of radius R equal to one-tenth the radius of earth has the same mass density as earth. Scientists R dig a well of depth on it and lower a wire of the 5 same length and of linear mass density 10 −3 kgm −1 into it. If the wire is not touching anywhere calculate the force applied at the top of the wire by a person holding it in place. Take the radius of earth to be 6 × 106 m and the acceleration due to gravity of earth to be 10 ms −2 .

SOLUTION

Given, mg = 100 N h⎞ ⎛ g′ = g ⎜ 1 − ⎟ ⎝ R⎠ h 1 = R 2

SOLUTION

⇒

1⎞ g ⎛ g′ = g ⎜ 1 − ⎟ = ⎝ 2⎠ 2

⇒

mg ′ =

mg 100 = = 50 N 2 2

ILLUSTRATION 17

What is the acceleration due to gravity of earth at the surface of moon if the distance between earth and moon is 3.8 × 10 5 km and radius of earth is 6.4 × 10 3 km ? SOLUTION

If M and R be the mass and radius of the earth then the acceleration due to gravity due to earth on the surface of earth i.e., g=

GM

…(1)

R2

Similarly, acceleration due to gravity at a distance r ( > R ) of the earth i.e., g′ =

GM

…(2)

r2

If r be the distance between earth and moon then g ′ will give you the value of acceleration due to gravity on the moon due to earth. Therefore, from equation (1) and (2)

( 6.4 × 10 ) ( 3.8 × 10 )

Rearth 10 Mearth Since, density ρ = 4 3 π Rearth 3 Mplanet Also, ρ = 4 3 π Rplanet 3 M Me ⇒ Mplanet = earth = 3 1000 10 Given, Rplanet = R =

Let the acceleration due to gravity at surface of planet and at the surface of earth be g p and g e respectively. Then gp = ⇒

⇒

5 2

g = 0.00275 ms

Mechanics II_Chapter 4_Part 1.indd 15

−2

g

{Q

g = 9.8 ms

}

GMe ( 10 )

( 10 )3 Re2

2

=

GMe

10Re2

ge 10

So, total force acting on wire is R

F=

∫ ( λ dx ) g

4R 5

⇒ −2

2 Rplanet

=

The value of g inside the planet at a distance x from centre of the planet is ⎛ x⎞ ⎛ x⎞ ginside = gsurface of planet ⎜ ⎟ = g p ⎜ ⎟ ⎝ R⎠ ⎝ R⎠

3 2

g′ =

gp =

GMplanet

λ gp ⎛ x2 ⎞ F= ⎟ ⎜ R ⎝ 2 ⎠

p

⎛ x⎞ ⎜⎝ ⎟⎠ R

R 4 R/5

Substituting the given values, we get F = 108 N

09-Feb-21 6:28:09 PM

4.16

JEE Advanced Physics: Mechanics – II

ILLUSTRATION 19

Two equal masses m and m are hung from a balance whose scale pans differ in vertical height by h. Calculate the error in weighing, if any, in terms of density of earth ρ . SOLUTION

Since g varies with height as g ′ = For h

R , we have

gR2

( R + h )2

g R

⇒

ω=

⇒

2π = T

⇒

T = 2π

g R R g

Substituting the values, 6400 × 10 3 10 T= hr 3600 T = 1.4 hr 2π

2h ⎞ ⎛ g′ = g ⎜ 1 − ⎟ ⎝ R⎠ From the figure we see that h1 > h2 , so W1 will be lesser than W2 and hence

⇒

Thus, the new time period should be 1.4 hr instead of 24 hr for the weight of a body to be zero on the equator. ILLUSTRATION 21

h1 h2

⎛h h ⎞ W2 − W1 = mg 2 − mg1 = 2mg ⎜ 1 − 2 ⎟ ⎝ R R⎠ Since g =

GM R

2

and h1 − h2 = h , so we get

A straight smooth tunnel is dug through a spherical planet whose mass density ρ0 is constant. The tunnel passes through the centre of the planet and is perpendicular to the planet’s axis of rotation, which is fixed in space. The planet rotates with the angular velocity ω so that objects in the tunnel have no acceleration relative to the tunnel. Find ω . SOLUTION

For no acceleration, we have mg ′ − mrω 2 = 0

⎛ GM ⎞ h 2GMmh W2 − W1 = 2m ⎜ 2 ⎟ = ⎝ R ⎠R R3

ω

⎛4 ⎞ Since M = ⎜ π R3 ⎟ ρ ⎝3 ⎠ ⇒

W2 − W1 =

g′

2mhG ⎛ 4 ⎞ 8 π R3 ρ ⎟ = πρGmh 3 ⎜ ⎝ ⎠ 3 3 R

rω

2

r

ILLUSTRATION 20

Suppose the earth increases its speed of rotation. At what new time period will the weight of a body on the equator becomes zero? Take g = 10 ms −2 and radius of earth R = 6400 km .

⇒

g ′ = rω 2

⇒

⎛ GM ⎞ 2 ⎜⎝ 3 ⎟⎠ r = rω R

⇒

⎛4 ⎞ G ⎜ π R3 ρ0 ⎟ ⎝3 ⎠ = ω2 R3

⇒

ω=

SOLUTION

The weight will become zero when g′ = 0 ⇒

g − Rω 2 = 0

Mechanics II_Chapter 4_Part 1.indd 16

{on the equator g ′ = g − Rω 2 }

4 π Gρ0 3

09-Feb-21 6:28:14 PM

Chapter 4: Gravitation and Satellites ILLUSTRATION 22

Two concentric spherical shells have masses m1 and m2 and radii r1 and r2 ( r2 > r1 ) . What is the force exerted by this system on a particle of mass m3 if it is placed at a distance r ( r1 < r < r2 ) from the centre? SOLUTION

The outer shell will have no contribution in the gravitational field at point P , because P lies inside the outer shell. So, at P , the field will only be due to the inner shell of mass m1 . m2

where ω is the angular speed of the earth and W0 is the scale reading when the ship is at rest. Also explain the significance of plus and minus sign. SOLUTION

Let R be the radius of earth and ω be its angular speed. When the ship is at rest, we have W0 = mg − mω 2 R

W = mg −

r

⇒ r2

…(1)

When the ship moves with a speed in the sense of rotation of earth, then its effective speed is v + Rω , otherwise it is v − Rω . So, we have

m3 r1

m ( v ± Rω ) R

2

⎛ mv 2 ⎞ W = mg − ⎜ + mω 2 R ± 2mvω ⎟ ⎝ R ⎠

⎛ mv 2 ⎞ W = W0 − ⎜ ± 2mvω ⎟ ⎝ R ⎠ From (1), we get ⇒

⇒

EP =

Gm1

r2 Thus, force on mass m3 placed at P is,

m=

F = ( m3 EP ) ⇒

F=

Gm1 m3 r2

r r The field EP and the force F both are towards centre O. ILLUSTRATION 23

A body is suspended on a spring balance in a ship sailing along the equator with a speed v . Show that 2vω ⎞ ⎛ the scale reading will be very close to W0 ⎜ 1 ± g ⎟⎠ ⎝

4.17

W0 g − Rω

2

=

…(2)

W0 ⎛ Rω 2 ⎞ g⎜ 1− g ⎟⎠ ⎝

Since

Rω 2 1 ≈ g 291

⇒

m=

⇒

2vω ⎞ ⎛ W ≈ W0 ⎜ 1 ± g ⎟⎠ ⎝

1

W0 g

Test Your Concepts-I

Based on Acceleration Due to Gravity, Gravitational Field and Applications 1. Calculate the escape velocity from the surface of moon. The mass of the moon is 7.4 × 1022 kg and radius is 1.74 × 106 m . 2. A planet of mass m1 revolves round the sun of mass m2. The distance between the sun and the planet is r. Considering the motion of the sun, find the total energy of the system assuming the orbits to be circular.

Mechanics II_Chapter 4_Part 1.indd 17

(Solutions on page H.255) 3. Consider the ring-shaped body of mass M and radius a. A particle of mass m is placed a distance x from the centre of the ring, along the line through the centre of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy of this system. Take the potential energy to be zero when the two objects are far apart.

09-Feb-21 6:28:18 PM

4.18

JEE Advanced Physics: Mechanics – II

(b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. dU (c) Use Fx = − to calculate the force between dx the objects. (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What is the force when x = 0? Explain why this result makes sense. 4. In the arrangement shown, find the magnitude and direction of the net gravitational force acting on the central particle at O. M

2M

4M

y

7M M

5M x

O

3M 7M

d

M

2M

d/2

d

d

5. The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square of r, the acceleration of the body A is a. If the gravitational force follows an inverse fourth power law, then what will be the acceleration of the body A? 6. A particle of mass 20 g experiences a gravitational force of 4 N along positive x-direction. Find the gravitational field at that point. 7. The density of the core of a planet is ρ1 and that of the outer shell is ρ2. The radii of the core and that of the planet are R and 2R respectively. If the

Gravitational Potential Energy (U) Gravitational potential energy of a system of particles is defined as the external work required to assemble the particles from infinity to a given configuration (as that required in system). When particles/masses are lying at infinity, their potential energy is taken to be zero, because no potential exists between them. Since the gravitational force between the particles is attractive in nature, so the work will be done by the system and final potential energy of system will be negative.

Mechanics II_Chapter 4_Part 1.indd 18

gravitational acceleration at the surface of the ρ planet is same as at a depth R, find the ratio 1 . ρ2 R

2R

2

8. Calculate the gravitational force of attraction on a particle of mass m placed at the centre of a semicircular wire of length L and mass M. 9. Inside a solid sphere of radius R, the density ρ is ρR given by ρ = 0 , where ρ0 is the density at the r surface and r is the distance from the centre. Find the gravitational field due to this sphere at a distance 2R from its centre. 10. A system consists of a thin ring of radius R and a very long uniform wire oriented along axis of the ring with one of its ends coinciding with the centre of the ring. If mass of ring be M and the linear mass density of the wire be λ, then calculate the interaction force between the ring and the wire. 11. Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh 60% of his weight at the pole. Take R = 6400 km . 12. A thick spherical shell has an inner radius R1 and an outer radius R2. It has mass M and uniform density. Find the gravitational field Er as a function of r. 13. At what height from the surface of earth will the value of g be reduced by 36% in comparison to the value at the surface? Given R = 6400 km .

In giving these arguments please keep this thing in mind that work done by a conservative force equals the decrease in potential energy of the system. When all particles of a system are separated far apart by infinite distance, there exists no interaction between them. This state, we take as reference of Zero Potential Energy (ZPE). Gravitational potential energy is categorised in two ways.

09-Feb-21 6:28:20 PM

Chapter 4: Gravitation and Satellites

(a) Gravitational interaction energy ( U ) of a system of particles. (b) Gravitational self energy ( U s ) of a body.

GRAVITATIONAL POTENTIAL ENERGY OF A SYSTEM OF TWO PARTICLES Figure shows two masses, a source mass m and a test mass m0 separated by a distance r . The gravitational potential energy of this system is found by calculating the work done in bringing test mass m0 from infinity to the given point P at a distance r from the source mass m . Let, at any instant the test mass m0 be at a distance x from the source mass m . Let m0 be displaced through dx towards m . The gravitational force of attraction F acts on m0 towards m . x

P F

r

m

m0

dx m0

⇒

GRAVITATIONAL POTENTIAL ENERGY FOR A SYSTEM OF PARTICLES When more than two particles are there in a system, the potential energy can be given by sum of potential energy of all the pairs of particles with no pair repeated. For example, if a system of three particles having masses m1 , m2 and m3 is given as shown in figure. m1

r12

−2

dW = Gmm0 x dx

⇒

and m4 (having total number of six potential = 4 C2 ), we have

∫

r14

⎞ ⎟ ∞⎠ r

m4

m3

r34

m m m m ⎡m m U = −G ⎢ 1 2 + 2 3 + 3 4 + r r r34 23 ⎣ 12

W∞→ P = −

Mechanics II_Chapter 4_Part 1.indd 19

m2 r13

r24

⎛1 1⎞ W = −Gmm0 ⎜ − ⎟ ⎝ r ∞⎠

Gmm0 U=− r

r12

m1

Gmm0 =W r So, work done by the external force in bringing the test mass m0 from ∞ to the point P under the influence of source mass m is Gmm0 W = W∞→ P = − r This work done is stored in the form of gravitational potential energy ( U ) of the system. So, ⇒

⎤ ⎥ = U12 + U13 + U 23 ⎦

Similarly, for an assembly of four masses m1 , m2 , m3

0

⇒

m3

⎡m m m m m m U =−G⎢ 1 2 + 2 3 + 1 3 r23 r13 ⎣ r12

W = Gmm0 x −2 dx ⎛ x −2 +1 W = Gmm0 ⎜ ⎝ −2 + 1

r23

The total potential energy of this system can be given as

r

⇒

r13

m2

∞

If dW be the work done, then r r dW = F ⋅ dx Gmm0 dx cos ( 0° ) ⇒ dW = x2

4.19

m4 m1 m1 m3 m2 m4 ⎤ + + r14 r13 r24 ⎥⎦ ⇒

U = U12 + U 23 + U 34 + U 41 + U13 + U 24

From above we see that the total potential energy is simply the sum of contributions due to distinct pairs. Generalising to N masses, we get N

U = −G

N

∑∑

i =1 j =1 ( j>i )

mi m j rij

09-Feb-21 6:28:29 PM

4.20

JEE Advanced Physics: Mechanics – II

where j > i assures that the pairs are not repeated. Other way round, we can count every pair twice and 1 multiply the result by . So, 2 N

U = −G

N

∑∑

mi m j

i =1 j =1

rij

=

1 2

N

∑ i =1

⎛ mi ⎜ −G ⎜ ⎜⎝

N

∑ j =1 j≠i

⎛m ⎞ dm = ⎜ 0 ⎟ dx ⎝ l ⎠ a−

mj ⎞ ⎟ rij ⎟ ⎟⎠

m0

C

dU = −

⇒

m m m m ⎞ ⎛m m U = −G ⎜ 3 2 + 3 1 + 2 1 ⎟ r31 r21 ⎠ ⎝ r32

Substituting in above, we get ⎛ 3 × 2 3 ×1 2 ×1⎞ + + U = − ( 6.67 × 10 −11 ) ⎜ ⎟ ⎝ 1 1 1 ⎠ U = −7.337 × 10 −10 J

ILLUSTRATION 25

Find the gravitational potential energy of a point mass m0 and a thin uniform rod of mass m0 and length l , if they are located along a straight line at a distance a measured from the centre of the rod. m0

m0 a

SOLUTION

Let us consider an infinitesimal element of length dx, mass dm at a distance x from the point mass m0. Then,

Mechanics II_Chapter 4_Part 1.indd 20

U=

∫ dU = −

Gm02 l

l⎞ ⎛ ⎜⎝ a + ⎟⎠ 2

∫

l⎞ ⎛ ⎜⎝ a − ⎟⎠ 2

⇒

⇒

U=−

⇒

l⎞ ⎛ a+ Gm02 ⎜ 2⎟ U=− log e ⎜ l⎟ l ⎜⎝ a − ⎟⎠ 2

⇒

U=−

Gm02 l

dx x

l⎞ ⎛ a+ ⎝⎜ 2 ⎠⎟

Gm02 U=− log e x l

where, r32 = r31 = r21 = 1.0 m , m1 = 1 kg m2 = 2 kg and m3 = 3 kg

2

Gm02 dx Gm0 dm =− x l x

1 kg

3 kg

x

If dU be the gravitational potential energy between dm and m0 , then

SOLUTION

2 kg

m0

a a+

Three masses of 1 kg, 2 kg and 3 kg are placed at the vertices of an equilateral triangle of side 1 m. Find the gravitational potential energy of this system. Take G = 6.67 × 10 −11 Nm 2 kg −2 .

2

dx

ILLUSTRATION 24

⇒

dm

l⎞ ⎛ a− ⎝⎜ 2 ⎠⎟

l⎞ l⎞⎤ ⎡ ⎛ ⎛ ⎢ log e ⎜⎝ a + 2 ⎟⎠ − log e ⎜⎝ a − 2 ⎟⎠ ⎥ ⎦ ⎣

Gm02 ⎛ 2a + l ⎞ log e ⎜ ⎝ 2 a − l ⎟⎠ l

GRAVITATIONAL SELF ENERGY FOR A THIN UNIFORM SHELL To calculate the gravitational self energy of a body, it is supposed that initially the particles of the body are scattered at infinite distance from each other. Therefore, in the formation of a body some external agent has to do some work in assembling the body. This energy is stored in the body as gravitational potential energy and is known as gravitational self energy of the body. Let us consider a thin uniform shell of mass M and radius R whose gravitational self energy is to be calculated. Let the shell at any instant have a mass m .

09-Feb-21 6:28:35 PM

4.21

Chapter 4: Gravitation and Satellites

Since the shell is uniform and thin, so this mass is lying at its surface. Now let us bring additional mass dm to the surface of this shell, then the gravitational potential energy between m and dm is Gmdm dU = − R On integrating, the gravitational self energy is obtained as M

Us = −

∫ 0

Gmdm GM 2 =− 2R R

R 16π 2 ρ 2G 4 16π 2 ρ 2G ⎛ r 5 r dr = − ⎜ 3 3 ⎜⎝ 5

∫ 0

R

0

⎞ ⎟ ⎟⎠

M , so on substituting the value of ρ in 4 π R3 3 the above equation, we get But ρ =

Us = −

3 GM 2 5 R

ILLUSTRATION 26

GRAVITATIONAL SELF ENERGY FOR A UNIFORM SPHERE Let us consider a uniform sphere of mass M and radius R whose gravitational self energy is to be calculated. Let the sphere at any instant have a mass m and radius r . Since the sphere is uniform, so this mass m is distributed uniformly on the sphere. If the sphere has a uniform mass density ρ , then ⎛4 ⎞ m = ⎜ π r3 ⎟ ρ ⎝3 ⎠

A solid sphere of mass M and radius R is initially placed at a distance 5R from the centre of a point mass m as shown in figure.

Now it is shifted to a position at a distance 3R from the point mass. During displacement, it is also uniformly expanded to a radius 2R so that its density decreases uniformly throughout its volume. Find the work required in this process. SOLUTION

dr

In this process during displacement the size of the sphere is also changing. So, we have to also take into account the self energy of the sphere. Initial total energy of system is the sum of self energy of M and the interaction energy of M and m . If Em be the self energy of m , then

dm r R

Now, let us bring an additional layer of mass dm in the form of a thin spherical shell of inner radius r and outer radius r + dr to be placed on the sphere and repeat the process until it becomes a full fledged solid sphere of radius R . Then

(

Us = −

⎛ 3 GM 2 ⎞ ⎛ GMm ⎞ Ei = Em + ⎜ − + − ⎟ ⎝ 5 R ⎟⎠ ⎜⎝ 5R ⎠ Finally the radius of M becomes 2R and it is situated at a distance 3R from m as shown in figure.

)

dm = 4π r 2 dr ρ The gravitational potential energy between the sphere and this infinitesimal shell is given by

(

)

⎛4 ⎞ G ⎜ π r 3 ρ ⎟ 4π r 2 ρdr ⎝3 ⎠ Gmdm dU = − =− r r On integrating, the gravitational self energy is obtained as

Mechanics II_Chapter 4_Part 1.indd 21

So, final total energy of system is given by ⎛ 3 GM 2 ⎞ ⎛ GMm ⎞ E f = Em + ⎜ − + − ⎟ ⎝ 5 2R ⎟⎠ ⎜⎝ 3R ⎠

09-Feb-21 6:28:40 PM

4.22

JEE Advanced Physics: Mechanics – II

External work W required in the process is given by W = E f − Ei

Conceptual Note(s)

⇒

W=

3 GM 2 GMm ⎛ 1 1 ⎞ − ⎜ − ⎟ 10 R R ⎝ 3 5⎠

Let us find the difference in potential energy of a mass m in two positions shown in figure. The potential energy of the mass on the surface of earth (at B) is,

⇒

W=

GM ( 9 M − 4m ) 30 R

GMm R and potential energy of mass m at height h above the surface of earth (at A) is, UB = −

POTENTIAL ENERGY OF A PARTICLE ON EARTH’S SURFACE If M be the mass of earth, R be the radius of earth and m is the mass of particle on earth’s surface, then potential energy of particle on earth’s surface is U=−

GMm = − mgR R

GM ⎫ ⎧ ⎨Q g = 2 ⎬ R ⎭ ⎩

At height h ,

If a particle is taken from the surface of the earth to a height h close to the surface then h R . So,

Ufinal = −

GMm = Ui and R

⇒ ΔU = UA − UB = −

( UA > UB ) GMm ⎛ GMm ⎞ −⎜ − ⎟ R+h ⎝ R ⎠

1 ⎞ GMmh ⎛1 ⇒ ΔU = GMm ⎜ − = ⎝ R R + h ⎟⎠ R ( R + h ) ⎫ ⎧ GM ⎨ 2 = g⎬ ⎩R ⎭

GMmh h⎞ ⎛ R 2 ⎜ 1+ ⎟ ⎝ R⎠ mgh ⇒ ΔU = h 1+ R

For h  R , ΔU ≈ mgh

GMm = Uf R+ h

A

Hence, work done is equal to change in energy. ⇒

GMm GMm W = U f − Ui = − + R+ h R

⇒

1 ⎞ ⎛ 1 W = GMm ⎜ − ⎝ R R + h ⎟⎠

⇒

−1 GMm ⎡ ⎛ h⎞ ⎤ W= 1 − 1 + ⎢ ⎜ ⎟ ⎥ R ⎣ ⎝ R⎠ ⎦

⇒

W

{Q for x ⇒

W

GMm ⎛ h ⎞ ⎜ ⎟ R ⎝ R⎠

⇒

W

⎛ GM ⎞ m ⎜ 2 ⎟ h = mgh ⎝ R ⎠

h B R

M

Thus, what we read the mgh is actually the difference in potential energy (not the absolute potential energy), that too for h  R.

GMm ⎡ ⎛ h⎞⎤ 1− ⎜ 1− ⎟ ⎥ R ⎢⎣ ⎝ R⎠ ⎦

Mechanics II_Chapter 4_Part 1.indd 22

GMm R+h

⇒ ΔU =

GMm Uh = − R+ h

Uinitial = −

UA = −

GRAVITATIONAL POTENTIAL (V) 1; ( 1 + x )

n

1 + nx }

The gravitational potential at a point P due to a source mass is the work done per unit test mass in bringing the test mass ( m0 ) from infinity to the point P under the gravitational influence of source mass m. So, VP = Potential at point P =

W∞→ P m0

09-Feb-21 6:28:46 PM

Chapter 4: Gravitation and Satellites

Gmm0 r W∞→ P Gm VP = =− m0 r

V = V1 + V2 + V3 + ...

Since W∞→ P = − ⇒

N

r

⇒

P

(

−1

Gm r

GRAVITATIONAL POTENTIAL DUE TO AN ASSEMBLY OF MASSES Before moving further with this discussion, let us know this and keep in mind that gravitational potential is a scalar quantity. Let us now calculate the gravitational potential due to an assembly of masses at point P shown.

m2

r3 r4

m m ⎞ m ⎛m m V = −G ⎜ 1 + 2 + 3 + 4 + 5 ⎟ r2 r3 r4 r5 ⎠ ⎝ r1

x

dx

Q

L

M dx L

The potential dV due to this element at point P is dV = −

⇒

Gdm x

⎛ M⎞ dx G⎜ ⎝ L ⎟⎠ V=− x

Net gravitational potential at point P is obtained by integrating this expression. So ⇒

V=

∫

GM dV = − L

r+L

∫ r

dx x

r+L

Problem Solving Technique(s)

⇒

V=−

GM log e x L

(a) The gravitational potential at earth’s surface is −GMe V= = − gRe Re

⇒

V=−

GM ⎛ r+L⎞ log e ⎜ ⎝ r ⎟⎠ L

(b) For an assembly of masses m1, m2, m3, …, at distances r1, r2, r3,… from point P, net gravitational potential equals the algebraic sum of gravitational potential due to each of the mass at point P. So,

Mechanics II_Chapter 4_Part 1.indd 23

P r

For this we consider an infinitesimal element of length dx at a distance x from the point P . Mass on this element is

m4

The gravitational potential V at point P due to the assembly of masses shown is the algebraic sum of the potential due to each of the mass in the assembly. So,

i

Consider a thin rod of length L , having a uniform mass M . Let us find the gravitational potential at a point P due to the rod at a distance r from one end of the rod.

m3

m5

mi

GRAVITATIONAL POTENTIAL DUE TO A THIN ROD

dm =

r2 P

∑r i

m1 r1

V = −G

)

It is a scalar quantity. Its SI unit is joule/kg Jkg and its dimensional formula is M 0 L2 T −2 The gravitational potential due to a source mass m at a distance r from it is

r5

m ⎛m m ⎞ V = −G ⎜ 1 + 2 + 3 + ... ⎟ r r r ⎝ 1 ⎠ 2 3

⇒

m

V=−

4.23

r

If mass density of the rod is λ , then λ =

M . So, L

⎛ r+L⎞ V = −Gλ log e ⎜ ⎝ r ⎟⎠

09-Feb-21 6:28:52 PM

4.24

JEE Advanced Physics: Mechanics – II

GRAVITATIONAL POTENTIAL DUE TO A RING AT ITS CENTRE Let us first find potential dV at centre C due to an infinitesimal mass dm on ring which is dV = −

Gdm R

Total potential at C is V = ⇒

THE GRAVITATIONAL POTENTIAL AND FIELD STRENGTH DUE TO A THIN SPHERICAL SHELL Let M be the mass and R , the radius of thin spherical shell. ⎡ GM ⎢− r V=⎢ ⎢ GM ⎢⎣ − R

∫ dV

GM dm V = −G =− R R

∫

Since all the infinitesimal dm’s of the ring are situated at same distance R from the ring centre C so, we can directly say that the total gravitational potenGM tial at centre of ring is VC = − . R M

r≥R

( outside and at surface ) r ve , body again escapes but now the path is a hyperbola. Here, v0 = orbital speed ⎛ GM ⎞ ⎜⎝ ⎟ at A and ve = escape velocity at A. r ⎠ 0 < v < v0

v0 < v < ve

⇒

A

V0

23

⎛ T ⎞ =⎜ ⎝ ηT ⎟⎠

23

=

1

η

23

Now when they are closest to each other, then, the separation between them is minimum i.e., as shown in figure. v1 r1

s1

r2

So, ω 21 =

⇒ NOTE: (a) From case (i) to (iv) total energy of the body is negative. Hence, these are the closed orbits and are also called Bound Trajectories. For case (v) total energy is zero and for case (vi) total energy is positive. In these two cases orbits are open and are called Unbound Trajectories.

r1 ⎛ T1 ⎞ = r2 ⎜⎝ T2 ⎟⎠

⇒

⇒

s2 v2

vrel v2 + v1 = r2 − r1 r⊥

ω 21

1 r1 ⎞ 2π r2 2π r1 ⎛ 1 2π ⎜ + + ⎝ T2 T1 r2 ⎟⎠ T2 T1 = = r r2 − r1 1− 1 r2

ω 21

1 1 ⎞ ⎛ 1 2π ⎜ + ⎝ ηT T η2 3 ⎟⎠ = 1 1− 2 3 η

ω 21

2π −1 3 (η + 1) = T 23 (η − 1)

h

ILLUSTRATION 49

R

A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to r1 and r2 respectively. Find the angular momentum of this planet relative to the centre of the sun. Mass of the sun is M .

(b) If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.

SOLUTION ILLUSTRATION 48

Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the opposite sense. The periods of revolutions are T and ηT respectively. Find the angular speed of S2 as observed by an astronaut in S1 , when they are closest to each other.

Applying the Law of Conservation of Angular Momentum and Mechanical Energy at P and A , we get mv1 r1 sin 90° = mv2 r2 sin 90° ⇒

⎛r ⎞ v2 = ⎜ 1 ⎟ v1 ⎝ r2 ⎠

…(1)

Also,

1 GMm 1 GMm mv12 − = mv22 − 2 2 r1 r2

…(2)

SOLUTION

Since T 2 ∝ r 3 2

⇒

r ∝T3

Mechanics II_Chapter 4_Part 1.indd 44

09-Feb-21 6:30:45 PM

Chapter 4: Gravitation and Satellites

Substituting value of v2 from equation (1) in equation (2), we get 1 ⎛ m 2 ⎜⎝ ⇒ ⇒

v12

r12 r22

⎞ ⎛ 1 1⎞ − 1 ⎟ v12 = GMm ⎜ − ⎟ ⎝ r2 r1 ⎠ ⎠

⎛ r1 + r2 ⎞ 2GM ⎜⎝ r ⎟⎠ = r 2 1 2GMr2 r1 ( r1 + r2 )

v1 =

r1 90° S

r2

90° A

⇒

b 2 = a 2 − ( a − r1 ) = 2 ar1 − r12

⇒

⎛ r +r ⎞ b 2 = 2r1 ⎜ 1 2 ⎟ − r12 = r2 r1 ⎝ 2 ⎠

2

2r r b2 = 12 a r1 + r2

DOUBLE STAR SYSTEM OR BINARY STAR SYSTEM

Since, angular momentum of the planet about the sun is L = mv1 r1 L=m

r1 + r2 2 Also, we know that b 2 = a 2 ( 1 − e 2 ) and a − r1 = ae a=

rL =

v1

⇒

Here semi major axis is given by

So, from equation (1), we get v2

P

4.45

2GMr1 r2 ( r1 + r2 )

ILLUSTRATION 50

If a planet revolves around the sun in an elliptical orbit such that its minimum distance from sun is r1 and maximum distance is r2 . Calculate the distance of planet from sun when it is at a position where the line joining the planet and sun is perpendicular to the major axis of the ellipse.

In motion of a planet round the sun we have assumed the mass of the sun to be too large in comparison to the mass of the planet. Under such situation the sun remains stationary and the planet revolves round the sun. If however masses of sun and planet are comparable and motion of sun is also to be considered, then both of them revolve around their centre of mass with same angular velocity but different linear speeds in the circles of different radii. The centre of mass remains stationary.

m1

r1

m2

r2

SOLUTION

The situation is shown in Figure.

We use following equations under this condition. m1 r1 = m2 r2

r1

rL

m1 r1ω 2 = m2 r2ω 2 =

r2

…(1) Gm1 m2

…(2)

( r1 + r2 )2

Solving these two equations, we can find that 1 Since, rL = (latus rectum) 2 ⇒

rL =

1 ⎛ 2b 2 ⎞ b 2 ⎜ ⎟= 2⎝ a ⎠ a

Mechanics II_Chapter 4_Part 1.indd 45

ω= …(1)

G ( m1 + m2 ) r3

=

GM r3

or T =

2π r 3 2 GM

Here, M = m1 + m2 and r = r1 + r2

09-Feb-21 6:30:50 PM

4.46 JEE Advanced Physics: Mechanics – II

μ=

m2

m1m2 m1 + m2

m1

Further, angular momentum of the system about centre of mass ⎛ m1 m2 ⎞ 2 L = ( I1 + I 2 ) ω = ⎜ r ω = μ r 2ω ⎝ m1 + m2 ⎟⎠ Kinetic energy of system, K=

1 ⎛ m1 m2 ⎞ 2 2 1 2 2 r ω = μr ω 2 ⎜⎝ m1 + m2 ⎟⎠ 2

and moment of inertia of system,

ILLUSTRATION 51

A double star is a system of two stars moving around the centre of mass of the system due to gravitation. Find the distance between the components of the double star if the total mass equals M and time period is T . SOLUTION

Let d be the distance between the stars and let d1 and d2 be the distances of stars from centre of mass d d ⎛ ⎞ ⎛ ⎞ Therefore, d1 = ⎜ m2 and d2 = ⎜ ⎟ ⎟ m1 m + m m + m ⎝ 1 ⎝ 1 2 ⎠ 2 ⎠ Since, we have

d2 Also, m1 + m2 = M

⎛ m1 m2 ⎞ 2 I=⎜ r = μr 2 ⎟ m m + ⎝ 1 2 ⎠

⇒

m1 m2 Here, μ = = reduced mass. m1 + m2

⇒

Thus, the two bodies can be replaced by a single body whose mass is equal to reduced mass. This single body revolve in a circular orbit whose radius is equal to the distance between two bodies and centripetal force of circular motion is equal to force of potential between two bodies for actual separation.

Gm1 m2

d1 =

= m1ω 2 d1 = m2ω 2 d2 {given}

d d m2 and d2 = m1 M M

Gm1 ( M − m1 ) d

2

⇒

GM = ω 2 d 2

⇒

d3 =

⇒

⎛ GMT 2 ⎞ 3 d=⎜ ⎝ 4π 2 ⎟⎠

⎤ ⎡ d = m1ω 2 ⎢ ( M − m1 ) ⎥ M ⎦ ⎣

GMT 2 4π 2 1

Test Your Concepts-V

Based on Satellites, Kepler’s Laws and Applications 1. Two satellites A and B of the same mass are orbiting the earth at altitudes R and 3R respectively, where R is the radius of the earth. Taking their orbits to be circular obtain the ratios of their kinetic and potential energies. 2. A satellite of mass m is orbiting a planet of mass M at a radial distance R from the centre of a planet. The satellite explodes by expelling very rapidly a small amount of its mass Δm in an opposite direction to its orbital velocity. The immediate recoil velocity of the satellite is vr (additional to the velocity already possessed by the satellite).

Mechanics II_Chapter 4_Part 1.indd 46

(Solutions on page H.263) (a) Show that the largest possible value of vr for which the satellite remains within the gravitational field assuming Δm  m is, GM R (b) Deduce the result for vr in case of Δm not being negligible. 3. If a satellite is revolving close to a planet of density ρ with period T, show that the quantity ρT2 is a universal constant. vr = ( 2 − 1)

09-Feb-21 6:30:54 PM

Chapter 4: Gravitation and Satellites

4. A satellite is projected into space with a velocity v0 at a distance r0 from the centre of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle α with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite. v0 rmax

α r0 rmin

5. Two small dense stars rotate about their common centre of mass as a binary system with the period of 1 year for each. One star is of double the mass of ⎛ 1⎞ the other and the mass of the lighter one is of ⎜ ⎟ ⎝ 3⎠ the mass of the sun. Given the distance between the earth and the sun is R. If the distance between the two stars is r, then obtain the relation between r and R. 6. In a double star, two stars one of mass m1 and another of mass m2, with a separation d, rotate about their common centre of mass. Find: (a) an expression for their time period of revolution (b) the ratio of their kinetic energies (c) the ratio of their angular momenta about the centre of mass and (d) the total angular momentum of the system (e) the kinetic energy of the system. 7. Two satellites A and B revolve around a planet in two coplanar circular orbits in the same sense with radii 104 km and 2 × 104 km respectively. Time period of A is 28 hours. What is time period of

Mechanics II_Chapter 4_Part 1.indd 47

4.47

another satellite. Find the speed of B with respect to A when A and B are at farthest distance from each other. 8. A satellite is revolving round the earth in a circular orbit of radius r and velocity v0. A particle is projected from the satellite in forward direction with ⎛ 5 ⎞ relative velocity v = ⎜ − 1 v . Calculate its min⎝ 4 ⎟⎠ 0 imum and maximum distances from earth’s centre during subsequent motion of the particle. 9. Two identical stars each of mass M orbit around their centre of mass. Each orbit is circular and has radius R, so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) What minimum energy would be required to separate the two stars to infinity? 10. Prove that the velocity of a satellite travelling in an elliptical orbit when it reaches point C on the end g . Also show that this of the semi-minor axis is R a velocity is the same as that for a circular orbit of radius a. C

vc

b a

11. Binary stars of comparable masses m1 and m2 rotate under the influence of each other’s gravity with a time period T. If they are stopped suddenly in their motions, find their relative velocity when they collide with each other. The radii of the stars are R1 and R2 respectively. G is the universal constant of gravitation.

09-Feb-21 6:30:55 PM

4.48

JEE Advanced Physics: Mechanics – II

SOLVED PROBLEMS PROBLEM 1

Two identical thin rods of length 2a have equal mass M that is distributed uniformly along their lengths. The rods lie along the x-axis with their centres separated by a distance b > 2 a .

in the field of the first rod. For finding the force on this rod let us again consider an infinitesimal element of length dx at a distance x from the nearest end of the left rod. If dF be the force due to this rod then ⎡ GM ⎤ dF = Eg dm = ⎢ λ dx ⎣ x ( x + 2 a ) ⎥⎦ y

b

dx

x

Show that the magnitude of the force exerted by the left rod on the right one is given by

⇒

SOLUTION

This problem is a very special one where we shall learn to calculate the force between two extended bodies using the concept of gravitational field. For this let us divide the problem in two parts. PART I Here we shall imagine the rod to the right to be absent and then let us calculate the field due to the rod on the left at a hypothetical point P (say) at a distance d from one end of the rod. Let us consider an element of length dx at a distance x from the point P. If dEg be the gravitational field due to the rod at P, then

⇒

⇒ ⇒

Gdm x2

⎛ GM ⎞ −2 =⎜ x dx ⎝ 2 a ⎟⎠

⎞ ⎛ d+ 2a GM GM ⎛ x −2 +1 ⎞ ⎜ Eg = x −2 dx ⎟ = 2a ⎜ 2 a ⎜⎝ −2 + 1 ⎟⎠ ⎟⎠ ⎝ d

∫

Eg = −

GM ⎛ 1 ⎞ ⎜ ⎟ 2a ⎝ x ⎠

d+ 2a

=− d

d+ 2a

d

1⎞ GM ⎛ 1 − ⎟ ⎜ 2a ⎝ d + 2a d ⎠

GM Eg = d ( d + 2a )

PART II Now after we have calculated the field due to one rod, we actually observe that the second rod will be lying

Mechanics II_Chapter 4_Part 1.indd 48

⇒

⇒

dF = F=

F=

⇒

F=

⇒

F=

b–a

b+a

GM 2 ⎡ dx ⎤ ⎢ ⎥ ( ) 2a ⎣ x x + 2a ⎦

∫

F=

⇒

a

M 2a

Since λ =

⎛ ⎞ GM b2 F= log e⎜ 2 2 2 ⎝ b − 4 a ⎟⎠ 4a 2

dEg =

–a

GM 2 dF = 2a

b

∫

b −2a

dx

x ( x + 2a )

2

GM ⎡ 1 ⎛ 2a + x ⎞ ⎤ − log e ⎜ ⎝ x ⎟⎠ ⎥⎦ 2 a ⎢⎣ 2 a

b b −2a

2

GM ⎡ ⎛ 2a + b ⎞ ⎛ b ⎞⎤ − log e ⎜ ⎟⎠ + log e ⎜⎝ ⎟ 2 ⎢ ⎝ b − 2 a ⎠ ⎥⎦ b 4a ⎣ GM 2 4 a2 GM 2 4 a2

⎤ ⎡ b2 log e ⎢ ⎥ ( ) ( ) ⎣ b − 2a b + 2a ⎦ ⎛ ⎞ b2 log e ⎜ 2 ⎝ b − 4 a 2 ⎟⎠

PROBLEM 2

Calculate the gravitational field due to a thin uniform very large hemispherical shell at its centre. Assume the shell to have a radius R and a mass M. SOLUTION

Let us consider an infinitesimal elemental ring on its surface having angular width dθ at an angle θ from its axis as shown. The surface area dA of this ring is dA = ( 2π R sin θ )( Rdθ )

09-Feb-21 6:31:02 PM

Chapter 4: Gravitation and Satellites

r

⇒ dr

⎛ M ⎞ E = Gσπ = G ⎜ π ⎝ 2π R2 ⎟⎠

⇒

E=

xθ

R

dθ O r = Rsinθ x = Rcosθ

dm = σ dA = σ ( 2π R2 sin θ dθ ) where σ is the surface mass density of the shell given by M σ= 2π R2 Since the gravitational field strength due to a thin ring of mass dm , radius r at a point lying on its axis at a distance x from its centre is Now due to this ring, gravitational field strength at centre C is G ( dm ) x

Also, from the figure, we observe that x = R cos θ and r = R sin θ

SOLUTION

The gravitational self energy of the sun is given by Us = −

3 ( 6.67 × 10 −11 ) ( 2 × 10 30 ) Us = − 5 ( 7 × 108 ) ⇒

Gσ ( 2π R2 sin θ dθ ) R cos θ

⇒

dE =

⇒

dE = Gσπ ( 2 sin θ cos θ dθ ) = Gσπ sin ( 2θ ) dθ

R3

Net gravitational field at center is obtained by integrating the above expression for dE between the π limits zero to . So, 2 π 2

E=

∫ dE = Gσπ ∫ sin ( 2θ ) dθ 0

⇒

⎛ cos ( 2θ ) E = Gσπ ⎜ − ⎝ 2

⇒

⎛ 1 1⎞ E = Gσπ ⎜ + ⎟ ⎝ 2 2⎠

Mechanics II_Chapter 4_Part 1.indd 49

π 2 0

⎞ ⎟ ⎠

2

U s = −2.29 × 10 41 J

Now, rate of change of energy of the sun, is dU s 3 M 2G dR = dt 5 R2 dt

Gdm ( R cos θ )

( R2 sin 2 θ + R2 cos2 θ )3 2

3 GM 2 5 R

This is actually the self energy of the solid sphere. So,

Substituting these values in equation (1), we get dE =

2R 2

Calculate the self energy of the sun, taking its mass to be equal to 2 × 10 30 kg and its radius to be very nearly 7 × 108 metre . If its radius contracts by 1 km per year, without affecting its mass, calculate the rate at which it radiates out energy.

…(1)

( r 2 + x 2 )3 2

GM

PROBLEM 3

Mass on this elemental ring is

dE =

⇒

4.49

where,

dR 1000 = ms −1 dt 365 × 24 × 3600

So, the rate of energy radiated out by the sun, dU s 3 M 2G dR = dt 5 R2 dt

(

)( 2

)

30 6.67 × 10 −11 ⎛ ⎞ dU s 3 2 × 10 1000 ⇒ = ⎜ 2 dt 5 ⎝ 365 × 24 × 3600 ⎟⎠ 7 × 108

⇒

(

)

dU s = 1.03 × 10 28 Js −1 dt

PROBLEM 4

Halley’s comet has a period of 76 years and in the year 1986 the distance of closest approach of the comet to the sun was 8.9 × 1010 m . Calculate the comet’s farthest distance from the sun if the mass of sun is 2 × 10 30 kg and G = 6.67 × 10 −11 Nm 2 kg −2 and 1 yr = 3.15 × 107 s

09-Feb-21 6:31:09 PM

4.50

JEE Advanced Physics: Mechanics – II

r r r The force F varying with height y as F = −2mg ( 1 − ay ) where a is a positive r constant. Find the work performed by the force F over the first half of the ascent and the corresponding increase in the potential energy of the body in the earth’s gravity field, which is assumed to be uniform.

SOLUTION

SOLUTION

From the problem it is self-evident that the orbit of the comet is elliptical with sun being at one focus. For elliptical orbits, according to Kepler’s third law, we have ⎛ 4π 2 ⎞ 3 T2 = ⎜ a ⎝ GMs ⎟⎠

First let us find the total height of ascent. At the beginning and the end of the path, the velocity of the body is happens to be equal to zero and therefore, the change in kinetic energy of the body is also equal to zero. Therefore, from Work-Energy Theorem, we have Work Done by All the Forces = 0 h

1

⇒

⎛ GMT 2 ⎞ 3 a=⎜ ⎝ 4π 2 ⎟⎠

⇒

h

⇒

⎛ 6.67 × 10 −11 × 2 × 10 30 × ( 76 × 3.15 × 10 a=⎜ ⎝ 4π 2

⇒

a ≈ 2.7 × 1012 m

)

7 2

1 ⎞3

⎟ ⎠

For an ellipse, we know that 2a = rmin + rmax ⇒

rmax = 2 a − rmin

⇒

rmax = 2 ( 2.7 × 1012 ) − 8.9 × 1010

⇒

rmax ≈ 5.3 × 10

r

0

where T = 76 × 365 × 86400

12

r

∫ ( F + mg ) dy = 0

m

⇒

∫ mg ( 1 − 2ay ) dy = 0 0

⇒

mgh ( 1 − ah ) = 0

⇒

h=

1 a

r The work performed by the force F over the first half of the ascent is, W=

h2

1 2a

0

0

∫ Fdy = 2mg ∫ ( 1 − ay ) dy =

3 mg 4a

The corresponding increase in the potential energy is

PROBLEM 5

ΔU =

A body of mass m ascends from the earth’s surface with zero initial velocity due to the action of two forces as shown in figure. y

m mg

mgh mg = 2 2a

PROBLEM 6

Find the speeds of a planet of mass m in its perihelion and aphelion positions. The semi-major axis of its orbit is a , eccentricity is e and the mass of the sun is M . Also find the total energy of the planet in terms of the given parameters. SOLUTION

Let v1 and v2 be the speeds of the planet at perihelion

Mechanics II_Chapter 4_Part 1.indd 50

09-Feb-21 6:31:16 PM

Chapter 4: Gravitation and Satellites v2

SUN

P r2

r1

P

4.51

A

M

v0

S

A v1

SOLUTION

and aphelion positions, then r1 = a ( 1 − e ) and r2 = a ( 1 + e )

…(1)

Applying the Law of Conservation of Angular momentum of the planet at P (perihelion) and A (aphelion) about the sun, we get mv1 r1 sin 90° = mv2 r2 sin 90° ⇒

v1 r1 = v2 r2

…(2)

Applying the Law of Conservation of Mechanical Energy in these two positions, we get

( U + K )at P = ( U + K )at A ⇒

1 GMm GMm 1 mv12 − = mv22 − 2 2 r1 r2

…(3)

Solving equations (1), (2) and (3), we get GM ⎛ 1 + e ⎞ GM ⎛ 1 − e ⎞ ⎜⎝ ⎟⎠ and v2 = ⎜ ⎟ a 1− e a ⎝ 1+ e ⎠

v1 =

Further, total energy of the planet is 1 GMm GMm 1 E = mv12 − = mv22 − 2 2 r1 r2 ⇒

E=

1 ⎡ GM ⎛ 1 + e ⎞ ⎤ GMm m ⎜ ⎟ − 2 ⎢⎣ a ⎝ 1 − e ⎠ ⎥⎦ a ( 1 − e )

⇒

E=

GMm ⎡ ⎛ 1 + e ⎞ ⎤ ⎜ ⎟ − 1⎥ a ( 1 − e ) ⎢⎣ ⎝ 2 ⎠ ⎦

⇒

GMm ⎛ e − 1 ⎞ E= ⎜ ⎟ a(1 − e ) ⎝ 2 ⎠

⇒

E=−

GMm 2a

PROBLEM 7

A cosmic body A coming from infinity with a velocity v0 is approaching the Sun of mass M, with its line of motion at a distance l from the Sun, as shown in the figure. When it gets closest to the Sun i.e. at P, what will be its distance from the sun?

Mechanics II_Chapter 4_Part 1.indd 51

By Law of Conservation of Mechanical Energy, we have

( U + K )at ∞ = ( U + K )at P 1 GMm 1 0 + mv02 = − + mv 2 …(1) 2 r 2 where M is the mass of the sun m is the mass of the body r is the distance of closest approach and v is the velocity of the body at the point P ⇒

Since the torque due to the gravitational force on the body about the sun is zero, so the angular momentum of the body about the sun will remain conserved, therefore by Law of Conservation of Angular Momentum, we have

( mv0 ) l = ( mv ) r ⇒

v0 l = vr

⇒

v=

…(2)

v0 l r

Substituting this value of v in equation (1), we get ⇒

1 GMm 1 ⎛ v0 l ⎞ mv02 = − + m⎜ ⎟ 2 2 ⎝ r ⎠ r

⇒

v02

⇒

v02 r 2

=−

(

2GMr + v02 l 2 r

2

2

)

+ 2GMr − v02 l 2 = 0

Solving this quadratic in r , we get r=−

−2GM ± 4G 2 M 2 + 4v04 l 2 2v02

Since we are here calculating the distance of closest approach, i.e. the least distance between the sun and the body, so we will reject the negative sign. Hence, we have

09-Feb-21 6:31:24 PM

4.52

JEE Advanced Physics: Mechanics – II

r=−

⇒

SOLUTION

−2GM + 4G 2 M 2 + 4v04 l 2

Tangential force, Ft = m

2v02

dv dt Rearranging, we get

2 ⎞ ⎛ ⎛ v2l2 ⎞ GM = 2 ⎜ −1 + 1 + ⎜ 0 ⎟ ⎟ v0 ⎜ ⎝ GM ⎠ ⎟ ⎝ ⎠

rmin

⇒

− av 2 = m

t

∫

PROBLEM 8

A body is projected vertically upwards from the surface of earth with a velocity sufficient to carry it to infinity. Calculate the time taken by it to reach a height h .

0

m dt = − a

Let v be the velocity of the body at a distance r from the centre of earth. Applying Law of Conservation of Mechanical Energy, we get

where, ve = 2 gR and g =

v= Since v =

t=

t=

r dr , so we have dt

∫ dt = R 0

1

R+ h

r 2g ∫

12

dr

R

2 1 ⎡ 32 32 ⎣ ( R + h ) − R ⎤⎦ 3 R 2g

⇒

t=

⇒

32 ⎤ 1 2R ⎡ ⎛ h⎞ t= ⎢ ⎜⎝ 1 + ⎟⎠ − 1 ⎥ 3 g ⎣ R ⎦

Mechanics II_Chapter 4_Part 1.indd 52

vi

m =− 2 a v vf vi

vf

∫v

−2

dv

vi

m⎛ 1⎞ =− ⎜− ⎟ a ⎝ v⎠

vf vi

GM GM and v f = nR R

m ⎛ nR R ⎞ − ⎜ ⎟ a ⎝ GM GM ⎠ GM R2

, so we get

m R a GM

(

n − 1)

PROBLEM 10

A satellite is to be placed in an orbit just above the earth’s 3 times the speed for a atmosphere with a speed 2 circular orbit at that height. The initial velocity imparted is horizontal. Calculate the maximum distance of the satellite from the earth, when it is in the orbit. SOLUTION

Since the satellite is to be placed in an orbit just above the atmosphere i.e. it is to be placed in an orbit close to surface of earth. Also, it is given that vi =

PROBLEM 9

An artificial satellite of the earth (radius R and mass M) moves in an orbit whose radius is n times the radius of the earth. Assuming resistance to the motion to be proportional to the square of velocity that is F = av 2 . Find how long the satellite will take to fall on to the earth.

∫

dv

1⎞ m⎛ 1 − ⎟ ⎜ a ⎝ v f vt ⎠

Since g =

R 2g

t

t=

Also, vi =

GM

dr R 2 g = dt r ⇒

⇒

…(1)

R2 Substituting in equation (1), we get

vf

m ⎛ v −2 +1 ⎞ t=− ⎜ ⎟ a ⎝ −2 + 1 ⎠

SOLUTION

1 GMm 1 GMm mv 2 − = mve2 − 2 r 2 R

dv dt

3 vorbit 2

Close to surface of earth v0 ≈ gR ⇒

vi =

3 gR 2

09-Feb-21 6:31:30 PM

4.53

Chapter 4: Gravitation and Satellites

By Law of Conservation of Angular Momentum, we have mvi ri = mv f r f ⇒

⎛ 3 ⎞ m⎜ gR ⎟ R = mv f rmax ⎝ 2 ⎠

…(1)

SOLUTION

r Let us consider a point P inside the cavity. Let E1 be gravitational field due to sphere at P , then r ⎛4 ⎞r E1 = ⎜ π Gρ ⎟ r1 ⎝3 ⎠

By Laws of Conservation of Energy, we have 1 ⎛3 GMm ⎞ GMm 1 = mv 2f − m ⎜ gR ⎟ − ⎝ ⎠ 2 2 2 R rmax ⇒

P

So, equation (2) becomes ⎛ 1 1⎜ − gR = ⎜ 4 2⎝

⇒

R2 rmax 2

−

3 gR3 2 rmax

2

⎞ ⎟ gR2 ⎟ − rmax ⎠

4 R 1 + =0 3 rmax 3

4 16 4 4 2 ± − ± R 3 9 3 3 3 2 1 = = = ± 2 3 3 2 rmax R rmax

= 1 or

1 3

Since rmax ≠ R ⇒

rmax = 3 R

So, maximum distance of satellite from the earth surface is hmax = rmax − R ⇒

P

E2

r1

hmax = 3 R − R = 2R

r2

C2 C1

3 3 gR 2 rmax

vf =

⇒

C1

gR2 3 1 gR − gR = v 2f − 4 2 rmax

From equation (1), we have

⇒

E1

…(2)

C2

If E2 be the gravitational field due to cavity at P , then r 4 r E2 = π G ( − ρ ) r2 3 r 4 r ⇒ E2 = − π Gρr2 3 So, net field at P , due to sphere and the cavity is r r r E = E1 + E2 ⇒

r 4 r r E = π Gρ ( r1 − r2 ) 3

…(1)

However, in triangle PC1C2 , we have r r r r1 + l = r2 r r r ⇒ r1 − r2 = − l Substituting in (1), we get r r 4 E = − π Gρ l 3 r 4 E = π G ρl ⇒ 3 So, we observe that the gravitational field inside the cavity is independent of the location of the point P and just depends upon the distance of the cavity’s centre from the centre of sphere.

PROBLEM 11

PROBLEM 12

Inside a uniform sphere of density ρ there is a spherical cavity whose centre is at a distance l from the centre of the sphere. Find the strength of the gravitational field inside the cavity.

A meteorite of mass m collides perpendicularly with a satellite which was orbiting around a planet of mass M in a circular path of radius R . Due to collision, the meteorite sticks to the satellite of mass 10m and the satellite is seen to have gone into an orbit whose

Mechanics II_Chapter 4_Part 1.indd 53

09-Feb-21 6:31:39 PM

4.54

JEE Advanced Physics: Mechanics – II

R . Determine 2 the velocity v of the meteorite before collision. minimum distance from the planet is

SOLUTION

The situation before collision and after collision is shown in the figure.

From equation (2) and (3), we get

v 10m R

v′

⎛ 20 ⎞ v ′′ = 2v ′ cos θ = ⎜ v ⎝ 11 ⎟⎠ 0

Path before collision

−

M

2

R/

v″

Before collision, the speed of satellite is the orbital speed given by GM u0 = R

…(1)

After collision with meteorite, let the combined mass 11m moves at an angle θ with the orbit as shown in figure. If finally, 11m moves at speed v ′ then applying the Law of Conservation of Momentum along horizontal and vertical directions, we get mv = 11mv ′ sin θ ⇒

⎛ v⎞ v ′ sin θ = ⎜ ⎟ ⎝ 11 ⎠

…(2)

10 mu0 = 11mv ′ cos θ ⇒

⎛ 10v0 ⎞ v ′ cos θ = ⎜ ⎝ 11 ⎟⎠

…(7)

Substituting the values of v ′ 2 from (6) and v ′′ from (7) in equation (5), we get

u

θ

…(6)

From equation (4), we get

m Path after collision

2

2

v 2 + 100u02 ⎛ 10u0 ⎞ ⎛ v⎞ v′2 = ⎜ ⎟ + ⎜ = ⎟ ⎝ 11 ⎠ ⎝ 11 ⎠ 121

2 2GM 1 ⎛ 20 ⎞ GM 1 ⎛ v 2 + 100u02 ⎞ + ⎜ = − + u ⎟⎠ ⎜ 0⎟ R 2⎝ 121 R 2 ⎝ 11 ⎠

⇒

v2 GM ⎛ =− +⎜ R ⎝ 242

⇒

v2 GM 300 ⎛ GM ⎞ =− + ⎜ ⎟ 242 R 242 ⎝ R ⎠

⇒

v2 58GM = 242 242R

⇒

v=

(

1⎞ ⎛ 1 ⎞ 2 2 ⎟⎠ ⎜⎝ ⎟⎠ 400u0 − 100u0 2 121

)

⎧ GM ⎫ ⎬ ⎨Q u0 = R ⎭ ⎩

58GM R

PROBLEM 13

Two iron spheres, each of mass 1 kg and 100 mm in diameter, are released from rest with a centre-tocentre separation of 1 m . Assume an environment in space with no forces other than the force of mutual gravitational attraction and calculate the time t required for the spheres to contact each other and the absolute speed v of each sphere upon contact.

…(3)

After collision applying the Law of Conservation of Angular Momentum, we get R …(4) 2 where v ′′ is the speed at perigee. Applying the Law of Conservation of Energy, we have 11m ( v ′ cos θ ) R = 11m ( v ′′ )

−

SOLUTION

Let vr be the relative speed at separation r , then by Law of Conservation of Mechanical Energy, we have

( U + K )at 1 m = ( U + K )r

GM ( 11m ) 1 GM ( 11m ) 1 + ( 11m ) v ′ 2 = − + ( 11m ) v ′′ 2 R 2 (R 2) 2

⇒

2GM v ′′ 2 GM v ′ 2 − + =− + R 2 R 2

Mechanics II_Chapter 4_Part 1.indd 54

…(5)

− ⇒

GmM 1 2 GmM 1 2 + μ(0) = − + μ vr 1 2 r 2

1 2 ⎛1 ⎞ μvr = Gmm ⎜ − 1 ⎟ ⎝r ⎠ 2

…(1)

09-Feb-21 6:31:46 PM

Chapter 4: Gravitation and Satellites

where, μ is the reduced mass given by

μ=

From equation (1), relative speed at the time of collision,

m( m) m = m+m 2

Substituting in equation (1) with m = 1 kg , we get ⎛1 ⎞ vr = 2 G ⎜ − 1 ⎟ ⎝r ⎠

…(2)

⇒

dr ⎛1 ⎞ − = 2 G ⎜ − 1⎟ ⎝r ⎠ dt

⇒

dt = −

⇒

0.1

∫ dt = − 2 G ∫ 0

1.0

r dr 1− r

∫

SOLUTION

(a) By Law of Conservation of Energy

Substituting r = sin 2 θ , we get

( U + K )surface = ( U + K )orbit

dr = 2 sin θ cos θ dθ ⇒ ⇒ ⇒ ⇒

2

sin θ

∫ cos θ 2 sin θ cosθ dθ I = 2 sin θ dθ = ( 1 − cos 2θ ) dθ ∫ ∫ I=

⇒

2

sin ( 2θ ) 2 I = θ − sin θ cos θ I =θ−

I = sin −1 r − r ( 1 − r ) t=−

1 2 G

( sin −1

r − r (1 − r ) )

0.1

t=

1 ⎡ −1 −1 0.1 + ( 0.1 )( 0.9 ) ⎤⎦ ⎣ sin ( 1 ) − 0 − sin 2 G

⇒

t=

1 ⎛π ⎞ ⎜ − 0.32 + 0.3 ⎟⎠ 2 G⎝2

⇒

t=

1 2 G

GMm1 1 ⎛ GMm1 ⎞ m1v12 = − −⎜ − ⎟ R⎞ ⎝ 2 R ⎠ ⎛ 2⎜ R + ⎟ ⎝ 4⎠ 6 GM ⇒ v12 = 5 R GMm2 GMm2 1 Similarly, − + m2 v22 = − R⎞ R 2 ⎛ 2⎜ R + ⎟ ⎝ 6⎠ ⇒

8 GM 7 R From equations (1) and (2), we get v12 v22

t = 95 × 10 s

Mechanics II_Chapter 4_Part 1.indd 55

⇒

…(1)

GMm2 1 ⎛ GMm2 ⎞ m2 v22 = − −⎜ − ⎟ R⎞ ⎝ 2 R ⎠ ⎛ 2⎜ R + ⎟ ⎝ 6⎠

⇒ v22 =

( 1.55 ) 3

GMm1 1 GMm1 + m1v12 = − R⎞ 2 R ⎛ 2⎜ R + ⎟ ⎝ 4⎠

⇒

1

⇒

⇒

⇒ −

2

But sin 2 θ = r ⇒

vr = 2.45 × 10 −5 ms −1 2

(a) the ratio of the speeds of projection from the surface of the earth. (b) the angular speed of S2 as observed by an astronaut in S1 when their radius vectors are making an angle 120° with each other.

r dr 1− r

Let the integral be I=

So, the speed of each sphere is

Two satellites S1 and S2 are to be set into the orbits R R of and above the earth’s surface respectively. 4 6 They revolve around the earth in coplanar circular orbit in the opposite sense. Determine

⎛1 ⎞ 2 G ⎜ − 1⎟ ⎝r ⎠ 1

1 ⎞ − 1 = 4.9 × 10 −5 ms −1 ⎝ 0.1 ⎟⎠

( 6.67 × 10 −11 ) ⎛⎜

vr = 2

PROBLEM 14

dr

t

4.55

=

v1 = v2

…(2)

21 20 21 20

09-Feb-21 6:31:56 PM

4.56

JEE Advanced Physics: Mechanics – II

(b) r1 = R +

PROBLEM 15

R 5R R 7R = and r2 = R + = 4 4 6 6

P = ( 0 , r2 ) ⎛ 3 r1 r ⎞ and Q = ( r2 cos 30°, − r1 sin 30° ) = ⎜ , − 1⎟ ⎝ 2 2⎠ 2

3 r12 ⎛ r ⎞ + ⎜ r2 + 1 ⎟ = 2.1 R ⎝ 4 2⎠

⇒ PQ =

Applying Lami’s Theorem, we get

SOLUTION

Orbital speed of satellite is

r2 r1 PQ = = sin ( 90° − β ) sin ( 90° − α ) sin ( 120° )

GM …(1) a Applying the Law of Conservation of Angular Momentum at P and Q , we get v0 =

r ⇒ cos α = 1 sin ( 120° ) = 0.5 and PQ cos β = Now, ω r =

r2 sin ( 120° ) = 0.48 PQ

mav0 = mvr av0 …(2) r Applying the Law of Conservation of Mechanical Energy at P and Q , we get

v2 cos α + v1 cos β PQ

⇒

GM GM cos α + cos β r2 r1

⇒ ωr =

An earth satellite is revolving in a circular orbit of radius a with velocity v0 . A gun is in the satellite and is aimed directly towards the earth. A bullet is fired v from the gun with muzzle velocity 0 . Neglecting 2 resistance offered by cosmic dust and recoil of gun, calculate maximum and minimum distance of bullet from the centre of earth during its subsequent motion.

1 ⎛ 2 v02 ⎞ GMm 1 GMm m ⎜ v0 + ⎟ − = mv 2 − ⎝ ⎠ 2 4 2 a r

PQ gR2 cos α + r2

⇒ ωr =

gR2 cos β r1

2.1R

Since, r1 =

v=

…(3)

⇒

5 2 GM v 2 GM v0 − = − 8 a 2 r

5R 7R and r2 = 4 6

v

a

gR2 cos α = 3666 r2 ⎛ 9.81 × 6400 × 10 3 ⎞ gR2 cos β = ⎜ ⎟ 0.48 54 r1 ⎝ ⎠

⇒

Substituting these values in equation (3), we get

ωr =

3666 + 3402 2.1 × 6400 × 10

Mechanics II_Chapter 4_Part 1.indd 56

3

= 5.25 × 10 −4 rads −1

v0 2

P

Substituting values of v and v0 from equations (1) and (2), we get 5 GM GM a 2 − = 2 a 8 a r

gR2 cos β = 3402 r1

v0

r

⎛ 9.81 × 6400 × 10 3 ⎞ gR2 cos α = ⎜ ⎟ 0.5 76 r2 ⎝ ⎠ ⇒

vnet

Q

⎛ GM ⎞ GM ⋅⎜ − ⎝ 2 a ⎟⎠ r

a 1 3 = 2− 8 a 2r r

⇒

−

⇒

−3 r 2 = 4 a 2 − 8 ar

⇒

3 r 2 − 8 ar + 4 a 2 = 0

09-Feb-21 6:32:04 PM

Chapter 4: Gravitation and Satellites

⇒ ⇒

8 a ± 64 a 2 − 48 a 2 6 8a ± 4a r= 6 r=

Mechanics II_Chapter 4_Part 1.indd 57

4.57

2a 3 Hence, the maximum and minimum distances are 2a 2a respectively. and 3 ⇒

r = 2 a and

09-Feb-21 6:32:06 PM

4.58

JEE Advanced Physics: Mechanics – II

PRACTICE EXERCISES SINGLE CORRECT CHOICE TYPE QUESTIONS This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 8GM 2 2GM 2 1. The gravitational force between a point like mass M (A) (B) 2 R 3 R2 and an infinitely long, thin rod of linear mass density λ at perpendicular distance L from M is

2.

3.

(A)

MGλ L

(B)

(C)

2MGλ L

(D) infinite

(A)

r 3ω R

(B)

r 2ω 3 R

(C)

r 3ω 2 R2

(D)

r 2ω 2 R

Consider a particle of mass m released from a point lying on the axis of a ring of mass M and radius r and at a distance r from the centre of ring. Assuming that the particle is moving under the gravitational attraction of the ring to reach the centre of the ring, then the velocity of the particle at the centre of the ring is GM 2GM (B) (A) r r 2GM r( 2 + 1)

(D)

2GM ⎛ 1 ⎞ ⎜1− ⎟ r ⎝ 2⎠

A homogeneous bar of length L and mass M is at a distance h from a point mass m as shown. The force on m is F .

3GM 2 2R 2

GMm

( h + L )2

GMm (C) F = h( h + L ) 5.

(B)

F=

GMm h2

GMm (D) F = L2

A uniform ring of mass M and radius R is placed directly above a uniform sphere of mass 8M and of same radius R. The centre of the ring is at a distance of d = 3 R from the centre of the sphere. The gravitational attraction between the sphere and the ring is

Mechanics II_Chapter 4_Part 2.indd 58

3GM 2 R2

A hole is drilled from the surface of earth to its centre. A particle is dropped from the surface of the earth. The speed of the particle when it reaches the centre of the earth in terms of its escape velocity ( ve ) at the surface of earth is ν (A) e (B) ν e 2 νe (C) 2ν e (D) 2

7.

Two identical solid spheres of radius R placed in contact with each other, the gravitational attraction between them is proportional to (A) R2 (C) R 4

8.

(C) 8g 9.

(B) R −2 (D) R −4

A planet has a mass and density both equal to eight times the mass and average density of the earth. If g be the acceleration due to gravity at the earth’s surface, then acceleration due to gravity at the planet’s surface is (B) 4g (A) 2g (D) 16g

Gravitational force between two charged bodies is F1 and electrostatic force between them is F2 . They are at distance r from each other. Now the air between them is sucked out and distance is increased to 2r . Then the new forces F1′ and F2′ are such that F1 4 F2 (C) F2′ > 4

(A) F1′ = (A) F =

(D)

6.

The radius of a planet is R . A satellite revolves around it in a circle of radius r with angular speed ω . The acceleration due to gravity on planet’s surface will be

(C) 4.

1 MGλ 2 L

(C)

F2 4 F1 (D) F1′ > 4

(B)

F2′ =

10. If g h and g d be the accelerations due to gravity at height h and at depth d , above and below the surface of earth respectively . Assuming h R and d R and if g h = g d then, (A) (B) (C) (D)

d=h d = 2h h = 2d data is insufficient to arrive at a conclusion

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Chapter 4: Gravitation and Satellites 11. A hemispherical shell having uniform mass density is shown in Figure. The direction of gravitational field intensity at point P will be along

(A) a (C) c

(B) b (D) d

12. What should be the velocity of a body thrown vertically upwards from the surface of earth (mass M and radius R ) so that it may reach at a height of 7R from the surface ? (A)

20GM 11R

(B)

20GM 9R

(C)

17GM 9R

(D)

7 GM 4R

13. Rate of change of weight near the earth’s surface varies with height ( h ) as (B) h o (A) h (C) h −1

(D)

1 h2

14. A uniform spherical planet of radius R has acceleration due to gravity at its surface g . Points P and Q located inside and outside the planet respectively g have same acceleration due to gravity equal to . 4 Maximum possible separation between P and Q is 7R 9R (B) (A) 4 4 3R (C) (D) None of these 2 −2

15. The radius of the earth is 6400 km and g = 10 ms . In order that the body of 5 kg weighs zero at the equator, the angular speed of the earth should be (A) (C)

1 rads −1 80 1 rads −1 800

1 rads −1 400 1 (D) rads −1 1600 (B)

16. If the radius of the earth were increased by a factor of 2, keeping the mass constant, then the factor by which its density has to be changed to keep g the same is 1 1 (B) (A) 8 4 1 (C) (D) 4 2

Mechanics II_Chapter 4_Part 2.indd 59

4.59

17. If radius of a solid sphere is decreased to half, keeping density of sphere unchanged, the slope of E-r graph will (A) remain unchanged (B) become two times (C) become four times 1 (D) remain th 8 18. The potential energy of a body of mass m is U = ax + by the magnitude of acceleration of the body will be ab ⎛ a+b⎞ (B) ⎜ (A) ⎝ m ⎟⎠ m (C)

a2 + b 2 m

(D)

a2 + b 2 m

19. The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface of earth for h R (A) is constant (B) increases parabolically with h (C) increases linearly with h (D) decreases parabolically with h 20. The gravitational field at a distance r from the centre of the sphere (having radius R , density ρ ) inside it is (A)

4 ρGπ r 3

(B)

4 ρGr 2 3

4 ρGR3 ρGR3 (D) πr 3r 2 21. The ratio of the energy required to raise a satellite upto a height h above the earth (of radius R) to the kinetic energy of the satellite into the orbit there is (C)

(A) h : R (C) 2h : R

(B) R : 2 h (D) R : h

22. A mass m is raised Slowly from a distance 2R from surface of earth to 3R . Work done to do so against gravity will mgR mgR (B) (A) 10 11 mgR mgR (C) (D) 12 14 23. A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at a distance 3R from the centre R of the sphere. A spherical cavity of radius is now 2 made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle. The ratio

F2 is F1

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JEE Advanced Physics: Mechanics – II

R/2

26. A particle of mass M is placed at the centre of a uniform spherical shell of mass 3M and radius R . The gravitational potential at the surface of the shell is

A

R

3R

(A) (C)

9 50 3 50

41 50 11 (D) 25 (B)

R is made in a solid sphere 2 of radius R , mass M . The gravitational field at the centre of the hole due to the remaining mass is

24. A spherical hole of radius

(A) −

GM R

(B)

−

3GM R

(C) −

2GM R

(D) −

4GM R

27. Two identical thin rings each of radius R are coaxially placed at a distance R . If the rings have a uniform mass distribution and each has mass m1 and m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is (A) zero (C)

(C)

GM 8R2 GM (D) R2

(B)

GM 2R 2

25. A ring of radius R , mass m and a solid sphere of same mass m and same radius R are placed with their centres on positive x-axis. The observer is moving from some finite distance on negative x-axis towards positive x-axis and the plane of the ring is perpendicular to x-axis. If the observer moves only upto the surface of the solid sphere, then the net gravitational field varies with the distance moved along x-axis as m

m R

(D)

Gm ( m1 − m2 )

(

2 −1

)

2R Gmm1 ( 2 + 1 ) m2 R

28. An infinite number of masses, each of 1 kg, are placed on the positive x -axis at 1 m, 2 m, 4 m, 8 m, … from the origin. The value of the gravitational field at the origin due to this distribution is 4G (B) (A) 2G 3 3G (C) (D) ∞ 4

R

(A) zero

Gm 2 ( m1 + m2 ) R

(B)

29. For a given density of planet, the orbital period of a satellite near the surface of planet of radius R is proportional to 1

3

(B)

(A) R 2 (C) R

−

1 2

R2

(D) R0

30. For a planet revolving around sun in an elliptical path as shown

+ x-axis

R

(A) E

(B) E distance

distance

(C) E

(D) E distance

Mechanics II_Chapter 4_Part 2.indd 60

distance

(A) v1r1 = v2 r2

(B)

2v1r1 = v2 r2

(C) v1r1 = 2v2 r2

(D) 2v1r1 = 3v2 r2

31. The magnitude of gravitational potential energy of the earth-satellite system is U with zero potential energy at infinite separation. If mass of satellite  mass of earth and the kinetic energy of satellite is K , then U (B) K = (A) K = 2U 2 (C) K = U (D) K = 4U

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Chapter 4: Gravitation and Satellites 32. A particle on earth’s surface is given a velocity equal to its escape velocity. The total mechanical energy of the particle is (A) infinite (B) negative (C) positive (D) zero 33. A small ball of mass m is released at a height R above the earth surface, as shown in the given figure. The maximum depth of the ball to which it goes inside the earth through a narrow groove before coming to R . The groove, contains an ideal rest momentarily is 2 spring of spring constant k and natural length R . The value of k , if R is the radius of earth and M mass of earth, is

3GMm 6GMm (B) R3 R3 9GMm 12GMm (C) (D) R3 R3 34. A projectile is fired upwards from the surface of the earth with a velocity kve where ve is the escape velocity and k < 1 . If r is the maximum distance from the centre of the earth to which it rises and R is the radius of the earth, then r equals R R (A) 2 (B) k 1 − k2 2R 2R (C) (D) k2 1 − k2 (A)

35. The binding energy of a particle at the surface of earth is E . If a kinetic energy greater than E is given to this particle, then the total energy of particle will be (A) zero (B) infinite (D) < 0 (C) > 0 36. Two concentric spherical shells are as shown in figure. The V -r graph will be m2 m1

r1

Mechanics II_Chapter 4_Part 2.indd 61

r2

(A) v

(B) v r

r

(C) v

(D) v r

r

37. A star with a mass of more than three times that of the sun contracts so much upon cooling that it loses its capacity to radiate; neither material particles nor light are able to overcome its gravitational field (such an object is called a “black hole”). Taking mass of sun to be 2 × 10 30 kg , the radius of such an object could be (approximately) (A) 6 km (B) 7 km (C) 8 km (D) 9 km 38. Four particles of equal mass M move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is GM R

(A)

⎡ GM ( )⎤ ⎢⎣ R 2 2 + 1 ⎥⎦

(C)

(B)

GM ⎤ ⎡ ⎢⎣ 2 2 R ⎥⎦

(D)

⎡ GM ⎛ 2 2 + 1 ⎞ ⎤ ⎢ ⎜ ⎟⎥ 4 ⎠⎦ ⎣ R ⎝

39. The gravitational field in a region of space is given by ur E = 4$i + $j Nkg −1 . The work done by this field is zero when a particle is moved along the line

(

)

(A) y + 4 x = 2

(B)

4y + x = 6

(C) x + y = 5

(D) y − 4 x = 2

40. The speed of a planet in an elliptical orbit with semimajor axis a about sun of mass M at a distance r from sun is (A)

⎛ 2 1⎞ GM ⎜ − ⎟ ⎝ r a⎠

(B)

⎛ 1 1⎞ GM ⎜ − ⎟ ⎝ r a⎠

(C)

⎛ 1 2⎞ GM ⎜ − ⎟ ⎝ r a⎠

(D)

GMr 2a2

41. A body is projected up with a velocity equal to three fourth of the escape velocity from the surface of the earth. The maximum distance it reaches from the centre of the earth is (Radius of the earth is R )

09-Feb-21 6:28:55 PM

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JEE Advanced Physics: Mechanics – II 10 R 9 9R (C) 8

16 R 7 10 R (D) 3

(B)

(A)

42. The work done in slowly lifting a body from earth’s surface to a height R (radius of earth) is equal to two times the work done in lifting the same body from earth’s surface to a height h , then R R (B) h = (A) h = 4 6 R R (C) h = (D) h = 3 2 43. A particle of mass m starts moving from rest along the line y = b with constant acceleration a . The areal velocity of the position vector of the particle at time t is abt 2 a 2bt (D) 2m

(A) constant (C)

(B)

abt 2m

44. A particle is placed in a field characterised by a value of gravitational potential given by V = − kxy , where k r is a constant. If Eg is the gravitational field then, r (A) Eg = k xi$ + y $j and is conservative in nature. r (B) Eg = k yi$ + x $j and is conservative in nature. r (C) Eg = k xi$ + y $j and is non-conservative in nature. r (D) Eg = k yi$ + x $j and is non-conservative in nature.

( ( ( (

) ) ) )

45. The gravitational potential of two homogeneous spherical shells A and B of same surface density at their respective centres are in the ratio 3 : 4 . If the two shells coalesce into single one such that surface charge density remains same, then the ratio of potential at an internal point of the new shell to the potential inside shell A is (B) 4 : 3 (A) 3 : 2 (C) 5 : 3 (D) 5 : 6

R 2R

(A) −

3GM 7R

(B)

−

9GM 14 R

(C) −

5GM 7R

(D) −

7GM 14 R

48. A particle is projected vertically upwards with a velocity gR , where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is R (B) R (A) 2 5R (C) 2R (D) 4 49. Two identical spherical balls each of mass m are placed as shown in figure. The variation of gravitational intensity g along the x-axis is best represented in

(A)

(B)

(C)

(D)

46. A satellite is revolving round the earth in an orbit of radius r with time period T . If the satellite is revolving round the earth in an orbit of radius r + Δr (Δr  r) with time period T + ΔT (ΔT  T) then, (A) (C)

ΔT 3 Δr = T 2 r ΔT Δr = T r

ΔT 2 Δr = T 3 r ΔT Δr (D) =− T r

(B)

47. The gravitational potential at the centre of a sphere of radius 2R having a concentric spherical cavity of radius R and net mass M is

Mechanics II_Chapter 4_Part 2.indd 62

50. A mass is taken from surface to a height h . The change in potential energy in this process is equal to the change in potential energy if it is now taken from that point to infinity. Then, (B) h = 2R (A) h = R (C) h =

3R 2

(D) h = 4 R

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Chapter 4: Gravitation and Satellites 51. The time period of an earth satellite close to the surface of the earth is 83 minute. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be (A) 83 minute (B) 83 8 minute (C) 664 minute (D) 249 minute 52. A particle of mass m moves along a straight line with constant speed in the x direction at a distance b from the x-axis as shown and sweeps areas A1 and A2 , then

4.63

56. A particle is projected from the surface of earth with velocity equal to its escape velocity at 45° with horizontal. The angle of its velocity with horizontal at height h = R , where horizontal at some point means a line parallel to tangent on earth just below that point, is ⎛ 1⎞ cos −1 ⎜ ⎟ ⎝ 3⎠

(A) 30°

(B)

(C) 60°

⎛ 1⎞ (D) cos −1 ⎜ ⎟ ⎝ 4⎠

57. A shell of mass M and radius R has a point mass m placed at a distance r from its centre. The gravitational potential energy U ( r ) vs r will be (A) O

(A) (B) (C) (D)

A1 > A2 if t4 − t3 > t2 − t1 A1 > A2 if t4 − t3 < t2 − t1 A1 = A2 if t4 − t3 < t2 − t1 A1 is always equal to A2

53. Two particles each of mass m are revolving in circular orbits of radius r = 5R in opposite directions. They collide perfectly inelastically and fall to the ground. The speed of combined mass on striking the ground is (A) 2 2v0 (C) 2v0

(B) 2v0 (D) v0

54. The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is 4.4 × 1015 m 2s −1 . The area enclosed by the earth’s orbit is approximately (A) 1 × 10 22 m 2

(B)

3 × 10 22 m 2

(C) 5 × 10 22 m 2

(D) 7 × 10 22 m 2

M and length R is placed 2 normally on surface of earth as shown. The mass of earth is M and its radius is R . Then the magnitude of gravitational force exerted by earth on the rod is

55. A uniform thin rod of mass

r

U(r)

U(r)

(C)

R

(B) O

r

O

R

r

(D) O

R

r

U(r)

U(r)

58. In a double star system, the masses of the two stars are M and 3M. The orbit radius of the lighter star is R. The time period of each star is (A) 8π

R3 GM

(B) 16π

R3 GM

(C) 8π

R3 27GM

(D) None of these

59. A particle of mass m is projected upwards with velocity half the escape velocity. At the highest point, the potential energy of the particle is GMm GMm (B) − (A) − 2R 4R 2GMm 3GMm (D) − 3R 4R 60. A satellite of mass m is moving around the earth in a circular orbit of radius r . If the mass of the earth is M , then the energy required to take the satellite to an orbit of radius 3r is (C) −

M 2

(A)

GM 2 4R2

(B)

GM 2 2R 2

(A)

GMm r

(B)

GMm 2r

(C)

4GM 2 9R 2

(D)

GM 2 8R2

(C)

GMm 3r

(D)

GMm 6r

Mechanics II_Chapter 4_Part 2.indd 63

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JEE Advanced Physics: Mechanics – II

61. Three particles each having a mass of 100 g are placed on the vertices of an equilateral triangle of side 20 cm . The work that has to be done to increase the side of this triangle to 40 cm is G = 6.67 × 10 −11 Nm 2kg −2 (A) 5 × 10 −8 J (B)

(

)

4 × 10 −10 J

(C) 5 × 10 −12 J (D) 5 × 10 −15 J 62. A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in figure. The force F exerted by the sphere on a particle of mass m located at a distance r from the centre of sphere varies as (0 ≤ r ≤ ∞)

R1

(A)

R2

64. The gravitational field due to a mass distribution K is E = 3 in the x-direction, where K is a constant. x Taking the gravitational potential to be zero infinity, its value at a distance x is K K (A) (B) x 2x K K (C) (D) 2 x 2x 2 65. A system consists of n identical particles each of mass m . The total number of interactions possible is n( n + 1) 2 n( n − 1) (D) 2

(A) n ( n + 1 )

(B)

(C) n ( n − 1 )

66. A straight rod of length l extends from x = α to x = l + α . If the mass per unit length is ( a + bx 2 ) . The gravitational force it exerts on a point mass m placed at x = 0 is given by

(B) 1 ⎞ ⎛ ⎛ 1 ⎞ (A) Gm ⎜ a ⎜ − ⎟ + bl ⎟ ⎝ ⎝ α α +l⎠ ⎠

(C)

(D)

Gm ( a + bx 2 ) l2 1 ⎞ ⎛ ⎛1 ⎞ (C) Gm ⎜ α ⎜ − ⎟ + bl ⎟ ⎝ ⎝ a a+l⎠ ⎠

(B)

1⎞ ⎛ ⎛ 1 ⎞ − ⎟ + bl ⎟ (D) Gm ⎜ a ⎜ ⎝ ⎝α+l α⎠ ⎠ 63. Imagine the acceleration due to gravity on earth is 10 ms −2 and on mars is 4 ms −2 . A traveler of mass 60 kg goes from earth to mars by a rocket moving with constant velocity. If effect of other planets is assumed to be negligible, which one of the following graphs shown the variation of weight W ( in N ) of traveler with time t (in second)

(A) A (C) C

Mechanics II_Chapter 4_Part 2.indd 64

(B) B (D) D

67. A thin spherical shell of mass M and radius R has a small hole. A particle of mass m is released at the mouth of the hole. Then the particle m

R

M

(A) will execute simple harmonic motion inside the shell. (B) will oscillate inside the shell, but the oscillations are not simple harmonic. (C) will not oscillate, but the speed of the particle will go on increasing. (D) stays at rest in its equilibrium position.

09-Feb-21 6:29:23 PM

Chapter 4: Gravitation and Satellites 68. A small meteorite of mass m travelling towards the centre of earth strikes the earth at the equator. The earth is a uniform sphere of mass M and radius R. The length of the day was T before the meteorite struck. After the meteorite strikes the earth, the length of day increases by (A) (C)

5mT 2M 4 mT 5M

mT M M (D) 3 mT

(B)

69. A ring of radius R has a total mass M distributed non-uniformly over its circumference. A point mass m is placed at the centre ( C ) of the ring. The work done in taking away this point mass from centre to infinity is GMm GMm (B) (A) − R R GMm GMm (C) (D) − 2R 2R

4.65

74. A particle is projected upward from the surface of earth with a speed equal to the orbital speed of a satellite near the earth’s surface. If R be the radius of earth, then the height to which it would rise is R 2R (B) (A) 2 (C) R

(D) 2R

75. Two small balls of mass m each are suspended side by side by two equal threads of length L . If the distance between the upper ends of the threads be a, the angle θ that the threads will make with the vertical due to attraction between the balls is

70. The gravitational self energy of a spherical shell of mass M and radius R is (A) −

GM 2 4R

(B)

−

GM 2 R

(C) −

3 GM 2 5 R

(D) −

GM 2 2R

71. An artificial satellite is moving in a circular orbit around the earth at some height with a speed equal to half the magnitude of the escape velocity from the surface of earth. Suppose the satellite is stopped suddenly in its orbit and allowed to fall freely. On reaching earth its speed will be (A)

gR

(C) 3 gR

(B)

2 gR

(B)

⎡ ( a − x )2 g ⎤ (C) tan −1 ⎢ ⎥ ⎣ Gm ⎦

⎡ ( a2 − x 2 ) g ⎤ (D) tan −1 ⎢ ⎥ Gm ⎣ ⎦

⎡ Gm ⎤ tan −1 ⎢ 2 ⎥ ⎣ (a − x) g ⎦

76. A planet is moving in an elliptical orbit around the sun as shown in figure. At positions P and Q it has respective speeds v1 and v2 and respective distances v r1 and r2 from the sun, then 1 equals v2 V2

(D) 5 gR

72. The height above the surface of earth where the total energy of a satellite is equal to its potential energy at a height of 2R from surface of earth (if R is the radius of the earth) is R R (A) (B) 4 2 (C) 2R (D) 4R 73. Two particles having masses m1 and m2 start moving towards each other from the state of rest from infinite separation. Their relative velocity of approach when they are interacting gravitationally at a separation r will be (A)

G ( m1 + m2 ) r

(B)

2G ( m1 + m2 ) r

(C)

3G ( m1 + m2 ) r

(D)

4G ( m1 + m2 ) r

Mechanics II_Chapter 4_Part 2.indd 65

⎡ (a − x)g ⎤ (A) tan −1 ⎢ ⎣ Gm ⎥⎦

r1

P

S

r2

Q

V1

(A) (C)

r1 r2

(B) r1 r2

r2 r1

⎛r ⎞ (D) ⎜ 1 ⎟ ⎝ r2 ⎠

2

77. A body is projected away from the earth with a speed 3ve where ve is the escape velocity. The speed of the body at infinity will be (A) ve

(B)

(C) 2ve

(D) 2 2ve

2ve

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JEE Advanced Physics: Mechanics – II

78. Two air bubbles in water (A) repel each other (B) attract each other (C) do not exert any force on each other (D) may attract or repel depending upon the distance between them 79. The figure shows the motion of a planet around the sun in an elliptic orbit with the sun at one focus. The shaded areas A and B can be assumed to be equal. If t1 and t2 represent the times taken by the planet to move from a to b and from c to d respectively, then

83. The period of revolution of an earth satellite close to the surface of earth is 90 minutes. The time period of another satellite in an orbit at an altitude of three times the radius of earth is (A) 90 8 min (C) 360 min

84. Two particles of equal mass m go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

85. (A) t1 < t2 (C) t1 = t2

(B) t1 > t2 (D) Data Insufficient

(A) (B) (C) (D)

T

Gm 2R

(C) V =

1 Gm 2 R

(D) V =

4Gm R

The minimum energy required to launch a satellite of mass m from the surface of earth (of mass M, radius R) in a circular orbit at an altitude 2R is GmM 3R 2GmM (D) 3R

(B)

Consider a circular disc of mass M , radius R with surface mass density σ. The gravitational potential due to the disc at a point P lying on its axis at distance r from the centre is (A) −2πσ Gr −4πσ Gr

r ⎛ ⎞ (C) −2πσ Gr ⎜ 1 − ⎟ 2 ⎝ r + R2 ⎠

T mπ ( ra + rp ) ra rp

r ⎛ ⎞ (D) −4πσ Gr ⎜ 1 − 2 2 ⎟ ⎝ r +R ⎠

2T mπ ( ra + rp ) ra rp

(B)

m

(D) m0

82. The time period of an artificial satellite in a circular orbit of radius R is 2 days and its orbital velocity is v0 . If time period of another satellite in a circular orbit is 16 days, then (A) its radius of orbit is 4R and orbit velocity is v0 (B) its radius of orbit is 2R and orbital velocity is v0 v (C) its radius of orbit is 2R and orbital velocity is 0 2 v0 (D) its radius of orbit is 4R and orbital velocity is 2

Mechanics II_Chapter 4_Part 2.indd 66

(B) V =

(B)

4T 81. A satellite of mass m moves along an elliptical path around the earth. The areal velocity of the satellite is proportional to

(C)

86.

2mπ ( ra + rp ) ra rp

1 m2

1 1 2R Gm

GmM 2R 5GmM (C) 6R

mπ ( ra + rp ) ra rp

(A) m −1

(A) V =

(A)

80. A satellite of mass m orbits the earth in an elliptical orbit having aphelion distance ra and perihelion distance rp . The period of the orbit is T . The semi-major and semira + rp minor axes of the ellipse are and rp ra respec2 tively. The angular momentum of the satellite is

(B) 270 min (D) 720 min

87.

Mass M is uniformly distributed only on curved surface of a thin hemispherical shell. A , B and C are three points on the circular base of hemisphere, such that A is the centre. Let the gravitational potential at points A , B and C be VA , VB , VC respectively. Then

(A) VA > VB > VC (B) VC > VB > VA (C) VB > VA and VB > VC (D) VA = VB = VC

09-Feb-21 6:29:43 PM

Chapter 4: Gravitation and Satellites 88.

An electron moves in a circular orbit at a distance from a proton with kinetic energy E . To escape to infinity, the energy which must be supplied to the electron is (B) 2E (A) 0.5E (C) E

89.

90.

(D)

93.

A body of mass m is kept at a small height h above the ground. If the radius of the earth is R and its mass is M, then the potential energy of the body and earth system, with the reference position being taken at infinity, is

4G ( Me + Mm ) r

(B)

4G r

( Me + Mm )

(C)

2G ( Me + Mm ) r

(D)

2G r

( Me + Mm )

The potential energy of gravitational interaction of a point mass m and a thin uniform rod of mass M and length l , if they are located along a straight line at a distance a from each other is (A) U = −

GMm ⎛ a+l⎞ log e ⎜ ⎝ a ⎟⎠ a

GMm + mgh R

(B)

−GMm + mgh R

1 ⎞ ⎛1 (B) U = GMm ⎜ − ⎝ a a + l ⎟⎠

(C)

GMm − mgh R

(D)

−GMm − mgh R

(C) U = −

GMm ⎛ a+l⎞ log e ⎜ ⎝ a ⎟⎠ l

(D) U = −

GMm a

The gravitational field in a r I = 2iˆ + 3 ˆj Nkg −1 .The work ⎛ particle from ( 1, 1 ) m to ⎜ 2, ⎝ 3 y + 2x = 5 is

(

)

region is given by done in moving a 1⎞ ⎟ m along the line 3⎠

94.

+20 J (D) +18 J

(B)

(C) −15 J

Three solid spheres each of mass m and radius R are released from the position shown in figure. The speed of any one sphere at the time of collision would be

d

95.

d

(A)

⎛1 3⎞ Gm ⎜ − ⎟ ⎝ d R⎠

(B)

⎛3 1⎞ Gm ⎜ − ⎟ ⎝ d R⎠

(C)

⎛ 2 1⎞ Gm ⎜ − ⎟ ⎝ R d⎠

(D)

⎛ 1 2⎞ Gm ⎜ − ⎟ ⎝ R d⎠

The masses and the radii of earth and moon are respectively Me, Re and Mm , Rm. The distance between the centres is r. At what minimum velocity a particle of mass m be projected from the mid point of the distance between their centres so that it may escape into space?

Mechanics II_Chapter 4_Part 2.indd 67

An object of mass 100 kg starts to fall from a height equal to radius of earth ( R ) . Its speed when it has lost half the initial height is [ M is mass of earth] (A)

100G 3R

(B)

GM 6R

(C)

50G 3R

(D)

GM 3R

The magnitude of the potential energy per unit mass of the object at the surface of earth is E . Then the escape velocity of the object is (A)

2E

(B)

2 E

(C)

E

(D)

E 2

96.

If g is the acceleration due to gravity at the surface of the earth, then the energy required to launch a satellite of mass m from the surface of the earth into a circular orbit at an altitude of 2R , R being the radius of the earth, is mgR mgR (B) (A) 6 3 2mgR 5 (D) mgR (C) 3 6

97.

If the period of revolution of an artificial satellite just above the earth’s surface is T and the density of earth is ρ , then ρT 2 equals

d

92.

(A)

(A)

(A) Zero 91.

2E

4.67

(A) (C)

3π G 2π G

3π 2G π (D) G (B)

09-Feb-21 6:29:54 PM

4.68 98.

JEE Advanced Physics: Mechanics – II Consider a massur m0 enclosed by a closed imaginary surface S. Let E be the gravitational r field intensity due to m0 at the surface element dS directed as outward normal to it. The surface integral of the gravitational field over S is (A) −m0G (B) −4π m0G (C) −

99.

m0G 4π

(D) None of above

If Me is the mass of earth and Mm is the mass of moon ( Me = 81Mm ) . The potential energy of an object of mass m situated at a distance R from the centre of earth and r from the centre of moon, will be ⎛ R ⎞ 1 + r⎟ 2 (A) −GmMm ⎜ ⎝ 81 ⎠R

(B)

⎛ 81 1 ⎞ (C) −GmMm ⎜ + ⎝ R r ⎟⎠

⎛ 81 1 ⎞ (D) GmMm ⎜ − ⎝ R r ⎟⎠

(C) −3 Jkg

(D) −7 Jkg

−1

101. A research satellite of mass 200 kg circles the Earth in 3R , where R is the radius an orbit of average radius 2 of the Earth. Assuming gravitational pull on a mass 1 kg on Earth’s surface to be 10 N, the pull on the satellite will be (B) 889 N (A) 880 N (C) 890 N (D) 892 N 102. A satellite is revolving round the earth with orbital speed v0 . If it stops suddenly, the speed with which it will strike the surface of earth (if ve is the escape speed of the particle from the earth’s surface), would be v2 (B) v0 (A) e v0 (C)

ve2 − v02

(D)

ve2 − 2v02

103. A planet is moving in an elliptic orbit. If T , U , E and L stand respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the centre of force, then (A) T is conserved (B) U is always positive (C) E is always negative (D) magnitude of L is conserved but its direction changes continuously

Mechanics II_Chapter 4_Part 2.indd 68

T⎛ e⎞ ⎜ 1 − ⎟⎠ 2⎝ π

(B)

⎛1 e ⎞ T⎜ − ⎝ 4 2π ⎟⎠

⎛ e ⎞ (C) ⎜ − 1 ⎟ T ⎝π ⎠

(D)

Tπ e

(A)

105. Inside a uniform sphere of mass M and radius R, a R is made in the sphere as shown. cavity of radius 3 Select the correct statement.

⎛ 81 1 ⎞ −GmMe ⎜ + ⎝ r R ⎟⎠

100. A person brings a mass of 2 kg from A to B . The increase in kinetic energy of the mass is 4 J and the work done by the person on the mass is −10 J . The potential difference between A and B is ΔV , given by (B) 7 Jkg −1 (A) 4 Jkg −1 −1

104. A planet revolves around the sun in an elliptical orbit of eccentricity e . If T is the time period of the planet, then the time spent by the planet between the ends of the minor axis close to the sun is

A R

B R/3 C

(A) Gravitational field inside the cavity is uniform (B) Gravitational field inside the cavity is non-uniform (C) The escape velocity of a particle projected from 88GM 15R (D) Escape velocity is defined for earth and particle system only point A is

106. The angular momentum ( L ) of earth revolving round the sun in circular orbit is proportional to r n, where r is the orbital radius of the earth. The value of n is (A) 1 (B) 2 (C) −

1 2

(D)

1 2

107. Consider a thin uniform spherical layer of mass M and radius R. The potential energy of gravitational interaction of matter forming this shell is (A) −

GM 2 R

(B)

−

1 GM 2 2 R

(C) −

3 GM 2 2 R

(D) −

2 GM 2 3 R

108. By what percent the energy of a satellite has to be 3 increased to shift it from an orbit of radius r to r ? 2 (A) 15% (B) 20.3% (C) 33.3% (D) 66.7%

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4.69

Chapter 4: Gravitation and Satellites 109. For a spherical shell of mass M and radius R, the energy required to double its radius is (A)

GM 2 R

(B)

3GM 2 2R

(C)

GM 2 4R

(D)

GM 2 2R

110. A satellite is seen after every 6 hours over the equator. If it revolves from east to west, the angular velocity of the satellite about the centre of earth is π π (A) radhr −1 (B) radhr −1 2 3 π π radhr −1 (C) radhr −1 (D) 4 8 111. Consider a solid sphere of mass M and radius R, then the potential energy of gravitational interaction of matter forming this solid sphere is GM 2 (A) − R (C) −

3 GM 2 5 R

(B)

1 GM 2 − 2 R

(D) −

3 GM 2 2 R

112. A planet of mass m is revolving in an elliptical orbit about the sun with an orbital period T . If m Msun and A be the area of orbit, then the angular momentum of the planet is mA mA (B) (A) 2T 4T 2mA 4mA (C) (D) T T 113. An artificial satellite moving in circular orbit around the earth has a total ( kinetic + potential ) energy E0. Its potential energy and kinetic energy respectively are (A) 2E0 and −2E0 (B)

−2E0 and 3E0

(C) 2E0 and −E0 (D) −2E0 and −E0 114. A satellite revolving in a circular equatorial orbit from west to east appears over a certain point on the equator every 8 hours . Its time period is (A) 16 hr (B) 8 hr (C) 6 hr (D) 32 hr 115. A “double star” is a composite system of two stars rotating about their centre of mass under their mutual gravitational attraction. Let us consider such a “double star” which has two stars of masses m and 2m at separation l . If T is the time period of rotation about their centre of mass then,

Mechanics II_Chapter 4_Part 2.indd 69

(A) T = 2π

l3 mG

(B)

T = 2π

l3 2mG

(C) T = 2π

l3 3 mG

(D) T = 2π

l3 4 mG

116. What should be the angular velocity of earth about its own axis so that the weight of the body at equator 3 would become th of its present value 4 1 1 −1 (A) rads rads −1 (B) 400 800 1 1 rads −1 rads −1 (C) (D) 1600 3200 117. Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. G (m + M) d Gm (B) V = − d GM (C) V = − d 2 G (D) V = − ( m + M ) d

(A) V = −

(

)

118. A point P R 3 , 0 , 0 lies on the axis of a ring of mass M and radius R. The ring is located in y -z plane with its centre at origin O. A small particle of mass m starts from P and reaches O under gravitational attraction only. Its speed at O will be (A)

GM R

(B)

Gm R

(C)

GM 2R

(D)

Gm 2R

119. A point particle is held on the axis of a ring of mass m and radius r at a distance r from its center C . When released, it reaches C under the gravitational attraction of the ring. Its speed at C will be (A)

2Gm r

(B)

2Gm ( 2 − 1) r

(C)

2Gm ⎛ 1 ⎞ ⎜1− ⎟ ⎝ r 2⎠

(D)

Gm r

120. A particle of mass m is placed on the centre of a fixed uniform semi-circular ring of radius R and mass M as shown. Then work required to displace the particle slowly from centre of ring to infinity is (Assume only gravitational interaction of ring and particle)

09-Feb-21 6:30:17 PM

4.70

JEE Advanced Physics: Mechanics – II 123. Two identical spherical masses are kept at some distance as shown. The potential energy when a mass m is taken from the surface of one sphere to the other GMm R GMm (C) πR (A)

GMm R GMm (D) − πR (B)

m

−

121. Consider an infinite plane sheet of mass with surface mass density σ. The gravitational field intensity at a point P at perpendicular distance r from such a sheet is (A) zero (B) −σ G (C) −2πσ G (D) −4πσ G

(A) (B) (C) (D)

increases continuously decreases continuously first increases then decreases first decreases then increases

124. The radius and mass of earth are increased by 0.5%. Which of the following statements are true at the surface of the earth? (a) g will decrease (b) (c) (A) (C)

122. In PROBLEM 121, the gravitational potential at the point P is (A) zero (B) −σ Gr (C) −2πσ Gr (D) −4πσ Gr

Escape velocity will remain unchanged Potential energy will remain unchanged (a) only (B) both (a) and (b) (a), (b) and (c) (D) both (b) and (c)

MULTIPLE CORRECT CHOICE TYPE QUESTIONS This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

2.

3.

An object is revolving around the earth of radius Re at height h from earth’s surface. Select the correct statement(s). (A) Its time period is independent of h . (B) Its time period depends on h . (C) Orbital velocity of an object depends on h . (D) If h Re then the time period of this object revolving around the earth is 24 hrs. If, at the surface of earth, the potential energy of a particle is U and potential is V , then the change in potential energy and the change in potential at a height h = R are ΔU and ΔV , then U U (B) ΔU = (A) ΔU = − 2 2 V V (C) ΔV = (D) ΔV = − 2 2

(C) EP = 0 for r < r2 (D) EP ≠ 0 for r < r2 4.

m ⎞ ⎛m (C) VP = −G ⎜ 1 + 2 ⎟ for r < r1 and r > r2 r ⎠ ⎝ r1 m ⎞ ⎛m (D) VP = −G ⎜ 1 + 2 ⎟ for r < r2 r2 ⎠ ⎝ r1 5.

A satellite is to be stationed in an orbit such that it can be used for relay purposes (such a satellite is called a Geostationary satellite). The condition(s) to be fulfilled by such a satellite is/are (A) Its orbit must lie in equatorial plane (B) Its sense of revolution must be from west to east (C) Its orbital radius must be 42400 km (D) Its orbit must not be elliptical and should be circular

6.

Two concentric spherical shells are as shown in figure. The magnitude of gravitational potential ( V ) and field strength ( E ) vary with distance ( r ) from centre as

A shell of mass m2 , radius r2 lies inside and is concentric with a larger uniform shell of mass m1, radius r1 . If EP is the gravitational field at point P at distance r from the common centre then, ⎛ m +m ⎞ (A) EP = G ⎜ 1 2 2 ⎟ for r > r1 & r > r2 ⎝ ⎠ r m (B) EP = G 22 for r < r1 and r > r2 r

Mechanics II_Chapter 4_Part 2.indd 70

In PROBLEM 3, if VP is the gravitational potential at the point P then, ⎛ m + m2 ⎞ (A) VP = −G ⎜ 1 ⎟⎠ for r > r1 and r > r2 ⎝ r (B) VP = 0 for r < r1

09-Feb-21 6:30:26 PM

Chapter 4: Gravitation and Satellites

(A) g (C)

(A)

(B)

v

v

r

(C)

r

(D)

E

E

r

r

7.

8.

A body is imparted a velocity v from the surface of the earth. If vo is orbital velocity and ve be the escape velocity then for (A) v = vo , the body follows a circular track around the earth. (B) v > vo but < ve , the body follows elliptical path around the earth. (C) v < vo , the body follows elliptical path and returns to surface of earth. (D) v > ve , the body follows hyperbolic path and escapes the gravitational pull of the earth. A tunnel is dug along a chord of the earth at a perpenR dicular distance from the earth’s centre. The wall of 2 the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The force N exerted by the particle on the wall and the acceleration a of the particle varies with x (distance of the particle from the centre) according to (B) N

(A) N

R/2

R

x

(C) a

R

x

R

x

R/2

R

x

The gravitational potential on the surface of a planet of radius R mass M is

Mechanics II_Chapter 4_Part 2.indd 71

−GM R

gM R

(D) −gR

10. Let V and E be the gravitational potential and gravitational field. Then select the correct alternative(s) (A) the plot of V vs r is continuous for a spherical shell (B) the plot of E vs r (distance from centre) is discontinuous for a spherical shell (C) the plot of V vs r is continuous for a solid sphere (D) the plot of E vs r is discontinuous for a solid sphere 11. Total energy of planet of mass m moving around the sun of mass Ms along an elliptical orbit depends on (A) Mass of the sun (B) Mass of the planet (C) Semi major axis (D) Independent of mass of the sun 12. A planet is revolving round the sun. Its distance from the sun at Apogee is rA and that at Perigee is rP. The mass of planet and sun is m and M respectively, vA and vP is the velocity of planet at Apogee and Perigee respectively and T is the time period of revolution of planet round the sun. (A) T 2 =

π2 ( rA + rP )3 2Gm

π2 ( rA + rP )3 2GM (C) vA rA = vP rP (B)

T2 =

(D) vA < vP ; rA > rP 13. Two tunnels are dug from one side of the earth’s surface to the other side, one along a diameter and the other along a chord. Now two particles are dropped from one end of each of the tunnels. Both the particles oscillate simple harmonically along the tunnels. Let T1 and T2 be the time periods and v1 and v2 be the maximum speed of the particles in the two tunnels. Then (B) T1 = T2 (A) T1 > T2 (C) v1 = v2 (D) v1 > v2 14. In some region, the gravitational field is zero. The gravitational potential in this region (A) may be zero (B) cannot be zero (C) must be constant (D) may be variable

(D) a

R/2

9.

R/2

(B)

4.71

15. A planet is revolving round the sun in an elliptical orbit. The work done on the planet by the gravitational force of sun is zero

09-Feb-21 6:30:35 PM

4.72

JEE Advanced Physics: Mechanics – II (A) (B) (C) (D)

in some parts of the orbit in any part of the orbit in no part of the orbit in one complete revolution

(A) (B) (C) (D)

16. A point P is lying at a distance r ( < a ) from the centre of shell of radius a. If E and V be the gravitational field and potential at the point P then, GM (A) E = 0 (B) E = − 2 r GM (C) V = 0 (D) V = − a 17. Select the correct statement(s). (A) Acceleration due to gravity increases with increasing altitude. (B) Acceleration due to gravity decreases with increasing depth (assume the earth to be a sphere of uniform density) (C) Acceleration due to gravity is independent of mass of the body ⎛ 1 1⎞ (D) The formula −GMm ⎜ − ⎟ is more accurate ⎝ r2 r1 ⎠ than the formula mg ( r2 − r1 ) for the difference of potential energy between two points r2 and r1 distance away from the centre of earth

18. Two objects of masses m and 4m are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If G is the universal gravitational constant. Then, at separation r (A) the total energy of the two objects is zero (B) their relative velocity of approach is magnitude

10Gm in r

4Gm2 r (D) net angular momentum of both the particles is zero about any point

(C) the total kinetic energy of the objects is

19. Which of the following statement(s) is/are correct (A) An astronaut going from Earth to Moon will experience weightlessness once (B) When a thin uniform spherical shell gradually shrinks maintaining its shape, the gravitational potential at the centre decreases (C) In the case of spherical shell, the plot of potential versus distance from centre is continuous (D) In the case of spherical shell, the plot of gravitational field intensity I versus distance from centre is continuous 20. Two identical satellites are orbiting at distances R and 7R from the surface of the earth, R being the radius of the earth. The ratio of their

Mechanics II_Chapter 4_Part 2.indd 72

kinetic energies is 4 potential energies is 4 total energies is 4 total energies is 1

21. In circular orbit of a satellite of mass m, revolving around a planet of mass M in an orbit of radius r and having the period of revolution T, we have (A) orbital speed equals (B)

T 2 ∝ r3

GM r

GMm 2r GMm (D) potential energy equals − 2r (C) kinetic energy equals

22. Which (one or more) of the following quantities remain constant during planetary motion in an elliptic orbit? (A) angular speed (B) areal speed (C) angular momentum (D) energy 23. A planet of mass m is revolving round the sun (of r mass Ms ) in an elliptical orbit. If v is the velocity of r the planet when its position vector from the sun is r and the planet rotates in counter clockwise direction, then areal velocity has a direction (A) given by “Right Hand Thumb Rule”. (B) given by “Left Hand Thumb Rule”. (C) normal to the plane of orbit upwards. (D) normal to the plane of orbit downwards. 24. For a solid sphere, the magnitude of (A) gravitational potential is maximum at centre (B) gravitational potential is minimum at centre (C) field strength is maximum at centre (D) field strength is minimum at centre 25. Two concentric spherical shells have masses m1 and m2 and radii r1 and r2. Select the correct statement(s).

(A) Outer shell will have no contribution in gravitational field at point P (B) Force on P is directed towards O Gm1m2 (C) Force on P is r2 Gm1m3 (D) Force on P is r2

09-Feb-21 6:30:41 PM

Chapter 4: Gravitation and Satellites

4.73

26. For a satellite revolving in a circular orbit, let v0 be the orbital speed, T be the time period of revolution, U be its potential energy and K the kinetic energy. Now if the value of G is decreased, then (B) T will decrease (A) v0 will decrease (C) U will decrease (D) K will decrease

(A) Error ΔT in measuring T, the time period, is 0.05 s (B) Error ΔT in measuring T, the time period, is 1 s (C) Percentage error in the determination of g is 5% (D) Percentage error in the determination of g is 2.5%

27. A double star in a system of two stars of masses m and 2m , rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to

33. Three point masses each of mass m are at the corners of an equilateral triangle of side l. The system rotates about the centre of the triangle with the separation of masses not changing during rotation. If T is the time period of rotation then,

(A)

3 r2

(C)

1 m2

(B)

r

(D)

1 − m 2

28. The acceleration due to earth’s gravity decreases if (A) we go down from the surface of the earth towards its centre (B) we go up from the surface of the earth (C) we go from the equator towards of the poles (D) the rotational speed of the earth is increased 29. For a planet revolving round the sun in an elliptical orbit, the (A) potential energy is constant (B) kinetic energy is constant (C) total mechanical energy is constant (D) angular momentum about centre of sun is constant 30. A particle of mass m is moved from the surface of the earth to a height h. The work done by an external agency to do this is (A) mgh for h

R

1 (C) − mgR for h = R 2

(B)

1

(B)

1 (D) mgR for h = R 2

32. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

T ∝ l2

1

(C) T ∝ m 2

(D) T ∝ m

−

1 2

34. Three masses each of mass m are kept at corner of an equilateral triangle and are revolving under the effect of mutual gravitational force. Select the correct statement(s). a (A) Radius of circular path followed by mass is 2 Gm (B) Velocity of mass is a 3Gm2 (C) Binding energy of system is 2a (D) Time period of mass is

π a3 2Gm

35. Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Select the incorrect statement(s). Energy

mgh for all h

31. A satellite is orbiting round the earth’s surface in an orbit as close as possible to the surface of the earth. Then (A) The time period of revolution of satellite is independent of its mass and is minimum (B) The orbital speed of satellite is maximum (C) The total energy of the “earth plus satellite” system is minimum (D) The total energy of the “earth plus satellite” system is maximum

Mechanics II_Chapter 4_Part 2.indd 73

3

(A) T ∝ l 2

A r

C B

(A) C shows the total energy, B the kinetic energy and A the potential energy of the satellite (B) A shows the kinetic energy, B the total energy and C the potential energy of the satellite (C) A and B are the kinetic and potential energies and C the total energy of the satellite (D) C and A are the kinetic and potential energies respectively and B the total energy of the satellite 36. The potential at the surface of a planet of mass M and radius R is assumed to be zero. Choose the most appropriate option

09-Feb-21 6:30:49 PM

4.74

JEE Advanced Physics: Mechanics – II (A) The potential at infinity is zero 3GM (B) The potential at the centre of the planet is − 2R GM (C) The potential at infinity is R GM (D) The potential at the centre of planet is − 2R

37. A body of mass m is projected vertically upwards with velocity v from earth’s surface and attains height h. If ve is the escape velocity from the surface of earth then for (A) v = ve the body reaches infinity. (B) small v , maximum height h attained by the v2 . body is 2g

(C) h = R , the projection kinetic energy is R (D) h = , total energy is negative. 2

mgR . 2

38. A double star consists of two stars having masses m and 2m separated by a distance r. Which of the following statement(s) is/are incorrect? 2r (A) Radius of circular path of star of mass 2m is 3 (B) Kinetic energy of 2m mass star is double that of lighter star (C) Time period of revolution of both are not same (D) Angular momentum of lighter star is more

REASONING BASED QUESTIONS This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) Bubble (B) Bubble (C) Bubble (D)

If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

1.

Statement-1: Kepler’s Second Law can be understood by conservation of angular momentum principle. Statement-2: Kepler’s Second Law is related with areal velocity, which can further be proved to be based dA 1 2 = r ω. on conservation of angular momentum as dt 2

2.

Statement-1: For the planets orbiting around the sun, angular speed, linear speed, kinetic energy changes with time, but angular momentum remains constant. Statement-2: No torque is acting on the rotating planet, so its angular momentum is constant.

3.

Statement-1: Escape velocity is independent of the angle of projection. Statement-2: Escape velocity from the surface of earth is 2gR , where R is radius of earth.

4.

5.

6.

Statement-1: Rate of change of weight near the earth’s surface with height h is independent of h . Statement-2: Since gravitational potential is given by GM V=− . r

7.

Statement-1: The magnitude of gravitational potential at the surface of solid sphere is less than that of the centre of sphere. Statement-2: Due to solid sphere, gravitational potential is same within the sphere.

8.

Statement-1: For a satellite revolving very near to earth’s surface the time period of revolution is given by 1 hour 24 minutes. Statement-2: The period of revolution of a satellite depends only upon its height above the earth’s surface.

Statement-1: A spherically symmetric shell produces no gravitational field anywhere. Statement-2: The field due to various mass elements cancels out, everywhere inside the shell.

9.

Statement-1: Work done in gravitational field in cyclic process is zero. Statement-2: Work done in conservative field does not depend upon path.

Statement-1: Gravitational potential is zero inside a shell. Statement-2: Gravitational potential is equal to the work done in bringing a unit mass from infinity to a point inside gravitational field.

10. Statement-1: The value of acceleration due to gravity does not depend upon mass of the body. Statement-2: Acceleration due to gravity is a constant quantity.

Mechanics II_Chapter 4_Part 2.indd 74

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Chapter 4: Gravitation and Satellites

4.75

LINKED COMPREHENSION TYPE QUESTIONS This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1

5.

The mean radius of the earth’s orbit around the sun is 1.5 × 1011 m and that of the orbit of mercury is 6 × 1010 m. Based on above information, answer the following questions. 1.

The mercury will revolve around the sun in nearly (A)

2 year 5 2

⎛ 2⎞ (C) ⎜ ⎟ year ⎝ 5⎠ 2.

(B)

5 2

(C) 2 10

year

(B)

2 10

(D)

1 10 2

Comprehension 2 The radius of the earth is R and acceleration due to gravity at its surface is g. A body of mass m is sent to a height of R from the earth’s surface. Based on above information, 4 answer the following questions. 3.

4.

6.

32

The ratio of the orbital velocity of mercury to that of the earth is (A)

(A) 10 4 π 1 (C) × 10 4 π 2

2 year 5

⎛ 2⎞ (D) ⎜ ⎟ ⎝ 5⎠

(A)

1 gR 2

(C)

1 gR 5

(B)

gR

(D)

2 gR 5

Comprehension 3 Two satellites S1 and S2 revolve round a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hour, respectively. The radius of the orbit of S1 is 10 4 km. Based on above information, answer the following questions.

Mechanics II_Chapter 4_Part 2.indd 75

(B)

2 × 10 4 π

(D) 4 × 10 4 π

The angular speed of S2 as observed by an astronaut is S1 is π π (A) (B) 2 3 π π (C) (D) 4 6

Comprehension 4 A particle of mass m is located at a distance r from the centre of a shell/sphere of mass M and radius R. The following plots (I), (II), (III) and (IV) show the variation of force F ( r ) between the shell and mass or alternatively the variation of force between the sphere and the mass. Based on above information, answer the following questions. (II) F(r) (I) F(r)

O

The potential energy increases by mgR mgR (A) (B) 4 3 mgR mgR (C) (D) 5 16 The minimum speed with which the body must be thrown from the surface of the earth so as to reach the specified height is

The speed of S2 relative to S1, when they are closest, in kmh −1, is

(III)

R

r

(IV) F(r)

F(r)

O

r

O

R

r

O

R

r

7.

The appropriate variation of the force between the shell and the point mass is represented by (A) I (B) II (C) III (D) IV

8.

The appropriate variation of the force between the sphere and the point mass is represented by (A) I (B) II (C) III (D) IV

Comprehension 5 An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the surface of earth. Now the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth. Assuming

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the radius of the earth to be R, the acceleration due to gravity on earth’s surface to be g. Based on above information, answer the following questions. 9.

The height of the satellite above the surface of the earth must be R 2R (B) (A) 2 3 (C) R (D) 2R

10. The speed with which it hits the surface of the earth is (A)

Rg

(B)

(C)

Rg 2

(D)

2Rg Rg 2

Comprehension 6 A smooth tunnel is dug across a diameter of earth and a particle is released from the surface of earth, the particle oscillates simple harmonically along it. Based on above information, answer the following questions. 11. Time period of the particle is equal to (A) 2π

R3 GM

(C) 84.6 min

(B)

2π

R g

(D) 24 hr

12. Maximum speed attained by the particle is (A)

2GM R

(C)

3GM 2R

(B)

GM R

(D)

GM 2R

Comprehension 7 A particle of mass m is placed at a distance x from the centre of ring along the line through the centre of the ring and perpendicular to its plane.

14. The force, when x = 0 is (A) (C)

GMm x2 GMmx

( a2 + x 2 )3 2

(B)

GMmx a2 + x 2

(D) zero

15. When given a small displacement x a from the centre of the ring, the particle of mass m will (A) move away from the centre (B) perform oscillations about the centre of ring (C) move to infinity and never return (D) None of these

Comprehension 8 A satellite of mass 5000 kg is projected in space with an initial speed of 4000 ms −1 making an angle of 30° with the radial direction from a distance 3.6 × 107 m away from the centre of the earth. 16. The angular momentum of satellite (A) 3.6 × 107 Js

(B)

4.9 × 107 Js

(C) 9.2 × 107 Js

(D) 3.6 × 1014 Js

17. The energy of satellite is (A) 1.6 × 107 joule

(B)

4.9 × 107 joule

(C) 0.2 × 107 joule

(D) −1.5 × 1010 joule

18. The semi-major axis of the orbit of satellite (A) 6.6 × 107 m

(B) 14.9 × 107 m

(C) 19.2 × 107 m

(D) 1.6 × 10 4 m

19. Semi-minor axis of the orbit of satellite (A) 16.6 × 107 m

(B)

3.92 × 107 m

(C) 10.2 × 1017 m

(D) 2.6 × 10 4 m

20. The minimum distance of satellite from earth (A) 66.6 × 107 m 7

(C) 1.29 × 10 m

(B) 14.9 × 107 m (D) 1.6 × 10 4 m

21. The maximum distance of satellite from earth.

Based on above information, answer the following questions. 13. Gravitational potential energy of this system (A) − (C)

GMm a2 + x 2

GMm x

Mechanics II_Chapter 4_Part 2.indd 76

(B)

GMm a

(D) None of these

(A) 6.6 × 107 m

(B)

24.9 × 107 m

(C) 11.9 × 107 m

(D) 1.6 × 10 4 m

Comprehension 9 A solid sphere having uniform density and radius R exerts a gravitational force of attraction F1 on a particle placed at P. The distance of P from the centre of the sphere is 2R. R A spherical cavity of radius is now made in the sphere. 2 The sphere (with cavity) exerts a gravitational force F2 on the same particle at P.

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Chapter 4: Gravitation and Satellites

R/2

P

R 2R

4.77

28. If K L be the kinetic energy of light body and K H kinetic energy of heavy body then (A) K L + K H =

ml 2ω 2 3

(B)

(C) K L + K H =

2ml 2ω 2 3

(D) K L + K H =

KL + KH =

2ml 2ω 2 9 ml 2ω 2 9

Based on the above facts, answer the following questions.

Comprehension 11

22.

A solid sphere of mass M and radius R is surrounded by a spherical shell of same mass M and radius 2R as shown. A small particle of mass m is released from rest from a height h ( R ) above the shell. There is a hole in the shell.

F1 is GMm R2 GMm (C) 3R2 (A)

23.

GMm 2R 2 GMm (D) 4R2 (B)

m h

F2 is

A

(A)

7 GMm 13 R2

(B)

7 GMm 16 R2

(C)

7 GMm 32 R2

(D)

7 GMm 36 R2

F 24. The ratio 1 is F2 9 (A) 7 1 (C) 2

7 9 4 (D) 3 (B)

25. The system revolves about (A) the centre (B) the centre of mass (C) both (D) None of these 26. Heavier star revolves in an orbit of radius l 2 l l (C) (D) 3 4 27. The system revolves with angular velocity ω (about an axis specified above) given by (B)

Gm l3

(B)

(C)

3Gm l3

(D) 2

Mechanics II_Chapter 4_Part 2.indd 77

2R

29. The time taken by the particle just to enter the hole at A is

A binary star system consists of two stars of masses m and 2m . The separation between their centres is l . They revolve two about a point under their mutual gravitational interaction force. Based on the above facts, answer the following questions.

(A)

R

Based on above information, answer the following questions.

Comprehension 10

(A) l

B

2Gm l3 Gm l3

(A) 2 (C)

hR2 GM hR2 GM

2 hR2 GM

(B)

(D) None of these

30. The time taken by the particle to move from A to B is (A) =

R2 GMh

(B)

(C)


R2 GMh

31. The approximate speed with which the particle hits B is (A)

2GM R

(B)

GM 2R

(C)

3GM 2R

(D)

GM R

Comprehension 12 The minimum and maximum distances of a satellite from the centre of the earth are 2R and 4R, respectively, where R is the radius of earth and M is the mass of the

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earth. Based on above information, answer the following questions.

Radius

32. The minimum speed of the satellite is

Orbital Radius

Gravitational Acceleration at the Surface

Orbital Period

(A)

GM 9R

(B)

GM 5R

Moon I

(C)

GM 6R

(D)

GM 3R

5 × 107 m

Moon II

9 × 107 m 3 × 10 5 s

33. The maximum speed of the satellite is (A)

3GM 2R

(B)

2GM 3R

(C)

2GM R

(D)

5GM 2R

Moon III 2 × 10 5 m

0.20 ms −2

2 × 106 s

Based on above information, answer the following questions. 35. In table, the mass of the planet is closest to

34. Radius of curvature at the point of minimum distance is 8R 5R (B) (A) 3 3 4R 7R (C) (D) 3 3

(A) 1.2 × 10 24 kg (C) 2.4 × 10

24

kg

(B) 1.7 × 10 24 kg (D) 4.8 × 10 24 kg

36. In table, the mass of Moon III is closest to

Comprehension 13

(A) 2.4 × 1019 kg

(B)

4.8 × 1019 kg

(C) 1.2 × 10 20 kg

(D) 2.4 × 10 20 kg

37. In table, the centripetal acceleration of Moon II due to orbital motion is closest to (A) 0.02 ms −2 (B) 0.04 ms −2

An Earth station receives data transmitted back in time from a future intergalactic expedition. The table summarizes the data for the moons of a planet what will be discovered in a distance galaxy.

(C) 0.06 ms −2

(D) 0.08 ms −2

MATRIX MATCH/COLUMN MATCH TYPE QUESTIONS Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles will look like the following: A B C D

1.

p p p p p

q q q q q

Match the gravitational field and potential in COLUMN-I to the respective quantities in COLUMN-II. COLUMN-I

COLUMN-II

(A) Eg = 0

(p) Inside a spherical shell

(B) Eg ≠ 0

(q) At centre of a solid sphere

r

s

t

r r r r

s s s s

t t t t

COLUMN-I

COLUMN-II

(C) V = 0

(r) Inside the solid sphere (excluding centre)

(D) V ≠ 0

(s) At centre of a circular ring (t) None of these

(Continued)

Mechanics II_Chapter 4_Part 2.indd 78

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Chapter 4: Gravitation and Satellites 2.

3.

Density of a planet is two times the density of earth and radius of this planet is half the radius of earth. Match, the following. (As compared to earth).

If the angular speed of rotation of earth about its own axis decreases, then match the following COLUMN-I

COLUMN-II

(A) Value of g at pole

(p) remains same

(B) Value of g at equator

(q) increases

(q) Same

(C) Distance of geostationary satellite

(r) decreases

(C) Gravitational potential at centre

(r) Two times

(D) Energy of geostationary satellite

(s) None of these

(D) Gravitational strength at centre

(s) Four times

COLUMN-I

COLUMN-II

(A) Gravitational strength at the surface of this planet

(p) Half

(B) Gravitational potential on the surface

6.

In elliptical orbit of a planet, as the planet moves from aphelion position to perihelion position, match the following table:

Match the following COLUMN-I

COLUMN-II

(A) Gravitational force

(p) Law of Conservation of Angular Momentum

(B) Kepler’s First Law

(q) T 2 ∝ r 3 2

(C) Kepler’s Second Law

(r) Inverse square law

(D) Kepler’s Third Law

COLUMN-I

COLUMN-II

(A) distance of planet from centre of sun

(p) remains same

(B) speed of planet

(q) decreases

(C) potential energy

(r) increases

(s) Orbit of planet is elliptical

(D) angular momentum about centre of sun

(s) zero

(t) Conservative 7.

(E) change in mechanical energy 4.

5.

On the surface of earth the gravitational field is E and gravitational potential is V. Match the following COLUMN-I

COLUMN-II

(A) At height h = R, value of E

(p) decreases by a factor

1 4

(A) Time period of a body (p) Independent of orbiting the planet in mass of body circular orbit

R (B) At depth d = , 2 value of E

(q) decreases by a factor

1 2

(B) Orbital velocity of body

(q) Dependent of radius of orbit

(C) At height h = R, value of V

(r) increases by a factor

11 8

(C) Mechanical energy of body

(r) Dependent of mass of planet

(D) At depth d =

(D) Escape velocity of a body from surface of planet

(s) Constant in an orbit

Match the following COLUMN-I

Mechanics II_Chapter 4_Part 2.indd 79

COLUMN-II

value of V

R , 2

(s) increases by a factor 2

(t) decreases by a factor

11 8

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4.80 8.

JEE Advanced Physics: Mechanics – II Consider A planet of mass M and radius R and a satellite is launched with velocity V from its surface. Match the following COLUMN-I (A) V >

2GM R

(p) Ellipse

(B) V =

2GM R

(q) Hyperbola

(C) V
ve (D) ve > v > v0

(s) The path of satellite is parabolic

(D)

2π mab = constant T

(s) None of these

INTEGER/NUMERICAL ANSWER TYPE QUESTIONS In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1.

2.

Calculate the work done, in newton, by the gravitational field when a particle is moved along the line 3 y + 2x = 5 in the region of gravitational field r E = 2iˆ + 3 ˆj Nkg −1 , from ( 1, 1 ) m to ( −2, 3 ) m . g

3.

speed with which it will crosses the centre of ring is

Calculate the ratio of the orbital radius of a communication satellite to the radius of earth so that it can cover 75% of the surface area of earth during its revolution. 2 Take ( 2.65 ) ≈ 7 .

(

)

A circular ring of mass M and radius R is placed in YZ plane with centre at origin. A particle of mass m is released from rest at a point x = 2R . The

Mechanics II_Chapter 4_Part 2.indd 80

1

y2 1 ⎞ GM ⎤ z ⎡ ⎛ v = ⎢x⎜ 1 − ⎟ . Find . xz y ⎠ R ⎥⎦ ⎣ ⎝ 4.

A particle is fired vertically upward with a speed of 15 kms −1. Find the speed of the particle in kms −1, when it goes out of the earth’s gravitational pull.

5.

A particle is projected from point A (which lies at a distance 4R from the centre of the Earth), with a speed v0 in a direction making 30° with the line joining the centre of the Earth and point A, as shown.

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Chapter 4: Gravitation and Satellites

4.81

v0

α

R

9.

Calculate the speed v0 of particle, in ms −1, if particle passes grazing the surface of the earth. Consider that only the gravitational interaction between these two exists. GM = 6.4 × 107 m 2s −2 Given that R 6.

7.

A ring of radius R = 4 m made of a highly dense material has a mass m1 = 5.4 × 109 kg distributed uniformly over its circumference. A highly dense particle of mass m2 = 6 × 108 kg is placed on the axis of the ring at a distance x0 = 3 m from the centre of the ring. Neglecting all other forces, except mutual gravitational interaction of the two, calculate (a) displacement of the ring, in cm, when particle is at the centre of ring and (b) speed of the particle, in cms −1, at that instant. Two particles each of mass M are fixed at positions M ( 0, a ) and ( 0, − a ) . Another particle of mass is 2 thrown from origin along +z axis so that it is just able

to reach a point ( 0 , 0 , 2 3 a ) . The speed with which the particle was projected from the origin is calculated as v =

8.

rmax

If a satellite is revolving around the earth in a circular orbit in a plane containing earth’s axis of rotation. If the angular speed of satellite is equal to that of earth, calculate the time (in hour) taken by it to move from a point above north pole to a point above the equator.

10. A smooth tunnel is dug along the radius of earth that ends at centre. A ball is released from the surface of earth along tunnel. Coefficient of restitution for collision between the centre of the earth and ball is 0.5. Calculate the distance travelled by ball just before second collision at centre is xR. Given mass of the earth is M and radius of the earth is R, find x . 11. If the distance between the centres of Earth and Moon is D and mass of Earth is 81 times that of Moon. The distance from the centre of earth where the gravitaxD tional field will be zero is , then calculate the value y of ( y − x ) . 12. If a planet was suddenly stopped in its orbit supposed to be circular, show that it would fall onto the sun in a 1 time times the period of the planet’s revolution. 2x Find x . 13. A large spherical mass M is fixed at one position and two identical masses m are kept on a line passing through the centre of M as shown in figure.

1 ⎞ αGM ⎛ 1− . Find α and β . a ⎜⎝ β ⎟⎠

A projectile of mass m is fired form the surface of the earth at an angle α = 60° from the vertical. The initial GM , where M and R speed of the projectile is v0 = R are mass and radius of earth. The projectile rises to a R , where ∗ is not readable. Find ∗ . Neglect height ∗ air resistance and the earth’s rotation.

Mechanics II_Chapter 4_Part 2.indd 81

The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l ⎛ M ⎞ from M the tension in the rod is zero for m = k ⎜ . ⎝ 288 ⎟⎠ Find the value of k .

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JEE Advanced Physics: Mechanics – II

14. Gravitational acceleration on the surface of a planet is 12 g, where g is the gravitational acceleration on the 11 surface of the earth. The average mass density of the 4 planet is times that of the earth. If the escape speed 3 on the surface of the earth is taken to be 11 kms −1, find the escape speed on the surface of the planet in kms −1. 15. A small point mass m is to be thrown with such a speed at a distance x from the axis of a long cylinder of radius R and density ρ, so that m starts revolving around the cylinder in a circular orbit of radius x with centre at the axis of cylinder. The speed with which 1

point mass is thrown is found to be v = R ( ∗Gρπ )∗ , where ∗ is not readable. Calculate ∗ . 16. A binary star consists of two stars A (mass 2.2 MS) and B (mass 11MS), where MS is the same mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. Find the ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass. 17. An artificial satellite is describing an equatorial orbit at 1600 km above the surface of the earth. Calculate its orbital speed, in ms −1 and the period of revolution, in seconds. If the satellite is travelling in the same direction as the rotation of the earth (i.e., from west to east), calculate the interval (in seconds) between two successive times at which it will appear vertically overhead to an observer at a fixed point on the equator. Radius of earth is 6400 km and g = π 2ms −2 at the surface of earth. 18. Inside a fixed sphere of radius R and uniform denR sity ρ , there is spherical cavity of radius such that 2 surface of the cavity passes through the centre of the sphere as shown in figure. A particle of mass m is released from rest at centre B of the cavity. The velocity with which particle strikes the centre A of the sphere is v . Neglect earth’s gravity. Initially sphere and particle are at rest. If the escape speed of the particle from 2v 2 surface of the outer bigger sphere is ve. Then find 2e . v

Mechanics II_Chapter 4_Part 2.indd 82

A

B

R

R/2

19. A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to 1 the planet’s gravity is th of its value at the surface 9 of the planet. If the escape velocity from the planet is vesc = v N , then the value of N is (ignore energy loss due to atmosphere) 20. A sphere of radius R has its centre at the origin. It has uniform mass density ρ0 except that there is a spheri1 1 cal hole of radius r = R whose centre is at x = R, as 2 2 shown in figure. The gravitational field at x = 2R is ⎛ a ⎞ ⎜⎝ ⎟ π Gρ0 R. Find a. a + 20 ⎠ y

r R

x

21. A spaceship approaches the moon (of mass M and radius R) along a parabolic path which is almost tangential to its surface. At the moment of maximum approach, the brake rocket is fired to convert the spaceship into a satellite of the moon. The change in speed is found to be

(

x − 1)

GM . Find x . R

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Chapter 4: Gravitation and Satellites

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ARCHIVE: JEE MAIN 1.

2.

[Online September 2020] K The mass density of a spherical galaxy varies as r over a large distance r from its center. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as 1 (A) T ∝ R (B) T 2 ∝ 3 R (C) T 2 ∝ R (D) T 2 ∝ R3 [Online September 2020] The height h at which the weight of a body will be the same as that at the same depth h from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected) 5R − R 5 R−R (A) (B) 2 2 (C)

3.

R 2

(D)

4.

5.

(B) r =

(C) r =

3 R 4

(D) r = R

Mechanics II_Chapter 4_Part 2.indd 83

)

A

(x

2

+ a2 )

32

(D) A ( x 2 + a 2 )

12

[Online September 2020] A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is (A) 1 (B) 2 1 (D) 2 (C) 2

7.

[Online September 2020] The value of the acceleration due to gravity is g1 at a R height h = (R = radius of the earth) from the surface 2 of the earth. It is again equal to g1 at a depth d below ⎛ d⎞ the surface of the earth. The ratio ⎜ ⎟ equals ⎝ R⎠

8.

5 R 9

[Online September 2020] On the x-axis and a distance x from the origin, the gravitational field due to a mass distribution is given Ax by in the x-direction. The magnitude of 2 ( x + a2 )3 2 gravitational potential on the x-axis at a distance x. Taking its value to be zero at infinity, is

+a

(B)

6.

[Online September 2020] The mass density of a planet of radius R varies with ⎛ r2 ⎞ the distance r from its center as ρ ( r ) = ρ0 ⎜ 1 − 2 ⎟ . ⎝ R ⎠ Then the gravitational field is maximum at 1 R 3

(x

2 12 32

(D) 2.5Re

(A) r =

A 2

(C) A ( x 2 + a 2 )

3R − R 2

[Online September 2020] A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius Re . By firing rockets attached to it, its speed is instantaneously increased in the direction 3 times larger. Due to of its motion so that is become 2 this, the farthest distance from the center of the earth that the satellite reaches is R, value of R is (A) 4Re (B) 3Re (C) 2Re

(A)

9.

(A)

7 9

(B)

4 9

(C)

1 3

(D)

5 9

[Online September 2020] The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω . An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is: (h R, where R is the radius of the earth) (A)

R2ω 2 8g

(B)

R2ω 2 4g

(C)

R2ω 2 g

(D)

R2ω 2 2g

[Online September 2020] A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is (B) 3 : 4 (A) 1 : 6 (C) 1 : 3 (D) 1 : 2

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JEE Advanced Physics: Mechanics – II

10. [Online September 2020] Two planets have masses M and 16M and their radii are a and 2a, respectively. The separation between the centers of the planets is 10a. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centers. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is (A) (C) 4

GM 2 ma GM a

(B)

3 5GM 2 a

(D) 2

GM a

11. [Online January 2020] A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms −2 at the north pole and the radius of the earth = 6400 km) (B) 194.66 N (A) 195.66 N (D) 195.32 N (C) 194.32 N 12. [Online January 2020] Consider two solid spheres of radii R1 = 1 m, R2 = 2 m and masses M1 and M2 , respectively. The gravitational field due to sphere (1) and (2) are shown. The value of M1 is M2

1 (A) 2 1 (C) 3

2 (B) 3 1 (D) 6

13. [Online January 2020] A body A of mass m is moving in a circular orbit of m radius R about a planet. Another body B of mass r ⎛ v⎞ 2 collides with A with a velocity which is half ⎜ ⎟ the ⎝ 2⎠ r instantaneous velocity v of A. The collision is completely inelastic. Then, the combined body (A) starts moving in an elliptical orbit around the planet. (B) continues to move in a circular orbit (C) Falls vertically downwards towards the planet (D) Escapes from the Planet’s Gravitational field.

Mechanics II_Chapter 4_Part 2.indd 84

14. [Online January 2020] Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are vA and vB, v n respectively, then A = . The value of n is vB 4 (A) 4 (B) 1 (C) 2 (D) 3 15. [Online January 2020] An asteroid is moving directly towards the center of the earth. When at a distance of 10R (R is the radius of the earth) from the earths center, it has a speed of 12 kms −1 . Neglecting the effect of earth’s atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 kms −1)? Give your answer to the nearest integer in kms −1. 16. [Online April 2019] Four identical particles of mass M are located at the corners of a square of side a . What should be their speed if each of them revolves under the influence of others’ gravitational field in a circular orbit circumscribing the square?

(A) 1.41

GM a

(B) 1.16

GM a

(C) 1.21

GM a

(D) 1.35

GM a

17. [Online April 2019] A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth’s volume is 64 times the volume of the moon. E E (B) (A) 64 4 E E (C) (D) 16 32 18. [Online April 2019] A solid sphere of mass M and radius a is surrounded by a uniform concentric spherical shell of thickness 2a

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Chapter 4: Gravitation and Satellites

and mass 2M. The gravitational field at distance 3a from the centre will be GM 2GM (B) (A) 2 9a 9a2 GM 2GM (C) (D) 3 a2 3 a2 19. [Online April 2019] A test particle is moving in a circular orbit in the gravitational field produced by a mass density K ρ ( r ) = 2 . Identify the correct relation between the r radius R of the particle’s orbit and its period T (A)

T is a constant R

(B)

(C)

T is a constant R2

T2 (D) is a constant R3

TR is a constant

20. [Online April 2019] The value of acceleration due to gravity at Earth’s surface is 9.8 ms −2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms −2, is close to (Radius of earth = 6.4 × 106 m ) (A) 9.0 × 106 m (B) 6.4 × 106 m 6 (C) 1.6 × 10 m (D) 2.6 × 106 m 21. [Online April 2019] A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? [Given:

Mass

of

planet 6

= 8 × 10 22 kg ,

planet = 2 × 10 m , Gravitational G = 6.67 × 10 −11 Nm 2kg −2] (A) 17 (B) 13 (C) 11 (D) 9 of

Radius constant

22. [Online April 2019] The ratio of the weights of a body on the Earth’s surface to that on the surface of a planet is 9 : 4 . The mass of 1 the planet is th of that of the Earth. If R is the radius 9 of the Earth, what is the radius of the planet? (Take the planets to have the same mass density) R R (B) (A) 4 3 R R (D) (C) 9 2

Mechanics II_Chapter 4_Part 2.indd 85

4.85

23. [Online January 2019] If the angular momentum of a planet of mass m , moving around the Sun in a circular orbit is L , about the center of the Sun, its areal velocity is (A) (C)

L m L 2m

4L m 2L (D) m (B)

24. [Online January 2019] The energy required to take a satellite to a height h above Earth surface (radius of Earth = 6.4 × 10 3 km ) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2 . The value of h for which E1 and E2 are equal, is (A) 3.2 × 10 3 km

(B) 1.6 × 10 3 km

(C) 1.28 × 10 4 km

(D) 6.4 × 10 3 km

25. [Online January 2019] A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is (A) 2mv 2 1 (C) mv 2 2

mv 2 3 (D) mv 2 2 (B)

26. [Online January 2019] Two stars of masses 3 × 10 31 kg each, and at distance 2 × 1011 m rotate in a plane about their common centre of mass O . A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is: (Take Gravitational constant G = 6.67 × 10 −11 Nm 2kg −2 ) (A) 2.8 × 10 5 ms −1

(B) 1.4 × 10 5 ms −1

(C) 2.4 × 10 4 ms −1

(D) 3.8 × 10 4 ms −1

27. [Online January 2019] A satellite is revolving in a circular orbit at a height h from the earth surface, such that h R where R is the radius of the earth. Assuming that the effect of earth’s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is (A)

2gR

(B)

gR ( 2 − 1 )

(C)

gR

(D)

gR 2

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JEE Advanced Physics: Mechanics – II

28. [Online January 2019] A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass m at x = 0, if the mass per unit length of the rod is A + Bx 2, is given by 1⎞ ⎤ ⎡ ⎛ 1 − ⎟ + BL ⎥ (A) Gm ⎢ A ⎜ ⎝ ⎠ a+L a ⎣ ⎦ 1⎞ ⎡ ⎛ 1 ⎤ − ⎟ − BL ⎥ (B) Gm ⎢ A ⎜ ⎣ ⎝ a+ L a⎠ ⎦ 1 ⎞ ⎤ ⎡ ⎛1 (C) Gm ⎢ A ⎜ − ⎟ + BL ⎥ ⎣ ⎝ a a+ L⎠ ⎦ 1 ⎞ ⎡ ⎛1 ⎤ (D) Gm ⎢ A ⎜ − ⎟⎠ − BL ⎥ ⎝ a a + L ⎣ ⎦ 29. [Online January 2019] A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be (A) such that it escapes to infinity (B) in a circular orbit of a different radius (C) in an elliptical orbit (D) in the same circular orbit of radius R 30. [Online January 2019] Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R and B is in a circular orbit of radius 2R around the earth. T The ratio of their kinetic energies, A is TB (A) 1

(B)

(C) 2

(D)

1 2 1 2

31. [2018] A particle is moving in a circular path of radius a k under the action of an attractive potential U = − 2 . 2 r Its total energy is (A) −

k 4 a2

(C) zero

k 2a2 3 k (D) − 2 2a (B)

32. [2018] A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then

Mechanics II_Chapter 4_Part 2.indd 86

3

n

(A) T ∝ R 2 for any n

(B)

T ∝ R2

( n+1)

(C) T ∝ R

2

+1

n

(D) T ∝ R 2

33. [Online 2018] A body of mass m is moving in a circular orbit of radius R about a planet of mass M . At some instant, it splits into two equal masses. The first mass moves R in a circular orbit of radius and the other mass, in 2 3R a circular orbit of radius . The difference between 2 the final and initial total energies is (A) +

GMm 6R

(B)

(C) −

GMm 2R

(D) −

GMm 2R GMm 6R

34. [Online 2018] Take the mean distance of the moon and the sun from the earth to be 0.4 × 106 km and 150 × 106 km respectively. Their masses are 8 × 10 22 kg and 2 × 10 30 kg respectively. The radius of the earth is 6400 km . Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on ΔF1 is the earth. Then, the number closest to ΔF2 (A) 2 (B) 0.6 (C) 6 (D) 10 −2 35. [Online 2018] Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence (A) there will be no change in weight anywhere on the earth. (B) weight of the object, everywhere on the earth, will increase. (C) except at poles, weight of the object on the earth will decrease. (D) weight of the object, everywhere on the earth, will decrease. 36. [Online 2018] Two particles of the same mass m are moving in circular −16 orbits because of force, given by F ( r ) = − r 3. r The first particle is at a distance r = 1 and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to

09-Feb-21 6:32:27 PM

Chapter 4: Gravitation and Satellites

(A) 6 × 10 −2 (C) 3 × 10 −3

(B) 10 −1 (D) 6 × 10 2

37. [2017] The variation of acceleration due to gravity g with distance d from centre of the Earth is best represented by ( R = Earth’s radius ) (A)

(C)

(B)

(D)

(A) 0.63 × 10 −3 rads −1

(B)

(C) 0.28 × 10 −3 rads −1

(D) 1.1 × 10 −3 rads −1

0.83 × 10 −3 rads −1

39. [Online 2017] The mass density of a spherical body is given by k ρ(r ) = for r ≤ R and ρ ( r ) = 0 for r > R , where r r is the distance from the centre. The correct graph that describes qualitatively the acceleration, a of a test particle as a function of r is (B)

40. [2016] A satellite is revolving in a circular orbit at a height h from the earth’s surface (radius of earth R; h R ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to (Neglect the effect of atmosphere) (A)

gR ( 2 − 1 )

(B)

2gR

(C)

gR

(D)

gR 2

41. [Online 2016] Figure shows elliptical path abcd of a planet around 1 the sun S such that the area of triangle csa is the 4 area of the ellipse. (See figure) With bd as the major axis, and ca as the minor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then

38. [Online 2017] If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so 3 that the person at the equator will weigh W. Radius 4 of the Earth is 6400 km and g = 10 ms −2.

(A)

(A) t1 = 4t2

(B)

(C) t1 = 3t2

(D) t1 = t2

Mechanics II_Chapter 4_Part 2.indd 87

(D)

t1 = 2t2

42. [Online 2016] An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R , while its mass is M . The gravitational pull FG on the astronaut is (A) Zero since astronaut feels weightless (B)

GMm

( R + h )2

(C) FG =

< FG
TB then RA > RB (B) If TA > TB then M A > MB 2

⎛T ⎞ ⎛R ⎞ (C) ⎜ A ⎟ = ⎜ A ⎟ ⎝ TB ⎠ ⎝ RB ⎠

3

(D) TA = TB 8.

[IIT-JEE 2002] A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km . Then, the time period of a spy satellite orbiting a few 100 km above the earth’s surface ( Re = 6400 km ) will approximately be 1 h (A) 2 (B) 1 h (C) 2 h (D) 4 h

9.

[IIT-JEE 2001] A simple pendulum has a time period T1 when on the earth’s surface and T2 when taken to a height R above the earth’s surface, where R is the radius of the T earth. The value of 2 is T1 2 (A) 1 (B) (C) 4 (D) 2

10. [IIT-JEE 1998] A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth (A) the acceleration of S is always directed towards the centre of the earth (B) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant (C) the total mechanical energy of S varies periodically with time (D) the linear momentum of S remains constant in magnitude 11. [IIT-JEE 1996] If the distance between the earth and the sun were half its present value, the number of days in a year would have been (A) 64.5 (B) 129 (C) 182.5 (D) 730

Mechanics II_Chapter 4_Part 2.indd 90

12. [IIT-JEE 1989] Imagine a light planet revolving around a very massive star in a circular orbit of radius R with period T. If the gravitational force of attraction between the planet and the star is proportional to R −5 2, the T 2 is proportional to R7 2

(A) R3

(B)

(C) R3 2

(D) R3.75

13. [IIT-JEE 1983] If g is the acceleration due to gravity at the earth’s surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is 1 mgR (B) 2 mgR (A) 2 1 (D) mgR (C) mgR 4 14. [IIT-JEE 1981] If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would (A) decrease by 1% (B) remain unchanged (C) increase by 2% (D) decrease by 2%

Multiple Correct Choice Type Problems In this section each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

[JEE (Advanced) 2015] A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P ( r ) is the pressure at r ( r < R ) , then the correct options is/are 3R ⎞ ⎛ P⎜ r = ⎟ ⎝ 4 ⎠ = 63 (B) (A) P ( r = 0 ) = 0 2R ⎞ 80 ⎛ P⎜ r = ⎟ ⎝ 3 ⎠ 3R ⎞ R⎞ ⎛ ⎛ P⎜ r = P⎜ r = ⎟ ⎟⎠ 16 ⎝ ⎝ 5 2 ⎠ = 20 = (D) (C) 2R ⎞ 21 R ⎞ 27 ⎛ ⎛ P⎜ r = P⎜ r = ⎟ ⎟ ⎝ ⎝ 5 ⎠ 3⎠

2.

[JEE (Advanced) 2013] Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are)

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Chapter 4: Gravitation and Satellites

4.91

(A) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is GM L (B) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4

GM L (C) The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2

(A) The gravitational force due to this object at the origin is ZERO. (B) The gravitational force at the point B ( 2, 0 , 0 ) is ZERO. (C) The gravitational potential is the same at all points of the circle y 2 + z 2 = 36 . (D) The gravitational potential is the same at all points on the circle y 2 + z 2 = 4 .

2GM L (D) The energy of the mass m remains constant 3.

[IIT-JEE 2012] Two spherical planets P and Q have the same uniform density ρ , masses MP and MQ and surface areas A and 4A , respectively. A spherical planet R also has uniform density ρ and its mass is ( MP + MQ ) . The escape velocities from the planets P , Q and R are VP , VQ and VR respectively. Then (A) VQ > VR > VP (C)

4.

5.

VR =3 VP

(B) VR > VQ > VP (D)

VP 1 = VQ 2

[IIT-JEE 1994] The magnitudes of gravitational field at distances r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 , respectively. Then (A)

F1 r1 = if r1 R F2 r12

(C)

F1 r1 = if r1 > R and r2 >R F2 r2

(D)

F1 r12 = if r1 < R and r2 < R F2 r22

[IIT-JEE 1993] A solid sphere of uniform density and radius 4 unit is located with its centre at the origin of coordinates, O. Two spheres of equal radii of 1 unit, with their centres at A ( −2, 0 , 0 ) and B ( 2, 0 , 0 ) respectively, are taken out of solid sphere, leaving behind spherical cavities as shown in figure.

Mechanics II_Chapter 4_Part 2.indd 91

Reasoning Based Questions This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE. 1.

[IIT-JEE 2008] Statement-1: An astronaut in an orbiting space station above the earth experiences weightlessness. Statement-2: An object moving around the earth under the influence of earth’s gravitational force is in a state of free-fall.

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after series of calculations based on the data provided in the question(s). 1.

[JEE (Advanced) 2015] A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to

09-Feb-21 6:32:59 PM

4.92

JEE Advanced Physics: Mechanics – II 1 th of its value at the surface 4 of the planet. If the escape velocity from the planet is

the planet’s gravity is

vesc = v N , then the value of N is (ignore energy loss due to atmosphere) 2.

[JEE (Advanced) 2015] A large spherical mass M is fixed at one position and two identical masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass

3.

[IIT-JEE 2010] Gravitational acceleration on the surface of a planet is 6 g , where g is the gravitational acceleration on the 11 surface of the earth. The average mass density of the 2 planet is times that of the earth. If the escape speed 3 on the surface of the earth is taken to be 11 kms −1, find the escape speed on the surface of the planet in kms −1.

nearer to M is at a distance r = 3l from M the tension ⎛ M ⎞ in the rod is zero for m = k ⎜ . The value of k is ⎝ 288 ⎟⎠

Mechanics II_Chapter 4_Part 2.indd 92

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Chapter 4: Gravitation and Satellites

4.93

ANSWER KEYS—TEST YOUR CONCEPTS AND PRACTICE EXERCISES Test Your Concepts-I (Based on Acceleration Due to Gravity, Gravitational Field and Applications) 1. 2.4 kms −1 Gm1m2 2r GMm 3. (a) − a2 + x 2 2. −

G ( m + M )2 d 3M 7. − R 6. −

8. (a) − (c) −

GMm ⋅ x

( x 2 + a2 )3 2

(e) 0

3GM 2 , along +x axis d2 a 5. 2 r 6. 200 Nkg −1 , along +x direction 7 3 2π GMm 8. L2 π RGρ0 9. 2

⎛ a+ L⎞ (b) −Gλ log e ⎜ ⎝ a − L ⎟⎠

Test Your Concepts-III (Based on Relation Between Gravitational Field and Potential) 2. − ( axy + byz ) + C 3. (a) 1 Jkg −1

(b) 1 Jkg −1 , conservative

(

)

4. − ( 6 xy ) iˆ − 3 x 2 + y 2 z ˆj − y 3 kˆ

(

5. (a) −2a xiˆ − yjˆ −4

3 GM 2 5 R

1. 3 Jkg −1

GMλ 10. R 11. 7.8 × 10

(b) −

⎛ l + l 2 + a2 ⎞ 9. (a) −2Gλ log e ⎜ ⎟⎠ ⎝ a

4.

7.

GM 2 2R

rads

−1

0 ⎡ ⎢ GM ⎛ r 3 − R13 ⎞ 12. ⎢⎢ 2 ⎜ 3 ⎟ r ⎝ R2 − R13 ⎠ ⎢ ⎢ GM ⎢ r2 ⎣

for

r < R1

for R1 < r < R2 for

r > R2

Test Your Concepts-II (Based on Gravitational Potential, Potential Energy and Applications) 1. −

Gm2 ⎛ 12 4 ⎞ 12 + + ⎟ a ⎜⎝ 2 3⎠

2. −

GMm0 ⎛ a+ L⎞ log e ⎜ ⎝ a ⎠⎟ L

GMm R GMm 4. − , remain the same r 3 GMm 5. − 2 R

Mechanics II_Chapter 4_Part 2.indd 93

(b) − ayiˆ − axjˆ

6. 5 2 Nkg −1

13. 1600 km

3.

)

7. 10 2 N

(

)

(

)

8. − 2 axiˆ + 2 ayjˆ + 2bzkˆ , 2 a 2 x 2 + y 2 + b 2 z 2 9. −2iˆ − 3 ˆj + kˆ

Test Your Concepts-IV (Based on Conservation Laws, Escape Velocity and Applications) ⎛ n ⎞ 1. (a) ⎜ mgR ⎝ n + 1 ⎟⎠

(b)

2ngR n+1

2. 11.2 kms −1 3. 2.1 × 10 −5 ms −1 , 4.2 × 10 −5 ms −1 4. 2 5. 6.

GM L

3 5GM 2 a

( (

) )

r0 1 + 1 − K ( 2 − K ) sin 2 ϕ , 2−K r0 1 − 1 − K ( 2 − K ) sin 2 ϕ 2−K

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JEE Advanced Physics: Mechanics – II

7. 0

9.

d3 G ( m1 + m2 )

6. (a) 2π

8. 42 kms −1 ⎛ 2 1⎞ 2G ( m1 + m2 ) ⎜ − ⎟ ⎝ a l⎠

(c)

m2 m1

(d) μω d 2

10. 9, 2

7. 56 2 hour , 3829 kmhr −1

Test Your Concepts-V (Based on Satellites, Kepler’s Laws and Applications)

8. r ,

5r 3

9. (a)

GM 2 4R2

1. 2, 2 4. ( 1 + sin α ) r0 , ( 1 − sin α ) r0

11.

5. r = R

(b)

m2 m1 1 (e) μω 2 d 2 2 (b)

GM R3 , 4 π GM 4R3

(c)

GM 2 4R

13 ⎡ ⎧⎪ ⎫⎪ ⎤ 1 4π 2 ⎥ −⎨ 2G ( m1 + m2 ) ⎢ ⎬ ⎢⎣ R1 + R2 ⎪⎩ G ( m1 + m2 ) T 2 ⎪⎭ ⎥⎦

Single Correct Choice Type Questions 1. C

2. C

3. D

4. C

5. D

6. D

7. C

8. C

9. C

10. B

11. C

12. D

13. B

14. B

15. C

16. C

17. A

18. C

19. C

20. A

21. C

22. C

23. B

24. C

25. B

26. D

27. B

28. B

29. D

30. C

31. B

32. D

33. D

34. B

35. C

36. D

37. D

38. D

39. A

40. A

41. B

42. C

43. B

44. B

45. C

46. A

47. B

48. B

49. A

50. A

51. C

52. B

53. A

54. D

55. A

56. C

57. C

58. C

59. C

60. C

61. C

62. C

63. C

64. D

65. D

66. A

67. D

68. A

69. B

70. D

71. A

72. B

73. B

74. C

75. B

76. B

77. D

78. B

79. C

80. A

81. D

82. D

83. D

84. C

85. C

86. C

87. D

88. C

89. B

90. A

91. D

92. A

93. C

94. D

95. A

96. D

97. A

98. B

99. C

100. D

101. B

102. D

103. C

104. B

105. A

106. D

107. B

108. C

109. C

110. C

111. C

112. C

113. C

114. C

115. C

116. C

117. D

118. A

119. C

120. A

121. C

122. C

123. C

124. C

Multiple Correct Choice Type Questions 1. B, C

2. A, D

3. A, B, C

4. A, C, D

6. A, C

7. A, B, C, D

8. B, C

9. C, D

5. A, B, C, D 10. A, B, C

11. A, B, C

12. B, C, D

13. B, D

14. A, C

15. A, D

16. A, D

17. B, D

18. A, B, C, D

19. A, B, C

20. A, B, C

21. A, B, C

22. B, C

23. A, C

24. A, D

25. A, B, D

26. A, D

27. A, D

28. A, B, D

29. C, D

30. A, D

31. A, B, C

32. A, C

33. A, D

34. B, C

35. A, B, D

36. C, D

37. A, B, C, D

38. A, B, C

Reasoning Based Questions 1. A

2. A

Mechanics II_Chapter 4_Part 2.indd 94

3. A

4. D

5. D

6. B

7. C

8. A

9. A

10. C

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Chapter 4: Gravitation and Satellites

Linked Comprehension Type Questions 1. D

2. A

3. C

4. D

5. A

6. B

7. A

8. C

9. C

10. A

11. A, B, C

12. B

13. A

14. D

15. B

16. D

17. D

18. A

19. B

20. C

21. C

22. D

23. D

24. A

25. B

26. C

27. C

28. A

29. A

30. C

31. D

32. C

33. B

34. A

35. D

36. C

37. B

Matrix Match/Column Match Type Questions 1. A → (p, q, r)

B → (r)

C → (t)

D → (p, q, r, s)

2. A → (q)

B → (p)

C → (p)

D → (q)

3. A → (q)

B → (r)

C → (q)

D → (p)

4. A → (p, q, r, s)

B → (p, q, r, s)

C → (q, r, s)

D → (p, r)

5. A → (p)

B → (q)

C → (q)

D → (q)

6. A → (r, t)

B → (s)

C → (p)

D → (q)

7. A → (p)

B → (q)

C → (s)

D → (t)

8. A → (q)

B → (s)

C → (p)

D → (r)

9. A → (r, q)

B → (s)

C → (p)

D → (r)

10. A → (s)

B → (r)

C → (p)

D → (p)

11. A → (r)

B → (q)

C → (p)

D → (r)

E → (s)

Integer/Numerical Answer Type Questions 1. 1.51 6. (a) 30

(b) 18

2. 0

3. 1.25

4. 10

7. α = 2 and β = 13

8. 2

9. 6

10. 2

5. 5656

11. 1

12. 5

13. 7

14. 3

15. 2

16. 6

17. 7155, 7071, 7702

18. 7

19. 1.5

20. 7

21. 2

ARCHIVE: JEE MAIN 1. C

2. A

3. B

4. B

5. A

6. C

7. D

8. D

9. A

10. B

11. D

12. D

13. A

14. A

15. 16

16. B

17. C

18. C

19. A

20. D

21. C

22. D

23. C

24. A

25. B

26. A

27. B

28. C

29. C

30. A

31. C

32. C

33. D

34. A

35. C

36. A

37. D

38. A

39. A

40. A

41. C

42. C

43. B

44. C

45. B

46. D

47. A

48. C

49. D

50. A

5. A

6. C

7. D

8. C

9. D

10. A

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1. B

2. C

3. B

4. B

11. B

12. B

13. A

14. C

Mechanics II_Chapter 4_Part 2.indd 95

09-Feb-21 6:33:09 PM

4.96

JEE Advanced Physics: Mechanics – II

Multiple Correct Choice Type Problems 1. B, C

2. B, D

3. B, D

4. A, B

5. A, C, D

Reasoning Based Questions 1. A

Matrix Match/Column Match Type Questions 1. 2

Mechanics II_Chapter 4_Part 2.indd 96

2. 7

3. 3

09-Feb-21 6:33:09 PM

HINTS AND EXPLANATIONS

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 1

2/9/2021 6:25:09 PM

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 2

2/9/2021 6:25:09 PM

CHAPTER 1: WORK, ENERGY, POWER AND CONSERVATION OF ENERGY

Test Your Concepts-I (Based on Work Done by a Constant and Variable Force)

⇒ ⇒

k

( gh ) − 0

a=

a=

= 2 ah

T

2

m θ

f = mg − ma

⇒

⎛ k2 ⎞ f = mg ⎜ 1 − ⎟ ⎝ 2 ⎠

mg x2

4.

k

2 1

− x22

sin

θ

)

x1

500 ⎡ 2 2 ⎣ ( 0.15 ) − ( 0.08 ) ⎤⎦ = 4.03 J 2

W1 = (b)

W2 = − ( mg sin θ ) ( x2 − x1 ) ⇒ W2 = − ( 6 )( 9.8 ) ( sin 15° )( 0.23 ) = −3.5 J ( 0.1, 0.4 )

5.

Method II: Since we know that

(a) W1 =

∫

  F ⋅ dr

( 0.1, 0 )   where F = −α x 2iˆ and dr = dxiˆ + dyjˆ + dzkˆ

Wnc = ΔU + Δk

2.

∫ ( −kx ) dx = 2 ( x

(a) W1 =

⎛ k2 ⎞ W f = − mgh ⎜ 1 − ⎟ ⎝ 2 ⎠

⇒

⎛1 ⎞ Wair drag = ( 0 − mgh ) + ⎜ mv 2 − 0 ⎟ ⎝2 ⎠

⇒

1 Wair drag = − mgh + mk 2 ( gh ) 2 Wair drag

mg

mg cos θ mg θ

Now W f = fh cos ( 180° )

⇒

N

ϕ

k g 2

⇒

}

Wmg = − mgl sin θ

k2g

mg − f = ma

π 2

{∵ θ = π }

Ff = − fl

2

If f be the air drag, then using Newton’s Second Law, we get

⇒

∵θ=

f

v 2 − u2 = 2 ah 2

{

WN = 0

Method I: Let a be the net downward acceleration of the ball. Then 2

WT = Tl cos ϕ

CHAPTER 1

1.

3.

( 0.1, 0.4 )

⇒ W1 =

∫ ) −α x dx = 0 ( 2

0.1, 0

( 0.3 , 0 )

⎛ k2 ⎞ = − mgh ⎜ 1 − ⎟ ⎝ 2 ⎠

(b)

W2 =

∫

−α x 2 dx = −

) α( 3)( α x = − ( 0.026 ) 3 3 ( 0.1, 0 )

−α x 2 dx = −

) α α( 3)( x = ( 0.026 ) 3 ( 0.3 , 0 ) 3

( 0.1, 0 )

WT = T  0

( 0.1, 0 )

(c)

N T

6.

mg

WN = 0

{∵ θ = 90° }

Wmg = 0

{∵ θ = 90° }

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 3

∫

( 0.3 , 0 )

f

W f = f  0 cos ( 180° ) = − f  0

W3 =

0.3 , 0

0.1, 0

Using our knowledge of constraints we have x = 8 cot θ ⇒

dx = −8cosec 2θ dθ F = 500 N θ

Fcos θ

θ x

y

2/9/2021 6:25:20 PM

H.4

JEE Advanced Physics: Mechanics – II y

60°

Since, W =

∫

( F cos θ ) dx

θ = 30° 60°

⇒

W=

∫

x=2

( 500 cos θ ) ( − 8cosec2θ ) dθ

( 2, 2 )

60°

⇒

W = −4000

∫ cosθ cosec θdθ 2

W2 =

30°

But ⇒

2

2

⇒ W2 = −

2, 2 2α 3 ( ) y 3 ( 2, 0 )

( )

⇒ W2 = −8 J

60°

{∵ α = 1.5 Nm −3 }

⇒

W = −4000 ( − cosec θ ) 30°

⇒

W = 4000 [ cosec ( 60° ) − cosec ( 30° ) ]

⇒

W = 4000 [ 1.155 − 2 ]

Since work done is equal to the area under a curve in F -x graph. So

⇒

W = 4000 ( −0.845 )

⇒

W = −3380 J

W=

W=

∫

(c) 9.

  F ⋅ dr

 r1

W=

W = 10 − 10 + 20 + 10

⇒

W = 30 J

1 1 W = ( 10 × 2 ) + ( 10 ) ( 4 − 2 ) + 0 + ( −5 ) ( 8 − 6 ) 2 2

∫ ( 3x iˆ + 2yjˆ ) ⋅ ( dxiˆ + dyjˆ + dzkˆ ) 2

⇒

 r1

 r2

⇒

W=

∫ ( 3x dx + 2ydy ) = ( x 2

3

+ y2

 r1

⇒

)

(a) W1 =

⇒ W0→ 3 =

( 2, 2 )

∫ −α y dy = ( ∫ ) −α y dy ( ) 2

W3→ 4 = 0

(c)

1 W4→7 = − × 2 × 1 = −1 J 2 W4→7 is negative, because force is in negative x-direction while displacement is towards positive x -axis.

0, 0

(Along the path x = y ) ( 2, 2 )

α 4 y = −4α ∵ α = 1.5 Nm −3 4 ( 0, 0 )

( )

⇒

W1 = −

⇒

W1 = −6 J

{

}

(b) Initially when we move along x-axis, the work done is zero because the force is acting along y-axis. Then along the path parallel to y -axis, we can put x = 2 to get

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 4

1 × 2 × 2 +1× 2 = 4 J 2

(b)

3

0, 0

W = 25 J

11. (a) W = Area under F -x graph

( 4, 6 ) ( 2, 3 )

W = 83 J ( 2, 2 )

1 ( −2 ) ( −10 ) + 1 ( −2 ) ( 10 ) + ( 10 ) ( 2 ) + 1 ( 2 ) ( 10 ) 2 2 2

⇒

 r2

⇒

As W1 ≠ W2 , force is non-conservative in nature.

10. The work from x = 0 to x = 8 m is the area under the curve.

 ˆ where dr = iˆ dx + ˆjdy + kdx

8.

−2α y dy ∫ ( ) 2, 0

∫ cosθ cosec θdθ = ∫ cosecθ cotθdθ ∫ cosecθ cotθdθ = − cosecθ

 r2

7.

x

O x=0

30°

(d) W0→7 = 4 + 0 − 1 = 3 J 12. Work done = Area under F -s graph 1 × 4 × 10 + ( 8 − 4 ) × 10 2

⇒

W=

⇒

W = 20 + 40

⇒

W = 60 J

2/9/2021 6:25:33 PM

Hints and Explanations   13. Since, dW = F ⋅ dr , where  dr = ( dx ) iˆ + ( dy ) ˆj

Work done by the gravitational force is Wmg = ( mg ) ( h ) cos ( 0° ) = mgh Work done by the tension force is

dW = − k ( ydx + xdy )

WT = Th cos ( 180° ) = −

Since, d ( xy ) = x dy + y dx ⇒ ⇒ ⇒

dW = − kd ( xy )

Test Your Concepts-II (Based on Kinetic Energy, Potential Energy and Power)

( a, a )

∫

W = dW = − k

∫

d ( xy )

( 0, 0 )

1.

P = Fv P v Now, we have

( a, a ) W = − k ( xy ) = − ka 2 − 0 = −ka 2 ( 0, 0 )

⇒

14. Since we know that work done is equal to the area under F -s graph.

…(1)

A1 A2 10

20

30 A3

⇒

s(m)

⇒

(a) Work done at the end of 10 m is the area under the curve from 0 to 10 m. ⇒ W = A1 =

⇒

(b) Work done at the end of 20 m is the area under the curve from 0 to 20 m .

⇒

⇒ W = A1 + A2 + A3 = 50 + 50 − 50 = 50 J 15. The FBD of the block placed in the lift is shown in Figure. T (by rope)

h

g 2

mg (weight)

∫ 0

Work done at the end of 30 m is the area under the curve from 0 to 30 m .

2.

mg − T = ma = T=

mg 2

mg 2

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 5

= f

P v

t

⎛ P⎞ vdv = ⎜ ⎟ dt ⎝ m⎠

∫ 0

2

⇒

⎛ P⎞ ⎛ 2P ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ ⎟t 2r m⎠

⇒

t=

Pm 8r 2

(a) P = constant ⇒ Fv = P ⇒ mv

{

dv =P dt

∵F=

dv dt

}

⇒ mvdv = Pdt v

According to Newton’s Second Law, we have

vmax

⎛ dv ⎞ P m⎜ = ⎝ dt ⎟⎠ v P 2r

⇒ W = A1 + A2 = 50 + 50 = 100 J

a=

P

P f

vmax =

Further, since F =

1 ( 10 )( 10 ) J = 50 J 2

m

F= f

So, from (1), we get

–10

⇒

dv =0 dt

At maximum speed,

10

(c)

F=

⎛ dv ⎞ F − f = ma = m ⎜ ⎝ dt ⎟⎠

F(N)

0

mgh 2

CHAPTER 1

⇒

H.5

t

∫

∫

⇒ m vdv = P dt 0

⇒

0

2

mv = Pt 2

⇒ v=

2Pt m

2/9/2021 6:25:45 PM

H.6

JEE Advanced Physics: Mechanics – II

(b) Since, v = ⇒

dx = dt

{

2P 1 2 t m

∵v=

dx dt

800 = −2000 passing 0.4 through the point ( x1y1 ) = ( 0.2, 800 ) is A straight line with slope m = −

}

y − y1 = m ( x − x1 )

2P 1 2 t dt m

⇒ dx = x

⇒

For t > 0.2 s

2Pt m

F(N)

t

2P 1 2 t dt m

∫ dx =

∫

0

{

⎛ 2 2P ⎞ 3 2 ⇒ x=⎜ t ⎝ 3 m ⎟⎠ ds ⎛ 2Pt ⎞ =⎜ ⎟ dt ⎝ m ⎠

12

i.e.,

∵

s

t

0

0

∫t

12

⎛ 2Pt ⎞ ⎟ m ⎠

∫ ds = ∫ ⎜⎝

dt =

2 32 t 3

}

0

Here x-axis is time axis and y-axis is force axis So, we get

12

dt

F − 800 = −2000 ( t − 0.2 )

which on integration gives ⎛ 2P ⎞ s=⎜ ⎝ m ⎟⎠ 3.

12

2 32 ⎛ 8P ⎞ t or s = ⎜ ⎝ 9m ⎟⎠ 3

12

t

32

W = ΔKE 1 Pt = m ( v 2 − u2 ) 2

{∵ W = Pt }

m( 2 v − u2 ) 2P Further, Fv = P ⇒

t=

⇒

⎛ dv ⎞ v=P m⎜ ⎝ ds ⎟⎠

⇒

kv 2

dx =P dt

v

⇒

∫

dv dv =v dt dx

}

⇒

P = Fv = ( −2000t + 1200 )( 50t ) watt

⇒

P = Fv = ( −20t + 12 ) 5t kW

Now, P =

dW dt

⇒

W = Pdt

∫

⇒

W=

0.2

v 2 dv =

∫

⇒

0

( v 3 − u3 ) = 3P x m

3x m = P v 3 − u3 Substituting in Equation (1), we get t=

3x ( u + v )

2 ( u2 + v 2 + uv )

5.

⇒

P = Fv = ( 40000 t ) watt = 40 t kW

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 6

∫ ( −100t

2

+ 60t ) dt

0.2

⎡ ⎛ t W = ⎢ 40 ⎜ ⎢⎣ ⎝ 2

2 0.2 ⎞ 0

⎛ t3 ⎟ − 100 ⎜ ⎠ ⎝ 3

0.4 ⎞

⎛ t2 ⎟ + 60 ⎜ ⎝ 2 0.2 ⎠

⎤ ⎟ ⎥ kJ 0.2 ⎠ ⎥ ⎦ 0.4 ⎞

⇒

100 ⎡ ( 0.056 ) + 30 ( 0.12 ) ⎥⎤ × 1000 J W = ⎢ 20 ( 0.04 ) − 3 ⎣ ⎦

⇒

W = ( 800 − 1867 + 3600 ) J

⇒

W = 2533 J

Since the collar is lifted with a constant speed of 0.6 ms−1. So F cos θ − mg = 0

For t ≤ 0.2 s ⎛ 20 ⎞ F = 800 N and v = ⎜ t = 50 t ⎝ 0.4 ⎟⎠

∫

0.4

40tdt +

0

P dx m

⇒

4.

F = −2000t + 1200

x

u

⇒

∵

F = 800 − 2000t + 400

⇒

for t ≤ 0.2 s ⎡ 40t So, P = ⎢ 2 ⎣ −100t + 60t for t > 0.2 s

…(1)

{

⇒

⎛ 20 ⎞ Further v = ⎜ t = 50 t ⎝ 0.4 ⎟⎠

From Work-Energy Theorem, we have

⇒

(0.2, 800) slope = –2000 t(s) 0.2 0.4

800

0

⇒

F cos θ = 5 ( 10 ) = 50

⇒

F=

50 cos θ

…(1)

2/9/2021 6:25:59 PM

Hints and Explanations   Now P = F ⋅ v ⇒

⎛ 50 ⎞ ( P=⎜ v cos θ ) ⎝ cos θ ⎟⎠

⇒

P = 50v

⇒

P = 30 W

W = ΔK =

{∵ v = 0.6 ms−1 }

Initial velocity u i.e. velocity at x = 0 is u = a( 0 ) = 0 Final velocity v i.e. velocity at x = 2 is

F = R = 3 × 10 4 N

( ) 3

v = 5 2 2 = 10 2 ms −1

Since we know that P = Fv ⇒ 7.

P = 3 × 10 4 × 40 = 1.2 × 106 watt

Since, F ∝ x −1 2 ⇒

dv a=v = kx −1 2 dx

∫

x

⇒

1 2x 2 1

8.

⇒

P∝

⇒

1 n=− 4

1 x2

1 1 − 2

= x4

=x

−

1 4

i.e., P ∝ x

−

1 4

 Here the only force acting is F = ( 2 + 3 x ) iˆ . So, body starts moving along x-axis. According to WorkEnergy Theorem,   W = F ⋅ dx = ΔK

∫

5

⇒

∫

1 2

⇒ ⇒ ⇒

⇒

dm ⎛ dV ⎞ = ρ⎜ = Avρ ⎝ dt ⎟⎠ dt

⇒

⎛ dm ⎞ F = v⎜ = v ( Avρ ) ⎝ dt ⎟⎠

⇒

F = Av 2 ρ

⇒

P = Fv = Av 3 ρ

⇒

P ∝ v3

So, n = 3 11.

ZPEL L

60°

L 30°

⎛ 3x ⎞ ⎜⎝ 2x + ⎟ 2 ⎠ 2

10 +

5

= 0

5 2 v 2

75 5 − 0 = v2 2 2

v = 19 ms

−1

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 7

m O L sin (30°) = A

( 2 + 3 x ) dx = ( 5 ) v 2 − 0

0

dm ⎛ dV ⎞ = ρ⎜ ⎝ dt ⎟⎠ dt

where A is the area of the generator through which wind flows

P = Fv 1

dm dt

dV ⎛ dx ⎞ = A⎜ = Av ⎝ dt ⎟⎠ dt

Since

⇒ v ∝ x4 Since power is defined as

x4

W = 50 J

where,

0

v = 2

⇒

1 ( 2 1 m v − u2 ) = ( 0.5 )( 200 ) 2 2

dV is the volume of wind flowing per second dt through the generator.

∫

2

W = ΔK =

If ρ be the density of wind, then

vdv = x −1 2 dx

0

⇒

10. Since P = Fv , where F = v

⇒ vdv = kx −1 2 dx Integrating both sides, we get v

1 ( 2 m v − u2 ) 2

Given that m = 0.5 kg and v = ax x = 5x 3 2

At constant speed, there is no acceleration, so the forces acting on the train are in equilibrium. ⇒

According to Work-Energy Theorem, we have

CHAPTER 1

6.

9.

H.7

L 2

reference line

Let the reference line pass through point of support as shown. ⇒

UO = 0

⇒

⎛ L⎞ U A = − mg ⎜ ⎟ and U B = − mgL ⎝ 2⎠

2/9/2021 6:26:13 PM

H.8

JEE Advanced Physics: Mechanics – II So, elastic energy stored in the final position is

mgL = U A − UB 2

⇒

UO − U A =

⇒

UO − U A =1 U A − UB

U=

1 2 1 2 kx2 = × 200 × ( 0.3 ) = 9 J 2 2

15.

12. Potential energy (PE) for a stretch of 20 cm is

(cm)f

1 2 U1 = k ( 0.2 ) = 30 J 2 ⇒

k=

(cm)i

30 × 2 = 1500 Nm −1 0.04

⇒

1 2 × 1500 × ( 0.6 ) = 1500 × 0.18 2

⎛ 7.5 ⎞ ΔU = mgh = ( 5 )( 10 ) ⎜ = 3.75 J ⎝ 100 ⎟⎠ 16.

U 2 = 270 J

0.25 m

60°

30° 0.5 m

Additional work required is

Final

W = U 2 − U1 = 270 − 30 = 240 J

B Initial

13. Fresistive = 20 vmax (in kN)

Vmax

Initially, the centre of mass of the metre scale lies 0.5 m below the hinge and finally 0.25 m below the hinge when the scale is turned to a position making 30° with horizontal. Therefore, centre of mass of the scale rises by

Fext

Power of engine is P = 50 HP = 50 × 746 = 37300 W

h = 0.5 − 0.25 = 0.25 m Hence work done by gravity is W = mgh cos180°

Let vmax be the maximum speed attained by the boat. At v = vmax , acceleration of boat is zero, so net force acting on boat is also zero. ⇒

Fext − Fresistive = 0

⇒

Fext = ( 20vmax ) 1000 N

Since, P = Fv

⇒

37300 = ( 20000vmax ) vmax

⇒

2 = vmax

⇒

vmax =

37300 20000

W = 0.1 × 10 × 0.25 × ( −1 ) = −0.25 J

Test Your Concepts-III (Based on Conservation of Energy and Work Energy Theorem) 1.

⇒

In this case three forces are acting on the object (i)

tension ( T )

(ii) weight ( mg ) and (iii) applied force ( F )

373 = 1.36 ms −1 = 4.9 kmh −1 200

h = l(1 – cos θ ) θ l T

14. When mass is suspended from the spring, then in equilibrium position, let extension in spring be x1. ⇒

mg = kx1

⇒

k=

10 cm

Since centre of mass of brick rises by 7.5 cm , so

PE acquired for a stretch of 20 + 40 = 60 cm is U2 =

7.5 cm 2.5 cm

mg 4 × 10 = = 200 Nm −1 ( x1 0.5 − 0.3 )

F mg

Using Work-Energy Theorem Wnet = ΔK .E.

Total stretch in the spring in the final state is x2 = 0.6 − 0.3 = 0.3 m

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 8

h

⇒

WT + Wmg + WF = 0

…(1)

2/9/2021 6:26:24 PM

Hints and Explanations ΔK .E. = 0 because K i = K f = 0

3.

From conservation of energy, 1 1 m2 gh2 = m1 gh1 + m1v12 + m2v22 2 2

Further, WT = 0 , as tension is always perpendicular to displacement. Wmg = − mgh ⇒

Wmg

2m

= − mg ( 1 − cos θ )

θ

1m

Substituting these values in Equation (1), we get

v2

WF = mg ( 1 − cos θ )

The initial and final configurations are shown in the figure. It is convenient to set U g = 0 at the floor. Initially, only m2 has potential energy. As it falls, it loses potential energy and gains kinetic energy. At the same time, m1 gains potential energy and kinetic energy. Just before m2 lands, it has only kinetic energy. Let v the final speed of each mass. Then, using the Law of Conservation of Mechanical Energy.

y v1 = 0 y1 = 0 Initial

x

O

v′2 = v y′2 = 0

⇒

1 ( m1 + m2 ) v2 + m1 gh = 0 + m2 gh 2

2

4.

⇒

2 ( m2 − m1 ) gh v = m1 + m2

Final

Substitute m1 = 3 kg, m2 = 5 kg, h = 5 m and g = 10 ms −2 . 2 ( 5 − 3 )( 10 )( 5 ) 5+3

⇒

v2 =

⇒

v = 5 ms −1

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 9

⇒

20 = 1.18 + 0.2v22 + v22

⇒

v2 = 3.96 ms −1 and v1 = v2 cos θ =

⇒

v1 = 3.54 ms −1

2 × 3.96 5

According to Work Energy Theorem, applied on A to B , we have ⎛ Work done by force ⎞ ⎛ Change in KE ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ from A to B ⎟⎠ from A to B

v′1 = v y′1 = h

2

2 × 10 × 1 = ( 0.5 )( 10 ) ( 5 − 2 ) + 1 1 ⎛ 2 ⎞ × 0.5 × v22 × ⎜ + × 2 × v22 ⎝ 5 ⎟⎠ 2 2

K f + U f = Ki + Ui ⇒

v1

v1 = v2 cos θ

Here the applied force F is variable. So, if we do not apply the work energy theorem, we will first find the magnitude of F at different locations and then inte  grate dW ( = F ⋅ dr ) with proper limits.

v2 = 0 y2 = h

m1

Using our knowledge of constraint relations, we get

Conceptual Note(s)

2.

√5 m

m2

θ

CHAPTER 1

as

H.9

(

1 m vB2 − vA2 2

)

⇒

WAB =

⇒

( 2T − mg sin θ ) ( AB ) = 2

⇒

5 ⎞ 1 ⎛ 2 ⎜⎝ 1150 − 130 × 10 × ⎟⎠ ( 3 ) = × 130 vB − 9 13 2

⇒

( 650 )( 3 ) = 65 ( vB2 − 9 )

⇒

vB2 − 9 = 30

⇒

vB = 39 = 6.25 ms −1

1

(

m vB2 − vA2

) (

)

While using work-energy theorem we are concerned with speed not the velocity. 5.

(a) For spring 1, X10 = For spring 2, X 20 = ⇒ U=

( m1 + m2 ) g = k1

30 = 0.02 m 1500

m2 g 10 = = 0.02 m k2 500

1 1 2 2 2 k1x01 + k2 x02 = ( 750 + 250 )( 0.02 ) 2 2

⇒ U = 0.4 J

2/9/2021 6:26:34 PM

H.10 JEE Advanced Physics: Mechanics – II (b) Let Δx1 and Δx2 be the additional elongation due to pulling m2 downwards by  = 0.08 m . Additional forces on m2 are equal in magnitude and opposite in direction.

7.

Let downward velocity of sphere be vs and horizontal velocity of wedge be vw . From our knowledge of constraint relations, the constraint equation for velocity can be written as, vs = tan α vw

⇒ k1Δx1 = k2 Δx2 ⇒

Δx1 1 = Δx2 3

Also, Δx1 + Δx2 = 0.08

…(1)

⇒

…(2)

When the sphere hits the ground, its centre will still be at a height R above the ground. Using Law of Conservation of Energy, we have

Solving Equation (1) and (2), we get Δx1 = 0.02 m and Δx2 = 0.06 m Wtotal = ΔK Since, m2 is pulled down slowly, so ΔK = 0 ⇒ Wmg + Wspring + WF = ΔKE = 0 …(3)

Also, Wspring = − ( U f − U i ) = U i − U f where, U i = 0.4 J and

⇒

(

)

1 1 2 k1 x01 + Δx12 + k2 ( x02 + Δx2 ) = 2.8 J 2 2

The lower block will bounce when extension in spring, kx = Mg

mgh − mgx =

⇒

2mgR = m ( 1 + tan 2 α ) vw2 = mvw2 sec 2 α

⇒

vw = 2 gR cos α and vs = vw tan α = 2 gR sin α

8.

(a) According to Work Energy Theorem, we have

⇒

1 1 mv 2 = WF + Wspring = Fx − kx 2 2 2

⇒

1 1 2 × 0.5 × v 2 = 20 × 0.25 − × 40 × ( 0.25 ) 2 2

⇒ v = 3.87 ms −1

Mg x= k

…(1)

By Law of Conservation of Energy, loss in gravitational potential energy (GPE) equals gain in elastic potential energy (EPE) of spring, so

⇒

⇒

Wtotal = ΔK

WF = − ( Wmg + Wspring ) = − ( 1.2 − 2.4 ) J = 1.2 J

⇒

mg ( 2R ) − mgR =

Wspring = −2.4 J

Since, from (3), we have

6.

1 1 mvs2 + mvw2 2 2

⇒

“Please note that the centre of mass of wedge stays at same distance from the ground both in the initial and final situations, so we have not accounted for change in potential energy of wedge, because that happens to be zero”.

Now, Wmg = m1 g Δx1 + m2 g ( Δx2 + Δx1 ) = 1.2 J

Uf =

…(1)

⎛ Loss in PE ⎞ ⎛ Gain in KE ⎞ ⎛ Gain in KE ⎞ ⎜⎝ of sphere ⎟⎠ = ⎜⎝ of sphere ⎟⎠ + ⎜⎝ of wedge ⎟⎠

According to Work Energy Theorem

⇒ WF = − ( Wmg + Wspring )

vs = vw tan α

2

M 2 g 2 Mmg 2 + M 2 g Mg k h = 2k = + mg k 2km

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 10

E=

1 2 1 kx + mv 2 = Fx = 20 × 0.25 = 5 J 2 2

If x f be the final maximum extension in the spring, then

1 2 kx 2

Mg ⎞ 1 ⎛ Mg ⎞ ⎛ mg ⎜ h − ⎟ = k⎜ ⎟ ⎝ k ⎠ 2 ⎝ k ⎠

(b) Total mechanical energy at point B is also equal to work done by the external force F . So, we have

1 2 kx f = 5 J 2 ⇒ xf =

10 = k

10 = 0.5 m 40

Hence, closest distance from wall is d = 0.6 − x f = 0.1 m

2/9/2021 6:26:46 PM

Hints and Explanations

11. When the block descends 12 mm, spring will further stretch 24 m. So total extension in the spring is ( 76 + 24 ) mm i.e., 100 mm

dθ

⎛ Decrease ⎞ ⎛ Increase ⎞ ⎛ Increase in ⎞ ⎜ in PE ⎟ = ⎜ in KE ⎟ + ⎜ Elastic PE ⎟ ⎜ of Block ⎟ ⎜ of Block ⎟ ⎜ of spring ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

θ

h=0 Final

Initial

⇒

1 2

( 50 )( 10 )( 0.012 ) = ( 1000 ) ⎡⎣ ( 0.1 )2 − ( 0.076 )2 ⎤⎦ + 1 ( 50 ) v 2 2

Initial potential energy is θ =π 2

∫

Ui =

( rdθ )( λ ) ( g ) ( r cos θ )

θ = 0°

(

U i = λ gr

2

) sin θ

π 2 0

6 = 25v 2 + 2.5

⇒

v = 0.374 ms −1

12. (a) From Work Energy Theorem, we have = λ gr

2

π (r 2) Finally, the centre of mass of chain is below 2 the last link or the slot. So, final potential energy is given by

⇒

⇒

2

ΔK = W =





∫ F ⋅ dr 0

 where dr = dxiˆ + dyjˆ + dzkˆ 2

∫ ( 2.5 − x )dx = 2.33 J

π 2r 2λ g ⎛ πr 2 ⎞ ⎛ πr ⎞ Uf = ⎜ × λ ⎟ ( g )⎜ − ⎟=− ⎝ ⎠ ⎝ 2 ⎠ 2 8

⇒ W=

⎛ π2 ⎞ ΔU = U f − U i = − r 2 λ g ⎜ 1 + ⎟ ⎝ 8 ⎠

⇒ W = ΔK = 2.33 J

By Law of Conservation of Energy Loss in PE = Gain in KE

2

0

(b) Speed of block at x is given by 1 × 1.5 × v 2 = 2

x

∫ ( 2.5 − x ) dx 2

⇒

− ΔU = ΔK

⇒

⎛ π 2 ⎞ 1 ⎛ πr ⎞ 2 r 2λ g ⎜ 1 + ⎟ ( λ )v ⎟= ⎜ ⎝ 8 ⎠ 2⎝ 2 ⎠

⎛ x3 ⎞ ⇒ v 2 = 1.33 ⎜ 2.5x − ⎟ ⎝ 3 ⎠

⇒

⎛ 1 π⎞ ⎛ 4 π⎞ v = 4 rg ⎜ + ⎟ = rg ⎜ + ⎟ ⎝π 8⎠ ⎝π 2⎠

For maximum K.E.,

10. Let T be the tension in the cable, then T − mg = ma T − 80 g = 80 a ⇒

T = 800 + 80 = 880 N

(a)

WT = Tx cos ( 0° ) = ( 880 )( 15 ) = 13.2 kJ

(b)

Wmg = mgx cos ( 180° ) = − ( 880 )( 15 ) = −12 kJ

(c)

KE = Wtotal = 13.2 − 12 = 1.2 kJ

(d)

1 mv 2 = 1200 2

⇒

40v 2 = 1200

⇒

v = 5.47 ms −1

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 11

CHAPTER 1

9. θ

H.11

⇒

0

dv 2 =0 dx

x3 ⎞ d ⎛ ⎜⎝ 2.5x − ⎟⎠ = 0 dx 3

⇒ 2.5 − x 2 = 0 ⇒ x 2 = 2.5 ⇒ x = 1.58 m ⇒ K max =

⎛ ( 1.58 )3 ⎞ 1 × 1.5 × 1.33 ⎜ 2.5 × 1.58 − ⎟ ⎝ 2 3 ⎠

⇒ K max = 2.635 J x

13. (a) U ( x ) = −Wspring =

αx ∫ ( α x + βx2 ) dx = 0

2

2

+

βx3 3

2/9/2021 6:26:58 PM

H.12 JEE Advanced Physics: Mechanics – II (b) By Law of Conservation of Energy, we get ⎛ Gain in KE ⎞ ⎛ Loss in PE ⎞ ⎜⎝ of Mass ⎟⎠ = ⎜⎝ of Spring ⎟⎠ ⇒

1 mv 2 = U i − U f = U ( 1 m ) − U ( 0.5 m ) 2

Since U ( x ) =

by xmax sin θ ⇒

α x2 βx3 + 2 3

⇒ U (1 m ) =

2

Thus, xmax = 3

80 ( 0.5 ) 24 ( 0.5 ) + = 11 J 2 3

1 ( 2 ) v 2 = 48 − 11 2

⇒ v 2 = 48 − 11 ⇒ v = 37 ms −1 ≈ 6 ms −1

1 1 ( m1 + m2 ) v2 + ( − m2 gx + m1 gx sin θ ) + 2 kx 2 = 0 2

kx = mg sin θ + μ s gm cos θ x=

⇒

U=

2g ( m2 − m1 sin θ ) = 0.98 m k

(b) In this case the change in kinetic energy is 1 ΔK = ( m1 + m2 ) v 2 . 2 The change in potential energy has the same form as in part (a), but with xmax replaced by x = 0.5 m , so

14. Block will begin to move when,

⇒

ΔK + ΔU g + ΔU s = 0

1 ⇒ 0 + ( − m2 gxmax + m1 gxmax sin θ ) + kx 2max = 0 2

80 ( 1 ) 24 ( 1 ) + = 48 J 2 3

⇒ U ( 0.5 m ) = ⇒

(a) At the maximum extension xmax , the blocks come to rest, and thus ΔK = 0 . Next, we must find the changes in Ug and Us. When m2 falls by xmax , the spring extends by xmax and m1 rises

Substituting values we get v = 1.39 ms−1.

mg ( sin θ + μ s cos θ )

17. When the block is released, the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural length. Thereafter, the block loses contact with the spring and moves with constant velocity.

k 1 2 ⎡⎣ mg ( sin θ + μ s cos θ ) ⎤⎦ kx = 2 2k

2

15. Using our knowledge of constrained motion of connected bodies, we observe that if x A = h (say), then xB = 2 h. Also, vA = x A = 2 ms −1 ⇒

v

vB = x B = 2vA = 4 ms −1 x

Using Law of Conservation of Energy, we get ⎛ Decrease ⎞ ⎛ Increase ⎞ ⎛ Increase in KE ⎞ ⎜⎝ in PE of A ⎟⎠ = ⎜⎝ in PE of B ⎟⎠ + ⎜⎝ of both A and B ⎟⎠ 1 2

1 2

⇒

( 30 )( 10 ) h = ( 5 )( 10 )( 2h ) + ( 30 ) ( 2 )2 + ( 5 ) ( 4 )2

⇒

300 h = 100 h + 60 + 40

⇒

200 h = 100

⇒

h = 0.5 m

16. To use E f = Ei we would need to assign the initial heights of the blocks arbitrary values h1 and h2 . The corresponding potential energies, m1 gh1 and m2 gh2 would appear in both Ei and E f and hence would cancel. We avoid this process by using the form ΔK + ΔU = 0 instead, since it does not require U = 0 reference level.

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 12

L0 . When 2 the distance of the block from the wall becomes x , where x < L0 , then compression is ( L0 − x ) . Using the Law of Conservation of Energy, Initially, the compression in the spring is

2

1 ⎛ L0 ⎞ 1 1 2 k ⎜ ⎟ = k ( L0 − x ) + mv 2 2 ⎝ 2 ⎠ 2 2 ⇒

v=

k ⎡ L20 2⎤ ⎢ − ( L0 − x ) ⎥ m⎣ 4 ⎦

When the spring acquires its natural length i.e., x = L0 , ⎛ k ⎞ L0 . Thereafter, the then velocity becomes v = ⎜ ⎝ m ⎟⎠ 2 block continues to move with this velocity.

2/9/2021 6:27:12 PM

Hints and Explanations 18. Since all surfaces are smooth. According to Law of Conservation of Mechanical Energy, we have

1 2 kxo and 2 E f = mgd sin θ . From Work Energy Theorem,

(a) Ki and Kf both are zero, so Ei =

⎛ Decrease in P.E. of ⎞ ⎛ Increase in K.E. of ⎞ ⎜⎝ both the blocks ⎟⎠ = ⎜⎝ both the blocks ⎟⎠

⇒

v=

Wnc = E f − Ei = ΔU + ΔK 1 ⇒ mgd sin θ − kx02 = − fd 2 ⇒ f = 0.82 N

2 ( 1 + π ) gr 3

(b) Initial energy mechanical Ei is still the same as above, but final mechanical energy at x = 0 is

Test Your Concepts-IV (Based on Work Energy Theorem for Non-conservative Systems) 1.

1 mv 2 + mgxo sin θ 2 From Work Energy Theorem, we have Ef =

(a) When the box is first placed on the belt there will be slipping between the two. But the force of friction on the box and its displacement are in the same direction. Consequently, the work done by kinetic friction is positive. Since the final speed of the box is v,

Wnc = E f − Ei = ΔU + ΔK ⇒

⇒ v = 2.45 ms −1 3.

N fk mg

1 ⎛ ⎞ (d) Total work done on the block = − ⎜ μ Mg + k 2 ⎟ ⎝ ⎠ 2 (e) According to Work Energy Theorem Wtotal = ΔK

…(1)

f = μ k N = μ k mg and W f = + fd , so from equation (1), we get

(b) The force of friction is 1 + μ k mgd = + mv 2 2 ⇒ (c)

(a) Work done by friction = − μ Mg 1 (b) Work done by the spring force = − k 2 2 (c) Gravitational force and normal reaction of the table do not work as they act in a direction perpendicular to displacement.

v

1 W f = ΔK = + mv 2 2

1 1 mv 2 + mgx0 sin θ − kx02 = − fx0 2 2

CHAPTER 1

⇒

1 πr ( mgr ) + ( 2mg ) ⎛⎜⎝ 2 ⎞⎟⎠ = 2 ( m + 2m ) v2

H.13

⇒ WC + WnC = ΔK

…(2)

⇒

v2 d= 2μk g

− ΔU + WnC = ΔK

⇒ 0−

If the box takes a time t to reach speed v , then v = at where a is the acceleration of box. In this 1 1 time it will move d = at 2 = vt. Since the belt’s 2 2 speed is fixed, in time t it moves a distance vt = 2d . The belt moves twice as far as the box while the box is accelerating.

1 1 ⎛ ⎞ Mv02 = − ⎜ μ Mg + k 2 ⎟ ⎝ ⎠ 2 2

⇒ 2 + ⇒ = 4.

2 μ Mg Mv02 − =0 k k

1⎡ 2 2 2 μ M g + Mkv02 − μ Mg ⎤⎦ k⎣

Net retarding force F = kx + μ N ⇒

F = kx + Mgbx

If a be the net retardation, then

2. d

a=−

x x=0 Us = 0 Ug = 0

0

⇒

∫

v0

vdv F ⎛ k + Mgb ⎞ = =⎜ ⎟x dx m ⎝ M ⎠ x

⎛ k + Mgb ⎞ vdv = − ⎜ ⎟ xdx ⎝ M ⎠

∫ 0

37°

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 13

2/9/2021 6:27:23 PM

H.14 JEE Advanced Physics: Mechanics – II 2 v02 ⎛ k + Mgb ⎞ x = −⎜ ⎟⎠ ⎝ 2 2 M

⇒

−

⇒

x = v0

Since

M k + Mgb

⇒

− ΔE = ( U + K )initial − ( U + K )final

1 ⎞ ⎛ ⎞ ⎛1 ⇒ Loss = ⎜ 0 + Mv02 ⎟ − ⎜ kx 2 + 0 ⎟ ⎠ ⎝ ⎠ ⎝2 2

5.

⇒ 7.

)

vmax = 0.82 ms −1

From Work Energy Theorem, we have ⇒

Wair = ΔKE − Wmg

⎛ M 2v02 ⎞ ⎛ gb ⎞ ⇒ Loss = ⎜ ⎝ 2 ⎟⎠ ⎜⎝ k + Mgb ⎟⎠

⇒

Wair =

From B to C

Negative sign indicates that work is done against the push of the air.

0 = ( 4 ) − 2 ( μK g ) ( 4 ) 2

⇒

8.

16 μK = = 0.2 ( 2 ) ( 10 ) ( 4 )

Block m2 will shift when kx = μm2 g m1

…(1) F

μ

For m1 , using Work Energy Theorem, we get

Wnc = ΔU + ΔK

1 2 kx + μm1xg 2 kx = 2 F − 2 μm1 g Fx =

Wfriction

1 = − mgR + mvB2 2

⇒

Wfriction

1 = − ( 0.2 )( 10 ) ( 2 ) + ( 0.2 )( 16 ) 2

⇒

Wfriction = −4 + 1.6 = −2.4 J

⇒

1 2 × 5 × ( 10 ) − ( 5 )( 10 )( 20 ) = −750 J 2

m2

From A to B Since we know that

6.

(

k 2 x − x02 − 2 μ K g ( x − x0 ) m

Wmg + Wair = ΔKE

M 1 1 Mv02 − kx 2 , where x = v0 2 2 k + Mgb

⇒ Loss =

vmax =

)

Substituting, k = 2000 Nm −1, m = 5 kg, x = 0.05 m, x0 = 0.009 m, μ k = 0.36

Loss in Mechanical Energy is − ΔE = Ei − E f ⇒

(

1 1 2 mvmax + μ K mg ( x − x0 ) = k x 2 − x02 2 2

⇒

(a) 100 = kx = 2000 x

Equating Equation (1) and (2), we get m ⎞ ⎛ F = ⎜ m1 + 2 ⎟ μ g ⎝ 2 ⎠ 9.

For horizontal portion, retardation a1 = μ g = 1.5 ms −2

⇒ x = 0.05 m

⇒

1 1 Since, mv 2 + μ k mgx = kx 2 2 2

v 2 = 2 a1s1 = 2 × 1.5 × 0.5 = 1.5 m 2s −2

⇒

v = 1.22 ms −1

⇒ v=

k 2 x − 2 μ K gx m

For inclined portion, acceleration is given by ⎛1 3⎞ a2 = g sin θ − μ g cos θ = ⎜ − 0.15 × ⎟ × 10 ⎝2 2 ⎠

⎛ 2000 ⎞ ( 2 ⇒ v= ⎜ 0.05 ) − 2 ( 0.36 )( 10 )( 0.05 ) ⎝ 5 ⎟⎠

⇒

a2 = 3.7 ms −2

⇒ v = 0.8 ms −1

⇒

s2 =

(b) Velocity is maximum where kx0 = μ K mg kx0

⇒ x0 =

1.5 v2 = = 0.2 m 2 a2 2 × 3.7

Work performed by friction forces is Wfriction = Wcase I + Wcase II

μ kmg

μ K mg 0.36 × 5 × 10 = = 0.009 m k 2000

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 14

…(2)

⇒

W f = − μmgs1 − ( μmg cos θ ) s2

⇒

⎛ 3⎞ W f = −0.15 × 0.05 × 10 ⎜ 0.5 + 0.2 × ⎟ = −0.044 J ⎝ 2 ⎠

2/9/2021 6:27:38 PM

Hints and Explanations 3.

vA = 2vB Therefore, vA = 2 ( 0.3 ) = 0.6 ms −1

⇒

as vB = 0.3 ms −1

{given}

⇒

Since Wnc = ΔU + ΔK

x A = 2xB = 2 m,

⇒

since

xB = 1 m,

⇒ mA = 4 kg,

⇒

μ = 0.115

⇒

W = μmgL

⇒

W = 0.5 × 1 × 10 × 1 = 5 J

1.

 We use F = −∇U = − 2 yiˆ + ( 2x + z ) ˆj + ykˆ

2.

  ⎛ ∂U ˆ ∂U ˆ ∂U ˆ ⎞ i+ j+ k Since, F = −∇U = − ⎜ ∂y ∂z ⎟⎠ ⎝ ∂x

(

( x, y )

)

4.

∫ ( 2axy dx + ax dy ) − ay dy ( ) 2

∫

⎧ ⎛ y3 ⎞ ⎫ ⎨ d ax 2 y − d ⎜ ⎬ ⎝ 3 ⎟⎠ ⎭ ⎩

(

)

2

⎛ y3 ⎞ d ⎜ ax 2 y − ⎟ ⎝ 3 ⎠

⇒

( x, y ) ⎤ ⎡⎛ ay 3 ⎞ ⎥ U − U 0 = ⎢ ⎜ − ax 2 y + ⎢⎝ ⎥ 3 ⎟⎠ ( 0 , 0 ) ⎥⎦ ⎢⎣

⇒

U = U 0 − ax 2 y +

ay 3 3

  Since ΔU = − F ⋅ dl

∫

where, dl = i dx + j dy + k dz

∂U = + aU 0 sin ( ax ) exp ( − az ) ∂x

( 2, 3 )

⇒

ΔU =

4 ( yi + x j ) ⋅ ( i dx + j dy + k dz ) ∫ ( ) 0,0

( 2, 3 )

⇒

Since, U has no dependence on y, so

U − ( −4 ) = 4

ydx + xdy ∫ ( ) 0,0

∂U =0 ∂y

Since, we know that d ( xy ) = xdy + ydx

∂U ∂ = −U 0 cos ( ax ) [exp ( − az )] Fz = − ∂z ∂z

( 2, 3 )

⇒

∂U = aU 0 cos ( ax ) exp ( − az ) ∂z

⇒

Fz = −

⇒

F = Fx iˆ + Fz kˆ

⇒

 F = aU 0 exp ( − az ) ⎡⎣ sin ( ax ) iˆ + cos ( ax ) kˆ ⎤⎦

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 15

U − U0 = −

( 0, 0 )

∂U ∂ = −U 0 exp ( − az ) [ cos ( ax ) ] ∂x ∂x

iˆ

∫

( 0, 0 )

Test Your Concepts-V (Based on Relation Between Conservative Force and Potential Energy and Types of Equilibrium)

Fy =

U( x , y ) − U( 0 , 0 ) = −

( x, y )

⇒

Fx =

)

0, 0

U − U0 = −

⇒

)

− y 2 dy

Let U( x , y ) = U and U( 0 , 0 ) = U 0 , then we have

W = μmg × horizontal distance covered

⇒

2

0, 0

11. Since the applied force is always parallel to the track, normal reaction will always be mg cos θ . In this case, work done against friction is

Fx = −

∫ dU = − ( ∫ ) 2axydx + a ( x ( ) ( x, y )

−80 μ = −10 + 0.72 + 0.045 80 μ = 9.235

( x, y )

0, 0

mB = 1 kg , so we get ⇒

) ( dU = − ⎡⎣ 2 axydx + a ( x 2 − y 2 ) dy ⎤⎦ (

dU = − ⎡⎣ 2 axy iˆ + a x 2 − y 2 ˆj ⎤⎦ ⋅ dx iˆ + dy ˆj + dz kˆ

( x, y )

1 1 − μmA gx A = − mB gxB + mAvA2 + mBvB2 2 2

⇒

Since we know that   dU = − F ⋅ dl

CHAPTER 1

10. From constraint relations, we can see that

H.15

U − ( −4 ) = 4

∫

d ( xy )

( 0,0 )

{∵ Fy = 0 }

( 2, 3 ) ( 0,0 )

⇒

U − ( −4 ) = 4 ( xy )

⇒

U − ( −4 ) = 4 [ ( 2 ) ( 3 ) − ( 0 )( 0 ) ] = 24

⇒

U = 24 − 4 = 20 J

2/9/2021 6:27:53 PM

H.16 JEE Advanced Physics: Mechanics – II 5.

Potential energy of the particle is given as 2

U = 2 − 20 x + 5x Therefore, its potential energy at x = −3 is

…(1)

Test Your Concepts-VI (Based on Vertical Circle) 1.

2

U1 = 2 − 20 × ( −3 ) + 5 ( −3 ) = 2 + 60 + 45 = 107 J Since the particle is released from x = −3 m , its kinetic energy at this position is zero and therefore its total energy is 107 + 0 = 107 J.

Now from Law of Conservation of Energy, we get mgh =

Particle will again come to rest at a position where its potential energy once again becomes 107 J. Substituting U = 107 J in equation (1), we get 2 − 20 x + 5x 2 = 107 ⇒

5x 2 − 20 x − 105 = 0

⇒

x 2 − 4 x − 21 = 0

⇒

x = −3 m and x = 7 m

Since F = −

2.

dU dx

⇒

dU = − Fdx

⇒

U(x) = −

x

∫

U(x) =

( − kx + ax 3 ) dx

=

U( x )

⇒

∫ 0

⇒

h≥

5 R 2

5 R 2

Let T be the tension in the string at the lowest point of circle. Then the net force on the stone is ( T − mg ), which is directed towards the centre of the circle and provides necessary centripetal force to the particle to revolve in a circle of radius r . So

O T

2k a 2k a

mg

2k , U ( x ) first increases and then a

decreases. From the given function we can see that F = 0 at x = 0 i.e., slope of U -x graph is zero at x = 0 . Therefore, the most appropriate OPTION is (D). Since, F = −

⇒

kx 2 ⎛ a ⎞⎛ a ⎞ x⎟ ⎜ 1 − x⎟ ⎜⎝ 1 + ⎠ ⎝ 2 2k 2k ⎠

Also, U ( x ) is negative for x >

7.

v 2 = 2 gh ≥ 5 gR

mv 2 r where v is the speed of the stone at the lowest point of the circle of radius r.

kx 2 ax 4 − 2 4

Now, U ( x ) = 0 at x = 0 and x =

For 0 < x
0 k

4u2 − 20 gr = u2 + gr

⇒

3u2 = 21gr

⇒

u = 7 gr

2E m Hence, the correct answer is (B).

Since v2 = u2 − 2gh

P = ( Fnet ) v

⇒

v = 7 gr − 2 g ( 2r ) = 3gr

Fnet = mg + 0.1mg + 0.02mg

⇒

⎛ 10 ⎞ v = 3 ( 10 ) ⎜ ⎟ = 10 ms −1 ⎝ 3 ⎠

⇒

Fnet = 1.12mg

Hence, the correct answer is (D).

⇒

P = ( 1.12 ) ( 1000 )( 10 )( 10 )

⇒

84.

1 mv 2 = E 2 v=

where

88.

⇒ P = 112 kW Hence, the correct answer is (A). 85.

)

(

(

It will rise further to a height h (say), then by Law of Conservation of Energy, we get

)

mgh = ⇒

)

⇒

W = 10 3 iˆ + ˆj + kˆ ⋅ ( 2iˆ + kˆ )

⇒

W = 10 3 ( 2 + 1 ) = 30 3 J

⇒

π 2

89.

Wext = WF0 =

0

x = 0.1 m Tension is maximum at the lowest point, so

(

m 2 u + gr TL = r

π F0l 1 2 = mv1 + mgl 2 2

)

l ( π F0 − 2mg ) m Hence, the correct answer is (A). ⇒

Tension is minimum at the highest point, so TH =

(

m 2 u − 5 gr r

Given that

)

TL =4 TH

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 32

∫

  F0 ⋅ dr

πF l ⇒ WF0 = 0 2 Applying Law of Conservation of Mechanical Energy, we get

Hence, the correct answer is (A). 87.

( L −  ) ( 1 − cos θ ) = L ( 1 − cos α )

⎛ L cos α −  ⎞ θ = cos −1 ⎜ ⎝ L −  ⎟⎠ Hence, the correct answer is (C).

⎛ Loss in Gravitational ⎞ ⎛ Gain in Elastic ⎞ ⎜⎝ Potential Energy ⎟⎠ = ⎜⎝ Potential Energy ⎟⎠ 1 2 kx 2 Putting values to solve the Quadratic Equation, we get

v2 = L ( 1 − cos α ) 2g

⇒

By Law of Conservation of Energy, we have

mg ( h + x ) =

h=

1 mv 2 2

Now h = ( L −  ) ( 1 − cos θ )

Hence, the correct answer is (B).

⇒

Velocity of bob at the lowest point just before hitting the pin is v 2 = 2 gL ( 1 − cos α )

iˆ + ˆj + kˆ Unit vector corresponding to iˆ + ˆj + kˆ is 3  30 ˆ ˆ ˆ ˆ ˆ ˆ i + j + k = 10 3 i + j + k So, F = 3      Since, W = F ⋅ Δr = F ⋅ ( r2 − r1 )

(

86.

) )

⇒

So, Total Energy = Kinetic Energy = E ⇒

( (

⇒

m 2 u + gr TL = r =4 TH m u2 − 5 gr r

90.

v1 =

In this case, the particle has both tangential and centripetal accelerations. Given that aT = α t

2/9/2021 6:30:52 PM

Hints and Explanations

⇒

⇒

93.

dv = αt dt v

t

0

0

1 1 mv 2 = kx 2 2 2

∫ dv = α ∫ tdt v=

⇒

αt 2 2 94.

Since atotal is equally inclined to ac and aT, so β = 45°

⇒

tan β = tan 45° =

aT =1 aC

αt αt × 4 × 2 =1 = ⎛ v2 ⎞ α 2t 4 ⎜⎝ ⎟⎠ r

a = g ( sin 30 − μ cos θ ) ⇒

⎛1 3⎞ a = 9.8 ⎜ − 0.2 ⎟ ⎝2 2 ⎠

⇒

a = 9.8 ( 0.3268 )

⇒

a = 3.2 ms −2 2

Since v 2 − ( 10 ) = 2 ( 3.2 )( 10 )

At t = 2 s , we have

α (2) × 4 × 2 =1 4 α2 (2) ⇒

⇒

v 2 = 164

⇒

v ≅ 13 ms −1

Hence, the correct answer is (B). 95.

α = 1 ms −3

Speed will be maximum at equilibrium, i.e., when F = kx

Hence, the correct answer is (B). 91.

⇒

Since the car is moving with a constant velocity, so Applied Force = Force of Friction ⇒

F = 50 N

⇒

W = Fs

⇒

W = ( 50 )( 1000 )

⇒

4

W = 5 × 10 J

mg = kd

…(1)

Further when it is allowed to fall suddenly, and if x is the maximum stretching in the spring, then by Law of Conservation of Energy ⎛ Loss in Gravitational ⎞ ⎛ Gain in Elastic ⎞ ⎜ Potential Energy ⎟ = ⎜ Potential Energy ⎟ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ of Body of Spring mgx = mg =

x = 2d Hence, the correct answer is (B).

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 33

…(2)

1 2 1 kx + mv 2 2 2

⇒

Fx =

⇒

1 ⎛ F⎞ 1 ⎛ F⎞ F ⎜ ⎟ = k ⎜ ⎟ + mv 2 ⎝ k⎠ 2 ⎝ k⎠ 2

⇒

F2 1 mv 2 = 2 2k

⇒

v=

2

F mk

Hence, the correct answer is (A). 96.

1 2 kx 2

1 kx 2 Equating (1) and (2), we get

⇒

F k

Wext = ΔU + ΔK

When it is lowered gradually then to attain the equilibrium, we have

⇒

x=

Applying MWET , we get

Hence, the correct answer is (A). 92.

1 ( 4 ) ( 1.6 )2 = 1 ( 2 ) x 2 2 2

⇒ x = 2.26 m Hence, the correct answer is (C).

Since, atotal = aC2 + aT2

⇒

By Law of Conservation of Energy

CHAPTER 1

⇒

H.33

⎛ Loss in Kinetic ⎞ ⎛ Gain in Gravitational ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ Potential Energy ⎟⎠ Energy ⇒

1 1 mV02 − mV 2 = mgh 2 2

⇒

V 2 = V02 − 2 gh

⇒

V = V02 − 2 gh

Hence, the correct answer is (C).

2/9/2021 6:31:00 PM

H.34 JEE Advanced Physics: Mechanics – II 97.

Also, we know that the work done by a conservative force equals the decrease in potential energy, so

Potential energy of cube at position 1, U1 = Mg ( 4 R ) = 4 MgR Applying Law of Conservation of Energy between 1 and 2, we get

⇒

4 MgR = 2 MgR +

⇒

4 MgR = Mv 2

⇒

v 2 = 4 gR

2.

1 Mv 2 2

N=

cos ( 37° ) =

h l

5h h −h= 4 4 By Law of Conservation of Energy, we have ⎛ Loss in EPE ⎞ ⎛ Gain in KE ⎞ ⎜⎝ of springs ⎟⎠ = ⎜⎝ of Bead ⎟⎠

⇒

⎛1 h2 ⎞ 1 2 ⎜ ( 1000 ) ⎟ = ( 5 ) v 2 ⎝2 16 ⎠ 2

⇒

v 2 = 25 h 2

⇒

v = 5h

⇒

T = mg +

⇒

T = 2mg

2mg ⎛ 1⎞ ⎜ 1 − ⎟⎠  ⎝ 2

⇒

μ ( 4 mg ) ≥ 2mg

⇒

μ ≥ 0.5

Since work done by tension is zero therefore work is done only by the gravitational force. So, (D) is also correct. Hence, (A), (B), (C) and (D) are correct. 3.

P = (Total Frictional Force) (velocity)

Speed of ball is maximum at equilibrium, so mg xeq = k

⇒

P = ( mf ) v

⇒

Pextra = mg ( sin θ ) v =

Hence, the correct answer is (A).

Hence, (A), (B) and (C) are correct.

Multiple Correct Choice Type Questions 1.

v 2 = 2 g ( 1 − cos ( 60° ) )

Wtotal = ΔK

Hence, the correct answer is (A). 99.

⇒

Hence, (A) is correct When m moves from A to B it has vertical displacement downwards, i.e., along the direction of gravitational force. So, (B) is correct. Tension is always perpendicular to the velocity, so power delivered by the tension is zero. So, (C) is correct. According to Work Energy Theorem,

x=l−h=

1 2 1 2 1 kx + kx = mv 2 2 2 2

mv 2 

For mass 4m to be stationary, f ≥ T

h 5h = ⇒ l= cos ( 37° ) 4 Extension x in spring is

⇒

Tension in the string will be maximum when mass m is at point B , so

1 mv 2 = mgh = mg ( 1 − cos θ ) 2

Let l be the length of spring at the situation shown in Figure. ⇒

{OPTION (A)}

By Law of Conservation of Energy,

M ( 4Rg ) − Mg = 3 Mg R Hence, the correct answer is (A). 98.

Wnc − ΔU = Δk

T = mg +

Mv 2 − Mg Since, N = R ⇒

{OPTION (B)}

Hence, (A), (B) and (C) are correct.

K 1 + U1 = K 2 + U 2 ⇒

Wc = −ΔU

According to Work-Energy Theorem for a non-conservative system, we have Wc + Wnc = Δk

Mechanics II_Chapter 1_Part 4_Hints and explanation.indd 34

4.

mghv s

Using the concepts of Laws of Motion and Kinematics we can draw following diagrams and then using the definition of work, we can find corresponding work done.

{OPTION (C)}

2/9/2021 6:31:09 PM

Hints and Explanations 40 N 3

T=

S2 = 0.3 m

S1 = 0.15 m

1 kg

 F1 is conservative because it is always directed towards a fixed point P2 . Therefore, W1 can be directly calculated as

20 N 3

W1 = F1 ( P1P2 ) = ( 20 ) ( 6 2 ) = 120 2 J

2 kg

Similarly, W2 = F2 ( OP2 ) = ( 30 )( 6 ) = 180 J 6( π 2 )

10 N

20 N

40 = 20 s

⇒

s=2m

0

3π

Hence, (A), (C) and (D) are correct. 7.

Since W = 40 J = Fs where F = 40 − 20 N = 20 N ⇒

∫ 15ds

W3 = 15s 0 = 45π J

For a light string pulled by a constant Force F = 40 N , tension (T) in the string is also 40 N . So, T = 40 N

3π

F3 ds =

0

Hence, (A), (B), (C) and (D) are correct. 5.

∫

and W3 =

According to Work Energy Theorem, work done by all the forces is equal to change in kinetic energy, so Wtotal = ΔK ⇒

Wext + Wnc = ΔU + ΔK N

Since acceleration is constant, so v ∝ t and hence P = Fv ⇒

P ∝t

Also, v = 2 ax ⇒

6.

ma

Mg

Since, Wnc = 0 , because friction is absent

P2

W1 =

∫ F cosθds

So, Wext = ΔU + ΔK

1

P1

Where, ds = ( 6 ) d ( −2θ ) = −12dθ and F1 = 20 N 0

W1 = −240

∫ cosθdθ

⎛π⎞ W1 = 240 sin ⎜ ⎟ = 120 2 J ⎝ 4⎠ O

P2

2θ

8.

F1

F3 θ

F2

1 2

⇒

2 ( Ma ) xmax = kxmax

⇒

xmax =

2 Ma k

The block will execute SHM and its initial position will be the one of the extreme positions i.e., block returns to its initial position after maximum elongation in the spring and compression in the spring is zero. So, (D) is also correct. Hence, (B) and (D) are correct.

π 4

⇒

Kx

P∝ x.

So, all options (A), (B), (C) and (D) are correct. Hence, (A), (B), (C) and (D) are correct.  Work done by F1 is

⇒

CHAPTER 1

2T =

H.35



ˆ + ˆjdy ) ∫ F ⋅ dr = ∫ ( iFˆ + ˆjF ) ⋅ ( idx ˆ + ˆjdy ) W = ( 7 iˆ − 6 ˆj ) ⋅ ( idx ∫ W=



x

−3

P1

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 35

⇒

∫

y

4

∫

W = 7 dx − 6 dy = −21 − 24 = −45 J 0

0

2/9/2021 6:31:49 PM

H.36 JEE Advanced Physics: Mechanics – II

9.

Displacements along the direction of X and Y axes are opposite to the direction of force acting along the respective directions, hence the particle must have some velocity at ( 0 , 0 ) .

K.E. increasing with time means v > u

Hence, (B) and (D) are correct.

OR P =

In region AB , s ∝ t ⇒ v = constant

W≠0 Since, K.E. is increasing, so P.E. must be decreasing and hence height above the ground must be decreasing. Hence, (A), (C), (D) and (E) are correct.

⇒

ΔK = 0

⇒

WA→ B = ΔK = 0

Similarly, from B to C , slope of s-t graph is decreasing, so v is also decreasing and hence W = ΔK = NEGATIVE Hence, (B) and (C) are correct.

mg k

…(1)

On application of F work done by F and gravity is used to increase the elastic potential energy of spring. So, 1 1 …(2) ( F + mg ) x0 = 2 k ( xi + x0 )2 − 2 kxi2 From (1) and (2) we get, 2F x0 = k Work done by applied force F is W = Fx0 Hence, (C) and (D) are correct.   11. Since P = F ⋅ v . At 1 s and 3 s, P is positive. Hence, angle is acute. At 7 s power is negative. Hence, angle is obtuse. Area under P-t graph is the work done which equals the change in kinetic energy. Hence, (A), (C) and (D) are correct. 12. In a conservative field, the work done WC by the conservative force is always equal to the decrease in potential energy. ⇒

Wc = U initial − U final = − ΔU

⇒

Wc = U A − U B

Either θ = 0°

OR 0 < θ ≤ 90° i.e. θ is acute W ≠ 0 , because t

15. Force on the particle is given by  F = −∇U ⇒

10. Let m be the mass of the block Initial elongation of the spring will be xi =

⇒

 ∂U ˆ ∂U ˆ F=− i− j = − 6 xy 2 + 6 iˆ ∂x ∂y

(

)

Now, for initial acceleration i.e. t = 0 and x = 1, y = 1, we have   F a= = 12iˆ − 6 ˆj M  ⇒ a = 6 5 ms −2 Since particle is at rest at x = 1 , y = 1 so, the total mechanical energy ( E ) is E=U+K ⇒

U ( 1, 1 ) + K ( 1, 1 ) = E

⇒

E=9J

According to modified work-Energy Theorem, we have Wext = ΔU + ΔK Since the particle is moving slowly, so ΔK = 0 ⇒

Wext = ΔU = U ( 0 , 0 ) − U ( 1, 1 )

⇒

Wext = −9 J

Hence, (A) and (C) are correct. 16. Free body diagram of block is as shown in figure. T = F = 40 N

2 kg

Decrease in potential energy implies U A > U B Hence, (B) and (C) are correct. 13. Conceptual (B) and (D) are correct. 14. By Work-Energy Theorem   1 1 mv 2 − mu2 = F ⋅ s 2 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 36

mg = 20 N

From Work Energy Theorem Wnet = ΔKE

2/9/2021 6:31:56 PM

Hints and Explanations

( 40 − 20 ) s = 40

⇒ s=2m Work done by gravity is Wg = −20 × 2 = −40 J

Hence, (A), (B) and (D) are correct. 17. From 0 to t1 , slope is increasing, so speed is increasing. Hence, work done is positive. From t1 to t2 , slope is decreasing, so speed is decreasing. Hence, work done is negative. From t2 to t3 , slope of s-t graph is zero, so speed is zero. Hence, work done is zero. From t3 to t4 , slope is constant, so speed is constant. Hence, work done is zero. Hence, (A), (B) and (C) are correct. 18. As soon as the block hits the wall, the suspension point B comes to a stop, whereas the particle C keeps moving with a velocity v0 towards left. In order that it complete a full circle, it must have kinetic energy enough to take it to the top of the circle.

⇒

v0 = 4 gl

Hence, (A) and (C) are correct. 19. Conceptual (B), (C) and (D) are correct. 20.

⇒

F∝

1 v

v 2 = 2 gR ⇒

a=

 7 24 ˆ a = iˆ − j 5 5  ⇒ a = 5 ms −2   Also, a ⋅ u = 0   ⇒ a⊥u ⇒

Further we have    v = u + at  24 ˆ ⎞ ⎛7 ⇒ v = 14.4iˆ + 4.2 ˆj + ⎜ iˆ − j⎟ 4 ⎝5 5 ⎠  ⎛ 28 ˆ 96 ˆ ⎞ i− j⎟ ⇒ v = 14.4iˆ + 4.2 ˆj + ⎜ ⎝ 5 5 ⎠  ⇒ v = 14.4iˆ + 4.2 ˆj + 5.6iˆ − 19.2 ˆj  ⇒ v = 20iˆ − 15 ˆj  ⇒ v = 20 2 + 152 = 25 ms −1 iˆ

1 mv 2 = kt 2

iˆ

⇒

v=

⇒

a=

2k 1 2 t m dv = dt

2k ⎛ 1 ⎞ m ⎜⎝ 2 t ⎟⎠

Since F = ma ⇒

1 F∝ t

From (1), we get 1 = t ⇒

a=

2k 1 m v 2k ⎛ 1 2k 1 ⎞ 2k ⎛ 1 ⎞ ⎛ 1 ⎞ = ⎜ ⎟⎜ ⎟ m ⎝ 2⎠ ⎝ v⎠ m ⎜⎝ 2 m v ⎟⎠

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 37

…(1)

v2 = 2g R

Hence, (A) and (C) are correct.   22. F = −∇U  ⇒ F = 7 iˆ − 24 ˆj  ⇒ ma = 7 iˆ − 24 ˆj

iˆ

K .E. ∝ t ⇒

k mv

21. Motion of block is non-uniform circular motion, so, acceleration is not constant throughout. At B:

WT = 40 × 2 = 80 J

1 mv02 = mg ( 2l ) 2

a=

Hence, (B) and (D) are correct.

and work done by tension is

⇒

⇒

CHAPTER 1

⇒

H.37

(

)

(

)

(

) (

)

Hence, (A), (B) and (C) are correct. 23. By Law of Conservation of Energy, we have 1 mu2 + 0 = 0 + mg ( 1 − cos θ ) 2 ⎛θ⎞ Since 1 − cos θ = 2 sin 2 ⎜ ⎟ ⎝ 2⎠ ⇒

⎡ ⎛θ⎞⎤ u2 = 2 g ⎢ 2 sin 2 ⎜ ⎟ ⎥ ⎝ 2⎠ ⎦ ⎣

⇒

θ ⎛ u ⎞ = sin −1 ⎜⎝ 2 g ⎟⎠ 2

2/9/2021 6:32:10 PM

H.38 JEE Advanced Physics: Mechanics – II

⇒

θ = 2 sin −1

⎛ u ⎞ ⎜⎝ 2 g ⎟⎠

Hence, (B) and (C) are correct. 24. At the highest point, if T ′ and v′ be the respective tension and velocity, then mg + T ′ =

mv′ 2 

(

Since T ′ = 2mg ⇒

2

mv′ = mg + 2mg  v ′ = 3 g

By Law of Conservation of Energy 1 1 mv′ 2 + mg ( 2 ) = mv 2 2 2 where v is the velocity at the lowest point ⇒

v = 7 g

Hence, (B) and (D) are correct. 25.

(

1 WSp = U i − U f = − k x22 − x12 2

)

1 2 kx 2 So, x2 = 0 and x1 = x i.e., either the spring was initially stretched by a distance x and finally was in its natural length or the spring was initially compressed by a distance x and finally was in its natural length. Since WSp =

Hence, (A) and (C) are correct. 26. In equilibrium F = 0 dU Since, F = − dr dU =0 dr Either U is zero or it is a non-zero constant. So, for equilibrium, we have ⇒

Either F = 0 , U = 0 or

force and force of gravity. When spring is detached only gravity is left i.e., retardation has reduced and so the block rises to a height greater than from where it was released more height. Hence, (A), (C) and (D) are correct.   F 1 ⎛ ∂U ˆ ∂U ˆ ⎞ i+ j {m = 1 kg} 28. a = = − ⎜ m m ⎝ ∂x ∂y ⎟⎠   F a = = − 3iˆ + 4 ˆj = constant m

F = 0 , U is a constant (but not zero).

Hence, (A) and (C) are correct. 27. In simple harmonic motion, the body is accelerated at all points except at mean position. At the starting point, spring force is zero. The only force is mg (downwards). Hence, acceleration is maximum, i.e., g (downwards). From mean position to initial position, retardation is due to forces, spring

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 38

)

⇒

ax = −3 ms −2 and ay = −4 ms −2

⇒

a = ax2 + ay2 = 5 ms −2 = constant

Hence, (A) is correct vx = −3t and vy = −4t x=6+

1 2 3 ax t = 6 − t 2 2 2

1 2 ay t = 4 − 2t 2 2 Particle crosses the y-axis when and y = 4 +

x = 0 or t = 2 s speed of particle at this moment v = at = 5 × 2 = 10 ms −1 co-ordinates of particle at t = 1 s are

( x , y ) = ( 4.5, 2 ) Hence (C) and (D) are also correct. Hence, (A), (C) and (D) are correct. 29. Work done against friction on ice is zero and work done against friction on the road is ( μmg )  . So, average work done is 0 + ( μmg )  2

= ( μmg )

 2

Thus, indicating that the effective length of the sledge that has to be dragged so that it just gets completely on  the road is . 2 Distance covered by the sledge on the road before coming to rest is

v02 . 2μ g

{∵ 02 − v02 = 2 ( − μ g ) s }

So total distance moved by the sledge is ⎛ v02 ⎞ ⎜ 2μ g + 2 ⎟ ⎝ ⎠

2/9/2021 6:32:16 PM

Hints and Explanations Distance covered by the sledge on the road is

⇒

⎛ ⎞ ⎛ ⎞ −⎜ + ⎟ =⎜ − ⎟ ⎝ 2μ g 2 ⎠ ⎝ 2μ g 2 ⎠ v02

x , Fnet = 0 2 When x ′ = x , then

30. In uniformly accelerated motion ⇒

2

a = g (upwards)

v = u2 + 2 as

34.

So, power, P = Fv = F ( u + at )

  ⎛ ∂U  ∂U  ∂U ⎞ F = −∇U = − ⎜ i +j +k ∂y ∂z ⎟⎠ ⎝ ∂x  ⇒ F = k yi + x j

(

also P = F u2 + 2 as i.e., power varies linearly with time and parabolically with displacement. So, (B) and (C) are incorrect Hence, (B) and (C) are correct. 31.

Fa

Hence, (A) and (C) are correct.   35. Power, P = F ⋅ v

  Between O and A angle between F ( = mg ) and is greater than 90° . Hence, power is negative. At angle is 90° . Hence, power is zero. From A to angle is less than 90° therefore, power is positive.

Fb

x=0

)

Work done by this force field is independent of the path followed between any two points and hence this force field is conservative in nature.

μ x = –a

x=b

v

1 1 1 W = ka 2 − kb 2 = k ( a 2 − b 2 ) 2 2 2 Since ΔK = 0

⇒

μmg ( a + b ) =

⇒

μ=

1 ( 2 k b − a2 ) 2

1 ( 2 k a − b2 ) 2

k(a − b) 2mg

Hence, (B) and (C) are correct. 1 32. Work done by spring in parts (A) and (C) is kx 2 and 2 1 work done by spring in parts (B) and (D) is − kx 2 . 2 Hence, (A) and (C) are correct. 33. At maximum extension, x , we have mgx =

1 2 kx 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 39

A

θ

According to MWET, we have Wnc = ΔU + ΔK − μmg ( a + b ) =

 v A B

y

Work done by spring on block is

⇒

a

mg

Hence, (B), (C) and (D) are correct.

and v = u + 2 as

k

m

kx − mg = ma

v = u + at

⇒

kx = 2 mg

2mg k

When x ′ =

Hence, (B), (C) and (D) are correct.

2

x=

CHAPTER 1

v02

H.39

O

B

x

At any instant, t  F = − mg ˆj  v = ( u cos θ ) iˆ + ( u sin θ − gt ) ˆj   ⇒ P = F ⋅ v = mg ( gt − u cos θ ) iˆ

⇒

P = mg ( gt − u cos θ )

i.e., P varies linearly with t Therefore, OPTIONS (B) and (D) are correct Hence, (B) and (D) are correct. 36. Momentum, p = 2mK Therefore, without knowing the mass we cannot compare the momentum. From Work Energy Theorem, more work will be done to stop B.

2/9/2021 6:32:22 PM

H.40 JEE Advanced Physics: Mechanics – II v2 , where Distance travelled before stopping is s = 2a a = μg .

2.

W = Fx =

Again, distance cannot be compared unless v is known and for v mass should be known to us.

38. For x > 0 , F = −ve

3.

Since mcar < mtruck , so retardation on the car is more than the truck and hence the car will stop earlier in lesser distance. Hence, the correct answer is (D).

4.

F=−

For x < 0 , F = +ve From x = 3 to x = 1 , displacement is in negative direction. From x = −1 to x = −3 also displacement is in negative direction. Now work done is positive if force and displacement are parallel to each other and work done is negative if they are antiparallel. Hence, (A) and (D) are correct. 39.

1 ( dm ) v 2 2 1 ( A dx ρ ) v 1 Power Loss = 2 = = ρ A v3 dt 2 dt 2

40.

Hence, the correct answer is (D). 5.

point is 5gR . At the highest point velocity is zero, however acceleration is non-zero equal to g . Hence, the correct answer is (D). 6.

1 mv 2 = mgh1 2

1 ρ A v2 2 Hence, (A) and (C) are correct.

Also, ⇒

1 2 kx0 2 7.

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 40

v2 2g

1 ( 2m ) v 2 = ( 2m ) gh2 2 h2 =

v2 2g

Work done and power developed is zero in uniform   W circular motion, because F ⊥ r . Also P = , so t Statement-2 is the correct explanation to Statement-1. Hence, the correct answer is (A).

Hence, (C) and (D) are correct.

Work done by force of gravity is path independent near the surface of Earth, for one reason that the force of gravity is constant near the surface of Earth. Also work done by force of gravity is path independent at greater height as the force of gravity is a conservative force. Hence, the correct answer is (D).

h1 =

Hence, the correct answer is (A).

W = Fx0

Reasoning Based Questions

gR ( ≠ 0 ) ,

because to just complete the circle, velocity at lowest

F=

2F ⇒ x0 = k Work done by applied force is

1.

At the topmost point, the velocity is

⇒

Fx0 =

dU dx

F = 0 at points B and C .

1 Further, Power = Fv = ρ A v 3 2 ⇒

1 mv 2 = ΔK 2

Since for both the cars of different mass, ΔK will be different, so W is also different. Hence, the correct answer is (D).

Hence, (B) and (D) are correct. 37. In region OA particle is accelerated, in region AB particle has uniform velocity while in region BC particle is decelerating. Therefore, work done is positive in region OA , zero in region AB and negative in region BC . Hence, (B) and (C) are correct.

According to Work-Energy Theorem, work done equals change in kinetic energy, so

8.

W ( mg ) + W ( sp ) + W ( Man ) = 0 ⇒

− ΔU + E + W = 0

⇒

ΔU = E + W

Work done by spring force is positive when a compressed spring is released or stretched spring released. Hence, the correct answer is (C).

2/9/2021 6:32:28 PM

Hints and Explanations By Law of Conservation of Momentum, we have

3.

At maximum extension (situation of momentary rest),

O = m1v1 − m2v2

vA = vB = zero By Law of Conservation of Energy

and K1 + K 2 = E the energy released in explosion ⇒

K1 m2 = K 2 m1

⇒

P = 2mE

⎛ Loss of Gravitational ⎞ ⎛ Gain in Elastic ⎞ ⎜ Potential Energy ⎟ = ⎜ Potential Energy ⎟ ⎟ ⎟ ⎜ ⎜ of block B ⎠ ⎝ of the spring ⎠ ⎝

Hence, the correct answer is (C). 10. Statement-1 is false because normal force cannot do any work in the absence of any displacement. The Statement-2 is correct because an external force alone can shift the position of centre of mass. Hence, the correct answer is (D).

1 2

⇒

2 ( 2m ) gxmax = kxmax

⇒

xmax =

4 mg k

Hence, the correct answer is (D). 4.

At x =

xmax 2 mg , we have = 2 k

Linked Comprehension Type Questions

vA = vB = v (say)

1.

{Because A and B are connected by a rigid string} Then by Law of Conservation of Energy

Just before the cord comes in contact with the peg, we have

⎛ Decrease ⎞ ⎛ Increase in ⎞ ⎛ Increase ⎞ ⎜ in GPE ⎟ = ⎜ EPE of ⎟ + ⎜ in KE of ⎟ ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ of B ⎠ ⎝ the spring ⎠ ⎝ both blocks ⎠

0.8 m 30°

T1 Peg

h = 0.8 sin (30°)

T1

mg

v = 2 gh

So, T1 − mg sin ( 30° ) =

mv 2 R1

{ R1 = 0.8 m }

mg m ( 2 g ) ( 0.4 ) 3 mg + = 2 0.8 2 Hence, the correct answer is (C). ⇒

2.

⇒

( 2 mg ) ⎛⎜⎝

⇒

8 m2 g 2 4 m2 g 2 = + 3 mv 2 k k

⇒

3 mv 2 =

h = 0.8 sin ( 30° ) = 0.4 ⇒

4 m2 g 2 k

5.

v = 2g

xmax mg = , we draw the free body diagrams 4 k for the blocks A and B, below

At x =

aA ( = a)

N

T

2

v = 2 gh

So, T2 − mg sin ( 30° ) =

mv 2 R2

T2 =

aB ( = a) kx

A

T

B

m ( 2 g ) ( 0.4 )

mg + 2 0.4 5mg ⇒ T2 = 2 Hence, the correct answer is (A). ⇒

2

2 mg ⎞ 1 ⎛ 2 mg ⎞ 1 2 ⎟ = k⎜ ⎟ + ( 3 m )v k ⎠ 2 ⎝ k ⎠ 2

m 3k Hence, the correct answer is (D). ⇒

T1 =

Just after the cord comes in contact with the peg, we have

1 2

( 2m ) gx = kx 2 + ( m + 2m ) v 2

h = 0.8 sin 30° = 0.4 m 2

1 2

⇒

mg v

⇒

CHAPTER 1

9.

H.41

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 41

mg

2 mg

Since A and B are connected with a rigid thread, so at any instant aA = aB = a (say)

2/9/2021 6:32:33 PM

H.42 JEE Advanced Physics: Mechanics – II ⇒ ⇒

T − kx = ma T − mg = ma

…(1)

Also, 2mg − T = 2ma

{

mg ∵x= k

}

…(2)

Hence, the correct answer is (A).   12. dW = F ⋅ d 

By Law of Conservation of Energy 1 mv 2 = mgH − mgh 2 v = 2g ( H − h )

Hence, the correct answer is (D).

h=

1 2 gt 2

t=

2h g

⎡ ⎛ y3 ⎞ ⎤ ⎛ x3 ⎞ dW = k ⎢ y 3 d ⎜ ⎟ + x 3 d ⎜ ⎝ ⎠ ⎝ 3 ⎟⎠ ⎥⎦ 3 ⎣

⇒

⎡ ⎛ x3 y 3 ⎞ ⎤ dW = k ⎢ d ⎜ ⎟⎥ ⎣ ⎝ 3 ⎠⎦

⇒

W = dW =

13.

⇒

2h s = 2g ( H − h ) g

⇒

s = 2 h( H − h )

d ( s2 ) So, =0 dh ⇒ ⇒

H − 2h = 0

H 2 Hence, the correct answer is (B).

smax = H

)

( a, a ) 9 k W( a , 0 )→( a , a ) = x 3 y 3 = ( a6 − 0 ) = 3 a6 3 ( a, 0 ) 3

(

…(1)

14.

)

( a, a ) k k ka6 W( 0 , 0 )→( a , a ) = x 3 y 3 = ( a6 − 0 ) = 3 3 ( 0, 0 ) 3

(

)

Hence, the correct answer is (C). ka6 . So, work done due to this force 3 is independent of the path followed between ( 0 , 0 ) to  ( a, a ) . Hence F is a conservative force.

paths is same i.e.,

Hence, the correct answer is (A). 16.

Wmg = mgh cos180° ⇒

Wmg = ( 2 ) ( 10 ) ( 2 ) ( −1 ) = −40 J

From Work Energy Theorem, we get Wmg + Wupthrust = ΔKE

Hence, the correct answer is (D). 10. Total height h = 2 +

)

Hence, the correct answer is (C).

h=

⇒

(

k 2 3 x y 3

15. Since we observe that the work done in all the three

d [ 4h ( H − h ) ] = 0 dh

smax

∫

Hence, the correct answer is (A).

For s to be MAXIMUM, s2 has to be MAXIMUM.

9.

⇒

(

s = vt

⎛ H⎞⎛ H⎞ =2 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠

)

dW = k x 2 y 3 dx + x 3 y 2 dy

( a , 0 ) 24 k ( 0 − 0 ) = zero W( 0 , 0 )→( a , 0 ) = x 3 y 3 = 3 3 ( 0, 0 )

Hence, the correct answer is (B).

⇒

(

⇒

From concepts of projectile motion, we have

⇒

8.

h = 2 + 4 sin 2 θ

11. Since, h = 2 + 4 sin 2 θ , so at θ = 30° , h = 3 m

g ⇒ a= 3 Hence, the correct answer is (C).

7.

⇒

Hence, the correct answer is (B).

mg = 3 ma

⇒

h = 2+

We can see that h = 6 m only when θ = 90°

Solving these two equations, we get

6.

2 g ( hA − hB ) sin 2 θ 2g

⇒

uB2 sin 2 θ 2g

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 42

⇒

−40 J + Wupthrust = 16

⇒

Wupthrust = 56 J

Hence, the correct answer is (D).

2/9/2021 6:32:41 PM

Hints and Explanations

56 = ( upthrust force ) × ( displacement ) × cos 0° ⇒

2 ⎛ m⎞ 56 = ⎜ ⎟ ρw g × 2 = × 1000 × 10 × 2 ρ ⎝ ρ⎠

Also, we observe that at r =

⇒

⎛ 40 ⎞ ρ = ⎜ ⎟ × 10 3 kgm −3 ⎝ 56 ⎠

E=

5 × 10 3 kgm −3 7 Hence, the correct answer is (A). ⇒

ρ=

18. At equilibrium r0 F= ⇒

F

dU =0 dr

r = r0

=

dU dr

r = r0

⇒ ⇒

d 2U dr 2

⇒

d 2U dr 2

r0 =

2A B

−

B ⎛ 2A ⎞ ⎜⎝ ⎟ B ⎠

⇒

E=−

3B2 16 A

Hence, the correct answer is (C).

d 2U 6 A 2B = 4 − 3 dr 2 r r 2A B

2

2r0 4 A = , we have 3 B

B2 B2 − 16 A 2 A

E=

=−

2A B + =0 r03 r02

22. In equilibrium ( Fnet = 0 ) T = kx 2T +

=

…(2)

kx

4

−

x 2

v 2

v 3

So, U is minimum at equilibrium position r0 . So, the equilibrium is stable. Hence, the correct answer is (A).

mg

Solving these equations, we have 2mg 2 × 1 × 10 = = 0.4 m 5k 5 × 10 Hence, the correct answer is (B). x=

23. Let v be the speed of block placed horizontally, then by Law of Conservation of Energy, we have

20. Work done by a conservative force equals the decrease in potential energy of the particle. So, W = U i − U f = U at equilibrium − U at ∞ 2

A B B − =− 4A r02 r0

and U ∞ = 0

B2 , so that the 4A particle is taken from equilibrium position to infinity.

So, work done by external agent will be Hence, the correct answer is (B).

21. For velocity to be zero, i.e., kinetic energy has to be zero, so the total energy must be purely potential.

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 43

T

2B ⎛ 2A ⎞ ⎜⎝ ⎟ B ⎠

= A POSITIVE VALUE

Now U i = U at r0 =

kx = mg 2

x

6A ⎛ 2A ⎞ ⎜⎝ ⎟ B ⎠

…(1)

2T + kx 2

2A B dU =− 3 + 2 dr r r

r0 =

A ⎛ 4A ⎞ ⎜⎝ ⎟ B ⎠

⇒

2A ⇒ r0 = B Hence, the correct answer is (B). 19. Since,

E=−

3B2 A B = − 16 A r 2 r

⇒

CHAPTER 1

17. Let ρ be the density of ball, then

H.43

2

2

1 1 ⎛ v⎞ 1 1 ⎛ x⎞ ⎛ x⎞ mv 2 + m ⎜ ⎟ + kx 2 + k ⎜ ⎟ = mg ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ 2⎠ 2 2 2 2 2 2 ⇒

mgx 3 2 3 mv 2 = − kx 8 2 8

⇒

v=

8 ⎛ gx 3 kx 2 ⎞ − ⎜ ⎟ 3⎝ 2 8m ⎠

Substituting the values, we have v=

8 ⎛ 10 × 0.4 3 × 10 × 0.16 ⎞ − ⎜ ⎟⎠ 3⎝ 2 8×1

⇒ v = 1.93 ms −1 Hence, the correct answer is (B).

2/9/2021 6:32:46 PM

H.44 JEE Advanced Physics: Mechanics – II

24.

KE = ⇒

1 ( 4 ) ( 20 )2 2

⇒

KE = 800 J

⇒

h ≤ 3R

⇒

h ≤ 1.5 m

Hence, the correct answer is (C). 25. Work done from OAB

Hence, the correct answer is (D).

1 ( 4 )( 12 ) = 24 J 2 Since the force is non-conservative from 0 to A, so KE must decrease and hence KE at x = 4 m is ( 800 − 24 ) J

Area =

⇒

KE

x= 4 m

31. Since, g sin θ − μ g cos θ = 2 ms −2

Block will never stop on rough inclined surface. So, it will cross point C infinite number of times. Hence, the correct answer is (A).

= 776 J

26. Gain in KE from BCDE is 60 J

32. Potential energy of the projectile at any instant will be

So, new KE is ( 776 + 60 ) J = 836 J Hence, the correct answer is (B).

28.

A → O is a non conservative force, so again negative work will be done while going back. Hence, the correct answer is (C).

U=

1 1 mv02 − mv 2 2 2

⇒

dU dv = 0 − mv dt dt

Since, m = 1 kg,

at = g sin 60° = 5 3 ms −2

  dU = a⋅v dt Hence, the correct answer is (B).

a = at2 + an2 = 18.2 ms −2

33. At height h , we have

Hence, the correct answer is (C).

K=

29. Height from bottom at when θ = 60° is h1 = R ( 1 − cos 60° ) =

⇒

v 2 16 = = 0.8 m 2 g 20

30.

a = g sin θ − μ g cos θ a = 10 ×

3 4 − 0.5 × 10 × = 2 ms −2 5 5

K v2 = 0 − 1 = constant U 2 gh

K 1 versus graph is a straight line not passU h ing through the origin. Hence, the correct answer is (B).

Hence,

Now since h1 + h2 = 1.3 m > R The bob will rise to θ = 90° , but there string will slack and total height will be less than 1.3 m. Hence, the correct answer is (C).

1 mv02 − mgh 2

and U = mgh

R = 0.5 m 2

Further height raised, h2 =

  a⋅v dv = at = = tangential acceleration dt v

⇒

v2 an = = 16 ms −2 R ⇒

g sin θ > μ g cos θ

⇒

Hence, the correct answer is (C).

27.

20 h ≤ 2 gR 3

34.

1 ( 2 )( 4 )2 = 16 J 2

KE

x= 5

=

PE

x= 5

= 20 + ( 5 − 2 ) = 29 J

2

⇒ Total Energy = Mechanical Energy So, E = U + K = 16 + 29 = 45 J

Velocity of block at B or at C is

⇒

20 h h v = 2 as = 2 × 2 × = sin θ 3 For the block, not to leave contact anywhere

Hence, the correct answer is (C).

v ≤ 2 gR

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 44

E = 45 J

35. KE possessed by the particle will be MAXIMUM, when PE is MINIMUM i.e. at x = 2 m So, PE

MIN

= 20 J

2/9/2021 6:32:53 PM

Hints and Explanations KE

MAX

⇒

= 45 − 20 = 25 J

Hence, the correct answer is (B).

Hence, the correct answer is (D).

36. The particle will be in equilibrium i.e. F = 0 ⇒

dU =0 dx

⇒

x0 − 2 = 0

⇒

x0 = 2 m

40.

Pav = Pinst ⇒

W = Fvinst t

⇒

v 2f − vi2 1 3 m = ( mav ) t 2 4

⇒

( 3 + 2t )2 − ( 3 )2 = atv

⇒

9 + 4t 2 + 12t − 9 =

⇒

4t 2 + 12t = 9t + 6t 2

⇒

2t 2 = 3t

⇒

t = 1.5 s

Hence, the correct answer is (B). 37. Let lower spring compresses maximum by x metre, then by Law of Conservation of Mechanical Energy, we have ⎛ Decreases in ⎞ ⎛ Increases in elastic ⎞ ⎜ potential energy ⎟ = ⎜ potential energy of ⎟ ⎜⎝ ⎟⎠ ⎜⎝ both the springs ⎟⎠ of block ⇒

( 2 ) ( 10 ) ( x + 1 ) = 1 × 10 × x 2 + 1 × 10 × ( x + 1 )2

⇒

20 x + 20 = 5x 2 + 5x 2 + 5 + 10 x

⇒ ⇒

2

2

(

3 2

T2 − mg =

2x − 2x − 3 = 0 x=

xmax = 1 + x = 2.82 m

3 ( 2 ) ( t ) ( 3 + 2t ) 2

T1 = mg cos θ

10 x − 10 x − 15 = 0 2

)

Hence, the correct answer is (A). 41.

2

2 ± 4 + 24 = 1.82 m 4 So, maximum extension in the upper spring is ⇒

W = 40 J

…(1)

mv 2 m = ( 2 gh )   2mg ( 1 − cos θ )  

⇒

T2 − mg =

⇒

T2 − mg = 2mg ( 1 − cos θ )

Hence, the correct answer is (C).

T1

38. At equilibrium position net force on the block should be zero. So, let it is at distance y from where it was released. Then,

θ

T1

20 = 10 y + 10 ( y − 1 )

⇒

20 y = 30

⇒

y = 1.5 m

h = l (1 – cos θ )

⇒

T2 = mg + 2mg ( 1 − cos θ )

…(2)

Given T2 = 2T1

Hence, the correct answer is (D). 39. Equation can be written as, v = ( 9 + 4s ) 2

Comparing this with v 2 = u2 + 2 as , we get u = 3 ms −1 and a = 2 ms −1 At t = 0 , velocity is 3 ms −1 and at t = 2 s , velocity is 7 ms −1 . From Work Energy Theorem, we have W = K f − Ki =

T2 T2

mg = Ky + K ( y − 1 ) ⇒

CHAPTER 1

⇒

H.45

(

1 m v 2f − vi2 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 45

)

=

1 ( 2 ) ( 49 − 9 ) 2

⇒

mg + 2mg ( 1 − cos θ ) = 2mg cos θ

⇒

cos θ =

3 4

⎛ 3⎞ θ = cos −1 ⎜ ⎟ ⎝ 4⎠ Hence, the correct answer is (D). ⇒

42. Tension is maximum at bottommost point, so amax =

2mg − mg =g m

2/9/2021 6:33:02 PM

H.46 JEE Advanced Physics: Mechanics – II Tmax = 2 mg

⇒

v2 =g 

⇒

2 ( 1 − cos θ ) =1 

⇒

2 − 2 cos θ = 1

⇒

2 cos θ = 1

⇒

⎛ 1⎞ π θ = cos −1 ⎜ ⎟ = ⎝ 2⎠ 3

45°

R amax

x′

mg

Since, − R = v′ sin ( 45° ) t −

Hence, the correct answer is (D). 43. Let t be the time taken by the ball from B to C.

⇒

1 R = gt 2 2 ⇒

t=

t=

1 2 gt 2

2R (1 + 2 ) g

x ′ = ( v′ cos ( 45° ) ) t

2R g

⇒

x′ = 2 ( 1 + 2 ) R

Hence, the correct answer is (C).

Since, x = vt

46. By Law of Conservation of Energy, we get

2R g

⇒

2R = v

⇒

v = 2 gR

1 7 gL 1 m = mgL + mv12 2 2 2

Applying Energy Conservation at points A and B , we get

⇒

7 gL = 4 gL + 2v12 y

1 1 mv02 = mv 2 + mg ( 2R ) 2 2 ⇒

v1

v0 = 6 gR

x

O

Hence, the correct answer is (D). 44.

vy = gt ⇒

L

vx

2R vy = g = 2 gR g

θ u=

and vx = 2 gR

v12 =

⇒

⎞  ⎛ 3 ⎞ˆ ⎛ 7 Δv = ⎜ gL ⎟ j − ⎜ gL ⎟ iˆ ⎝ 2 ⎠ ⎝ 2 ⎠

⇒

 Δv =

v′ = vx2 + vy2 = 2 gR + 2 gR

⇒

 Δv = 5 gL

v ’ = 2 gR

Hence, the correct answer is (D).

vy vx

=1

v

vy

θ = 45°

Hence, the correct answer is (B). 45. Since all surfaces are frictionless, so collision is elastic, hence

⇒

3 gL 2    Δv = v1 − u

⇒

tan θ = ⇒

7gL 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 46

2 gL 3 gL + 2 2

2/9/2021 6:33:08 PM

Hints and Explanations 1 ⎛ 7 gL ⎞ 1 2 m⎜ ⎟⎠ = mgL ( 1 + sin θ ) + mv ⎝ 2 2 2

⇒

7 gL = 4 gL ( 1 + sin θ ) + 2v 2

⇒

2v = gL ( 3 − 4 sin θ ) 2

y

⇒

The normal force N B at B is mvB2 = 4 mg R Hence, the correct answer is (C). NB =

…(1)

53.

v

O

θ

m

θ

x

mg

⇒

54.

vD = 2 g ( 2R ) = 2 gR Retardation offered beyond D is

Also, when the motion ceases to be circular, then T = 0 ⇒

mv 2 = mg sin θ L

⇒

v 2 = gL sin θ

⇒

2 gL sin θ = gL ( 3 − 4 sin θ )

⇒

6 sin θ = 3

a = − ⎡⎣ g sin ( 30° ) + μ k cos ( 30° ) ⎤⎦ ⇒

⇒

gL 2

a=−

(

g 1 + 3 μk 2

)

2 = 2 ax Since v 2 − vD

⇒ θ = 30° Hence, the correct answer is (A).

⇒

m ( 6 gR )

NC = mg +

Let the final velocity when its stops be v = 0 .

7gL 2

v 2 = gL sin ( 30° ) =

⇒

0 2 − ( 4 gR ) = −

⇒

x=

(

4R 1 + 3 μk

Hence, the correct answer is (C).

 v = − v cos θ iˆ + v sin θ ˆj

(

gL − 3iˆ + ˆj 2

)

2g 1 + 3 μk x 2

55.

x

)

x1

Hence, the correct answer is (C).

30°

49. Work done by all the force = change in KE Hence, the correct answer is (D). 50. Gravitational force is constant. Hence, the correct answer is (C).

37°

2m

x = x1 − x2 ⇒

x=

52. Velocity at B is

⇒

x=

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 47

x2

F

Let us consider the motion of a point on the free portion of string on which F is being applied. Then

51. Work done by kinetic friction in this case is negative because force of friction is acting in the opposite direction of motion. Hence, the correct answer is (C). vB = 2 gh = 2 g ( 2R )

2

1

iˆ

 v=

mvC2 R

= 7 mg R Hence, the correct answer is (D).

L

48.

vC = 6 gR

The normal force NC at C is NC = mg + h = L (1 + sin θ )

u=

vC = 2 gh′ = 2 g ( 3 R ) ⇒

L

in gs

vB = 2 gR

CHAPTER 1

47. Since

H.47

2 2 − sin ( 30° ) sin ( 37° ) ⎛ ⎜⎝

2 2 10 − = 4− 1⎞ ⎛ 3⎞ 3 ⎟ ⎜ ⎟ 2⎠ ⎝ 5⎠

2/9/2021 6:33:15 PM

H.48 JEE Advanced Physics: Mechanics – II

⇒

x=

59. The path followed is shown in the Figure.

2 m 3

y

So, work done is (a, a)

⎛ 2 ⎞ 100 W = Fx = 50 ⎜ ⎟ = J ⎝ 3⎠ 3 Hence, the correct answer is (C). (0, 0)

56. From Work-Energy Theorem, we have W = ΔK ⇒

100 1 = mv 2 = 5v 2 3 2

⇒

v=

For the path between dy = 0 . ⇒

20 ms −1 3

Hence, the correct answer is (C). 57. Initial Acceleration is a1 = Final Acceleration is a2 = So, ratio ⇒

F cos ( 30° ) m F cos ( 37° ) m

a1 cos ( 30° ) = a2 cos ( 37° )

dy = dx

W( a , 0 )→( a , a ) = W2 =

a

∫ ady = a

2

Hence, the correct answer is (A).   60. Since, dW = F ⋅ d  = ydx + xdy Also, we know that d ( xy ) = ydx + xdy ⇒

dW = d ( xy )

∫

) ( dxiˆ + dyjˆ )

Taking y = x , dy = dx and the limits from ( 0 , 0 ) to

( a, a ) , we get a

∫

⇒

∫ ydx + xdy

W = d ( xy )

∫ ∫( W = ( ydx + xdy ) ∫ W = 2xdx = ( x

W( a , 0 )→( a , a ) = W2 =

Integrating, we get

Since, work done is   W = F ⋅ d  = yiˆ + xjˆ ⇒

⇒

( a, 0 ) to ( a, a ) , x = a , so

W = W1 + W2 = a 2

58. As the particle moves straight from ( 0 , 0 ) to ( a, a ) , then it must move along the line y=x

⇒

For the path between dx = 0.

∫ ydx + xdy = 0

0

Hence, the correct answer is (B).

dy =1 dx

W( 0 , 0 )→( a , 0 ) = W1 =

( 0, 0 ) to ( a, 0 ) , y = 0 , so

So, the total work done is

a1 5 3 = 8 a2

⇒

x

(a, 0)

2

)0=a a

2

0

Hence, the correct answer is (A).

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 48

( x2 , y 2 ) ( x1 , y1 )

⇒

W = ( xy )

⇒

W = x2 y 2 − x1y1

In this case, we observe that the work done by the force depends only on ( x1 , y1 ) , the initial position and ( x2 , y 2 ) , the final position. So, the work done by the force depends only on initial and final values of x and y . Also, when the object returns to the original position, then we have ( x2 , y 2 ) = ( x1 , y1 ) , and hence W=0. Hence, (A) and (D) are correct.

2/9/2021 6:33:23 PM

H.49

Hints and Explanations

Matrix Match/Column Match Type Questions A → (q) B → (s) C → (r) (D) → (r)

θ

θ

mg = 10 N

and Wext = ΔU + ΔK = ( U f + K f ) − ( U i + K i )

From Work-Energy Theorem, we have

A → (s) B → (r) C → (p) D → (q)

Work done by all the forces = ΔKE = 32 J

vB2 = uA2 − 2 ghAB = ( 9 g ) − ( 2 g ) = 7 g

Writing equation of motion, we have,

Work done by gravity = − mgh = − ( 1 ) ( 10 )( 16 ) = −160 J

⇒

vB = 7 g

⇒

vB2 =7 g

⇒

⇒

Again,

=

vA2

⇒

vC = 5 g

⇒

vC2 =5 g

TC = 4 mg

⇒

TC =4 mg

…(1)

x

⇒

N sin 30° = f cos 30°

⇒

N = 3f

− 2 ghAC = ( 9 g ) − 2 g ( 2 ) = 5 g

…(2)

f =6N and N =

18 =6 3 N 3

⎛ 3⎞ ( 16 ) = 144 J So, WN = ( N cos θ )( s ) = ( 6 3 ) ⎜ ⎝ 2 ⎟⎠ mvC2 

A → (t) B → (p) C → (s) D → (q)

⎛ 1⎞ and W f = ( f sin θ ) ( s ) = ( 6 ) ⎜ ⎟ ( 16 ) = 48 J ⎝ 2⎠ 4.

A → (r) B → (q) C → (p) D → (t) v f − vi = area under a-x graph = 12 ms −1

At t = 4 s , v = at = 8 ms −1 and s =

3 N + f = 24

Solving equations (1) and (2), we have

Further, TC + mg = ⇒

y

∑F = 0

7 TB = 2mg 2 vC2

∑ F = ma

N cos 30° + f sin 30° − 10 = ma = 2

mvB2 Further, TB = = 7 mg 

3.

y

a

x

Since Wtotal = ΔK and WC = −ΔU

2.

f

θ

CHAPTER 1

1.

N

1 2 at = 16 m 2

KE =

1 mv 2 = 32 J 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 49

⇒

v f = 12 + 4 = 16 ms −1

⇒

ΔKE =

(

)

1 m v 2f − vi2 = 120 J 2

Work done by all the forces = ΔKE = 120 J Kf =

1 mv 2f = 128 J 2

Work done by conservative forces = U i − U f = 240 J

2/9/2021 6:33:29 PM

H.50 JEE Advanced Physics: Mechanics – II Work done by external force i.e. Wext is given by ⎛ Total ⎞ ⎛ Work done by ⎞ Wext = ⎜ work ⎟ − ⎜ conservative ⎟ ⎜⎝ done ⎟⎠ ⎜⎝ ⎟⎠ forces ⇒ 5.

6.

7.

Wext = −112 J

A → (p, q) B → (t) C → (r) D → (s) Angle between net force and the string can never be obtuse. It is 90° at A , 0° at B and acute in between. A → (p, q, r) B → (p, q, r) C → (q, r) D → (p, r) Conceptual

⇒

225 15 = = 7.5 N 4 2   P = F⋅v F=

⇒

P = ( Ft ) ( v )

⇒

P = ( 6 ) ( 12 ) = 72 watt Pav =

⇒

W t

Pav = Ft ( vav ) 1

1

But vav =

∫

∫

12 tdt

vdt

0 1

0

=

∫

1

dt

0

⇒

A → (r) B → (p) C → (q, s) D → (q, s) Conceptual

⇒

vav =

12t 2 2

1

= 6 ms −1 0

Pav = ( 6 )( 6 ) = 36 watt

10. A → (s) B → (r) C → (q) D → (p) Just after the vehicle is stopped, the bob swings out to an angle of 60° . So, by Law of Conservation of Mechanical Energy, we have

8.

A → (p, s) B → (q, r) C → (p, s) D → (q, r) Conceptual

9.

A → (s) B → (r) C → (p) D → (q)

⇒

v2 = 9t 2 R

1 mv02 = mg ( 1 − cos θ ) 2

⇒

1⎞ ⎛ v0 = 2 ( 10 )( 5 ) ⎜ 1 − ⎟ = 5 2 ms −1 ⎝ 2⎠

mgh =

⇒

v 2 = ( 16 ) ( 9t 2 )

⇒

v = 12t

⇒

at =

⇒

Ft = mat =

θ

dv = 12 dt

v2 =9 R 9 Fn = man = N 2

⇒

F=

Ft2

+

T T

1 ( 12 ) = 6 N 2

At t = 1 s , an = ⇒

1 mv02 2

Fn2

81 = 36 + 4

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 50

mg sin θ

mg cos θ mg

So, net force on the bob at the lowest point is F=

mv02 ( 2 ) ( 50 ) = = 20 N  5

2/9/2021 6:33:35 PM

Hints and Explanations 3.

Acceleration of the bob at lowest point is a=

v02

F = = 10 ms −2 m 

H.51

(a) The forces acting on the block are shown in the figure. N

F

At the highest point, net force is F = mg sin ( 60° ) = 10 3 N

53° f

So, acceleration at highest point is F = 5 3 ms −2 m

mg

Clearly, WN = 0 and Wg = 0, whereas WF = Fs cos θ

Integer/Numerical Answer Type Questions 1.

W f = − fs = − μ k Ns where N = mg − F sin θ

Since the block does not slide on the wedge. So f = mg sin ( 45° ) =

( 1 ) ( 10 ) 2

=

According to Work Energy Theorem,

10 N 2

ΔK = Wnet = WF + W f

v = 2 ms1

⇒

ΔK = Fscosθ − μk (mg − Fsinθ )s

⇒

ΔK = ( 30 )( 2 )( 0.6 ) −

(b) Now ΔK = N

g

si

n(

45

°)

s = vt = 2m

m

mg cos(45°)

therefore, v f = 5 ms −1 = 500 cms −1 f=

mg √2

4.

(a) By Law of Conservation of Energy, we get 0 + mg ( 4 R ) =

45° 45°

⇒

  Now W f = f ⋅ s

2.

⇒

W f = ( f cos 45° ) ( 2 )

⇒

⎛ 10 ⎞ ⎛ 1 ⎞ ( ) Wf = ⎜ 2 ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

⇒

W f = 10 J

Since, P = Fv ⇒

F=

P 8 × 10 5 = = 4 × 10 4 N v 20

At constant speed, the forces acting on the train are in equilibrium. Resolving the forces parallel to the hill ⎛ 1 ⎞ F = R + ( 2 × 10 5 ) × g × ⎜ ⎝ 50 ⎟⎠ ⇒

4 × 10 4 = R + 39200

⇒

R = 800 N

1 1 mv 2f − mvi2 = 32 J 2 2

Since vi = 3 ms −1

f

mg

1 ( 40 − 24 )( 2 ) = 32 J 8

CHAPTER 1

a=

4 gR =

1 mv 2 2

v2 2

At Q , we have N=

mv 2 m ( 8 gR ) = = 8 mg R R

So, net force at Q is FQ = N 2 + ( mg ) ⇒

2

FQ = 64 m2 g 2 + m2 g 2 = 65mg

⇒ x = 65 (b) For block to exert force (on the top of track) equal to weight, we have N = mg Since mg + N = ⇒

mv 2 R

mv 2 = 2mg R

Therefore, the resistance is 800 N.

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 51

2/9/2021 6:33:41 PM

H.52 JEE Advanced Physics: Mechanics – II 6.

⇒ v 2 = 2Rg

Let v be the velocity of the particle at the highest point. v

Applying Law of Conservation of Energy, we get mg ( h ) = ⇒

mg

1 m ( 2Rg ) + mg ( 2R ) 2

H

N R

h = 3R

⇒ ∗= 3 5.

Let m be the mass of B . From its free-body diagram

According to Newton’s Second Law, the net force

T − μN = m ( 0 ) = 0

Fnet = N + mg provides the centripetal force to the body to revolve in a circle of radius R . So

where T is the tension in the string and N = mg ⇒

mv 2 R The body does not fall at the uppermost point of the loop if N ≥ 0. In the limiting case, N = 0 . So,

T = μ ( mg )

N + mg =

From the free-body diagram of the spring T − T′ = 0 where T ′ is the force exerted by A on the spring ⇒

mg =

T = T ′ = μmg μ N′

T

μN

B

T

(a)

N′

⇒ T′

From (1), v 2 = gR , so, we get 1 mgH = mg ( 2R ) + m ( gR ) 2

From the free-body diagram of A ⇒

2 g − ( T ′ + μN ′ ) = 2 ( 0 ) = 0

⇒

2g = T ′ = μmg

⇒

m=

T = T ′ = μmg = 0.2 × 10 × 9.8 = 19.6 N Now, for a spring, we have F = kx

⇒

x=

⇒ 7.

1 m 100

Since Wnc + Wext = ΔU + ΔK

⇒

fx cos ( 180° ) + 0 =

1 2 1 kx + ( − mgx sin θ ) + mv 2 2 2

− ( μ k mg cos θ ) x =

1 2 1 kx − mgx sin θ + mv 2 2 2 N

f mg

2

1 2 1 ⎛ 1 ⎞ kx = × 1960 × ⎜ = 0.098 J ⎝ 100 ⎟⎠ 2 2

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 52

H = ( 2.5 )( 40 ) = 100 cm

Fs

⇒ Energy stored in the spring is given by U=

5 R = 2.5R 2

Since R = 40 cm , we get

Tensile force on the spring is given by

19.6 = 1960 x

H=

Let the block move down the incline through a distance x , then

2g 2 = = 10 kg μ g 0.2

⇒

…(1)

1 mgH = mg ( 2R ) + mv 2 2

A

where N ′ is the normal reaction of the vertical wall of C on A and N ′ = 2 ( 0 ) = 0 (because there is no horizontal acceleration of A )

v 2 = gR

Applying Law of Conservation of Energy at the initial and highest point of the loop, we get

2g (c)

T′ (b)

mv 2 R

θ

2/9/2021 6:33:49 PM

Hints and Explanations

(b) Maximum kinetic energy of the block is the mechanical energy of spring block system which actually is the energy dissipated as heat = 80 J 1 1 (c) 80 = kx 2 = ( 640 ) x 2 2 2

x

x sin θ

θ

−1 = 1 − 6 + v 2

⇒

v2 = 4

⇒

v = 2 ms −1

⇒

10. Let  0 be the unstretched length of the spring, then by Law of Conservation of Energy, we have

( U + K )at A = ( U + K )at B 1 1 1 1 2 2 k ( 3 −  0 ) + mvA2 = k ( 5 −  0 ) + mvB2 2 2 2 2

Let u be the speed of ice block when it leaves the table top. Then

But vB = 0

1 1 mu2 = kx 2 2 2 ⇒

x = 0.5 m = 50 cm

⎛ k⎞ u2 = ⎜ ⎟ x 2 ⎝ m⎠ u A h

Table B

v

If v be its speed when it reaches the floor, then By Law of Conservation of Energy, we have

( U + K )at A = ( U + K )at B ⇒

1 1 mu2 + mgh = mv 2 + 0 2 2

⇒

v = u2 + 2 gh =

⇒

( 3 −  0 )2 + 4 = ( 5 −  0 )2

⇒

(9 + 

⇒

16 − 4 0 = 4

⇒

0 = 3 m

2 0

)

(

− 6 0 + 4 = 25 +  20 − 10 0

Before we make use of the above relation, we must observe that (i) A move down by xA, so its potential energy decreases by mAghA where hA = xA sin θ = ( 1 ) ⎛⎜ 3 ⎞⎟ = 3 m. ⎝ 5⎠ 5

kx 2 + 2 gh m

xA

(ii) B moves up by xB = hB , so its potential energy as well as kinetic energy increases.

m = 120 g = 0.12 kg h = 2.5 m

⇒ 2

+ 2 ( 10 )( 2.5 )

⇒

v=

⇒

v = 50 + 50 = 10 ms −1

0.12

hA

θ

k = 2400 Nm −1

( 2400 )( 0.05 )

)

11. Using our knowledge of constraint relations, we observe that at any instant xB = 2x A and vB = 2vA. So, let vA = v after x A = 1 m . Using Work Energy Theorem for a non-conservative system, we get Wnc = ΔU + ΔK .

Now, x = 5 cm = 0.05 m

9.

CHAPTER 1

8.

⇒

H.53

(a) Increase in thermal energy of block-floor system is the work done against friction. ⇒ Wfriction = μ K mgs = ( 0.25 ) ( 4 ) ( 10 )( 8 ) ⇒ Wfriction ≈ 80 J

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 53

f K x A cos ( 180° ) = ΔU A + ΔK A + ΔU B + ΔK B

4 , x A = 1 m, 5 xB = 2 m, hA = 0.6 m , mA = 30 kg, mB = 5 kg,

where

f K = μ K mA g cos θ ,

cos θ =

g = 10 ms −2 . ⇒

− ( μ K mA g cos θ ) x A = − mA ghA + mB ghB + 1 1 2 mAv 2 + mB ( 2v ) 2 2

2/9/2021 6:33:57 PM

H.54 JEE Advanced Physics: Mechanics – II Substituting the values, we get 1⎞ ⎛ 4⎞ ⎛ 3⎞ ⎟ ( 300 ) ⎜⎝ ⎟⎠ = − ( 300 ) ⎜⎝ ⎟⎠ + ( 50 ) ( 2 ) + 5⎠ 5 5

⎛ −⎜ ⎝

1 1 ( 30 ) v 2 + ( 5 )( 2v )2 2 2 ⇒

−48 = −180 + 100 + 25 v 2

⇒ v=

32 4 2 = 25 5

⇒ ∗= 2 12.

h =  sin θ = 2 sin θ

H=

⇒

H = 0.96 m = 96 cm

⇒

H = 24 x = 96

⇒

x=4

v = 2 gh , where h = R ( 1 − cos θ ) ⇒

mv 2 

v = 2 gR ( 1 − cos θ )

Also, at angle θ , we have N + mg cos θ =

⇒

2mg − mg sin θ =

mv 2 

⇒

2 ( 10 ) − 10 sin θ =

25 + 40 sin θ 2

⇒

sin θ =

⇒

1⎞ θ = sin ⎜ ⎟ ⎝ 4⎠

mv 2 R

⇒

N + mg cos θ = 2mg ( 1 − cos θ )

⇒

N = 2mg − 3 mg cos θ

…(1)

When the tube just breaks its contact with the ground then

1 4

2N cos θ = Mg

−1 ⎛

Substituting the value of N from (1), we get 4 mg cos θ − 6 mg cos 2 θ = Mg

So, ∗ = 4 sin θ =

⎝ 25 ⎠ 24 = 2 ( 10 ) 25

H=

Given, T = 2mg

13.

( 30 ) ⎛⎜ 16 ⎞⎟

⇒

v 2 = 25 + 40 sin θ

Further, T − mg sin θ =

v 2 sin 2 ( 90° − θ ) v 2 cos 2 θ = 2g 2g

14. Speed of each particle at angle θ using Law of Conservation of Energy is

v 2 = v02 + 2 gh = 25 + 2 × 10 × 2 sin θ ⇒

The maximum height of the projectile is

Substituting, θ = 60° , we get

3 5

At Q , T = 0 and hence the particle will become a projectile following a parabolic path.

2mg − ⇒

v 90°– θ

3 mg = Mg 2

m =2 M

Conceptual Note(s)

Q 3m

θ

O

5m

Initially, the normal reaction on each ball will be radially outwards and later it will be radially inwards, so that the net normal reaction on the tube is radially outwards to break it off from the ground.

P

⇒

mv 2 = mg sin θ r

⇒

⎛ 3⎞ v = rg sin θ = ( 5 )( 10 ) ⎜ ⎟ = 30 ms −1 ⎝ 5⎠

⇒

∗ = 30

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 54

15. Let v be the speed of ball at point B . Then by Law of Conservation of Energy applied between A and B , we get U A + K A = UB + KB ⇒

1 1 2 0 + ( 1 ) ( 5 ) = ( 1 ) ( 10 ) ( 1 ) + ( 1 ) v 2 2 2

2/9/2021 6:34:05 PM

Hints and Explanations v 2 25 = − 10 2 2

⇒

v2 = 5

Force acting on particle at B is ⎛ mv ⎞ ⎟ r ⎠ 2

( mg )2 + ⎜⎝

F=

2

⇒

⎛ 1× 5⎞ 2 F = ( 10 ) + ⎜ ⎝ 1 ⎟⎠

⇒

F = 125

⇒

F=5 5 N

⇒

x=5

2

ARCHIVE: JEE MAIN 1.

After reading the problem, we conclude that

3.

Since, P = constant . From Work-Energy Theorem

for 0 ≤ x ≤ 15 , F = 200 N and for 15 ≤ x < 30 , F = 200 − ⇒

100 ( x − 15 ) 15

200 , for 0 ≤ x ≤ 15 100 ⎢ 300 − x , for 15 ≤ x < 30 15 ⎣

⎡

W=

∫ Fdx

15

30

0

15

∫ 200dx + ∫

100 ⎞ ⎛ x ⎟ dx ⎜⎝ 300 − 15 ⎠

W = 3000 + 4500 − 2250 = 5250 J

Given that, U = − ⇒

F=−

A B + 12 6 r r

At equilibrium, F = 0 6 A 12B − 13 r7 r

0=

⇒

6A 1 = 12B r 6

⇒

⎛ 2B ⎞ r=⎜ ⎝ A ⎟⎠

So, U eq = U

⇒

s = ds =

ds = dt

∫

2Pt = m 2P m

2P 1 2 t m t

∫t

12

0

dt =

2 2P 3 2 t 3 m

dx = dt

⇒

v=

⇒

x = dx =

⇒

x=

∫

2Pt = m 2P m

2P 1 2 t m t

∫t

12

0

dt =

2 2P 3 2 t 3 m

2 2(1) 2 ( 9 )3 2 = ( 1 )( 27 ) = 18 m 3 2 3

Hence, the correct answer is (18).

⎛ 2B ⎞ r=⎜ ⎝ A ⎟⎠

U eq = −

v=

16

2

16

=−

A B + 2 2 B A ) 4B A 2 (

2

(

2

A A A + =− 2B 4B 4B Hence, the correct answer is (C). ⇒

⇒

1 mv 2 2

Since, P = constant , so from Work-Energy Theorem, 1 we get W = mv 2 . Also, W = Pt 2 2Pt 1 ⇒ K = Pt = mv 2 i.e., v = m 2

dU = − A ( −6 r −7 ) + B ( −12r −13 ) dr

⇒

K = Pt =

4.

Hence, the correct answer is (B). 2.

⇒

Hence, the correct answer is (D).

⎛ 100 ⎞ ⎛ (30 2 − 152 ) ⎞ ⎟⎜ ⇒ W = ( 200 )( 15 ) + ( 300 )( 15 ) − ⎜ ⎟ ⎝ 15 ⎠ ⎝ 2 ⎠ ⇒

1 mv 2 2

Also, we know that W = Pt

F ( in N ) = ⎢

Since, W = ⇒

W=

CHAPTER 1

⇒

H.55

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 55

)

5.

⇒

1 mv 2 = mgh 2 Fs = mgh

⇒

F ( 0.2 ) = ( 0.15 )( 10 )( 20 )

⇒

F = 150 N

Since, WF =

Hence, the correct answer is (150).

2/9/2021 6:34:13 PM

H.56 JEE Advanced Physics: Mechanics – II 6.

T = 2000 g + f = 2000 + 4000

Applying Work-Energy Theorem, we get

⇒ T = 24000 N Power of the electric motor of the elevator is P = Fv = Tv ⇒

60 × 746 = 1.865 ≈ 1.9 ms −1 24000 Hence, the correct answer is (1.9). ⇒

mg sin θ ( AC + 2 AC ) − ( μmg cos θ ) AC = 0 ⇒

μ = 3 tan θ

⇒

k=3

v=

10. Applying Law of Conservation of Mechanical Energy between A and B , we get

Hence, the correct answer is (3). 7.

60 × 746 = ( 24000 ) v

U A + K A = UB + KB

Since the elevator is moving with a constant speed, so we have

⇒

mg ( 2 ) = mg ( 1 ) + K 2

⇒

K 2 = mg = 10 J

Hence, the correct answer is (10). 11.

W = ΔK = Area under F -x graph ⇒

ΔK = W =

⇒

ΔK = 6.5 J

1 × ( 3 + 2 ) × ( 3 − 2 ) + 2 × 2 = 2.5 + 4 2

Hence, the correct answer is (D). 12. Mass of hanging portion is

T = 6800 + 9200 + 6000 = 22000 N

m1 =

So, power delivered by the motor is P = Tv = 22000 × 3 = 66000 W

M n

Hence, the correct answer is (C).  rg

8.

W=

∫  ri

⇒

  F ⋅ dr =

0

∫ 1

1

L/n 0

⎛ y2 ⎞ ⎛ x2 ⎞ − xdx + ydy = ⎜ − ⎟ + ⎜ ⎝ 2 ⎠ 1 ⎝ 2 ⎟⎠

∫ 0

1 0

⎛ 0 2 12 ⎞ ⎛ 12 0 2 ⎞ W = −⎜ − ⎟ +⎜ − ⎟ =1J ⎝ 2 2⎠ ⎝ 2 2 ⎠

Hence, the correct answer is (B). 9.

Let the elevator be moving up with a constant speed v. Then frictional force f will be downwards. So, tension in the cable is

So, to pull the complete hanging portion up, the centre of mass of hanging portion has to be pulled through L . 2n ⇒

⎛ Mg ⎞ ⎛ L ⎞ W=⎜ ⎝ n ⎟⎠ ⎜⎝ 2n ⎟⎠

MgL 2n2 Hence, the correct answer is (C). ⇒

W=

13. Initial equilibrium position of block is at x = 0 Speed of the block will be maximum at equilibrium position i.e., when F = kxeq ⇒

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 56

xeq =

F k

2/9/2021 6:34:19 PM

Hints and Explanations Applying Law of Conservation of Energy, we get

By Law of Conservation of Energy,

( U + K )i = ( U + K ) f Fxeq =

⎛ Loss in GPE ⎞ ⎛ Gain in KE ⎞ ⎜⎝ of pendulum ⎟⎠ = ⎜⎝ of pendulum ⎟⎠

1 2 1 kxeq + mv 2 2 2 2

mgh =

K1 = mgh = mgL ( 1 − cos θ )

⇒

1 ⎛ F⎞ 1 ⎛ F⎞ F ⎜ ⎟ = k ⎜ ⎟ + mv 2 ⎝ k⎠ 2 ⎝ k⎠ 2

⇒

⇒

F2 1 mv 2 = 2 2k

Now, when L is doubled, keeping θ same, then Kinetic energy is also doubled. So

⇒

v=

K 2 = 2K1

F mk

Hence, the correct answer is (A).

Hence, the correct answer is (D). 14.

1 mv 2 2

⇒

v=

18. Force on particle in the given potential field is given as  ⎛ dU ⎞ ˆ k ˆ F = −⎜ r= 3r ⎝ dr ⎟⎠ r

dx = 6t dt

u = v(t = 0 ) = 0 v = v ( t = 5 s ) = 30 ms −1

CHAPTER 1

⇒

H.57

Fc

1 ⇒ W = ΔKE = (2)(30)2 = 900 J 2 Hence, the correct answer is (B). 15.

N − mg = ⇒

N=

mg 2

For body to move in a circle, we have FC =

3 mg 2

1⎛ g⎞ Since, s = ⎜ ⎟ t 2 2⎝ 2⎠ So, work done by normal reaction is

k r2 ⇒ The kinetic energy of particle is ⇒

16.

)

W = K f − K i = 3iˆ − 12 ˆj ⋅ 4iˆ = 12 J ⇒

k k − 2 =0 2 2r 2r Hence, the correct answer is (C). ⇒

W=

(

k 1 mv 2 = 2 2 2r

Total energy of particle TE = KE + PE

⎛ 3 mg ⎞ ⎛ g 2 ⎞ W=⎜ t ⎝ 2 ⎟⎠ ⎜⎝ 4 ⎟⎠

3 mg 2t 2 8 Hence, the correct answer is (A). ⇒

mv 2 =

KE =

WN = Ns cos ( 0° ) = Ns ⇒

k mv 2 = r r3

19. For particle to move at speed v in a circular orbit, we have FC =

K f = 12 + K i = 12 + 3 = 15 J

Hence, the correct answer is (A). 17. θ

E=

⇒

16 mv 2 + r3 = r r

⎛ 16 ⎞ mv 2 = ⎜ + r 3 ⎟ ( r ) = 16 + r 4 ⎝ r ⎠ v

L

Fc r

m h

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 57

O

GPE = 0

2/9/2021 6:34:25 PM

H.58 JEE Advanced Physics: Mechanics – II So, kinetic energy of particle is ⇒

1 1 mv 2 = ( 16 + r 4 ) 2 2 Ratio of kinetic energy of the two particles is given by K=

1( ( )4 ) 16 + 1 K1 2 16 + 1 = = 1 4 K2 ( 16 + ( 4 ) ) 16 + 256 2 ⇒

K1 17 = = 0.06 K 2 272

Hence, the correct answer is (B). 20.

v=a s ⇒

ds =a s dt s

⇒

∫ 0

⇒

ds = s

22.

∫ adt

v=

⇒

−

⇒

k=

m v0t0

⇒

k=

10 −2 10 × 10

⇒

k = 10 −4 kgm −1

k 1 = − t0 v0 m

F = 6t = m

dv where m = 1 kg dt 1

v

∫ ∫

dv = 6t dt

0

2 2

⇒

2

0

⎛ t2 ⎞ v = 6⎜ ⎟ ⎝ 2⎠

1 ( 1 ) ( 9 ) = 4.5 J 2 Hence, the correct answer is (A). W = ΔKE =

1 ⎛ a 4t 2 ⎞ 1 4 2 1 mv 2 − 0 = m ⎜ ⎟ = ma t 2 2 ⎝ 4 ⎠ 8

23.

P

Hence, the correct answer is (C). 1 mv02 Kf 1 8 21. Since, = = 1 Ki mv02 4 2

⇒

vf

4m

h = 2m

30° O

Energy lost over path PQ is

v v vf = i = 0 2 2

Energy lost over path QR is

v0 2

∫

v0

R Q x

1 = vi 2

Since, F = − kv 2 =

⇒

0

⇒ v = 3 ms According to Work Energy Theorem, we have

W = ΔK

⇒

1

−1

ds a t = 2 dt

W=

k t0 m

k 1 2 − = − t0 v0 v0 m

⇒

Applying Work Energy theorem, we get

⇒

=−

v0

⇒

0

2 s = at s=

v0 2

Hence, the correct answer is (C).

t

at 4 Velocity at any time t is given by ⇒

⎛ 1⎞ ⎜⎝ − ⎟⎠ v

E1 = μmg cos θ × 4

mdv dt

t0

− kdt dv = m v2

∫ 0

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 58

E2 = μmgx Since E1 = E2 ⇒

μmgx = μmg cos θ × 4

⇒

x = cos θ × 4 = 4 cos ( 30° )

⇒

x = 2 3 = 3.45 m

2/9/2021 6:34:32 PM

Hints and Explanations From Q to R energy loss is half of the total energy loss, so

⇒

1 ( mgh ) 2

So, if then before reaching top, ingredients fall and mix proper. Hence, the correct answer is (A).

μ = 0.29

Hence, the correct answer is (D). 24. Work done against gravity in lifting the mass 1000 times is W = ( 1000 ) ( mgh )

⎛ 100 ⎞ ⎛ 1 ⎞ 26. Inclination of road, θ = tan −1 ⎜ = tan −1 ⎜ ⎟ ⎝ 1000 ⎟⎠ ⎝ 10 ⎠ 1 ⇒ tan θ = 10 1 (for very small value of θ ) 10 When car is moving uphill with a speed of u = 10 ms −1 , then F = W sin θ + f ⇒

⇒

W = 10 × 9.8 × 10 3

⇒

W = 9.8 × 10 4 Joule

Let M kg of fat be used up in the process. Then energy supplied by fat is E = ( 3.8 × 107 M )

tan θ ≈ sin θ =

N

20 J 100

This energy supplied by the fat must equal to the work done by him against gravity. So,

100 m

θ

f θ

W W cos θ 1 km

M = 12.89 × 10 −3 kg

Hence, the correct answer is (A). 25. In the frame of rotating drum, at the topmost point the mixture will not loose contact with the drum wall if

⇒

P = Fu = ( W sin θ + f ) u

⇒

⎛W W⎞ P=⎜ + × 10 ⎝ 10 20 ⎟⎠

3W 3W × 10 = 20 2 When car is moving downhill with a constant speed, say v , then ⇒

mv 2 ≥ mg r ⇒

u

W sin θ

20 ( 3.8 × 107 M ) = 9.8 × 10 4 100 ⇒

60 10 ≈ 27 rpm 2π 1.25

CHAPTER 1

μmgx =

ω in rpm is ω RPM =

H.59

v ≥ gR

P=

F ′ = W sin θ − f

mv2 R v mg v

⇒

P = F ′v = ( W sin θ − f ) v 2

⇒

3W ⎛ W W ⎞ =⎜ − v ⎝ 10 20 ⎟⎠ 4

ω

N

f

v W sin θ θ

If v ≥ gR , then proper mixing will not take place Thus, for proper mixing v < gR ⇒

ω=

v = R

g 10 rads −1 = R 1.25

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 59

θ

100 m

W W cosθ 1 km

⇒

3W W = v 4 20

⇒

v = 15 ms −1

Hence, the correct answer is (C).

2/9/2021 6:34:38 PM

H.60 JEE Advanced Physics: Mechanics – II 27.

–1

v(ms )

⇒

50

Ui =

2 ( 0.1 )( 9 ) = 0.6 J 3

Hence, the correct answer is (B). 30. Centripetal force F = α r 2

v

mv 2 = αr2 r So, kinetic energy of the particle is ⇒

(0, 0)

2

10

t(s)

1 αr3 mv 2 = 2 2 Potential energy of the particle is

Given that, m = 10 kg , t = 2 s

K=

u = 50 ms −1 at t = 0 s Since, a =

Δv 50 − 0 = = −5 ms −2 Δt 0 − 10

r

U=

So, speed of the body at t = 2 s is given by

E = K +U =

1 1 1 1 2 2 mv 2 − mu2 = ( 10 )( 40 ) − ( 10 )( 50 ) 2 2 2 2

⇒

W = 5 ( 40 2 − 50 2 ) = 5 × ( 40 − 50 ) ( 40 + 50 )

⇒

W = −4500 J

vT2 = n2 R2t 2

⇒

vT = nRt

5 E = αr3 6 Hence, the correct answer is (D). 31.

W=

vT2

L

∫ Fdx =∫ ( ax + bx ) dx 2

0

R

Tangential force on the particle is

αr3 αr3 + 2 3

⇒

L

Hence, the correct answer is (C).

⇒

0

αr3 3

Total energy of the particle is

Using Work-Energy theorem i.e. W = ΔK , we get

28. Centripetal acceleration is ac = n2 Rt 2 =

∫

0

v = u + at = 50 + ( −5 ) × 2 = 40 ms −1

W=

∫

r

Fdr = α r 2 dr =

⇒

W=

0

2

aL bL3 + 2 3

Hence, the correct answer is (C). 32.

dv FT = M T = MnR dt Power delivered to the particle is

X

N=0

⇒ P = Mn2 R2t Hence, the correct answer is (B).

R cos θ

ϕ θ

v

R sin ϕ

ϕ

29. The block comes momentarily to rest means that its velocity at that point was also 3 ms −1 . ⎛ Loss in EPE ⎞ ⎛ Gain in KE ⎞ ⎜⎝ of Spring ⎟⎠ = ⎜⎝ of Block ⎟⎠

h = R (cos θ – sin ϕ )

Y

P = FT vT = ( MnR )( nRt )

O

Applying Energy Theorem, we get mgh =

1 mv 2 2

⇒

1 ⎛ 2 x2 ⎞ 1 k ⎜ x − ⎟ = mv 2 2 ⎝ 4 ⎠ 2

⇒

⇒

1 2⎛ 1⎞ 1 kx ⎜ 1 − ⎟ = mv 2 ⎝ 2 4⎠ 2

At point Y, we have along radial direction in circular motion

⇒

1 2 4 2 kx = U i = mv 2 = mv 2 2 6 3

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 60

v = 2 gR(cos θ − sin ϕ )

mv 2 = mg sin ϕ R

2/9/2021 6:34:45 PM

Hints and Explanations ⇒

2mg ( cos θ − sin ϕ ) = mg sin ϕ

⇒

2 cos θ = 3 sin ϕ

If ρ be the density of air, then mass of wind intercepting the blades per second is m = ρ Av

Hence, the correct answer is (C). 33. To find KE of spring, we consider an elemental part of spring as shown. m Mass of element is dm = dx 

dK 1 1 = mv 2 = ρ Av 3 dt 2 2

CHAPTER 1

Hence, the correct answer is (D).

l

dx

Kinetic energy of wind mass intercepting the section of blades per second is

⇒ Electrical power ∝ v 3

v′

x

35. If plank thickness is x , then, we have m

H.61

Plank

v

As spring is uniformly stretched, the speed v′ of element is

u

u–u n

a

⎛ v⎞ v′ = ⎜ ⎟ x ⎝ ⎠ Kinetic energy of element is dK = ⇒

1⎛ m ⎞ ⎛ v ⎞ dK = ⎜ dx ⎟ ⎜ x ⎟ 2⎝  ⎠ ⎝  ⎠

⇒

1 mv 2 2 dK = x dx 2 3

∫

1 dmv′ 2 2

2

1⎞ ⎛ u2 ⎜ 1 − ⎟ = u2 − 2 ax ⎝ n⎠

2



∫ 0

Total kinetic energy of spring is given as K=

x

1 mv 2 ⎛  3 ⎞ 1 2 ⎜ ⎟ = mv 2 3 ⎝ 3 ⎠ 6

⇒

2 ⎡ ⎛ n − 1⎞ ⎤ 2 ax = u2 ⎢ 1 − ⎜ ⎟ ⎝ n ⎠ ⎥⎦ ⎣

⇒

⎡ 2n − 1 ⎤ 2 ax = u2 ⎢ ⎣ n2 ⎥⎦

Using constant retardation, we have 1

Hence, the correct answer is (D). 34. Wind volume flowing through the cross-sectional area of blades (rotating) of wind mill per unit time is given by

2

3

–

u

–

N

vf = 0

V = Av v metre

Nx

area = A Volume flow in 1 s

0 = u2 − 2 a ( Nx )

v Wind speed

⇒

⎡ 2n − 1 ⎤ u2 = Nu2 ⎢ ⎣ n2 ⎥⎦

⇒

N=

n2 2n − 1

Hence, the correct answer is (A).

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 61

2/9/2021 6:34:49 PM

H.62 JEE Advanced Physics: Mechanics – II 36. Applying Work-Energy Theorem from point A to C, we get A

v0

…(1)

B

C

W1 =

vf = 0

1 ( k x ) F2 k2 x22 = 2 2 = 2 2k 2 2k 2 2

W2 = ⇒

Hence, the correct answer is (B). 37. Maximum energy loss is

v = 2 ah1 = 2 gh2

v

g

h2 = 1 m

h1 = 0.5 m

u=0

F

F W = mg = 700 N

gh2 10 × 1 = = 20 ms −2 h1 0.5

Velocity at take-off is v = 2 × 10 × 1 = 20 ms −1 Power delivered by muscles is P = Fv = mav P = 70 × 20 × 20

Mechanics II_Chapter 1_Part 5_Hints and explanation.indd 62

W1 k2 = W2 k1

⇒

k2 > k1

⇒

k1 < k2

Statement 2 is true.

38. If a is upward acceleration of man before leaping, then

vf = 0

{using (1)}

Since, W1 > W2

1 Mm M ⎛1 2⎞ ( v − 0 )2 = ⎜ mv ⎟⎠ 2M+m M + m⎝ 2 So, Statement-I is wrong. Hence, the correct answer is (D).

take off point

{using (1)}

Work done on spring S2 is

v2 h L= 0 + 2μ g μ

a=

1 ( k x ) F2 k1x12 = 1 1 = 2 2k 1 2k1 2

v

1 mv02 + mgh − μmgL = 0 2

⇒

Hence, the correct answer is (B).

Work done on spring S1 is Smooth

⇒

P = 6.26 × 10 3 W

39. For the same force, F = k1x1 = k2 x2

h

⇒

⇒

For the same extension, i.e. for x1 = x2 = x Work done on spring S1 is W1 =

1 1 k1x12 = k1x 2 2 2

Work done on spring S2 is W2 =

1 1 k2 x22 = k2 x 2 2 2

⇒

a

…(2)

W1 k1 = W2 k2

Since, k1 < k2 ⇒

W1 < W2

Statement 1 is false. Hence, the correct answer is (D). 40. By Work Energy Theorem   W = F ⋅ Δr = ΔK

(

)(

⇒

ΔK = 7 iˆ + 4 ˆj + 3 kˆ ⋅ 2iˆ + 3 ˆj + 4 kˆ

⇒

ΔK = 14 + 12 + 12 = 38 J

)

Hence, the correct answer is (A). 41. As x → ∞ , U ( x ) = 0 At equilibrium, force on atoms is F = 0 ⇒

F=−

⇒

x6 =

dU 12 a 6b = − =0 dx x13 x 7

2a b

2/9/2021 6:34:55 PM

Hints and Explanations 1

Potential energy at equilibrium is given by a

U eqbm =

12 2a ⎞ 6

U eqbm =

a ⎛ 4a ⎞ ⎜⎝ 2 ⎟⎠ b 2

6

−

U eqbm =

⇒

⎛ b2 ⎞ b2 D = U ∞ − U eqbm = 0 − ⎜ − ⎟ = ⎝ 4a ⎠ 4a

Hence, the correct answer is (D).

⎛ 2a ⎞ 6 ⎜⎝ ⎟⎠ b

⎛ ⎜⎝ ⎟⎠ b ⇒

b

−

b2 b2 b2 − =− 4 a 2a 4a

⇒

b 2a b

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1.

2 and normal force N 3 inwards on bead, when cos θ >

Let at any instant, the bead make an angle θ with the vertical, as shown in Figure. N

cos θ


m2 g k

{∵ kx = m2 g } v=0

12. Consider the two blocks plus the spring to be the system. No external force acts on this system in horizontal direction. Hence, the linear momentum will remain constant. Suppose, the block of mass M moves with a speed V and the other block with a speed v after losing contact with the spring. From Law of Conservation of Linear Momentum in horizontal direction, we get

m1 x m1

m2

Substituting, x =

⇒

v = 2gH

m2

m2 g in Equation (1), we get k 2

m2 g ⎛ m2 + 2m1 ⎞ k ⎜⎝ 2m1 ⎟⎠

H=

⇒

H min =

…(1)

1 The initial energy of the system is Ei = kx 2 2 The final energy of the system is 1 1 E f = mv 2 + MV 2 2 2 Since there is no friction so, mechanical energy will remain conserved. 1 1 1 ⇒ mv 2 + MV 2 = kx 2 …(2) 2 2 2 Solving equations (1) and (2), we get

⎛ m g⎞ ⎛ m g⎞ 2m1 gH = k ⎜ 2 ⎟ + 2m1 g ⎜ 2 ⎟ ⎝ k ⎠ ⎝ k ⎠ ⇒

MV − mv = 0 m V= v M

m2 g ⎛ m2 + 2m1 ⎞ k ⎜⎝ 2m1 ⎟⎠

11. Let us assume the final velocities of the particles m and 2m be v1 and v2 , respectively, as shown in Figure.

v=x

kM and m( M + m )

V=x

km M( M + m)

Test Your Concepts-VII (Based on Oblique Collisions) 1.

In perfectly inelastic collision with the horizontal surface, the component parallel to the surface will remain unchanged. P θ θ

m ( 2v ) + 2m ( − v ) = m ( v1 ) + 2m ( v2 ) ⇒

θ D

vccosθ

0 = mv1 + 2mv2

v2 − v1 = 2v − ( − v ) v2 − v1 = 3v

…(2)

Solving equations (1) and (2), we get v1 = −2v and v2 = v This simply means that both masses reverse their direction of motion after the collision as shown in Figure.

C

H L By Law of Conservation of Energy, we have cos θ =

vc = 2 gH Similarly, when the string becomes taut again, the component perpendicular to its length will remain unchanged. So, vc cos 2 θ =

(

2 gH

2

) HL2

=v

{say}

Again, using Law of Conservation of Energy, we get h=

Mechanics II_Chapter 2_Hints and Explanation_1.indd 88

θ vc

⇒ v1 + 2v2 = 0 …(1) Since collision is elastic, so relative velocity of approach of bodies equals the relative velocity of separation of bodies. Hence, we get ⇒

L H vccosθ

vccos2θ

Applying conservation of linear momentum, we get

B

2

v = 2g

H4 5 L4 = H 4 2g L

( 2 gH )

2/9/2021 6:32:38 PM

Hints and Explanations  Let v be the velocity of the sphere after impact. To find  v we must separate the velocity components parallel and perpendicular to the wall. Since, after impact the component of velocity parallel to the wall remains unchanged while component perpendicular to the wall becomes e times in opposite direction. So, we have 3  v = − iˆ + ˆj 2 Therefore, the velocity of the sphere after impact is 3 − iˆ + ˆj 2 The loss in kinetic energy is

To find the coefficient of restitution we require the velocity components, before and after impact, in the direction of impulse i.e., along the line of impact, i.e., in the direction −3iˆ + 4 ˆj . The unit vector in the direction  1 of J is −3iˆ + 4 ˆj . So, the magnitudes of the velocity 5 components in this direction just before impact is

(

( 4iˆ − ˆj ) ⋅ 15 ( −3iˆ + 4 ˆj ) = − 165 and just after impact is

( iˆ + 3 ˆj ) ⋅ 51 ( −3iˆ + 4 ˆj ) = 95

16 is sim5 ply an indication that this component is in a direction  opposite to that of J . 16 The speed of approach to the wall is therefore, and 5 9 the speed of separation is . Since 5 The significance of the negative sign in −

− ΔK = K i − K f 2 ⎞ 27 m 1 ⎛ 2 2 ⎛ 3⎞ J m ⎜ 3 + 1 − ⎜ ⎟ − 12 ⎟ = ⎝ ⎠ ⎝ ⎠ 2 2 8   Since impulse J = Δp , so we have  ⎛ 3 ⎞ ⎛ 9m ⎞ ˆ J = m ⎜ − iˆ + ˆj ⎟ − m 3iˆ + ˆj = − ⎜ i ⎝ 2 ⎠ ⎝ 2 ⎟⎠

⇒

− ΔK =

(

3.

)

Since the collision is perfectly inelastic, so the putty wedge system of mass ( M + m ) moves as a single body with a velocity V. Applying the conservation of linear momentum, we get mv0 cos θ = ( M + m )V

6.

1 1 mv02 − ( M + m )V 2 2 2 mv0 cos θ where, V = M+m ⇒

− ΔK =

⇒ ⇒ 4.

1 1 ⎛ mv0 cos θ ⎞ mv02 − ( M + m ) ⎜ ⎝ M + m ⎟⎠ 2 2 mv02 ⎛ m cos 2 θ ⎞ − ΔK = ⎜⎝ 1 − ⎟ 2 M+m ⎠ − ΔK =

Speed of Separation Speed of Approach

e=

95 9 = 16 5 16

 v′ = u 2 + v 2

Method I: By using conservation of linear momentum If we take wedge and ball as a system, then no external force is acting along the horizontal direction, so linear momentum is conserved along the horizontal direction. Since the ball is moving along the normal direction, so it has no component of velocity along the inclined surface, i.e. the common tangent direction or along the t-line.

2

mv02 ⎛ M + m sin 2 θ ⎞ ⎜ ⎟ 2 ⎝ M+m ⎠

Since impulse equals change in momentum, so we have  J = m iˆ + 3 ˆj − m 4iˆ − ˆj  ⇒ J = m −3iˆ + 4 ˆj

( (

e=

Applying Law of Conservation of Linear Momentum in vector form, we get  muiˆ = 0 + mvjˆ + mv′  ⇒ v′ = uiˆ − vjˆ ⇒

Loss = − ΔK = K i − K f − ΔK =

⇒ 5.

mv0 cos θ ⇒ V= M+m The loss in kinetic energy of the system is

⇒

)

CHAPTER 2

2.

H.89

)

)

(

Mechanics II_Chapter 2_Hints and Explanation_1.indd 89

)

Let velocity of the ball and wedge after collision be vn and V respectively, then by conservation of linear momentum applied along the horizontal direction, we have

2/9/2021 6:32:46 PM

H.90 JEE Advanced Physics: Mechanics – II mv0 sin θ = MV + mvn sin θ

…(1)

V sin θ − vn = e ( v0 − 0 )

…(2)

y

Also, v2 − v1 = e ( u1 − u2 ) ⇒

x

Solving equations (1) and (2), we get

30°

⎡ ( 1 + e ) m sin θ ⎤ V=⎢ ⎥ v0 ⎣ M + m sin 2 θ ⎦

Squaring and solving, we get R ≈ 2 m and t ≈ 1 s

⎛ m sin 2 θ − Me ⎞ and vn = V sin θ − ev0 = ⎜ v ⎝ M + m sin 2 θ ⎟⎠ 0

Now, ux = 2 cos ( 30° ) = 3 ms −1 and uy = 2 sin ( 30° ) = 1 ms −2

Method 2: By using the impulse method The impulse acting on the ball and wedge is shown in Figure.

ax = g sin ( 30° ) = 4.9 ms −2 and ay = − g cos 30° = −4.9 3 ms −2 Speed after bouncing is v = vx2 + ( evy )

Since impulse equals the change in momentum of the body, so for the wedge, we have J sin θ = MV …(1) for the ball, we have − J = mvn − mv0 …(2) Also, v2 − v1 = e ( u1 − u2 ) ⇒

V sin θ − vn = e ( v0 − 0 )

9.

(

2

(

3 + 4.9 × 1 ) + 0.6 ( 1 − 4.9 3 × 1 )

⇒

v=

⇒

v ≅ 8 ms −1

2

)2

If after collision, the particle moves along the track, then this means that there is no normal component of velocity and hence e = 0. Just before collision, the components of velocity along the normal and along the tangent are un = u cos θ and ut = u sin θ

…(3)

Solving equations (1), (2) and (3), we get

When the velocity of particle becomes horizontal just after collision with the track, then we have

⎡ ( 1 + e ) m sin θ ⎤ V=⎢ ⎥ v0 and ⎣ M + m sin 2 θ ⎦ ⎛ m sin 2 θ − Me ⎞ vn = ⎜ v ⎝ M + m sin 2 θ ⎟⎠ 0 7.

Vertical component of velocity and hence the time of flight becomes e times. Since, Range = uH T where, uH is the horizontal component of velocity which remains unchanged due to collision. Hence the new range will become e times i.e., L2 = eL1

8.

t=

2h = g

⎛ R ⎞ 4+⎜ ⎝ 3 ⎟⎠ g

Since, R = vAt ⇒

⎛ R ⎞ 4+⎜ ⎝ 3 ⎟⎠ R=2 g

Mechanics II_Chapter 2_Hints and Explanation_1.indd 90

The component of velocity along the common tangent direction will remain unchanged because in this direction there is no impulse. Hence vt = ut = u sin θ

…(1)

Also, speed of the particle just after impact along the n-line is e times the speed of particle before collision along the n-line. vn = eun = eu cos θ

…(2)

Since after collision, the particle moves in horizontal direction, so vertical component of particle velocity should be zero. For this condition to be obeyed, we have

2/9/2021 6:32:54 PM

Hints and Explanations eun cos θ = vt sin θ

…(3)

Solving equations (1), (2) and (3), we get ( eu cos θ ) cos θ = ( u sin θ ) sin θ

Single Correct Choice Type Questions 1.

⎛π⎞ 1 e = tan 2 θ = tan 2 ⎜ ⎟ = ⎝ 6⎠ 3

and uy = 20 2 sin 45° = 20 ms −1

10. (a) ur = 0 , ar = g

After 1 s, horizontal component remains unchanged while the vertical component becomes vy = uy − gt

⇒ vr = 2 gh1 After collision relative velocity vr′ = e 2 gh1 and relative retardation is still g (downwards). So,

⇒

( v′ ) h2 = r = e 2 h1 2g 2

(b)

g 5g = 4 4 So, just before collision, we have

H = h1 + h2

⎛ 5g ⎞ vr = 2 ⎜ h ⎝ 4 ⎟⎠ 1

Where, h1 be the height attained in 1 s.

Just after collision, vr′ = evr

1 2 h1 = ( 20 ) ( 1 ) − ( 10 ) ( 1 ) = 15 m 2 and h2 be the height attained after 1 s. ⇒

5g Relative retardation is still 4 2 vr′ ) ( 2 Hence, h2 = = e h1 ⎛ 5g ⎞ 2⎜ ⎝ 4 ⎟⎠ 11. The horizontal component of the velocity of ball during the path OAB is ucos α while in its return journey BCO it is eucos α . The time of flight T also remains unchanged. Hence,

⇒

2.

B

a

2u sin α a a = − eu cos α g u cos α

⇒

2u2 sin α cos α − ag a = gu cos α eu cos α

⇒

e=

⇒

e=

ag 2u sin α cos α − ag

Mechanics II_Chapter 2_Hints and Explanation_1.indd 91

( 20 )2 2 × 10

= 20 m

Let speed of block be V. Then by Law of Conservation of Linear Momentum (applied along the horizontal direction), velocity of cylinder will be v in opposite direction, then mv − MV = 0

10 ms −1 3 Hence, the correct answer is (A). ⇒

2

1 ⎛ u2 sin ( 2α ) ⎞ − 1⎟ ⎜⎝ ag ⎠

2g

=

Mv = MV 2 ⇒ v = 2V By Law of Conservation of Mechanical Energy, we have 1 1 2 mgh = MV 2 + m ( 2V ) 2 2 where, h = R − r = 1 m Substituting the values, we get ( 1 ) ( 10 ) ( 1 ) = 1 ( 2 ) ( V 2 ) + 1 ( 1 ) ( 4V 2 ) 2 2 ⇒ 3v 2 = 10

O

⇒

v′y2

⇒

A

2u sin α a a = + g u cos α eu cos α

h2 =

⇒ H = 20 + 15 = 35 m Hence, the correct answer is (B).

T = tOAB + tBCO

⇒

vy = 20 − ( 10 ) ( 1 ) = 10 ms −1

Due to explosion one part comes to rest. Hence, from Conservation of Linear Momentum, vertical component of second part will become v′y = 20 ms −1. Therefore, maximum height attained by the second part will be

ur = 0 , ar = g +

C

Horizontal and vertical components of initial velocity are ux = 20 2 cos 45° = 20 ms −1

CHAPTER 2

⇒

H.91

3.

v=

Since centre of mass of 1 g, 2 g and 3 g is at ( 2, 2, 2 ) ⇒

a mass of ( 1 + 2 + 3 ) g = 6 g is placed at ( 2, 2, 2 ).

2/9/2021 6:33:07 PM

H.92 JEE Advanced Physics: Mechanics – II  Let the new mass of 4 g be placed at r, such that the new centre of mass is new at origin ( 0, 0, 0 ). Hence  6 ( 2, 2, 2 ) + 4 r ( 0, 0, 0 ) = 10  ⇒ r = ( −3 , − 3 , − 3 ) Hence, the correct answer is (C). 4.

7.

where, v0 = 2 gh and for mass 2m, we have J = 2mv From equations (1) and (2), we get

Let the displacement of wedge be x (leftwards). Horizontal displacement of A and B with respect to wedge is 10 cos 45° or 5 2 cm (rightwards) or the horizontal displacement of A and B with respect to ground is ( 5 2 − x ) cm rightwards. The centre of mass of the whole system will not move in horizontal direction. So,

( 2m ) x = ( 5 2 − x ) ( m + 2m )

⇒

3 Hence, the correct answer is (D). 8.

5mx = 15 2 m

m = m0 e − λ t ⇒

⎛ dm ⎞ − λt ⎜⎝ − ⎟ = m0 λ e dt ⎠

⎛ dm ⎞ Now thrust force F = u ⎜ − = um0 λ e − λ t ⎝ dt ⎟⎠ ⇒ ⇒ ⇒ ⇒

⎛ dv ⎞ m⎜ = um0 λ e − λ t ⎝ dt ⎟⎠ dv m0 e − λ t = um0 λ e − λ t dt dv = uλ dt

(

)

v

t

0

0

∫ dv = uλ ∫ dt

⇒ v = uλt Hence, the correct answer is (B). 6.

By Work-Energy Theorem 1 ( 2 m v − u2 ) = Fs …(1) 2 For the second case 1 ( 2 m v′ − u′ 2 ) = Fs …(2) 2 where F is the retarding force offered by the plank to the bullet. Equating (1) and (2), we get

9.

Before explosion, particle was moving along x-axis, i.e., it has no y-component of velocity. Therefore, the centre of mass will not move in y-direction ⇒ ycm = 0

( 150 )2 − ( 125 )2 = v′ 2 − ( 90 )2

⇒

v′ 2 = ( 35 )

2

⇒ v′ = 35 ms −1 Hence, the correct answer is (B).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 92

m1y1 + m2 y 2 m1 + m2

⇒

ycm =

⇒

⎛ 3m ⎞ ⎛ m⎞( ⎟(y) ⎜⎝ ⎟⎠ +15 ) + ⎜⎝ 4 ⎠ 0= 4 ⎛ m 3m ⎞ ⎜⎝ + ⎟ 4 4 ⎠

⇒ y = −5 cm Hence, the correct answer is (A).  Since Fext ≠ 0   acm = ao, vcm = vo at any time t.  F  Since acm = ext so a0 ≠ 0 M v0 = 0 and a0 ≠ 0 Hence, the correct answer is (B).

10. Speed of ball before collision is ⎛ v = u cos 60° = ( 20 ) ⎜ ⎝

1⎞ −1 ⎟ = 10 ms 2⎠

Since, collision is perfectly inelastic ( e = 0 ), the ball will not bounce. It will move along the plane with velocity v′ = v cos 30° =

10 3 = 5 3 ms −1 2 v′

u = 20 ms–1

v 2 − u 2 = v ′ 2 − u′ 2 ⇒

…(2)

2 gh

v=

⇒ x = 3 2 cm Hence, the correct answer is (B). 5.

Let J be the impulse and v the common speed. Since, Impulse = Change in Linear Momentum So, for mass m, we have J = mv0 − mv …(1)

60°

30° v = u cos θ 30°

Maximum height attained by the ball is H=

u2 sin 2 60° v′ 2 + 2g 2g

2/9/2021 6:33:18 PM

Hints and Explanations 2

The position vectors of points G1 and G2 are h  r1 = Ri + j and 4 3R  3R   − j = Ri − r2 = Ri + j 8 8 According to the problem the centre of mass lies at C  and hence rcm = Ri   M r 1 + M2 r 2  Since rcm = 1 M1 + M2

( 5 3 )2

2 + 20 20 300 75 375 ⇒ H= + = = 18.75 m 20 20 20 Hence, the correct answer is (B). ⇒

H=

( )

11. Retardation due to friction a = μ g = ( 0.25 )( 10 ) = 2.5 ms −2 Since collision is elastic, i.e., after collision first block comes to rest and the second block acquires the velocity of first block. Distance travelled by it will be 2

(5) v2 = =5m 2 a ( 2 ) ( 2.5 ) So, final separation x is x= 5−2= 3 m Hence, the correct answer is (C). s=

12. In perfectly inelastic collision between two particles, linear momentum is conserved. If θ be the angle between the velocities of the two particles before collision, then p 2 = p12 + p22 + 2 p1p2 cos θ ⇒

h = 3 R Hence, the correct answer is (A). Substituting values to get

2

v⎞ ⎛ 2 2 ⎜⎝ 2m ⎟⎠ = ( mv ) + ( mv ) + 2 ( mv )( mv ) cos θ 2

1 = 1 + 1 + 2 cos θ 1 ⇒ cos θ = − 2 ⇒ θ = 120° Hence, the correct answer is (D). ⇒

13. In one dimensional collision of two particles, the velocities are interchanged when collision is elastic and masses are equal. Hence, the correct answer is (A). 1 14. M1 = Mass of cone = π R2 hρ 3 2 M2 = Mass of hemisphere = π R3 ρ 3 h 3h Since CG1 = therefore AG1 = 4 4 3R Also CG2 = 8

15.

⎛ KE of different ⎞ ⎛ Kinetic energy ⎞ ⎛ KE of ⎞ ⎜ particles in the ⎟ ⎜ of a system ⎟ = ⎜ centre of ⎟ + ⎜ ⎟ ⎜⎝ of particles ⎟⎠ ⎜⎝ mass ⎟⎠ ⎜ frame of reference ⎟ ⎝ of centre of mass ⎠

CHAPTER 2

⎛ ⎞ ( 20 )2 ⎜ 3 ⎟ ⎝ ⎠

H.93

1 mv 2 2 1 So, KE of the system of particles ≥ mv 2 2 Hence, the correct answer is (C). Here, KE of centre of mass is

16. This problem can be done by using the concept of “Negative Mass being added to a system”.

rcm =

( 784 k ) ( 0 ) + ( −441 k ) 7 784 k + ( −441 k )

⇒ rcm = 9 cm Hence, the correct answer is (B). ⎛ m − m2 ⎞ 17. Fraction of energy retained = ⎜ 1 ⎝ m1 + m2 ⎟⎠ ⇒

− mn ⎞ ⎛m Fraction = ⎜ nucleus ⎝ mnucleus + mn ⎟⎠ 2

2

⎛ Amn − mn ⎞ ⎛ A − 1⎞ Fraction = ⎜ =⎜ ⎟ ⎝ A + 1 ⎟⎠ ⎝ Amn + mn ⎠ Hence, the correct answer is (B). ⇒

2

2

18. Let v be the velocity of ball w.r.t. wedge and V be the velocity of wedge.

Mechanics II_Chapter 2_Hints and Explanation_1.indd 93

2/9/2021 6:33:26 PM

H.94 JEE Advanced Physics: Mechanics – II Applying conservation of linear momentum, we get mV = m ( v cos 45° − V ) v ⇒ V= 2 2 ⇒ v = 2 2V Applying conservation of energy, we get 1 ⎡ 2 2 m ⎣ ( v cos 45° − V ) + ( v sin 45° ) + V 2 ⎤⎦ = mgh 2 2 2 ⎛ v ⎞ ⎛ v ⎞ ⇒ ⎜ −V⎟ + ⎜ + V 2 = 2 gh ⎟ ⎝ 2 ⎠ ⎝ 2⎠ 2

2

⇒

⎛ 2 2V ⎞ ⎛ 2 2V ⎞ ⎛ g ⎞ 2 −V⎟ + ⎜ ⎜⎝ ⎟ + V = 2 g ⎜⎝ ⎟ ⎠ ⎝ 2 2 ⎠ 2⎠

⇒

V 2 + 4V 2 + V 2 = gR 2

⇒

6V 2 = gR 2

⇒

V=

v1′ − v2′ v1 − v2 = u2 − u1 u − 3u

⇒

2eu = v2 − v1

Hence, the correct answer is (C).   m v + m2v2  22. vcm = 1 1 m1 + m2

…(1)

along y-axis mv = 2mV sin θ v ⇒ V sin θ = …(2) 2 Squaring (1) and (2) and adding, we get v V= 2 Hence, the correct answer is (C).

v M

P

B

A

J J

B

Before collision

A

P2

P2 = P − J

Coefficient of restitution e =

B

P1

After collision

Mechanics II_Chapter 2_Hints and Explanation_1.indd 94

P1 − P2 P

Man (m) Ladder (M – m)

Since, Impulse = Change in Linear Momentum, so we get For B: J = P1 …(1) For A: J = P − P2 ⇒

vr – v

+ve

20. Let P1 and P2 be the momenta of A and B after collision. A

)

23. The rope tension is the same both on the left and right hand side at every instant, and consequently momentum of both sides are equal

along x-axis mv = 2mV cos θ v 2

)

(

V=

V cos θ =

e=

(

gR 2 6

19. By Law of Conservation of Momentum

⇒

21.

e=

Since m1 = m2 = m {say}   v1 + v2  ⇒ vcm = { m1 = m2 } 2  ⇒ vcm = iˆ + ˆj ms −1   a +a 3  Similarly, acm = 1 2 = iˆ + ˆj ms −2 2 2   Since, vcm is parallel to acm, so the path will be a straight line. Hence, the correct answer is (C).

gR 3 2 Hence, the correct answer is (A). ⇒

P1 − P + J P J − P + J 2J ⇒ e= = −1 P P Hence, the correct answer is (B). ⇒

…(2)

v

Mv = ( M − m ) ( − v ) + m ( vr − v ) m vr ⇒ v= 2M Momentum of the centre of mass is ⇒

P = P1 + P2 ⇒

2Mvcom = Mv + Mv m ⇒ vcom = v = vr 2M Hence, the correct answer is (B).

2/9/2021 6:33:36 PM

Hints and Explanations  m ( 0 ) + ma  acm = m+m  a  acm = 2

⎛ Loss in Elastic ⎞ ⎛ Gain in Kinetic ⎞ ⎜⎝ Potential Energy ⎟⎠ = ⎜⎝ Energy of m ⎟⎠ 2

a

m1 = m

m2 = m

⇒ v = 1 ms −1 If x be the maximum extension in the spring, then by Law of Conservation of Mechanical Energy, we get 1 1 1 1 ( 3 ) ( 1 )2 + ( 6 ) ( 2 )2 = ( 200 ) x 2 + ( 9 ) ( 1 )2 2 2 2 2 ⇒ 3 + 24 = 200 x 2 + 9

30.

27. For completely inelastic collision, the ball will stick to the pendulum ⇒ mv = ( m + m )V v ⇒ V= 2 V 2 v2 = 2g 8g Hence, the correct answer is (D). ⇒

h=

1 4 ⎞ 1 2 ( 5 ) ⎛⎜ ⎟⎠ = ( 5 ) v2 ⎝ 2 100 2

⇒

v2 = 0.04 ms −1

2

⎛ m − m2 ⎞ v1 = ⎜ 1 u1 ⎝ m1 + m2 ⎟⎠ Since, m1 =

4 3 4 π r1 and m2 = π r23 3 3

⎛ r3 − r3 ⎞ ⇒ v1 = ⎜ 13 23 ⎟ u1 ⎝ r1 + r2 ⎠

⇒

⎛ 2r 3 ⎞ v2 = ⎜ 3 1 3 ⎟ u1 ⎝ r1 + r2 ⎠

…(2)

Put r1 = 4 cm and r2 = 2 cm in (1) and (2), we get ⎛ 64 − 8 ⎞ v1 = ⎜ 81 = 63 cms −1 ⎝ 64 + 8 ⎟⎠ 2 ( 64 ) × 81 = 144 cms −1 64 + 8 Hence, the correct answer is (A). v2 =

31.

( 1 )( 12 ) + ( 2 ) ( −24 ) = 1v1 + 2v2 ⇒

v1 + 2v2 = −36

…(1)

⎛ v − v1 ⎞ Further e = − ⎜ 2 ⎝ u2 − u1 ⎟⎠ 2 ⎛ v − v1 ⎞ = −⎜ 2 ⎝ −24 − 12 ⎟⎠ 3 2 v2 − v1 ⇒ = 3 36 ⇒ v2 − v1 = 24 Add (1) and (2) ⇒

28. In an elastic oblique collision if two particles are of equal masses and the second particle is at rest, then after collisions the particles scatter at right angles. Hence, the correct answer is (A).

⇒

v2 = −4 ms −1

⇒

v1 = −28 ms −1

29. On breaking off from the wall the speed of m1 is zero and the spring is unstretched. If m2 has a speed v2 at that instant, then by Law of Conservation of Energy.

⇒ Loss = Total Initial − Total Final Since 1 1 Total Initial= ( 1 )( 144 ) + ( 2 ) ( 576 ) 2 2

Mechanics II_Chapter 2_Hints and Explanation_1.indd 95

…(1)

⎛ 2m1 ⎞ Similarly, v2 = ⎜ u1 ⎝ m1 + m2 ⎟⎠

x=

26. Since, both have equal mass and collision is elastic, so pendulum will have a velocity v after collision By Law of Conservation of Energy 1 mv 2 = mgh 2 v2 ⇒ h= 2g Hence, the correct answer is (B).

⇒

So, vCM =

18 200 ⇒ x = 0.3 m ⇒ x = 30 cm Hence, the correct answer is (D). ⇒

1 2 1 kx = m2v22 2 2

m1v1 + m2v2 0 + 5 ( 0.04 ) 0.04 = = ms −1 m1 + m2 10 + 5 3 4 ⇒ vCM = cms −1 3 Hence, the correct answer is (B).

Hence, the correct answer is (B). 25. At maximum extension, velocity of both the blocks will be same. Let v be the common velocity of the blocks (towards right). By Law of Conservation of Linear Momentum, we get 6 ( 2 ) + 3 ( −1 ) = ( 3 + 6 ) v

⇒

CHAPTER 2

24.

H.95

…(2)

3v2 = −12

2/9/2021 6:33:49 PM

H.96 JEE Advanced Physics: Mechanics – II ⇒

Ei = 72 + 576 = 648 J

⇒

1 1 Similarly, E f = ( 1 ) ( 784 ) + ( 2 ) ( 16 ) 2 2 ⇒ E f = 392 + 16 ⇒

3 mx = mL ( 1 − cos θ ) L ⇒ x = ( 1 − cos θ ) 3 Hence, the correct answer is (D). ⇒

E f = 408 J

⇒ Loss = 648 − 408 ⇒ Loss = 240 J Hence, the correct answer is (C).

2mx = m [ L ( 1 − cos θ ) − x ]

36.

32. When A and C exchange their position, then displacements of A, B, C and plank ( P ) are shown in Figure. 37.

⎛ m − m2 ⎞ v1 = ⎜ 1 u1 ⎝ m1 + m2 ⎟⎠ ⇒ v1 = 0 Hence, the correct answer is (D). ⎛ Relative ⎞ ⎛ Relative ⎞ ⎜ velocity of ⎟ = ⎜ velocity of ⎟ = v ⎜⎝ separation ⎟⎠ ⎜⎝ approach ⎟⎠ 2π r ⇒ Time of next collision = v Hence, the correct answer is (B).

{ as e = 1}

38. By Law of Conservation of Momentum 2.9 ( 0 ) + 0.1( 150 ) = ( 2.9 + 0.1 ) v ⇒

Since xcm system remains fixed, so Δxcm = 0 ⇒

mA Δx A + mB ΔxB + mC ΔxC + mP ΔxP = 0

40 ( 4 + x ) + 50 x − 60 ( 4 − x ) + 90 x = 0 240 x = 80 1 ⇒ x=+ m 3 1 m , rightwards So, B shifts 3 Hence, the correct answer is (B). ⇒ ⇒

33.

4 Mx = M ( 10 R − x ) ⇒ 4 Mx = 10 MR − Mx ⇒ x = 2R Hence, the correct answer is (B).

34. In an inelastic collision, only the momentum of system (ball and earth) may remain conserved. Some energy can be lost in the form of heat, sound etc. Hence, the correct answer is (C). 35. Let x be the displacement of bead. Displacement of particle with respect to bead is L ( 1 − cos θ ), i.e., displacement of particle with respect to ground will be L ( 1 − cos θ ) − x. Since net force in horizontal direction on the system is zero. Therefore, the centre of mass will not move in horizontal direction.

Mechanics II_Chapter 2_Hints and Explanation_1.indd 96

v=

( 0.1 )( 150 ) 3

⇒ v = 5 ms −1 By Law of Conservation of Energy ⎛ Loss in K.E. of ⎞ ⎛ Gain in P.E. of ⎞ ⎜⎝ combined system ⎟⎠ = ⎜⎝ combined system ⎟⎠ ⇒ ⇒

1 ( M + m ) v 2 = ( M + m ) gh 2 v 2 = 2 gh

But h =  ( 1 − cos θ ) ⇒

v 2 = 2 g ( 1 − cos θ )

25 = 2 ( 10 ) ( 2 ⋅ 5 ) ( 1 − cos θ ) 1 ⇒ cos θ = 2 ⇒ θ = 60° Hence, the correct answer is (C). ⇒

39. Fraction Transferred 4 m1m2 Fraction Lost = m ( 1 + m2 )2 ⇒ Fraction =

4n

( n + 1 )2

Hence, the correct answer is (D). 40. At maximum extension 1 2 kx = mgx 2 2mg ⇒ x= …(1) k Free body diagram of both the blocks at this instant is shown in Figure.

2/9/2021 6:34:01 PM

Hints and Explanations kx

A

B

mg

mg

46. When the string becomes taut both particles begin to move with velocity components v in the direction AB. Applying Law of Conservation of Linear Momentum in the direction AB, we get mu cos 30° = mv + mv

2kx − 2mg kx − mg So, acm = = 2m m ⇒ acm = g

⇒

…(2)

A A

⎛ R⎞ m ⎜ ⎟ + 2m ( x ) ⎝ 2⎠ 41. xcm = m + 2m R + 2x ⇒ R= 2 3 5 ⇒ x= R 4 Hence, the correct answer is (C). v1 = 14 ms–1 m1 = 10 kg

vcm =

vcm =

60°

⇒

acm = g sin θ

Hence, the correct answer is (D).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 97

mu M

⎛ mu ⎞ h mh d=⎜ = ⎝ M ⎟⎠ u M So, total distance of man from floor when ball reaches the floor is m⎞ ⎛ s = d + h = h⎜ 1 + ⎟ ⎝ M⎠ ⇒

{down the plane} {down the plane}

v=

is the velocity of the man in the upward direction. Since, no gravity is existing hence ball will reach the h floor in a time t = u During this time the man will move up by a distance d = vt

45. Let mA = m, then mB = 2m . If f be the friction between the two blocks, then net force on block A is

⇒

v

0 = mu + M ( − v )

44. The system is stationary in air implies equilibrium, i.e. net force is zero. If the man now climbs up to the balloon using the rope, the centre of mass of the ‘man plus balloon’ system will remain stationary Hence, the correct answer is (A).

acm

u

48. By Law of Conservation of Momentum

Hence, the correct answer is (C).

⇒

30°

30°

47. Since this situation is satisfied only for an elastic collision, so both momentum and energy are conserved. Hence, the correct answer is (D).

m1v1 + m2v2 10 × 14 + 4 × 0 = = 10 ms −1 m1 + m2 10 + 4

3 mg sin θ F +F = 1 2 = 3m m + 2m

B

Hence, the velocity of ball A just after the string becomes u 3 taut is 4 Hence, the correct answer is (A).

m1v1 + m2v2 m1 + m2

Where, f = μmg cos θ

J

2

B

m2 = 4 kg

Similarly, net force on block B is F2 = 2mg sin θ + f

usin30°

B

v2 = 0

F1 = mg sin θ − f

v J



10 × 14 + 4 × 0 10 × 14 = = 10 ms −1 ⇒ vcm = 10 + 4 14 Hence, the correct answer is (C). 43.

u 3 4 A

Hence, the correct answer is (A).

42.

v=

CHAPTER 2

kx

H.97

Hence, the correct answer is (A). 49.

⎛ m − m2 ⎞ v1 = ⎜ 1 u1 ⎝ m1 + m2 ⎟⎠ ⇒

2 m1 − m2 = 3 m1 + m2

⇒

2m1 + 2m2 = 3 m1 − 3 m2

⇒

5m2 = m1

2/9/2021 6:34:09 PM

H.98 JEE Advanced Physics: Mechanics – II 51. If the shape of the body is spherical with its centre at origin then Rcm = ( 0 , 0 )

m1 5 = m2 1 Hence, the correct answer is (C). ⇒

50. Let a be the acceleration of wedge leftwards and ar the relative acceleration of block down the plane. Then absolute acceleration of block in horizontal direction will be ( ar cos θ − a ) towards right. Net force on the system in horizontal direction is zero. Therefore, acceleration of COM in horizontal direction will be zero or acceleration of wedge towards left is equal to the acceleration of block towards right.

So, ar cos θ − a = a ⇒ 2a = ar cos θ

and Rcm Rcm

1 ri dmi M =R

∫

⇒ Rcm ≤ R Hence, the correct answer is (B). 52. Let the origin O be located just below the top position of m. Let the centre of mass of M be located at point ( x , y ) before m is released. When m comes down to the floor, let the centre of mass of M move to the left by . The centre of mass of m has travelled through ( h cot α −  ) to the right.

…(1)

Now let N be the normal reaction between the block and the wedge. Then free body diagram of wedge gives N sin θ = ma

…(2)

Free body diagram of block with respect to wedge is

Net force on block perpendicular to plane is zero. Hence, N + ma sin θ = mg cos θ

…(3)

Solving equations (1), (2) and (3), we get 2 g sin θ ar = 1 + sin 2 θ x

y

Acceleration of block vertically downwards is ay = ar sin θ 2 g sin 2 θ 1 + sin 2 θ So, acceleration of COM is ⇒

ay =

acm =

ay 2

=

g sin 2 θ

( 1 + sin 2 θ )

Hence, the correct answer is (C).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 98

Mx + 0 Mx = …(1) M+m M+m M ( x −  ) + m ( h cot α −  ) …(2) Finally, ( xcm )final = M+m Further no external force is acting on the system and the system ( M plus m ) is initially at rest, so we must conclude that the position of centre of mass remains fixed. Mx M ( x −  ) + m ( h cot α −  ) ⇒ = M+m M+m M m ( h cot α −  ) ⇒ 0=− + M+m M+m mh cot α ⇒ = M+m Hence, the correct answer is (C). Initially, ( xcm )initial =

53. When the car C accelerates from rest to a velocity v0 relative to the double-boat system, the two boats accelerate to the left, so we have vC ( to right ) + vA ( to left ) = v0 By conservation of momentum, we get mvC = 2 MvA Solving, we get mv0 2 Mv0 vA = , vC = m + 2M m + 2M After the car applies brakes to come to rest, then the tension in the string connecting A, B also becomes zero. Applying conservation of momentum to A and C, we get

2/9/2021 6:34:18 PM

Hints and Explanations mvC − MvA = ( m + M ) v′A

⇒

58.

Hence, the correct answer is (A). 2

ρ ( av ) a 3 where, ρ = 10 kgm −3, av = 20 × 10 −6 m 3 s −1 π d 2 π ( −2 )2 and a = = 10 4 4 Substituting the values, we get

( 103 ) ( 20 × 10 −6 )2

π⎞ ⎛ 55. Angle θ = ⎜ π − ⎟ ⎝ n⎠ π ⇒ π −θ = n   Impulse = change in linear momentum = p f − pi θ pf = mv

pi = mv

pi

2

+

⇒

2v

e=

2

1 ⎞ 2d ⎛ = ⎜ 1+ ⎟ ⎝ e⎠ v

2v sin ( 45° ) g

2 g

d ev

1 ⎞ 2d ⎛ = ⎜ 1+ ⎟ ⎝ e⎠ v

gd v − gd 2

Hence, the correct answer is (C).  59. F = macm  F acm = m

h

π –θ pf

⇒

2 2 Δp = ( mv ) + ( mv ) − 2 ( mv )( mv ) cos ( π − θ )

⇒

⎛θ⎞ Δp = mv 2 ( 1 + cos θ ) = 2mv cos ⎜ ⎟ ⎝ 2⎠

which is independent of h . Hence, the correct answer is (D). 60. Let v be the velocity of ball after an elastic collision with the wall, so e=1

⎛π π ⎞ ⎛ π ⎞ Δp = 2mv cos ⎜ − = 2mv sin ⎜ ⎝ 2 2n ⎟⎠ ⎝ 2n ⎟⎠ Hence, the correct answer is (D). ⇒

4 ms–1

57. Let v′ be the velocity of block. Then by Law of Conservation of Linear Momentum, we have

v 1

56. When two bodies of equal mass collide elastically head-on, then they exchange their velocity. Hence, the correct answer is (A).

⇒

d v

where T =

⇒

= 5.1 × 10 −3 N π −4 × 10 4 Hence, the correct answer is (D). F=

T=

CHAPTER 2

mMv0

( m + M )( m + 2M )

54. F = ρav 2 =

(1 + n )v − u

n Hence, the correct answer is (C).

So, velocity of A (to right) is v′A =

vr =

ms–1

Before Collision

1 ms–1

After Collision

⇒

⎛ Relative velocity ⎞ ⎛ Relative velocity ⎞ ⎜⎝ of separation ⎟⎠ = ⎜⎝ of approach ⎟⎠

mu = mv + mnv′

⇒

v−1= 4+1

⎛ u−v⎞ v′ = ⎜ ⎝ n ⎟⎠

⇒

v = 6 ms −1

Velocity of bullet relative to block is ⎛ u−v⎞ vr = v − v ′ = v − ⎜ ⎝ n ⎟⎠

Mechanics II_Chapter 2_Hints and Explanation_1.indd 99

H.99

{away from the wall}

Hence, the correct answer is (D). 61. Let vr be the speed of bullet with respect to gun and v the velocity of gun. Then from the two figures it is clear that ϕ > 45°.

2/9/2021 6:34:25 PM

H.100

JEE Advanced Physics: Mechanics – II

vr 2

vr 2

45° velocity of bullet with respect to gun

vr –v 2 velocity of bullet with respect to ground

ϕ

vr 2

Hence, the correct answer is (C).

0=

m ( 30 2 ) + 3m v 5 5

⇒

v = −10 2 ms −1

1 mv 2 = 3 mgh 2

⇒

v = 6 gh

So, hB = h + ⇒

v 2 sin 2 ( 60 ) 2g

hB = h +

9 h 13 h = 4 4

4 hA = hB 13 Hence, the correct answer is (C). ⇒

62. If mass is non-uniformly distributed, then centre of mass of ring may lie from origin to circumference. Hence, 0 ≤ b ≤ a . Hence, the correct answer is (D). 63.

⇒

65. Let σ be surface mass density. Then mdisc = ( π R2 )σ mcavity = − 2σ

C

C1

R/2

x

Hence, the correct answer is (B). 64. After collision balls exchange their velocities, so

where  is length of square. Now, R  cos ( 45° ) = 2 R ⇒ = 2 ⇒

vA2 =h 2g

mcavity = −

 R  ( πR2σ ) 0 −  R σ   2

vcos60° = v 2

rCM remainder =

v 60°

A

vB

M

B

⇒ hB h

N

But path of B will be first straight line and then parabolic as shown in figure. For N to B, Loss in KE = Gain in PE 1 1 ⇒ mvB2 + mv 2 = mgh 2 2 1 1 ⇒ mv 2 = m ( 8 gh ) − mgh 2 2

Mechanics II_Chapter 2_Hints and Explanation_1.indd 100

C2

R 2σ 2 Now assume origin to be at centre of disc, then

vA = 2 gh and vB = 2 g ( 4 h ) = 2 2 gh Height attained by A will be hA =



rCM remainder = −

 2  2  R 2σ πR σ − 2 2

R 2 ( 2 π − 1)

Negative sign indicates that CM of remained is to the left of centre of disc i.e., origin. Hence, the correct answer is (C). 66. Thrust = v

dm dt

⇒

 2    Thrust = ( 2 )  1000  5

⇒

Thrust = 8 × 10 −4 N

Hence, the correct answer is (D).

2/9/2021 6:34:31 PM

Hints and Explanations 67. By Law of Conservation of Momentum mu − mv = m ( 0 ) + mv′ v′ = u − v

L 2 L ⇒ x= 4 So, maximum offset is L  L 3L Lmax =  − x  + = 4 2  2 For two bricks this can be generalised as L 1 Lmax =  1 +  2 2 For three bricks, we have L 1 1  11L Lmax =  1 + +  = 2 2 3  12 For four bricks, we have L 1 1 1 Lmax =  1 + + +  and so on. 2 2 3 4 Hence, the correct answer is (D). ⇒

…(1)

Further

⇒

 v − v1  e = − 2   u2 − u1   v′ − 0  e = −   −v − u 

⇒

v′ = e ( v + u )

…(2)

Equate (1) and (2), we get u − v = ev + eu ⇒ (1 − e ) u = (1 + e ) v u 1+ e = v 1− e Hence, the correct answer is (A). ⇒

68. At the highest point velocity before explosion is v cos 60 along + x -axis. By Law of Conservation of Momentum m m mv′ ( mv cos 60 ) iˆ = ( 100 ˆj ) + ( −100 ˆj ) + 3 3 3 ⇒ ⇒

3v ˆ 3 ( 200 ) ˆ i= i 2 2 v′ = 300iˆ ms −1

70.

69. Let us first consider maximum offset/overhang of one brick as shown in Figure.

So, for one brick, we have L L Lmax = ( 1) = 2 2 Now let us consider two bricks and calculate maximum offset (or overhang) for combination.

By Law of Conservation of Moments, we have L  mgx = mg  − x  2 

Mechanics II_Chapter 2_Hints and Explanation_1.indd 101

p 2 = px2 + py2 ⇒

p 2 = 122 + 52

⇒

p = 13 kgms −1

Hence, the correct answer is (C).

v′ =

Hence, the correct answer is (B).

2x =

CHAPTER 2

⇒

H.101

71.

dx = 6t 2 dt Impulse ( J ) = ∆p = m ( v f − vi ) v=

⇒

J = 2 ( 6 × 1 − 0 ) = 12 Ns

Hence, the correct answer is (B). 72. When the man moves forward with a velocity v with respect to the plank, then the plank moves backwards with a velocity V with respect to ground. So, velocity of the man with respect to ground in the forward direction is ( v − V ) . Applying conservation of linear momentum, we get MV = m ( v − V ) ⇒

V=

mv  m  L  =   m + M  m + M  t 

Hence, the correct answer is (B). 73. By Law of Conservation of Mechanical Energy 1 2 1 2 kx = µvr 2 2 Here, µ = reduced mass of the blocks, so ( m )( 2m ) 2 µ= = m m + 2m 3 and vr = relative velocity of the two blocks Substituting in equation (1), we get kx 2 =

…(1)

2 mvr2 3

2/9/2021 6:34:33 PM

H.102

⇒

JEE Advanced Physics: Mechanics – II  3k  vr =  x  2m 

Hence, the correct answer is (A).   74. vb , w = v′b , w = 2 2 ms −1    So, v′b = v′b , w + vw ⇒ ⇒ ⇒

79. By Law of Conservation of Linear Momentum Hence, the correct answer is (C). 80.

2 g − 1g 10 = ms −2 2+1 3 ( 2) ( a) − (1) ( a) ⇒ acm = 2+1 a=

2  2 v′b = ( 2 ) + ( 2 2 ) + 2 ( 2 ) ( 2 2 ) cos 45°  v′b = 4 + 8 + 8 = 20  v′b = 2 5 ms −1

a

a

Hence, the correct answer is (D). 75.

M0 g = Thrust = v

dM dt

⇒

dM M0 g = dt v

⇒

⇒

dM 60000 = = 60 kgs −1 1000 dt

Since, scm

Hence, the correct answer is (B). 76.

2 kg

F − Mg = Ma ⇒

F = M ( g + a)

⇒

F = (6000 ) (10 + 20 )

⇒

F = 18 × 10 N 4

Further F=v

dM dt

dM ⇒ 18 × 10 4 = (1000 ) dt dM −1 = 180 kgs ⇒ dt Hence, the correct answer is (D). 77. Since the system is free from external force, hence acm = 0 and since initially they are at rest, so Vcm = 0

Hence, the correct answer is (A). 1+ e  78. After collision v2 =  u  2  1− e  and v1 =  u  2 

Here, u = initial speed of ball A Now, v2 = 2v1 ⇒

1 e= 3

Hence, the correct answer is (B).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 102

⇒

acm =

scm =

1 kg

a 10 = ms −2 3 9 1 = acmt 2 2

{downwards}

1 ⎛ 10 ⎞ 2 20 m = 2.22 m ⎜ ⎟ ( 2) = 2⎝ 9 ⎠ 9

Hence, the correct answer is (B). 81. At the highest point velocity of the shell is v cos θ. At the highest point, it explodes into two pieces of equal masses out of which one piece retraces the path i.e. has a velocity −v cos θ. M v cos θ

v cos θ M/2

Just Before Explosion

M/2

v′

Just After Explosion

So, by Law of Conservation of Momentum M Mv′ M ( v cos θ ) = ( −v cos θ ) + 2 2 ⇒ v′ = 3v cos θ Hence, the correct answer is (A). 82. Since Impulse = Change in linear momentum    ⇒ F∆t = m ( v f − vi ) ⇒

( 2iˆ + ˆj + 3kˆ ) ( 2 ) = (1) ( v − ( 2iˆ + ˆj ) ) f

⇒

 v f = 6iˆ + 3 ˆj + 6 kˆ

⇒

 2 2 2 v f = ( 6 ) + ( 3 ) + ( 6 ) = 9 ms −1

Hence, the correct answer is (C). 83. When the block just reaches the top of the wedge then the velocity of block with respect to wedge at the top of the wedge is zero. Let v be the horizontal velocity of

2/9/2021 6:34:36 PM

Hints and Explanations both at this instant. By Law of Conservation of Linear Momentum, we have

⇒

(2m + m) v = mu ⇒

v=

v1 =

H.103

v 2 gl = 3 3

Hence, the correct answer is (B).

u 3

By Law of Conservation of Mechanical Energy, we get 1 1 mu 2 = ( 3 m) v 2 + mgh 2 2

87.

y cm

  m  M ( ) mL =   = = y initial (m + M) m+M

Finally, when ice melts y final = zero mL 2(m + M) Hence, the correct answer is (C).

v

u

h

2m

2m

θ

⇒ ⇒ ⇒

θ

m

88. If vcm be the velocity of the centre of mass, then vcm =

 u2  u = 3   + 2 gh  9  2 2 u = 2 gh 3 2

 Gain in K.E. of  Loss in   =  K.E. of m   centre of mass    of system

Hence, the correct answer is (B). 4 m1m2

( m1 + m2 )

mu + 0 u = m + ηm 1 + η

Further by Law of Conservation of Energy

u = 3 gh

84. Fraction Lost =

CHAPTER 2

Shift = ∆y = −

2

=

8 9

Hence, the correct answer is (D). 85. The centre of mass cannot move towards left. It  will always move towards right. Since wedge has a tendency to move towards left so the only external force on the system in horizontal direction is friction which acts towards right. Hence, the correct answer is (D).

  +  Gain in    P.E. of m  

1 1 2 mu 2 = ( m + ηm ) vcm + mgh 2 2

⇒

1  u = 2 gh  1 +  η 

Hence, the correct answer is (C). 89. Let v1 and v2 be the velocities of the two masses after collision in the same direction. Then m2

m1

m1

Before Collision

86. Velocity of mass m just before string becomes taut is

v1

v2

After Collision

m1v = m1v1 + m2v2

v = 2 gh = 4 gl = 2 gl

m2

and v2 − v1 = ev

…(1) …(2)

Solving equations (1) and (2) we get, v1 =

v ( m1 − em2 ) m1 + m2

For v1 to be positive m1 > em2 ⇒

Since, impulse e quals the change in momentum, so we have for mass 2m , J = 2mv1 and for mass m , J = mv − mv1 ⇒ mv − J = mv1 ⇒ mv − 2mv1 = mv1 ⇒

3mv1 = mv

Mechanics II_Chapter 2_Hints and Explanation_1.indd 103

m1 >e m2

Hence, the correct answer is (B). 90. Since, the lower end moves towards positive x-axis the force of friction will be along negative x-direction. Therefore, centre of mass of the rod will finally fall at  x < 0. Hence, the lower end will be at x < . 2 Hence, the correct answer is (C).

2/9/2021 6:34:38 PM

H.104

JEE Advanced Physics: Mechanics – II

   F12   F21    2 + M M     Ma1 + ( 2 M ) a2   2M  = M 91. aCM = 3M 3M   F + F21  =0 ⇒ aCM = 12 3M Hence, the correct answer is (A).  92. In the figure let, v12 = velocity of ball w.r.t. wedge before collision and  v12 = velocity of ball w.r.t. wedge after collision, which must be in vertically upward direction as shown. M v v′12 12 α α v1 30° O N – v2 30°

  ′ will make equal angle In elastic collision v12 and v12 (say α ) with the normal to the plane. We can show that α = 30° ⇒ ∠MON = 30°

v2 = cot 30° = 3 v1 Hence, the correct answer is (A). 93.

∫ x dm ∫ dm rcm =

 x2  ∫0 x  m0 L  dx L

m0 x 2 dx L 0

∫

⇒ V=−

4v A−4

Negative sign indicates the recoil speed. Hence, the correct answer is (A). 96. If m be mass of each block, then acceleration of system is mg sin ( 60° ) − mg sin ( 30° ) a= 2m  3 −1 a= g  4       ma1 + ma2 a1 + a2 = Now acm = 2m 2 ⇒

 acm =

a2 + a2  3 − 1  = g 2  4 2 

Hence, the correct answer is (A). L4 = 43 L 3

97. Let h be the height of each step. Then 2 gh v 1 = = u 2 2 g ( 2h ) Hence, the correct answer is (B). e=

3L 4 Hence, the correct answer is (D).

⇒

…(2)

(0 ) A = 4v + ( A − 4 ) V

⇒

L

⇒

95.

v2 =

 3 −1   Here, a1 and a2 are   g directed at right angles to  4  each other.

Now,

rcm =

3 mv 2 ( M + m) From equations (1) and (2), we get v1 2 = v2 3 Hence, the correct answer is (B). ⇒

rcm =

98. Let Origin be placed at B and x-axis along BC. By Law of Conservation of Momentum

94. Let mass of gun to be M and that of shell to be m . The two cases are shown in figure as below

Here, v1 and v2 are the recoil speeds of the gun in two cases. Applying Law of Conservation of Linear Momentum in horizontal direction in two cases, we get CASE-1: m (v − v1 ) = Mv1 mv ⇒ v1 = M+m

CASE-2: m ( v cos 30° − v2 ) = Mv2

Mechanics II_Chapter 2_Hints and Explanation_1.indd 104

along x-axis 0 + mv cos 30 − mv cos 30 = 0 − mv cos 30 + mvx

…(1)

⇒

vx = v cos 30 along y-axis

…(1)

− mv + mv sin 30 + mv sin 30 = 0 − mv sin 30 + mvy

2/9/2021 6:34:40 PM

Hints and Explanations ⇒

vy = v sin 30

…(2)

H.105

104.

Squaring (1) and (2) and adding, we get vnet = vx2 + vy2 = v at

m

m

m

COM

x

R/2

The centre of mass will follow the same path, so R mx = m + mR 2 3 R 3 × 100 = = 150 m ⇒ x= 2 2

Hence, the particle C will have the same magnitude of velocity but directed along the original direction (direction before collision) of particle B . Hence, the correct answer is (C). 99. By Law of Conservation of Momentum 0 = 8 ( 6 ) + 4v ⇒

v = −12 ms −1

Negative sign indicates the recoil of second piece 1 2 KE = ( 4 )( 12 ) = 288 J 2 Hence, the correct answer is (D).  dp  100. F = = 2Bt dt     When a and u are at 45°, F and p will also be inclined at 45°. This will happen at t =

105.

  A F = 2B B Hence, the correct answer is (C).

101. Net external force on the two blocks (whether they move with same retardation or not) is Fext = (0.2) ( 2 + 1) (10 ) = 6 N Fext = 2 ms −2 2+1 Hence, the correct answer is (B). acm =

2

 m − m2  102. K.E. Retained =  1  E  m1 + m2 

6 (9) = (12 + 6 ) v ⇒ v = 3 ms −1 Hence, the correct answer is (A).

106.

A B

⇒

⇒

Hence, the desired distance is (100 + 150 ) m = 250 m Hence, the correct answer is (C).

ˆj component, i.e., component of velocity parallel to wall remains unchanged while iˆ component will  1 become  −  ( 2iˆ ) = −iˆ. Therefore, velocity vector of  2 the sphere after it hits the wall is −iˆ + 2 ˆj . Hence, the correct answer is (B).

107. Since ∆xcm =

Hence, the correct answer is (C).   103. Change in linear momentum ∆p = F∆t  Here, F is weight, i.e., mg  ⇒ Δp = ( mg ) ( Δt ) = (1) (10 ) (1) = 10 kgms −1 Hence, the correct answer is (C).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 105

m1∆x1 + m2 ∆x2 =0 m1 + m2

M ( 4 L − x ) = 5 Mx 2 ⇒ x = L to left 3 Hence, the correct answer is (D).  d2 R  108. M 2 = Mg dt  d2 R  ⇒ =g dt 2 Hence, the correct answer is (C). ⇒

2

⇒ K.E. Retained =  8 − 2  E   8+2 ⇒ K.E. Retained = 0.36 E

CHAPTER 2

R

109.

sin θ = ⇒

r 1 = 2r 2

θ = 30° v

θ v

2/9/2021 6:34:43 PM

H.106

JEE Advanced Physics: Mechanics – II

By Law of Conservation of Linear Momentum mu = 2mv cos ( 30° ) ⇒

v=

u 3

…(1)

⇒

1 2 1 1 m22v02 kx = m2v02 − ( m1 + m2 ) 2 2 2 2 ( m1 + m2 )

⇒

kx 2 = m2v02 −

 ( v2 )n − ( v1 )n  Now e = −    ( u2 )n − ( u1 )n 

kx 2 =

u v 2 ⇒ e= = 3 = u cos ( 30° ) u 3 3 2 Hence, the correct answer is (C). 110. Unless m1 = m3 the COM of all the four particles can never be at the centre of the square. Hence, the correct answer is (D).

⇒

2

m1m2v02 + m22v02 − m22v02  m1m2  2 =  v0 m1 + m2  m1 + m2  m1m2

x = v0

( m1 + m2 ) k

Hence, the correct answer is (D). 114. Velocity of man with respect to ground is 1 ms −1 in opposite direction. ⇒

111. Let u be the velocity of ball before collision. Speed of ball after collision will become.

vcm =

m1v1 + m2v2 m1 + m2

40 × 2 − 80 × 1 =0 40 + 80 Hence, the correct answer is (A). ⇒

2

5  u   u  v=   +  = 8u  2 2 2

vcm =

115. Impulse = Change in Linear Momentum

1 1 mu 2 − mv 2 2 2 Fraction of KE lost in collision = 1 mu 2 2

eu cosα u sin α

2

⇒ Fraction Lost = 1 −  v  = 1 − 5 = 3   8 8 u Hence, the correct answer is (C).      112. ( m1v1′ + m2v2′ ) − ( m1v1 + m2v2 ) = Δp  ⇒ Δp = change in momentum of the two particles ⇒ ⇒ ⇒

 External force on Δp = × time interval the system  Δp = ( m1 + m2 ) g ( 2t0 )  Δp = 2 ( m1 + m2 ) gt0

u cosα Before collision

⇒

u sin α

After collision

Ft = m ( eu cos α + u cos α )

mu cos α (1 + e ) t Hence, the correct answer is (D). ⇒

116.

F=

 m − m2   2m2  v1 =  1  u1 +  m + m  u2 m m + 2  2   1  1 For m1 clockwise torque about P . In critical case these two torques are equal 50 – x

x P

100g

⇒

100 g ( 50 − x ) = 5 gx

⇒

105 x = 50 × 100

⇒

x = 47.6 cm

Hence, the correct answer is (D).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 107

vr cos θ – v

vr

CHAPTER 2

⇒

2

H.107

m ( x ) + 2m ( x ′ ) m + 2m

Ft 2 x + 2 x′ = 6m 3

Ft 2 x − 4m 2 Hence, the correct answer is (B). ⇒

126.

x′ =

MV = m (0 ) + ( M − m) v MV M−m Hence, the correct answer is (C). ⇒

v=

u 2 sin 2θ ( 10 ) sin 90° = = 10 m g 10 2

127. Range of stone = 5g

Net force on system in horizontal direction is zero. Therefore, centre of mass will remain stationary in horizontal direction. Hence,

(60 + 40 ) x = (1) (10 ) Where, x is the displacement of boy + platform

2/9/2021 6:34:49 PM

H.108

JEE Advanced Physics: Mechanics – II

⇒

x=

Force of friction on the two blocks before the blocks reach a common velocity is as shown in Figure.

10 = 0.1 m = 10 cm 100

a1

Hence, the correct answer is (B).

m/2

128. Since both particles move under the influence of gravity, so relative acceleration of one as seen by other is zero. Therefore, the relative motion between the two is uniform. Relative velocity of the two is vr = (10 + 5) ms −1 = 15 ms −1 in horizontal direction. Therefore, the collision will take place after a time 60 =4s 15 Net linear momentum in horizontal direction of the two particles is zero. Therefore, after collision the combined mass will fall vertically downwards. The desired distance from A would be

f=

a1 = µg and a2 =

3 ar = µg 2 Hence, (A), (B) and (C) are correct.

3.

x = vr t = 10 × 4 = 40 m

Multiple Correct Choice Type Questions Horizontal velocity of the ball B at the time when it strikes the ground is vB, then 10 = vB × t

⇒

vB = 10 ms −1

⇒

(10 )

2

mv02 mv02 mv02 mv02 − 2mgh = − = 2 2 4 4 Hence, (A) and (D) are correct. K system =

…(1) Since collision is perfectly elastic and mass of ball A and that of B are same, therefore velocity of ball A, before collision is also 10 ms −1. So, OPTION (C) is correct. 1 1 KE = mv 2 = m  2 2 

When the string becomes taut, speed of balls is given by mv0 v0 = V= 2m 2 If CM having velocity v0 is raised up by height h , then we have 2 1 v (2m) gh = (2m) ⎛⎜ 0 ⎞⎟ ⎝ 2⎠ 2 v02 ⇒ h= 8g When CM is at highest point, then

2× 5 10 = vB = vB g 10

10 = vB ×

µ g 2

⇒

Hence, the correct answer is (A).

⇒

m a2

t=

1.

μ mg 2

f = μ mg 2

+ ( 10 )   2

4.

Since collision is elastic so, the kinetic energy will be conserved. Let v be the speed of ball after collision. Then, 1 (1) (10 )2 = 1 ( 4 )( 4 )2 + 1 (1) v 2 2 2 2

2

⇒

1 KE = m × 2 × 100 = ( m × 100 ) J 2

v = 6 ms −1

So, OPTION (B) is also correct. Hence, (B) and (C) are correct. 2.

By Law of Conservation of Linear Momentum, m m   u =  m + v 2 2 

Let J be the impulse between the ball and wedge during collision. Since Impulse = Change in Linear Momentum

⇒

⇒

Work done against friction = Ei − E f = W f ⇒

1 m 1  3 m  u  Wf =   u2 −    2 2  2  2  3 

⇒

1 2 1  W f = mu 2 =  mu 2  6 3 4 

Mechanics II_Chapter 2_Hints and Explanation_1.indd 108

J sin 30° = ( 4 )( 4 )

⇒ J = 32 Ns Hence, (A) and (D) are correct.

2

5.

Since initially the system is at rest so the centre of mass remains fixed. xcm =

mx1 + Mx2 m+ M

2/9/2021 6:34:51 PM

Hints and Explanations m∆x1 + M∆x2 =0 M+m

∆xcm =

⇒

m∆x1 + M∆x2 = 0

⇒

m ( L − D ) + MD = 0

M m Now speed of ball 1 after impact is one third its initial speed, so u v1 = ± 3 Substituting this in equation (1), we get M 1 x= = OR 2 m 2 Hence, (B) and (C) are correct.

Where, x =

mL 1 s, we have 1( −2) + 2 ( 4 ) = 2 ms −1 1+ 2 Hence, (B), (C) and (D) are correct. vcm =

Mechanics II_Chapter 2_Hints and Explanation_1.indd 112

2/9/2021 6:35:04 PM

Hints and Explanations 26. Velocity of ball just before striking the ground is After collision, its speed becomes v1 = ev0

y v

Since the ball rises to a height of 2.5 m after the first collision, so we have v 2 e 2v02 e 2 ( 200 ) h= 1 = = = 2.5 2g 2g 20 ⇒

e 2 = 0.25

⇒

e = 0.5

Impulse equals the change in momentum, so we have

θ

⇒

T ∝ uy

⇒

a=

J = Δp = m (v1 − v0 ) = mv0 (1 + e ) ⇒

⎛ 10 ⎞ ( J=⎜ 10 2 ) (1 + 0.5) ≈ 2.1 kgms −1 ⎝ 100 ⎟⎠

Hence, (A), (B) and (C) are correct. ⎡ (1 + e ) m2 ⎤ ⎛ m − em1 ⎞ 27. Since, v1 = ⎜ 2 u1 + ⎢ ⎥ u2 ⎝ m1 + m2 ⎟⎠ ⎣ m1 + m2 ⎦ ⇒

v m −m 2 v1 =  1 v + 0 = 4  m+m 

 ( 1 + e ) m1   m1 − em2  Also, v2 =   u1 +  m + m  u2 m m + 2  2   1  1 ⇒

v2 =

( 1 + 1 2 ) m = 3v 2m

4

Impulse = ∆p1 = ∆p2 ⇒

 3  3 mv Impulse = m  v  = 4 4 

T1 uy 1 = = T2 u′y e

H=

2 u 2 sin 2 θ uy = 2g 2g

⇒

H ∝ uy2

⇒

b=

⇒

2u 2 sin θ cos θ 2ux uy = g g R ∝ uy

2 1 H1 uy = 2 = 2 H 2 u′y e

R=

R1 uy 1 = = R2 u′y e Hence, (A), (B) and (D) are correct. ⇒

c=

30. Just after collision, we have 2  m − 5m  vm =  2 gl  2 gl = − 3  m + 5m  2 gl  2× m  and v5 m =   2 gl = 3  m + 5m 

Also, T − mg = Loss in kinetic energy is K i − K f = −∆K ⇒ ⇒

−∆K = −∆K =

 1  3v  2 1  v  2  1 mv 2 −  m   + m    2 2  4  2  4   3 mv 2 16

Hence, (B) and (C) are correct. 28. If A and B are free to move, no external forces are acting and hence p and E both are conserved. But when B is fixed, (with the help of an external force) then E is conserved but p is not conserved. Hence, (A) and (C) are correct. 29. As discussed and studied, the horizontal component ux remains unchanged while the vertical component uy becomes euy. Since

Mechanics II_Chapter 2_Hints and Explanation_1.indd 113

x

CHAPTER 2

v0 = 2 gh = 10 2 ms

2u sin θ 2uy = g g

T=

−1

H.113

mvm2 m  8 gl  =   l l  9 

17 mg 9 By Law of Conservation of Energy, we get ⇒

T=

hm =

vm2 4l = 2g 9

Hence, (B), (C) and (D) are correct. 31. Momentum of the system is always conserved. Potential energy of the system is maximum when both the particles move with same velocity i.e., mass becomes double while momentum is constant. 1 P2 or K ∝ K= m 2m i.e., at this moment KE becomes half the original i.e., K . The rest is in the form of elastic potential energy. 2

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H.114

JEE Advanced Physics: Mechanics – II

Therefore, minimum kinetic energy of system is K 2 and at the same moment, elastic potential energy of K . the system is maximum i.e., 2 Hence, (B) and (D) are correct. 32.

acm = acm =

m1a1 + m2 a2 m1 + m2

( m) ( g ) + ( m) ( g )

=g 2m i.e., acceleration of centre of mass of particles is g downwards

Horizontal and vertical components of velocity of centre of mass will be 10 ms −1 each. So, H=

vy2 2g

Where, vy is the vertical component of velocity of centre of mass ⇒

H=

( 10 )2 2 × 10

=5m

33. For elastic collision oblique collision in which bodies are of equal mass and target body is at rest, then θ + φ = 90°

θ = 60°

⇒

φ = 30°

3 2

Mechanics II_Chapter 2_Hints and Explanation_1.indd 114

So, OPTION (D) is correct. Hence, (A) and (D) are correct. 35. Out of two blocks, one block of mass m is also moving in vertical direction (downwards). Therefore, CM is moving vertically downwards and momentum of the system is not conserved in vertical direction because of application of force. However, since no force is acting on system in horizontal direction, so momentum of system is conserved along the horizontal direction. Hence, (C) and (D) are correct. mv + mv = v and 2m

mg + mg =g 2m Electrostatic forces between the particles are equal and opposite. On the lower mass it will be downwards and on upper mass it will be upwards. Therefore, both the particles will always lie on a vertical line and the horizontal displacement of the lower particle will be less and the upper particle will be more than the value which would had been in the absence of charges on them because time of flight of lower mass will decrease while that of upper mass will increase. Hence, (A), (B), (C) and (D) are correct. acm =

MV = m (v − V ) mv m+M So, work done by man is Wman given by ⇒

Hence, (B) and (C) are correct. 34. Let after collision, the velocity of 2m and m be respectively, then v − v1 e= 2 v ⇒ v2 − v1 = ev

t=

37. If the man walks along the rails with velocity v relative to car, then some velocity say V is also imparted to car. If M be the mass of car, then by Law of Conservation of Linear Momentum, we get

From figure, we see that

⇒

2πR { ∵ collision is elastic e = 1 } v So, OPTION (A) is correct. 2πR , Time after which next collision takes place is v which is independent of mass. ⇒

36. Since, vcm =

Therefore, maximum height of centre of mass from the ground will be ( 20 + 5) or 25 m. Hence, (B) and (C) are correct.

sin θ =

Relative velocity vr = v2 − v1 Time after which the next collision takes place is 2πR 2πR 2πR t= = = vr v2 − v1 ev

and v1

V=

Wman =

1 1 2 m ( v − V ) + mV 2 2 2

1  mM  2 1 2 Wman =   v < mv 2 m+ M  2 …(1)

Hence, OPTION (B) is correct.

2/9/2021 6:35:08 PM

Hints and Explanations

Reasoning Based Questions 1.

p2 2m During firing of bullet from gun, momentum is conserved. So, p = constant

6.

So, centre of mass may be at rest or may move with a constant velocity. Hence, the correct answer is (D). 7.

Net force acting on the wedge and block system is gravity therefore centre of mass is accelerated in downward direction. If external force acting on the system is zero, centre of mass may be in rest or moving with constant velocity. Hence, the correct answer is (C).

8.

If m be mass of man, then Mladder = M − m

E=

⇒ ⇒

E∝

1 m

Egun Ebullet

=

Δycm =

Centre of mass is the property of mass distribution and it is possible that no mass lies at the centre of mass e.g., a ring, a shell. Hence, the correct answer is (D).

3.

N

During collision force exerted by wedge on particle is perpendicular to the inclined face. So linear momentum of wedge is conserved along the face of wedge, because the impact force has got no component along the wedge surface. Hence, the correct answer is (A). 4.

9.

10. The momentum of a system of particles from any frame is given by   p = mvcm   From the centre of mass frame, vcm = 0   ⇒ p=0 Hence, the correct answer is (A).

Linked Comprehension Type Questions 1.

Since the collision is elastic velocity of the centre of mass will remain unchanged. Mv0 ⇒ vcm = M+m Hence, the correct answer is (B).

2.

Since m remains stationary wrt ground, so L t1 = v0 Hence, the correct answer is (A).

3.

Applying conservation of linear momentum, we get

Hence, the correct answer is (C). 5.

When man moves 4 m w.r.t. boat, let boat move x in opposite direction. So, net displacement of man w.r.t. ground is ( 4 − x ). So, 60 ( 40 − x ) = 60 x ⇒

240 − 120 = 0

⇒ x = 2 m is distance moved by boat. Hence, the correct answer is (D).

Mechanics II_Chapter 2_Hints and Explanation_1.indd 115

( Fext ) Δt = Impulse = Δp Internal forces are always zero, in a system because they always form an action-reaction pair, so no need to say “must be zero”. Hence, the correct answer is (C).

By Law of Conservation of Linear Momentum and Kinetic Energy, we have 2u1  2m1  v2 =  u1 = m2    m1 + m2   1 + m  1

My + ( M − m ) ( − y ) + m ( h − y )

2M mh ⇒ Δycm = (upward) 2M Hence, the correct answer is (D).

mbullet mgun

Hence, the correct answer is (A). 2.

Net external force is zero,   ⇒ acm = 0

CHAPTER 2

If the man moves normal to the rails, then car will not move. Hence, work done by him in this case will just 1 be mv 2 and hence OPTION (C) is also correct. 2 Hence, (B) and (C) are correct.

H.115

Mv0 = Mv1 + mv2 Since the collision is elastic, so we have v2 − v1 =1 v0 ⇒ v2 − v1 = v0 ⇒ v1 = v2 − v0

…(1) …(2)

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H.116

JEE Advanced Physics: Mechanics – II

⇒ Mv0 = M ( v2 − v0 ) + mv2

⇒

2 Mv0 M+m Mv0 Since vcm = , so velocity of bead in the centre of m+ M mass frame is Mv0 v = v2 − vcm = m+ M Hence, the correct answer is (B).

⇒ v2 ≈ 5 ms −1 Hence, the correct answer is (C).

⇒ v2 =

4.

v22 = 24.7

6.

Let v1 be the velocity of block 2 kg just before collision, v2 be the velocity of block 2 kg just after collision and v3 be the velocity of block M just after collision. Applying work energy theorem, according to which, change in kinetic energy equals the work done by all forces at different stages as shown in Figure.

Since, ΔK = Wfriction + Wgravity

(

)

⇒

1 M 0 − v32 = − ( 0.5 )( μ )( M ) g cos θ − Mgh2 2

⇒

− v32 = − μ g cos θ − 2 gh2

⇒

v32 = ( 0.25 )( 10 )( 0.99 ) + 2 ( 10 )( 0.025 )

⇒

v32 = 2.975

⇒ v3 ≈ 1.72 ms −1 Hence, the correct answer is (C). 7.

Coefficient of restitution e is given by e=

Relative velocity of separation Relative velocity of approach

⇒

e=

v2 + v3 5 + 1.72 6.72 = = v1 8 8

⇒

e ≈ 0.84

ΔKE = Wfriction + Wgravity

(

)

⎡1 2 ⎤ 2 ⎢⎣ 2 m v1 − ( 10 ) ⎥⎦ = −6 μmg cos θ − mgh1 Since, m = 2 kg, so we get ⇒

v12 − 100 = −2 ⎡⎣ 6 μ g cos θ + gh1 ⎤⎦ where, cos θ = 1 − sin 2 θ = 1 − ( 0.05 ) ≈ 0.99

Hence, the correct answer is (D).

2

⇒

v12 = 100 − 2 ⎡⎣ ( 6 )( 0.25 )( 10 )( 0.99 ) + ( 10 )( 0.3 ) ⎤⎦

8.

2v1 = Mv3 − 2v2

⇒ v1 ≈ 8 ms −1 Hence, the correct answer is (B). 5.

Since, ΔK = Wfriction + Wgravity ⇒

Applying conservation of linear momentum before and after collision, we get

( )

1 ⎡ 2 m ( 1 ) − v22 ⎤ = −6 μmg cos θ + mgh1 ⎦ 2 ⎣

2 ( v1 + v2 ) 2 ( 8 + 5 ) 26 = = 1.72 1.72 v3

⇒

M=

⇒

M ≈ 15.12 kg

Hence, the correct answer is (D). 9.

⎛ m − m2 ⎞ ⎛ 2m2 ⎞ Since, v1 = ⎜ 1 u1 + ⎜ u2 ⎟ ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎟⎠ ⎛ 2m1 ⎞ ⎛ m − m1 ⎞ u1 + ⎜ 2 u2 and v2 = ⎜ ⎟ ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎟⎠ where m1 = m

⇒

1 − v22 = 2 ( −6 μ g cos θ + gh1 )

⇒

1 − v22 = 2 ⎡⎣ ( −6 )( 0.25 )( 10 )( 0.99 ) + ( 10 )( 0.3 ) ⎤⎦

u1 = − 2 gH

⇒

1 − v22

u2 = 2 gH

= −23.7

Mechanics II_Chapter 2_Hints and Explanation_1.indd 116

m2 = 2m

2/9/2021 6:35:24 PM

Hints and Explanations 15.

2 ( 2m ) ⎛ m − 2m ⎞ v1 = ⎜ − 2 gH + 2 gH ⎝ 3 m ⎟⎠ 3m

(

)

m v

5 ⇒ v1 = 2 gH 3 Similarly

⇒

)

m ( 3r ) =r 3m Hence, the correct answer is (A). rcm =

2 1 2 gH + 2 gH 3 3 16.

1 ( 2 gH ) H 9 h2 = = = 2g 2g 9

17.

v2 = −

v22

Hence, the correct answer is (A). 25 ( 2 gH ) 25H v12 = 9 = 11. h1 = 2g 9 2g 12. Since the centre of mass of system remains at rest, so we have m ( 5l − x ) = Mx 5ml ⇒ x= M+m Hence, the correct answer is (D). 13. By Conservation of Linear Momentum, we get MvM − mvm = 0 By Conservation of Energy, we get 1 2 1 1 2 2 kl = mvm + MvM 2 2 2 Solving equations (1) and (2), we get km

M( M + m)

Hence, the correct answer is (C).

vm =

M m

km l= M( M + m)

kM l m( M + m )

kMl 2 1 2 mvm = 2 2( M + m ) Hence, the correct answer is (B). ⇒

2 2 2 2 2 mr + m ( 2r ) + ( 2m ) ( 2r ) + 2mr 2 5 5 18 2 48 2 mr + 6 mr 2 = mr ⇒ I cm = 5 5 Hence, the correct answer is (C). I cm =

…(1) …(2)

Combined solution to 18, 19, 20   m v + m2v2  Since vcm = 1 1 m1 + m2 So, velocity of 1 and 2 w.r.t. centre of mass frame are given by   m ( v − v2 )    v1 cm = v1 − vcm = 2 1 m1 + m2   m1 ( v2 − v1 )    v2 cm = v2 − vcm = m1 + m2  Total momentum of system ( psystem ) equals    psystem = m1v1 cm + m2v2 cm     m m ( v − v2 ) m2 m1 ( v2 − v1 )  + =0 ⇒ psystem = 1 2 1 m1 + m2 m1 + m2 Please note that the total momentum of a system in centre of mass reference frame, i.e. C-frame is always zero 1 1 m1v12 cm and K 2 cm = m2v22 cm 2 2 In C-frame, we have K1 cm =

⎛ M⎞ 14. Since vm = ⎜ ⎟ vM ⎝ m⎠ ⇒

mv v = 3m 3 Hence, the correct answer is (D). vcm =

18-20. The correct answer is 18(D), 19(C) and 20(C).

Hence, the correct answer is (C).

vM = l

K=

1 m1m22   2 1 m2 m12   v − v2 ) + ( v2 − v1 )2 2( 1 2 ( m1 + m2 ) 2 ( m1 + m2 )2

⇒

K=

KE =

Mechanics II_Chapter 2_Hints and Explanation_1.indd 117

2r 2r

1 2 gH 3 Hence, the correct answer is (B). ⇒

10.

v2 = −

r

O

⎛ 2m − m ⎞ ⎛ 2m ⎞ v2 = ⎜ 2 gH − 2 gH + ⎜ ⎝ 3 m ⎟⎠ ⎝ 3 m ⎟⎠

(

2m

r

CHAPTER 2

⇒

H.117

1 ⎛ m1m2 ⎞   2 ( v1 − v2 ) 2 ⎜⎝ m1 + m2 ⎟⎠

2/9/2021 6:35:32 PM

H.118

JEE Advanced Physics: Mechanics – II

Consider two particles of mass m1 and m2 having   velocity v1 and v2 respectively in ground frame. Let  velocity of centre of mass equals vcm in ground frame, then    v1 cm = v1 − vcm    v2 cm = v2 − vcm Kinetic energy in ground frame K= ⇒

K=

1 1 2 2 m1 ( v1 cm ) + m2 ( v2 cm ) 2 2

(

)

1   2 m1 v12 + vcm − 2v1 ⋅ vcm + 2 1   2 m2 v22 + vcm − 2v2 ⋅ vcm 2

(

⇒

K=

)

1 1 1 2 m1v12 + m2v22 + ( m1 + m2 ) vcm − 2 2 2    ( m1v1 + m2v2 ) ⋅ vcm

1 1 1 2 m1v12 + m2v22 − ( m1 + m2 ) vcm 2 2 2 1 2 ⇒ K in C-frame = K in ground frame − ( m1 + m2 ) vcm 2 Since m1 + m2 = m and vcm = v ⇒

K=

⇒

1 K in ground frame = K in cm frame + mv 2 2

Applying Law of Conservation of Linear Momentum, we get

40v + 4 × 5 = 0

10 5 = ms −1 22 11

24. If v1 and v2 are the velocities of object of mass m and block of mass 4m, just after collision then by conservation of momentum, mv = mv1 + 4 mv2 , i.e., v = v1 + 4v2 …(1) Further, as collision is elastic 1 1 1 mv 2 = mv12 + 4 mv22 , i.e., v 2 = v12 + 4v22 2 2 2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 118

s > 2d

…(3)

(Toppling will take place if line of action of weight does not pass through the base area in contact) However, the motion of B is retarded by frictional force f = μ ( 4 m + 2m ) g between table and its lower surface. So, the distance moved by B till it stops

}

25. If v = 2vo = 5 6 μ gd , the object will rebound with speed 3 v = 3 6 μ gd 5 and as time taken by it to fall down v1 =

2h 2d = {∵ as h = d } g g The horizontal distance moved by it to the left of P in this time

20 = 44v v=

2 v 5 3 Substituting in equation (1), v1 = v 5 When v2 = 0 , v1 = v2 , but it is physically unacceptable. Now, after collision the block B will start moving with velocity v2 to the right. Since, there is no friction between blocks A and B the upper block A will stay at its position and will topple if B moves a distance s such that Therefore, v2 =

t=

1 v = − ms −1 2  v = 0.5 ms −1

Applying Law of Conservation of Linear Momentum, we get ⇒

2 v 5

{

Combined solution to 21, 22, 23

⇒

v2 = 0 or v2 =

v2 ⎛ 6 μmg ⎞ 0 = v22 − 2 ⎜ s , i.e., s = 2 ⎟ ⎝ 4m ⎠ 3μ g Substituting this value of s in equation (3), we find that for toppling of A 2 2v v22 > 6 μ gd or v > 6 μ gd as v2 = 5 5 5 5 i.e., v > 6 μ gd or vmin = vo = 6 μ gd 2 2 Hence, the correct answer is (C).

21-23. The correct answer is 21(B), 22(C) and 23(D).

⇒

Solving, these two equations we get either

…(2)

x = v1t = 6 d 3 μ Hence, the correct answer is (B). mux 26. mux = + Mvx 2 mu cos θ ⇒ = Mvx 2 mu cos θ ⇒ vx = 2M Hence, the correct answer is (C). 27. Loss in kinetic energy of bullet block system is − ΔK =

⎤ ⎡ 1 ⎛ ux ⎞ 2 1 1 2 mux2 − ⎢ m ⎜ ⎟⎠ + Mvx ⎥ ⎝ 2 2 2 ⎣2 ⎦

2/9/2021 6:27:08 PM

Hints and Explanations

⇒ ⇒

dK 1 2 = μv dt 2 Hence, the correct answer is (D).

1 1 1 mux2 − mux2 − Mvx2 2 8 2 3 1 2 2 − ΔK = mux − Mvx 8 2 2 3 mu cos 2 θ 1 ⎛ m2u2 cos 2 θ ⎞ − ΔK = − M⎜ ⎟⎠ 8 2 ⎝ 4M2

⇒

− ΔK =

⇒

3 mu2 cos 2 θ m2u2 cos 2 θ − ΔK = − 8 8M

⇒

3 mMu2 cos 2 θ mu2 cos 2 θ − ΔK = − 8M 8M

34. No additional force is required. Hence, the correct answer is (C). 35.

⎛ 2m1 ⎞ v2 = ⎜ u1 ⎝ m1 + m2 ⎟⎠ For m1  m2

m ( 3 M − m ) u2 cos 2 θ 8M Hence, the correct answer is (C). ⇒

28.

− ΔK =

v1  u1 v2 ≅ 2u1 Hence, the correct answer is (C).

Δvx = v0 cos θ

36. If the two bodies are of equal mass and second body is at rest, then after impact they exchange their velocities i.e. body one comes to rest and body two moves with velocity of body 1 and hence the maximum transfer of momentum takes place when m1 = m2 and m2 is at rest ⇒ 3 − x2 + x = 1

Hence, the correct answer is (C). 29. Since collision is elastic, so for the particle Δvx = v0 cos θ Hence, the correct answer is (C). 30. During collision, r⊥ = 0 , so L = mvr⊥ = 0 Hence, the correct answer is (B). 31. By Newton’s Law F − μv = ( M + μt) v

⇒

∫ 0

⇒

dv = F − μv

t

dv dt

dt

∫ M + μt 0

Ft v= M0 + μt

Hence, the correct answer is (D). dv =0 dt So, from (1), we get F = μv

32. For constant velocity,

⇒ P = Fv = μv 2 Hence, the correct answer is (C). 33.

⎛ m − m2 ⎞ v1 = ⎜ 1 u1 ⎝ m1 + m2 ⎟⎠

1 Mv 2 2 For constant velocity, v = constant K=

…(1)

⇒

x2 − x − 2 = 0

⇒

x 2 − 2x + x − 2 = 0

⇒

x ( x − 2 ) + 1( x − 2 ) = 0

⇒

( x + 1 )( x − 2 ) = 0

CHAPTER 2

⇒

H.119

⇒ x = −1 or x = 2 Hence, the correct answer is (B, D). 1 1 m1u12 − m1v12 2 2 37. Fraction of energy lost by m1 is f = 1 m1u12 2 ⎛ m − m2 ⎞ Put v1 = ⎜ 1 u1, we get ⎝ m1 + m2 ⎟⎠ f =

4 m1m2

( m1 + m2 )2

Hence, the correct answer is (A). 38. The normal reaction by the wall provides external force to the system of blocks A and B. The block B will not move but the block A will move due to initial compression of spring.

dK v 2 dM = dt 2 dt Since M = M0 + μt ⇒

⇒

dM =μ dt

Mechanics II_Chapter 2_Hints and Explanation_2.indd 119

2/9/2021 6:27:28 PM

H.120

JEE Advanced Physics: Mechanics – II 9m 2 g 2 2k 3 mg ⇒ x= k 2 Hence, the correct answer is (C).

Acceleration of block A just after release is

⇒

k ( 3 mg k ) = m m ⇒ aA = 3 g , towards right aA =

Fsp

and acceleration of block B is zero. Hence acceleration of the system is acm

m a + mB aB m ( 3 g ) + m ( 0 ) 3 = A A = = g (m + m) mA + mB 2

41.

kx 2 =

Line of impact

60° B

Hence, the correct answer is (C).

B

30

°

39. The block B will lose contact with wall when block is moving towards right and spring attains its natural length.

60cos(30°)

Line of tangent 60 ms–1

Before

After

Let speed of block just before it strikes the second inclined plane be v then, For velocity of block A, using ΔK + ΔU = 0

⇒

1 2⎞ ⎛1 ⎞ ⎛ 2 ⎜⎝ mvA − 0 ⎟⎠ + ⎜⎝ 0 − kx0 ⎟⎠ = 0 2 2 k vA2 = x02 m ⎛ k ⎞ ⎛ 3 mg ⎞ k x0 = ⎜ ⎜ ⎟ ⎝ m ⎟⎠ ⎝ k ⎠ m

⇒

vA =

⇒

m vA = 3 g k

The velocity of block B just at the time of losing contact with wall will be zero. So, we have vcm

⎛ m⎞ m⎜ 3g ⎟ ⎝ m v + mBvB k ⎠ 3g m = A A = = mA + mB m+m 2 k

Hence, the correct answer is (C). 40. When both the blocks move with same velocity, the extension spring will be maximum. The common velocity of the blocks is equal to the velocity of CM. Applying Conservation of Mechanical Energy to the initial and the final state when extension is maximum, we get ΔK + ΔU = 0 ⇒

⎛ 1( ⎞ ⎛ 1 2 1 2⎞ 2 ⎜⎝ m + m ) vcm − 0 ⎟⎠ + ⎜⎝ kx − kx0 ⎟⎠ = 0 2 2 2

⇒

⎛ 9 g 2 m ⎞ 1 ⎛ 2 9m2 g 2 ⎞ 1 ( 2m ) ⎜ + k x − ⎟ =0 ⎝ 4 k ⎟⎠ 2 ⎜⎝ 2 k2 ⎠

⇒

9m 2 g 2 ⎛ 1 2 9m 2 g 2 ⎞ + ⎜ kx − ⎟ =0 ⎝2 4k 2k ⎠

Mechanics II_Chapter 2_Hints and Explanation_2.indd 120

1 mv 2 = mg ( 3 tan 60° ) 2 ⇒

v = 60 ms −1

Speed of block immediately after it strikes the second incline is 45 ms −1. because in perfectly inelastic collision the component of velocity along line of impact becomes zero. Hence, the correct answer is (B). 42. By Conservation of Mechanical Energy 2 1 1 mvC2 = m ( 45 ) + mg ( 3 3 tan 30° ) 2 2

⇒

vC2 = 45 + 60 = 105

⇒ vC = 105 ms −1 Hence, the correct answer is (B). 43. If collision is completely elastic, then vertical component of velocity becomes, 45 15 × 3 45 sin 30° − 15 sin 60° = − =0 2 2 15 ms–1 60° 30° 45 ms–1

Hence, the correct answer is (C). 44. Maximum potential energy is stored when the relative velocities of the balls become zero (i.e., both have the same velocities) and the potential energy is stored at the cost of kinetic energy.

2/9/2021 6:27:41 PM

Hints and Explanations Since, in both the cases, when the relative velocity becomes zero, the balls have same velocities. Hence, the correct answer is (C). 45. Since, there is no external force, center of mass will move with uniform speed all the time. Hence, the correct answer is (B).

H.121

∑m x = ∑m x R L

L L

⇒

4d = 1( 4 − d )

⇒

d = 0.8 m

47. Since collision is elastic, so angle of reflection is same as that of incidence angle. In frame of big sphere, small sphere will rebound at an angle 2α with vertical as shown in Figure.

i.e., m is already at x = 3.2 m at this instant. Time taken by m to fall to ground from this instant will be t=

2h = g

2×2 = 10

2 s 5

CHAPTER 2

46. Since, the collision is inelastic kinetic energy will not be conserved but the velocity of center of mass will not change because there is no external force. Hence, the correct answer is (B).

The x-coordinate of m where it strikes the ground will be x = 3.2 + v1t = 3.2 + 4 2 × Hence, the correct answer is (B). 48. In frame of larger sphere, the smaller sphere is travelling with 2v0 before second collision. Since larger sphere is massive in comparison to smaller sphere, so the smaller sphere will rebound with same velocity 2v0 as collision is elastic and surface is frictionless. The small sphere will reflect at same angle. Hence, the correct answer is (C).

Hence, the correct answer is (C). 51. Let N be normal reaction between M and m. Then free body diagram of m will be as shown below. N1 30°

49. At an instant when m is just leaving M, let v1 and v2 be the velocities of m and M (both horizontal). v2

M

m

CAUSE

By Law of Conservation of Linear Momentum, we have …(1)

By Law of Conservation of Mechanical Energy, we have

( 1 ) ( 10 ) ( 4 − 2 ) = 1 ( 1 ) v12 + 1 ( 4 ) v22 2

⇒

2

v12 + 4v22 = 40

…(2)

Solving these two equations, we get v2 = 2 ms −1 and v1 = 4 2 ms −1 Hence, the correct answer is (D). 50. Let d be the displacement of M, when m just leaves M, then position of centre of mass remains fixed, so

Mechanics II_Chapter 2_Hints and Explanation_2.indd 121

60°

W = 10 N

v1

1 × v1 = 4v2

2 ≈ 6.8 m 5

∑F

H

⇒ ⇒

a = 5 3 ms–2 EFFECT

= maH

N cos 30° = ( 1 ) 5 3 cos ( 60° ) N = 5 Newton

Hence, the correct answer is (B). 52. Since, N ′ = Mg + N sin ( 30° ) ⇒ N ′ = 40 + 2.5 ⇒ N ′ = 42.5 N Hence, the correct answer is (C). 53. Relative acceleration of the bullet is zero Relative velocity of the bullet is 100 ms −1 ⇒

t=

Initial Separation 100 = =1s Relative Velocity of Approach 100

Hence, the correct answer is (A).

2/9/2021 6:27:53 PM

H.122 54.

JEE Advanced Physics: Mechanics – II

 Δp = F Δt = ( m1 + m2 ) g Δt  ⎛ 30 + 20 ⎞ ( 10 ) ( 10 −4 ) Δp = ⎜ ⎝ 1000 ⎟⎠  ⇒ Δp = 5 × 10 −5 Ns Hence, the correct answer is (B). ⇒

60. For the following collision line of impact is as shown in the Figure. V sin θ

θ

55. Since collision is perfectly inelastic, so

V cos θ

( 20 )( 90 ) − ( 30 )( 10 ) = ( 20 + 30 ) v ⇒ v = 30 ms −1 Hence, the correct answer is (D). 56. The block is not free to move along horizontal or vertical. It can move along the inclined surface when the bullet collides, jerk or impulse will be generated by the inclined surface. As the surface is smooth, only normal impulse will be generated. So, the forces along the inclined will not be impulsive and conservation of linear momentum is applicable along the inclined. Hence, the correct answer is (D).

Line of impact

So, velocity of approach along line of impact is vn = V cos θ

…(1)

Hence, the correct answer is (C). 61. Let Vcm be the velocity of cm of the disc after collision and V ′ be velocity of ball after collision, then e=

Vcm − V V cos θ

…(2)

V sin θ

57. As discussed above, the impulse by inclined plane will be along normal to surface. The impulsive force between bullet and block will be internal. So net impulse will be along normal to the incline. Hence, the correct answer is (C).

θ V cos θ

Line of impact

58. By impulse momentum theorem, I = Δp Conserving Linear Momentum along the line of impact, we get MVcosθ = MVcm + MV ′ From equation (2), we get MV cos θ = MVcm + M ( − eV cos θ + Vcm ) Initial momentum along the normal is pi = − mu sin θ

V cos θ ( 1 + e ) 2 Hence, the correct answer is (D). ⇒

Also, p f = 0 ⇒

I = Δp = mu sin θ

62. Speed of ball after collision is

Hence, the correct answer is (A).

u′ =

Matrix Match/Column Match Type Questions

mu cos θ m+ M

Stopping distance s = ⇒ ⇒

u2 2a

2

1 ⎛ mu cos θ ⎞ s=⎜ × ⎝ m + M ⎟⎠ 2 g sin θ s=

2 2

2

cos 2 θ ( 1 − e ) + sin 2 θ 4 Hence, the correct answer is (B). V′ = V

59. Applying conservation of linear momentum along the incline, we get mu cos θ = ( m + M ) u′ ⇒

Vcm =

A → (p); B → (r); C → (s); D → (r) (A) K =

p2 2m

K′ =

2

m u cos θ 2 g sin θ ( m + M )

1.

2

Hence, the correct answer is (C).

Mechanics II_Chapter 2_Hints and Explanation_2.indd 122

( 3 p )2 = 9 ⎛ 2m

p2 ⎞ = 9K ⎝⎜ 2m ⎠⎟

% increase in K = 800% = 8 ( 100% )

2/9/2021 6:28:06 PM

Hints and Explanations So, at t = 5 s, we have

p = 2 mK p′ = 2 ( 4 K ) m = 2 2Km = 2 p

aCM =

% increase in p = 100% = 1( 100% ) p2 2m For small percentage changes, % increase in K = 2 (% increase in p) = 2% = 4 ( 0.5% )

vCM =

(C) K =

4.

%

increase

in

= 0.5% = 1( 0.5% ) 2.

(%

increase

in

K)

A → (p); B → (p); C → (q); D → (r) 1 mv 2 = μmgx 2 ⇒

v = 2 Mgx = 2 × 0.5 × 10 × 2.5 = 5 ms −1 ⇒

1 mv 2 = mgR ( 1 − cos θ ) 2

⇒

v = 5 ms −1 2

⇒ 3.

v = 25 − 2 × μ g × 1.6 = 3 ms

0

  ⎛8 ⎞ and r2 = v2 dt = ⎜ ˆj ⎟ ⎝3 ⎠

∫ 0

The time of flight of the particle launched from the ground is 2u 2 × 20 = =4s 10 g The time taken by the second particle dropped from a height of 180 m is t1 =

2 ( 180 ) =6s 10

Taking the downward direction as positive, we have at t=0

vCM =

1 80  16 + 64 = unit vcm = 3 3  m vcm = 80 unit

2

A → (q); B → (q); C → (p); D → (s)

aCM =

( )

∫

⇒

8   ( 1 ) ( 4iˆ ) + ( 2 ) ⎛⎜ ˆj ⎞⎟ ⎝3 ⎠ m1r1 + m2 r2  = rcm = 3 m1 + m2

⇒

16 ˆ ⎞  ⎛4 rcm = ⎜ iˆ + j⎟ ⎝3 9 ⎠

⇒

16 256 20  + = rcm = 9 81 9

e=1

t2 =

)

( 1 ) ( 4iˆ ) + ( 2 ) 4 ˆj  4iˆ + 8 ˆj = Vcm = 3 3

−1

v2 − v1 u1 − u2

2h = g

= 25 ms −1

2

2

Since, e =

2m

  Now, r1 = v1dt = ( 4iˆ )

2

Since, v = u + 2 as ⇒

( m )( 0 ) + ( m )( 50 )

= 5 ms −2

A → (q); B → (s); C → (r); D → (p)    Fcm = F1 + F2    Fcm = m1a1 + m2 a2 = 2iˆ + 8 ˆj  ⇒ Fcm = 4 + 64 = 68 unit    m v + m2v2 Vcm = 1 1 m1 + m2 ⇒

Apply Law of Conservation of Energy, we get

⇒

2m

(

(D) p = 2mK 1 p= 2

( m )( 0 ) + ( m )( 10 )

CHAPTER 2

(B)

H.123

m1a1 + m2 a2 ( m )( 10 ) + ( m )( 10 ) = = 10 ms −2 m1 + m2 2m

m1v1 + m2v2 ( m ) ( −20 ) + ( m )( 0 ) = = −10 ms −1 m1 + m2 2m

At t = 5 s, the first particle will come to rest whereas the second particle will still be moving with a downward velocity v2 = gt = 50 ms −1

Mechanics II_Chapter 2_Hints and Explanation_2.indd 123

5.

A → (p, r, s); B → (p, r, s); C → (p); D → (p, q) Since, vcm = ⇒ vcm =

Total momentum Σp = Total mass Σm

2×3 2 = ms −1 = constant 3+6 3

At maximum deformation in the spring, both the blocks move together with the velocity that equals the velocity of centre of mass, so we have 2 ms −1 3 Up to the maximum extension in spring, the spring force on 6 kg is towards left. So 6 kg block will accelerate and its velocity will be maximum. v3 kg = v6 kg = vcm =

2/9/2021 6:28:25 PM

H.124 6.

JEE Advanced Physics: Mechanics – II

A → (r); B → (p); C → (q); D → (s) p1 + p2 = p

According to Impulse - Momentum Theorem, we have   J = P2 − P1

…(1)

Further, K1 + K 2 = K p12

⇒

2m

+

p22 4m

=

⇒

p 2m

m

p, K

2m

v2

…(2) p1

m

Before Collision

2m

and J = 3 × v2

p2

v2 = 1.6 ms −1

After Collision

A → (r); B → (q); C → (s); D → (p) Let the spring be compressed to state P from its natural state at Q as shown in Figure.

When released, the block B goes from P to Q, so we have kx aB = mB and aCM =

mB aB ( mB ) ( kx ) = mA + mB mA + mB

When the block goes from P to Q, we see that the compression x decreases, so aCM decreases. After Q, we observe that A leaves contact with the wall and spring comes to its natural length, hence net force on system is zero. Therefore, aCM is zero. Also, we see that from P to Q, the velocity of B increases and hence velocity of centre of mass also increases. After this the acceleration of centre of mass, i.e. aCM becomes zero and hence vCM becomes constant. 8.

A → (q); B → (r); C → (p); D → (p)

…(2)

Solving (1) and (2), we get

Solving these two equations we get, 4 p2 = p 3 p K and p1 = − , K1 = 3 9 8K and K 2 = 9 7.

…(1)

J

2 p12 + p22 = 2 p 2

⇒

− J = ( 0.5 + 1.5 ) v2 − 0.5 × 16

2

Since, a = ⇒

( m2 − m1 ) g = ⎛ 3 − 2 ⎞ g = g ⎜ ⎟ ( m1 + m2 ) ⎝ 3 + 2 ⎠ 5

a = 2 ms −2

Also, hmax =

v 2 1.6 2 = = 0.64 m 2a 2 × 2

24 = 0.64 s a A → (p, r); B → (p, r); C → (q); D → (q) The velocity of the centre of mass is Total momentum Σp vcm = = Total mass Σm ( 1 ) ( 10 ) + ( 2 ) ( −5 ) = 0 = constant ⇒ vcm = 3 ⇒ pcm = Mvcm = 0 = constant

and t = 9.

Finally, both blocks the will stop. 10. A → (q); B → (p, r); C → (p, r); D → (r, s)  Mdvcm  Since, F = = Macm dt   dvcm  = Macm = 0 ⇒ M dt   ⇒ acm = 0 So, (A) → (q)   dv  But acm = cm = 0 dt  ⇒ vcm = constant and this constant value may be zero also. So, (B) → (p, r) and similarly (C) → (p, r) No comments can be given about velocities of individual particles. So, (D) → (r, s) 11. A → (s); B → (p); C → (p); D → (r) (a) When A moves x towards right, then B and C will move towards left as shown in Figure.

Mechanics II_Chapter 2_Hints and Explanation_2.indd 124

2/9/2021 6:28:42 PM

Hints and Explanations

x 3 (b) When B moves x towards left, then A and C will move towards right as shown in Figure ⇒

x1 =

For perfectly inelastic collision, e = 0, so vA = 1 3v For inelastic collision having e = , vA = 2 4 5v 1 For inelastic collision having e = , vA = 8 4 14. A → (q); B → (q); C → (q); D → (p) vcm =

Under these conditions the centre of mass of the system will remain fixed, so we have 60 x = 60 x2 ⇒ (c)

x2 = x

When A moves x towards right and B moves x towards left, then C will move towards right as shown in Figure.

⇒

x3 = x

(d) When A and B both move x towards right, then C will move towards left as shown in Figure.

( 2 )( 2 ) + ( 2 ) ( 0 ) = ( 2 + 2 ) vcombined ⇒ vcombined = 1 ms −1 So, (B) → (q) Now, U max = K i − K f 1 1 2 2 × 2 × ( 2 ) − × 4 × (1) = 2 J 2 2 So, (D) → (p) 1 2 =2J Since, Kxmax 2 ⇒ xmax = 1 m So, (C) → (q)

T = m2 g = 200 N For equilibrium of ( M + m1 ) , we have 3T = Wman + m1 g

x4 = 3x

12. A → (s); B → (s); C → (p, q, r); D → (p, q, r) Conceptual 13. A → (r); B → (s); C → (p); D → (q) By conservation of linear momentum, we have mv = mvA + mvB ⇒ vA + vB = v Also, applying the definition of e, we get ⎛ v − vB ⎞ e = −⎜ A ⎝ 0 − v ⎟⎠ ⇒ vA − vB = ev From equations (1) and (2), we get ⎛ 1− e⎞ ⎛ 1+ e ⎞ vB = ⎜ v and vA = ⎜ v ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ For elastic collision, e = 1, so vA = v

Mechanics II_Chapter 2_Hints and Explanation_2.indd 125

U max =

15. A → (r); B → (r, s); C → (p, q); D → (p) (a) Let T be the tension in string connecting m1 to m2, then tension in the string held by the man is 2T. For equilibrium of m2, we have

Under these conditions the centre of mass of the system will remain fixed, so we have 90 x = 30 x 4 ⇒

m1v1 + m2v2 = 1 ms −1 m1 + m2

So, (A) → (q) At maximum compression by Law of Conservation of Linear Momentum, we get

⇒

Under these conditions the centre of mass of the system will remain fixed, so we have 30 x + 30 x3 = 60 x

v 2

CHAPTER 2

Under these conditions the centre of mass of the system will remain fixed, so we have 30 x = 90 x1

H.125

⇒

600 = Wman + 100

⇒

Wman = 500 N

(b) The net upward force acting on the system is 4T and the net downward force acting on the system is the weight of all. For the centre of mass to accelerate upwards, we must have 4T > 200 + 100 + 500

…(1) (c)

⇒ T > 200 N For the centre of mass to accelerate downwards, we must have

…(2)

4T < 200 + 100 + 500 ⇒

T < 200 N

(d) When in equilibrium, the forces acting on the man are shown in Figure.

2/9/2021 6:29:03 PM

H.126

JEE Advanced Physics: Mechanics – II v m m

v 2

v Just Before Impact

Since, ΣF = 0 ⇒ N + 400 = 500 ⇒ N = 100 N

17. A → (q, r); B → (q, r); C → (p, s); D → (p) In the absence of external force, the state of centre of mass will remain unchanged and the linear momentum of system remains conserved e.g. a man walking on a platform which lies on a frictionless horizontal surface. Centre of mass of a body can lie inside or outside it. Centre of mass of a solid cube will lie inside the cube.

2.

⇒

ϕ = 2θ = 120°

⇒

ϕ = 6° 20

⎛ m − mB ⎞ ⎛ 3 − 2⎞( ) v′A = ⎜ A vA = ⎜ 2 = 0.4 ms −1 ⎝ 3 + 2 ⎟⎠ ⎝ mA + mB ⎟⎠ ⎛ 2mA ⎞ ⎛ 6 ⎞( ) vB′ = ⎜ vB = ⎜ 2 = 2.4 ms −1 ⎝ 3 + 2 ⎟⎠ ⎝ mA + mB ⎟⎠ Retardation of both the blocks will be, a = μ K g = 0.4 × 10 = 4 ms −2

18. A → (r); B → (r); C → (q); D → (p)

sA =

For (A), p = 2mK = 2 × 0.5 × 4 = 2 kgms −1 For (C), i.e. at maximum compression the blocks move with a common velocity and will have momentum equal to half the momentum of centre of mass of the particles. p ⇒ pA = pB = CM = 1 kgms −1 2

v′A2 v′ 2 and sB = B 2a 2a

Therefore, the desired distance is,

For (B), pCM = pA + pB = 0 + 2 = 2 kgms −1

d = sB − sA = d= 3.

For (D), since we know that for an elastic collision between two equal masses, the velocities are interchanged. Hence, B comes to rest and A starts moving.

2 2 vB′ 2 − v′A2 ( 2.4 ) − ( 0.4 ) = 2a 8

5.6 = 0.7 m = 70 cm 8

For stable circular motion of centre of mass of rope, we use ⎛ l⎞ T = m⎜ ⎟ ω2 ⎝ 2⎠ For the rope not to break, we must have T ≤ 40 N

19. A → (q); B → (s); C → (p); D → (r) Using xcm

m x + .... + mn xn = 1 1 m1 + ... + mn

⇒

⎛ l⎞ m ⎜ ⎟ ω 2 ≤ 40 ⎝ 2⎠

and ycm =

m1y1 + .... + mn yn m1 + ..... + mn

⇒

2 ( 0.4 ) ⎛⎜ ⎞⎟ ω 2 ≤ 40

⇒

ω 2 ≤ 100

⇒

ω ≤ 10 rads −1

⇒

ω max = 10 rads −1

Integer/Numerical Answer Type Questions mv cos θ + mv cos θ = ( m + m ) 2mv cos θ = mv ⇒

cos θ =

⇒

θ = 60°

1 2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 126

Just After Impact

So, angle between the two before collision is

16. A → (q); B → (p); C → (r); D → (r) Apply conservation of linear momentum in horizontal direction and conservation of mechanical energy.

1.

θ θ

v 2

4.

⎝ 2⎠

Along common tangent direction, the speed of the particle remains unchanged. So, we have

2/9/2021 6:29:16 PM

Hints and Explanations

H.127

v0 cos α = v sin α ⇒ v = v0 cot α …(1) Also, we can conserve linear momentum of the ‘particle + wedge’ system along horizontal direction. So, mv0 + 0 = 2mV v0 2 Coefficient of restitution ⇒

V=

e=

6.

…(2)

( Relative velocity of separation )n-line ( Relative velocity of approach )n-line

V sin α + v cos α V v = + cot α v0 sin α v0 v0 Using (1) and (2), we get e=

⇒

⇒

10 gl

)

2

− 2 gl = 8 gl



B

A

v2



B

A

v2

 v2

m 8 g  + m ( 0 ) = ( m + m ) v2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 127

⇒

x=4

Since no external force acts in z direction, hence z coordinate of the centre of mass of the ball should be zero. To make z coordinate zero other ball should fall symmetrically with respect to z axis. Hence, z coordinate of other ball is z2 = −5 m. The balls do not have any external force in x direction. Hence in x direction the centre of mass should move with constant velocity. The x coordinate of centre of mass at t = 2.0 s is

B

v2

m1x1 + m2 x2 m1 + m2

20 × 250 + 20 x2 20 + 20

⇒

400 =

⇒

x2 = 800 − 250 = 550 m

7.

m1y1 + m2 y 2 m1 + m2

⇒

20 × 0 + 20 × y 2 20 + 20 y 2 = 20 m

⇒

x2 + y 2 + z2 = −5 + 20 + 20 = 35

⇒

Let v2 be the common velocities of both A and B, just after string becomes taut. Then from Law of Conservation of Linear Momentum, we have v1

v = 2 g = 4 g

Since, ycm =

v1 = 8 gl

A

⇒

Height fallen by centre of mass at t = 2 s is 1 2 h = × 10 × ( 2 ) = 20 m 2 Hence, y-coordinate of centre of mass is ycm = 30 − 20 = 10 m

String becomes taut when A moves upwards by a distance  under influence of gravity. Let v1 be the velocity of A at this moment, then

(

v 2 = 4 g

Since, xcm =

1 + cot 2 α 2 Given that tan α = 2 1 ⇒ cot α = 2 1 1 3 ⇒ e = + = = 0.75 2 4 4

v12 =

⇒

xcm = 200 × 2 = 400 m

e=

5.

v2 =

CHAPTER 2

8 gl v1 = 2 2 Both particles return to their original height with the same speed v2 and the string becomes loose as soon as B strikes the ground and the speed v with which A strikes the ground is, 8 g v 2 = v22 + 2 g = + 2 g 4 ⇒

10 =

Assuming wedge, block 1 and block 2 as a system. Since no external force acts on the system in horizontal direction, so both the linear momentum and mechanical energy will be conserved for the system. Final step: Let velocity of block 1 in vertical direction be v. Hence velocity of block 2 with respect to wedge will be v towards left. Let velocity of wedge towards right be V.

2/9/2021 6:29:35 PM

H.128

JEE Advanced Physics: Mechanics – II mv − J = 0

Hence velocity of block 2 towards left (with respect to ground) will be ( v − V ). Applying conservation of linear momentum for the system in horizontal direction, we get

( M + m )V − m ( v − V ) = 0 Since M = 3 m ⇒

v = 5V

…(1)

Applying conservation of mechanical energy for the system, we get ΔK + ΔU = 0 ⇒

8.

( M + m) 2

V2 +

m m ( v − V )2 + v 2 − 0 + ( − mgh ) = 0 2 2

⇒

1 1 1 ( 3 m + m )V 2 + m ( 5V − V )2 + m ( 5V )2 = mgh 2 2 2

⇒

V=

⇒

α = 45

(a)

ycm =

24 = ⇒

9.

⇒

J = 1 2 × 10 × 2 = 2 10 kgms −1 = 40 kgms −1

⇒

α = 40 vm = 4 ga and v2 m = 0

Also, we know that ⎛ m − em2 ⎞ v1 = ⎜ 1 u1 ⎝ m1 + m2 ⎟⎠

m1y1 + m2 y 2 m1 + m2 600 ⎞ ⎟ ( 80 ) 48 1000 ⎠ = m1 + 0.6 m1 + 0.6

( m1 ) ( 0 ) + ⎛⎜⎝

m1 = 1.4 kg

 acm at t = 3 s is ( 36 ms −2 ) ˆj   iˆ Fcm = ( m1 + m2 ) acm = ( 2 ) ( 36 ) ˆj = ( 72 N ) ˆj ⇒

The system will be in equilibrium about the centre of mass. Considering origin at the left end of the rod we get 25 × 0 + 8.5 × 30 + 10 × 50 + 5 × 80 + 5 × 90 53.5 1605 xc = = 30 cm 53.5 xc = ⇒

J = mv = m 2 gh

11. At the bottommost point, just before collision,

2 gh 45

So, total mass = ( 1.4 + 0.6 ) kg = 2 kg  (b) iˆ vcm = ( 6 ms −3 ) t 2 ˆj   dv ⇒iˆ acm = cm = ( 12 ms −3 ) tjˆ dt  acm t = 2 s = ( 12 )( 2 ) = 24 ms −2 (c)

⇒

10. The string will become taut when the particle will fall through a distance 2 m in downward direction. Applying impulse momentum equation, we get

Mechanics II_Chapter 2_Hints and Explanation_2.indd 128

⎡ ( 1 + e ) m1 ⎤ and v2 = ⎢ ⎥ u1 ⎣ m1 + m2 ⎦ ⇒

⇒

1 ⎛ ⎞ m − × 2m ⎜ ⎟ 2 4 ga = 0 vm ′ =⎜ ⎝ m + 2m ⎟⎠ 1⎞ ⎛⎛ ⎞ ⎜ ⎝⎜ 1 + 2 ⎠⎟ m ⎟ 1 4 ga = 4 ga v2′ m = ⎜ ⎝ m + 2m ⎟⎠ 2

After second impact, 1⎞ ⎛⎛ ⎞ 4 ga 4 ga ⎜ ⎜⎝ 1 + 2 ⎟⎠ 2m ⎟ ⎛ − 4 ga ⎞ vm =− ′′ = ⎜ ⎟=− ⎝ m + 2m ⎟⎠ ⎜⎝ 2 ⎠ 2 2 1 ⎛ ⎞ 2m − × m ⎛ 4 ga ⎞ −4 ga ⎜ ⎟ 2 v2′′m = ⎜ − ⎟⎠ = − ⎝ 2m + m ⎟⎠ ⎜⎝ 2 4

( vm′′ )2 = a

So,

hm =

⇒

x = 2 and y = 8

2g

2

and h2 m =

( v2′′m )2 = a 2g

8

12. When chain leaves the vertex, then P.E. is decreased by the same factor by which K.E. is increased hf sin ( 45° ) = 10 2 ⇒

h f = 10 m

2/9/2021 6:29:53 PM

Hints and Explanations

H.129

90° g aA = √2

⇒

U f = −1000 J

Also, U i = − mghi = −10 × 10 × 5 ⇒

U i = −500 J

Since ΔU + ΔK = 0

⇒

1 × 10 × v 2 = 0 2 v 2 = 100

⇒

v = 10 ms −1

⇒

−500 +

aA + aB = g

Also, mA = mB g ˆ ˆ 1    ⇒ acm = ( aA + aB ) = i + j , downwards 2 2 2

(

⇒

m ( 0 ) + m ( 40 ) = 20 m 2m

Initial velocity of centre of mass is ucm =

m ( 50 ) + m ( 30 ) = 40 ms −1 2m

Acceleration of centre of mass is

)

g  ⎛ g ⎞( acm = ⎜ 2 ) = = 5 ms −2 ⎝ 2 2 ⎟⎠ 2

15. When bock reaches ground and if B is displaced toward right by x, the displacement of centre of mass of system in horizontal direction must be zero hence we use 4 M ( l − x ) = 20 Mx

13. Let us consider ball A and B as a system, then the initial height of centre of mass from ground is

( ycm )i =

g √2

CHAPTER 2

Since, U f = − mgh f = −10 × 10 × 10

aB =

x=

l 12 = =2m 6 6

16. Velocity of disc A in tangential direction is u …(1) ( v1 )t = u sin 45° = 2 The disc B will move along common normal direction only.

acm = − g = −10 ms −2 2 2 Since vcm − ucm = 2 acm Δycm

where Δycm is displacement of centre of mass along y direction (taken vertically upwards). ⇒

0 2 − 40 2 = −2 × 10 × Δycm

⇒

Δycm =

⇒

( ycm ) f − ( ycm )i = 80 ( ycm ) f = 20 + 80 = 100 m

⇒

1600 = 80 m 20

Hence, maximum height reach by centre of mass from ground level will be hmax = 100 m 14. Acceleration of both the blocks will be g a1 = a2 = g sin 45° = 2 at right angles to each other. Since,   m a + mB aB  acm = A A mA + mB

Mechanics II_Chapter 2_Hints and Explanation_2.indd 129

Since both the discs move perpendicular to each other, so disc A should move along common tangent direction. So, vA =

u , along common tangent direction. 2

Applying conservation of linear momentum along common normal direction, we get

⇒

⎛ u ⎞ + 0 = m ( 0 ) + ( 2m ) ( v2 )n m⎜ ⎝ 2 ⎟⎠ u ( v2 )n = 2 2

2/9/2021 6:30:10 PM

H.130

JEE Advanced Physics: Mechanics – II

Hence, velocity of disc B along common normal is Since,

( Relative velocity of separation )n-line e= ( Relative velocity of approach )n-line ( v2 )n − 0 u 2 2 1

⇒

e=

⇒

1 =2 e

u cos 45°

=

u

2

=

u 2 2

.

2

2

x − 9x + 20 = 0

⇒

2 × 10 e=

= 3.2

3 = 0.6 5

⎛2 π R3 ⎞ 7π R 3 ρ m2 = ⎜ π R3 − ρ= ⎟ ⎝3 12 ⎠ 12 Let its centre of mass be at a height y Since common centre of mass of m1 + m2 (i.e. a hemi3R sphere) is at a height so we get 8 ⎛ π R3 ⎞ R ⎛ 7 ⎞ + ρ ⎜ π R3 ⎟ y ρ⎜ ⎝ 12 ⎠ ⎝ 12 ⎟⎠ 4 3R = 8 ⎛ 2 3⎞ ρ ⎜ πR ⎟ ⎝3 ⎠ ⇒

2 ⎛ 3R ⎞ R 7 y + ⎜ ⎟= 3 ⎝ 8 ⎠ 48 12

⇒

7 y R R 11R = − = 12 4 48 48

⇒

y=

11R 11 ( 28 ) = 11 cm = 28 28

19. Velocity of B after collision is

( 1 + e ) m1u1 m1 + m2

⎡ (1 + e )m ⎤ ( 10 ) =⎢ ⎣ m + m ⎥⎦

v2 = 5 ( 1 + e )

Height to which B rises is h =

Mechanics II_Chapter 2_Hints and Explanation_2.indd 130

v

m0 2

0

m0

∫ mdv = ∫ ⇒

udm

v = u ln 2 = 2 ln ( 2 ) ms −1 ≈ 1.4 ms −1 1 1 mV 2 = ( μmg ) ( 0.06 ) + kx 2 2 2

x = 4 and x = 5

v2 =

[ 5 ( 1 + e ) ]2

21. According to Work-Energy Theorem, we have

( x − 4 ) ( x − 5 ) = 09

18. Let ρ be the mass density, then mass of the removed cone is ⎛ 1 ⎛ R ⎞ 2 ⎞ π R3 ρ m1 = ρ ⎜ π ⎜ ⎟ R ⎟ = ⎝3 ⎝ 2⎠ ⎠ 12 R Also, centre of mass of cone is at height 4 Mass of remaining hemisphere is

⇒

⇒

20. Since gases are ejected out during motion we use

17. Maximum transfer of momentum takes place when both the masses are equal i.e., x 2 − 9x + 21 = 1 ⇒ ⇒ ⇒

v22 = 3.2 2g

Now,

⇒ ⇒ ⇒ ⇒

1 1 ( 0.18 )V 2 = ( 0.1 )( 0.18 )( 10 )( 0.06 ) + ( 2 ) ( 0.06 )2 2 2 V = 0.4 ms −1 N 0.4 = 10 N=4

22. Since no force acts along the horizontal direction and initial velocity of centre of mass along a horizontal direction is also zero, so x coordinate of centre of mass remains zero. However, ( acm )y = − g Initial velocity of centre of mass along y-direction is

( ucm )y = ⇒

m [ 10 sin ( 30° ) ] + m [ 10 sin ( 30° ) ] 2m

( ucm )y = 5 ms −1

At t = 10 s 1 ( acm )y t 2 2 1 2 ⇒ ycm = ( 5 )( 10 ) + ( −10 )( 10 ) 2 ⇒ ycm = 50 − 500 = −450 m ycm = ( ucm )y t +

⇒

x = 0 and y = −450

⇒

x − y = 450

23. Initial state at t = 0 and final state at time t are shown in Figure.

v22 2g

2/9/2021 6:30:32 PM

Hints and Explanations

H.131

Since no external forces are acting on the system, the displacement of the centre of mass of the system should be zero. Hence Δxcm =

Applying conservation of momentum at time t , we get mv0 = mv′ + 2mvy ⇒

⇒

v0 = v′ + 2vy

At the time of collision θ = 0 and v′ = vy v ⇒ v′ = v y = 0 3 Applying conservation of energy, we get

(

25.

)

⎛ v02 v2 ⎞ + 2 ⎜ vx2 + 0 ⎟ ⎝ 9 9 ⎠

v02 =

⇒

1 2⎞ ⎛ v02 ⎜ 1 − − ⎟ = 2vx2 ⎝ 9 9⎠

⇒

⎛ 9−3⎞ v02 ⎜ = 2vx2 ⎝ 9 ⎟⎠

v0 3 So, net velocity of A at the time of collision is ⇒

( 20 )2 × 3 u2 sin 2 θ 4 = 15 m H= = 2g 2 × 10 i.e., the shell strikes the ball at highest point of its trajectory. Velocity of (ball + shell) just after collision, using Law of Conservation of Linear Momentum is u cos 60° v= 2 20 ⇒ v= = 5 ms −1 2×2 At the highest point, combined mass is at rest relative to the trolley. Let v be the velocity of trolley at this instant. Then by Law of Conservation of Momentum, we get

1 1 ⎡1 ⎤ mv02 = mv′ 2 + 2 ⎢ m vx2 + vy2 ⎥ 2 2 ⎣2 ⎦ ⇒

40 x + 50 ( 2 + x ) − 60 ( 2 − x ) =0 150 2 x= m ≈ 13 cm 15

CHAPTER 2

⇒

mP ΔxP + mR ΔxR + mM Δx M =0 mP + mR + mM

2 × 5 = ( 2 + 18 ) v

vx =

⇒

v=

1 ms −1 2

vA = vx2 + vy2 ⇒ ⇒

vA = vA =

θ

4 2v0 v02 v02 + = v0 = 3 9 9 3

v vr = 0

2v0 ⎛ 2 ⎞ = ⎜ ⎟ ( 9 ) = 6 ms −1 ⎝ 3⎠ 3

24. Let displacement of the platform be Δx p = x , towards right as shown in Figure.

The displacement of Rahul is ΔxR = ( 2 + x ), rightwards

The displacement of Mohit is Δx M = ( 2 − x ), leftwards

Mechanics II_Chapter 2_Hints and Explanation_2.indd 131

Using Law of Conservation of Mechanical Energy, we get ⎛ ⎜⎝

1⎞ ⎛ ⎛ 1⎞ 2 ⎟ ( 2 ) ( 5 ) − ⎜⎝ ⎟⎠ × ( 2 + 18 ) ⎜⎝ 2⎠ 2

2

1⎞ ⎟ = ( 2 ) ( 10 ) ( 1 ) ( 1 − cos θ ) 2⎠

⇒ ⇒ ⇒

25 − 2.5 = 20 ( 1 − cos θ ) 1 − cos θ = 1.125 cos θ = 0.125

⇒

cos θ =

⇒

∗=8

1 8

26. Let v be velocity of the ball after collision along the normal, v′ be velocity of the ball after collision along incline and J be impulse on ball, then J = v − ( −2 cos 30° ) = v + 3

2/9/2021 6:30:51 PM

H.132

JEE Advanced Physics: Mechanics – II ⎡ ( vCM )y ⎤ ⎦ = 36 = 1.8 m =⎣ 2 ( aCM )y 2 × 10 2

⇒

v+ 3 = 2v1 2 v = 4v1 − 3

…(1)

Coefficient of restitution is defined as ⎛ v − v1 ⎞ e = −⎜ 2 ⎝ u2 − u1 ⎟⎠ ⇒

( hCM )max

⇒

( hCM )y = 180 cm

28. Before we can draw working diagrams, the velocity components, before and after impact, parallel and perpendicular to the line of centres must be found. To do this we need unit vectors in these two directions. Let the unit vector along the line of centres i.e., along ˆ where n-line be n,

Impulse on wedge is J sin 30° = Mv1 = 2v1 ⇒

⇒

(

)

(

)

1 ˆ ˆ i−j 2 and a unit vector perpendicular to the line of centres i.e., along t-line be tˆ where nˆ =

1 ˆ ˆ tˆ = i+j 2

v ⎞ ⎛ v+ 1⎟ 1 ⎜⎝ 2⎠ = 2 2 cos 30°

3 v1 ⇒ v= − 2 2 Solving (1) and (2), we get

…(2)

1 1 v1 = ms −1 and v = ms −1 3 3 For the ball, velocity along incline remains constant, so v′ = 2 sin ( 30° ) = 1 ms

The velocity components along these directions are calculated in table. Magnitude of component Sphere

Parallel to nˆ

( iˆ + 2 ˆj ) ⋅ 12 ( iˆ − ˆj )

−1

Hence final velocity of ball is

A

=−

2

2 ⎛ 1 ⎞ v2 = 12 + ⎜ = ms −1 ⎝ 3 ⎟⎠ 3 ⇒

27. If all the particles of the system have same accelera tion, a say, then acceleration of centre of mass will also  be equal to a. Now, in this problem since both particles are projected in the air, so both particles have acceleration g downwards.  ⇒ aCM = − gjˆ  Now, if aCM remains constant, then equation of motion can be applied to centre of mass. In this question, maximum height attained by the centre of mass is asked, thus we need not to think about the vertical motion of centre of mass.

⇒

( vCM )y =

5 ( 10 sin 30° ) + 2 ( 17 sin 30° ) 5+2

( vCM )y =

25 + 17 42 = = 6 ms −1 7 7

Also, ( aCM )y = −10 ms −2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 132

( iˆ + 2 ˆj ) ⋅ 12 ( iˆ + ˆj )

1 2

=

( −iˆ + 3 ˆj ) ⋅ 12 ( iˆ − ˆj )

B

α=2

Parallel to tˆ

3 2

( −i + 3 ˆj ) ⋅ 12 ( iˆ + ˆj )

= −2 2

= 2

Using Law of Conservation of Linear Momentum and the Law of Restitution along the line of impact i.e., n-line, we get ⎛ 1 ⎞ + 2m ( 2 2 ) = mu + 2mv m⎜ ⎝ 2 ⎟⎠ 1⎛ 1 ⎞ ⎜⎝ 2 2 − ⎟ =u−v 2 2⎠ Solving, we get 5 u = 2 2 and v = 2 4 The velocity of A after impact is and

(

)

(

3 ˆ 1 ˆ ˆ ⎛ 3 ⎞ 1 ˆ ˆ  i+j vA = − unˆ + t = ( −2 2 ) i − j +⎜ ⎝ 2 ⎟⎠ 2 2 2 1  vA = − iˆ + 7 ˆj 2

(

)

)

2/9/2021 6:31:11 PM

Hints and Explanations

(

(

)

(

⇒

)

)

⇒

100 100 + ms −1 16 15 1⎞ ⎛ 1 v2 = 100 ⎜ + ms −1 ≈ 13 ms −1 ⎝ 16 15 ⎟⎠ v2 =

31. (a) Let v be velocity of block when the bullet emerges from it. Applying Law of Conservation of Linear Momentum, we get

( 1 ) v + ( 4 × 10 −3 ) ( 100 ) = ( 4 × 10 −3 ) ( 600 )

29. Taking four monkeys together as system, the external forces acting on the system is tension and gravitational force.

v = 2 ms −1

⇒

CHAPTER 2

⇒ α = 2 , x1 = 1 and y1 = 7 The velocity of B after impact is 1 ˆ ˆ  ⎛ 5 ⎞ 1 ˆ ˆ ( 2⎟ i+j vB = − vnˆ + 2tˆ = ⎜ − i − j + 2) ⎝ 4 ⎠ 2 2 1 ˆ  ⇒ vB = − i + 9 ˆj 4 ⇒ β = 4, x2 = 1 and y 2 = 9

H.133

Using v 2 = u2 − 2 as , we get 0 = ( 2 ) − 2 ( μk g ) s 2

0 = 4 − 2 μ k ( 10 )( 0.4 ) 1 ⇒ μ k = 0.5 = 2 ⇒ ∗= 2 (b) Decrease in kinetic energy of bullet is 1 2 2 − ΔK = ( 4 × 10 −3 ) ⎡⎣ ( 600 ) − ( 100 ) ⎤⎦ = 700 J 2 (c) Kinetic energy of block 1 2 K = × 1× (2) = 2 J 2 32. Just before collision, the arrangement is shown in Figure. Acceleration of centre of mass of system is m a + m2 a2 + m3 a3 + m4 a4 acm = 1 1 m1 + m2 + m3 + m4 Taking upward direction is positive, we get 15 ( 1 ) + 25 ( −3 ) + 20 ( 0 ) + 40 ( 0 ) acm = 15 + 25 + 20 + 40 60 6 = ms −2, downwards ⇒ acm = − 100 10 Since, ∑ Fext = Macm ⇒

Mg − T = Ma

⇒ ⇒

( 100 )( 9.8 ) − T = 100 ( 0.6 ) T = 980 − 60 = 920 N

30. Mass of gun is 80 m − 5m = 75m After first shot, if speed of car is v1, then we have 75mv = 5m ( 100 − v ) 500 100 = ms −1 ⇒ v1 = 80 6 After second shot, if speed of car is v2 , then we have ⎛ 100 ⎞ 75m ⎜ = 70 mv2 − 5m ( 100 − v2 ) ⎝ 16 ⎟⎠ 75 × 100 + 500 ⇒ 75v2 = 16

Mechanics II_Chapter 2_Hints and Explanation_2.indd 133

2 2R 2 2 = 2R + R 3 Let initial velocity of A be u and final velocity of each small balls be v Applying conservation of linear momentum, we get mu = 2mv cos θ So, cos θ =

Since, e = ⇒

e=

⇒

e=

( Relative velocity of separation )n-line ( Relative velocity of approach )n-line

v u cos θ 1 1 9 = = ≈ 0.56 2 cos 2 θ 2 ⎛ 8 ⎞ 16 ⎜⎝ ⎟⎠ 9

33. If final velocity of shuttle is v, then by conservation of momentum, we get M ⎛ 4M ⎞ M ( 4000 ) = ⎜ v + ( 3880 ) ⎝ 5 ⎟⎠ 5

2/9/2021 6:31:30 PM

H.134

⇒ ⇒

JEE Advanced Physics: Mechanics – II ⇒

3880 4v = 4000 − = 3224 5 5 v = 4030 kph

⇒

vB = 6 ms −1

⇒

vC =

( 2m )( 6 )

3m vC = 4 ms −1

35. Applying conservation of momentum along x and y direction. Along x direction, …(1) mBv = mAvA cos θ If A moves at an angle θ to initial direction of B which is taken as x axis, then along y direction v mB = mAvA sin θ …(2) 2 Dividing (2) by (1), we get 1 tan θ = 2 ⎛ 1⎞ ⇒ θ = tan −1 ⎜ ⎟ ⎝ 2⎠ ⇒ α=2 Since collision is elastic, so we have ⎛ v2 ⎞ 1 1 1 mBv 2 = mB ⎜ ⎟ + mAvA2 ⎝ 4 ⎠ 2 2 2 Using equation (1), we get

( mAvA cosθ ) v = ( mAvA cosθ ) 1 ⇒ v cos θ = v cos θ + VA 4 3 ⇒ VA = v cos θ 4 2 1 Since, tan θ = , so cos θ = 2 5 ⇒

⎛ 3v ⎛ 3⎞⎛ 2 ⎞ VA = ⎜ ⎟ ⎜ v= =⎜ ⎝ 4 ⎠ ⎝ 5 ⎟⎠ 2 5 ⎝

⇒

β = 0.45

v + mAvA2 4

1 1 1 2 mBv02 = ( mA + mB ) v 2 + kxm 2 2 2 1 1 1 2 2 2 × ( 1 ) × ( 2 ) = × ( 1 + 1 ) × ( 1 ) + × ( 200 ) × xm 2 2 2 ⇒

2 2 = 1 + 100 xm

⇒

xm = 0.1 m = 10 cm

37. Let smaller block be moving with speed v1 relative to bigger block when it reaches the bottommost position and at this instant, bigger block is moving at v2 (say). Applying conservation of momentum in horizontal direction we get

6 mv2 = m ( v1 − v2 ) Applying conservation of energy, we get mga =

…(2)

Solving equations (1) and (2), we get 2 ga = 36v22 + 6v22 ga 21

⇒

v2 =

⇒

∗ = 21

38. Let K be the initial kinetic energy of mass m and K1 and K 2 the kinetic energies after collision in the direction shown in figure. Initial Momentum of the particle is 2mK . So,

m

9 ⎞ ⎟ v = 0.45v 4×5⎠

…(1)

1 1 2 m ( v1 − v2 ) + ( 6 m ) v22 2 2

y

K1

36. At maximum compression ( xm ), velocity of both the blocks is same, say it is v. Applying Law of Conservation of Linear Momentum, we get ( mA + mB ) v = mBv0

Mechanics II_Chapter 2_Hints and Explanation_2.indd 134

v=

Substituting the values, we get

Now, for completely inelastic collision between B and C , we have 2m ( vB ) + m ( 0 ) vC = 2m + m ⇒

v0 2 = = 1 ms -1 2 2 By Law of Conservation of Mechanical Energy, we get ⇒

34. For elastic collision between A and B, we have ⎛ 2m ⎞ vB = ⎜ 9 ⎝ 2m + m ⎟⎠

( 1 + 1 ) v = ( 1 ) v0

K

m 5m

x

30° K2

Applying conservation of linear momentum in vector form, we get,

(

2mK iˆ = 2mK1 ˆj + 2 MK 2 cos 30iˆ − sin 30 ˆj

)

2/9/2021 6:31:52 PM

Hints and Explanations we have

3 and 2 1 2mK1 = 10 mK 2 2 Solving, we get 2mK = 10 mK 2

vcm = 50 = ⇒

4 K K 2 = K and K1 = 15 3 ⇒

K1 + K 2 =

v2 =

( 3 )( 30 ) + 2v2 5

( 5 )( 50 ) − ( 3 )( 30 ) 2

160 = 80 ms −1 2

39. After time t, let us calculate velocity of centre of mass of system by using impulse momentum equation. As spring force is an internal force of system, so we have

9 K = 0.6 K 15

t

∫

psystem = ( 2 + 3 ) vcm = 5tdt =

So, loss in KE = K − 0.6 K = 0.4 K Hence, KE will be decreased by 40%.

0

t

∫

psystem = ( 2 + 3 ) vcm = 5tdt = 0

5t 2 2

5t 2 2

…(1)

At t = 10 s, from equation (1) we get

39. After time t, let us calculate velocity of centre of mass of system by using impulse momentum equation. As spring force is an internal force of system, so we have

2

vcm =

5 ( 10 ) 2 = 50 ms −1 5

If at this instant, let velocity of 2 kg mass be v2 , then we have

…(1)

vcm = 50 =

At t = 10 s, from equation (1) we get 2

vcm =

=

5 ( 10 ) 2 = 50 ms −1 5

CHAPTER 2

⇒

H.135

( 3 )( 30 ) + 2v2 5

( 5 )( 50 ) − ( 3 )( 30 )

v2 =

⇒

1 1 1 1 2 mu2 + ( 3 m )( 0 ) = mv 2 + ( 3 m ) v12 2 2 2 2

2

=

160 = 80 ms −1 2

⇒

If at this instant, let velocity of 2 kg mass be v2 , then

ARCHIVE: JEE MAIN 1.

Applying Law of Conservation of Momentum, we get   pi = p f

⇒ u2 = v 2 + 3v12 Substituting value of v1 from equation (1), we get ⎛ u2 + v 2 ⎞ u2 = v 2 + 3 ⎜ ⎟ ⎝ 9 ⎠ ⇒

3u 2 = 3v 2 + u 2 + v 2

2u2 = 4v 2 u ⇒ v= 2 Hence, the correct answer is (D). ⇒



⇒

( m ) ( uiˆ ) + ( 3 m ) ( 0 ) = ( m ) ( vjˆ ) + ( 3 m ) v1

⇒

mui − mvj = 3 mv1

⇒

uiˆ − vjˆ  v1 = 3

⇒

 v1 =

u2 + v 2 3

2.

Applying momentum conservation along y-axis, we get

…(1)

u2 + v 2 9 Since collision is perfectly elastic, so we have ⇒

v12 =

ΣK initial = ΣK final

Mechanics II_Chapter 2_Hints and Explanation_2.indd 135

2/9/2021 6:32:08 PM

H.136

JEE Advanced Physics: Mechanics – II m1v1 sin θ1 = m2v2 sin θ 2

⇒

mv1 sin θ1 = 10 mv2 sin θ 2

⇒

v1 sin θ1 = 10v2 sin θ 2

⇒

( K f )m

So, nearest integer value of X = 23 Hence, the correct answer is (23.00).

…(1)

Since it is given that

4.

1 = ( K i )m 2

⇒

1 1⎛ 1 ⎞ mv12 = ⎜ mu2 ⎟ ⎠ 2 2⎝ 2

⇒

v1 =

u 2

⇒

1 1 1 mu2 = mv12 + ( 10 m ) v22 2 2 2

⇒

1 1⎛ 1 ⎞ 1 mu2 = ⎜ mu2 ⎟ + ( 10 m ) v22 ⎠ 2 2 2⎝ 2

⇒

1 1 mu2 = ( 10 m ) v22 4 2

⇒

v2 =

⇒

0.1 × 20 = 2v

⇒ v = 1 ms −1 So, final kinetic energy is 1 K f = mgh + mv 2 = 21 J 2 Hence, the correct answer is (A). 5.

u 20

…(3)

Substituting equations (2) and (3) in equation (1), we get u u sin θ1 = 10 sin θ 2 2 20

Since all collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So, if the final velocity be V, then on applying momentum conservation, we get

sin θ1 = 10 sin θ 2

⇒ n = 10 Hence, the correct answer is (10.00). 3.

Since, pi = p f

…(2)

Also, collision is elastic, so we have K f = K i

⇒

X = 2 ( 4π − 1 ) = ( 8π − 2 ) = 23.12

Since xcm

mv = 16 mV ⇒

m x − m2 x2 = 1 1 m1 − m2

V=

v 16

Now initial energy is Ei = Final energy is E f = ⇒

Ef =

1 mv 2 2

v 1 ( 16 m ) ⎛⎜ ⎞⎟ ⎝ 16 ⎠ 2

2

mv 2 32

Energy loss is Ei − E f given by

where, m1 is mass of complete disc and m2 is removed mass. Let σ be the surface mass density of disc material, then xcm = ⇒

xcm =

2(

(

)

σπ a 0 ) − σ a 4 d 2

(

2

2

σπ a − σ a 4

)

−d a =− 4π − 1 2 ( 4π − 1 )

Mechanics II_Chapter 2_Hints and Explanation_2.indd 136

=

2

−a d 4 2

(

π a − a2 4

)

⇒

Ei − E f =

1 1 v2 mv 2 − m 2 2 16

Ei − E f =

1 1 ⎞ ⎛ mv 2 ⎜ 1 − ⎟ ⎝ 2 16 ⎠

1 ⎛ 15 ⎞ mv 2 ⎜ ⎟ ⎝ 16 ⎠ 2 So, percentage loss in KE is ⇒

Ei − E f =

Energy loss ΔK × 100% × 100% Original energy K

2/9/2021 6:32:27 PM

Hints and Explanations

⇒ ⇒

So, from equations (1) and (2), we get

1 ⎛ 15 ⎞ mv 2 ⎜ ⎟ ⎝ 16 ⎠ ΔK 2 × 100% = × 100 = 93.75% 1 K mv 2 2 p ≈ 94

θ = 105° Hence, the correct answer is (D). 8.

3R = 3 cm 8

Since, x =

Hence, the correct answer is (D). Momentum conservation along x direction, we get

⇒

x=3

Hence, the correct answer is (3). 9. 2mv0 cos θ = ( 2m ) ⇒

cos θ =

v0 2

Let 1 kg be taken as origin and xy axis is shown in Figure.

1 2

CHAPTER 2

6.

H.137

⇒ θ = 60° Angle is 2θ = 120° Hence, the correct answer is (120.00). 7.

Given that the respective initial velocity of A and B be    u1 = 3iˆ + ˆj ms −1 -and u2 = 0

(

)

Also, m1 = 2m2 and after collision, we have   v1 = iˆ + 3 ˆj ms −1 and v2 = ?

(

So, xcm =

)

Applying Law of Conservation of Linear Momentum, we get     m1u1 + m2u2 = m1v1 + m2v2  ⇒ 2m2 3iˆ + ˆj + 0 = 2m2 iˆ + 3 ˆj + m2v2

(

⇒ ⇒ ⇒

(  v2 = 2 (

)

) ( 3iˆ − ˆj ) + 2 ( i −  v2 = 2 ( 3 − 1 ) ( iˆ − ˆj )  v2 = 2

(

) 3 ˆj )

)

1( 0 ) + 1.5 ( 3 ) + 2.5 ( 0 ) = 0.9 cm and 5

1( 0 ) + 1.5 ( 0 ) + 2.5 ( 4 ) = 2 cm 5 Hence, the correct answer is (B). ycm =

10. Let lamina be divided in two parts such that m1 = 3 kg and m2 = 1 kg

3iˆ + ˆj − 2 iˆ + 3 ˆj

  If θ be the angle between v1 and v2 , then we have   v ⋅ v 2 ( 3 − 1 )( 1 − 3 ) cos θ = 1  2 = v1 v2 ( 2 ) ⎡⎣ 2 2 ( 3 − 1 ) ⎤⎦ 1− 3 2 2 Since we can write, cos ( 105° ) = cos ( 60° + 45° ) ⇒

cos θ =

⇒

cos ( 105° ) = cos 60° cos 45° − sin 60° sin 45°

⇒

⎛ 1⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ cos ( 105° ) = ⎜ ⎟ ⎜ − ⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠

⇒

1− 3 cos ( 105° ) = 2 2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 137

…(1)

…(2)

Mass of plate-1 is assumed to be concentrated at ( 0.5, 1,.5 ) and mass of plate-2 is assumed to be concentrated at ( 1.5, 2.5 ) , so we have xcm =

m1x1 + m2 x2 3 × 0.5 + 1 × 1.5 = = 0.75 and m1 + m2 4

ycm =

m1y1 + m2 y 2 3 × 1.5 + 1 × 2.5 = = 1.75 m1 + m2 4

Hence, the correct answer is (D).

2/9/2021 6:32:44 PM

H.138

JEE Advanced Physics: Mechanics – II

11. By conservation of linear momentum, we have

( )

(

)

 iˆ ( 0.1 ) ( 3iˆ ) + ( 0.1 ) 5 ˆj = ( 0.1 ) ( 4 ) iˆ + ˆj + ( 0.1 ) v  ⇒ v = − iˆ + ˆj  Speed of B after collision is v = 2 Now, kinetic energy of B is KB =

1 1 1 mv 2 = ( 0.1 ) ( 2 ) = 2 2 10

⇒ x=1 Hence, the correct answer is (1.00). 12. Applying Law of Conservation of Moments about the centre C of the full disc, we have

⇒

( R2 + R + 1 ) ( 2 − R ) = 1

Hence, the correct answer is (C). 13. Applying Law of Conservation of Momentum, we get ⎛ iˆ + ˆj ⎞  ( mu ) iˆ + mu ⎜ = ( m + m )v ⎝ 2 ⎟⎠ u  3 v = uiˆ + ˆj 4 4  u ⇒ v = 10 4 So, final kinetic energy is ⇒

2

1 5 ⎛u ⎞ 2m ⎜ 10 ⎟ = mu2 ⎝4 ⎠ 2 8 and initial kinetic energy is K final =

1 1 ⎛ u 2⎞ 6 2 mu2 + m ⎜ ⎟ = mu 2 2 ⎝ 2 ⎠ 8 So, loss in kinetic energy is Ki =

1 mu2 8 Hence, the correct answer is (B). − ΔK = K i − K f =

m1r1 = m2 r2 , where 4 m1 = mremainder = π ( R3 − ( 13 ) ) ρ , r1 = 2 − R and 3 4 3 m2 = mcavity = π ( 1 ) ρ , r1 = R − 1 3 ⇒

mremainder ( 2 − R ) = mcavity ( R − 1 )

⇒

4 ⎡4 3 ( )3 ⎤ ( ) ⎡ 4 ( )3 ⎤ ( ) ⎢⎣ 3 π R ρ − 3 π 1 ρ ⎥⎦ 2 − R = ⎢⎣ 3 π 1 ρ ⎥⎦ R − 1

⇒

( R3 − 1 ) ( 2 − R ) = R − 1 ( R2 + R + 1 ) ( 2 − R ) = 1

L

⇒

⇒ Alternatively, we can also get the desired result by applying the formula xcm =

mfull disc x1 − mremoved disc x2 mfull disc − mremoved disc

Considering origin to be at the centre of the full disc, we have xcm = − ( 2 − R ) , x1 = 0 and x2 = R − 1 ⇒

4 3 0 − π (1) ρ ( R − 1) − ( R − 1) 3 −(2 − R) = = 3 3 4 3 4 3 πR ρ − π (1) ρ R − (1) 3 3

⇒

− ( 2 − R ) ( R3 − 1 ) = − ( R − 1 )

⇒

( 2 − R ) ( R − 1 ) ( R2 + R + 1 ) = ( R − 1 )

Mechanics II_Chapter 2_Hints and Explanation_2.indd 138

∫ xdm = ∫ ( λdx )x ∫ dm ∫ dm

14. Since, xcm =

xcm

∫

⎛ bx 2 ⎞ ⎜⎝ a + 2 ⎟⎠ xdx L

aL 2 b L4 + 2 L 4 = 0L = 2 2⎞ L3 b ⎛ bx aL + 2 ⎜⎝ a + 2 ⎟⎠ dx L 3 L

∫ 0

( 4 a + 2b ) ⇒

xcm =

8

( 3a + b )

=

3 ⎛ 2a + b ⎞ ⎜ ⎟L 4 ⎝ 3a + b ⎠

3 Hence, the correct answer is (D). 15. Applying Law of Conservation of Momentum along x direction, we get

2/9/2021 6:33:01 PM

Hints and Explanations Substituting (2) in (1), we get

mu + mu = 2mv′ 2 3u ⇒ v′ = 4 Range after collision is 3u 2 H ⎛ 3u ⎞ R = v′t = ⎜ t= ⎝ 4 ⎟⎠ g 4

⇒

R=

{

∵H=

1 2 gt 2

}

3V V⎞ ⎛ = m2 ⎜ V + ⎟ ⎝ 2 4⎠ 6 ⇒ m2 = kg 5 Hence, the correct answer is (D). 20. Applying momentum conservation, we get mv = ( 4 m + m ) v′

u2 sin 2 ( 60° ) 2g

v 5 Applying energy conservation, we get

So, common speed is v′ =

3u 2u2 sin 2 ( 60° ) ⎛ 3u ⎞ ⎛ u 3 ⎞ =⎜ ⎝ 4 ⎟⎠ ⎜⎝ 2 g ⎟⎠ g ( 2g ) 4

⎛ v2 ⎞ 1 1 mgh + 5m ⎜ ⎟ = mv 2 ⎝ 25 ⎠ 2 2

3 3u 2 8g Hence, the correct answer is (C). ⇒

R=

⇒

( )

mgh =

1⎞ 1 4 1 ⎛ mv 2 ⎜ 1 − ⎟ = mv 2 ⎝ ⎠ 2 5 2 5

2mv 2 2v 2 = 5 × mg 5 g Hence, the correct answer is (B).

( 2m ) ajˆ + 3 m × aiˆ + ma ( − iˆ ) + 4 m × a − ˆj  16. aCM = 2m + 3 m + 4 m + m ˆ 2 ai − 2 ajˆ a ˆ ˆ  ⇒ aCM = = i−j 10 5 Hence, the correct answer is (D).

(

CHAPTER 2

where, H =

H.139

⇒

)

h=

21. Initial momentum is pi = 2mv + 2mv = 4 mv Let v′ be the speed of each splitted particle, then by momentum conservation, we get

17. Applying Conservation of Linear Momentum, we get m1v1 + m2v2 = m1v3 + m2v4 ⇒

m1v1 + 0.5m1v2 = 0.5m1v1 + 0.5m1v4

⇒

v1 = v4 − v2

Hence, the correct answer is (B). 18.

X-coordinate of CM of remaining sheet is xcm =

⇒

xcm

MX − mx M−m 22

5b 12

V = V2 −

V 4

Mechanics II_Chapter 2_Hints and Explanation_2.indd 139

v ′ = 2 2v

Applying conservation of linear momentum along x and y directions. Along x direction, initial and final momentum are

Since, ( Σpx )initial = ( Σpx )final

19. Applying Conservation of Linear Momentum, we get

⇒

⇒

( Σpx )initial = M ( 10 cos 30° ) + 2 M ( 5 cos 45° ) ( Σpx )final = 2 M ( v1 cos 30° ) + M ( v2 cos 45° )

⎛ 5 a 5b ⎞ So, centre of mass is located at ⎜ , ⎝ 12 12 ⎟⎠ Hence, the correct answer is (B). 2V 2V = + m2V2 4 ⎛ v − v1 ⎞ Since e = − ⎜ 2 =1 ⎝ u2 − u1 ⎟⎠

2

Hence, the correct answer is (D).

( 4 m ) × ⎛⎜ a ⎞⎟ − M ⎛⎜ 3 a ⎞⎟ ⎝ 4 ⎠ 5a ⎝ 2⎠ = = 4m − m 12

Similarly, ycm =

mv′ = 4 mv 2

⇒

…(1)

…(2)

10 v = 3v1 + 2 2 2 v ⇒ 5 3 + 5 2 = 3v1 + 2 …(1) 2 Along y direction, initial and final momentum are ⇒

5 3+

( Σpy )initial = 2M ( 5 sin 45° ) − M ( 10 sin 30° )

2/9/2021 6:33:22 PM

H.140

JEE Advanced Physics: Mechanics – II

( Σpy )final = 2M ( v1 sin 30° ) − M ( v2 sin 45° ) = ( Σp y ) Since, ( Σpy ) initial final ⇒

v 5 2 − 5 = v1 − 2 2

⇒

⇒

3 1 mgL sin θ = mgL cos θ 2 2 1 ⇒ tan θ = 3 Hence, the correct answer is (C). ⇒

…(2)

Solving equations (1) and (2), we get

(

⎛L ⎞ ⎛L ⎞ mg ⎜ sin θ ⎟ = mg ⎜ cos θ = L sin θ ⎟ ⎝2 ⎠ ⎝2 ⎠

3 + 1 ) v1 = 5 3 + 10 2 − 5

26. Since, ΔE1 =

v1 = 6.5 ms −1 and v2 = 6.3 ms −1

Hence, the correct answer is (D).

⇒

ΔE1 =

23. Applying momentum conservation, we get Also, V1 + V2 = 0.70

⇒

V1 = 0.20

Hence, the correct answer is (A).

y CM =

1 ⎛ 2mM ⎞ ⎛ ⎜ ⎟⎜ 2 ⎝ 2m + M ⎠ ⎝

v⎞ ⎟ 2⎠

v 2

2

5⎛ 1 2⎞ ⎜ mv ⎟⎠ 6⎝ 2

1⎛ m⎞ 2 ⎛ 1⎞ ⎛ m⎞ 2⎛ M ⎞ 5⎛ 1 2⎞ ⎜ ⎟ v + ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ v ⎜⎝ ⎟ = ⎜ mv ⎟⎠ 2⎝ 2 ⎠ 2 2 2m + M ⎠ 6 ⎝ 2 10 5 M ⇒ 1+ = = 2m + M 6 3 M =4 ⇒ m Hence, the correct answer is (B). ⇒

 3  0.5,  2  

( m )( 0 ) + ( 2m ) ( 1 ) + ( 3 m ) ⎛⎜ 1 ⎞⎟ 6m

ΔE2 =

Since, ΔE1 + ΔE2 =

24.

xCM =

1⎛ m⎞ 2 1 2 ⎜ ⎟ v = mv 2⎝ 2 ⎠ 4

Velocity after collision is V =

50V1 = 20V2 ⇒

1 2 μvrel 2

⎝ 2⎠ 7 = m 12

⎛ ⎞ ( m )( 0 ) + ( 2m )( 0 ) + ( 3 m ) ⎜ 3 ⎟ ⎝ ⎠ 2

6m

=

3 m 4

27. Let bullet meet the piece of wood in time t , then we have 1 100 − 10t 2 = 100t − 5t 2 2 ⇒ t=1s Applying Law of Conservation of Momentum, we get

Hence, the correct answer is (D). 25. Let mass of one rod be m, then balancing the torque about one end of the rod, we get

90 ( 0.02 ) − 10 ( 0.03 ) = v ( 0.05 ) ⇒

v=

1.8 − 0.3 1.5 150 = = = 30 ms −1 0.05 0.05 5

30 2 30 × 30 = = 45 m 2 × 10 2 × 10 So, maximum height above the building is ⇒

s2 =

hmax = 45 − 5 = 40 m Hence, the correct answer is (B). 3 × 10 k Since spring constant is high, so initial compression is low. Let v1 be velocity after collision, then by Conservation of Momentum, we have

28. Initial compression is xi =

mg ( C1P ) = mg ( C2 N )

Mechanics II_Chapter 2_Hints and Explanation_2.indd 140

4v1 = v0

…(1)

2/9/2021 6:33:41 PM

Hints and Explanations

Applying Conservation of Linear Momentum, we get

where, v0 = 2 gh = 2 g × 100

mv0 = mv1 + mv2

⇒

1 ( 4 ) v12 = 1 kx 2 2 2

⇒

⇒

x = 2 cm

According to the question, we have

*No given option is correct.

v1 + v2 = v0

Kf =

29. Since, u = ωθ 0 and v = ωθ1

…(1)

3 Ki 2

…(2)

1 1 3⎛ 1 ⎞ mv12 + mv22 = ⎜ mv02 ⎟ ⎠ 2 2 2⎝ 2 3 ⇒ v12 + v22 = v02 2 From equation (1), we have

u θ0 = v θ1

⇒

⎛ M −m⎞ Also, v = ⎜ u ⎝ M + m ⎟⎠ ⇒

M + m u θ0 = = M − m v θ1

⇒

M θ 0 + θ1 = m θ 0 − θ1

…(3)

( v1 + v2 )2 = v02 ⇒

v12 + v22 + 2v1v2 = v02

Using equation (3), we get

⎛ θ + θ1 ⎞ M = m⎜ 0 ⎝ θ 0 − θ1 ⎟⎠ Hence, the correct answer is (A). ⇒

30. Let assume linear mass density is λ . The corresponding coordinates of the rod are shown in Figure.

⇒ ⇒

CHAPTER 2

⇒

H.141

3 2v1v2 = v02 − v02 2 2 v 2v1v2 = − 0 2 4v1v2 = − v02

Since, ( v1 − v2 ) = ( v1 + v2 ) − 4v1v2 2

2

⇒

( v1 − v2 )2 = v02 − ( −v02 ) = 2v02

⇒

v1 − v2 = 2v0

Hence, the correct answer is (B). 32. In an elastic oblique 2-D collision, if two bodies are of equal masses and second body is at rest, then after collision, the bodies scatter at right angles. So, the bodies have equal mass i.e.

then, m1 = 2Lλ and r1 cm ≡ ( L, L ) L⎞ ⎛ m2 = Lλ and r2 cm ≡ ⎜ 2L, ⎟ and ⎝ 2⎠

munknown = m

⎛ 5L ⎞ m3 = Lλ and r3 cm ≡ ⎜ , 0⎟ ⎝ 2 ⎠ Since, xcm = ⇒

xcm =

and ycm ⇒

m1x1 + m2 x2 + m3 x3 m1 + m2 + m3

33. The kinetic energy of an object just after it hits the ground is 50% of K.E. of the object. So, we have

( 2m ) L + m ( 2L ) + m ( 5L 2 )

2m + m + m m1y1 + m2 y 2 + m3 y 3 = m1 + m2 + m3

xcm =

( 2m ) L + m ( L 2 ) + m ( 0 )

2m + m + m Hence, the correct answer is (C). 31.

Hence, the correct answer is (A).

v0

Mechanics II_Chapter 2_Hints and Explanation_2.indd 141

v1

=

1 1⎛ 1 ⎞ mv′ 2 = ⎜ mv 2 ⎟ ⎠ 2 2⎝ 2

13 L 8 ⇒

=

v′ =

v 2

By definition, we have

5 L 8 v2

⇒

e=

Relative velocity after collision Relative velocity beforee collision

e=

v′ 1 = v 2

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H.142

JEE Advanced Physics: Mechanics – II

When a ball dropped from a height h , total distance covered when

36. Consider a small segment dx of the rod at a distance x from A. Mass of this small segment bx ⎞ ⎛ dm = μdx = ⎜ a + ⎟ dx ⎝ L⎠

⎛ 1 + e2 ⎞ t → ∞ is  = h ⎜ ⎝ 1 − e 2 ⎟⎠ ⇒

1⎞ ⎛ 1+ ⎜ 2 ⎟ = 3h  = h⎜ 1⎟ ⎜⎝ 1 − ⎟⎠ 2

Then CM of the rod AB (from A ) is given by L

None of the given OPTIONS is correct.

∫ xdm = ∫ = ∫ dm ∫ μdx

( μdx ) x

34. Let the length AB = p and BC = q If λ be the linear mass density of the rod, then according to question, centre of mass of the rod lies vertically below point A, so xcm = p cos 60°

xCM

0

L

0

L

⇒

7 L= 12

∫

⎛ bx 2 ⎞ ⎜⎝ ax + ⎟ dx L ⎠

⎛ aL2 bL2 ⎞ + ⎜⎝ ⎟ 2 3 ⎠ 0 = L bL ⎞ ⎛ bx ⎞ ⎛ ⎜⎝ aL + ⎟⎠ ⎜⎝ a + ⎟⎠ dx 2 L

∫ 0

⇒

q p ( λ q ) ⎛⎜⎝ 2 ⎞⎟⎠ + ( λ p ) ⎛⎜⎝ 2 ⎞⎟⎠ cos 60° = p cos 60° = λ(p + q)

⇒

xCM

⇒

q2 p 2 + p 4 = 2 2 (p + q)

⇒

p 2 + pq = q2 +

⇒

1+

⇒

q 1 ⎛ q⎞ ⎜⎝ p ⎟⎠ − p − 2 = 0

⇒

Hence, the correct answer is (C). 37.

p2 2

q q2 1 = + p p2 2 2

2 ⎛ 1⎞ − ( −1 ) ± ( −1 ) − 4 ( 1 ) ⎜ − ⎟ ⎝ 2⎠

⇒

q = p

⇒

q 1+ 3 = = 1.366 ≈ 1.37 p 2

2×1

m ( 2v ) iˆ + ( 2m ) ˆj  vcombined = m + 2m

(

⇒

2v ˆ ˆ  i+j vcombined = 3

⇒

2 2v  vcombined = 3

)

Now loss in KE is − ΔE = Ei − E f 1± 3 = 2

Hence, the correct answer is (C). 35. The centre of mass of a uniform solid cone of height h h 3h lies at a distance of from base or from vertex. 4 4 Hence, the correct answer is (B).

Mechanics II_Chapter 2_Hints and Explanation_2.indd 142

a b + 7 2 3 = 12 a + b 2 b = 2a

⎛ 2 2v ⎞ 1 1 1 2 m ( 2v ) + ( 2m ) v 2 − ( 3 m ) ⎜ ⎝ 3 ⎟⎠ 2 2 2

⇒

Loss =

⇒

Loss = 3 mv 2 −

2

4 5 mv 2 = mv 2 3 3

5mv 2 3 × 100% = 56% 2mv 2 + mv 2 Hence, the correct answer is (C). ⇒

% Loss =

38. Hence, the correct answer is (B).

2/9/2021 6:34:11 PM

Hints and Explanations

H.143

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems

Since, H =

1.

u02 sin 2 α 2g

…(2)

From equations (1) and (2), we get

  Since both x and y components of v are equal, so v makes an angle of 45° with horizontal. Hence, the correct answer is (A).

⎛π⎞ h Since, cos ⎜ ⎟ = ⎝ n⎠ R ⇒

Δ = R−h=

5.

h 1 ⎡ ⎤ − h = h⎢ ⎥ cos π − cos ( π n ) n 1 ) ( ⎣ ⎦

Hence, the correct answer is (D).

first collision, they will exchange their velocities. So, after two collisions they will again reach at point A as shown in Figure.

2.

Since, K = ⇒

1 1 mv 2 = mg 2t 2 2 2

{∵ v = gt }

K ∝ t2

Therefore, K -t graph is parabola. However, during collision, retarding force is just like the spring force ( F ∝ x ) , therefore kinetic energy first decreases to elastic potential energy and then increases. Hence, the correct answer is (B). 3.

Hence, the correct answer is (C). 6.

Final momentum of object E Power × Time Pt p= = = c Speed of light c

4.

Before the first collision, the particle having speed 4π ⎞ ⎛ 2v will rotate 240° ⎜ or ⎟ while other particle ⎝ 3 ⎠ 2π ⎞ ⎛ having speed v will rotate 120° ⎜ or ⎟ . After ⎝ 3 ⎠

CHAPTER 2

 ⎛ u cos α ⎞ ˆ ⎛ u0 cos α ⎞ ˆ iˆ v = ⎜ 0 ⎟⎠ i + ⎜⎝ ⎟⎠ j ⎝ 2 2

30 × 10

−3

× 100 × 10 3 × 108

⇒

p=

⇒

p = 1.0 × 10 −17 kgms −1

−9

Since, R = u

2h g

⇒

20 = v1

2×5 2×5 and 100 = v2 10 10

⇒

v1 = 20 ms −1 , v2 = 100 ms −1

Applying momentum conservation just before and just after the collision, we get

( 0.01 )( v ) = ( 0.2 )( 20 ) + ( 0.01 )( 100 )

Hence, the correct answer is (B).

⇒

From momentum conservation, we have

Hence, the correct answer is (D). 7.

u 20 − 2gH

Since, yCM = ⇒

  Σpi = Σp f ⇒ iˆ m ( u0 cos α ) iˆ + m

(

)

 u02 − 2 gH ˆj = ( 2m ) v

Mechanics II_Chapter 2_Hints and Explanation_2.indd 143

…(1)

v = 500 ms −1

y CM =

m1y1 + m2 y 2 + m3 y 3 + m4 y 4 + m5 y 5 m1 + m2 + m3 + m4 + m5

( 6 m )( 0 ) + ( m )( a ) + m ( a ) + m ( 0 ) + m ( − a )

6m + m + m + m + m a ⇒ y CM = 10 Hence, the correct answer is (A).

2/9/2021 6:34:20 PM

H.144 8.

JEE Advanced Physics: Mechanics – II M

On a system of particles if, ∑ Fext. = 0 then,

v cos θ

psystem = constant  iˆ p = A cos ( kt ) iˆ − A sin ( kt ) ˆj Since,  dp ⇒ iˆ F = = − kA sin ( kt ) iˆ − kA cos ( kt ) ˆj dt   Since, F ⋅ p = 0   So, angle between F and p is 90° Hence, the correct answer is (D). 10.

−1

Since, vcm = ⇒

vcm

m1v1 + m2v2 m1 + m2

10 × 14 + 4 × 0 10 × 14 = = = 10 ms −1 10 + 4 14

Hence, the correct answer is (C). 11. Change in momentum of two particles is      Δp = ( m1v1′ + m2v2′ ) − ( m1v1 + m2v2 )  External force on × time interval Since, Δp = the system  ⇒ Δp = Fext Δt = ( m1 + m2 ) g ( 2t0 )  ⇒ Δp = 2 ( m1 + m2 ) gt0

⇒

ycm = 0

⇒

ycm =

⇒ ⇒

m1y1 + m2 y 2 m1 + m2

⎛ 3m ⎞ ⎛ m⎞( ⎟(y) ⎜⎝ ⎟⎠ +15 ) + ⎜⎝ 4 4 ⎠ 0= ⎛ m 3m ⎞ ⎜⎝ + ⎟ 4 4 ⎠ y = −5 cm

Hence, the correct answer is (A). 13. At the highest point velocity of the shell is v cos θ . At the highest point, it explodes into two pieces of equal masses out of which one piece retraces the path i.e., has a velocity −v cos θ .

Mechanics II_Chapter 2_Hints and Explanation_2.indd 144

v′

So, by Law of Conservation of Momentum M ( v cos θ ) =

M Mv′ ( −v cos θ ) + 2 2

⇒ v′ = 3v cos θ Hence, the correct answer is (A). 14. In an inelastic collision, only the momentum of system (ball and earth) may remain conserved. Some energy can be lost in the form of heat, sound etc. Hence, the correct answer is (C). 15. Since the system is free from external force, hence acm = 0 and since initially they are at rest, so Vcm = 0 Hence, the correct answer is (A).

Multiple Correct Choice Type Problems 1.

The rate of collision of particle with the piston is f =

1 v = 2L v 2L

The speed of the particle after collision (when it collides with a speed v ) with the piston is v + 2V . Assuming that the piston moves inwards by dL, then speed of the particle increases by

Hence, the correct answer is (C). 12. Before explosion, particle was moving along x-axis , i.e., it has no y-component of velocity. Therefore, the centre of mass will not move along the y-direction

M/2

Just After Explosion

Just Before Explosion

Hence, the correct answer is (A). 9.

v cos θ M/2

⇒

⎛ dL ⎞ ⎛ v ⎞ dv = 2V ⎜ ⎝ V ⎟⎠ ⎜⎝ 2L ⎟⎠ dv dL = v L

Since L is decreasing, so we have dv dL =− v L v

⇒

∫

v0

dv =− v

L0 2

∫

L0

dL L

⇒

( log e v ) v

⇒

⎛ v ⎞ ⎛ log e ⎜ ⎟ = − log e ⎜ ⎝ ⎝ v0 ⎠

⇒

v = 2v0

v 0

= − ( log e L )

L0 2 L0

1⎞ ⎟ 2⎠

1 So, kinetic energy at L0 is K L0 = mv02 and kinetic 2 energy at L0 2 is

2/9/2021 6:34:31 PM

Hints and Explanations

2

=

1 2 m ( 2v0 ) = 4 K L0 2

Total kinetic energy is K = ⇒

Hence, (B) and (C) are correct. 2.

Since, Δxcm of the block and point mass system is zero i.e., CM coil not move along x-direction . ⇒

Hence, (A) and (C) are correct. 4.

m ( x + R ) + Mx = 0

where, x is displacement of the block.

  Initial momentum of the system is Σpinitial = p1 + p2 = 0   So, final momentum of system Σpfinal = p1′ + p2′ = 0 OPTION (B) is allowed because if we put c1 = − c2 ≠ 0 ,   p1′ + p2′ will be zero. Similarly, we can check other options. Hence, (A) and (D) are correct.

mR ⇒ x=− M+m Applying conservation of momentum, we get 5.

0 = mv − MV ⇒

K = 4.5 J

1 × 1 × 32 2

mv = MV

Here, at maximum compression xmax , we introduce the concept of reduced mass μ of the system (details discussed in Rotational Dynamics) given by

Applying conservation of energy, we get 1 1 mv 2 + MV 2 2 2 Solving these two equations, we get mgR =

v=

2 gR m 1+ M

Hence, (A) and (B) are correct. 3. 1

u

5

2 ms−1

1

Before collision

1 1 1 = + μ m m

v

5

m 2 Further, by Law of Conservation of Energy we get ⇒

After collision

Since collision is elastic, so e = 1 ⇒ Velocity of Approach = Velocity of Separation ⇒

u=v+2

…(1) ⇒

( 1 ) u = ( 5 ) v − ( 1 )( 2 ) u = 5v − 2

xmax = v

So, K.E. =

…(2)

Equating (1) and (2), we get

m 2k

1 2 1⎛ m⎞ 2 μv = ⎜ ⎟ v 2 2⎝ 2 ⎠

mv 2 4 Hence, (B) and (D) are correct. ⇒

v + 2 = 5v − 2 ⇒

μ=

1 2 1 2 μv = kxmax 2 2

By Law of Conservation of Linear Momentum, we get ⇒

v = 1 ms −1 and u = 3 ms −1

K.E. =

Momentum of system psystem = ( 1 ) ( 3 ) = 3 kgms −1

Reasoning Based Questions

Momentum of 5 kg after collision is

1.

If a force is applied at centre of mass of a rigid body, its torque about centre of mass will be zero, but acceleration will be non-zero. Hence, velocity will change. Hence, the correct answer is (D).

2.

In case of elastic collision, coefficient of restitution e =1. Magnitude of relative velocity of approach equals magnitude of relative velocity of separation.

p2 = ( 5 ) ( 1 ) = 5 kgms −1 So, kinetic energy of centre of mass is K cm = ⇒

CHAPTER 2

K L0

H.145

2

mu 1 1 1× 3⎞ ( m1 + m2 ) ⎛⎜ 1 ⎞⎟ = ( 1 + 5 ) ⎛⎜⎝ ⎟ 2 2 6 ⎠ ⎝ m1 + m2 ⎠

K cm = 0.75 J

Mechanics II_Chapter 2_Hints and Explanation_2.indd 145

2

2/9/2021 6:34:41 PM

H.146

JEE Advanced Physics: Mechanics – II

But relative speed of approach is not equal to relative speed of separation. Hence, the correct answer is (D).

Linked Comprehension Type Questions 1.

⇒

vC2 = 45 + 60 = 105

⇒

vC = 105 ms −1

Hence, the correct answer is (B). 3.

Line of impact

If collision is completely elastic, then vertical component of velocity becomes, √15 ms−1

60° B 30

Line of tangent

B

60°

√60 cos(30°)

30°

°

√45 ms−1

√60 ms−1 Before

After

45 15 × 3 − =0 2 2 Hence, the correct answer is (C). 45 sin 30° − 15 sin 60° =

Let speed of block just before it strikes the second inclined plane be v then,

⇒

1 mv 2 = mg ( 3 tan 60° ) 2 v = 60 ms −1

Speed of block immediately after it strikes the second incline is 45 ms −1 . Because in perfectly inelastic collision the component of velocity along line of impact becomes zero. Hence, the correct answer is (B). 2.

Integer/Numerical Answer Type Questions 1.

For elastic collision between A and B , we have ⎛ 2m ⎞ vB = ⎜ 9 ⎝ 2m + m ⎟⎠ ⇒

Now, for completely inelastic collision between B and C , we have

By Conservation of Mechanical Energy 2 1 1 mvC2 = m ( 45 ) + mg ( 3 3 tan 30° ) 2 2

Mechanics II_Chapter 2_Hints and Explanation_2.indd 146

vB = 6 ms −1

vC = ⇒

2m ( vB ) + m ( 0 ) ( 2m )( 6 ) = 3m 2m + m

vC = 4 ms −1

2/9/2021 6:34:45 PM

CHAPTER 3: ROTATIONAL DYNAMICS

Test Your Concepts-I (Based on Moment of Inertia and Applications) 1.

4.

{∵ dm = λ dx }

(a) dI = x 2 ( dm ) = x 2 ( λ 0 x ) dx

dx

Axis A: I A = m1d12 + m2 d22 = ( 3 kg ) ( 1 m ) + 2

dm

CHAPTER 3

( 5 kg ) ( 2 m )2 = 23 kgm 2 Axis B: I B = m1 ( 0 ) + m2 ( d1 + d2 ) = 45 kgm 2 2

2L

Axis C: IC = m1 ( d1 + d2 ) + m2 ( 0 ) = 27 kgm 2 2

⇒

Axis D: I D = 0

⇒

For each mass we need its perpendicular distance from the axis. For each axis, two masses do not contribute to the moment of inertia. The other two are at the same distance 3sin ( 53° ) = 2.4 m I A = ( 4 kg ) ( 2.4 m ) + ( 2 kg ) ( 2.4 m ) 2

⇒

3

⎛ x4 I = λ0 ⎜ ⎝ 4

2L ⎞

⎟ ⎠

0

λ0 ( 16 L4 ) 4

⇒

I=

⇒

I = 4 λ 0 L4

…(1)

Now dm = λ dx = λ 0 xdx

2

I A = 34.6 kgm 2

⇒

I B = ( 1 kg ) ( 2.4 m ) + ( 3 kg ) ( 2.4 m 2 ) = 23 kgm 2 2

3.

∫ x dx 0

I D is zero because we treated the masses as point particles and perpendicular distances are equal to zero. 2.

I = λ0

Moment of inertia of rod 1 about axis P is ml 2 I1 = 3

⇒ ⇒

2L

∫

M = dm = λ 0

∫ xdx 0

⎛ x2 M = λ0 ⎜ ⎝ 2

2L ⎞ 0

⎟ ⎠

M = 2λ 0 L2

So, I = 2 ML2 (b) Similarly, in this case I=

Moment of inertia of rod 2 about axis P, I2 =

ml 2 l⎞ ⎛ + m⎜ 5 ⎟ ⎝ 12 2⎠

2

5.

1 ML2 12 Now, L = 2π r I1 =

⇒

So, moment of inertia of a system about axis P, ml 2 ml 2 l⎞ ⎛ I = I1 + I 2 = + + m⎜ 5 ⎟ ⎝ 3 12 2⎠ ⇒

ml 2 I= 3

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 147

1 ML2 3

r=

L 2π

I 2 = 2 Mr 2 =

2

⇒

…(1)

ML2 2π 2

I1 ⎛ ML2 ⎞ ⎛ 2π 2 ⎞ π 2 = =⎜ ⎟ 6 I 2 ⎝ 12 ⎠ ⎜⎝ ML2 ⎟⎠

2/9/2021 6:42:28 PM

H.148 6.

JEE Advanced Physics: Mechanics – II

Using parallel axis theorem I1 = IC + ma 2

…(1)

I 2 = IC + mb 2

…(2)

a

From (1) and (2)

b

I1 − I 2 = m ( a 2 − b 2 ) 7.

(a) I = 0 + m ( 2d ) + m ( 2d ) + m ( 2d ) 2

2

I = 4 md 2 + 2md 2 + 2md 2

⇒ ⇒

I = 8 md

2

σ=

I = I total = I can + 2I lids ⇒

⎛1 ⎞ I = MR2 + 2 ⎜ mR2 ⎟ ⎝2 ⎠

⇒

I = ( 2π Rhσ ) R2 + ( π R2σ ) R2

⇒

I = σπ R

3(

⇒

⇒ ⇒

mR2 ( 15 + 16 + 250 ) I= 20

⇒

I=

⇒

I ≅ 14 mR2

I = I sphere of − I sphere of radius a

⇒ ⇒

⎛ a2 ⎞ I cavity = π a 2σ ⎜ + b2 ⎟ ⎝ 2 ⎠

⇒

I=

radius b

2 ⎡⎛ 4 2 ⎡⎛ 4 ⎞ ⎤ ⎞ ⎤ I = ⎢ ⎜ π a3 ⎟ ρ ⎥ a2 − ⎢ ⎜ π b 3 ⎟ ρ ⎥ b 2 ⎠ ⎦ ⎠ ⎦ 5 ⎣⎝ 3 5 ⎣⎝ 3 ⎛ 2⎞ ⎛ 4 ⎞ I = ⎜ ⎟ ⎜ πρ ⎟ ( a 5 − b 5 ) ⎝ 5⎠ ⎝ 3 ⎠

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 148

2 1 ( π R2σ ) R2 − π a2σ ⎛⎜ a + b 2 ⎞⎟ ⎝ 2 ⎠ 2

1 I = πσ ( R 4 − a 4 − 2 a 2b 2 ) 2 Using (1), we get I=

4 ( 3 πρ a − b 3 ) 3

1 ( π a2σ ) a2 + ( π a2σ ) b 2 2

⇒

mR2 ( 281 ) 20

M=

1 ( π R2σ ) R2 and 2

⇒

4 ⎛4 ⎞ 10. Given M = ρ ⎜ π a 3 − π b 3 ⎟ ⎝3 ⎠ 3 ⇒

Idisc =

I cavity =

25 3 ⎛2 ⎞ mR2 + 2 ⎜ mR2 + mR2 ⎟ ⎝ ⎠ 4 5 4

⎛ 3 4 25 ⎞ I = mR2 ⎜ + + ⎟ ⎝4 5 2 ⎠

…(1)

By Parallel Axis Theorem, we have

2 ⎡2 1 ⎡ ⎛ 3R ⎞ ⎤ 2 2 + R⎟ ⎥ I= ⎣ m ( 3 R ) ⎤⎦ + 2 ⎢ mR + m ⎜⎝ ⎠ ⎦ 12 2 ⎣5

I=

M

π ( R2 − a2 )

Now, about the said axis, we have

2h + R )

⇒

2 ⎛ a5 − b 5 ⎞ M⎜ ⎟ 5 ⎝ a3 − b 3 ⎠

Also I = Idisc − I cavity

I = I total = I rod + I spheres ⇒

I=

4 M πρ = 3 3 a − b3

11. Let σ be mass per unit area i.e., surface mass density, then

I = 4 m 2

⇒

9.

⇒

2

(b) I = 0 + m 2 + m 2 + m ( 2 )

8.

From (1),

2

…(1)

1 ⎛ R 4 − a 4 − 2 a 2b 2 ⎞ M⎜ ⎟⎠ 2 ⎝ R2 − a2

12. Moment of inertia of a semi-circular disc about an axis passing through centre and perpendicular to plane of MR2 disc, I = 2 Using parallel axis theorem I = I cm + Md 2, d is the perpendicular distance between two parallel axis passing through centre C and CM. I= ⇒

4R MR2 ,d= 2 3π

MR2 ⎛ 4R ⎞ = I cm + M ⎜ ⎝ 3π ⎟⎠ 2

2

2/9/2021 6:42:46 PM

Hints and Explanations

⇒

2 ⎡ MR2 ⎛ 4R ⎞ ⎤ I cm = ⎢ − M⎜ ⎥ ⎟ ⎝ 3π ⎠ ⎦ ⎣ 2

⇒

⎛ 1 16 ⎞ I cm = MR2 ⎜ − 2 ⎟ ⎝ 2 9π ⎠

If dI be the moment of inertia of element, then dI = x 2 dm

2

⇒

I = 2π kR x 5 dx

⇒

⎛ R6 ⎞ π kR6 I = 2π k ⎜ = ⎝ 6 ⎟⎠ 3

But from (2) π kR 4 = 2 M

a

⇒

a

a 2

∫ 0

a2 I = ( m1 + m2 ) 4 m1

a 2

I=

16. So, I =

m2

⇒

14. k = R ⇒

dI = x 2 ( 2π kx 3 dx )

CHAPTER 3

⎛ a⎞ ⎛ a⎞ 2 13. I = m1 ( 0 ) + m2 ⎜ ⎟ + m3 ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

m2

⇒

R

2

⇒

H.149

2 MR2 3

mL2 mL2 mL2 + + 3 4 3

I=

11 2 mL 12

I = Mk 2 = MR2

z 2

mL 3

mL2 4 mL2 3

d

By Parallel Axis Theorem, we have

x

I = IG + Md 2 ⇒ ⇒ ⇒

1 MR2 + Md 2 2 1 Md 2 = MR2 2 R d= 2 MR2 =

17. Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC ( i.e., D ) is I1 =

15. dm = σ dA ⇒ ⇒

dm = ( kx 2 ) ( 2π xdx )

∫

…(1)

I 2 = I1 + mr 2

R

∫

⎛ R4 ⎞ M = 2π k ⎜ ⎝ 4 ⎟⎠

⇒

M=

⇒

0

π kR 4 2

m 2 12

According to Parallel Axis Theorem, moment of inertia of this rod about the asked axis is

M = dm = 2π k x 3 dx

⇒

y

I2 =

2

m 2 m 2 ⎛  ⎞ + m⎜ = ⎟ ⎝2 3⎠ 12 6

…(2)

x

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 149

x + dx

2/9/2021 6:43:00 PM

H.150

JEE Advanced Physics: Mechanics – II

So, moment of inertia of the system is ⎛ m 2 ⎞ I = 3I 2 = 3 ⎜ ⎝ 6 ⎟⎠ ⇒

I=

m 2 2

Since, I = ( Mtotal ) K 2 = ⇒

( 3m ) K 2 =

⇒

K=

m 2 2

m 2 2

 6

Test Your Concepts-II (Based on Rotational Kinematics, Combined Effect of Rotation and Translation Motion) 1.

Angular velocity ω = ⇒

The acceleration of point 1 is    a1 = a1C + aC

dθ d ( 4t − 3t 2 + t 3 ) = dt dt

   But a1C = ( a1C )tangential + ( a1C )radial

ω = 4 − 6t + 3t 2

(a) At t = 2 s

ω = 4 − ( 6 )( 2 ) + 3 ( 2 ) ⇒

ω = 4 rads

2

−1

⇒

2

⇒ (c)

28 − 4 4−2

  where, a3C = − ( Rω 2 ) iˆ − ( Rα ) ˆj and aC = aiˆ

Instantaneous angular acceleration is,

⇒

 a2 = ( a + Rα ) iˆ − ( Rω 2 ) ˆj

Acceleration for the point 3 is    a3 = a3C + aC

α av = 12 rads −2

α=

2.

t f − ti

=

 a1 = ( a + Rω 2 ) iˆ + ( Rα ) ˆj

⇒

(b) Average angular acceleration

ω f − ωi

⇒

  where, a2C = ( Rα ) iˆ − ( Rω 2 ) ˆj and aC = aiˆ

ω = 28 rads −1

α av =

  a1C = ( Rα ) ˆj + ( Rω 2 ) iˆ and aC = aiˆ

Similarly, the acceleration of point 2 is    a2 = a2C + aC

At t = 4 s

ω = 4 − ( 6 )( 4 ) + 3 ( 4 )

⇒

⇒

dω d ( = 4 − 6t + 3t 2 ) dt dt

 a3 = ( a − Rω 2 ) iˆ − ( Rα ) ˆj

Acceleration for the point 4 is    a4 = a4C + aC

α = −6 + 6t

At t = 2 s, α = −6 + ( 6 ) ( 2 ) = 6 rads −2

  where, a4C = − ( Rα ) iˆ + ( Rω 2 ) ˆj and aC = aiˆ

At t = 4 s, α = −6 + ( 6 ) ( 4 ) = 18 rads −2

⇒

The motion of disc is both translational and rotational, so the acceleration of any point P on the disc can be expressed by    aP = aPC + aC For all the specified points 1, 2, 3 and 4, the accelerations are shown in Figure.

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 150

3.

 a4 = ( a − Rα ) iˆ + ( Rω 2 ) ˆj

For particle at P, r = CP = 1 m ⇒

α=

dω d = ( 2t ) = 2 rads −2 dt dt

At t = 1 s, ω = 2 rads −1 and α = 2 rads −2 So, at = rα = 2 ms −2 , ar = rω 2 = 4 ms −2, a = 2 ms −2

2/9/2021 6:43:19 PM

Hints and Explanations

H.151

In order to calculate the radial acceleration, we first need to find the angular velocity at the given time. So,

Net acceleration of P is the vector sum of three terms a, ar and at as shown in Figure below.

ω = ω 0 + α t = 0 + ( 60 rads −2 ) ( 0.15 s ) ⇒

ω = 9 rads −1

Since aC = rω 2, so aC = ω 2 r = ( 81 rad2s −2 ) ( 0.2 m ) = 16.2 ms −2

iˆ

( −4 sin 37°iˆ − 4 cos 37° ˆj )

⇒

 aP = 2iˆ + 1.6iˆ − 1.2 ˆj − 2.4iˆ − 3.2 ˆj

a = aC2 + aT2 = 20.2 ms −2

(

1 1 (b) Since θ = α t 2 = 60 rads −2 2 2 ⇒ θ = 1.88 rad

 aP = ( 1.2iˆ − 4.4 ˆj ) ms −2

Here α = π rads

⇒

−2

ω0 = 0

7.

t=4s (a) ω = 0 + ( π rads −2 ) × 4 s = 4π rads −1

5.

n ( 2π ) rad = 8π rad

⇒

n=4

n=

)( 0.25 s )

2

θ 1.88 = = 0.3 rev 2π 2π

 Given that ω 0 = 0.2 rads −1, R = 10 cm = 0.1 m . Since r rotates relative to O with constant angular velocity ω 0 . So angular velocity relative to C is

ω c = 2ω 0 = 0.4 rads −1  Modulus of velocity v = Rω c = ( 0.1 )( 0.4 )  ⇒ v = 0.04 ms −1 = 4 cms −1 Modulus of total acceleration is  a = Rω c2

1 (b) θ = ω 0t + ( π rads −2 ) × ( 16 s 2 ) = 8π rad 2 (c) Let the number of turns be n ⇒

CHAPTER 3

The magnitude of the net linear acceleration is

)

 aP = 2iˆ + 2 cos 37°iˆ − 2 sin 37° ˆj +

iˆ ⇒ 4.

(

⇒

The motion of the rod can be considered as super position of pure translation and pure rotation.

⇒ ⇒

 2 a = ( 0.1 )( 0.4 ) = 0.016 ms −2  a = 1.6 cms −2

The direction of its total acceleration (centripetal acceleration) will be towards centre C. 8.

Let the angular acceleration of the bobbin is α as shown in Figure.

As end A, the rod slide along horizontal surface, the component of its velocity normal to ground should be zero. So, we get

ωl cos θ = 0 2 ωl vC = cos θ 2 − vC +

⇒ 6.

(a) The tangential acceleration is constant and given by aT = α r = ( 60 rads −2 ) ( 0.2 m ) = 12 ms −2

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 151

Point B is directly connected with roof through string. It means net acceleration of point B should be zero.

2/9/2021 6:43:34 PM

H.152

JEE Advanced Physics: Mechanics – II    aB = aBC + aC

⇒

0 = −α r + A

⇒

α=

where, ω =

A r

lω ⎛ l ⎞ ⎛ 2v ⎞ =⎜ ⎟⎜ ⎟ =v 2 ⎝ 2⎠ ⎝ l ⎠ So, the resultant velocity of A is given by vAC =

…(1)

Similarly, point A on the bobbin is directly connected with block. This means that the acceleration of point A should be same as the acceleration of the block. So, we have    aA = aAC + aC ⇒

a = αr + A

2 vA = vAC + vC2 + 2vAC vC cos θ

where, θ = 90° + 30° = 120° ⇒

⎛ vAC sin θ ⎞ ϕ = tan −1 ⎜ ⎝ vAC cos θ + vC ⎟⎠

⎛ A⎞ a= ⎜ ⎟R+ A ⎝ r⎠

9.

⇒

A r = a R+r

At time t the bottommost point will rotate an angle θ = ω 0t with respect to the centre of the disc C. The centre C will travel a distance s = v0t .

R

ω0

P

C

θ

⇒

Net velocity of P is the vector sum of v and rω as shown in Figure.

v0 x

s = v0t

(

 vP = 2iˆ + 5 cos 53°iˆ − 5 sin 53° ˆj

From the figure, PQ = R sin θ = R sin ( ω 0t ) and CQ = R cos θ = R cos ( ω 0t )

Coordinates of point P at time t are x = OM − PQ = v0t − R sin ω 0t and y = CM − CQ = R − R cos ω 0t So, P ( x , y ) ≡ ( v0t − R sin ( ω 0t ) , R − R cos ( ω 0t ) ) 10. The point A is translating as well as rotating, so the velocity of the end A is the combined effect of rotation and translation. The combined velocity of end A is given by    vA = vAC + vC

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 152

rω = ( 1 ) ( 5 ) = 5 ms −1

Q

M

O

⎛ v sin 120° ⎞ π ϕ = tan −1 ⎜ = ⎝ v cos 120° + v ⎟⎠ 3

11. For particle P, r = CP = 1 m

y

C

vA = v 2 + v 2 + 2v 2 cos 120° = v

  If vA makes angle ϕ with the direction of vC , then

…(2)

From equations (1) and (2), we get

⇒

2v , vC = vcm = v and l

⇒

(

)

)

 vP = 2iˆ + 3iˆ − 4 ˆj = 5iˆ − 4 ˆj ms −1

Test Your Concepts-III (Based on Instantaneous Axis of Rotation, Pure Rolling and Conservation of Energy) 1.

For pure rolling, we have KR k 2 2 = = KT r 2 5 So, translational kinetic energy at topmost point is 5 KT = mg ( h − 2R ) 7 1 5 ⇒ mv 2 = mg ( h − 2R ) 2 7 10 ⇒ v2 = g ( h − 2R ) …(1) 7 For the ball not to leave the track, we have mv 2 mg = …(2) R

2/9/2021 6:43:50 PM

Hints and Explanations

h = 2.7 R In case of frictionless track, there will be no rotational kinetic energy, so 1 mv 2 = mg ( h − 2R ) 2 ⇒ v 2 = 2 g ( h − 2R ) …(3) Solving equation (2) and (3), we get

4.

⇒

K system =

⇒

K system =

⇒

K system

mv 2 = mg cos θ R+r

If m be the mass of the ball, then total kinetic energy at B is EB = mgh

KR r 2 2 = = KT k 2 5 KR =

r

2 5 mgh and KT = mgh 7 7

mg

mgh =

mgH = KT ⇒

mgH =

⇒

H=

5 mgh 7

where Ι =

Since speed of rod is v, so speed of centre of mass of the cylinder and the sphere is v 2. For a body rolling without slipping, the kinetic energy is R2 ⎞ 1 2 ⎛ K = KT + K R = mvcm ⎜⎝ 1 + 2 ⎟⎠ 2 K So, for the rolling sphere and the cylinder, we have K cylinder =

1 ⎛ m2 ⎜ 2 ⎝

2

v⎞ ⎛ ⎟ ⎜ 1+ 2⎠ ⎝

1⎞ 3 m2v 2 ⎟= 2 ⎠ 16

2

2⎞ 7 1 ⎛ v⎞ ⎛ m1 ⎜ ⎟ ⎜ 1 + ⎟ = m1v 2 2 ⎝ 2⎠ ⎝ 5 ⎠ 40 The motion of the rod is purely translational, so kinetic energy of rod is 1 K rod = mv 2 2 The total kinetic energy of the system is K system = K rod + K cylinder + K sphere ⇒

K system =

mv 2 Ιω 2 + 2 2

2 2 mr , v = rω 5

and h = ( R + r ) ( 1 − cos θ ) From these equations we get

5 h 7

and K sphere =

h R θ θ

In portion BC , friction is absent, so rotational kinetic energy will remain constant and translational kinetic energy will get converted to potential energy. So, if H be the height to which ball climbs in BC , then

3.

…(1)

where v is the speed of the centre of the sphere at that moment and θ is the corresponding angle. The speed v can be found by using the Law of Conservation of Energy, according to which

The ratio of rotational to translational kinetic energy would be,

⇒

1 3 7 mv 2 + mv 2 + mv 2 2 4 8 3 7 ⎞ 17 1 2⎛ = mv ⎜ 1 + + ⎟ = mv 2 ⎝ 2 2 4⎠ 8

The equation of motion for the centre of the sphere at the moment of breaking off, N = 0 is

h = 2.5R 2.

7 1 3 mv 2 + ( 4 m ) v 2 + ( 5m ) v 2 2 16 40

CHAPTER 3

Solving, equation (1) and (2), we have

H.153

1 3 7 mv 2 + m2v 2 + m1v 2 2 16 40

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 153

ω= 5.

10 g ( R + r ) 17 r 2

When the bead comes to its lowest position, then change in gravitational potential energy of the bead is ΔU = − mgh = − mg ( 2R ) = −2mgR

…(1)

Similarly, change in kinetic energy will take place both for the bead and the disc, so we have ΔK = ΔK bead + ΔKdisc ⇒

ΔK =

1 ⎛ mR2 ⎞ 2 1 mv 2 + ⎜ ⎟ω 2 2⎝ 2 ⎠

v R mv 2 mv 2 ⇒ ΔK = + 2 4 3 ⇒ ΔK = mv 2 4 By Law of Conservation of Energy, we have ΔU + ΔK = 0 Since ω =

…(2)

2/9/2021 6:44:06 PM

H.154

6.

JEE Advanced Physics: Mechanics – II

⇒

−2mgR +

⇒

v=

Angular velocity of the rod can also be directly calculated as

3 mv 2 = 0 4

Angular velocity will be maximum, when the rod is in the vertical position. Then by Law of Conservation of Energy, we have

⇒

⎛ Loss in Gravitational ⎞ ⎛ Gain in Rotational ⎞ ⎜ Potential Energy of ⎟ ⎜ Kinetic Energy ⎟ ⎟ ⎜ Centre of Mass of ⎟ = ⎜ of the ⎟ ⎜ ⎟ ⎜ Rod th he Rod ⎠ ⎝ ⎠ ⎝ ω max

7.

2

1 ⎛ ml 2 ⎞ ⎛ 3v ⎞ 1 ⎛ v⎞ m⎜ ⎟ + ⎜ ⎟⎜ ⎟ 2 ⎝ 2⎠ 2 ⎝ 12 ⎠ ⎝ l ⎠

2

1 mv 2 2 Alternatively, we can calculate angular velocity of the rod and velocity of centre of mass by IC method. Let the IC of the rod be at a distance x from the centre of mass of the rod as shown in Figure. ⇒

2

K=

2 = 2 2

⇒

mg (  ) =

1 ( m ) ( 2 ) 2 ω max 2 3

⇒

ω max =

3g 2

By Law of Conservation of Energy, we have ⎛ Loss in Gravitational ⎞ ⎛ Gain in Rotational ⎞ ⎜ Potential Energy ⎟ = ⎜ Kinetic Energy of ⎟ ⎜ of the point mass ⎟ ⎜ po int mass + Disc ⎟ ⎠ ⎝ ⎠ ⎝

8.

K=

1 2 Ιω max 2

mgh = where, h =

vrel 3v = r⊥ l Since, we know that the kinetic energy 1 1 2 K = KT + K R = mvcm + I cmω 2 2 2

ω=

8 gR 3

⇒

mg ( R ) =

⇒

ω=

1⎛ 1 2 2⎞ 2 ⎜ MR + mR ⎟⎠ ω 2⎝ 2 4 mg

( 2m + M ) R

If vcm be the velocity of the centre of mass of the rod, then for end A we have ⎛ l⎞ vA = v = ⎜ ⎟ ω − vcm ⎝ 2⎠

…(1)

Similarly, for end B, we have ⎛ vB = 2v = vcm + ⎜ ⎝

l⎞ ⎟ω 2⎠

Since the rod will be in pure rotation about the IC so for end A, we have ⎛ l ⎞ v = ⎜ − x⎟ ω …(3) ⎝2 ⎠ and for end B, we have ⎛ l ⎞ 2v = ⎜ + x ⎟ ω …(4) ⎝2 ⎠ From equations (3) and (4), we get l 3v x = and ω = 6 l Also, the velocity of centre of mass is given by ⎛ l ⎞ ⎛ 3v ⎞ v vC = xω = ⎜ ⎟ ⎜ = ⎝ 6 ⎠ ⎝ l ⎟⎠ 2 Rotational kinetic energy of the rod is KR =

Translational kinetic energy of the rod is

…(2)

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 154

2

1 1 ⎛ v⎞ 1 2 mvcm = m ⎜ ⎟ = mv 2 2 2 ⎝ 2⎠ 8 Total kinetic energy of the rod is KT =

Solving Equations (1) and (2), we have v 3v ω= and vC = 2 l

2 3 1 1 ⎛ ml 2 ⎞ ⎛ 3v ⎞ I cmω 2 = ⎜ = mv 2 ⎟ ⎜ ⎟ 8 2 2 ⎝ 12 ⎠ ⎝ l ⎠

K = K R + KT ⇒

K=

3 1 1 mv 2 + mv 2 = mv 2 8 8 2

2/9/2021 6:44:18 PM

Hints and Explanations Total kinetic energy of the rod also equals the rotational kinetic energy of the rod about the instantaneous centre of zero velocity (IC) and is given by

11. By Law of Conservation of Energy, we get 1 2 1 1 kx = mv 2 + Ιω 02 2 2 2

)

1 1 I ICω 2 = I cm + mx 2 ω 2 2 2

⇒

K=

1 ⎡ ml 2 ⎛ + m⎜ ⎢ ⎝ 2 ⎣ 12

⇒

K=

1 mv 2 2

Since v = Rω 0 and Ι =

2 2 l ⎞ ⎤ ⎛ 3v ⎞ ⎟⎠ ⎟⎠ ⎥ ⎜⎝ 6 ⎦ l

(

v 3 ga 3 g ω= = = 2a 2a 2 a

ω a m

ω

kx 2 =

⇒

x = Rω 0

m

h 2

⎛ Work done by ⎞ ⎛ Change in KE of ⎞ ⎜⎝ Applied Force ⎟⎠ = ⎜⎝ Cradle and Disks ⎟⎠

h

⇒

a

m ZPEL

By Law of Conservation of Mechanical Energy, we have ⎛ Loss in ⎞ ⎛ Gain in ⎞ ⎛ Gain in ⎞ ⎜ RKE of ⎟ = ⎜ GPE of ⎟ + ⎜ GPE of ⎟ ⎜ System ⎟ ⎜ Particle ⎟ ⎜ CM of Ring ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2

1 ( m ( 2a )2 + 2ma2 ) ⎛⎜⎝ 3 g ⎞⎟⎠ = mg ( 2a + h ) + 2 2 a h⎞ ⎛ mg ⎜ a + ⎟ ⎝ 2⎠

⇒ ⇒

3 mgh 27 mga = 3 mga + 4 2 h = 2.5 a

10. mgh = K R + KT =

3 mv 2 4

where h = s sin θ ⇒

gs sin θ =

⇒

s=

⇒

7.5 = 13.5v 2

⇒

v = 0.745 ms −1

Test Your Concepts-IV (Based on Torque and Applications) 1.

Decrease in the height of centre of mass of the rod  h = ( cos 37° − cos 60° ) 2  ( 0.8 − 0.5 ) = 0.15 2 By Law of Conservation of Mechanical Energy, we get ⇒

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 155

h=

mgh =

1 1 ⎛ m 2 ⎞ 2 IOω 2 = ⎜ ⎟ω 2 2⎝ 3 ⎠

⇒

⎛ m 2 ⎞ 2 0.15mg = ⎜ ω ⎝ 6 ⎟⎠

⇒

an = ω 2 = 0.9 g

…(1)

Angular acceleration, ⎞ ⎛ mg ⎟ sin ( 37° ) ⎠ τ O ⎜⎝ 2 α= = 1 2 ΙO m 3

3 2 v 4

( 3 )( 4 ) 3v 2 = = 0.6 m 4 g sin θ ( ) ( ) ⎛ 1 ⎞ 4 10 ⎜ ⎟ ⎝ 2⎠

1 2

( 30 )( 0.25 ) = ( 9 ) ( v 2 ) + 2 ⎛1 1⎛ 1 ⎞v ⎞ 2 ⎜ ( 6 ) ( v2 ) + ⎜ × 6 × r 2 ⎟ 2 ⎟ ⎠r ⎠ ⎝2 2⎝ 2

2a

B

⇒

3m 2k

12. In rolling without sliding on a stationary ground, work done by friction is zero. Hence

CM a

3 mR2ω 02 2

⇒

m CM

)

⎛1 ⎞ kx 2 = m R2ω 02 + ⎜ mR2 ⎟ ω 02 ⎝2 ⎠

The initial angular velocity imparted to the system is

B

1 mR2 , so we get 2

CHAPTER 3

9.

(

K=

H.155

⇒

at = α = 0.9 g

…(2)

2/9/2021 6:44:30 PM

H.156

JEE Advanced Physics: Mechanics – II for pulley, we get

O

α=

60° 37°

⇒

Fn

Solving the above equations, we have a = 1.87 ms −2 , T1 = 233.5 N and

Tangential force, Ft = ( dm ) α = 0.9 gdm = 9dm

T2 = 238.2 N

τ = F cos θ Since, α =

τ F cos θ 3 F cos θ = = I m m 2 3 4.

θ

⇒

dω 3 F cos θ ω = dθ m

⇒

ω dω = ω

⇒

∫ 0

⇒ 3.

3 F cos θ dθ m ϕ

3F ω dω = cos θ dθ m

ω=

∫ 0

6F sin ϕ m

⇒

ω=

v 2.73 = = 27.3 rads −1 R 0.1

⇒

t=

2s = a

2×2 = 1.47 s 1.87

At angle θ , we have by Law of Conservation of Energy

…(1)

…(2)

3g  ( 1 − cos θ ) an = ω 2 = 2 2  3 and at = rα = α = g sin θ 2 4 ⇒

T1

T2

20 kg

30 kg

20 g

v = 2 as = 2 × 1.87 × 2 = 2.73 ms −1

Since, an = rω 2

The free body diagrams of the bodies and the pulley are shown in Figure, then

a

⇒

 1 2 Ιω = mg ( 1 − cos θ ) 2 2 1 m 2 2 mg ( 1 − cos θ ) ω = ⇒ 2 3 2 3g ( 1 − cos θ ) ⇒ ω2 =  τ Since α = I ⎛ ⎞ mg ⎜ sin θ ⎟ ⎝2 ⎠ 3 g sin θ = ⇒ α= 2 m 2 3



α

…(3) …(4)

Normal force, Fn = ( dm ) ω 2 = 0.9 gdm = 9dm

So, net force, F = Fn2 + Ft2 = 9 2dm 2.

1 mR mR2 2 2 ( T2 − T1 ) 0.4 ( T2 − T1 ) α= = R 5R

For no slipping, a = Rα

dm

Ft

( T2 − T1 ) R = 2 ( T2 − T1 )

30 g

f = max = m ( at cos θ − an sin θ )

α a

T1

T2

⇒

3 g sin θ ⎛3 ( 1 − cos θ ) ⎞⎟ f = m ⎜ g sin θ cos θ − ⎝4 ⎠ 2

⇒

f =

3 ⎛3 ⎞ mg sin θ ⎜ cos θ − 1 ⎟ ⎝2 ⎠ 2

for 20 kg mass, we get T1 − 20 g = 20 a

…(1)

30 g − T2 = 30 a

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 156

c

θ

for 30 kg mass, we get …(2)

an

x

at y

2/9/2021 6:44:44 PM

Hints and Explanations

⇒

N = m ( g − ay )

⇒

N = m ⎡⎣ g − ( at sin θ + an cos θ ) ⎤⎦

⇒ ⇒ ⇒

3 g cos θ 3 ⎡ ( 1 − cos θ ) ⎤⎥ N = m ⎢ g − g sin 2 θ − 4 2 ⎣ ⎦ mg ( 4 − 3 sin 2 θ − 6 cosθ + 6 cos2 θ ) N= 4 mg ( 1 − 3 cos θ )2 N= 4

8.

m1 T1

T2

m1g

m2

( T2 − T1 ) R

1 mR2 2 For no slipping condition, we have a = Rα Solving these equations, we get

α= and 6.

2 ( m2 − m1 ) g 2 m ( 1 + 2m2 + m ) R

T1 m1 ( m + 4 m2 ) = T2 m2 ( m + 4 m1 )

TR 2T = 1 m 2 1R m1R 2

…(1) …(2) …(3)

⎛ 2m2 g sin θ ⎞ ⎛ m m g sin θ ⎞ a=⎜ and T = ⎜ 1 2 ⎝ 2m2 + m1 ⎟⎠ ⎝ 2m2 + m1 ⎟⎠ Total energy of system when m2 is at a height h is mgh. When m2 reaches the bottom total energy, due to energy conservation will still be mgh.

m2g

…(1)

However, this mgh gets distributed to pulley as ⎛ 1 ⎞ rotational kinetic energy ⎜ = Iω 2 ⎟ and to mass m2 ⎝ 2 ⎠ ⎛ 1 ⎞ as translation kinetic energy ⎜ = m2v 2 ⎟ , so speed at ⎝ 2 ⎠ bottom is

…(2)

…(3)

⎛ 4 m2 g sin θ ⎞ ⎛ h ⎞ v = 2 as = ⎜ ⎜ ⎟ ⎝ 2m2 + m1 ⎟⎠ ⎝ sin θ ⎠

…(4)

⇒ 9.

v=

2 gh 1 + ( m1 2m2 )

Let a be the acceleration of truck. The free body diagram of hemisphere in the frame of reference of the truck is shown in Figure.

(a) F cos θ = Mg sin θ ⇒ (b)

7.

Equations of motion are,

a

For cylinder

α=

2g a

Solving these equations, we get

For mass m2 m2 g − T2 = m2 a

4 g 2a a g

and a = Rα

For mass m1 T1 − m1 g = m1a

{∵  = 4a }

2a g

ω=4

α=

T2 a

t=

4a 1 2 2g 2 = αt = t a 2 a

m2 g sin θ − T = m2 a

Let α be the angular acceleration of the cylinder and a be the linear acceleration of two bodies T1

⇒

⇒

1⎞ So, N = 0 at θ = cos ⎜ ⎟ ⎝ 3⎠ Hence, it will certainly slip beyond that.

α

θ=

Since, ω = α t =

−1 ⎛

5.

⇒

CHAPTER 3

Further, mg − N = may

H.157

F = Mg tan θ

f =0

2mga 4 g = 1 2 a ma 2 Since  = rθ

α=

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 157

The centre of gravity is at G such that 3r OG = 8

2/9/2021 6:45:00 PM

H.158

JEE Advanced Physics: Mechanics – II

The hemisphere is in equilibrium in this frame of reference. Hence, net moment of all the forces about bottommost point must be zero. ⇒ ⇒

3r ⎛ 3r ⎞ ⎛ ⎞ mg ⎜ sin θ ⎟ = ma ⎜ r − cos θ ⎟ ⎝ 8 ⎠ ⎝ ⎠ 8 3 g sin θ a= 8 − 3 cos θ

τ 10. α = = I

12. α =

Since ω = ω 0 + α t ⇒

C

A α

B

⇒

mg

11. From the free body diagram, we get T T

α

a

Mg − T M TR 2T = 1 MR 2 MR 2

a = Rα Solving these three equations, we get

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 158

1 τθ = Ιω 2 2

⇒

( 40 × 0.25 ) ⎛⎜

⇒

3ω 2 = 200

⇒

ω = 8.16 rads −1

5 ⎞ 1 = × 4 × ω2 ⎝ 0.25 ⎟⎠ 2

⇒

mgh =

1 1 mv 2 + Ιω 2 2 2

⇒

mgh =

1 1 2 m ( Rω ) + Ιω 2 2 2

⇒

( 4 ) ( 10 )( 5 ) = 1 ( 4 ) ⎛⎜ 1 ⎞⎟ ω 2 + 1 ( 6 ) ω 2 2

⎝ 16 ⎠

2

2

ω + 3ω 2 8

⇒

200 =

⇒

ω 2 = 64

⇒

ω = 8 rads −1

The difference in answers is due to the fact that in the first case T = 40 N and in the second case T < 40 N .

Mg

2g Mg a= and T = 3 3

⇒

⎛ Loss in ⎞ ⎛ Gain in ⎞ ⎛ Gain in ⎞ ⎜ GPE of ⎟ = ⎜ KE of ⎟ + ⎜ RKE ⎟ ⎜ 4 kg mass ⎟ ⎜ 4 kg mass ⎟ ⎜ of pulley ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ 3g ⎞ mg − T = m ⎜ ⎝ 4 ⎟⎠ mg T= 4

α=

{∵ ω 0 = 0 }

40 = 4 kg 10 By Law of Conservation of Energy, we have

aB = α =

a=

32 g a

(b) Mass of the block is m =

3g 2 3g  (b) aC = α = 2 4 (c) Since mg − T = maC ⇒

ω = αt =

13. (a) W = ΔK

⎞ ⎟ 2⎠ = 3 g 2 m 2 3

⎛ mg ⎜ ⎝

T

(a)

τ 4 mga 8 g = = I 1 ma 2 a 2

14. Equations of motion for the discs are, …(1)

a=

…(2)

…(3)

mg − T m

α2 =

2T TR = 2 mR 2 mR

…(1) …(2)

Similarly, we get

α1 = ⇒

2T TR = mR2 2 mR

…(3)

α1 = α 2

2/9/2021 6:45:15 PM

Hints and Explanations

a

α2

T

⇒

2x 2 =

⇒

mg

For no slipping, a = Rα 2

…(4)

Further, a − Rα 2 = Rα 1

…(5)

Solving these equations, we get mg 2g 4g T= ,a= and α 2 = 5 5 5R 15. By Law of Conservation of Mechanical Energy, ⎛ Loss in ⎞ ⎛ Gain in ⎞ Loss in ⎞ ⎜ GPE of ⎟ ⎛ ⎜ RKE of ⎟ ⎜ CM of ⎟ + ⎜ GPE of ⎟ = ⎜ Ring and ⎟ ⎟ ⎜ ⎟ ⎜⎝ Particle ⎟⎠ ⎜ ⎝ Ring ⎠ ⎝ Particle ⎠ ⇒

2x 2 =1 k + x2

⇒

2

α max = α

⇒

ω=

 2 3

=

g 3 

1. a

α

a=

1 ⎛ 3 mr 2 2⎞ + m ( 2r ) ⎟ ω 2 ⎜⎝ ⎠ 2 2 3 3 mgr 11 2 2 = mr ω 2 4

x=

Test Your Concepts-V (Based on Uniform and Accelerated Pure Rolling)

mgr sin 60° + mg ( 2r sin 60° ) =

⇒

⎧ 2 2 ⎫ ⎨∵ k = ⎬ 12 ⎭ ⎩

2 + x2 12  x= 2 3

CHAPTER 3

B

α1

⇒

H.159

and α =

F − μ k Mg F = − μk g M M

τ ( F − μ k Mg ) R 2 ⎛ F ⎞ = = ⎜ − μk g ⎟ ⎝M ⎠ 1 Ι R 2 MR 2

2.

6g 3 11r

Since, τ = mgr cos ( 60° ) + mg 2r cos ( 60° ) ⇒ ⇒

⇒

τ=

3 mgr 2

mg sin θ − T …(1) m Tr α= …(2) Ι a = rα …(3) Solving these three equations, we get g sin θ a= Ι 1+ mr 2 NOTE: Compare the above result with rolling on a rough ground and think what are the similarities between the two. a=

3 mgr 2

τ = I ⎛ 3 mr 2 2⎞ + m ( 2r ) ⎟ ⎜⎝ ⎠ 2 3 mgr 2 3 g α= = 11mr 2 2 11r α=

16. Let us calculated the torque about O, so τ 0 = I 0α ⇒

⎛ M 2 ⎞ Mgx = ⎜ + Mx 2 ⎟ α ⎝ 12 ⎠

⇒

α=

3.

gx 2 , where k 2 = 2 2 12 k +x

α1

For α to be maximum, we have dα =0 dx ⇒

Let us first draw the free body diagrams for cylinder, pulley and mass. Then

R1

x ( −1 ) ( k + x 2

)

2 −2 (

2x ) + ( k + x

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 159

2

)

2 −1 (

1) = 0

M1

T1 a1 f

T2

α2 T1 M2

a2

R2 T2

5g

2/9/2021 6:45:29 PM

H.160

JEE Advanced Physics: Mechanics – II

for 5 kg mass, 5 g − T2 = 5 a2 for pulley,

( T2 − T1 ) R2 1 M2 R22 2

…(1)

= α2

…(2)

for cylinder, T1 + f = M1a1

( T1 − f ) R1 = α 1 M1R12 2

…(4)

1

…(5)

a2 = R2α 2 = a1 + R1α 1

…(6)

Solving above equations, we get 4 a2 = g 11 4 gt ⇒ v = a2 t = 11

⇒

0.8 F − μ s N ( R ) ( μ s N ) ( R ) 2 μ s N = = 1 M M MR2 2

⇒

0.8 F = 3 ( 0.4 ) ( Mg − 0.6 F )

⇒

F = 0.79 Mg

Therefore, maximum value of F is 0.79 Mg 6.

For thin walled lawn roller, we have

Ι 2 = = 0.4 for sphere mR2 5 Ι 1 = = 0.5 for disc and = 1 for hoop 2 2 mR 2 s= =4m sin ( 30° ) For sphere, we have

Ι = MR2

g sin θ a= = Ι 1+ mR2

a

f

F = MR2

⇒

For pure rolling to take place, we have a = Rα Also F − f = Ma

f ⎛ fR ⎞ a = R⎜ = ⎝ MR2 ⎟⎠ M F ⇒ f = 2 F− f F and a = = M 2M 5.

( 9.8 ) ⎛⎜ 1 ⎞⎟

⎝ 2⎠ = 3.5 ms −1 1 + 0.4

v = 2 as = 2 × 3.5 × 4 = 5.29 ms −1 1 mg sin θ 3 × 9.8 × 2 f = = = 4.2 N ⎛ 1 ⎞ mR2 1+ ⎜ ⎟ 1+ ⎝ 0.4 ⎠ Ι

2s 2×4 = = 1.51 s a 3.5 Similarly, the values for disc and hoop can be obtained. t=

⎛τ⎞ where a = Rα = R ⎜ ⎟ ⎝ Ι⎠ ⇒

3F = Mg − 0.6 F 5

For pure rolling a = Rα

…(3)

Also, a1 = R1α 1

4.

N = Mg −

7.

Force of friction in this case will be backwards. For pure rolling, a = Rα ⇒

1 ⎛ μ MgR − Fb ⎞ F − μ Mg ) = R ( ⎜ 1 ⎟ M MR2 ⎟ ⎜⎝ ⎠ 2

The free body diagram for cylindrical wheel is shown in Figure (b).

Solving this equation, we get 3 μ MgR F= R + 2b Thus, maximum value of F can be 3 μ MgR Fmax = R + 2b

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 160

2/9/2021 6:45:43 PM

H.161

Hints and Explanations 8.

The free body diagrams for plank and cylinder are shown in Figure. Also

for plank, we have F − T − f = Ma

…(1)

for cylinder, we have T − f = ma fR 2f = 1 mR 2 mR 2 for no slipping at A, we have

α=

…(1)

for cylinder, we have f1 − f 2 = Ma2

…(2)

( f + f )R α= 1 2

…(3)

1 MR2 2 a2 = Rα

Rα − a = a ⇒ Rα = 2 a Solving these equations, we get a=

…(4)

and a1 = a2 + Rα

…(5)

9.

F 3m + M

h = ( R − r ) ( 1 − cos θ )

…(1)

By Law of Conservation of Energy, we have

4 F cos θ 3 MF cos θ , f1 = 3 M + 8m 3 M + 8m

⎛ Loss in KE of ⎞ ⎛ Gain in ⎞ = ⎜⎝ Rolling ⎟⎠ ⎜⎝ GPE ⎟⎠

MF cos θ 3 M + 8m

Equation of motion for the block is mg sin θ − T = ma

⇒

2 1 ⎛1 2 2⎞⎛ k ⎞ ⎜⎝ mv0 − mv ⎟⎠ ⎜⎝ 2 ⎟⎠ = mgh 2 2 R

⇒ ( 1 ) ( 9.8 ) sin 30° − T = ( 1 ) a ⇒ T + a = 4.9 Equation of motion for disk is TR =α 1 MR2 2 2T ⇒ α= MR For no slipping, we have 2T Rα = a = M 2T ⇒ a= 5 ⇒ T = 2.5 a Substituting in equation (1), we get 3.5 a = 4.9

⇒

1 ⎛1 2 2⎞⎛ ⎜⎝ mv0 − mv ⎟⎠ ⎜⎝ 1 + 2 2

⇒

…(1) ⇒ ⇒

R

Equation of motion at angle θ is, N − mg cos θ = N=

m

⎛

( R − r ) ⎜⎝

mv 2 (R − r)

v02 −

10 ⎞ gh ⎟ + mg cos θ ⎠ 7

Substitute value of h from (1), we get ⎞ ⎛ m ⎞ ⎛ 2 10 g ( N=⎜ v − R − r ) ( 1 − cos θ ) ⎟ + ⎠ ⎝ R − r ⎟⎠ ⎜⎝ 0 7

α

mg cos θ

f T

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 161

θ

mg cos θ

10. Free body diagram for cylinder and plank are shown below, then

a

N

h

⇒

f

2⎞ ⎟ = mgh 5⎠

1 1 5 mv02 − mv 2 = mgh 2 2 7 10 v 2 = v02 − mgh 7

v

a = 1.4 ms −2

A

…(4)

11. At P, we have

Solving these equations, we get

and f 2 =

…(3)

CHAPTER 3

for plank, we have F cos θ − f1 = ma1

a2 =

…(2)

a

⇒

Fn = N =

mg mv02 ( 17 cos θ − 10 ) + (R − r) 7

2/9/2021 6:45:58 PM

H.162

JEE Advanced Physics: Mechanics – II

Tangential force is the force of friction given by f = ⇒

mg sin θ

(

1 + k2 r2

)

mvL 9mL2 = ω 12 24

⇒

mg sin θ = 1+ (5 2)

2v 9L (b) The desired ratio is,

ω=

⇒

2 Ft = f = mg sin θ 7

Test Your Concepts-VI (Based on Angular Momentum and its Conservation) 1.

K system K bullet

Let v0 be the initial velocity and v its velocity in the π position when the thread forms an angle with the 2 vertical. The ball experiences two forces, the gravitational force and the tension of the thread. Moment of these forces is zero relative to the vertical axis passing through the point O. Consequently, the angular momentum of the ball relative to the given axis is constant. So, we have mv0 sin θ = mv

3.

⇒ 4.

The ball moves in the Earth’s gravitational field under the influence of an external force, the tension of the thread. That force is always perpendicular to the velocity vector of the sphere and, therefore, does not perform any work. It follows that mechanical energy of the ball remains constant

v0 = 2 g sec θ 2.

(a) Applying the Law of Conservation of Angular Momentum about the pivot, we get

∑ mvr

⊥

1 9

ω2 =

I1ω 1 I2

where, I1 = I , ω 1 = ω 0 , I 2 = I + mR2

θ

Solving these two equations, we get

=

Since, net external torque on the system is zero. Therefore, angular momentum will remain conserved. So,

⇒

O

…(2)

K bullet

I1ω 1 = I 2ω 2

…(1)

mv02 mv 2 = + mg cos θ 2 2

K system

⇒

⎛ 9mL2 ⎞ ⎛ 2v ⎞ 2 1 2 Iω ⎜⎝ ⎟⎜ ⎟ 24 ⎠ ⎝ 9L ⎠ = 2 = 1⎛ m⎞ 2 ⎛ mv 2 ⎞ ⎜ ⎟v ⎜⎝ ⎟ 2⎝ 6 ⎠ 6 ⎠

5.

ω2 =

Iω 0 I + mR2

By Law of Conservation of Angular Momentum about centre of disc, we have I1ω 1 = I 2ω 2 ⇒

1 2 ⎡ ma ⎢ ⎛ I1 ⎞ 2 ω 2 = ⎜ ⎟ ω1 = ⎢ ⎝ I2 ⎠ ⎢ 1 ma 2 + 2m ⎛ ⎜⎝ ⎢⎣ 2

⇒

ω2 =

⎤ ⎥ ω 2 ⎥ a⎞ ⎥ ⎟⎠ ⎥ 2 ⎦

ω 2

Applying Law of Conservation of Angular Momentum about axis of rotation, we get mvr = ( mrω ) r + Iω where, I =

1 ( 2m ) r 2 2

= ( I system ) ω

2 2 ⎛ L⎞ ⎡m L ⎛ L⎞ ⎤ mbullet v ⎜ ⎟ = ⎢ rod + mbullet ⎜ ⎟ ⎥ ω ⎝ 2⎠ ⎣ 3 ⎝ 2⎠ ⎦ If m be the mass of the rod, then m m mbullet = rod = 6 6 2 2 ⎛ m ⎞ ⎛ L ⎞ ⎛ mL mL ⎞ ⇒ ⎜ ⎟ v⎜ ⎟ = ⎜ + ⎟ω ⎝ 6 ⎠ ⎝ 2⎠ ⎝ 3 24 ⎠

⇒

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 162

⇒

( 4m

)

1 gr r = ( mrω ) r + ( 2m ) r 2ω = 2mr 2ω 2

2/9/2021 6:46:10 PM

Hints and Explanations

ω=2

Since friction is absent, so, mechanical energy will remain conserved

g r

⇒

Considering the sudden change in the linear momentum of the particle caused by the impulsive tension J , we have, by

v0 = 2 gr sec θ 0 ⇒

Impulse = Change in Momentum

6.

(

J = − mrω − −4 m gr

⇒

J = 2m gr

)

8.

⇒

⇒

2m ⎞ ⎛ ω2 = ⎜ 1 + ⎟ ω0 ⎝ M⎠

(b) W = E f − Ei = ⇒

⇒ 7.

2 gr sec θ 0

J  = ( I system ) ω =

I1ω 1 = I 2ω 2

⇒

( v0 )min =

According to Angular Impulse − Angular Momentum Theorem, we have

(a) By Law of Conservation of Angular Momentum, we get

⎛1 2 2⎞ ⎜⎝ MR + mR ⎠⎟ ⎛ I1 ⎞ 2 ω2 = ⎜ ⎟ ω0 = ω0 ⎛1 ⎝ I2 ⎠ 2⎞ ⎜⎝ MR ⎟⎠ 2

…(2)

Solving equation (1) and (2), we get

Impulse − Momentum Theorem, ⇒

v 2 = v02 − 2 gh

9.

J=

2

m ( 2 ) ω 3

4 mω 3

By Law of Conservation of Angular Momentum, we have Li = L f {about bottom most point}

⇒

1 1 I 2ω 22 − I1ω 12 2 2

⇒

2m ⎞ 2 1⎛ 1 ⎞⎛ W = ⎜ MR2 ⎟ ⎜ 1 + ⎟ ω0 − ⎝ ⎠ ⎝ 2 2 M⎠

Iω 0 = 2 ( Iω + mRv )

⇒

1⎛ 1 2 2⎞ 2 ⎜ MR + mR ⎟⎠ ω 0 2⎝ 2

⎛1 ⎛1 ⎞ 2⎞ 2 ⎜⎝ mR ⎟⎠ ω 0 = 2 ⎜⎝ mR ω + mR ( ω R ) ⎟⎠ 2 2

⇒

ω=

⇒

v = ωR =

2

W=

1 2m ⎞ ⎛ mω 02 R2 ⎜ 1 + ⎟ ⎝ M⎠ 2

The following two forces are acting on the particle. (i) Normal Reaction ( N ) (ii) Weight ( mg )

ω0 6

∑ mvr

⊥

θ0

N

⇒

h = rcos θ0

mg

The torque due to the normal reaction N about centre O is zero. The torque due to the weight is not zero, but its component along a vertical axis passing through O is zero. Hence, angular momentum of the particle about the same axis will remain conserved. So, we have mv0 r sin θ 0 = mvr

…(1)

where, v is the speed of the particle at the topmost position.

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 163

ω0R 6

10. Applying Law of Conservation of Angular Momentum, Angular Velocity of disc is

O r

CHAPTER 3

⇒

H.163

= ( I system ) ω

⎛1 ⎞ m1vR = ⎜ m2 R2 + m1R2 ⎟ ω ⎝2 ⎠

m1R ⎛ 2m1 ⎞ ⎛ v ⎞ v=⎜ ⎜ ⎟ 1 2 2 ⎝ m2 + 2m1 ⎟⎠ ⎝ R ⎠ m2 R + m1R 2 where v is the relative speed of man with respect to disc. v So, = ω r (the relative angular speed of man) R 2m1 ⇒ ω= ⋅ω ( m2 + 2m1 ) r ⇒

ω=

⇒

dϕ ⎛ 2m1 ⎞ dθ = dt ⎜⎝ m2 + 2m1 ⎟⎠ dt

2/9/2021 6:46:25 PM

H.164

JEE Advanced Physics: Mechanics – II 13. Linear momentum, angular momentum and kinetic energy are conserved in the process. Applying Law of Conservation of Linear Momentum, we get

Integrating we get, ⎛ 2m1 ⎞ ϕ=⎜ θ ⎝ m2 + 2m1 ⎟⎠ 11. Applying conservation of angular momentum, we get ⎛v ⎞ Mv1R = I ⎜ 2 ⎟ + Mv2 R ⎝ R⎠ ⇒

v2 =

⇒

ω=

⇒

v1 1+

I ⎞ ⎛ R⎜ 1 + ⎟ ⎝ MR2 ⎠

J = mv0

…(1)

Also, on applying Angular Impulse − Angular Momentum theorem, we get m 2 ⎞ ω ⎟⎠ = Iω = 2 12

…(1)

…(2)

6v0 

⎛ M 2 ⎞ mvd = ⎜ ω ⎝ 12 ⎟⎠

⇒

⎛ 12mvd ⎞ ω=⎜ ⎝ M 2 ⎟⎠

⇒

v0

…(2)

⎛ Relative Speed ⎞ ⎛ Relative Speed ⎞ ⎜⎝ of Approach ⎟⎠ = ⎜⎝ of Separation ⎟⎠

⇒ v = V + ωd Substituting the values, we get v=

12mvd 2 m v+ M M 2

m=

M 2 12d 2 +  2

A

ω

)ω

Since the collision is elastic, so e = 1

⇒

C

system

⇒

Solving these two equations, we get

ω=

m v M

∑ mvr = ( I

v1

12. Let J be the linear impulse applied at the end B and ω the angular speed of rod, then by Impulse − Momentum Theorem, we have

⎛ J⎜ ⎝

V=

Applying Law of Conservation of Angular Momentum about the centre of the rod, we get

I MR2

v2 = R

MV = mv

14. Let ω be the angular velocity of rod just after the impulse is applied. Applying Angular Impulse Angular Momentum Theorem, we get A a

D O J

B

a J

C

Linear speed of D (mid-point of CB i.e., centre of mass of lower half) relative to C is

2a

⎛ ⎞ 3 v = ω ⎜ ⎟ = v0 ⎝ 4⎠ 2 Force exerted by upper half on the lower half,

B

⎛ m ( 4 a )2 ⎞ Ja = Iω = ⎜ + m ( a2 ) ⎟ ω ⎝ 12 ⎠

mv 2 2 F= l4 3 Substituting v = v0 , we get 2 9mv02 F= 2

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 164

⇒

ω=

3J 7 ma

For describing the complete revolution 1 2 Iω > mgh, where h = 2 a 2

2/9/2021 6:46:37 PM

Hints and Explanations ⇒

2

1 ⎛ 7 2 ⎞ ⎛ 3J ⎞ ⎜ ma ⎟⎠ ⎜⎝ ⎟ > mg ( 2 a ) 2⎝ 3 7 ma ⎠

⇒

7 ga J > 2m 3

( ω 0 − αt ) r = 2 ( a1t )

Substituting the values, we get 2rω 0 t= 9μ g

15. In the process of motion of the given system the kinetic energy and the angular momentum relative to the axis of rotation do not vary. So, by Law of Conservation of Energy, we have 1 2 1 2 1 Iω 0 = Iω + mv 2 2 2 2

⇒ 2.

s=

1 2 2ω 02 r 2 a2 t = 2 81μ g

Initially there is forward slipping so, the friction is backwards and maximum. v0

v vr

O

⇒

r

μ mg

m

Iω 02 = Iω 2 + mv 2

Let velocity becomes zero in time t1 and angular velocity becomes zero in time t2 . Then, 0 = v0 − at1 v v ⇒ t1 = 0 = 0 …(1) a μg

…(1)

By Law of Conservation of Angular Momentum, we have Iω 0 = ( I + mr 2 ) ω

where v 2 = vr2 + ω 02 r 2

…(2)

and 0 = ω 0 − α t2

…(3)

⇒

t2 =

Solving these equations, we get

ω0r

vr =

1+

⇒

Test Your Concepts-VII (Based on Rolling with Slipping) 1.

μmg a1 = a2 = = μg m 5μ g μmg α= = 2 2r 2mr 5

ω0 μmg

3.

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 165

ω0R 2μ g

…(2)

⇒

ω 0 R v0 > 2μ g μ g

⇒

ω0 >

⇒

( ω 0 )min =

2v0 R 2v0 R

The situation given in the problem is shown in Figure.

μmg

Let pure rolling start after time t, then for pure rolling to occur, we have ωr − v = v

ω r = 2v

t2 =

μmgR 2 μ g = R mR2 2

Disk will return when t2 > t1

Let, a1 be the linear acceleration of sphere (towards right), a2 be the linear acceleration of plank (towards left) and α be the angular retardation of sphere, then

⇒

ω0 α

where, α =

mr 2 I

CHAPTER 3

⇒

H.165

In this case sliding friction acts on cylinder in forward direction which increases its linear speed and decreases its angular speed. Let pure rolling start after time t, when final velocity be v f and final angular velocity be ω f , then we must have

2/9/2021 6:46:50 PM

H.166

JEE Advanced Physics: Mechanics – II v f = Rω f

(as pure rolling starts)

1 mω 02 R2 …(5) 12 According to Work Energy theorem, work done by friction is ⇒

For translational motion, applying the impulse momentum equation, we get ft = mv f − 0

Wfriction = E f − Ei

Initial momentum of the cylinder is taken as zero as it does not have any translational speed. ⇒ μmgt = mv f …(1)

Substituting the values of Ei and E f from equations (3) and (5), we get

For rotational motion, applying the angular impulse angular momentum equation, we get

Wfriction =

Iω 0 − fRt = Iω f ⇒

⇒

v ⎛1 ⎛1 2⎞ 2⎞ f ⎜⎝ mR ⎟⎠ ω 0 − μmgRt = ⎜⎝ mR ⎟⎠ 2 2 R

…(2)

4.

Dividing equations (1) and (2), we get μ gt =1 Rω 0 − 2 μ gt ⇒

μ gt = Rω 0 − 2 μ gt

⇒

ω R t= 0 3μ g

Ef =

1 Wfriction = − mω 02 R2 6

Applying Law of Conservation of Angular Momentum about bottommost point, we get Li = L f ⇒ ⇒ ⇒

During sliding, friction is the only force acting on the cylinder and we know that the work done by the friction is always negative or loss in kinetic energy of system.

⎛2 ⎞ mv0 R = mvR + ⎜ mR2 ⎟ ω ⎝5 ⎠

s

1 1 E f = mv 2f + Iω 2f 2 2

1 ω0R 3

At this moment pure rolling starts, thus its angular velocity at this instant is R

=

μ k mg = μ k g and v 2 = v02 − 2 as m 25 v02 − v02 v02 − v 2 49 s= = 2a 2μk g

Since, a =

⇒

…(4)

During sliding the acceleration of cylinder due to f friction ( f = μmg ) is a = = μ g, thus after time t it m gains a velocity

vf

}

ω

Ei =

ωf =

v R

v

v0

1 2 Iω 0 2

v f = μ gt =

∵ω=

a

1⎛ 1 1 2⎞ 2 2 2 …(3) ⎜ mR ⎟⎠ ω 0 = mR ω 0 2⎝ 2 4 Finally, when pure rolling starts, kinetic energy of cylinder is ⇒

{

2 mv0 R = mvR + mvR 5 5 v = v0 7

At t = 0, the kinetic energy of the cylinder is Ei =

1 1 mω 02 R2 − mω 02 R2 12 4

⇒ 5.

s=

12v02 49 μ k g

Till the pure rolling starts, friction ( f1 ) is upwards and it is maximum. Linear acceleration for μ = 1 a=

3 is

μmg cos 30° − mg sin 30° =0 m

μ gt 1 = ω0 R 3

From equation (4), final kinetic energy is Ef =

2

1 ⎛ Rω 0 ⎞ 1⎛ 1 2 ⎞ ⎛ ω0 ⎞ m⎜ ⎟ + ⎜⎝ mR ⎟⎠ ⎜⎝ ⎟ 3 ⎠ 2 ⎝ 3 ⎠ 2 2

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 166

2

2/9/2021 6:47:04 PM

Hints and Explanations i.e., the cylinder will keep on rotating at the same place till its angular velocity becomes zero. Angular 1 retardation, for μ = and θ = 30° is 3

( μmg cosθ ) R = mR2 2

t1 =

g R

ω0 ω0R = α g

3R

30°

⇒

⎛ 5μ g ⎞ 0 = ω0 + ⎜ − t ⎝ 2R ⎟⎠

⇒

t=

g sin θ

(

1 + I mR2

)

=

v2 ⎛ v ⎞ 1 s = v0 ⎜ 0 ⎟ − ( μ g ) 2 0 2 μ g ⎝ μg ⎠ 2

⇒

s=

2s 12R R = =6 a g 3 g

So, total time t = t1 + t2

6.

a=

v02 2μ g

v0 2Rω 0 = μg 5μ g

g sin 30° g = 1 + (1 2 ) 3

Let t2 be the time taken by the cylinder to reach the bottom, then

⇒

…(2)

Equating (1) and (2), we get

Also, s = 3 R cosec ( 30° ) = 6 R

t2 =

2Rω 0 5μ g

⇒

Once, the angular velocity becomes zero, cylinder starts rolling downwards with an acceleration a=

…(1)

Distance travelled is 1 s = v0t + ( − μ g ) t 2 2

…(1)

s

v0 μg

Also, ω = ω 0 + α t

If t1 be the time when the angular velocity of cylinder becomes zero, then 0 = ω 0 − α t1 ⇒

t=

⇒ 7.

ω0 =

5v0 2R

From conservation of angular momentum about bottommost point. Li = L f ⇒

2 mv0 R = mvR + mR2 ⋅ ω 5

⇒

2 mv0 R = mvR + mvR 5

ω R R t= 0 +6 g g f = μg m

v0

μ ( Mg ) R fR τ = = 2 I 2 MR2 MR2 5 5 5μ g α= 2R

∵ω=

v R

}

f at t = 0

Since v = v0 + ( − μ g ) t 0 = v0 − ( μ g ) t

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 167

v

s

⇒

v=

5 v0 7

Since, a =

v0

⇒

{

a

α=

⇒

CHAPTER 3

α=

⇒

H.167

μ k mg = μk g m

So, from v 2 = v02 − 2 as v2 − v2 We get s = 0 = 2a ⇒

s=

25 2 v0 49 2μk g

v02 −

12v02 49 μ k g

2/9/2021 6:47:20 PM

H.168 8.

JEE Advanced Physics: Mechanics – II

Since the cylinder does not slip over the plank, so it is the case of pure rolling and hence we can take friction on cylinder in any direction, say towards right as shown in Figure.

From equation (4), we get a1 =

3F 3M + m

Since, a2 is acceleration of cylinder relative to the plank, so net acceleration of the cylinder is anet = a1 − a2 =

F 3M + m

Please note that in the equation a2 = Rα , a2 is the acceleration of cylinder with respect to plank. So, Since friction on cylinder is acting towards right, so it must act on the plank towards left. Let plank moves toward right with an acceleration a1 , due to which the cylinder will experience a pseudo force ma1 towards left, and hence it will roll towards left with respect to plank with an acceleration a2. Since we have used the concept of pseudo force, so we must say that the acceleration a2 must be with respect to the plank. Let the angular acceleration of the cylinder during rolling be α , so we have a2 = Rα For translational motion of plank, we have F − f = Ma1

…(1)

For translational motion of cylinder with respect to plank we have ma1 − f = ma2

⎛1 ⎞⎛ a ⎞ fR = ⎜ mR2 ⎟ ⎜ 2 ⎟ ⎝2 ⎠⎝ R ⎠

1 ma2 2 From equation (2) and (3), we get ⇒

f =

…(3)

1 ma1 − ma2 = ma2 2 3 a2 2 Using equations (1), (2) and (3), we get ⇒

a1 =

1 3 F − ma2 = Ma2 2 2 ⇒

a2 =

2F 3M + m

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 168

9.

a1 − a2 F = R R( 3M + m )

Let us draw the situations at t = 0 and at time t. ω0

v0

ω

at t = 0

Taking forward direction as positive and CW sense as positive, we get a = μg

τ μ MgR 5 μ g = = 2R I 2 MR2 5 Now, at time t, we have α=

…(2)

For rotational motion of cylinder with respect to plank, we have fR = Iα ⇒

α=

v = v0 − ( μ g ) t ⎛ 5μ g ⎞ and ω = −ω 0 + ⎜ t ⎝ 2R ⎟⎠ Since at time t, we have v = Rω ⇒

⎡ ⎛ 5μ g ⎞ ⎤ v0 − ( μ g ) t = R ⎢ −ω 0 + ⎜ t ⎝ 2R ⎟⎠ ⎥⎦ ⎣

⇒

t=

…(1) …(2)

2 ⎛ v0 + Rω 0 ⎞ 7 ⎜⎝ μ g ⎟⎠

So, speed of ball at time t is v = v0 − ( μ g ) t

…(4) ⇒

v = v0 −

⇒

v=

2 ( v0 + Rω 0 ) 7

1 ( 5v0 + 2Rω 0 ) 7

10. Given μ > tan α ⇒

μmg cos α > mg sin α

2/9/2021 6:47:30 PM

Hints and Explanations

Single Correct Choice Type Questions

Since, a = ( μ g cos α − g sin α ) Also, α =

H.169

τ ( μmg cos α ) r 2 μ g cos α = = 1 2 Ι r mr 2

1.

Since axis passes through ends of two rods and is perpendicular to their length, so I 1P = I 3 P =

1 2 m 3 P

Slipping will stop say at time t, then v = rω ⇒

at = r ( ω 0 − α t )

⇒

rω 0 rω 0 ⎛ ⎞ t= =⎜ a + rα ⎝ 3 μ g cos α − g sin α ⎟⎠

2

1 m 2 12 3 , so by Parallel Axis and point P is at a distance 2 Theorem Moment of inertia of rod 2 about it CG is

If d1 is the distance travelled during this time t, then 1 2 r 2ω 02 ( μ cos α − sin α ) at = 2 2 2 g ( 3 μ cos α − sin α )

d1 =

Also, v = at =

rω 0 ( μ cos α − sin α ) ( 3 μ cos α − sin α )

Once slipping has stopped, pure rolling continues if

μ>

tan α

(

1 + mR2 I

⇒

μ>

tan α 1+ 2

⇒

μ>

tan α 3

)

Since, in the question it is already given that μ > tan α , so the retardation of cylinder is given by a′ =

g sin α

(

1 + I mr

2

)

=

g sin α 2 = g sin α 1+1 2 3

3 r 2ω 02 ( μ cos α − sin α ) v2 = 2 a′ ( 3 μ cos α − sin α )2 ( 4 g sin α ) 2

So, dmax = d1 + d2 ⇒

⎛ r 2ω 02 ( μ cos α − sin α ) ⎞ dmax = ⎜ × 2⎟ ⎝ 2 g ( 3 μ cos α − sin α ) ⎠ 3 ( μ cos α − sin α ) ⎞ ⎛ ⎜⎝ 1 + ⎟⎠ 2 sin α

⇒

dmax =

r 2ω 02 ( μ cos α − sin α ) 4 g sin α ( 3 μ cos α − sin α )

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 169

I 2P =

⎛ 3 ⎞ 1 m 2 + m ⎜  ⎝ 2 ⎟⎠ 12

⇒

I 2P =

1 3 m 2 + m 2 12 4

⇒

I 2P =

10 2 m 12

⇒

⎛ 10 1 1 ⎞ 2 I total = ⎜ + + m ⎝ 12 3 3 ⎟⎠

⇒

⎛ 10 + 4 + 4 ⎞ 2 I total = ⎜ ⎟⎠ m ⎝ 12

⇒

I total =

2

3 2 m 2

Further I total = ( 3 m ) k 2

If it travels a further distance d2 under this new retardation a′ , then d2 =

3

CHAPTER 3

1

3 2

⇒

( 3 m ) k 2 = m 2

⇒

k=

 2

Hence, the correct answer is (C). 2.

If we assume complete disc to be present, then it would have a mass 4 times the mass of the sector. Then, moment of inertia of the complete disc is Idisc =

1 1 Mdisc R2 = ( 4 Msector ) R2 2 2

Hence, I sector = ⇒

I sector =

Ι disc 4

1 MR2 2

2/9/2021 6:47:39 PM

H.170

JEE Advanced Physics: Mechanics – II

Calculus Method:

J

m

A

v

CM

m

Consider an element of mass dm, thickness dx and radius x of the sector. dm =

⇒

M 2π xdx 2 M = 2 xdx ⎛ π R2 ⎞ 4 R ⎜⎝ ⎟ 4 ⎠

⎛ 2M ⎞ dI = x 2 ⎜ 2 x dx ⎟ ⎠ ⎝ R

⇒

2M dI = 2 x 3 dx R

(

J J J + = 2m 2m m Hence, the correct answer is (C). ⇒

4.

I=

)

I=

Work done W =

1 2 Ιω 2

I = ( 0.3 ) x 2 + ( 0.7 ) ( 1.4 − x )

2M x 3 dx R2

2

For work to be minimum, the moment of inertia ( I ) should be minimum dI ⇒ =0 dx

∫ 0

⇒

vA =

If x is the distance of mass 0.3 kg from the centre of mass, we will have,

R

⇒

J J = 2m 2 2m

So, vA = v + ω

If dI be the moment of inertia of the element, then dI = x 2 dm ⇒

ω=

B

1 MR2 2

General Funda:

⇒

2 ( 0.3 x ) − 2 ( 0.7 ) ( 1.4 − x ) = 0

⇒

( 0.3 ) x = ( 0.7 ) ( 1.4 − x ) ( 0.7 ) ( 1.4 )

= 0.98 m 0.3 + 0.7 Hence, the correct answer is (C). ⇒

5.

x=

Assuming mass of body to be m, the mechanical energy is conserved in both the cases, so 2

⎛ Moment of Inertia ⎞ ⎛ Moment of Inertia ⎞ of a SECTOR of ⎜ of a DISC of mass ⎟ ≡ ⎜ ⎟ ⎜⎝ M with radius R ⎟⎠ ⎜⎝ mass M with radius R ⎟⎠ Hence, the correct answer is (A). 3.

Since, Impulse = Change in Momentum, so

Hence, ⇒

J 2m ⎛ Angular ⎞ ⎛ Change in Angular ⎞ = Also, ⎜ ⎟⎠ Momentum ⎝ Impulse ⎟⎠ ⎜⎝

I ⎛v ⎞ 25 2 v0 = v02 + ⎜ 0 ⎟ m⎝ R ⎠ 16

2

If K be the radius of gyration, then I = mK 2, so

J = ( 2m ) v ⇒

1 ⎛5 ⎞ 1 1 m ⎜ v0 ⎟ = mv02 + Ιω 2 2 ⎝4 ⎠ 2 2

mK 2 25 = 1+ 16 mR2

v=

⇒

⇒

J  = ( I system ) ω

⇒

J  = ( 2m 2 ) ω

9 2 R 16 3 ⇒ K= R 4 Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 170

⇒

K2 =

2/9/2021 6:47:47 PM

Hints and Explanations 6.

H.171

So, from (1), we get

Moment of inertia about the axis passing through x = 2R , y = 0 shown by ⊗ direction in figure is

⇒

1 2 I1 = mR2 + m ( 2R ) 2

v′ =

v J = 2 2m

Hence, the correct answer is (B).

9 ⇒ I1 = mR2 …(1) 2 Moment of inertia about the axis passing through y = d, z = 0, shown as dotted line in figure is

8.

Since at t = 0, P is at the top of rim, so ⎛ π +θ ⎞ vP = 2v0 sin ⎜ ⎝ 2 ⎟⎠

y

CHAPTER 3

y = d, z = 0

x x = 2R, y=0 2R

I2 =

1 mR2 + md 2 4

…(2)

⇒

⎛θ⎞ ⎛ v t⎞ vp = 2v0 cos ⎜ ⎟ = 2v0 cos ⎜ 0 ⎟ ⎝ 2⎠ ⎝ 2R ⎠

⇒

⎛ v t⎞ vP2 = 4v02 cos 2 ⎜ 0 ⎟ ⎝ 2R ⎠

Equating (1) and (2), we get

At t = 0, vP2 = 4v02

1 9 mR2 + md 2 = mR2 4 2 17 ⇒ d= R 2 Hence, the correct answer is (B). 7.

Also, vP = 0 at point of contact i.e., when P reaches the T πR bottom i.e., in a time t = = 2 v0

Let v be the velocity of centre of mass of ring just after the impulse is applied and v′ be its velocity when pure rolling starts. Angular velocity ω of the ring at this v′ instant will be ω = r v

ω

v′

Hence, the correct answer is (B). 9.

Let L be the length of the rod and F1 the magnitude of other force. If C is the centre of the rod, then F1 − F = ma ⇒

F1 = F + ma

…(1)

Since the rod moves translationally, so the net torque about C is zero.

Since, Impulse = Change in Linear Momentum, so we have

F F1

J = mv J …(1) m Between the two positions shown in figure, force of friction on the ring acts backwards. Angular momentum of the ring about bottommost point will be conserved, so ⇒

L

C

v=

Li = L f ⇒

mvr = mv′r + Ιω

⇒

mvr = mv′r + ( mr 2 )

v′ = 2mv′r r

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 171

⇒

⎛ L⎞ ⎛L ⎞ F ⎜ ⎟ = F1 ⎜ −  ⎟ ⎝ 2⎠ ⎝2 ⎠

⇒

⎛ L⎞ ⎛L ⎞ F ⎜ ⎟ = ( F + ma ) ⎜ −  ⎟ ⎝ 2⎠ ⎝2 ⎠

⇒

⎛ L⎞ F ⎜ ⎟ = ( F + ma )  ⎝ 2⎠

2/9/2021 6:47:55 PM

H.172

⇒

JEE Advanced Physics: Mechanics – II

L=

2 ( F + ma )  ma

F ⎞ ⎛ L = 2 ⎜ 1 + ⎟ ⎝ ma ⎠ Hence, the correct answer is (B). ⇒

10. The free body diagram of arrangement is shown in Figure.

mg = N1 Also, μ N1 = N 2 ⇒

N 2 = μmg

⇒

 3 Fx a= α = 2 2 m

⇒

a∝x

So, a-x graph is a straight line passing through origin 3F with slope . 2m Hence, the correct answer is (B).  13. The angular momentum of a body L may be expressed as the sum of two parts, (a) one arising from the motion of the centre of mass of the body and (b) the other from the motion of the body with respect to its centre of mass.     i.e. Ltotal = LC. M . + rC. M . × p     ⇒ Ltotal = LC. M . + M ( rC. M . × vC. M . ) For this Problem

Taking moment of forces about O, we get

⇒ ⇒

LC. M . = Iω =

l μmgl sin θ = mg cos θ 2 1 tan θ = 2μ

  M ( rC. M . × vC. M . ) = MRvCM = MR ( Rω ) ⇒

⎛ 1 ⎞ θ = tan −1 ⎜ ⎝ 2 μ ⎟⎠

11. Given, that I 0 is the moment of inertia of table and gun and m the mass of bullet. Initial angular momentum of system about centre

(

Li = I 0 + mr

)ω 0

  M ( rC. M . × vC. M . ) = MR2ω

1 3 MR2ω + MR2ω = MR2ω 2 2 Hence, the correct answer is (B). ⇒

Hence, the correct answer is (D).

2

1 MR2ω and 2

Ltotal =

14. Total Revolutions is equal to area under the curve. Area= Ar ΔABF + Ar ΔBCD + Ar Rectangle BDEF

…(1)

Let ω be the angular velocity of table after the bullet is fired. Final angular momentum L f = I 0ω − m ( v − rω ) r

…(2)

where ( v − rω ) is absolute velocity of bullet to the right, then From (1) and (2), we get mvr I 0 + mr 2 This is also the increase in angular velocity Hence, the correct answer is (B).

( ω − ω0 ) =

12. The rod will rotate about point A with angular acceleration given by

α=

τ Fx 3 Fx = = 2 I ⎛ m ⎞ m 2 ⎝⎜ 3 ⎟⎠

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 172

⇒

Area =

( 600 ) 1 ( 1800 ) 1 600 ( 8 ) + ( 16 ) +8× 2 60 2 60 60

7200 + 4800 + 4800 = 280 60 Hence, the correct answer is (B). ⇒

Area =

15. Applying conservation of energy, we get 2 1 1 ⎛ mv ⎞ 1 ⎛ ML2 ⎞ 2 mv 2 = M ⎜ + ⎜ ⎟ (ω ) ⎟ 2 2 ⎝ M⎠ 2 ⎝ 12 ⎠

…(1)

2/9/2021 6:48:03 PM

Hints and Explanations

H.173

Applying conservation of angular momentum, we get 2 ⎛ L ⎞ ML mv ⎜ ⎟ = ω ⎝ 2⎠ 12 6mv ⇒ ω= ML From (1) and (2), we get

…(2)

⎛ 36 m2v 2 ⎞ 1 1 m2v 2 1 mv 2 = + ML2 ⎜ ⎝ M 2 L2 ⎟⎠ 2 2 M 24

⇒

m 3m + M M 4m 1= M 1=

⎛b ⎞ 0 = v2 − 2 ⎜ g ⎟ l ⎝h ⎠ hv 2 2bg Hence, the correct answer is (B).

M 4 Hence, the correct answer is (B). ⇒

a≤

CHAPTER 3

⇒

b g h Final velocity of truck is zero, so we have ⇒

⇒

m=

l=

19. Since, for pure rolling, we have

16.

ω=

v R y

r θ

I AB = I A ′B′ = I and ICD = IC ′D ′ If I 0 be the moment of inertia of the square plate about an axis passing through O and perpendicular to the plate, then by perpendicular axis theorem I 0 = I AB + I A ′B′ = 2I AB

…(1)

OR I 0 = ICD + IC ′D′ = 2ICD

…(2)

From (1) and (2) ICD = I AB = I

{OPTION (A)}

Hence, the correct answer is (A). 17. Net torque about O due to weights acting at midpoints of the respective rods should be zero. Hence, ⎛ ⎞ ⎛ ⎞ mg ⎜ sin 60° ⎟ = Mg ⎜ sin 30° ⎟ ⎝2 ⎠ ⎝2 ⎠ M sin 60° = = 3 m sin 30° Hence, the correct answer is (C). ⇒

18. Let a be the acceleration of the truck, then from FBD of the block, taking torque about right edge from where normal reaction is passing, we get ⎛ h⎞ ⎛ b⎞ ma ⎜ ⎟ ≤ mg ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 173

P

θ

v

rω = vr R x

Velocity of point P is resultant of two velocity vectors as shown in figure. So,  vP = vx iˆ + vy ˆj ⇒iˆ ⇒ iˆ

 vr sin θ ⎞ ˆ ⎛ vr cos θ ⎞ ˆ ⎛ vP = ⎜ v + ⎟i − ⎜ ⎟j R ⎠ ⎝ R ⎠ ⎝  vP = ω ⎡⎣ ( R + r sin θ ) iˆ − ( r cos θ ) ˆj ⎤⎦

Hence, the correct answer is (C). 20. Moment of inertia of both the rods about an axis passing through their point of intersection O and perpendicular to the plane of the rods is I0 =

m 2 m 2 m 2 + = 12 12 6

Now axis 1 and axis 2 are two mutually perpendicular axes passing through Olying in the plane of two rods. Then by symmetry, we have I1 = I 2 = I

{say}

2/9/2021 6:48:09 PM

H.174

JEE Advanced Physics: Mechanics – II C 1

⇒

2

⇒ A

B O

⇒

Now I solid sphere < I hollow

D

Now, according to Perpendicular Axis Theorem, we have I1 + I 2 = I 0 ⇒

2I1 = I 0 =

⎛ v⎞ Iω 0 = I ⎜ ⎟ + mvR ⎝ R⎠ Iω 0 v= I + mR R ω0 v= 1 mR + R I

⇒

vsolid < vhollow

⇒

v1 < v2

Hence, the correct answer is (C).

m 6

2

23. Considering an elemental mass dm on ring as shown in Figure. Then

m 2 12 Hence, the correct answer is (C). ⇒

I1 =

21. Let OA = x and OB = y as shown in figure

B

C

α

Mdθ ⎛ M ⎞( dm = ⎜ Rdθ ) = ⎝ 2π R ⎟⎠ 2π Velocity of this elemental mass is

v

y

β O

A

x

v = 2v0 cos ϕ

Then BC + CA =  ⇒

y cos α + x cos β = 

…(1)

Differentiating both sides of (1) w.r.t. time t, we get

⇒

Kinetic energy of element is

⎛ dy ⎞ ⎛ dx ⎞ ⎜⎝ ⎟⎠ cos α + ⎜⎝ ⎟ cos β = 0 dt dt ⎠ Since,

dK = ⇒

v cos β ⎛ dy ⎞ So, ⎜ = vB = − ⎝ dt ⎟⎠ cos α Here, negative sign implies that y decreases as t increases. Hence, the correct answer is (D). 22. Applying Conservation of Angular Momentum about point of contact, we get Iω 0 = Iω + mvR

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 174

θ 1 1 ( dm ) v 2 = ( dm ) ⎛⎜ 4v02 sin 2 ⎞⎟ ⎝ 2 2 2⎠

Mv02 ( 1 − cos θ ) dθ 2π Total kinetic energy of segment ACB is

dx = vA = v dt

ω

⎛θ⎞ v = 2v0 cos ϕ = 2v0 sin ⎜ ⎟ ⎝ 2⎠

ω

v

dK =

K = K ACB =

Mv02 2π

+

π 2

∫ ( 1 − cosθ )dθ

−

π 2

Mv02 Mv02 − 2 π Hence, the correct answer is (A). ⇒

24. τ = ⇒

K=

dL d = ( Ιω ) dt dt

τ=

⎞ d ⎛ ML2 + mx 2 ⎟ ω ⎜ ⎠ dt ⎝ 3

2/9/2021 6:48:17 PM

Hints and Explanations

τ = 2mx

dx ω dt

1 ⎛1 ⎞ 27. ⎜ MR2 + Mb R2 ⎟ ω = MR2ω ′ ⎝2 ⎠ 2

Now, x = vt ⇒

(Since the boy reaches the centre, so final angular momentum of boy is zero).

τ ∝t

Finally, torque becomes zero. Hence, the correct answer is (B). 25. No external torque is acting on the system, so angular momentum is conserved. Further there exists no nonconservative forces in the system, so total energy is also conserved. Hence, the correct answer is (B). 26. By Law of Conservation of Linear Momentum, we have mu = mv + mV ⇒

u = v+V m

u

m

v a

a

3ω = 15 rpm 2 Hence, the correct answer is (C). ⇒

( KE )Plank = 1 m ( 2v )2 2

( KE )Plank = 2mv 2

2

Just After Impact

By Law of Conservation of Angular Momentum about centre of rod, we have ⎛ ma 2 ⎞ mua = mva + ⎜ ω ⎝ 3 ⎟⎠ aω …(2) 3 Further applying the definition of coefficient of restitution ( e = 1 ) at the point of impact, we get Relative speed of approach = Relative speed of separation ⇒ u = ( V + aω ) − v …(3) u=v+

Solving these three equations (1), (2) and (3) we get, 2 6u u and ω = 5 5a 1 1 ma 2 ⎛ 6u ⎞ ⎛2 ⎞ × m × ⎜ u⎟ + × ×⎜ ⎝5 ⎠ ⎝ 5 a ⎟⎠ 2 2 3

8 mu2 25 Hence, the correct answer is (D). K=

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 175

R

⇒

( KE )Cylinder = 1 mv 2 ⎛⎜ 1 + 1 ⎞⎟ = mv 2 ⎝ ⎠

⇒

( KE )Plank 2 = ( KE )Cylinder 1

2

1

Hence, the correct answer is (B). 29. Let M be the mass of the square plate before the holes are cut. If m be the mass of one hole, then

πM ⎛ M ⎞ m=⎜ π R2 = ⎝ 16 R2 ⎟⎠ 16 Moment of inertia of remaining portion is I = I square − 4 I hole ⇒

I=

⎡ mR2 ⎤ M( 16 R2 + 16 R2 ) − 4 ⎢ + m ( 2R 2 ) ⎥ 12 ⎣ 2 ⎦

⇒

I=

8 MR2 − 10 mR2 3

⎛ 8 10π ⎞ I=⎜ − MR2 ⎝ 3 16 ⎟⎠ Hence, the correct answer is (C). ⇒

So, kinetic energy of rod is 2

ω′ =

28. For the plank P

ω

m

Just Before Impact

⇒

( 100 + 50 ) ω = 100ω ′

2⎞ ⎛ ( KE )Cylinder = 1 mv 2 ⎜ 1 + K 2 ⎟ ⎝ ⎠

a

m

K=

⇒

For the cylinder

V

V=

1 ⎛1 ⎞ ⎜⎝ M + Mb ⎟⎠ ω = Mω ′ 2 2

…(1)

a

⇒

⇒

CHAPTER 3

⇒

H.175

2

30. Conceptual the correct answer is (B). 31. Applying angular momentum conservation, we get ⎡ m ( 2l )2 ⎤ 2⎢ ⎥ ω 0 = I ′ω ′ ⎣ 12 ⎦

…(1)

2/9/2021 6:48:24 PM

H.176

JEE Advanced Physics: Mechanics – II 34. v C

ω v

where, I ′ =

2 ⎡ m ( 2l )2 m ( 2l ) 2⎤ +⎢ + m ( 2l ) ⎥ 12 ⎣ 12 ⎦

ml 2 13 ml 2 14 ml 2 ⇒ I′ = + = 3 3 3 So, equation (1) becomes 2 2 14 2 ml ω 0 = ml ω ′ 3 3

ω ⇒ ω′ = 0 7 Hence, the correct answer is (B).

2 ( mR2 ) ( 2π − θ ) = ⇒

4π m − 2mθ =

M θ 2

1 MR2θ 2

8π m 4m + M Hence, the correct answer is (D). ⇒

θ=

35. At the critical condition, normal reaction N will pass through point P. In this condition

τN = 0 = τ f

(about P) N

32. By Law of Conservation of Angular Momentum

F

L/2 f

(

)

⎛ a⎞ mv ⎜ ⎟ = Ι system ω ⎝ 2⎠ about O 2 ⎛ a⎞ ⎡1 ⎛ a ⎞ ⎤ mv ⎜ ⎟ = ⎢ Ma 2 + M ⎜ ⎥ω ⎝ 2⎠ ⎣6 ⎝ 2 ⎟⎠ ⎦

3v 4a Hence, the correct answer is (A). ⇒

ω=

33. By Law of Conservation of Energy Loss in Potential energy = Gain in Kinetic energy mg ( 1 − cos θ ) =

1 mv 2 2

mv 2 = 2mg ( 1 − cos θ )  At point of breaking ⇒

mg cos θ = ⇒

the block will topple when

τ F > τ mg ⇒

FL > ( mg )

⇒

F>

L 2

mg 2

Therefore, the minimum force required to topple the bock is mg Fmin = 2 Hence, the correct answer is (B). 36. Relative velocity is   vP − vQ = 2Rω cos 30° = 3 Rω

mv 2 

mg cos θ = 2mg ( 1 − cos θ )

2 3 Hence, the correct answer is (B). ⇒

mg

cos θ =

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 176

Hence, the correct answer is (D).

2/9/2021 6:48:31 PM

Hints and Explanations

37. When m is placed in left pan, m1 is placed in right pan. So,

H.177

The intersection of 2 y + 3 x = 16 and 3 y − 4 x = −1 is ⎛ 50 61 ⎞ , ⎜⎝ ⎟ 17 17 ⎠ Hence, the correct answer is (D).

…(1)

2

When m is placed in right pan, m2 is placed in left pan. …(2) So, m2 gx = mg ( L − x ) Divide (1) by (2), we get

2

m = m1m2 m = m1m2

Hence, the correct answer is (B). 38.

f = μmg , a = μ g

α=

τ μmgR 5 g μ = = I 2 mR2 2 R 5

a

a v0 = α ω0

⇒

2R v0 = ω0 5

⇒ 5v0 = 2Rω 0 Hence, the correct answer is (B).

⇒

mv 2 r 16 =

Solving equations (1) and (2), we get θ θ2 = 1 3 n So, the disc will make more rotations before it comes 3 to rest.

42.  =

2π R 4

⇒

R=

2 π

v0 ω 0 = a α

⇒

39. T =

…(2)

Hence, the correct answer is (B). α

Now, t =

…(1)

⎛ω ⎞ and 0 = ⎜ 0 ⎟ − 2αθ 2 ⎝ 2 ⎠

2

⇒

⎛ ω0 ⎞ 2 ⎜⎝ ⎟ = ω 0 − 2αθ1 2 ⎠

CHAPTER 3

mgx = m1 g ( L − x )

41. Since, retarding torque is constant, so, angular retardation say α will also be constant. Applying ω 2 = ω 02 − 2αθ we get,

( 16 ) v 2 144

⇒ v = 12 ms −1 Hence, the correct answer is (D). 40. Axis of pure rotation will be the point of intersection of the lines perpendicular to both the straight lines. A line perpendicular to 2x − 3 y = 2 and passing through ( 4 , 2 ) is 2 y + 3 x = 16 and a line perpendicular to 3 x + 4 y = 7 and passing through ( 1, 1 ) is 3 y − 4 x = −1

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 177

⎛ 2 ⎞ Since, I = mR2 = m ⎜ ⎟ ⎝ π ⎠

2

2m 2 5 Hence, the correct answer is (D). ⇒

I=

{∵ π 2 ≈ 10 }

⎛ MΙ of ⎞ ⎛ MΙ of Complete ⎞ ⎛ MΙ of Removed ⎞ ⎟ − ⎜ Portion of ⎟ Disc of 43. ⎜ Given ⎟ = ⎜ ⎟ ⎜ Radius R 3 ⎟ ⎜ Shape ⎟ ⎜ Radius R ⎠ ⎝ ⎝ ⎠ ⎝ ⎠ ⇒

I remaining = I whole − I removed

⇒

I=

2 2 ⎡1 R 1 1 2R ⎞ ⎤ ( 9 M ) ( R )2 − ⎢ m ⎛⎜ ⎞⎟ + m ⎛⎜ ⎥ ⎟ 2 2 ⎝ 3 ⎠ ⎦ ⎣2 ⎝ 3⎠

…(1)

2

9M ⎛ R⎞ ×π⎜ ⎟ = M ⎝ 3⎠ π R2 Substituting in equation (1), we get I = 4 MR2 Hence, the correct answer is (A).

where, m = σ a =

2/9/2021 6:48:43 PM

H.178

JEE Advanced Physics: Mechanics – II

44. If m be the mass of cube and a be the side of cube, then the cube will slide when F > μmg

…(1)

and it will topple if torque due to F about P, is greater than torque due to mg about P i.e., ⎛ a⎞ Fa > ⎜ ⎟ mg ⎝ 2⎠ ⇒

F>

47. When the string is cut the weight Mg of rod acting at centre produces the torque about P. ⇒

Also, τ = Iα ⇒

1 mg 2

…(2) N = mg

⎛ L⎞ τ = Mg ⎜ ⎟ ⎝ 2⎠ 2 ⎛ L ⎞ ⎛ ML ⎞ Mg ⎜ ⎟ = ⎜ α ⎝ 2 ⎠ ⎝ 3 ⎟⎠

3g 2L Hence, the correct answer is (D). ⇒

α=

48. T sin θ = m ( L sin θ ) ω 2 a force of friction = f

θ mg a 2

P Tcosθ

θ

From equations (1) and (2), we observe that cube will topple before sliding if 1 2 Hence, the correct answer is (D).

mrω

2

45. Mass of cotton pad after time t is m = μt Applying conservation of angular momentum, we get ⎛ m0 r 2 ⎞ ⎛ m0 r 2 ⎞ω + μtr 2 ⎟ ⎟⎠ ω = ⎜⎝ ⎜⎝ ⎠ 2 2 2 m0 r 2 =

r Tsinθ

mg

μ>

⇒

T

⇒

324 = ( 0.5 )( 0.5 ) ω 2

⇒

ω2 =

324 0.5 × 0.5

⇒

ω=

324 0.5 × 0.5

18 = 36 rads −1 0.5 Hence, the correct answer is (B). ⇒

m0 r 2 + μtr 2 2

ω=

49. For equilibrium of rod (considering limiting friction), we have

m ⇒ t= 0 2μ Hence, the correct answer is (B).

f1 = N 2 ⇒

46.

μ N1 = N 2

…(1)

Also, −W + N1 + f 2 = 0 ⇒

N1 + μ N 2 = W

⇒

N1 + μ 2 N1 = W

…(2) f2

⎛ 3 ⎞ 2 2 I = m( 0 ) + m( 0 ) + m⎜  ⎝ 2 ⎟⎠ ⇒

I=

2

3 m 2 4

Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 178

2/9/2021 6:48:51 PM

Hints and Explanations

⇒

2 ( 3μ 2 + 4μ ) = 3 + 3μ 2

⇒

6μ 2 + 8μ = 3 + 3μ 2

⇒

3μ 2 + 8μ − 3 = 0

⇒

3 μ 2 + 9μ − μ − 3 = 0

⇒

3 μ ( μ + 3 ) − 1( μ + 3 ) = 0

⇒

( 3μ − 1 )( μ + 3 ) = 0

Since μ ≠ −3 1 3 Hence, the correct answer is (B). ⇒

μ=

where α =

⇒

( μg ) ω 2μ g R

0

=

rω 0 2

rω 0 i.e., v0 < rω 0 , so f acts such that it 2 decreases v0 and hence acts opposite to v0 i.e., backwards. Also, the point of contact of disc with the ground has non-zero velocity as v0 < rω 0 . Since v0 =

Hence, the correct answer is (B). 53. At the topmost point of the loop minimum value of linear speed of centre of sphere is v = gR So, translational kinetic energy at topmost point is 1 1 mv 2 = mgR 2 2 In case of pure rolling of a solid sphere the ratio of rotational to translational kinetic energy is KT =

KR k 2 2 = = KT r 2 5

50. By Law of Conservation of Angular Momentum

( MR2 )ω = MR2ω ′ + 2mR2ω ′

v0 =

τ fR ( μmg ) R 2 μ g = = = 1 I I R mR2 2

CHAPTER 3

Balancing all torques about point A, we get ⎛ 3⎞ ⇒ −W ⎜ ⎟ + f 2 ( 3 ) + N 2 ( 4 ) = 0 ⎝ 2⎠ 3W ⇒ 3 f2 + 4N2 = 2 3W ⇒ 3μN2 + 4N2 = 2 3 ⇒ μ N1 ( 3 μ + 4 ) = N1 ( 1 + μ 2 ) 2

H.179

So, total kinetic energy at topmost point is 7⎛ 1 ⎛ 5+ 2⎞ ⎞ 7 K=⎜ K = mgR ⎟ = mgR ⎠ 10 ⎝ 5 ⎟⎠ T 5 ⎝⎜ 2 Now by Law of Conservation of Mechanical Energy, we have

⎛ M ⎞ ω′ = ⎜ ω ⎝ M + 2m ⎟⎠ Hence, the correct answer is (B). ⇒

51. By Law of Conservation of Angular Momentum

∑ mv r = ( I

⇒

system

)ω

2  ( 2 m ) ( 2 ) 2m ( 4 2 ) mv = ω= ω 2 12 12

3v ( anticlockwise ) 4 Hence, the correct answer is (A). ⇒

52. t = ⇒

ω=

v0 ω 0 = a α ⎛ a⎞ v0 = ⎜ ⎟ ω 0 ⎝α⎠

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 179

7 mgR = mg ( h − 2R ) 10 ⇒ h = 2.7 R Hence, the correct answer is (D). 54. For pure rolling, acm = Rα , where α = ⇒ ⇒

F ⎡ F( h − R) ⎤ = R⎢ ⎥ 2 m mR2 ⎥ ⎢ ⎣ 5 ⎦

τ F( h − R) = 2 I mR2 5

h − R = 0.4 R

⇒ h = 1.4 R Hence, the correct answer is (C). 55. For a given perimeter area of a circle is maximum, i.e., distribution of mass is at a maximum distance from the axis in case of a circle. Hence, the correct answer is (A).

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H.180

JEE Advanced Physics: Mechanics – II

56. Mass of the loop = M = Lρ Further if r is the radius of the loop, then 2π r = L L ⇒ r= 2π 3 Moment of Inertia about XX′ is I = Mr 2 2 3 L2 3 ρL3 ⇒ I = ( Lρ ) = 2 ( 2π )2 8π 2 Hence, the correct answer is (D).

62.

By Law of Conservation of Moments, ⎛ Total Clockwise ⎞ ⎛ Total Anticlockwise ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ Moments Moments ⇒

57. By Law of Conservation of Angular Momentum

⇒

mvr = ( I + mR2 ) ω

mvr I + mR2 Hence, the correct answer is (A). ⇒

ω=

M dx 58. Mass of the element dx is m = L This element needs centripetal force for rotation, so ⎛M 2 ⎞ xω dx ⎟ dF = mxω 2 = ⎜ ⎝ L ⎠ L

⇒

∫

F = dF = 0

L

0

This is the force exerted by the liquid at the other end. Hence, the correct answer is (A). 59. Friction is not sufficient for pure rolling. Therefore, maximum friction will act upwards in all the three bodies. So, linear acceleration a of all the three bodies will be same and equal to ( g sin θ − μ g cos θ ). Therefore, time taken by all the three will be same. Hence, the correct answer is (D). 60. Linear velocity of all three will be same. However, ring having maximum moment of inertia will have least angular acceleration and hence least rotational kinetic energy. Hence, the correct answer is (A). 61. At any angle θ , mv 2 + mg cos θ r At mean position θ = 0° T=

And v = 2 gr ⇒

T=

m ( 2 gr )

+ mg r ⇒ T = 3 mg Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 180

63.

x=

1 2 1 Iω = Mv 2 2 2 ⇒

1 1 ( 3 )( 9 ) = ( 27 ) v 2 2 2

⇒

v = 1 ms −1

Hence, the correct answer is (A). 64. Since L = Iω

2

m 2 Mω L ω xdx = 2 L

∫

L W ⎛ 3W ⎞ = (0) + ⎜ x ⎝ 4 ⎟⎠ 2 4

2L 3 Hence, the correct answer is (C).

mvr = ( I system ) ω

⇒

W

⇒

L = ( mr 2 ) ω

r L when r becomes , L becomes . 2 4 Hence, the correct answer is (A). 65. If v be the linear velocity of centre of mass of the spherical body and ω be its angular velocity about centre of mass, then v ω= 2R KE of spherical rolling body is

⇒

K1 =

1 1 mv 2 + Ιω 2 2 2

K1 =

⎛ v2 ⎞ 1 2 1( mν + 2mR2 ) ⎜ ⎝ 4 R2 ⎠⎟ 2 2

3 mν 2 4 Speed of the block will be v′ = vcm + Rω = 2Rω + Rω = 3 Rω ⇒

K1 =

…(1)

⎛ v ⎞ 3 v′ = 3 Rω = ( 3 R ) ⎜ = v ⎝ 2R ⎟⎠ 2 1 So, KE of block is K 2 = mv′ 2 2 ⇒

⇒

K2 =

2

1 ⎛3 ⎞ 9 m ⎜ v ⎟ = mv 2 2 ⎝2 ⎠ 8

…(2)

2/9/2021 6:49:09 PM

Hints and Explanations From equations (1) and (2), we get

69. M1 = M2

K1 2 = K2 3 Hence, the correct answer is (B).

⇒

mv 2 r Further, by Law of Conservation of Angular Momentum L = mvr = constant Ar n =

⇒

v=

⇒

Ar n =

m⎛ L ⎞ ⎜ ⎟ r ⎝ mr ⎠

⇒

Ar n =

L2 −3 r m

⇒

n = −3

)

(

)

where t is thickness of both discs.

n

⇒

(

⎡ π R12 t ⎤ d1 = ⎡ π R22 t ⎤ d2 ⎣ ⎦ ⎣ ⎦

L mr 2

⇒

R12 d2 = R22 d1

1 MR12 d 2 ⇒ = 2 1 2 MR2 d1 2 Hence, the correct answer is (B). 70. r⊥ = ( b − a ) sin 30° =

b−a 2

y

⎫ ⎧ L2 ⎨∵ = constant ⎬ ⎩ m ⎭

v0

Hence, the correct answer is (D). 67. According to Angular Impulse – Angular Momentum Theorem, we have   τ av Δt = ΔL …(1) where, Δt = time of flight =

⇒ ⇒ ⇒ ⇒

 ( mu sin θ ) ( u2 sin 2θ ) ΔL = g  mu3 sin θ sin 2θ ΔL = g   g ⎞ ΔL ⎛ mu3 sin θ sin 2θ ⎞ ⎛ τ aν = = ⎟ ⎜⎝ 2u sin θ ⎟⎠ Δt ⎜⎝ g ⎠

 mu2 sin ( 2θ ) τ aν = 2 Hence, the correct answer is (C). 68. Equilibrium of m gives {T =Tension in string}

Net torque about point of contact of spool should be zero. Hence, ( 2R ) ( Mg sin α ) = TR 2Mg sin α = mg

⇒ m = 2 M sin α Hence, the correct answer is (A).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 181

O

30°

a

30°

r⊥

a/2

x

mν 0 ( b − a ) 2 Hence, the correct answer is (D). ⇒

L = mv0 r⊥ =

71. Mg = mω 2

⇒

⇒

30° b−a

2u sin θ g

   ΔL = L f − Li about point of projection  ΔL = ( mu sin θ ) ( range ) − 0

T = mg

CHAPTER 3

66. T = Ar

H.181

⇒

f =

1 2π

Mg m

Hence, the correct answer is (A). 72. a =

g I ⎞ ⎛ ⎜⎝ 1 + ⎟ mr 2 ⎠

Since v 2 − u2 = 2 as ⇒

v 2 = 2 ah

⇒

v2 =

⇒

v=r

(

)

2 mgr 2 h I + mr 2

2 mgh I + mr 2

2 mgh v = r I + mr 2 Hence, the correct answer is (B). ⇒

ω=

2/9/2021 6:49:19 PM

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JEE Advanced Physics: Mechanics – II

73. The combined effect of translation and rotation is equivalent to a single effect of rotation about the point having zero velocity i.e., IC just like the hoop being pinned at IC and rotating about IC.

θ P

f

v0 R

O(IC)

mg sin θ

θ v0 R

So, this can be assumed as a pure rotation about point v of contact say O (i.e., IC) with angular velocity ω = , R where R is the radius of hoop. Speed of P will be

θ⎞ ⎛ vP = ( OP ) ω = ⎜ 2R sin ⎟ ω ⎝ 2⎠ ⇒

76. Initially the spring force kx is less than mg sin θ , i.e., the cylinder is accelerated downward or force of friction f is upwards. It will reverse its direction when kx > mg sin θ .

Hence, the correct answer is (D). v and vc = 0 R Hence, the correct answer is (C).

77. ω =

78. Let friction act on cylinder in backward direction

⎛θ⎞ vP = ( 2Rω ) sin ⎜ ⎟ ⎝ 2⎠

⎛θ⎞ vP = 2v0 sin ⎜ ⎟ ⎝ 2⎠ Hence, the correct answer is (B). ⇒

74. I1 = mr 2

Since, a = Rα

I 2 =moment of inertia about any one of the diameters

⇒

…(1)

mg sin 60° − f = Ma

…(2)

Since fR = Iα

1 = mr 2 2 3 ⇒ I = I1 + I 2 = mr 2 2 Hence, the correct answer is (D).    75. L = m ( r × v )

⇒

iˆ ˆj kˆ  L=2 1 1 0 2 −2 2  L = 2 2iˆ − 2 ˆj − 4 kˆ  L = 4iˆ − 4 ˆj − 8 kˆ

⇒

Lz = −8 kgm 2s −1

⇒

(

)

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 182

MR2 ⎛ a ⎞ ⎜ ⎟ 2 ⎝ R⎠

⇒

fR =

⇒

f =

⇒

Mg sin 60° =

⇒

a=

Ma 2

…(3) 3 Ma 2

g 2 g sin 60° = 3 3

Mg 2 3 For sliding, we have ⇒

f =

⇒

f = μ Mg cos 60°

⇒

Mg ⎛ 2 − 3 x ⎞ ⎛ 1 ⎞ =⎜ ⎟⎜ ⎟ 2 3 ⎝ 3 ⎠ ⎝ 2⎠

⇒

2 − 3x = 1

⇒

3x = 1

1 m 3 Hence, the correct answer is (A). ⇒

x=

2/9/2021 6:49:28 PM

Hints and Explanations

80. Consider the origin at x = 0. Let the equilibrium be established at a distance r from origin. Then by Law of Conservation of Moments ⎡⎣ 12 + 22 + 3 2 + ....... + ( 100 )2 ⎤⎦ = ( 1 + 2 + 3 + ....... + 100 ) r ⇒

( 100 ) ( 100 + 1 ) ( 200 + 1 ) 6

⇒

201 r= 3

⇒

r = 67 cm

=

( 100 ) ( 100 + 1 ) 2

r

π m , the rod will rotate an After the given time t = 12 J angle ⎛ 6 J ⎞ ⎛ π m ⎞ π θ = ωt = ⎜ = ⎝ m ⎟⎠ ⎜⎝ 12 J ⎟⎠ 2 ⇒

J  v P = 2v = 2 m Hence, the correct answer is (C). ⇒

83. Net external torque is zero. Therefore, angular momentum of system will remain conserved, i.e., Li = L f Initial angular momentum Li = 0, so final angular momentum should also be zero Angular ⎛ ⎞ ⎛ Angular Momentum ⎞ ⎜ Momentum of ⎟ + ⎜ ⎟ =0 of Platform in ⎜ Man in CW sense ⎟ ⎜ ⎟ CCW sense ⎝ ⎠ ⎝ ⎠

Hence, the correct answer is (B). 81. Angular momentum of system cannot remain conserved as some external unbalanced torque is present due to forces at axles. Kinetic energy is not conserved, because slipping is there and work is done against friction. Hence, the correct answer is (C). 82. Let v and ω be the linear and angular speeds of the rod after applying an impulse J at B. Then Impulse = Change in Momentum

P

J

C

⇒

mvr − ( I platform ) ω = 0

⇒

ω=

⇒ ω = 0.7 rads −1 Hence, the correct answer is (B). 84. Angular velocity of man relative to platform is 1 v0 = 0.7 + = 1.2 rads −1 r 2 If t be the time taken by man to complete one round around the platform, then 2π t= ωr

ω

ω 6

ν

A

2ν P

ν

B

2π s 1.2 Angle rotated with respect ground in this time is ⇒

B

B

⇒

mv = J

⇒

v=

J m

…(1)

⎛ Angular ⎞ ⎛ Change in Angular ⎞ = Also, ⎜ ⎟⎠ Momentum ⎝ Impulse ⎟⎠ ⎜⎝ ⇒

⎛ ⎞ Iω = J ⎜ ⎟ ⎝ 2⎠

⇒

m 2 ⎛ ⎞ ω = J⎜ ⎟ ⎝ 2⎠ 12

⇒

ω=

mv0 r ( 70 ) ( 1 )( 2 ) = I 200

ωr = ω +

A

C

⎛ ⎞ ⎛  ⎞ ⎛ 6J ⎞ J v = ⎜ ⎟ω = ⎜ ⎟⎜ = ⎝ 6⎠ ⎝ 6 ⎠ ⎝ m ⎟⎠ m

CHAPTER 3

79. Since spheres are smooth, so no transfer of angular momentum takes place from A to B. However, sphere A only transfers its linear velocity v to sphere B and stops. Hence, we conclude that A stops but continues to rotate with same angular speed ω and B moves with speed of A but with zero angular speed. Hence, the correct answer is (B).

H.183

6J m

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 183

⎛v ⎞ ⎛ 1 ⎞ ⎛ 2π ⎞ 5π θ =⎜ 0 ⎟t=⎜ ⎟⎜ = ⎝ r ⎠ ⎝ 2 ⎠ ⎝ 1.2 ⎟⎠ 6 Hence, the correct answer is (B). τ μmgR μ g = 85. Angular retardation α = = R I mR2 Since, ω = ω 0 − α t

ω0 − ω α ω ω0 − 0 ω R 2 = 0 ⇒ t= μg 2μ g R Hence, the correct answer is (D). ⇒

…(2)

t=

t=

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H.184

JEE Advanced Physics: Mechanics – II

86. Length of each side is 2L, so if r is the perpendicular distance of wire from the centre of mass (in this case, the geometrical centre), then

89. Let μ be the friction coefficient between sphere and each wall. Free body diagram of sphere is

N2

O

mg

N1 P

1

1

30° 30°

r

Net force on the sphere in horizontal direction is zero, so N1 cos 60° + μ N 2 cos 60° = N 2 cos 30° + μ N1 cos 30°

60°

The moment of inertia about O is

⇒

I = 6 ( I one side ) ⇒

ω

Q

r = 3L

⎛ m ( 2L ) ⎞ I = 6⎜ + mr 2 ⎟ ⎝ 12 ⎠

⇒

N1 ( 1 − 3 μ ) = N 2 ( 3 − μ )

⇒

N1 3−μ = N2 1 − 3μ

2

⎛ mL2 ⎞ I = 6⎜ + 3 mL2 ⎟ = 20 mL2 ⎝ 3 ⎠ Hence, the correct answer is (B).

N1 + μ N 2 = 3 ( N 2 + μ N1 )

1 MR 2 2 Perpendicular distance on y = x + c from origin is 0−0+c c r⊥ = d = = 2 2

1 we get, 3 1 3− N1 3 3 −1 4 3 = = = 1+ N2 3 3− 3 3 1− 3 f 4 μ N1 = 1+ Since, 1 = f2 μN2 3

According to Parallel Axis Theorem, we have

Hence, the correct answer is (B).

⇒

Substituting μ =

87. Moment of inertia about z-axis is

I = IG + Md ⇒

90. For Plank, Mg sin θ + f = Ma1 For Sphere, mg sin θ − f = ma2

2

1 1 ⎛ c ⎞ MR2 = MR2 + M ⎜ ⎝ 2 ⎟⎠ 2 4

2

where a2 − Rα = a1

c=

⇒

f

Mg sin θ

⎛2 ⎞ I system = 4 ⎜ Mr 2 ⎟ + 2 Ma 2 ⎝5 ⎠ 2 I system = M ( 4 r 2 + 5 a 2 ) 5

M

α mg sin θ

⎛2 ⎞ Since, fR = ⎜ mR2 ⎟ α ⎝5 ⎠ Solving equations (1), (2), (3) and (4), we get

…(4)

f =0 Hence, the correct answer is (A). r M

91. Since τ = Iα ⇒

M

M a

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 184

a2

f

88. I system = 4 I sphere + 2 Ma 2 ⇒

…(3)

a1

R 2 Hence, the correct answer is (A). ⇒

…(1) …(2)

⎡ m ( 2 ) 2 ⎤ ⎛ ω ⎞ τ=⎢ ⎥⎜ ⎟ ⎣ 12 ⎦ ⎝ t ⎠

m 2ω 3t Hence, the correct answer is (B). ⇒

τ=

2/9/2021 6:49:44 PM

Hints and Explanations

(

93. I =

98. The rod will rotate about A. Therefore, by Law of Conservation of Mechanical Energy, we get ⎛ Decrease in ⎞ ⎛ Increase in Rotational ⎞ ⎟ ⎜ Gravitational ⎟ = ⎜ Kinetic Energy ⎟ ⎜ Potential Energy ⎟ ⎜ about A ⎠ ⎝ ⎠ ⎝

)

3 Ma 2 10

Hence, the correct answer is (D).

⇒

⎛ mg ⎜ ⎝

⎞ 1 2 ⎟ = I Aω 2⎠ 2

⇒

⎛ mg ⎜ ⎝

 ⎞ 1 ⎛ m 2 ⎞ 2 ⎟= ⎜ ⎟ω 2⎠ 2⎝ 3 ⎠

⇒

ω2 =

3g 

…(1) ω

94. From conservation of angular momentum

( Iω = constant ) , angular velocity will remain half. As, 1 2 Iω 2 The rotational kinetic energy will become half. Hence, the correct answer is (B). K=

95. Moment of inertia of semi-circular portions about x and y axes are same. But moment of inertia of straight portions about x-axis is zero. So, we have Ix < I y ⇒

Ix BC > AB A

107. Applying Law of Conservation of Angular momentum about A, we get A

H.187

K=

h1 14 = h2 15 Hence, the correct answer is (C). ⇒

109. The rod will rotate about point A. Let a be the linear acceleration of centre of mass of the rod and α the angular acceleration of the rod about A. Then

112.

∫ τ dt = ΔL Since, τ = Fr⊥ = ( 2t ) ( R + r )

T

t

⇒

A α

∫

L = 2t ( R + r ) dt = ( R + r ) t 2 0

Hence, the correct answer is (C).

mg

mg − T = ma  τ mg 2 3 g Also, α = = = 2 I m 2 3  ⇒ a= α 2

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 187

…(1)

…(2)

113. The desired moment of inertia is xsinα = r⊥

α x

dx

…(3)

2/9/2021 6:50:09 PM

H.188

JEE Advanced Physics: Mechanics – II x =+ 

I=

∫



dI =

I=

2

−

x = − 

∫

dm ( r⊥ )

⎛ m

⎞

∫ ⎜⎝ 2 dx ⎟⎠ ( x sin α )

m 2 I= sin 2 α 3 Hence, the correct answer is (B). v1

⇒

x=

v1  v1 + v2

118.

⇒

n=

⇒

n  16

θ 100 = 2π 2π

I2 =4 I1 M2 R22 =4 M1R12

⇒

R23 =4 R13

⇒

R2 = 43 R1

1

T 24 = n2 n2 So, here we formulate a rule to calculate the new time 1 period. If Radius of earth shrinks by ( n > 1 ) then n T 24 new time period is 2 i.e. 2 . So new time period of n n rotation also decreases. Hence, the correct answer is (B). T′ =

5μ g 2R

θ = 100 rads −1

M2 = ( 2π R2 ) λ

2π 2π = n2 T′ T

α=

⇒

M1 = ( 2π R1 ) λ and

R I Since R goes to , so Ι goes to 2 , hence ω goes to n n n2ω {∵ Iω = constant } ⇒ ω ′ = n2ω

⇒

1 2 θ = ( 2 ) ( 10 ) 2

Let λ be the linear mass density. Then

Iω = constant

116. Angular retardation α =

⇒

⇒

115. By Law of Conservation of Angular Momentum

⇒

ω 0 40 = = 16 s α 2.5

Hence, the correct answer is (B).

Hence, the correct answer is (C).

⇒

t=

1 117. Since θ = α t 2 2

−x

v1 v = 2 x −x

α = 2.5 rads −2

Hence, the correct answer is (C).

IC

ω=

⇒

⇒

x

ω

α=

Now, 0 = ω 0 − α t

2

−

114.

5 × 0.1 × 10 2×1

⇒

τ ( μmgR ) = 2 I mR2 5

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 188

Hence, the correct answer is (B). 119.

ma cosθ

θ ma (pseudo force)

mg sinθ

θ

The sphere will continue pure rolling when ma cos θ = mg sin θ ⇒

a = g tan θ

Hence, the correct answer is (D).

2/9/2021 6:50:18 PM

Hints and Explanations 120. In case of rolling without slipping, the point of contact P on the ground is at rest. So, the cylinder will rotate about P with angular acceleration

H.189

Moment of inertia when the tortoise is at A MR2 2

I1 = mR2 +

and moment of inertia when the tortoise is at B I 2 = mr 2 +

Here, r 2 = a 2 + ⎡⎣ R2 − a 2 − νt ⎤⎦

P α

⇒

α=

2

From conservation of angular momentum

ω 0 I1 = ω ( t ) I 2

FR τP = I P 3 MR2 2

Substituting the values, we can see that variation of ω ( t )is non-linear. Hence, the correct answer is (B).

2F 3 MR

123. (0, 4)

4

Hence, the correct answer is (B).

CHAPTER 3

α=

MR2 2

=

x

+

121. If m be the mass of sphere, then

y

Ι ( x ) = Ι = Ι c + mx 2 Ι = Ι c at x = 0

(0, 0)

(−4, 0)

Therefore, Ι versus x graph is a parabola with minimum value of Ι = Ι c at x = 0. Therefore, the correct graph is (B).

L = mvr⊥ where r⊥ =

x

⇒

⎛ 4 ⎞ L = ( 5 )( 3 2 )⎜ ⎝ 2 ⎟⎠

⇒

L = 60 cgs unit

C

x

Hence, the correct answer is (D).

x

Hence, the correct answer is (B). 122. Since, there is no external torque, angular momentum will remain conserved. The moment of inertia will first decrease till the tortoise moves from A to C and then increase as it moves from C and D. Therefore, ω will initially increase and then decrease.

0−0−4 4 = 2 2

124. Conceptual the correct answer is (C). 125.

AB =2 BC ⇒

AB = DC =

  and BC = AD = 3 6

A

B

D

C

O a A

B C

D

vt

Let R be the radius of platform, m the mass of disc and M is the mass of platform.

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 189

Similarly, mAB = mDC = and mBC = mAD =

m 3

m 6

2/9/2021 6:50:24 PM

H.190

JEE Advanced Physics: Mechanics – II

Now I = 2I AB + I AD + I BC ⇒

⎧⎛ 1 ⎞ ⎛ m ⎞ ⎛  ⎞ 2 ⎫ ⎛ m ⎞ ⎛  ⎞ 2 I = 2⎨⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎬ + ⎜ ⎟ ⎜ ⎟ + 0 ⎩⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎭ ⎝ 6 ⎠ ⎝ 3 ⎠

189 7 m 2 = m 2 4374 162 Hence, the correct answer is (D). ⇒

I=

126. Let ω be the angular velocity of the rod. Applying, angular impulse equals change in angular momentum about centre of mass of the system, we get ω M

M

J = Mv

⎛ L⎞ J ⎜ ⎟ = Ι cω ⎝ 2⎠ ⇒

⎛ ML2 ⎞ L ( Mv ) ⎛⎜ ⎞⎟ = ( 2 ) ⎜ ⎟⎠ ω ⎝ ⎠ ⎝ 2

4

ω=

127. mg sin θ component is always down the plane whether it is rolling up or rolling down. Therefore, for no slipping, sense of angular acceleration should also be same in both the cases. Therefore, force of friction f always act upwards. Hence, the correct answer is (B). 128.

Hence, the correct answer is (B). 130. Let m be the mass of the disc. Then translational kinetic energy of the disc is 1 KT = mv 2 …(1) 2 As it ascends on a smooth track its rotational kinetic energy will remain same while translational kinetic energy will go on decreasing. At highest point. KT = mgh ⇒

1 mv 2 = mgh 2

⇒

h=

2

(6) v2 = 1.8 m = 2 g 2 × 10

Hence, the correct answer is (B).

v L Hence, the correct answer is (A). ⇒

Since, τ t = ΔL ⇒ ( mgR sin θ ) t = L

131. Since, net force on centre of mass is zero so, the centre of mass will not move at all. Hence, the body can only rotate about centre of mass. Hence, the correct answer is (B). 132. In case of pure rolling bottommost point is the instantaneous centre of zero velocity. Q

C P O ω

Velocity of any point on the disc, v = rω , where r is the distance of point from O. rQ > rC > rP

′

⇒

vQ > vC > vP

Hence, the correct answer is (A).

By Law of Conservation of Angular Momentum ⎛1 ⎛1 2⎞ 2 ⎜⎝ MR ⎟⎠ ω = ⎜ MR + ⎝2 2 ⇒

1⎛ M⎞ 2⎞ ⎜ ⎟ R ⎟ ω′ 2⎝ 4 ⎠ ⎠

5 ω = ω′ 4

4ω 5 Hence, the correct answer is (B). ⇒

133. At t = 2 s, A and B are at ( 0 , 2 m ) and ( 0 , 0 ) as shown. Component of relative velocity perpendicular to AB is 2 ms −1 . So,

ω=

2 = 1 rads −1 2 v1 = j A 2m

ω′ =

129. Force of friction passes through point P. Hence, its torque about P will be zero. Hence, only mg sin θ will provide torque to ring about P.

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 190

B

v2 = 2i

Hence, the correct answer is (B).

2/9/2021 6:50:31 PM

Hints and Explanations

F a

P α

f

Point of contact P is momentarily at rest, about Instantaneous Centre of Zero Velocity IC (located at P), i.e., ring will rotate about P, so

α=

F τ P F ( 2R ) = = I P 2 MR2 MR

For translational motion of ring, we have F + f = Ma = MRα = F ⇒

7 1⎛ 1 ⎛ 1⎛ M⎞ 2 ⎞ ⎞⎛ v ⎞ MgR = ⎜ M ′R′ 2 ⎟ ⎜ cm ⎟ + ⎜ ⎜ ⎟ vcm ⎟⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2⎝ 4 ⎠ 8 2 2 R′

⇒

7 1⎛ M⎞ 2 ⎛ M⎞ 2 MgR = ⎜ ⎟ vcm + ⎜ ⎟ vcm ⎝ 8 ⎠ 8 2⎝ 8 ⎠

⇒

7 ⎛ M⎞⎛ 3⎞ 2 MgR = ⎜ ⎟ ⎜ ⎟ vcm ⎝ 8 ⎠⎝ 2⎠ 8

⇒

vcm =

137. Tangential Force = FT = ma = m ( α L ) = N ∴ ⇒

Limiting value of friction ( f s )max = μ N = μ FT

( fs )max = μ N = μmα L

…(1)

Further if ω is the angular velocity at time t, then

ω = αt Also, the centripetal force is

…(2)

FC = mLω 2 = mL ( α 2t 2 )

135. Since moment of inertia of a rod about an axis passing through one end, making angle θ with it is I=

14 gR 3

Hence, the correct answer is (B).

f =0

Hence, the correct answer is (B).

1 2 m sin 2 θ 3

…(3)

For bead to slide FC > ( f s )max ⇒

mL ( α 2t 2 ) > μmα L

⇒

t>

μ α

So, the minimum time after which the bead begins to μ start is α Hence, the correct answer is (A).

y

θ

⎫ m 2 ⎧ m 2 So, I x = I y = 2 ⎨ sin 2 45° ⎬ = 3 ⎩ 3 ⎭ ⎛ m 2 ⎞ 2 2 Iz = 2 ⎜ = m ⎝ 3 ⎟⎠ 3 Hence, I x = I y < I z Hence, the correct answer is (B). Loss in Gain in ⎞ ⎛ ⎞ ⎛ Gain in ⎞ ⎛ Gravitational ⎟ ⎜ Rotational ⎟ ⎜ Translational ⎟ ⎜ 136. = + P.E. of K.E. of ⎜ ⎟ ⎜ K.E. of ⎟ ⎜ ⎟ ⎜⎝ ⎟⎠ ⎜⎝ C.M. ⎟⎠ ⎜⎝ ⎟⎠ C.M. C.M. ⇒

2

⇒

CHAPTER 3

134. Let f be the friction on the ring towards right, a be the linear acceleration and α be the angular acceleration of the ring about centre of mass.

H.191

⎛1 ⎞ 1 2 MgR − M ′gR′ = ⎜ I ′ω 2 ⎟ + M ′vcm ⎝2 ⎠ 2

⎧ R π R 2 ρ M ⎫ = ⎬ ⎨ where R′ = & M ′ = π R′ 2ρ = 4 ⎭ 2 4 ⎩

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 191

138. In pure rolling mechanical energy remains conserved, therefore speed will be same in both the cases. Acceleration of the sphere, a, down the plane is proportional to sin θ . So, a ∝ sin θ i.e., acceleration and hence, time of descend will be different. Hence, the correct answer is (B).    139. v = ω × r  ⇒ v = ( 2kˆ ) × 2iˆ + 2 ˆj = 4 ˆj − iˆ

(

) (

)

Hence, the correct answer is (B). 140. In uniform circular motion the only force acting on the particle is centripetal (towards centre). Torque of this force about the centre is zero. Hence, angular momentum about centre remain conserved. Hence, the correct answer is (A).

2/9/2021 6:50:38 PM

H.192

JEE Advanced Physics: Mechanics – II

  141. For Angular Momentum to be conserved τ ext = 0 a 3 6 ⇒ = = 2 −6 −12 ⇒ a = −1 Hence, the correct answer is (B).

The cube will topple if torque due to mg sin θ about P is greater than torque due to mg cos θ about P i.e.,

( mg sin θ ) ⎛⎜⎝ 2 ⎞⎟⎠ > ( mg cosθ ) ⎛⎜⎝ 2 ⎞⎟⎠ a

a

Since f , the force of friction passes through P, so torque due to f about P is zero.

142. Torque τ = TR

⇒

tan θ > 1

…(2)

From equations (1) and (2), we observe that cube slides before toppling if

μ μmg cos θ tan θ > μ

I cm =

The

143. τ = FR = Iα

⇒

2 MR2 5

K=R

147. By Law of Conservation of Energy

…(1)

Mgh =

K2 ⎞ 1 2 ⎛ MvCM ⎜⎝ 1 + 2 ⎟⎠ 2 R

Hence, the correct answer is (C). mg

sin

os

c mg

148. R.K .E. = ⇒

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 192

1 2 Iω = 1500 2

1 2 I ( α t ) = 1500 2

2/9/2021 6:50:48 PM

Hints and Explanations 2 152. I RIM = 2 MR

( 1.2 ) ( 25 )2 t 2 = 3000

⇒ t2 = 4 ⇒ t=2 Hence, the correct answer is (B).

I DIAMETER =

149. Since, Impulse = Change in Momentum, so we get J = mv

…(1) A

ω

1 MR2 2

⇒ Required Ratio= 4 : 1 Hence, the correct answer is (B). 153. Conceptual the correct answer is (D). 154. Pd = I Aω where I A =

 ω Since the distance from A to C is , we get vC = . 2 2 ⎛ ω ⎞ ⎤ ⎡1 d = ⎢ m 2 ⎥ ω ⇒ m⎜ ⎝ 2 ⎟⎠ ⎣3 ⎦

x v J

where v is the linear speed of centre of mass of rod. Also, Angular Impulse = Change in Angular Momentum ⎛ m 2 ⎞ …(2) Jx = Iω = ⎜ ω ⎝ 3 ⎟⎠  ⇒ v= ω …(3) 2 where, ω is the angular speed of rod about point A. Solving these three equations, we get 2 x=  3 Hence, the correct answer is (B). ⇒

150. Let the extension produced in the spring due to rotation be x. Then, for the body to be in equilibrium Fspring = Fcentrifugal ⇒

2  3 Hence, the correct answer is (B).

⇒

d=

155. sin θ =

h d d

Also, condition for banking is tan θ = ⇒

2 ⎡ ⎛ h⎞ ⎤ v tan ⎢ sin −1 ⎜ ⎟ ⎥ = ⎝ d ⎠ ⎦ rg ⎣

156. In case of pure rolling upto B, we have for a cylinder

x=

151. For the cube to topple about P, we must have τ F > τ mg

KT =2 KR where KT =

F 3a

v2 rg

Hence, the correct answer is (B).

kx = m (  + x ) ω 2

mω 2 k − mω 2 Hence, the correct answer is (B). ⇒

1 2 m 3

CHAPTER 3

⇒

H.193

A

mg a 2

P

a ⎛ 3a ⎞ ⇒ F ⎜ ⎟ > ( mg ) ⎝ 4 ⎠ 2 2 ⇒ F > mg 3 2 So, the minimum value of F is mg 3 Hence, the correct answer is (C).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 193

2 1 mgh and K R = mgh 3 3

h B h C

After B: Rotational kinetic energy is constant (because no torque is provided due to absence of friction) however translational kinetic energy increases.

2/9/2021 6:50:57 PM

H.194

JEE Advanced Physics: Mechanics – II

At C: KT =

2 5 mgh + mgh = mgh 3 3

while K R =

1 mgh 3

160. Let ω 1 and ω 2 be the final angular velocities when the slipping has ceased. Then v1 v2

KT =5 KR Hence, the correct answer is (B).    157. aP = aP0 + a0  Here, aP0 = acceleration of P with respect to O, so So,

   aP0 = aP0 t + aP0 n ⇒

ω2

ω1

v1 = v2 ⇒

ω 1R = ω 2 ( 2 R )

ω1 …(1) 2 ⎛ Angular ⎞ ⎛ Change in Angular ⎞ = Since, ⎜ ⎟⎠ , so Momentum ⎝ Impulse ⎟⎠ ⎜⎝ ⇒

    aP = ( aP0 t + aP0 n ) + a0

ω2 =

JR = Iω 1

…(2)

and J 2R = 4 I ( ω 0 − ω 2 )

rω 2

ap

  where, aP0 t =tangential component of aP0   and aP0 n =normal component of aP0 So,

  a0 + aP0 t = a + rα and  aP0 n = rω 2

⇒

2  2 aP = ( a + rα ) + ( rω 2 )

Hence, the correct answer is (A). 158. In case (i) work done by friction is zero, while in case (ii) it is non-zero. Hence, the correct answer is (B). 159.

P

2v v

O

where, J is the linear impulse due to friction which acts tangentially and is equal for both the cylinders. Solving equations (1), (2) and (3), we get

ω0 2 Hence, the correct answer is (B). ω 1 = ω 0 and ω 2 =

161. The frictional force μmg is the only horizontal force acting on the two bodies. So, each body has an μmg acceleration = μ g in opposite direction. So, m relative acceleration is 2 μ g. Hence, the correct answer is (B). g 162. a = M 1+ 2m So, v 2 = 2 ah ⇒ v is independent of R. Hence, the correct answer is (D). 163. I = Mk 2 = ( 40 )( 0.5 ) ⇒

⇒

ar =

vr2 4v 2 2v 2 = = 2R 2R R

Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 194

⇒

2

I = 40 ( 0.25 ) = 10 kgm 2

ω 0 = 1800 ×

Q

vr = vPQ = 2v

…(3)

α=

2π = 60π rads −1 60

ω 0 60π = = 2π rads −2 t 30

Also, τ = Iα ⇒ τ = 20π Hence, the correct answer is (B).

2/9/2021 6:51:05 PM

Hints and Explanations 164. By Law of Conservation of Energy

167. Since L = mvr⊥ and as r⊥ is constant, so L is constant. Hence, the correct answer is (B).

K2 ⎞ 1 2 ⎛ Mgh = MvCM ⎜⎝ 1 + 2 ⎟⎠ 2 R

168. Let J be the linear impulse imparted to the ball, then

K2 1 For a cylinder 2 = 2 R vCM =

v0

h

4 gh 3

ω0

Hence, the correct answer is (B).

Impulse =Change in Momentum

165. Let V be the linear velocity of centre of mass of rod just after collision and ω be its angular velocity, then by Law of Conservation of Linear Momentum, we have mv0 = Mv

⇒

J = mv0

…(1)

⎛ Angular ⎞ ⎛ Change in Angular ⎞ Also, ⎜ = ⎟⎠ Momentum ⎝ Impulse ⎟⎠ ⎜⎝

…(1) B

Jh = Iω 0 =

⇒

2 2 mr ω 0 5

…(2)

From equations (1) and (2), we get V

ω0 =

ω

CHAPTER 3

⇒

H.195

5 v0 h 2 r2

Hence, the correct answer is (B). 169. The point P will have a velocity v0 and a tangential

A

v ⎛ r⎞ velocity vT = ⎜ ⎟ ω = 0 inclined to each other at 60°. ⎝ 2⎠ 2

By Law of Conservation of Angular Momentum about centre of rod, we get ⎛ ML2 ⎞ mv0 x = ⎜ ω ⎝ 12 ⎟⎠ For A to be at rest just after collision, we have L ω =V 2

60° vT =

O

…(3)

So, vP2 = v02 +

Solving equations (1), (2) and (3), we get

v02 4

2

⎛v ⎞ + 2 ( v0 ) ⎜ 0 ⎟ cos ( 60° ) ⎝ 2⎠

v0 7 2 Hence, the correct answer is (C). ⇒

L 6 Hence, the correct answer is (B). x=

166. From the concepts of Projectile Motion.

P

r 2

…(2)

vP =

170. ϕ ϕ

ω = 1 2 gt 2 4r t= g 2r =

⇒

Hence, the correct answer is (D).

Since x = AB = ( 2v ) t ⇒

AB = 4v

gϕ 2ϕ Δϕ 2ϕ = = = 2 v sin ϕ Δt T ⎛ ⎞ v sin ϕ ⎜⎝ g ⎟⎠

r g

Hence, the correct answer is (C).

Mechanics II_Chapter 3_Part 7 Hints and Explanation_1.indd 195

171. Angular momentum, L = Iω 1 For the said axis, I = m 2 3 m 2 ⇒ L= ω 3 Hence, the correct answer is (B).

2/9/2021 6:51:12 PM

H.196

JEE Advanced Physics: Mechanics – II

Multiple Correct Choice Type Questions 1.

I1 =

1 1 MR2 + M 2 2 12

⇒

1 1 I1 = MR2 + M ( 4 R2 ) 2 12

⇒

1 1 5 I1 = MR2 + MR2 = MR2 2 3 6

Also, I 2 =

3.

( )

⇒

1 4 11 MR2 + MR2 = MR2 2 3 6

I2 =

⇒

I 2 > I1 and I 2 − I1 = MR2

Since the net linear momentum imparted to the triangular wedge is along x-axis and is non-zero, so the centre of mass of wedge ABC will move along x-axis. Hence, the correct answer is (B). 4.

In the frame of rod, the small vertical rods will experience centrifugal forces which forms a couple in clockwise direction in the state given in problem. To balance this couple force by hinge at A on the rod must be downward and the force by hinge at B must be upward. The angular momenta of the vertical rod particles about point O will be inclined to rod hence option (D) is also correct. Hence, (A), (B) and (D) are correct.

5.

Vc =

Hence, (A) and (B) are correct. 2.

Due to Law of Conservation of Angular Momentum  L =constant   ⇒ L ⋅ L = constant d ( ) L⋅L = 0 ⇒ dt   dL =0 ⇒ 2L ⋅ dt   dL ⇒ L⊥ dt    Since τ = A × L  dL   ⇒ = A×L dt    dL ⇒ must be perpendicular to A as well as L dt     A ⋅ L Further component of L along A is = x(say). Since   A  dL  dA A⊥ and =0 dt dt   d (   )  dL  dA ⇒ + L⋅ =0 A⋅L = A⋅ dt dt dt   ⇒ A ⋅ L = constant   A⋅L ⇒ = x = constant A    dL (or τ ) is perpendicular to L , hence it cannot Since  dt change magnitude of L but can surely change direction of L . Hence, (A), (B) and (C) are correct.

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 196

Wedge ABC

 p Triangular = ( mv ) iˆ Wedge ABC

1 1 MR2 + M ( 4 R2 ) 2 3

⇒

By Law of Conservation of Linear Momentum, we have    mviˆ + mvjˆ + mv − ˆj + 0 = p Triangular + 0

2m ( − v ) + m ( 2v ) + 8 m ( 0 ) =0 2m + m + 8 m

{OPTION (A)}

Further, by Law of Conservation of Angular Momentum

(I

system about O

)ω = ∑ mv r

⊥

system about O

⎡ 1 ⎤ 2 2 I system = ⎢ ( 8 m )( 6 a ) + m ( 2a ) + 2ma 2 ⎥ = 30 ma 2 12 ⎣ ⎦ about O

∑ mvr

⊥

= ( 8 m )( 0 )( 0 ) + 2m ( − v ) ( − a ) +

system about O

m ( 2v )( 2 a ) ⇒

∑ mvr

⊥

= 6 mva

system about O

⇒

( 30ma2 )ω = 6 mva

⇒

ω=

v 5a

{OPTION (C)}

Further total energy of the system E is 1 3 E = Ιω 2 = mv 2 {OPTION (D)} 2 5 Hence, (A), (C) and (D) are correct.

2/9/2021 6:41:43 PM

Hints and Explanations

λ0 g , then for equilibrium of the rod, we have 2 Στ = 0 and ΣF = 0.

L = mv0 r sin θ

If N =

L = mv0b = constant y

ΣF = 0 gives, Fapplied = Fpseudo = Ma

v0dt r1

b

x

dA mv0 r sin θ L = = 2m 2m dt

Hence, (B) and (D) are correct. C

VC = (2v)i

ω VB = vi

The mass M of the rod is l

λ ⎛λ ⎞ M = λ dr = ⎜ 20 r ⎟ dr = 0 ⎝ l ⎠ 2

∫

∫

0

0

Position of centre of mass of rod from end lying on the surface is l

rcm =

∫ 0

⎛ λ0 ⎞ ⎜⎝ 2 rdr ⎟⎠ r l l

λ0

∫l

2

rdr

9.

tan θ =

g 3 = a 4

⇒ θ = 37° Hence, (A), (B) and (C) are correct.  Fi = 3 ma {∵ Net force= ( Total mass )( acceleration ) } 1 F   i a = a0 = 3m Since τ = I 0α ⇒

Fb = ( 3 mr 2 ) α Fb ⇒ α= 3 mr 2 Hence, (A) and (C) are correct.

Hence, (B) and (C) are correct. l

4 g 3

⇒

A(VA = 0)

8.

a=

⇒

1 Since, dA = ( v0 dt ) r sin θ 2

B

⇒

Similarly, Στ = 0 gives F sin θ = N cos θ

  Also, r1 ≈ r2 = r

7.

2λ 0 g λ 0 a = 3 2

r2

θ

⇒

⇒

CHAPTER 3

6.

H.197

10. L1 = Iω = MK 2ω L2 = Iω + MvR

…(1)

⇒

L2 = MK 2ω + MR ( ω R )

⇒

L2 = Mω ( K 2 + R2 )

{∵ v = Rω } …(2)

From equations (1) and (2), we can see that L2 = 2L1, when K = R and L2 > 2L1, when K > R Hence, (B) and (D) are correct.

=

2l 3

0

11. Conceptual (C) and (D) are correct. 12. ω = ω 0 − aϕ ⇒

Free body diagram of rod is shown in Figure.

dϕ = ω 0 − aϕ dt ϕ

⇒

∫ 0

4W 4 Mg 2λ 0 g Given that F = = = 3 3 3

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 197

t

dϕ = dt ω 0 − aϕ

∫ 0

ϕ

⇒

1 − log e ( ω 0 − aϕ ) = t a 0

⇒

log e ( ω 0 − aϕ ) − log e ω 0 = − at

⇒

ω 0 − aϕ = e − at ω0

2/9/2021 6:41:53 PM

H.198

JEE Advanced Physics: Mechanics – II

ω0 ( 1 − e − at ) a dϕ ω 0 d [ 1 − e − at ] Further, ω = = dt a dt ω ⇒ ω = 0 [ − e − at ( − a ) ] a ⇒ ω = ω 0 e − at ⇒

ϕ=

{OPTION (A)}

⎛ v ⎞ α = cos −1 ⎜ ⎝ Rω ⎟⎠ So, the required angle θ is ⇒

{OPTION (C)}

Hence, (A) and (C) are correct. 13. When cylinder comes down, at the point where string leaves contact with the cylinder is point of instantaneous rest, thus string does zero work. 2 1 ⎛ mR2 ⎞ ⎛ v ⎞ 1 2 ω I ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ R⎠ KR 2 C 1 = = 2 = 1 1 KT 2 2 2 mv mv 2 2 Hence, (A) and (C) are correct.

14. The farther the mass is distributed from the Axis of Rotation (AOR), the more the Moment of Inertia. Here C must be towards heavier side and hence lies between O and B. ⇒

Rω cos α = v

I A > IO > I C > I B

Hence, (A) and (C) are correct. 15. Since L = J  Where J = mv is the impulse. Also, L = Ιω L J ⇒ ω= = 1 I m 2 3 3J ⇒ ω= m L2 3 J 2 = 2I 2m  3J and vC = ω = 2 2m Hence, (A), (B), (C) and (D) are correct. Kinetic energy =

16. Particle P will have a velocity in vertical direction, when

⎛ v ⎞ ⎛ v ⎞ θ = π − cos −1 ⎜ and θ = π + cos −1 ⎜ ⎝ Rω ⎟⎠ ⎝ Rω ⎟⎠ ⎛ v ⎞ Point P corresponds to θ = π − cos −1 ⎜ and it has ⎝ Rω ⎟⎠ velocity in vertically upward direction, while Q cor⎛ v ⎞ responds to θ = π + cos −1 ⎜ and it has velocity in ⎝ Rω ⎟⎠ vertically downward direction. Hence, (C) and (D) are correct. 17. If vcm be the velocity of centre of mass of the cylinder, ω be the angular velocity of the cylinder, then for no sliding between plank A and cylinder C, we have Rω − vcm = v

…(1)

Similarly, for no sliding between plank B and cylinder C, we have vcm + Rω = 2v

…(2)

Solving equations (1) and (2), we get 3v v and ω = , clockwise 2 2R Below the centre of the cylinder at a distance x (say), we observe a point which happens to be at instantaneous rest and this point is called the instantaneous centre (IC) of zero velocity and the axis passing through it is called the Instantaneous Axis of Rotation. At this point, we have vcm =

vcm − xω = 0 ⇒

x=

v2 vcm R = = ω 3v 2R 3

The translational kinetic energy of the system is KT = K A + K B + ( KC )T ⇒ ⇒

1 1 1 ⎛ v⎞ 2 m ( 2v ) + ( 2m ) v 2 + ( 8 m ) ⎜ ⎟ ⎝ 2⎠ 2 2 2 2 KT = 4 mv KT =

2

Since the cylinder is in pure rolling, so the kinetic energy of the system is KT = K A + K B + ( KC )rolling The kinetic energy in rolling is the sum of translational kinetic energy and rotational kinetic energy and is given by K rolling =

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 198

K2 ⎞ 1 2 ⎛ Mvcm ⎜⎝ 1 + 2 ⎟⎠ 2 R

2/9/2021 6:42:02 PM

Hints and Explanations 2

v 1 1 3 ( 8 m ) ⎛⎜ ⎞⎟ ⎛⎜ 1 + ⎞⎟ = mv 2 ⎝ 2⎠ ⎝ 2 2⎠ 2

1 1 3 2 m ( 2v ) + ( 2m ) v 2 + mv 2 2 2 2 9 ⇒ KT = mv 2 2 Hence, (A), (B), (C) and (D) are correct. ⇒

K=

fs Rough

Now F − f s = ma

18. Since, the broken away piece of mass m (say) has the same linear speed as it had before detaching from the disc, the angular speed of the remaining disc will remain stay unchanged. Further, the linear speed of the piece will first decrease and then increase due to gravity. So, its angular speed about an axis passing through O will first decrease and then increase.

m O

ω

R

O

Hence, (C) and (D) are correct. ⎛ Angular ⎞ ⎛ Change in Angular ⎞ 19. Since ⎜ = ⎟⎠ , so Momentum ⎝ Impulse ⎟⎠ ⎜⎝ L = τt ⇒

cm

L = F ( 2R ) t

…(1)

⎛1 ⎞ a Also, τ = f s R = I cmα = ⎜ mR2 ⎟ ⎝2 ⎠R 1 ⇒ f s = ma 2 Solving (1) and (2), we get 2F F a= and f s = 3M 3

…(2)

Hence, (B), (C) and (D) are correct. 24. Point of contact P is at rest. Taking torque due to all the forces about point P we observe that Torque is clockwise if F1 is applied i.e., spool will rotate clockwise. Torque due to F2 is zero i.e., spool will not rotate if F2 is applied, because F2 passes through P.

CHAPTER 3

So, ( KC )rolling =

H.199

Torque due to F3 and F4 is anticlockwise i.e., spool will rotate anticlockwise if only F3 or F4 is applied. Hence, (B) and (D) are correct. 25. Since surface is frictionless, the centre of mass of the rod will fall along the straight line

i.e., L varies linearly with time. Hence, (B) and (C) are correct.    20. Velocity of particle is v = ω × r and centripetal acceleration is       ac = ω × v = ω × ( ω × r )

CM y /2

θ

/2

CM

Hence, (B) and (D) are correct. 21. On a smooth horizontal surface, it can roll without slipping if v = Rω and no external force is acting on it. Hence, (A), (C) and (D) are correct. 22. On melting, moment of inertia increases and hence ω decreases. Hence, (B) and (D) are correct. 23. Since cylinder will roll towards right and required torque about centre of mass can be provided by friction if friction acts in the backward direction.

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 199

ycm =

 ( 1 − cos θ ) 2

⇒

dy ⎛  ⎞ dθ ω = ⎜ sin θ ⎟ sin θ = ⎠ dt dt ⎝ 2 2

⇒

a=

d2 y  2 ⎛ ⎞ = ω cos θ + ⎜ sin θ ⎟ α 2 ⎝ ⎠ 2 2 dt

α ω 2 cos θ + sin θ 2 2 Hence, (A) and (D) are correct. ⇒

a=

2/9/2021 6:42:08 PM

H.200

JEE Advanced Physics: Mechanics – II

26. Since the pulley has mass, so tension in both branches of string must be different.

29. From symmetry we have I1 = I 2 and I 4 = I 3

…(1)

Also, by perpendicular Axis Theorem I 0 = I1 + I 2 = I 4 + I 3

…(2)

From (2), we get

{∵

2I1 = 2I 3 ⇒

I1 = I 3

⇒

I 0 = I1 + I 2 = I 4 + I 3 = I1 + I 3

of ( 1 ) }

Hence, (A), (B) and (C) are correct.    30. Ltotal = Labout cm + Lcm about O Assume T1 to be absent then T2 provides a torque T2 R (in anticlockwise sense). Now assume T2 to be absent then T1 provides a torque T1R (in clockwise sense). So, Net Torque

⇒

CCW

Iα = ( T1 − T2 ) R

⇒

⎛ a⎞ I ⎜ ⎟ = ( T1 − T2 ) R ⎝ R⎠

{∵ τ = Iα }

{

∵α=

Ia ⇒ ( T1 − T2 ) = 2 R Further for m1

a R

}

…(1)

m1 g − T1 = m1a

…(2)

CW

Taking CW as positive, we get

τ = T1R − T2 R ⇒

 LO = ( I cm ) ω + Mvcm ( r⊥ cm from O )  

⇒

 ⎛1 ⎞ LO = − ⎜ MR2 ⎟ ω + M ( Rω )( 3 R ) ⎝2 ⎠  ⎛5 ⎞ LO = ⎜ MR2 ⎟ ω ⎝2 ⎠

Similarly, angular momentum about A is    LA = Labout cm + Lcm about A ⇒

 ⎛1 ⎞ ⎛1 ⎞ LA = ⎜ mR2 ⎟ ω + 0 = ⎜ mR2 ⎟ ω ⎝2 ⎠ ⎝2 ⎠

Hence, (A) and (C) are correct.

and for m2 T2 − m2 g = m2 a

…(3)

Adding (2) and (3) and using (1), we get m1 − m2 ⎡ ⎤ a=⎢ g 2 ⎥ m + m + I R 1 2 ⎣ ⎦

(

)

31. When the body is on the verge of toppling, then we can apply the equilibrium condition to get the desired results. Also, if the body topples about the point P, then the normal reaction will shift to this point as shown in Figure.

Hence, (A), (B), (C) and (D) are correct. 27. When the centre of mass moves a distance S, then the distance covered by the point of application of force F is 2S, therefore work done is W = F ( 2S ) From Work Energy Theorem, we have W = 2 FS = ⇒

v=

1 1 mv 2 + Iω 2 2 2

20 FS 7M

Hence, (A) and (B) are correct. 28. Moment of inertia will decrease but radius of gyration will increase because the distance of remaining mass from the axis has been increased. Radius of gyration depends on the distribution of mass from the axis. Hence, (A), (B) and (D) are correct.

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 200

So, we have ΣF = 0 and Στ = 0 For translational equilibrium of the body N = mg and F = fl = μ N = μmg For rotational equilibrium of the body, taking torque about P, we get ⎛ a⎞ F 3 = mg ⎜ ⎟ ⎝ 2⎠ ⇒

F=

mg 2 3

2/9/2021 6:42:20 PM

Hints and Explanations For rotational equilibrium of the body, taking torque about O, we get ⎛ 3a ⎞ mg ⎛ 3 a ⎞ ⎛ a⎞ ⎜ ⎟ + μmg ⎜⎝ ⎟ = mg ⎜⎝ ⎟⎠ 2 ⎠ 2 2 3⎝ 2 ⎠

μ=

2 3

When μ = 2 μmin P, we get

α=

a ⎞ ⎛ 17 2 ⎞ ma ⎟ α ⎟ =⎜ ⎠ 2 ⎠ ⎝ 12

6g 17 a

Hence, (A) and (C) are correct. 32. Relative velocity between A and B along AB should be zero. Hence vA sin θ = vB cos θ ⇒

α=

1 1 = for ring and 1+1 2

β=

2 2 = for solid sphere 2+5 7

Hence, (A) and (D) are correct. mg 1 = and F = , taking torque about 3 3

mg ( 3 a ) − mg ⎛⎜⎝ 3 ⇒

Therefore, fraction of its total energy associated with rotation is

vA = vB cot θ =

4v0 3 5

3

37° 4

⎛ component of relative velocity ⎞ ⎜⎝ ⎟⎠ perpendicular to AB Since, ω = AB B vA cos θ + vB sin θ 

⇒

ω=

⇒

⎛ 4v0 ⎞ ⎛ 4 ⎞ ⎛ 3⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ + v0 ⎜⎝ ⎟⎠ 5v 3 5 5 = 0 ω=  3

Hence, (B) and (D) are correct. 33. Conceptual (A), (B), (C) and (D) are correct. 34. In case of pure rolling,

KR k2 = 2 , so KT k + r 2

35. If F1 or F2 is applied, the centre of mass of spool will move towards right and it will rotate anticlockwise about centre of mass. Hence, friction will be towards left. If F3 is applied, centre of mass of spool will move towards left and it will rotate anticlockwise about centre of mass. Hence, friction will be towards left. If F4 is applied, then the centre of mass of spool will move towards left and it will rotate counter clockwise and F4 in itself has a tendency of both. So, friction may be zero, leftwards or rightwards depending upon the moment of inertia, mass and radius. Hence, (A), (B) and (D) are correct.

CHAPTER 3

⇒

1

H.201

36. Since no external torque is acting on the system hence  L is conserved.    ( Lstudent + stool )initial + ( Lwheel )initial = ( Lstudent + stool )final +  ( Lwheel )final    ⇒ 0 + L0 = ( Lstudent + stool )final + ( − L0 )   ⇒ ( Lstudent + stool )final = 2L0 Hence, (A) and (C) are correct. 37. Speed of the bottommost point is zero but acceleration is not zero. Friction force may be there if it is an accelerated motion but work done by friction is always zero. Hence, (A) and (D) are correct. 38. This figure makes us conclude that section AB and BC have equal kinetic energy. Also, section ABC has greater kinetic energy than section ADC and BC has greater kinetic energy than CD.

KR K 2 = 1 for a ring and R = for a solid sphere. KT KT 5 where, K R =rotational kinetic energy and KT = translational kinetic energy

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 201

Hence, (A) and (B) are correct.

2/9/2021 6:42:33 PM

H.202

JEE Advanced Physics: Mechanics – II

39. L = mvx h ⇒

⎛ v 2 sin 2 45 ⎞ L = m ( v cos 45 ) ⎜ ⎟ 2g ⎝ ⎠

vx = v cos 45°

v 45°

⇒

L=

h

v 2 sin 2 45 ⎪⎫ ⎪⎧ ⎬ ⎨∵ h 2g ⎭⎪ ⎩⎪

mv 3 4 2g

Also, from h = ⇒

2 Ki 3 Obviously, the lost energy must change to heat and actually this energy is lost due to the presence of friction which helps the two bodies attain a common angular velocity. Hence, (A), (B) and (D) are correct. ⇒

Loss =

42. Initially, when rod is vertical, magnitude of normal reaction is mg and only gravitational force is acting on the centre of mass of the rod in the vertical downward direction.

v2 4g

mv Also, L = h 2 L= ⇒

θ

(

)

m 2 gh h 2

2

Hence, (B) and (D) are correct. 40. When the net force on a rigid body is zero, then the summation of moments about any point is constant. Hence, (A) and (C) are correct. 41. By Law of Conservation of Angular Momentum ⎛1 ⎛1 2⎞ 2 2⎞ ⎜⎝ MR ⎟⎠ ω = ⎜⎝ MR + MR ⎟⎠ ω ′ 2 2 ω ⇒ ω′ = 3 Loss in Kinetic energy= K i − K f 1 2 1 Iω = MR2ω 2 2 4

⇒ ⇒ ⇒

⎛ M 2 ⎞ ⎛ ⎞ Also, N ⎜ cos θ ⎟ = Ια = ⎜ α ⎝2 ⎠ ⎝ 12 ⎟⎠ ⇒

⇒

α

 = acos θ 2

…(1) …(2) …(3)

After solving (1), (2) and (3), we get Mg 4 Hence, (A), (B) and (D) are correct. N=

43. When seen from the frame attached to the plank, net torque about P is zero, so we have R R ⎛ ⎞ ma ⎜ R − cos θ ⎟ = mg sin θ ⎝ ⎠ 2 2

1 ( I total )ω ′ 2 2 1⎛ 3 ⎞⎛ 1 ⎞ K f = ⎜ MR2 ⎟ ⎜ ω 2 ⎟ ⎠⎝ 9 ⎠ 2⎝ 2 1 Kf = MR2ω 2 12 ⎛1 1⎞ Loss = ⎜ − ⎟ MR2ω 2 ⎝ 4 12 ⎠ 1 MR2ω 2 6 2⎛ 1 ⎞ Loss = ⎜ MR2ω 2 ⎟ ⎠ 3⎝ 4

cosθ

Mg − N = Ma

Kf =

⇒

mg

Now consider the situation in the figure, then

L = m 2 gh 3

Where K i =

CM

N

v = 2 gh

Loss =

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 202

⇒

a=

g sin θ 2 − cos θ

2/9/2021 6:42:45 PM

Hints and Explanations

Reasoning Based Questions

Also, f = μ N = ma

μmg = ma

…(1)

1.

a 1 = g 2 For the plank, we have ⇒

μmin =

T ∝ I ∝ R2 it means if size of earth changes then it’s moment of inertia changes. If radius is half so time T 24 period will become = = 6 hr. 4 4 Hence, the correct answer is (A).

F − f − f ′ = ma ⇒

F − μmg − 2 μmg = ma

…(2)

Adding equations (1) and (2), we get F − 2 μmg = 2ma 2.

⇒ F = 100 N Hence, (B) and (D) are correct. 44. T cos θ = mg and T sin θ =

3.

⇒ ⇒

v=

4.

5.

θ T

Tcosθ

θ O

r

Q

6.

Now, PO =  cos θ

LH = ( m )( v )( PO )

7.

Since net torque is zero angular velocity remains constant. Hence, the correct answer is (A).

8.

Moment of inertia is then sum of mr 2 terms. We cannot change all the r’s, keep m’s the same, and expect Σmi ri2 to remain unchanged. Hence, the correct answer is (A).

9.

Velocity of point of contact

LH = ( 2 )( 1.7 ) ( 0.86 ) = 2.9 kgm 2s −1

Vertical component of angular momentum is LV = ( m )( v ) ( OQ ) LV = ( 2 )( 1.7 ) ( 0.5 ) = 1.7 kgm 2s −1   dL Also, = τ = mgr = ( 2 ) ( 10 )( 0.5 ) dt  dL ⇒ = 10 kgm 2s −2 dt Hence, (A), (B) and (C) are correct.

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 203

As the person climbs up, normal reaction and friction between the ladder and the wall both increases. This decreases normal reaction from the floor, decreasing limiting value of friction there. This increases the possibility of the ladder slipping. Hence, the correct answer is (C).

PO = ( 1 ) ( 0.86 ) = 0.86 m

⎛ 1⎞ Also, r =  sin θ = ( 1 ) ⎜ ⎟ = 0.5 m ⎝ 2⎠ Horizontal component of angular momentum is

If system has only rotational kinetic energy than momentum may be zero. Hence, the correct answer is (D).

mg

⇒

If angular velocity is constant then frictional force acting on sphere is zero. In case of pure rolling velocity of contact point is zero. Hence, the correct answer is (B).

P

⇒

Perpendicular axis theorem is not valid for a sphere. Hence, the correct answer is (D).

0.5 × 10 ≈ 1.7 ms −1 1.731

rg = 3

In the case of smooth surface, f r = 0, whereas in pure rolling motion friction can be present or absent. Hence, the correct answer is (B).

mv 2 r

1 v2 = tan θ = tan ( 30° ) = rg 3

When earth shrinks its angular momentum remains constant i.e., 2 2π L = Ιω = mR2 × = constant 5 T

CHAPTER 3

When θ = 37° , we get a = 5 ms −2 ⇒

H.203

v = vcm − Rω When pure rolling occurs vcm = Rω ⇒

V=0

Also, frictional force provides torque which further helps in achieving the pure rolling condition. Hence, the correct answer is (B).

2/9/2021 6:42:56 PM

H.204

JEE Advanced Physics: Mechanics – II  Since τ f ≠ 0  ⇒ τ N ≠ 0 and torque by friction and normal reaction will be in opposite. Hence, the correct answer is (A).

10. ω

vCM, acm rω 2

rω

v

15. I1ω 1 = I 2ω 2

During the motion of rolling there is radial acceleration toward the centre. Hence contact point moves vertically upward.

If I 2 < I1 ⇒ ω 2 > ω1 The diver does work by pulling his limps and thus, ω increases or rotational kinetic energy increases. Hence, the correct answer is (A).

F = 2R 5

Linked Comprehension Type Questions

f=0

1. Hence, the correct answer is (D). 11.

By Law of Conservation of Mechanical Energy, we have 2 1 ⎡ ( 3 m ) ( 4 ) ⎤ 2 2 2 ⎥ ω = 8 m ω 2⎣ 3 ⎦

1 2

( 3 m ) ( g ) ( 2 ) = Iω 2 = ⎢

ω

⇒

ω=

r

1 3g 2  A

tion

ω 3

e ntan

a

Inst

a ω f rot is o x a ous

C

Hence, the correct answer is (A). 12. In sliding down, the entire potential energy is converted into kinetic energy. While in rolling down same part of potential energy is converted into kinetic energy of rotation, therefore linear velocity acquired is less. Hence, the correct answer is (A).

B

⎛ Change in ⎞ ⎛ Angular ⎞ ⎜ Angular ⎟ , we get = Applying, ⎜ ⎟ ⎝ Impulse ⎠ ⎜ ⎟ ⎝ Momentum ⎠

13. Frictional force on an inclined plane, for a disc is 1 f = g sin α 3 Hence, the correct answer is (B).

J1 ( 3 ) = Iω

14. As the block remains stationary Σfx = 0 i.e., F = N Σfy = 0 i.e., f = mg and Στ = 0   ⇒ τ f + τN = 0

⇒

⎛ 1 3g ⎞ 3 J1 = ( 16 m 2 ) ⎜ ⎝ 2  ⎟⎠

⇒

J1 =

3g 8 m 3 

⇒

J1 =

g 8 m 3 g = 8 m 3 3

Hence, the correct answer is (C). 2.

f

Let ω ′ be the angular speed in opposite direction. Again, applying Law of Conservation of Mechanical Energy, we get 1 2

( 3 m ) ( g ) (  ) = I ( ω ′ )2 = 8 m 2 ( ω ′ )2

F N mg

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 204

⇒

ω′ =

1 2 2

3g 

2/9/2021 6:43:06 PM

Hints and Explanations 6.

H.205

By Law of Conservation of Angular Momentum, we have

ω′ 60°

( 2m )( 2u ) (  ) =

( 3 m ) ( 2 ) 2 3

ω − ( 2m ) ( v1 ) (  )

A

A

ω

J 2 ( 3 ) = I ( ω + ω ′ ) = ( 16 m 2 ) ⇒

3.

2m 2u

1 3g ⎛ 1 ⎞ ⎜⎝ 1 + ⎟ 2  2⎠

C

B Just After Impact

⇒

4u = 4ω − 2v1

Hence, the correct answer is (B).

⇒

2u = 2ω − v1

Torque about point A is

Since, for an elastic collision we have, e = 1, so

⇒

⎛ 3 mR2 ⎞ F ( 2R ) = ⎜ α ⎝ 2 ⎟⎠

⇒

4F α= 3 mR

⇒

Hence, the correct answer is (C).

4F 3m

7.

In this case, the component of velocity along AC remains unchanged. Proceeding in the same manner, we can just replace 2u by 2u cos ( 30° ) = 3u and find v2 .

Hence, the correct answer is (D). 4.

…(2)

2u 3

v1 =

at = Rα at =

2u = v1 + ω

From equation (1) and (2), we get

So tangential acceleration of center of mass is

⇒

…(1)

⎛ Relative ⎞ ⎛ Relative ⎞ ⎜ speed of ⎟ = ⎜ speed of ⎟ , at point C ⎜ approach ⎟ ⎜ separation ⎟ ⎠ ⎝ ⎠ ⎝

τ = F ( 2 R ) = Iα

A

F 2u

R

A

A

ω

2m

N2

60°

3u

C

N1

C

u B

mg

N 2 = mg 5.

2m

B Just Before Impact

2 g 4 m 6 g ( 2 + 1 ) = 4 m ( 2 + 1 ) 3 3

J2 =

v

C

u ⎛ 2⎞ 3 Hence, v = ⎜ ⎟ u= ⎝ 3⎠ 2 3

F + N1 = mat

Speed of particle after impact is

N1 =

F 3

Hence, the correct answer is (C).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 205

{

∵ at =

4F 3m

}

v2 = v 2 + u 2 =

v

C u

B Just Before Imapct

Hence, the correct answer is (B).

⇒

CHAPTER 3

⎛ Angular ⎞ ⎛ Change in Angular ⎞ Since, ⎜ = ⎟⎠ , so Momentum ⎝ Impulse ⎟⎠ ⎜⎝

B Just After Imapct

2 u2 + u2 = u 3 3

Hence, the correct answer is (B).

2/9/2021 6:43:17 PM

H.206 8.

JEE Advanced Physics: Mechanics – II ω

μmg

v2, a2 m

v1, a1

μmg

m

11. Let v be the velocity of centre of mass (also at C) of rod and two particles and ω the angular velocity of the system. 10 ms–1

μmg a1 = = μg m

C

μmg = μg m

a2 =

6

If v1 and v2 be the velocities of plank and cylinder at time t, then relative velocity of centre of mass of sphere w.r.t. plank is ( v1 + v2 ) . So, when pure rolling begins, then

ω=

v1 + v2 r

where v1 = ( μ g ) t and v2 = ( μ g ) t

τ ( μmg ) r Now, α = = 2 2 I mr 5 ⇒

A

ms–1

A C

B

B

By Law of Conservation of Linear Momentum, we have ( 0.08 ) ( 10 + 6 ) = ( 0.08 + 0.08 + 0.16 ) v ⇒ v = 4 ms −1 Hence, the correct answer is (D). 12. Also, AB = CB = 0.5 m Similarly applying Conservation Momentum about the point C, we get

where, I system = I rod + I two particles =

( 1.6 ) (

v1 + v2 = rω

( 2μ g ) t = r ( ω 0 − αt )

⇒

( 2μ g ) t = rω 0 − ⎛⎜⎝ 2 μ g ⎞⎟⎠ t

Substituting in equation (1), we get

⇒

2 ⎛ rω ⎞ t= ⎜ 0⎟ 9 ⎝ μg ⎠

Hence, the correct answer is (C).

⇒

5

Hence, the correct answer is (B). Since v2 = ( μ g ) t v2 =

1 2 a1t 2

⇒

xP =

1 2 rω ( μ g ) ⎛⎜ 9 μ g0 ⎞⎟ 2 ⎝ ⎠

⇒

xP =

2 ⎛ r 2ω 02 ⎞ 81 ⎜⎝ μ g ⎟⎠

12

3

)

…(1)

2

+ 2

I system = 0.08 kgm 2

ω = 2 rads −1 13. Loss of kinetic energy = − ΔK = K i − K f 1 1 ( 0.08 )( 10 )2 + ( 0.08 )( 6 )2 − 2 2

⇒

− ΔK =

⇒

1 1 ( 0.08 + 0.08 + 0.16 )( 4 )2 − ( 0.08 )( 2 )2 2 2 − ΔK = 4 + 1.44 − 2.56 − 0.16

2 rω 0 9

Hence, the correct answer is (A). 10. xP =

Angular

2 ( 0.08 )( 0.5 )

⇒

⇒

of

( 0.08 )( 10 )( 0.5 ) − ( 0.08 )( 6 )( 0.5 ) = I systemω

5 ⎛ μg ⎞ α= ⎜ ⎟ 2⎝ r ⎠

when pure rolling begins at time t, then

9.

v

ω

⇒ − ΔK = 2.72 J Hence, the correct answer is (B). 2

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 206

14. By Law of Conservation of Mechanical Energy, we have 1⎛ 1 ⎞ mg ( 2R ) = ⎜ MR2 + mR2 ⎟ ω 2 ⎠ 2⎝ 2 ⇒

ω=

8 mg

( 2m + M ) R

2/9/2021 6:43:32 PM

Hints and Explanations

17. Just after hinge B breaks (and F is removed too) the rod will not be in equilibrium. Let the new reaction at A be N , then

m O

ω=

M , so we have 2 2g R

15. Let F be the force exerted by disk on the particle (upwards). Then,

( 2 g ) 2 = ⎜⎝

⇒

⎛ 2 2 ⎞ g = ⎜ α ⎝ 3 ⎟⎠

3g 2 ⎛ ⎞ Since a = ⎜ ⎟ α ⎝ 2⎠

α=

F − mg = mRω 2

N

2

8m g

⇒

F = mg +

⇒

mg ( 10 m + M ) F= ( 2m + M )

( 2m + M )

A

 ⎛ 3g ⎞ ⎜ ⎟ 2 ⎝ 2 ⎠ 3g ⇒ a= 4 So, from (1), we get ⎛ 3g ⎞ 20 − N = 2 ⎜ ⎝ 4 ⎟⎠

M , so we have 2

3 Mg 2 Hence, the correct answer is (D). 16. Since the rod is in equilibrium, so N1 + N 2 = ( 2 ) ( 10 ) + 4 N1 + N 2 = 24 N

…(1)

N1

N2  2

 4

⇒

a=

⇒

20 − N =

⇒ N=5N Hence, the correct answer is (D). ⎛ Loss in GPE ⎞ ⎛ Gain in RKE ⎞ ⎜⎝ of CM of Rod ⎟⎠ = ⎜⎝ of Rod ⎟⎠

B 2g

4N

⇒

Taking torque due to all forces about A, we get ⎛ ⎞ ⎛ 3 ⎞ N1 ( 0 ) + ( 20 ) ⎜ ⎟ + 4 ⎜ ⎟ + N 2 = 0 ⎝ 2⎠ ⎝ 4 ⎠     CCW ⇒

10 + 3 − N 2 = 0

⇒

N 2 = 13 N N1 = 11 N

Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 207

⎛( ) 2⎞ ( 2 ) g ⎛⎜  ⎞⎟ = 1 ⎜ 2  ⎟ ω 2 ⎝ ⎠ ⎝ ⎠ 2

2

3

3g  So, acceleration is ⇒

CW

From (1), we get

60 4

18. By Law of Conservation of Energy

 4

A

CW

B 2g

F=

⇒

⎛ ( 2 ) 2 ⎞ ⎟α 3 ⎠



⇒

⇒

Hence, the correct answer is (C).

Since, m =

…(1)

τ = Iα

m Final

Initial

Since, m =

2 g − N = 2a

ω

CHAPTER 3

R

O M

H.207

ω=

an = ω 2 = 3 g = 30 ms −2 …(2)

Hence, the correct answer is (C). 19. Applying Conservation of Angular Momentum, we get

∑ mvr

⊥

= ( I system ) ω

2/9/2021 6:43:44 PM

H.208

JEE Advanced Physics: Mechanics – II 1 ⎝3

⇒

( mu ) L = ⎛⎜ ML2 + mL2 ⎞⎟ ω

⇒

ω=

But u =

⎠

3 mu

( M + 3m ) L

Since m  M , so, we get 3mu ω≅ ML

…(1)

Hence, the correct answer is (C). 20. By Law of Conservation of Energy, we have

M 2 1 ⎛θ⎞ ( m ) ⎛⎜ ⎞⎟ gL sin ⎜ ⎟ ⎝ 2⎠ ⎝ m⎠ 3 2

⇒

Δp ≅

⇒

Δp ≅ M

gL ⎛θ⎞ sin ⎜ ⎟ ⎝ 2⎠ 6

Hence, the correct answer is (D). 22. Applying Law of Conservation of Angular Momentum about the hinge we get ⎛ ML2 ⎞ mvx = Iω = ⎜ ω ⎝ 3 ⎟⎠

⎛ Loss in RKE ⎞ ⎛ Gain in ⎞ ⎛ Gain in ⎞ ⎜ of Ball-Rod ⎟ = ⎜ GPE of ⎟ = ⎜ GPE of ⎟ ⎜ System ⎟ ⎜ CM of Rod ⎟ ⎜ Ball ⎟ ⎠ ⎝ ⎝ ⎠ ⎝ ⎠ ⇒

M 2 ⎛θ⎞ gL sin ⎜ ⎟ ⎝ 2⎠ m 3

O

x

mgL ( 1 − cos θ )

3g ( 1 − cos θ ) L

⇒

ω≅

⇒

3 mu ≅ ML

⇒

u≅

v

m

( M + 3 m ) Lω 2 = 3 g ( 1 − cos θ ) ( M + 2m ) M + 3 m ≅ M and M + 2m ≅ M

O

ω

1⎛ 1 ⎛ L⎞ 2 2⎞ 2 ⎜ ML + mL ⎟⎠ ω = Mg ⎜⎝ ⎟⎠ ( 1 − cos θ ) + 2 2⎝ 3

Since M  m , so

…(1)

Also, by Law of Conservation of Linear Momentum, we get Pi = Pf

M 2 ⎛θ⎞ gL sin ⎜ ⎟ ⎝ 2⎠ m 3

⎛L ⎞ mv = Mvc = M ⎜ ω ⎟ ⎝2 ⎠ From equations (1) and (2), we get 2L x= 3

Hence, the correct answer is (C).

Hence, the correct answer is (C).

3g ⎡ ⎛θ⎞⎤ 2 sin 2 ⎜ ⎟ ⎥ ⎝ 2⎠ ⎦ L ⎢⎣

21. Since angular momentum is conserved about the hinge so there will be no change in angular momentum of system, however the rod is hinged at support and due to this the linear momentum changes. So, Δp = p f − pi ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

⎡ ⎛ Lω ⎞ ⎤ Δp = ⎢ m ( Lω ) + M ⎜ − mu ⎝ 2 ⎟⎠ ⎥⎦ ⎣ M⎞ ⎛ Δp = ⎜ m + ⎟ Lω − mu ⎝ 2 ⎠ M ⎞ ⎛ 3 mu ⎞ ⎛ Δp = ⎜ m + ⎟ ⎜ ⎟ − mu ⎝ 2 ⎠⎝ M ⎠ ⎛ m 1⎞ Δp = ⎜ + 3 mu − mu ⎝ M 2 ⎟⎠ 3 Δp  mu − mu 2 1 Δp ≅ mu 2

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 208

⎛ ⎞ vC = Lω ⎝ 2⎠

⇒

23. I = ⇒

…(2)

m 2 3 m 2 = 3 I

…(1)

Further, let R be the radius of ring, then ⇒

 = 2π R  R= 2π

⎛ 2 ⎞ ⎛ 3I ⎞ I1 = mR2 = m ⎜ 2 ⎟ = ⎜ 2 ⎟ ⎝ 4π ⎠ ⎝ 4π ⎠ Hence, the correct answer is (B). 24.  = 2 ( 2π R ) ⇒

R′ =

 4π 2

I 2 = 2 m ( R′ ) =

2m 2 ⎛ 3 ⎞ =⎜ ⎟I 16π 2 ⎝ 8π 2 ⎠

Hence, the correct answer is (D).

2/9/2021 6:44:00 PM

Hints and Explanations 25. For block

R1 7 = R2 5

v ( θ ) = 2 gR1 cos θ

⇒

⎛ dθ ⎞ R1 ⎜ − ⎟ = 2 gR1 cos θ ⎝ dt ⎠

Hence, the correct answer is (C). ΔL 1 = = 0.5 Nm Δt 2 Hence, the correct answer is (C).

27. τ av =

R1 R1 θ h=R1cosθ

⎛ ω + ω0 ⎞ ⎛ L + L0 ⎞ t=⎜ t 28. θ = ⎜ ⎝ 2 ⎟⎠ ⎝ 2I ⎟⎠

CHAPTER 3

⇒

H.209

⎡ 3 + 2 ⎤( ) θ=⎢ 2 = 12.5 rad ⎣ 2 ( 0.4 ) ⎥⎦ So, number of revolutions is θ 12.5 N= = = 1.99 ≈ 2 2π 2 ( 3.14 ) Hence, the correct answer is (A). ⇒

⇒

− dθ = cos θ

⇒

2g dt = R1

⇒

2g dt R1

t1

0°

0

π 2

∫ ∫

t1 =

R1 2g

0°

∫

π 2

− dθ cos θ − dθ cos θ

…(1)

E = mgh , where h = gR2 cos θ 10 gR2 cos θ 7 Now again proceeding in the similar manner, we get v (θ ) =

7 R2 10 g

0°

∫

π 2

− dθ cos θ

30. C is the centre of mass of rod and O the combined centre of mass of system (ball + rod) after collision. If we consider the origin at the centre of the rod, C, then the combined centre of mass of ball + rod is at a distance x from C, as shown. Conserving angular momentum 2 about point O. 2 ⎡ ⎛ x ⎞ 2 m 2 ⎛ x⎞ ⎛ x⎞ ⎤ mv0 ⎜ ⎟ = I 0ω = ⎢ m ⎜ ⎟ + + m⎜ ⎟ ⎥ω ⎝ 2⎠ ⎝ 2⎠ ⎦ 12 ⎣ ⎝ 2⎠

…(2)

From equations (1) and (2), on dividing and using R1 = 2R2, we get t1 10 = t2 7 ⇒

t1 > t2

So, the ball will reach the bottom of the track first. Hence, the correct answer is (B).

⇒

⎛ x⎞ v0 ⎜ ⎟ ⎝ 2⎠ 6v0 x ω= 2 = ⎛  ⎞ ⎛ x 2 ⎞ ( 2 + 6x 2 ) ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ 12 2

⇒

ω=

d ⎛ 2 ⎞ ⎜⎝ 6 x + ⎟⎠ = 0 dx x

t1 = t2 R1 7 R2 = 2 10

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 209

6v0 ⎛ 2 ⎞ ⎜⎝ 6 x + ⎟⎠ x

⎛ 2 ⎞ For ω to be maximum or minimum quantity ⎜ 6 x + ⎟ ⎝ x⎠ should be minimum or maximum

26. From equations (1) and (2), already calculated, when

then

W = 6.25 J

Hence, the correct answer is (D).

where E is the total mechanical energy, so

t2 =

⇒ ⇒

2 K For the ball, we have, R = KT 5 1 5 ⇒ mv 2 = E 2 7

⇒

L2 L20 − 2I 2I 1 ( 4 − 9 ) = −6.25 J W= 2 ( 0.4 )

29. W = ΔK =

⇒

6−

2 =0 x2

2/9/2021 6:44:15 PM

H.210

JEE Advanced Physics: Mechanics – II v0

m

ω O

x/2 C

C

⇒

⎛ Jh ⎞ J R⎜ ⎟ > ⎝ Ι ⎠ m

⇒

h>

⇒ Just Before Impact

⇒

Just After Impact

⇒ h > 0.4 R i.e., in CASE-2 Hence, the correct answer is (D).

 x= 6 34.

⎛ 2 ⎞ d2 ⎜ 6x + ⎟ ⎝ x⎠

B

 Further is positive, hence at at x = 6 dx 2  x= , the denominator is minimum and hence ω is 6 maximum

mg

Further at x = 0, ω = 0, therefore ω will first increase and then it will decrease. Hence, the correct answer is (B).

J = pi − p f ( as pi > p f

L τ 0 mg 2 + mgL 9 g α= = = I0 8L mL2 + mL2 3 Hence, the correct answer is (B).

Now, J will be maximum when p f or v f for ball is

So, J

max

v ⎞ mv0 ⎛ = m ⎜ v0 − 0 ⎟ = ⎝ 2⎠ 2

Hence, the correct answer is (A). 32. Translational kinetic energy in both cases will be equal but rotational kinetic energy in CASE-1 will be less. This is because perpendicular distance of linear impulse from the centre is less. So, angular impulse imparted to the sphere will be less or ω will be less. Hence, the correct answer is (C). 33. Friction will be in forwards direction in the case of backward slipping just after being hit by the cue, i.e., Rω > v By Angular Impulse − Angular Momentum Theorem, we have Jh = Ιω where J is the linear impulse

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 210

α

mg

)

least. v f will be least when the ball strikes at centre v of rod, because in that case v f = 0 (by Law of 2 Conservation of Linear Momentum). In all other cases, one more term rω , will be added to this.

O

A

31. Impulse J = Δp = p f − pi ⇒

Ι mR 2 mR2 5 h> mR

35. Let ω be the angular speed just before striking with ground. B

B

O

mg 3L 4 O

A mg Initially

L 30° 4

60°

α

A Finally

⎛ Decrease in ⎞ ⎛ Increase in ⎞ ⎜ Gravitational ⎟ = ⎜ Rotational ⎟ ⎜ Potential Energy ⎟ ⎜ Kinetic Energy ⎟ ⎝ ⎠ ⎝ ⎠ ⇒

⎞ ⎛ 3 L ⎞ 1 ⎛ mL2 ⎛ L L⎞ mg ⎜ − ⎟ + mg ⎜ L − + mL2 ⎟ ω 2 ⎟⎠ = ⎜⎝ ⎝ 2 4⎠ ⎝ ⎠ 4 2 3

Solving this equation, we get, g L ⇒ vA = ( OA ) ω = Lω = 1 ⋅ 1 gL Hence, the correct answer is (D).

ω = 1⋅ 1

2/9/2021 6:44:27 PM

Hints and Explanations 36. The distance of centre of mass of the system from point A is  3

Fy = Centripetal Force

Therefore, the magnitude of horizontal force exerted by the hinge on the body is F =centripetal force ⇒

F = ( 3 m ) rω

⇒

⎛  ⎞ 2 F = ( 3m ) ⎜ ω ⎝ 3 ⎠⎟

⇒

37. Angular acceleration of system about point A is

τ α= A = IA

⎛ 3 ⎞ ⎟ ⎝ 2 ⎠

( F )⎜

2m

2

=

3F 4 m

Now, acceleration of centre of mass along x-axis is ⎛  ⎞⎛ 3F⎞ ax = rα = ⎜ ⎝ 3 ⎠⎟ ⎜⎝ 4 m ⎟⎠ ax =



τ 2t 2 2I 0 where τ = F ( 2R ) − F ( R ) = FR ⇒

K=

⇒

K=

Substituting the values, we get

K=

x

Fx 3 2

⇒

K=

⇒

K=

⇒

K=

CM 

B

C

Now, let Fx be the force applied by the hinge along x-axis. Then, Fx + F = ( 3 m ) ax ⇒

⎛ F ⎞ Fx + F = ( 3 m ) ⎜ ⎝ 4 m ⎟⎠

⇒

Fx + F =

⇒

Fx = −

⇒

Fx =

3 F 4

F 4

F , along negative x-axis 4

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 211

F 2 R 2t 2 2I 0

( 10 )2 ( 1 )2 ( 2 )2

40. In this case the body will rotate about bottommost point, so

y



F

2

1 ⎛ τ ⎞ 2 t I0 2 ⎜⎝ I 0 ⎟⎠

= 50 J 2×4 Hence, the correct answer is (B).

Fy

ω, α

K=

K=

F 4m

A

⇒ Fy = 3 mω 2 Hence, the correct answer is (C). 39. If ground is smooth the body will rotate about O, so 1 K = I 0ω 2 2 1 2 ⇒ K = I0 ( αt ) 2

2

⇒ F = 3 mω 2 Hence, the correct answer is (C).

⇒

38. If Fy be the force applied by the hinge along y-axis. Then,

CHAPTER 3

r=

H.211

τ 2t 2 2I

[ F ( 4R ) − F ( 3R ) ]2 t 2 2 2 ⎡⎣ I 0 + m ( 2R ) ⎤⎦

F 2 R 2t 2 2 ⎡⎣ I 0 + 4 mR2 ⎤⎦

( 10 )2 ( 1 )2 ( 6 )2 2[ 4 + 8 ]

⇒ K = 150 J Hence, the correct answer is (D). 41. Since both spheres are of equal masses and the second sphere is at rest, so just after collision the first sphere comes to rest and it will have only ω i.e., it will slip backwards. So, friction will be maximum and in forward direction. Let ν ′ be its linear speed and ω ′ its angular speed when it again starts pure rolling. Friction is passing through its bottommost point, so we can conserve angular momentum about an axis passing through its bottommost point and perpendicular to plane of motion. So,

2/9/2021 6:44:41 PM

H.212

JEE Advanced Physics: Mechanics – II ω

ω′

44. Minimum value of μ required in this case is v′

μmg

Just After Impact

When Pure Rolling Beging

⇒

2( 2 mR2 ) ω = ( mR2 ) ω ′ + mv′R 5 5

a

45. M =

2( 2⎛ ⎞ ⎛ v′ ⎞ mR2 ) ω = ⎜ mR2 ⎜ ⎟ + mv′R ⎟ ⎝ R⎠ ⎠ 5 5⎝

∫ 0

ρ0 ( a + x ) ρ ⎛ x2 ⎞ dx = 0 ⎜ ax + ⎟ 2 ⎠ a a ⎝

v′ =

a

42. Thinking on the same concept as discussed above, we again have

46. xcm =

∫

a

( dm ) x

0

ω″ v

v″

μmg

⇒

xcm

M

=

2 ⇒ mvR = ( mR2 ) ω ′′ + m ( ω ′′R ) R 5 5v ⇒ ω ′′ = 7R Now, angular impulse imparted to the sphere is ⎛2 ⎞⎛ 5 v⎞ ΔL = Ιω ′′ = ⎜ MR2 ⎟ ⎜ ⎝5 ⎠ ⎝ 7 R ⎟⎠ 2 ⇒ ΔL = mvR 7 Hence, the correct answer is (A). 43. In case of pure rolling g sin θ a= I 1+ mR2 a is minimum when I is maximum which is for ring

∫ 0

a

= 0

2s = a

2×5 =2s ⎛ 10 ⎞ ⎜⎝ ⎟⎠ 4

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 212

5a 9

Hence, the correct answer is (C). a

a

∫ 0

∫ 0

⇒

IA =

7 ρ0 a 3 12

⇒

IA =

7 Ma 2 18

ρ0 ( a + x ) 2 ρ ⎛ ax 3 x 4 ⎞ x dx = 0 ⎜ + ⎟ 4 ⎠ a a ⎝ 3

{

∵ M=

a 0

3 aρ0 2

}

Hence, the correct answer is (A). ⎛ Angular ⎞ ⎛ Change in Angular ⎞ = 48. Since, ⎜ ⎟⎠ , so we get Momentum ⎝ Impulse ⎟⎠ ⎜⎝ Ja = I Aω = ⇒

ω=

7 Ma 2 ω 18

18 J 7 Ma A

5a 9

g sin θ 10 = = ms −2 2 4

tmax =

0

3 aρ0 2

3 aρ0 2

equal to mR2. So,

⇒

=

ρ0 ( a + x ) xdx a

2 ⎛ ax 2 x 3 ⎞ = + ⎟ ⎜ 3a ⎝ 2 2 ⎠

47. I A = ( dm ) x 2 =

Li = L f

amin

a

Hence, the correct answer is (D).

2 Rω 7 Hence, the correct answer is (B). ⇒

tan θ mR2 1+ I

I is maximum for ring, hence minimum value of μ required for pure rolling will have maximum value for ring. So, it will be the first to start slipping. Hence, the correct answer is (A).

Li = L f ⇒

μmin =

C J

B

2/9/2021 6:44:53 PM

Hints and Explanations

1 I Aω 2 > Mg ( 2 AC ) 2

⇒

s=

v2 2μ g

⇒

s=

r 2ω 02 32 μ g

2

⇒

1 ⎛ 7 Ma 2 ⎞ ⎛ 18 J ⎞ ⎛ 5a ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ > 2 Mg ⎜⎝ ⎟⎠ 2 18 7 Ma 9

⇒

J>

⇒

52. Since 0 2 − v 2 = 2 ( − μ g ) s

M 70 ag 9 M J min = 70 ag 9

Hence, the correct answer is (D). 53. v =

J 10 = = 5 ms −1 m 2

CHAPTER 3

B will pass through a point vertically above A when

H.213

Hence, the correct answer is (B). 49. At the time of pure rolling, we have − v = v − rω ⇒

r 2ω 02 = 2v 2 8

rω 0 4 Hence, the correct answer is (A). ⇒

x

rω v= 2

Also, we have 1⎛ 1 1 1 ⎛1 2 2⎞ 2⎞ 2 ⎜⎝ I cmω 0 ⎟⎠ = mv + ⎜⎝ mv + I cmω ⎟⎠ 2 2 2 2 2 ⇒

v

ω

v=

50. When slipping ceases, we have 0 = v − ( μ g )t ⇒

v = ( μ g )t

⇒

t=

rω 0 4μ g

⎛ ⎞ J⎜ ⎟ ⎝ 2⎠ τ J 6J ω= = = = Ι Ι ⎛ m 2 ⎞ m 2⎜ ⎝ 12 ⎟⎠ ⇒

ω=

6 × 10 = 15 rads −1 2×2

At point P, which is say at distance r from centre where if, v and rω cancel each other then that point will be at rest just after impact. v 5 1 Therefore, r = = = m ω 15 3 Hence, the correct answer is (C). 54. In the given time angle through which the rod rotates is ⎛ π ⎞ π θ = ωt = 15 ⎜ ⎟ = = 60° ⎝ 45 ⎠ 3 y

Hence, the correct answer is (D). ⎛1 ⎞ 51. Li = ⎜ mr 2 ⎟ ω 0 ⎝2 ⎠

1m

⎛1 ⎞ L f = ⎜ mr 2 ⎟ ω ⎝2 ⎠ Since ω =

x

O 1m

ω0 , so 4

ΔL = L f − Li =

A

60°

πm 9

3 2 mr ω 0 8

Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 213

The displacement of centre of mass from origin O is ⎛ π ⎞ π s = vt = ( 5 ) ⎜ ⎟ = m ⎝ 45 ⎠ 9

2/9/2021 6:45:06 PM

H.214

JEE Advanced Physics: Mechanics – II

⇒

xA =

⎛π π 3⎞ + ( 1 ) sin ( 60° ) = ⎜ + ⎟ m ⎝9 9 2 ⎠

57. In process A, T = 5 N In process B, mg − T ′ = ma

1 ⇒ y A = 1 cos ( 60° ) = m 2 Hence, the correct answer is (D).

⇒

55. Acceleration of block aB = g sin ( 45° ) − μ g cos ( 45° ) ⇒

aB =

So, tangential acceleration will be greater in process A than that in B. Hence, the correct answer is (B). 58. The mechanical energy is conserved in both Aand B. Hence, the correct answer is (A).

g g g − = 2 2 2 2 2

and acceleration of cylinder aC = 1 where, I = ( 2m ) R2 2

g sin 45° Ι 1+ 2mR2

…(1)

59. Let the a be the acceleration of C.M. w.r.t. plank, then 2 μmg = ma ⇒

a = 2μ g

⇒

Substituting in equation (1), we get 2g 3 Hence, the correct answer is (D).

α=

μmg

α

5°

)

gc FBD of Cylinder

…(1) …(2)

μ N1 + N 3 = 2mg cos ( 45° )

…(3)

2mg sin ( 45° ) − N1 − f = 2ma

…(4)

1 ( 2m ) R 2 2 For no slipping, we have a = Rα Solving these equations, we get 3g a= 5 2 Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 214

⇒

a′ = 4 μ g

⇒

a′ = 4 ms −2

Hence, the correct answer is (C).

mg sin ( 45° ) + N1 − μ N 2 = ma

( f − μ N1 ) R Also, α =

…(2)

a′ = a + a0

2m

gs 2m

45°

a0 = 2 μ g

Acceleration a′ of point of contact w.r.t. the plank is

os

(4 in

(4 os gc m

Equations of motion are, N 2 = mg cos ( 45° ) + μ N1

Let a0 be the acceleration of point of contact due to rotation. a0 = rα

(4

5°

) 5°

) 5° (4 in gs m

FBD of Block

μmg

⇒

f

)

μ N2

N

3

1

μ

N

N

2

N

1

N

1

a

56. The free body diagrams of the block and the cylinder are as shown. a

mr 2 α 2

2μ g r

Acceleration of cylinder aC =

45°

…(1)

Further, τ = μ ( mg ) r =

I 1 = 2 2 2mR

⇒

T ′ = ( 5 − ma ) < T

…(5)

…(6)

60. t =

v v = a′ 4 μ g

⇒ t = 2.5 s Hence, the correct answer is (B). 61. Distance travelled by the cylinder w.r.t. plank in 2.5 s is 1 s = − vt + ( 2 μ g ) t 2 2 ⇒ s = –18.75 m ⇒ s = 18.75 m After t = 2.5 s the velocity of cylinder is v′ = − v + ( 2 μ g ) ( 2.5 ) ⇒

v′ = −5 ms −1

Remaining distance = 40 − 18.75 = 21.25 m

2/9/2021 6:45:22 PM

Hints and Explanations Time taken to cover 21.25 m is 21.25 t′ = = 4.25 s 5

⇒

64. Let N be normal force between the wall and sphere at the time of impact and Δt the time of impact, then

62. Let vcm be the velocity of centre of mass, then

N Δt = Δp = 8 m

Rω − 2

{∵ Δv = 8 ms −1 } ⎛2 ⎞ Further R ( μ N Δt ) = Δ ( Ιω ) = ⎜ mR2 ⎟ ( 5 ) ⎝5 ⎠

Rω Rω vcm = = 2 2 2

vP = Rω −

…(1)

{∵ Δω = 5 rads−1 }

Since vP = 2 ms −1 ⇒

vcm = Rω = 4 ms −1

⇒

ω = 4 rads −1

…(2)

ω

{∵ R = 1 m }

v = 4 ms-1

Now, acceleration of point P is

CHAPTER 3

⇒

⎛ 2⎞ F = ( 180 ) ⎜ ⎟ = 120 N ⎝ 3⎠

Hence, the correct answer is (C).

So, total time T = 2.5 + 4.25 = 6.75 s Hence, the correct answer is (D).

vP = vcm

H.215

2 aP = 10 = aT2 + aN

aT2

R⎞ ⎛ + ⎜ ω2 ⎟ ⎝ 2⎠

⇒

10 =

⇒

aT = 36 = 6 ms −2

Dividing equation (2) by (1), we get 1 μ= 4 Hence, the correct answer is (A).

2

If acm be the acceleration of the centre of mass then Rα =6 2

aT = acm −

⇒

acm −

⇒

acm = 12 ms −2

J net =

(

)

1 + μ2 J μJ

acm =6 2

⇒

65. Net linear impulse is J net = J 2 + μ 2 J 2

Hence, the correct answer is (C). J

63. If f be the force of friction, then ⎛ R⎞ ⎛1 ⎞ F ⎜ ⎟ − fR = Iα = ⎜ mR2 ⎟ α ⎝ 2⎠ ⎝2 ⎠ ⇒

where, J = ΔP = 16 Ns

F − f = ( 5 ) ( 12 ) = 60 2

…(1)

⎛ 1 ⎞ So, net impulse is J net = ⎜ 1 + ⎟ ( 16 ) = 4 17 Ns ⎝ 16 ⎠ Hence, the correct answer is (C).

F R/2 R

f

Also, F + f = macm = 10 × 12 = 120 From (1) and (2), we get 3F = 180 2

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 215

1 2 Iω K2 Rotational KE 2 66. = = 2 ∵ I = MK 2 , v = Rω Translational KE 1 Mv 2 R 2 Hence, the correct answer is (D). 1 2 Iω RKE K2 2 67. = = 2 2 TE 1 2⎛ R2 ⎞ R + K Iω ⎜ 1 + 2 ⎟ ⎝ 2 K ⎠ Hence, the correct answer is (D).

{

CM

…(2)

}

2/9/2021 6:45:36 PM

H.216

JEE Advanced Physics: Mechanics – II

1 mv 2 TKE R2 2 68. = = 2 2 2 TE ⎛ K ⎞ R +K 1 mv 2 ⎜ 1 + 2 ⎟ ⎝ 2 R ⎠

71. Direction of velocities of two points of the rod A and B are tangential as shown.

Hence, the correct answer is (B).

O ω

69. At maximum height, both the sphere and wedge have same horizontal velocity say v, then by Impulse − Momentum Theorem, we have

vA

Total Change in ⎞ ⎛ ( Im pulse ) = ⎜⎝ Momentum of Wedge ⎟⎠ + ( sphere ) ⇒

P = ( 4m ) v + ( m ) v

⇒

P = 5 mv

1 1 1 ⎛1 ⎞ 1 mv02 + I cmω 02 = ⎜ mv 2 + I cmω 02 ⎟ + ( 4 m ) v 2 ⎝2 ⎠ 2 2 2 2 1 1 mv02 = ( 5m ) v 2 + mgh 2 2

⇒

h=

h

4P 2 2P 2 = 10 m2 g 5m2 g

⇒

KE =

⎛2 ⎞ P⎜ r⎟ ⎝5 ⎠ Ph ω0 = = 2 2 I cm mr 5

ω0 =

⎞ 1 1 ⎛ m 2 I ΙAORω 2 = ⎜ + mh 2 ⎟ ω 2 ⎝ ⎠ 2 2 12

⇒

⎛ 1 ( 4 )( 4 ) ⎞ 4 × 10 × 3 = ⎜ + ( 4 )( 3 )⎟ ω 2 ⎝ 2 12 ⎠

⇒

ω = 3.2 rads −1

Hence, the correct answer is (A). 72. vc = hω = ( 3 ) ( 3.2 ) = 5.5 ms −1 Hence, the correct answer is (B). 73-76. The correct answer is 73(A), 74(C), 75(B) and 76(A). Combined solution to 73, 74, 75, 76 a1

α F

P mr

f

So, from (1), we get

⇒

mgh =

where, h = 4 − 1 = 3 m

1 1 ( 5m ) v 2 + I cmω 02 …(1) 2 2 Now, by Angular Impulse − Angular Momentum Theorem, we have 2 Ph = I cmω 0 , where h = 0.4 r = r 5

⇒

O

1 mC

1 ⎛1 ⎞ 1 70. KE = ⎜ mv 2 + I cmω 02 ⎟ + ( 4 m ) v 2 ⎝2 ⎠ 2 2

⇒

vB

C

Hence, the correct answer is (A).

⇒

B

When we draw perpendicular on them, they meet at O, the centre of sphere. So, the rod can be assumed in pure rotation about an axis passing through O and perpendicular to plane of paper (also called IAOR). Since, the surface is smooth mechanical energy will remain conserved. So, decrease in gravitational potential energy of CM of rod = Increase in rotational kinetic energy about IAOR

Just after sharp impulse is given to the sphere, let ω 0 be the angular velocity and v0 be the velocity of centre of mass of sphere, then by Law of Conservation of Energy, we have

⇒

A

2

KE =

P ⎞ 1⎛ 2 1 ⎞⎛ P ⎞ ( 5m ) ⎛⎜ + ⎜ mr 2 ⎟ ⎜ ⎝ 5m ⎟⎠ ⎠ ⎝ mr ⎟⎠ 2 2⎝ 5

KE =

3P 2 10 m

Hence, the correct answer is (C).

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 216

2

f

a2

F − f = ma1

…(1)

f = ma2

…(2)

2f fR τ = = I 1 mR2 mR 2 and a1 − Rα = a2 Also, α =

…(3) …(4)

2/9/2021 6:45:51 PM

Hints and Explanations Solving equations, (1), (2), (3) and (4), we get

greater than zero. Since, 4ω 2 = 4 kg − 5 g ( 1 − cos θ ) has a real ω at θ = π , when

F F F 3F a1 = , a2 = , f = ,α = 4m 4m 4 2mR

4kg − 5 g ( 1 − cos π ) > 0

77. Moment of inertia of the system about the given axis, 2

The particle will therefore, describe complete circle, only when

+ m ( 2 ) = 8 m …(1) 3 By Law of Conservation of Mechanical Energy in the two positions shown in figure, we get I=

2

k>

5 2 Hence, the correct answer is (B). ⇒

2

1 ⎛ kg ⎞ 1 2 I⎜ ⎟⎠ = Iω + mg 2 ( 1 − cos θ ) + ⎝  2 2 3 mg ( 1 − cos θ ) Substituting the value of I from equation (1), we get 4ω 2 = 4 kg − 5 g ( 1 − cos θ ) A

Also, T ( 1 ) =

A

ω θ B

m

B

Initial Position

General Position

When the rod comes to instantaneous rest, ω = 0, so, we get 4 kg − 5 g ( 1 − cos θ ) = 0 3 Substituting, k = 2, we get cos θ = − 5

π –θ A

T=a

⇒

1g − 1a = 1a

⇒

a=

⇒

T=a=5N

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 217

g = 5 ms −2 2

{∵ R = 1 m }

81. W = τθ = ( 5 )( 40 ) = 200 J Hence, the correct answer is (D). 1 2 1⎛ 1 2⎞ 2 Iω = ⎜ ( 2 )( 1 ) ⎟ ( 5 × 4 ) ⎠ 2 2⎝ 2

⇒ ΔK = 200 J Hence, the correct answer is (C). 2⎫ ⎧ M ( R 2 )2 ⎛ R ⎞ ⎪ ⎪ 2 + M⎜ 83. I = 4 ⎨ ⎬ + mR ⎟ ⎝ ⎠ 12 2 ⎪⎩ ⎪⎭

⇒

I=

8 MR2 + mR2 = 20 kgm 2 3

Hence, the correct answer is (B). 84.

c

f

Hence, the correct answer is (C). 78. Since the particle is attached to a rigid rod, its motion is restricted to a circular path. It will therefore, describe complete circles if its speed at the highest point is

…(2)

1 1 80. θ = α t 2 = ( 5 )( 16 ) = 40 rad 2 2 Hence, the correct answer is (A).

In this position, θ is an obtuse angle, so the height h of B above A is given by, ⎛ 3 ⎞ 6 h = 2 cos ( π − θ ) = 2 ⎜ ⎟ = ⎝ 5⎠ 5 6 Hence, B rises to above the level of A. 5

1 ( 2)a 2

Hence, the correct answer is (B).

h

θ

…(1)

⇒

82. ΔK =

B 2

kmin =

79. 1g − T = 1a

…(2)

ω0

5 2

CHAPTER 3

( 3 m ) ( 2 ) 2

H.217

θ

( 4 M + m ) g sin θ − f = ( 4 M + m ) a

…(1)

2/9/2021 6:46:05 PM

H.218

JEE Advanced Physics: Mechanics – II

⎛ a⎞ Also, τ = fR = I ⎜ ⎟ ⎝ R⎠ Solving (1) and (2)

⇒

…(2)

v0 2 Hence, the correct answer is (B). ⇒

7g 24 Hence, the correct answer is (D). a=

f = 20 a = μ × 28 g cos ( 30° )

μ=

5 12 3

⇒

Mv − Mv0 = − μ ( Mg ) t

⇒

v − v0 = − μ gt

Now, since pure rolling begins at t =

Hence, the correct answer is (D). 86.

ω0

ω v0

v

f at t = 0

at time t, when pure rolling starts

Iω − I ( −ω 0 ) = ( fR ) t ⇒

Iω + Iω 0 = ( fR ) t

⇒

1 ⎛v v ⎞ MR2 ⎜ + 0 ⎟ = fRt ⎝R R⎠ 2

⇒

Mv + Mv0 = 2 ft

…(1)

where f = μ Mg Also, we have Mv − Mv0 = − ft

⇒

⎛ 2v ⎞ v − v0 = − μ g ⎜ 0 ⎟ ⎝ 3μ g ⎠

⇒

v = v0 −

⇒

v=

…(1) 2v0 3μ g

2v0 3

v0 3

This result can also be obtained by using Law of Conservation of Angular Momentum, according to which Mv0R − Icmω 0 = MvR + Icmω

Let pure rolling start at time t, then

…(2)

So, from (1) and (2), we get 2v t= 0 3μ g Hence, the correct answer is (C). 87. Applying Law of Conservation of Angular Momentum about the point of contact, because net torque on the disc about the point of contact is zero, so Linitial = Lfinal Since, for a body in combined effect of rotation and translation, we have     L = Lcm + m ( r × v )

⇒

⎛1 ⎞v ⎛1 ⎞v Mv0R − ⎜ MR2 ⎟ 0 = MvR + ⎜ MR2 ⎟ ⎝2 ⎠ R ⎝2 ⎠R

⇒

Mv0R −

⇒ ⇒

Mv0R MvR = MvR + 2 2 Mv0R 3MvR = 2 2 v v= 0 3

Hence, the correct answer is (C). 89. L = mvr sin θ ⇒

L = ( 2 )( 4 ) ( 3 ) sin 30

⇒

L = 12 kgm 2s −1

Direction is found by Right Hand Thumb Rule Hence, the correct answer is (A). 90. τ = Fr sin θ ⇒

⎛ 1⎞ τ = ( 2 )( 3 )⎜ ⎟ ⎝ 2⎠

Lfinal = MvR + I cm ( 0 )

⇒

τ = 3 Nm

⎛1 ⎞v Mv0 R − ⎜ MR2 ⎟ 0 = MvR ⎝2 ⎠ R

Direction is found by Right Hand Thumb Rule. Hence, the correct answer is (D).

So, Linitial = Mv0 R − I cmω 0 and

⇒

v=

88. Since Mv − Mv0 = − ft

85. From above we get

⇒

Mv0 R = MvR 2

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 218

2/9/2021 6:46:20 PM

Hints and Explanations

1.

A → (r) B → (p) C → (p) D → (s)

∫

⇒

1 mv 2 Final KE of Ball 9 = 2 = Initial KE of Ball 1 mv 2 25 0 2

⇒

Impulse Delivered to Rod ( J ) 2 = Initial Momentum of Ball ( mv0 ) 5

mv − mv0 = − Fdt = − J

…(1)

mV = J

…(2)

( Li )ball about CM

…(3)

( Lf

( L f )rod

=

mv0 = mv + mV ⇒

v0 = v + V

⇒

ω ⎞ ⎡⎛ ⎤ ⎢ ⎜⎝ V + 2 ⎟⎠ − v ⎥ e = 1= −⎢ ⎥ 0 − v0 ⎢⎣ ⎥⎦ ω − v = v0 2

⇒

V+

⇒

v + v0 = V +

⎛ Also, mv0 ⎜ ⎝ ⇒

w 2

⎞ ⎛ ⎟ = mv ⎜⎝ 2⎠

2.

ω v0 = v + 6

…(5)

From (3) and (5), we get V=

ω 6

⇒

⇒

( Krod )rotation ( Krod )translation

 2ω 2 =3 12V 2

A → (q, r) B → (p) C → (s) D → (q) From graph in question, we see that F − f ⎛τ⎞ Since a = Rα = R ⎜ ⎟ , where a = ⎝I⎠ m

2 ω 3

…(6)

⇒

Solving (5) and (6), we get

ω=

=

F = 10t

So, from (4), we get v + v0 =

⎛ m 2 ⎞ ⎜⎝ ⎟⎠ ω 2 = 12 = ⎛ ⎞ 5 mv0 ⎜ ⎟ ⎝ 2⎠

⎛ m 2 ⎞ 2 1 2 ω Iω ⎝⎜ ⎠⎟ 2 = 12 2 1 mV mV 2 2

…(4)  ⎞ ⎛ m 2 ⎞ ⎟ +⎜ ⎟ω 2 ⎠ ⎝ 12 ⎠

)rod

( Li )ball

Iω ⎛ ⎞ mv0 ⎜ ⎟ ⎝ 2⎠

CHAPTER 3

Matrix Match/Column Match Type Questions

H.219

F − 20 ⎛ 20 × 1 ⎞ = (1) ⎜1 10 2⎟ ⎜⎝ × 10 × 1 ⎟⎠ 2

12v0 5

v0

F

v

F

ω

⇒ V

F = 60 N

So, when F = 60 N, friction reaches its maximum value, i.e. slipping will start and F becomes 60 N at 6 s. When f = 10 N, then

Just Before Impact

During Impact

Just After Impact

F − 10 ⎛ 10 × 1 ⎞ = (1) ⎜1 10 2⎟ ⎜⎝ × 10 × 1 ⎟⎠ 2

So, we get v=

3v0 2v ,V = 0 5 5

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 219

⇒

F = 30 N

Since F = 10t , so F becomes 30 N at 3 s

2/9/2021 6:46:33 PM

H.220

JEE Advanced Physics: Mechanics – II Since, I z = I x + I y

3.

4.

A → (r, s) B → (p, q) C → (s) D → (p, q, r, s) Conceptual

⇒

A → (s) B → (q) C → (p) D → (r) (A) Moment of inertia of the frame about an axis passing through AC

Iz =

4 ml 2 3

Moment of inertia about an axis passing through D ⎛ ID = I z + 4m ⎜ ⎝ 5.

2

l⎞ 4 2 7 ml 2 2 ⎟⎠ = ml + ml = 2 3 3

A → (q) B → (p) C → (s) D → (r) ⎛θ⎞ In general, as vP = 2v0 sin ⎜ ⎟ ⎝ 2⎠

6. ⎛1 ⎞ 2 I AC = 4 ⎜ ml 2 sin 2 45° ⎟ = ml 2 ⎝3 ⎠ 3 Using parallel axis theorem, we get I PP ′ = I AC ⇒

I PP′

(B) IQQ′ =

2

2 ⎛ l ⎞ + 4m ⎜ = ml 2 + 2ml 2 ⎝ 2 ⎟⎠ 3

8 = ml 2 3 2

2ml 5ml + ml 2 = 3 3

2

7.

1 2 (C) I RR′ = 4 × ml 2 sin 2 45° = ml 2 3 3

A → (r) B → (q) C → (t) D → (p) For (A), I =

5 MR2 4

For (B), I =

7 MR2 5

For (C), I =

3 MR2 2

For (D), I =

1 MR2 2

A → (s) B → (r) C → (q) D → (p) For rolling without slipping, acceleration of point P at bottom of cylinder equals the acceleration of plank, i.e. 2 ms −2 . So, we get ⇒

Rα − a = 2

…(1)

For rolling without slipping, acceleration of point A at the top of cylinder equals acceleration of the block B, i.e. 6 ms −2 . So, we get ⇒ ⎡ 1 ⎛ (D) I x = I y = 2 ⎢ ml 2 + m ⎜ ⎝ ⎣ 12

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 220

2 l ⎞ ⎤ 2ml 2 ⎟⎠ ⎥ = 2 ⎦ 3

Rα + a = 6

…(2)

where a is the acceleration of centre of mass of cylinder towards left.

2/9/2021 6:46:43 PM

Hints and Explanations

H.221

⎛ m 2 ⎞ ⎛ 5⎞ I2 = 0 + 2 ⎜ + m 2 = ⎜ ⎟ m 2 ⎝ 3 ⎟⎠ ⎝ 3⎠ ⎡ m 2 ⎤ 2 I3 = 4 ⎢ sin 2 45° ⎥ = m 2 = I1 ⎣ 3 ⎦ 3 m2 3

D

a = 2 ms −2 and α = 1 rads −2 ⇒

2 a = 4 ms

and α = 1 rads

m2 3

−2

Further we have, for B,

T = ( mcyl ) a = 2mcyl

⇒

2mcyl = 4 mB

⇒

mcyl mB

⇒

=2

1 2 a′t 2 where a′ is the acceleration at which the thread unwraps from the cylinder, so a′ = Rα 1 2 ⇒  = 20 + ( Rα ) ( 2 ) 2 ⇒  = 20 + 8 ⇒  = 28 m ⇒ Δ =  −  0 = 28 − 20 = 8 m

9.

A → (p, s) B → (p, r) C → (q, r) D → (p, r) For the case of pure rolling (upwards or downwards), the required value of friction acts in upward direction and in case of slip (forward or backward) maximum friction will act in backward or forward direction. A → (q) B → (s) C → (q) D → (r) ⎛ m 2 ⎞ ⎛ I1 = 2 ⎜ + 2( m )⎜ ⎝ ⎝ 12 ⎟⎠

2

⎛ 5 ⎞ m 2 4 2 + m⎜ ⎟⎠ = m ⎝ 12 2 3

( I AB )4 = ( IBC )4 =

Also,  =  0 +

8.

{∵ of symmetry}

MI of rods AB and BC about D i.e., 4 is

For cylinder, we have

mB g − 2mcyl = 6 mB

B

Note that I1 = I 3

mB g − T = mB ( 6 )

⇒

G

A

mB g − T = mB aB ⇒

5 2

⎛ m 2 ⎞ ⎛4 ⎞ 10 2 I4 = 2 ⎜ m + 2 ⎜ m 2 ⎟ = ⎝ 3 ⎟⎠ ⎝3 ⎠ 3

10. A → (p, q, r) B → (p, q, r) C → (p, q, r) D → (s) Conceptual 11. A → (r) B → (p) C → (p) D → (q) If total mass of disc be M, then mass of cut out portion M is m1 = 4 2

1 1 ⎛ R⎞ 15 MR2 − m1 ⎜ ⎟ = MR2 2 2 ⎝ 2⎠ 32 Also, by parallel axis theorem, we have

So, I1 =

I 2 = I1 +

By perpendicular axis theorem, we have I3 + I4 = I2

2

⇒

3M ⎛ R ⎞ ⎜ ⎟ 4 ⎝ 2⎠

I1 3 M ⎛ R ⎞ + ⎜ ⎟ 2 4 ⎝ 2⎠ I I2 − I3 = 1 2

and I 3 =

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 221

2

3M ⎛ R ⎞ 21 MR2 ⎜ ⎟ = 4 ⎝ 2⎠ 32

Since, I 2 = I1 +

⎞ 2 2 ⎟ = m 2⎠ 3

CHAPTER 3

Solving (1) and (2), we get −2

C 5 2

2

2

{

∵ Idiameter =

I1 2

}

2/9/2021 6:46:57 PM

H.222

JEE Advanced Physics: Mechanics – II

12. A → (p) B → (s) C → (p) D → (r) For all cases if J is the linear impulse, then

Since  = ⇒

J m Jr Also, ω = ⊥ IE v=

where, r⊥ is the perpendicular distance from centre E and v is rightwards. For pure rolling ω should be clockwise and hence J should be applied at A. If it is below A, angular velocity is anticlockwise and it will cause forward slip. 13. A → (q) B → (r) C → (p) D → (r) L2 and Iω = constant. 2I Insect first moves away from the axis, then towards it. Hence, I will first increase and then decrease.

Since, τ = 0, so L = constant, K =

14. A → (t) B → (s) C → (p) D → (p) g sin θ I = 1 for a thin walled cylindrical but 2 I mR 1+ mR2 shell a=

⇒ ⇒

g sin θ g = 2 4 g a g sin θ α= = = R 2R 4R a=

KT = ⇒

KT = KR =

⇒

KR =

1 1 1 ⎛ g sin θ ⎞ 2 mv 2 = m ( at ) = m ⎜ ⎟ 2 2 2 ⎝ 2 ⎠ 2 2

2

2

( t2 )

2 2

mg t sin θ mg t = 2 8

1 2 1 1 ⎛ g sin θ ⎞ 2 Ιω = Ι ( α t ) = ( mR2 ) ⎜ ⎝ 2R ⎟⎠ 2 2 2 2 2

2

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 222

Moment of inertia I of hollow sphere is maximum, hence a is minimum and t is maximum Total kinetic energy equals mgh for all KR 2 = , for solid sphere KT 5 KR 2 = , for hollow sphere KT 3 KR 1 = , for cylinder KT 2 So, rotational kinetic energy is maximum for hollow sphere. K Since T is maximum for solid sphere, hence it has KR maximum translational kinetic energy. 16. A → (p, r, s) B → (q, r, s) C → (q, r, s) D → (p, r, s) Conceptual 17. A → (q) B → (s) C → (p) D → (s) For A: At 2, K Total = mgh = E (say) 2 K Since, R = KT 5 2 2 ⇒ K R = E = mgh 7 7 ⎛ h⎞ For B: At 3, K Total = mg ⎜ ⎟ ⎝ 2⎠ Since, ⇒

( t2 )

g sin θ 2S , where a = I a 1+ mR2

KR 2 = KT 5

KT =

h⎞ 5 5 5⎛ E = ⎜ mg ⎟ = mgh 7 7⎝ 2 ⎠ 14

For C: From 3 to 4, K R will be constant, so

2 2

mg t sin θ mg t = 2 8

15. A → (q) B → (s) C → (q) D → (p)

2

t=

1 2 at 2

At 4, K R =

h⎞ 1 2⎛ ⎜⎝ mg ⎟⎠ = mgh 7 2 7

For D: At 4, KT = K Total − K R ⇒

1 6 KT = mgh − mgh = mgh 7 7

2/9/2021 6:47:12 PM

Hints and Explanations 18. A → (p, r) B → (p, r) C → (q, s) D → (p, s) Conceptual

r⊥

O

P

30° T

1m T

19. A → (s) B → (p) C → (p) D → (q, r) Conceptual

30°

Since the tension passes through O, so torque due to tension about point O is zero. So, τ mg = F × r⊥

⇒

    a1 = a1 cm + acm where acm = ( Rα ) iˆ

⇒

 a1 = rα iˆ − ω 2 rjˆ + Rα iˆ = ( R + r ) α iˆ − ω 2 rjˆ

Similarly, we have  a2 = − rα ˆj − ω 2 riˆ + Rα iˆ = ( Rα − ω 2 r ) iˆ − rα ˆj

where, r⊥ = OP = 1 sin ( 30° ) = 0.5 m

2.

⇒

τ mg = ( mg ) ( 0.5 )

⇒

τ mg = ( 1 ) ( 10 )( 0.5 )

⇒

τ mg = 5 Nm

If v be the linear velocity of rod after impact (upwards), ω be the angular velocity of rod and J be the linear impulse at A during impact, then by v

 a3 = − rα iˆ + ω 2 rjˆ + Rα iˆ = ( Rα − rα ) iˆ + ω 2 rjˆ

ω

 a4 = rα ˆj + ω 2 riˆ + Rα iˆ = ( Rα + ω 2 r ) iˆ + rα ˆj 21. A → (q) B → (s) C → (s) D → (p)

J

θ v0 A During impact

A

θ After impact

Impulse − Momentum Theorem, we have J = Δp = p f − pi

I m 2

For axis 1, K =

2 m ( 2a ) = a 3 3

For axis 2, K =

m ( 2a ) a = 12 3

For axis 3, K =

ma 2 a = 3 3

For axis 4, K =

ma 2 a = 12 12

2

⇒

J = mv − ( − mv0 )

⇒

J = m ( v + v0 )

Two forces are acting on the ball. (i) tension ( T ) (ii) weight ( mg )

Mechanics II_Chapter 3_Part 8 Hints and Explanation_2.indd 223

…(1)

⎛ Angular ⎞ ⎛ Angular ⎞ ⎜ Further by ⎜ − Momentum ⎟ , we have ⎟ ⎝ Impulse ⎠ ⎜ ⎟ ⎝ Theorem ⎠ m 2 ⎛ ⎞ J ⎜ cos θ ⎟ = Iω = ω …(2) ⎝2 ⎠ 12 Since the collision is elastic, so at the point of impact, e = 1. v ω 2

Integer/Numerical Answer Type Questions 1.

CHAPTER 3

mg

20. A → (q) B → (r) C → (p) D → (s)  Acceleration of 1 w.r.t. centre of mass is a1 cm = rα iˆ − ω 2 rjˆ    Since a1 cm − = a1 − acm

Since K =

H.223

θ

⇒

θ

⎛ Relative speed ⎞ ⎛ Relative speed ⎞ ⎜⎝ of approach ⎟⎠ = ⎜⎝ of separation ⎟⎠

2/9/2021 6:47:24 PM

H.224

JEE Advanced Physics: Mechanics – II

l v0 = v + ω cos θ 2 Solving equations (1), (2) and (3), we get ⇒

ω=

where v is the speed of the centre of the sphere at that moment and θ is the corresponding angle. The speed v can be found by using the Law of Conservation of Energy, according to which

…(3)

6v0 cos θ

l ( 1 + 3 cos 2 θ )

mgh =

For ω to be maximum, we have

mv 2 Iω 2 + 2 2

dω =0 dθ ⇒

6v0 d ⎛ cos θ ⎞ ⎜ ⎟ =0 l dθ ⎝ 1 + 3 cos 2 θ ⎠

⇒

( 1 + 3 cos2 θ ) ( − sin θ ) − cosθ ( −6 sin θ cosθ ) =0 ( 1 + 3 cos2 θ )2

⇒

1 + 3 cos 2 θ = 6 cos 2 θ

⇒

3 cos 2 θ = 1 1 cos θ = 3

⇒ ⇒ 3.

2 2 mr , v = rω and h = ( R + r ) ( 1 − cos θ ) 5 From these equations we get where I =

10 g ( R + r ) 17 r 2

ω=

∗= 3

For the block to be in equilibrium F = f = μmg

…(1)

R⎞ ⎛ 10 g ⎜ R + ⎟ ⎝ 16 ⎠

⇒

ω=

⇒

ω=

⇒

ω = 40 rads −1

⎛ R⎞ 17 ⎜ ⎟ ⎝ 16 ⎠

For block not to topple, we have ⎛ F⎜ b + ⎝

a⎞ a ⎟⎠ < mg 2 2

…(2)

5.

⇒

a⎞ ⎛ ⎛ a⎞ μmg ⎜ b + ⎟ < mg ⎜ ⎟ ⎝ ⎝ 2⎠ 2⎠

⇒

a⎞ a ⎛ ⎜⎝ b + ⎟⎠ μ < 2 2

⇒

0.4 = μ
torque due to N about G

Mg

mg

F.B.D. of hoop

F.B.D. of block of mass m

Solving, we get a=

Taking the limiting value, we get τ f = τN ⇒ ⇒

⇒

( f ) ( h ) = ( N )( b ) ( ma )( h ) = ( mg ) ( b )

⎛ b⎞ ⎛ 0.6 ⎞ ( ) a=⎜ ⎟g=⎜ 10 = 5 ms −2 ⎝ 1.2 ⎟⎠ ⎝ h⎠ So, if the acceleration of truck exceeds this value, the block will topple about A. The maximum acceleration/retardation at which the block is not disturbed, is the smaller of the two values, obtained above, i.e., 4 ms −2

(c)

=

2 = 1 ms −2 2

T =1N

and α =

⇒

mg

( M + 2m )

Tr T = = 5 rads −2 I Mr

16. Between A and B, there is forward slipping. Therefore, friction will be maximum and backwards (rightwards). At point B where v = Rω , ball starts rolling without slipping and force of friction becomes zero.

CHAPTER 3

N = mg

H.227

Hence, the maximum retardation can be 4 ms −2 Since, u = 72 kmhr −1 = 20 ms −1

{Q

So, applying v 2 = u2 − 2 as 2

⇒

s=

u 2a

⇒

s=

( 20 )2 = 50 m ( 2 )( 4 )

v = 0}

{Q a = 4 ms−2 }

14. Li = L f ⇒

⎛ ⎛ 200 R2 ⎞ 200 R2 ⎞ 2 ⎜⎝ 80 R + ⎟⎠ ω = ⎜⎝ 0 + ⎟ ω1 2 ⎠ 2

⇒

ω 1 = 1.8ω 0 = ( 1.8 )( 5 )

⇒

ω 1 = 9 rpm

15. If α be the angular acceleration of the hoop and a be the acceleration of its centre, acceleration of m would be a + rα . If I be the moment of inertia of the hoop, then I = Mr 2 Since, τ = Tr = Iα ⇒

Tr T α= = I Mr

and T = Ma For m, we have mg − T = m ( a + rα )

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 227

By Law of Conservation of Angular Momentum between points A and B about bottommost point (because torque of friction about this point is zero), we get LA = LB ⇒

m ( 0.7 v0 ) R − Ιω 0 = mvR + Ιω

Substituting ω 0 = v=

2 v0 v , ω = and I = mR2, we get R R 5

3 ⎛ 3 ⎞ v0 = ⎜ ⎟ ( 7 ) ms −1 = 1.5 ms −1 ⎝ 14 ⎠ 14

17. FBD of plate w.r.t. truck is shown in figure. Fy A Pseudo force ma

Fx

α C

x y

mg

Equations of motion for centre of mass of plate is ma − Fx ax = m

09-Feb-21 6:38:56 PM

H.228

⇒

JEE Advanced Physics: Mechanics – II ay = 0

v = 2 as = ( 2 )( 4 )( 2 ) = 4 ms −1

Fy = mg = 100 N

ω=

v 4 = = 40 rads −1 R 0.1

t=

2s = a

Angular acceleration about A is maR 2a = 3 3 R 2 mR 2 Since ax = Rα

α=

⇒

19. If moment of inertia of bigger disc is I =

ma − Fx 2 = a m 3 Fx =

a

T2

10 kg

30 kg

10 g

30 g

2

I I Iω 1 + ( 0 ) = Iω 2 + ω 2 4 4 ⇒

ω2 =

4ω 1 5

Initial kinetic energy is K i =

18. The free body diagrams of the bodies and the pulley are shown in figure, then T1

M(R 2)

MR2 2

I = 2 4 By conservation of angular momentum, we have

So, MI of small disc is I 2 =

1 1 ⎛ 100 ⎞ ma = ⎜ ⎟ ( 0.9 ) = 3 N 3 3 ⎝ 10 ⎠ Absolute acceleration of centre of mass is F 3 aC = x = = 0.3 ms −2 = 30 cms −2 m 10 2 a 2 ⎛ 0.9 ⎞ -2 Since, α = = ⎜ ⎟ = 1 rads 3 R 3 ⎝ 0.6 ⎠ ⇒

2×2 = 1 second 4

1 2 Iω 1 2

Final kinetic energy K f is Kf =

a

⇒

2

I ⎞ ⎛ 4ω 1 ⎞ 1 2⎛ 4⎞ 1⎛ ⎜ I + ⎟⎠ ⎜⎝ ⎟ = Iω 1 ⎜⎝ ⎟⎠ 2⎝ 4 5 ⎠ 2 5

⎛ Ki − K f p% = ⎜ ⎝ Ki

4 1− ⎞ 5 ⎟⎠ 100% = 1 × 100 = 20%

20. h = ( 48 − 20 ) m = 28 m

for 10 kg mass, we have

Total kinetic energy of the sphere at B is

T1 − 10 g = 10 a

…(1)

K = mgh

for 30 kg mass, we have

A

30 g − T2 = 30 a

…(2) α

h B

v

20 m T1

T2

d

for the pulley, we have

( T − T1 ) R = 2 ( T2 − T1 ) α= 2 1 mR2 2

⇒

α=

In case of pure rolling, we have

mR

2 ( T2 − T1 ) 0.1( T2 − T1 ) = 20 R R

For no slipping, a = Rα Solving the above equations, we get −2

a = 4 ms , T1 = 500 N and T2 = 180 N

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 228

…(3)

KT r 2 5 = = , for a sphere KR k 2 2

⇒

KT =

1 5 mv 2 = mgh 2 7

⇒

v=

10 ⎛ 10 ⎞ gh = ⎜ ⎟ ( 10 )( 28 ) = 20 ms −1 ⎝ 7 ⎠ 7

…(4) Now, t =

2h g

09-Feb-21 6:39:12 PM

Hints and Explanations 2 ( 20 ) =2s 10

Let moment of inertia of triangular plate ADE about its centroid ( G ′ ) be I1 and its mass be m1

t=

⇒

d = vt = ( 20 ) ( 2 ) = 40 m

m1 =

21. ⇒

( 4 m ) ( 2 R )2 2

(

2

3a 4

m1 ⎛ ⎜ 12 ⎝

)

2

m 3 ⎛ a⎞ ⎜ ⎟ = 4 ⎝ 2⎠ 4

×

2

a⎞ m a 2 ma 2 = ⎟⎠ = 2 4 × 12 4 192 a 2 3

a 3

3 − mR2 2

⇒

3⎞ ⎛ IO = mR2 ⎜ 8 − ⎟ ⎝ 2⎠

⇒

13 IO = mR2 2 IP =

I1 =

m

CHAPTER 3

⇒

IO =

a a 2 − = 3 2 3 2 3

Distance GG ′ =

So, MI of part ADE about centroid G is

⎛ mR ⎞ 3 ( 4 m ) ( 2 R )2 − ⎜ + md 2 ⎟ ⎝ 2 ⎠ 2 2

where d 2 = R2 + ( 2R )

H.229

2 ma 2 m ⎛ a 2 ⎞ ⎛ a ⎞ = + ⎜ ⎟ I 2 = I1 + m1 ⎜ ⎟ ⎝2 3⎠ 192 4 ⎝ 12 ⎠

2

5ma 2 192 So, MI of remaining part is ⇒

I2 =

= ⇒ 11 mR2 2

⇒

I P = 24 mR2 −

⇒

IP =

⇒

37 IP 37 = 2 = ≈3 IO 13 13 2

37 mR2 2

22. Let the side of triangle be a and its mass be m

23.

ma 2 5ma 2 11ma 2 11I 0 − = = 12 192 12 × 16 16

N = 11

1 ⎛l ⎞ I 0ω 2 = mg ⎜ sin θ ⎟ ⎝4 ⎠ 2 ⇒

2 2 1 ⎛ m ( 4a ) ⎛ l⎞ ⎞ 2 ⎛l ⎞ + m ω = mg ⎜ sin θ ⎟ ⎜ ⎟ ⎜ ⎝ 4 ⎠ ⎟⎠ ⎝4 ⎠ 2 ⎝ 12

⇒

7⎛ l⎞ 2 ⎜ ⎟ ω = g sin θ 6⎝ 4⎠

⇒

ω=

24 g sin θ 7l

τ Since, α = = I0

MI of plate ABC about centroid G is ma 2 I= 12 a Triangle ADE is also an equilateral triangular of side . 2

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 229

⎛l ⎞ mg ⎜ cos θ ⎟ ⎝4 ⎠ 12 g cos θ = 2 7l 7 ⎛ l⎞ m⎜ ⎟ 3 ⎝ 4⎠

μN θ

/4

N /4

O an

C

at θ mg

y

x

09-Feb-21 6:39:28 PM

H.230

JEE Advanced Physics: Mechanics – II

Now

∑ F = ma y

25. Equation of motion for the block is mg sin θ − T = ma

y

⇒

mg cos θ − N = mat

⇒

N = mg cos θ − mat

⇒

⎛l ⎞ N = mg cos θ − m ⎜ α ⎟ ⎝4 ⎠

⇒

⎛ 3 g cos θ ⎞ 4 N = mg cos θ − m ⎜ ⎟⎠ = mg cos θ ⎝ 7 7

⇒

{

Q at =

l α 4

}

⇒ ⇒

4 ⎛l ⎞ μmg cos θ − mg sin θ = m ⎜ ω 2 ⎟ ⎝4 ⎠ 7 4 μmg cos θ 6 mg sin θ − mg sin θ = 7 7 4 13 μmg cos θ = mg sin θ 7 7 4μ tan θ = 13

⇒

{

2 mv0 R = mvR + mvR 5 5 v = v0 7 a

Qω=

v = Rω v0

v

ω μk mg

s

μ k mg = μ k g and v 2 = v02 − 2 as m 25 v02 − v02 v02 − v 2 49 s= = 2a 2μk g

Since, a =

⇒ ⇒

a=

⇒

T = 1.5 a

Substituting in equation (1), we have 2.5 a = 5

24. Applying Law of Conservation of Angular Momentum about bottommost point, we get Li = L f

⇒

v R

a = 2 ms −2 2

⎛ a⎞ 26. Since, I = 0 + m ⎜ ⎟ + ma 2 ⎝ 2⎠

} 5 2 25 2 ma = ma 4 20

⇒

I=

⇒

N = 25

27. (a) By Law of Conservation of Mechanical Energy, we have Gain in Loss in ⎞ ⎛ ⎞ ⎛ ⎟ ⎜ Gravitational ⎟ ⎜ Rotational ⎜ Potential Energy ⎟ = ⎜ Kinetic Energy of ⎟ ⎟ ⎜ ⎟ ⎜ of Particle ⎝ ⎠ ⎝ disc + Particle ⎠ ⇒

mg ( 2R ) =

s=

12v02 Substituting the values, we get 49 μ k g

⇒

ω=

s=

( 12 )( 12 )2 = 4.8 m ( 49 )( 0.75 )( 9.8 )

⇒

ω=

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 230

2T M

2T 3

⇒

⇒

⎛2 ⎞ mv0 R = mvR + ⎜ mR2 ⎟ ω ⎝5 ⎠

α=

Rα = a =

So, ∗ = 4

⇒

…(1)

TR =α 1 MR2 2 ⇒

μ N − mg sin θ = man

⇒

⇒ T+a=5 Equation of motion for disk is

2T MR For no slipping, we get

Rod begins to slip, when

⇒

( 1 ) ( 10 ) sin 30° − T = ( 1 ) a

1⎛ 1 2 2⎞ 2 ⎜ MR + mR ⎟⎠ ω 2⎝ 2

8 mg

( 2m + M ) R ( 8 )( 5 )( 10 ) = 4 rads −1 ( 10 + 40 ) 0.5

09-Feb-21 6:39:44 PM

Hints and Explanations (b) If F be the force exerted by disk on the particle (upwards), then F − mg = mRω 2

1 1 (b) θ = α t 2 = × ( 2.618 )( 100 ) = 131 rad 2 2 O ω

(c) W = ⇒

m Final

Initial

30. Just after the thread is burnt, the forces acting on the rod are as shown in Figure.

2

8m g

F = mg +

( 2m + M ) mg ( 10 m + M ) F= ( 2m + M ) ( 5 )( 10 ) ( 50 + 40 ) F= ( 10 + 40 )

⇒ ⇒

F 0.1 m

C O

28. Equations of motion for the discs are, mg − T a= m

Angular acceleration about O is given by

α=

T

⇒

Similarly, α 1 = ⇒

TR 1 mR2 2

=

…(3)

α1 = α 2

For no slipping a = Rα 2

…(4)

Further, a − Rα 2 = Rα 1

…(5)

Solving these equations, we get T= ⇒

mg 2g 4 , a = g and α 2 = 5 5 5R

T = 4 N, a = 8 ms −2 , α 2 = 8 rads −2

29. (a) ω =

250 × 2π = 26.18 rads −1 60

1 2 × 30 × ( 0.2 ) = 0.6 kgm 2 2 Now, ω = α t Ι=

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 231

α ≅ 30 rads −2 ac = 3 ms −2

{downwards}

Since W + kx − F = maC

mg

2T mR

τ 5.6 = Ι ⎛ ( 2 )( 1 )2 2⎞ + ( 2 ) ( 0.1 ) ⎟ ⎜⎝ ⎠ 12

Now, ac = rα = ( 0.1 ) α

a

⇒ T

F = kx = 6 N

τ = ( 20 )( 0.1 ) + ( 6 )( 0.6 ) = 5.6 N

…(1) …(2)

B

W = 20 N

Torque due to forces about O is

TR 2T = 1 mR 2 mR 2

A

0.6 m

α

F = 90 N

α2 =

1 1 2 × Ι × ω 2 = × ( 0.6 ) × ( 26.18 ) 2 2 W ≈ 206 J

CHAPTER 3

R O M

⇒

ω = 2.618 rads −2 t

and τ = Ια = 1.57 Nm

m

⇒

α=

⇒

H.231

⇒

F = W + kx − maC

⇒ ⇒

F = 20 + 6 − ( 2 ) ( 3 ) F = 20 N

31. ω 0 = 600 rpm = 600 × ⇒

ω 0 = 20π rads −1

2π rads −1 60 {revolutions per minute}

Let α be the constant angular retardation, then ω = ω 0 − αt 0 = ( 20π ) − 3 ( α ) 20 ⇒ α= π rads −2 3 τ Further, α = I If R be the radius of fly wheel and F be the tangential force exerted by the brake lining on fly wheel, then ⇒

09-Feb-21 6:40:00 PM

H.232

JEE Advanced Physics: Mechanics – II 34. (a) By Law of Conservation of Angular Momentum, we get I1ω 2 = I 2ω 2

τ = μ FR 1 where I = mR2 2 From the above equations, we get μ FR 2μF 20 α= π= = 1 3 mR 2 mR 2 10π mR ⇒ F= 3μ Substituting the values, we get 10 × 22 × 20 × 0.12 F= 3 × 7 × 0.1 ⇒ F = 251.43 N

where ω 1 = ( 0.5 )( 2π ) rads −1 2

I1 = 1.6 + 2 × 4 × ( 0.9 ) = 8.08 kgm 2 2

I 2 = 1.6 + 2 × 4 × ( 0.15 ) = 1.78 kgm 2 and ω 1 = 0.5 rads −1 ⇒

⎛ 2π ⎞ ⎛ 2π ⎞ I1 ⎜ = ( I1 + ΔI ) ⎜ ⎝ T1 ⎟⎠ ⎝ T2 ⎟⎠

⇒

T2 I1 + ΔI ΔI = = 1+ T1 I1 I1

⇒

⎛ ΔI ⎞ ΔT = ( T2 − T1 ) = T1 ⎜ ⎝ I1 ⎟⎠

where, I1 =

1 1 2 I1ω 12 = ( 8.08 ) ( 2π × 0.5 ) ≈ 40 J 2 2 1 1 2 E f = I 2ω 22 = ( 1.78 ) ( 2π × 2.27 ) = 181 J 2 2 (c) W = E f − Ei ≈ 141 J (b) Ei =

32. By Law of Conservation of Angular Momentum, we get I1ω 1 = I 2ω 2 ⇒

⎛ 8.08 ⎞ ( )( ) ω2 = ⎜ 0.5 2π = 14.3 rads −1 ⎝ 1.78 ⎟⎠

35. For equilibrium of cylinder in horizontal direction, we have N1 = μ N 2

…(1)

For equilibrium of cylinder in vertical direction, we have …(1)

2 2 MR2 and ΔI = mR2 5 3

where, M is the mass of earth, m is the mass of ice. Substituting the values in equation (1), we get

N 2 + μ N1 = F + Mg

…(2)

Solving these two equations with F = 40 N, M = 2 kg 1 and μ = , we get 3 N1 = 18 N and N 2 = 54 N

( 24 × 3600 ) ⎛⎜ 2 × 3 × 1019 ⎞⎟ ⎝3 ⎠ = 0.7 s ΔT = ⎛2 ⎞ 24 ⎜⎝ × 6 × 10 ⎟⎠ 5 r r r r 33. Since, τ = ( r2 − r1 ) × F r r ⇒ τ = ⎡⎣ 4iˆ + 3 ˆj − kˆ − iˆ + 2 ˆj + kˆ ⎤⎦ × F r ⇒ τ = 3iˆ + ˆj − 2kˆ × iˆ + 2 ˆj + 3 kˆ

(

) ( ) (

(

⇒ ⇒ ⇒

iˆ ˆj r τ= 3 1 1 2 r τ = 195 x = 195

kˆ −2 = 7 iˆ − 11ˆj + 5kˆ 3 Nm

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 232

) )

Since, Fd = μ ( N1 + N 2 ) r ⇒ ⇒

⎛ 1⎞( 18 + 54 )( 0.1 ) μ ( N1 + N 2 ) r ⎜⎝ 3 ⎟⎠ d= = F 40 d = 0.06 m = 6 cm

12-Feb-21 4:38:03 PM

Hints and Explanations

H.233

ARCHIVE: JEE MAIN 1.

For equilibrium, τ B = 0 (torque about point B is zero)

4.

Angular momentum conservation gives ⎛ Ml 2 ⎞ mvl = ⎜ ω + ml 2ω ⎝ 3 ⎟⎠

( 1 )( 6 )( 1 )

1 ⎛ Ml 2 ⎞ 2 1 ( 2 ) 2 ⎜ ⎟ ω + ml ω = ( m + M ) rcm ( 1 − cos θ ) 2⎝ 3 ⎠ 2 ml + ( Ml 2 ) where, rcm = m+ M Ml ⎞ ⎛ ml + l 2ω 2 ⎛ M ⎜ ⎞ ( 2 ⎟ g ( 1 − cos θ ) ) ⇒ + m⎟ = m + M ⎜ ⎜ ⎝ m + M ⎟⎠ ⎠ 2 ⎝ 3

TA ( 100 ) − ( mg ) ( 50 ) − ( 2mg ) ( 25 ) = 0 ⇒

100TA = 100 mg

⇒ TA = 1mg Hence, the correct answer is (D). 2.

For equilibrium, ( τ )P = 0 ⇒

FR − mgx = 0

⇒

⎛ x⎞ F = mg ⎜ ⎟ ⎝ R⎠

⇒

⎛ ⎜⎝

2

5 ⎞ ⎛ 18 ⎞ ⎟ ⎜ ⎟ = 20 ( 1 − cos θ ) 6⎠⎝ 5 ⎠

⎛ 18 ⎞ ⎛ 3 ⎞ 1 − cos θ = ⎜ ⎟ ⎜ ⎝ 5 ⎠ ⎝ 20 ⎟⎠ 27 23 ⇒ cos θ = 1 − = 50 50 ⇒ θ ≈ 63° Hence, the correct answer is (B). ⇒

5. Since, x = R2 − ( R − a )

ω=

CHAPTER 3

18 rads −1 = 2 5 +1 3 Energy conservation gives ⇒

2

From free body diagram, we see that FV = mg and ml 2 FH = ω sin θ 2

2

⎛ R− a⎞ F = mg 1 − ⎜ ⎝ R ⎟⎠ Hence, the correct answer is (D). ⇒

3.

Since, both discs are rotating in same sense, so applying conservation of an angular momentum to the system, we get Linitial = Lfinal ⇒

I1ω 1 + I 2ω 2 = ( I1 + I 2 ) ω f

⇒

( 0.1 )( 10 ) + ( 0.2 )( 5 ) = ( 0.1 + 0.2 ) ( ω f

)

20 ⇒ ωf = 3 Kinetic energy of combined disc system is RKE =

20 1 1 ( I1 + I 2 )ω 2f = ( 0.1 + 0.2 ) ⎛⎜⎝ ⎞⎟⎠ 2 2 3

⎛ 0.3 ⎞ ⎛ 400 ⎞ 120 20 RKE = ⎜ J = = ⎝ 2 ⎟⎠ ⎜⎝ 9 ⎟⎠ 18 3 Hence, the correct answer is (D). ⇒

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 233

2

Net torque about CM due to FV and FH is ⎛l ⎞ ⎛l ⎞ τ net = FV ⎜ sin θ ⎟ − FH ⎜ cos θ ⎟ ⎝2 ⎠ ⎝2 ⎠ 2 2 ml ω According to problem, τ net = sin θ cos θ 12 l ⎛l ⎞ ⎛ ⎞⎛ l ⎞ ⇒ mg ⎜ sin θ ⎟ − ⎜ mω 2 sin θ ⎟ ⎜ cos θ ⎟ = ⎝2 ⎠ ⎝ ⎠⎝ 2 ⎠ 2 ml 2ω 2 sin θ cos θ 12 3g 2ω 2l Hence, the correct answer is (B). ⇒

cos θ =

…(2)

09-Feb-21 6:40:26 PM

H.234

6.

JEE Advanced Physics: Mechanics – II

I1 =

MR2 ρ ( π R2 ) tR2 = 2 2

⇒

I ∝ R4

⇒

I1 R14 1 = = I 2 R24 16

10. The moment of inertia of system about the specified axis is 2

2 ⎛ l ⎞ 2 + m ( 2l ) I = m( 0 ) + ( 2 )m⎜ ⎝ 2 ⎟⎠

R1 1 = R2 2 Hence, the correct answer is (B). ⇒

7.

Moment of inertia about O is IO =

M( 2 M( 2 L + B2 ) = 80 + 60 2 ) 12 12

2 2

2ml 2 + 2ml 2 = 3 ml 2 2 Angular momentum of system about the specified axis is ⇒

I=

L = Iω = ( 3 ml 2 ) ω Hence, the correct answer is (B).

Moment of inertia about O′ is obtained by applying parallel axis theorem, so we have

11. Consider an infinitesimal element of mass dm, length dx at a distance x from the specified axis of rotation, then

∫

M( 2 2 80 + 60 2 ) + M ( 50 ) 12 M( 2 80 + 60 2 ) 1 IO 12 = = ⇒ IO′ M ( 80 2 + 60 2 ) + M ( 50 )2 4 12 Hence, the correct answer is (B). ⇒

8.

IO′ =

Loss in KE is given by 1⎛ I I ⎞ 2 Loss = − ΔK = ⎜ 1 2 ⎟ ( ω 1 − ω 2 ) 2 ⎝ I1 + I 2 ⎠ 1 ⎡ ( I ) ( 3I ) ⎤

⇒

Loss =

⇒

⎛1 ⎞3 Loss = ⎜ Iω 2 ⎟ ⎝2 ⎠4

2 ⎢⎣

4I

⎥⎦ ω

2

Loss 3 = ⇒ Ki 4 Hence, the correct answer is (C). 9.

∫

I = r 2 dm = x 2 λ dx

IO′ = I 0 + Md 2

Moment of inertia of the cone about the specified axis is MR2 I= 2 Hence, the correct answer is (A).

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 234

L

⇒

x⎞ ⎛ I = x 2 λ 0 ⎜ 1 + ⎟ dx ⎝ L⎠

∫ 0

L

⇒

⎛ L3 L3 ⎞ ⎛ x3 ⎞ I = λ 0 ⎜ x 2 + ⎟ dx = λ ⎜ + ⎟ ⎝ ⎝ 3 4⎠ L⎠

⇒

7 L3 λ 0 I= 12

∫ 0

L

…(1) L

x⎞ ⎛ Since M = λ dx = λ 0 ⎜⎝ 1 + ⎟⎠ dx L

∫ 0

⇒ ⇒

∫ 0

L ⎞ 3λ L ⎛ M = λ0 ⎜ L + ⎟ = 0 ⎝ 2⎠ 2 2 λ0 L = M 3

…(2)

From (1) and (2), we get 7 ⎛ 2 ⎞ 2 7 ML2 ⎜ M⎟ L = 12 ⎝ 3 ⎠ 18 Hence, the correct answer is (D). I=

09-Feb-21 6:40:37 PM

Hints and Explanations

⇒

2 1 1⎛ 1 ⎞⎛ v ⎞ 3 mv 2 + ⎜ mr 2 ⎟ ⎜ 2 ⎟ = mv 2 ⎠⎝ r ⎠ 4 2 2⎝ 2

u=

4 gh 3

u 1 4 gh = r r 3 Hence, the correct answer is (C). ⇒

l2 l2 + m = mk 2 12 16

⇒

m

⇒

7l2 = k2 48

Hence, the correct answer is (A). 17. Since a =

⇒

ω=

v = R

2 ( m1 − m2 ) gh ( m1 + m2 ) R2 + I

18. Mass of plate is

k=l

R

∫

M = ρ0 r ( 2π rdr ) = 0

⇒ ⇒ dI = σ ( 2π rdr ) r 2

⇒

dI = 2π ( A + Br ) r 3 dr a

∫ dI = 2π ∫ ( Ar

3

R

∫

∫

IC = ( dm ) r 2 = ρ0 r ( 2π rdr ) r 2 = 0

2πρ0 R5 5

5 ⎛ 1 1 ⎞ 16πρ0 R I = IC + MR2 = 2πρ0 R5 ⎜ + ⎟ = ⎝ 3 5⎠ 15 8 8⎛ 2 3⎞ 2 2 I = ⎜ πρ0 R ⎟ R = MR ⎠ 5⎝ 3 5

19. If α be initial angular acceleration (before it slips off), then ⎛ ml 2 ⎞ ⎛ l⎞ mg ⎜ ⎟ = Ια = ⎜ α ⎝ 2⎠ ⎝ 3 ⎟⎠

+ Br 4 ) dr

0

⎛ A aB ⎞ I = 2π a 4 ⎜ + ⎟ ⎝ 4 5 ⎠ Hence, the correct answer is (A). ⇒

15. Rotational KE is RKE =

2πρ0 R3 3

Hence, the correct answer is (D).

⇒

3g 2l Angular speed acquire by the box in time τ = 0.01 s is ⇒

⎛ K2 ⎞ 1 mv 2 ⎜ 1 + 2 ⎟ = 8.75 × 10 −4 J ⎝ 2 R ⎠ Hence, the correct answer is (A). 16. From parallel axis theorem, we get 2 ⎡ 2 ⎛ d⎞2 ⎛ d ⎞ ⎤ 13 I0 = 3 × ⎢ M ⎜ ⎟ + M ⎜ Md 2 ⎥= ⎟ ⎝ ⎠ ⎝ ⎠ 5 2 10 3 ⎣ ⎦ 2

13 ⎛ d ⎞ = Md 2 + Md 2 Now, I A = I 0 + 3 M ⎜ ⎝ 3 ⎠⎟ 10 23 Md 2 10

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 235

α=

3 × 10 × 0.01 1 ⎛ 3g ⎞ ( ω = αt = ⎜ 0.01 ) = = rads −1 ⎝ 2l ⎟⎠ 2 × 0.3 2

1 1 mv 2 + Iω 2 2 2

RKE =

IA =

I R2

( m1 − m2 ) gR2 ( m1 + m2 ) R2 + I

⎡ ( m1 − m2 ) gR2 ⎤ v = 2 ah 2 ⎢ ⎥h 2 ⎢⎣ ( m1 + m2 ) R + J ⎥⎦

⇒

⇒

m1 + m2 +

=

⇒

14. Since dI = r 2 dm

⇒

( m1 − m2 ) g

Hence, the correct answer is (B).

7 48 Hence, the correct answer is (B).

⇒

I 0 13 = I A 23

ω=

13. I = I cm + md 2

⇒

⇒

CHAPTER 3

12. mgh =

H.235

The angle by which it would rotate when it hits the ground is θ = ωt ′ , where, ω = constant and t ′ = time of fall =

2H = 1 sec g

1 radian 2 Hence, the correct answer is (D). ⇒

θ=

20. For sphere, we have mghsph = ⇒

mghsph =

7 mv 2 10

1⎛ 2 1 ⎞⎛ v⎞ mv 2 + ⎜ mR2 ⎟ ⎜ ⎟ ⎠⎝ R⎠ 2 2⎝ 5

2

…(1)

09-Feb-21 6:40:52 PM

H.236

JEE Advanced Physics: Mechanics – II

For cylinder, we have mghcylinder = ⇒

hsph hcylinder

2

1 ⎛ mR ⎞ ⎛ v ⎞ 3 1 2 mv 2 + ⎜ ⎟ ⎜ ⎟ = mv 2 2⎝ 2 ⎠ ⎝ R⎠ 4 2

7 × 4 14 = = 10 × 3 15

Hence, the correct answer is (D). 21. Since, τ = ⇒

dE dθ

For ring, h1 =

5ω 1 6

I1ω 12 24 Also, loss in kinetic energy is ⇒

ΔKE = −

Loss = − ΔK =

α=

22. Since, mgh =

ωc =

1 ⎛1 ⎞ 1 Loss in KE = ⎜ I1ω 12 + I 2ω 22 ⎟ − ( I1 + I 2 ) ω c2 ⎝2 ⎠ 2 2

2kθ = Iα

2kθ I Hence, the correct answer is (A). ⇒

⇒

1 I pω 2 2

2 1⎛ 3 3 ⎛ v2 ⎞ 2⎞ v = ⎜ ⎟ ⎜⎝ mr ′ ⎟⎠ 2 2 2 4⎝ g ⎠ ( r′ )

− ΔK =

1 ⎛ I1 ⎞ ⎛ ω 12 ⎞ I1ω 12 ⎟= ⎜ ⎟⎜ 24 2⎝ 3 ⎠ ⎝ 4 ⎠

I1ω 12 24 Hence, the correct answer is (D). ⇒

1 ( 2mr ′ 2 ) v 2 v2 = 2 2 mg ( r ′ ) g

For cylinder, h2 =

⇒

1 ⎛ I 1I 2 ⎞ ( ω 1 − ω 2 )2 2 ⎜⎝ I1 + I 2 ⎟⎠

ΔE = ΔK = −

26. Since, x = x0 + a cos ω 1t and y = y0 + b sin ω 2t ⇒

vx = − aω 1 sin ( ω 1t ), vy = bω 2 cos ( ω 2t )

⇒

ax = − aω 12 cos ( ω 1t ), ay = − bω 22 sin ( ω 2t )

7 2 m ( r′ ) v2 1 5 7 ⎛ v2 ⎞ For sphere, h3 = × = ⎜ ⎟ 2 g 2 10 ⎝ g ⎠ ( r′ )

At t = 0, x = x0 + a , y = y0

⇒ h : h : h = 2 : 3 : 14 = 20 : 15 : 14 1 2 3 2 10 No given option is correct.

⇒

23. Applying conservation of angular momentum, we get ⎛ ML2 ML mL2 ⎞ ω0 = ⎜ +2 ⎟ω ⎝ 12 12 4 ⎠ 2

Mω 0 M + 6m Hence, the correct answer is (C). ⇒

ω=

24. Since, ω = α t Also, E =

1 2 Iω = 1200 2

25. Applying conservation of angular momentum, we get

( I1 + I 2 )ω common = I1ω1 + I 2ω 2 ω common = ω c =

⇒

r τ = m ( − aω 12 ) × y0 ( − kˆ ) r τ = my0 aω 12 kˆ

(

)

Hence, the correct answer is (C). 27. Surface mass density is σ = kr 2 R

Mass of disc, M =

∫ ( kr

2

) 2π rdr

0

R 4 π kR 4 = 4 2 Moment of inertia of the disc about the axis is ⇒

M = 2π k

∫ ∫ ∫ I = ( Kr ) ( 2π rdr ) r ∫

I = dI = ( dm ) r 2 = σ dAr 2

1 2 ⇒ × 1.5 ( 20t ) = 1200 J 2 ⇒ t=2s Hence, the correct answer is (B).

⇒

ax = − aω 12 and ay = 0

I1ω 1 4 = ⎛ 5 × 2⎞ω ⎜⎝ ⎟ 1 I1 4 3⎠ I1 + 2

I1ω 1 +

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 236

⇒

2

R

⇒

∫

I = 2π kr 5 dr = 0

2

π kR6 2 = MR2 3 3

Hence, the correct answer is (D). r 28. Since, r = 2tiˆ − 3t 2 ˆj r r dr = 2iˆ − 6tjˆ ⇒ v= dt r r At t = 2 s, r = 4iˆ − 12 ˆj and v = 2iˆ − 12 ˆj

09-Feb-21 6:41:10 PM

Hints and Explanations r r r L = m ( r × v ) = 2 4iˆ − 12 ˆj × 2iˆ − 12 ˆj = −48 kˆ

(

) (

)

32. Applying conservation of mechanical energy, we get

Hence, the correct answer is (D). 29. Since τ = Iα and ω = ω 0 + α t ⇒ ⇒

25 × 2π = ( α ) 5 α = 10π

⇒

⎛5 ⎞ ⎛ 5⎞ τ = ⎜ mR2 ⎟ α ≈ ⎜ ⎟ ( 5 × 10 −3 )( 10 −4 ) 10π ⎝4 ⎠ ⎝ 4⎠

⇒ τ = 2.0 × 10 −5 Nm Hence, the correct answer is (C). M 4 3 ⎧ ⎛ R⎞ ⎫ πr ρ = ⎨Q r = ⎜⎝ ⎟⎠ ⎬ 3 8 2 ⎭ ⎩ ⎛ 7 M ⎞ ( )2 1 14 = MR2 2R For sphere, I1 = ⎜ ⎝ 8 ⎟⎠ 2 8

30. For small sphere,

⇒ ⇒

2 ⎛l ⎞ 1 ⎛ ml ⎞ 2 mg ⎜ sin 30° ⎟ = ⎜ ω ⎝2 ⎠ 2 ⎝ 3 ⎟⎠ 3 g 30 = ω2 = 2l 1

ω = 30 rads −1

Hence, the correct answer is (B). 33. The force per unit area P is F P= π R2

2⎛ M⎞ 2 ⎜ ⎟r 5⎝ 8 ⎠

For disc, I 2 =

CHAPTER 3

⇒

H.237

Torque due to frictional force is R

∫

τ = μ ( dN ) x =

2 ⎛ M ⎞ ⎛ R2 ⎞ MR2 ⎟= ⎜ ⎟⎜ 5⎝ 8 ⎠ ⎝ 4 ⎠ 80 I1 14 × 80 = = 140 ⇒ (8) I2 Hence, the correct answer is (A). σ 31. Since, σ = 0 r ⇒ ⇒

I2 =

dm =

σ0 2π rdr r

∫

m = dm = σ 0 2π ( b − a ) b

⇒

I = 2πσ 0

∫ a

2πσ 0 ( 3 r dr = b − a3 ) 3

⇒ ⇒

2πσ 0 ( 3 2πσ 0 ( b − a ) k = b − a3 ) 3 1 k 2 = ( b 2 + ab + a 2 ) 3 2

1 ( b 3 − a3 ) 3 b−a Hence, the correct answer is (D). ⇒

k=

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 237

0

3

τ=

34. Since, F − f = Ma Also, fR =

⇒ ⇒

2

Since, I = mk 2

∫

μ F × 2π R 2 = μ FR 2 3 3π R Hence, the correct answer is (D). ⇒

⇒

μF 2π x 2 dx π R2

⇒

MR2 a 2 R

Ma 2 3 Ma F= 2 2F a= 3M f =

a 2F = R 3 MR Hence, the correct answer is (B). ⇒

α=

35. Since, τ = I pα ⇒

2 5 M0 gl − 4 M0 gl = ⎡⎣ 2 M0 ( 2l ) + 5 M0l 2 ⎤⎦ α

⇒

M0 gl = 13 M0 l 2α

10-Feb-21 3:26:11 PM

H.238

JEE Advanced Physics: Mechanics – II 41. Since I = I1 + I 2 + I 3

g 13l Hence, the correct answer is (C). ⇒

α=

⎛ MR2 ⎞ MR2 + 2⎜ + MR2 ⎟ = 3 MR2 ⎝ 4 ⎠ 2 Hence, the correct answer is (A). ⇒

36. Since, I = I spheres + I rod where, I rod = I spheres ⇒

M × 4 R2 MR2 = 12 3

42. Since,

⎛2 ⎞ 44 = 2 ⎜ MR2 + 4 MR2 ⎟ = MR2 ⎝5 ⎠ 5

44. Since, L = mv0 r

2

Also,

15 15 KML2 = I0 16 16 Hence, the correct answer is (B). r r r 38. Since, τ = τ 1 + τ 2 r where, τ 1 = 2iˆ + 3 ˆj × Fkˆ = F 3iˆ − 2 ˆj I=

)

(

(

)

(

r τ = F 3iˆ − 2 ˆj + 3 kˆ

1 1 mv02 = mv12 + mgh 2 2

⇒

v02 = 25 + 2 × 10 × 10 = 225

⇒

v0 = 15 ms −1

⇒

L = 20 × 10 −3 × 15 × 20 = 6 kgm 2s −1

Hence, the correct answer is (D).

)

r τ 2 = 6 ˆj × F − sin 30°iˆ − cos 30° ˆj = 3 Fkˆ ⇒

100 + 400 R12 + R22 = = 250 2 2

Hence, the correct answer is (C).

M⎛ L⎞ ML2 I1 = K ⎜ ⎟ = K 4 ⎝ 2⎠ 16

(

R=

I ( x ) = I 0 + mx 2

So, I = I 0 − I1 ⇒

)

43. According to parallel axis theorem, we have

I=

37. Let I 0 = KML2 ⇒

⇒

(

M 2 R1 + R22 = MR2 2

⇒ R ≈ 16 cm Hence, the correct answer is (C).

⎛ 44 1 ⎞ I=⎜ + MR2 ⎝ 5 3 ⎟⎠

137 MR2 15 Hence, the correct answer is (D). ⇒

I=

45. Moment of inertia of one of the outer disc about an axis passing through point O and perpendicular to the plane

)

Hence, the correct answer is (A). 39. Since τ p = I pα ⇒

F ( 2R ) = 2 MR2α

⇒

F 40 α= = MR 0.5 × 5

⇒ α = 16 rads −2 Hence, the correct answer is (A). 40. Since τ = Fr sin θ ⇒ 2.5 = 1 × 5 sin θ ⇒ sin θ = 0.5 π ⇒ θ= 6 Hence, the correct answer is (B).

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 238

I1 =

9 1 2 MR2 + M ( 2R ) = MR2 2 2

Moment of inertia of the system about point O, IO =

1 1 ⎛9 ⎞ 55 MR2 + 6 I1 = MR2 + 6 ⎜ MR2 ⎟ = MR2 ⎝2 ⎠ 2 2 2

Required moment of inertia of the system about point P is 2

I P = IO + 7 M ( 3 R ) =

55 181 MR2 + 63 MR2 = MR2 2 2

Hence, the correct answer is (D).

09-Feb-21 6:41:41 PM

Hints and Explanations 9M π R2 2

9M ⎛ R⎞ Mass of removed portion is M ′ = ×π⎜ ⎟ = M ⎝ 3⎠ π R2 Let moment of inertia of removed portion be I1 2

3

M⎛ R⎞ MR2 ⎛ 2R ⎞ ⎜⎝ ⎟⎠ + M ⎜⎝ ⎟⎠ = 2 3 3 2 Let I 2 be the moment of inertia of the whole disc, so ⇒

I1 =

9 MR2 2 Moment of inertia of remaining disc is I = I 2 − I1

⇒

μmg = mω 2 r

⇒

μ=

2

2

9 MR MR 8 MR = = 4 MR2 − 2 2 2 Hence, the correct answer is (D). ⇒

I=

47. Moment of force will be maximum when line of action of force is perpendicular to line AB.

2 1 tan θ = = 4 2 Hence, the correct answer is (A).

mgh = mgl sin α = ⇒

6 gl sin α = v 2

⇒

v = 6 gl sin α

l − Mx 2

⇒

mx = M

⇒

⎛ l⎞ 1 m=⎜M ⎟ −M ⎝ 2⎠ x

1 This is equation of straight line with variables m and . x Hence, the correct answer is (D). 49. As coin is at rest on rotating disc, centripetal force is provided by the friction force between the coin and disc. f = mω 2 R

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 239

1 ⎛ ml 2 ⎞ 2 ⎜ ⎟ω 2⎝ 3 ⎠

⇒ v ∝ sin α Hence, the correct answer is (D). 51. Applying law of conservation of angular momentum, i.e. Li = L f , we get

⎛ L⎞ ⎛ L⎞ m ( 2v ) ⎜ ⎟ + 2m ( v ) ⎜ ⎟ = Iω ⎝ 3⎠ ⎝ 6⎠ ⇒

2 2 ⎡ 1 ⎛ L⎞ ⎛ L⎞ ⎤ mvL = ⎢ ( 8 m ) L2 + m ⎜ ⎟ + 2m ⎜ ⎟ ⎥ ω ⎝ 3⎠ ⎝ 6⎠ ⎦ ⎣ 12

⇒

5 ⎛2 1 1 ⎞ v = L ⎜ + + ⎟ ω = ωL ⎝ 3 9 18 ⎠ 6

6v 5L Hence, the correct answer is (A). ⇒

⎛l ⎞ mgx = Mg ⎜ − x ⎟ ⎝2 ⎠

μ=

50. Applying conservation of energy, we get

48.

Balancing torque about point of suspension X, we get

2

4π 2 ( 3.5 ) × 1.25 × 10 −2 = 604 × 10 −3 ≈ 0.6 10 Hence, the correct answer is (D). ⇒

I2 =

2

2 ω 2 r ( 2πν ) r = g g

CHAPTER 3

46. Mass per unit area of disc is σ =

H.239

ω=

52. Moment of inertia about z-axis is I z =

mR2 2

Moment of inertia about z′ axis is I z ′ = I z + mR2 = ⇒

3 mR2 2

I z : I z′ = 1 : 3

Hence, the correct answer is (B). 53. Since, I = ⇒

I=

mR2 ml 2 + 4 12

m ⎛ 2 l2 ⎞ ⎜ R + ⎟⎠ 4⎝ 3

If V be volume of cylinder, then V = ( π R2 ) l

09-Feb-21 6:41:58 PM

H.240

⇒

JEE Advanced Physics: Mechanics – II

I=

m ⎛ V l2 ⎞ + ⎟ ⎜ 4 ⎝ πl 3 ⎠

For I to be minimum, we have

55. From figure, we conclude mg − T = ma Moment of inertia of a uniform disc is I = acceleration is a = α R

…(1) 2

MR and an 2

Since, τ = TR = Iα

dI m ⎛ −V 2l ⎞ = ⎜ + ⎟ =0 dl 4 ⎝ π l 2 3 ⎠ ⇒ ⇒

⎛ MR2 ⎞ ⎛ a ⎞ TR = ⎜ ⎝ 2 ⎟⎠ ⎜⎝ R ⎟⎠

⇒

T=

Ma 2

V 2l = 2 3 πl 2π l 3 V= 3

⇒

π R 2l =

⇒

l2 3 = R2 2

2π l 3 3

3 l ⇒ = R 2 Hence, the correct answer is (D). 54. Torque at angle θ is ⎛l ⎞ τ = Fr⊥ = Mg ⎜ sin θ ⎟ ⎝2 ⎠ Since, τ = Iα ⇒

⇒

Iα =

Mgl Ml 2 sin θ , where I = 2 3

Substituting this value in equation (1) Ma mg − = ma 2 ⇒

M⎞ ⎛ mg = a ⎜ m + ⎟ ⎝ 2 ⎠

2mg M + 2m Hence, the correct answer is (D). ⇒

a=

1 l Hence, the correct answer is (A).

56. W1l 1 = W2l 2, i.e. W ∝

57. Moment of inertia of the disc about the given axis ID =

MR2 2

⇒

⎛ Ml 2 ⎞ Mgl sin θ ⎟⎠ α = ⎜⎝ 3 2

⇒

lα ⎛ sin θ ⎞ = g⎜ ⎝ 2 ⎟⎠ 3

3 g sin θ 2l Hence, the correct answer is (A). ⇒

α=

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 240

2

M M ⎛ R⎞ ×π⎜ ⎟ = ⎝ 4⎠ 16 π R2 Moment of inertia of removed portion about the given axis, using parallel axes theorem is Mass of removed portion is M ′ =

IR =

1 ⎛ M ⎞ ⎛ R2 ⎞ ⎛ M ⎞ ⎛ 9R2 ⎞ 19 MR2 ⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟= 2 ⎝ 16 ⎠ ⎝ 16 ⎠ ⎝ 16 ⎠ ⎝ 16 ⎠ 512

Required moment of inertia 1 19 MR2 237 MR2 − = MR2 2 512 512 Hence, the correct answer is (D). I = ID − IR =

09-Feb-21 6:42:11 PM

Hints and Explanations r r r iˆ L = r × p = rp sin θ nˆ 58. Since, r ⇒ L = pr⊥ nˆ

10 60 10 rads −1 = rpm = 27 rpm 1.25 2π 1.25 Hence, the correct answer is (A). ⇒

62.

For A to B r R mv ( − kˆ ) L= 2 For C to D

r

ω

So, vleft point = ω r ′ < ω r = vright point So, the roller will turn to left. Hence, the correct answer is (D). 60. Since no external torque acts on the system, therefore total angular momentum of the system about point O remains constant. So, Li = L f ⎛ a⎞ mv ⎜ ⎟ = Iω ⎝ 2⎠

mva ⇒ ω= 2I where, I is the moment of inertia of cube about its edge 2

⎛ 2a ⎞ a2 ma 2 ma 2 2ma 2 I = m + m⎜ = + = ⎟ ⎝ 2 ⎠ 6 6 2 3

mva × 3 3v 3×2 = = = 5 rads −1 2 × 2ma 2 4 a 4 × 0.3 Hence, the correct answer is (D). ⇒

⇒

2R = 3 a

⇒

a=

2R 3

⇒

m=

3 M ⎛ 2R ⎞ ⎜ ⎟ 4π R3 ⎝ 3 ⎠

⇒

m=

2M 3π

ω=

⎛ M ⎞ 3 a ⎜ 4 3⎟ ⎜⎝ π R ⎟⎠ 3

3

Since MI of cube is I =

59. Let the distance of central line from instantaneous axis of rotation be r, then r from the point on left becomes lesser than that for right.

⇒

( Diameter of Sphere ) = ( Diagonal of the Cube )

Mass of Cube = m = ρa 3 =

r ⎛ R ⎞ L=⎜ + a ⎟ mv ( kˆ ) ⎝ 2 ⎠ For B to C r ⎛ R ⎞ L=⎜ + a ⎟ mv ( kˆ ) ⎝ 2 ⎠ Hence, (A) and (C) are correct.

⇒

ω=

CHAPTER 3

For D to A r R mv ( − kˆ ) L= 2

r

H.241

1 2 ma 6

4 MR2 9 3π Hence, the correct answer is (C). ⇒

I=

63. Equation of motion for solid cylinder is F − f = ma

…(1)

Also, τ = fR = Iα For pure rolling, we have a = α R mR2 a ⋅ 2 R ma ⇒ f = 2 From equations (1) and (2), we get ⇒

fR =

…(2)

ma = ma 2 3 ⇒ F = ma 2 Hence, the correct answer is (C). F−

64. For a thin uniform square sheet I1 = I 2 = I =

ma 2 12

61. For just one complete rotation, speed of the drum at top position, v = Rg , where R = 1.25 m Angular velocity of the drum is

ω=

v = R

g R

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 241

Hence, the correct answer is (C).

09-Feb-21 6:42:27 PM

H.242

JEE Advanced Physics: Mechanics – II

2 2 65. R = ( 0.8 ) + ( 0.6 ) = 1 m, v = rω = ( 0.6 ) ω

⇒

dω ( = 4t − t 2 ) dt dt

⇒ dω = ( 4t − t 2 ) dt On integrating, we get t3 3 At direction of reversal of motion, ω = 0 ⇒ t=6s dθ t3 Since, ω = = 2t 2 − dt 3 3 ⎛ t ⎞ ⇒ dθ = ⎜ 2t 2 − ⎟ dt ⎝ 3⎠

ω = 2t 2 −

Angular momentum of the particle about point O is L = mvR sin 90° = m ( rω ) R ⇒ L = ( 2 ) ( 0.6 ) ( 12 )( 1 ) = 14.4 kgm 2s −1 Hence, the correct answer is (D).

On integrating, we get

66. In this case L changes in direction but not in magnitude. Hence, the correct answer is (C). 67. Since, mg − T = ma

…(1)

Also, τ = TR = Ια ⇒

⎛ a⎞ TR = ( mR2 ) ⎜ ⎟ ⎝ R⎠

2t 3 t 4 − 3 12 At, t = 6 s, θ = 36 rad

θ=

⇒

36 μmg cos θ

Since P has most of its mass concentrated at surface, so

⇒

I P > IQ

⇒ tan θ > 3 Condition of toppling is

g sin θ I 1+ mR2 aQ > aP (as its moment of inertia is less)

In case of pure rolling, a = ⇒

⇒

8.

5.

vsinθ v

θ

v

v

vcos θ

vr = 2v sin θ ⇒

T/2

T

dL d = ( Iω ) dt dt

⎞ d ⎛ ML2 + mx 2 ⎟ ω ⎜ ⎠ dt ⎝ 3 dx ⇒ τ = 2mx ω dt Now, x = vt ⇒ τ ∝t Finally, torque becomes zero. Hence, the correct answer is (B). ⇒

7.

2 1 MR2 = Mr 2 + Mr 2 5 2

⇒

Hence, the correct answer is (D).

τ=

I=

2 3 MR2 = Mr 2 5 2

2 R 15 Hence, the correct answer is (A).

vr

6.

9.

⇒

vr = 2v sin ( ωt )

0

v

tan θ >

1 mR2 2 ⇒ Body is disc. Hence, the correct answer is (D).

vsinθ

θ vcos θ

15 ⎞ ⎛ 10 ⎞ ⎟⎠ > ( mg cos θ ) ⎜⎝ ⎟⎠ 2 2

2 ⎛ 3v 2 ⎞ 1 1 ⎛ v⎞ mv 2 + I ⎜ ⎟ = mg ⎜ ⎟ ⎝ ⎠ 2 2 R ⎝ 4g ⎠

⇒

θ

θ

( mg sin θ ) ⎛⎜⎝

2 …(2) 3 With increase in value of θ , condition of sliding is satisfied first. Hence, the correct answer is (D). ⇒

ω Q > ω P as vP > vQ

Hence, the correct answer is (D).

…(1)

( τ mg sin θ )about O > ( τ mg cosθ )about O

Therefore, Q reaches first with more linear speed and more translational kinetic energy. v Further, ω = R ⇒ ω ∝v ⇒

tan θ > μ

τ=

Condition of sliding is

t

r=

10. On smooth part BC , due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from B to C translational kinetic energy converts into gravitational potential energy. Hence, the correct answer is (D). MI of MI of ⎛ MI of ⎞ ⎛ ⎞ ⎛ ⎞ 11. ⎜ Given ⎟ = ⎜ Complete Disc ⎟ − ⎜ Removed Portion ⎟ ⎜ Shape ⎟ ⎜ of Radius R ⎟ ⎜ of Radius R 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⇒

I remaining = I whole − I removed

⇒

I=

2 2 ⎡1 R 1 1 2R ⎞ ⎤ ( 9 M ) ( R )2 − ⎢ m ⎛⎜ ⎞⎟ + m ⎛⎜ ⎥ ⎟ 2 2 ⎝ 3 ⎠ ⎦ ⎣2 ⎝ 3⎠

…(1)

2

9M ⎛ R⎞ ×π⎜ ⎟ = M 2 ⎝ 3⎠ πR Substituting in equation (1), we get

where, m = σ a =

I = 4 MR2 Hence, the correct answer is (D).

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 244

09-Feb-21 6:43:06 PM

Hints and Explanations

r r Direction of ( r × v ), hence the direction of angular momentum remains the same. Hence, the correct answer is (D).

increase as it moves from C and D. Therefore, ω will initially increase and then decrease.

13. From conservation of angular momentum ( Iω = constant ) , angular velocity will remain half. As,

O r

1 2 Iω 2 The rotational kinetic energy will become half. Hence, the correct answer is (B). K=

14. In case of pure rolling bottommost point is the instantaneous centre of zero velocity. Velocity of any point on the disc, v = rω , where r is the distance of point from O. C

Q

P

A

Let R be the radius of platform, m the mass of disc and M is the mass of platform. Moment of inertia when the tortoise is at A MR2 2 and moment of inertia when the tortoise is at B I1 = mR2 +

I 2 = mr 2 +

2

Here, r 2 = a 2 + ⎡⎣ R2 − a 2 − νt ⎤⎦

vQ > vC > vP

From conservation of angular momentum

15. In uniform circular motion the only force acting on the particle is centripetal (towards centre). Torque of this force about the centre is zero. Hence, angular momentum about centre remain conserved. Hence, the correct answer is (A). 16. Let ω be the angular velocity of the rod. Applying, angular impulse equals change in angular momentum about centre of mass of the system, we get ω M

ω 0 I1 = ω ( t ) I 2

M

Substituting the values, we can see that variation of ω ( t )is non-linear. Hence, the correct answer is (D). 18. mg sin θ component is always down the plane whether it is rolling up or rolling down. Therefore, for no slipping, sense of angular acceleration should also be same in both the cases. Therefore, force of friction f always act upwards. Hence, the correct answer is (B). 19. If we assume complete disc to be present, then it would have a mass 4 times the mass of the sector. Then, moment of inertia of the complete disc is

J = Mv

⎛ L⎞ J ⎜ ⎟ = Ι cω ⎝ 2⎠ ⎛ ML2 ⎞ L ( Mv ) ⎛⎜ ⎞⎟ = ( 2 ) ⎜ ⎟⎠ ω ⎝ ⎠ ⎝ 2

Idisc =

1 1 Mdisc R2 = ( 4 Msector ) R2 2 2

Hence, I sector = 4

v L Hence, the correct answer is (A). ⇒

MR2 2

rQ > rC > rP Hence, the correct answer is (A).

⇒

D

B C vt

O ω

⇒

a

ω=

CHAPTER 3

r r r 12. L = m ( r × v )

H.245

Idisc 4

1 MR2 2 Calculus Method: ⇒

I sector =

17. Since, there is no external torque, angular momentum will remain conserved. The moment of inertia will first decrease till the tortoise moves from A to C and then

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 245

09-Feb-21 6:43:15 PM

H.246

JEE Advanced Physics: Mechanics – II N

Consider an element of mass dm, thickness dx and radius x of the sector. dm =

F

2π xdx 2 M = 2 xdx ⎛ π R2 ⎞ 4 R ⎜⎝ ⎟⎠ 4 M

L L/2 f

If dI be the moment of inertia of the element, then dI = x 2 dm ⇒

⎛ 2M ⎞ dI = x 2 ⎜ 2 x dx ⎟ ⎝ R ⎠

⇒

dI =

(

2M 3 x dx R2

the block will topple when τ F > τ mg

)

⇒

I=

2M x 3 dx R2

∫ 0

⇒

I=

FL > ( mg )

1 MR2 2

General Funda:

⎛ Moment of Inertia ⎞ ⎛ Moment of Inertia ⎞ of a SECTOR of ⎜ of a DISC of mass ⎟ ≡ ⎜ ⎟ ⎜⎝ M with radius R ⎟⎠ ⎜⎝ mass M with radius R ⎟⎠ Hence, the correct answer is (A). 20. Mass of the loop = M = Lρ

F>

22. No external torque is acting on the system, so angular momentum is conserved. Further there exists no nonconservative forces in the system, so total energy is also conserved. Hence, the correct answer is (B). r 23. The angular momentum of a body L may be expressed as the sum of two parts, (a) one arising from the motion of the centre of mass of the body and (b) the other from the motion of the body with respect to its centre of mass. r r r r i.e. Ltotal = LC. M . + rC. M . × p r r r r ⇒ Ltotal = LC. M . + M ( rC. M . × vC. M . ) For this Problem

Further if r is the radius of the loop, then

r=

L 2π

Moment of Inertia about XX′ is I = ⇒

I=

⇒

1 3 MR2ω + MR2ω = MR2ω . 2 2 Hence, the correct answer is (C).

3 Mr 2 2

⇒

3 L2 3 ρL3 = ( Lρ ) 2 2 ( 2π ) 8π 2

Hence, the correct answer is (D). 21. At the critical condition, normal reaction N will pass through point P. In this condition

τN = 0 = τ f

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 246

1 MR2ω and 2 r r M ( rC. M . × vC. M . ) = MRvCM = MR ( Rω ) r r M ( rC. M . × vC. M . ) = MR2ω

LC. M . = Ιω =

2π r = L ⇒

L 2

mg 2 Therefore, the minimum force required to topple the bock is mg Fmin = 2 Hence, the correct answer is (D). ⇒

R

⇒

mg

(about P)

Ltotal =

24. Since spheres are smooth, so no transfer of angular momentum takes place from A to B. However, sphere A only transfers its linear velocity v to sphere B and stops. Hence, we conclude that A stops but continues to rotate with same angular speed ω and B moves with speed of A but with zero angular speed. Hence, the correct answer is (C).

09-Feb-21 6:43:24 PM

Hints and Explanations 25. By Law of Conservation of Angular Momentum ⎛ a⎞ mv ⎜ ⎟ = I system ω ⎝ 2⎠ about O

(

( 0.7 ) ( 1.4 )

= 0.98 m 0.3 + 0.7 Hence, the correct answer is (D). ⇒

)

x=

H.247

M dx L This element needs centripetal force for rotation, so

29. Mass of the element dx is m = ⎛M 2 ⎞ dF = mxω 2 = ⎜ xω dx ⎟ ⎝ L ⎠

⎛ mv ⎜ ⎝

26.

A′

L

⇒

L

m Mω 2 L F = dF = ω 2 xdx = 2 L

∫ 0

∫ 0

This is the force exerted by the liquid at the other end. Hence, the correct answer is (D).

C′

30. By Law of Conservation of Angular Momentum

( MR2 )ω = MR2ω ′ + 2mR2ω ′

CHAPTER 3

2 a⎞ ⎡1 ⎛ a ⎞ ⎤ 2 ⎥ω ⎟⎠ = ⎢ Ma + M ⎜ ⎝ 2 ⎟⎠ ⎦ 2 ⎣6 3v ⇒ ω= 4a Hence, the correct answer is (A).

⇒

ω′ D′

B′

I AB = I A ′B′ = I and ICD = IC ′D ′ If I 0 be the moment of inertia of the square plate about an axis passing through O and perpendicular to the plate, then by perpendicular axis theorem I 0 = I AB + I A ′B′ = 2I AB

…(1)

I 0 = ICD + IC ′D′ = 2ICD

…(2)

⎛ M ⎞ ω′ = ⎜ ω ⎝ M + 2m ⎟⎠ Hence, the correct answer is (C). ⇒

Multiple Correct Choice Type Problems

OR

1.

From (1) and (2) ICD = I AB = I

{OPTION (A)}

⎛ mL2 ⎞ mvx = ⎜ mx 2 + ⎟ω ⎝ 3 ⎠ 3vx ⇒ ω= 2 L + 3x 2 For ω to be maximum, we have dω =0 dx L ⇒ xM = 3

Hence, the correct answer is (A). 27. Since L = mvr⊥ and as r⊥ is constant, so L is constant. Hence, the correct answer is (B). 28. Work done W =

1 2 Iω 2

If x is the distance of mass 0.3 kg from the centre of mass, we will have, I = ( 0.3 ) x 2 + ( 0.7 ) ( 1.4 − x )

⇒

3v 2L Hence, (A), (C) and (D) are correct.

2

For work to be minimum, the moment of inertia ( I ) should be minimum dI ⇒ =0 dx ⇒ 2 ( 0.3 x ) − 2 ( 0.7 ) ( 1.4 − x ) = 0

( 0.3 ) x = ( 0.7 ) ( 1.4 − x )

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 247

Applying conservation of angular momentum about hinge, we get

⇒

2.

ωM =

Applying conservation of energy (as the friction is acting at the point of no slipping), we get MgL 1 ( 1 − cos 60° ) = Ι 0ω 2 2 2

09-Feb-21 6:43:34 PM

H.248

JEE Advanced Physics: Mechanics – II L 2 L 2

aR =

Since I 0 = ⇒

ω=

L L 2 aT = α 2 ω 2

aR =

3 3g 4L For CM of the rod, we have ⇒

⇒

3g 2L

L 2 3g ω = 2 4 L Also τ 0 = Mg sin ( 60° ) = Ι 0α 2 ⇒

Now, x =

ML2 3

α=

Mg − N = May

L sin θ and y = L cos θ 2

x2 ⎛ L⎞ ⎜⎝ ⎟⎠ 2

2

+

y2 =1 L2

So, path of A is an ellipse. Hence, (A), (C) and (D) are correct. 4.

Let Ω be the angular velocity of assembly perpendicular to axis of the rod, then for pure rolling, we have aω = lΩ aω ⇒ Ω= l

where, ay = aT sin ( 60° ) + aR cos ( 60° )

ω 2L L α sin ( 60° ) + cos ( 60° ) 2 2

⇒

ay =

⇒

ay =

⇒

15 g ⎛ 9 3⎞ ay = ⎜ + g= ⎝ 16 8 ⎟⎠ 16

So, component of Ω along z-axis is Ω cos θ

⇒

⎛ 15 g ⎞ Mg − N = M ⎜ ⎝ 16 ⎟⎠

⇒

3 ⎛ 3 3g ⎞ ⎛ 3g ⎞ ⎛ L ⎞ ⎛ 1 ⎞ ⎜ ⎟ +⎜ ⎟⎜ ⎟⎜ ⎟ 4 ⎝ 4 ⎠ ⎝ 2L ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

Mg 16 Hence, (B), (C) and (D) are correct. ⇒

3.

N=

When the bar makes an angle θ , the height of its CM L (mid-point) is cos θ 2 Displacement of CM is L y = ( 1 − cos θ ) 2 Since, force on CM is only along the vertical direction, hence CM is falling vertically downward. Instantaneous torque about point of contact is

⇒

⎛L ⎞ τ = mg ⎜ sin θ ⎟ ⎝2 ⎠ τ ∝ sin θ

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 248

Ω cos θ =

aω 24 lω ω = = l 5 5l 5

{Q l =

24 a

}

So, (A) is correct. Let L1 be the angular momentum of assembly about its centre of mass, then ⎡ ma 2 ( 4 m )( 2 a )2 ⎤ ⎛ 17 2 ⎞ + L1 = ⎢ ma ⎟ ω ⎥ ω = ⎜⎝ ⎠ 2 2 ⎣ 2 ⎦ So, (C) is correct If I 2 be the moment of inertia of assembly about and axis perpendicular to axis of rod (marked as 2), then ⎛ ma 2 ⎞ ⎡ ( 4 m )( 2 a )2 2⎤ I2 = ⎜ + ml 2 ⎟ + ⎢ + ( 4 m ) ( 2l ) ⎥ ⎝ 4 ⎠ ⎣ 4 ⎦ 17 2 17 2 ma + 17 ml 2 = ma + 17 m ( 24 a 2 ) 4 4 So, angular momentum of assembly about axis 2 is L2 = I 2Ω ⇒

I2 =

09-Feb-21 6:43:45 PM

Hints and Explanations ⇒

r r r vP = vPO + vO

⇒

⎛ Rω ˆ 3 Rω ˆ ⎞ r vP = ⎜ − k ⎟ + 3 Rω iˆ i+ ⎝ 4 ⎠ 4

H.249

11 3 r vP = Rω iˆ + Rω kˆ 4 4 Hence, (A) and (B) are correct. aω ⎠ 24 a

⎛1 ⎞ L2 = ( 17 ma 2 ) ⎜ + 24 ⎟ ⎝4

⇒ ⇒

6.

⎛ 97 ⎞ ω L2 = ( 17 ma 2 ) ⎜ ≅ 17 24 ma 2ω ⎝ 4 ⎟⎠ 24 ⎛ 24 ⎞ ( 17 )( 24 ) 2 L2 cos θ = ( 17 24 ma 2ω ) ⎜ ma ω = ⎝ 5 ⎟⎠ 5

⎛ 17 2 ⎞ ⎛ Also, L1 sin θ = ⎜ ma ω ⎟ ⎜ ⎝ 2 ⎠⎝ ⇒

LZ = L2 cos θ − L1 sin θ =

⇒

LZ ≅ 80 ma 2ω

( −2 × 1 ) + ( 0.1 × 20 ) = ( 0.1 × 0 ) + ( 2 × v )

Here, v is the velocity of CM of ring after impact. Solving the above equation, we get v = 0 So, CM becomes stationary. Correct OPTION is (A) Linear impulse during impact 1. Along horizontal direction is

1⎞ ⎟ 5⎠ 17 2 ⎛ 1⎞ ma ω ⎜ 24 − ⎟ ⎝ 5 2⎠

J1 = Δp = 0.1 × 20 = 2 Ns

So, (D) is not correct Angular momentum of CM of assembly about O is LCM about O where vcm

The data seems to be incomplete. Let us assume that friction from ground on ring is not impulsive during impact. From linear momentum conservation along horizontal direction, we get

CHAPTER 3

⇒

2.

⎛ 9l ⎞ = ( 5m ) ( vcm ) ⎜ ⎟ ⎝ 5⎠

Along vertical direction is J 2 = Δp = 0.1 × 10 = 1 Ns Writing the equation (about CM)

m ( aω ) + 4 m ( ( 2 a ) ω ) 9 aω = = 5m 5

⎛ 24 ⎞ ⎛ 9 aω ⎞ ⎛ 9l ⎞ LCM about O = ( 5m ) ⎜ = 81 ma 2ω ⎜ ⎝ 5 ⎟⎠ ⎜⎝ 5 ⎟⎠ ⎝ 5 ⎟⎠

5.

So, (B) is not correct Hence, (A) and (C) are correct. r Velocity of point O is vO = ( 3 Rω ) iˆ Rω r Also, vPO = in the direction shown in Figure. 2

Since, angular impulse equals change in angular momentum, so we get ⎛ 3 1⎞ 1 1 ⎞ 2⎛ 1× ⎜ × ⎟ − 2 × 0.5 × = 2 × ( 0.5 ) ⎜ ω − ⎟ ⎝ 2 ⎠ ⎝ 2 2 0.5 ⎠ Solving this equation, ω comes out to be positive i.e. ω is counter clockwise. So just after collision, rightwards slipping is taking place. Hence, friction is leftwards. Hence, (A) and (C) are correct. 7.

iˆ

In vector form, r Rω Rω sin 30°iˆ + cos 30°kˆ vPO = − 2 2 Rω ˆ 3 Rω ˆ r vPO = − i+ k 4 4 r r r Since, vPO = vP − vO ⇒

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 249

C

ω B

VC = (2v)i

VB = vi

A (VA = 0)

Hence, (B) and (C) are correct.

09-Feb-21 6:43:53 PM

H.250 8.

JEE Advanced Physics: Mechanics – II

In case of pure rolling, f =

⇒

mg sin θ mR2 1+ I

⇒ (upwards)

Therefore, as θ decreases force of friction will also decrease. Hence, (C) and (D) are correct. 9.

Hence, (A), (B) and (C) are correct. 11. Vc =

f ∝ sin θ

Due to Law of Conservation of Angular Momentum r L =constant r r ⇒ L ⋅ L = constant d (r r) L⋅L = 0 ⇒ dt r r dL =0 ⇒ 2L ⋅ dt r r dL ⇒ L⊥ dt r r r Since τ = A × L r dL r r ⇒ = A×L dt r r r dL ⇒ must be perpendicular to A as well as L dt ur r r ur A ⋅ L Further component of L along A is = x(say). Also A r ur d ( ur r ) ur dL r dA + L⋅ =0 A⋅L = A⋅ dt dt dt r ur ⎧ ur dL dA r ⎫ and = 0⎬ ⎨Q A ⊥ dt dt ⎩ ⎭ ur r ⇒ A ⋅ L = constant ur r A⋅L ⇒ = x = constant A r r r dL Since (or τ ) is perpendicular to L , hence it cannot dt r change magnitude of L but can surely change direction r of L .

I 0 = I1 + I 2 = I 4 + I 3 = I1 + I 3 2m ( − v ) + m ( 2v ) + 8 m ( 0 ) =0 2m + m + 8 m

Further, by Law of Conservation of Angular Momentum

(I

system about O

)ω = ∑ mv r

∑ mvr

= ( 8 m )( 0 )( 0 ) + 2m ( − v ) ( − a ) + m ( 2v )( 2 a )

⊥

system about O

⇒

∑ mvr

⊥

I1 = I 2 and I 4 = I 3

…(1)

I 0 = I1 + I 2 = I 4 + I 3 From (2), we get 2I1 = 2I 3 ⇒

I1 = I 3

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 250

…(2)

{Qof ( 1 ) }

= 6 mva

system about O

⇒

( 30ma2 )ω = 6 mva

⇒

ω=

v 5a

{OPTION (C)}

Further total energy of the system E is 1 2 3 Iω = mv 2 2 5 Hence, (A), (C) and (D) are correct. E=

{OPTION (D)}

12. L = mvx h ⇒

⎛ v 2 sin 2 45 ⎞ L = m ( v cos 45 ) ⎜ ⎟ 2g ⎝ ⎠

⇒

L=

v 2 sin 2 45 ⎪⎫ ⎪⎧ ⎬ ⎨Q h 2g ⎭⎪ ⎩⎪

mv 3 4 2g

vx = vcos 45°

v 45°

Also, from h =

Also, by perpendicular Axis Theorem

⊥

system about O

⎡ 1 ⎤ 2 2 I system = ⎢ ( 8 m )( 6 a ) + m ( 2a ) + 2ma 2 ⎥ = 30 ma 2 12 ⎣ ⎦ about O

Hence, (A), (B) and (C) are correct. 10. From symmetry we have

{OPTION (A)}

⇒

v = 2 gh

⇒

L= L=

h

v2 4g

mv h gives 2

(

)

m 2 gh h 2

09-Feb-21 6:44:03 PM

Hints and Explanations v

L = m 2 gh 3

⇒

Hence, (B) and (D) are correct.

v 2 ω 2 ⎛ 2 R2 ⎞ = ⎜r − ⎟ 2 2 ⎝ 4 ⎠

g sin θ a= I 1+ mR2

⇒

v = ω r2 −

I solid < I hollow

⇒

r

asolid > ahollow

( KE )Hollow = ( KE )Solid = decrease in PE = mgh

Hence, the correct answer is (D).

Linked Comprehension Type Questions 1.

∫ R 4

Solid cylinder will reach the bottom first. Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant. ⇒

R 2

⇒

In case of pure rolling on inclined plane,

⇒

∫

0

Reasoning Based Questions 1.

∫

r

vdv = ω 2 rdr

The point A will be IAOR (i.e. instantaneous axis of rotation). For no slipping, we have

R2 dr = 4 dt t

dr R2 r − 4

∫

= ω dt

2

0

⇒

⎛ R2 2 ⎜ r+ r − 4 ln ⎜ R ⎜ ⎜⎝ 2

⇒

r + r2 −

⇒

r2 −

⇒

⎛R ⎞ R2 R2 ⎞ − ⎟ ⎜ + ⎟ 4 4 ⎟ = ωt ⎟ − ln ⎜ 2 R ⎟ ⎜ ⎟ ⎟⎠ ⎜⎝ ⎟⎠ 4

CHAPTER 3

⇒

H.251

R2 R ωt = e 4 2

R 2 R 2 2ω t R = e + r 2 − 2r e ω t 4 4 2

R 2 2ω t R 2 e + 4 = R ( eω t + e −ω t ) r= 4 4 Reω t

Hence, the correct answer is (C). 4.

Rω = ( R − r ) ω 0 ⇒

r r Since, Frot = Fin + 2m vrot iˆ × ω kˆ + m ( ω kˆ × riˆ ) × ω kˆ

(

)

⇒

r mrω 2iˆ = Fin + 2mvrotω − ˆj + mω 2 riˆ

⇒

Fin = 2mvrω ˆj

( )

⎛ R−r⎞ ω = ω0 ⎜ ⎝ R ⎟⎠

1( 2 2mR2 ) ω 2 = mω 02 ( R − r ) 2 Hence, the correct answer is (A). ⇒

2.

KE =

If height of the cone is h

r

Then, μ N = mg ⇒

μm ( R − r ) ω 02 = mg

g μ(R − r ) Hence, the correct answer is (C). ⇒

3.

ω0 =

Force on block along slot is the centripetal force, so ⎛ vdv ⎞ mω 2 r = ma = m ⎜ ⎝ dr ⎠⎟

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 251

R ( ωt e + e −ω t ) 4 dr R = vr = ( ω e ω t − ω e − ω t ) 4 dt r Rω ( ω t Fin = 2m e − e −ω t ) ω ˆj 4 r mRω 2 ( ω t Fin = e − e −ω t ) ˆj 2

Since, r = ⇒ ⇒ ⇒

09-Feb-21 6:44:12 PM

H.252

JEE Advanced Physics: Mechanics – II

Also, reaction is due to the disc surface, so

⇒

r mRω 2 ( ω t Freaction = e − e −ω t ) ˆj + mgkˆ 2 Hence, the correct answer is (B).

⇒

5-6. The correct answer is 5(D), 6(A). Combined solution to 5, 6 Following points must be kept in mind. (i) Every particle of the disc is rotating in a horizontal circle. (ii) Actual velocity of any particle is horizontal. (iii) Magnitude of velocity of any particle is v = rω where, r is the perpendicular distance of that particle from actual axis of rotation (z-axis) (iv) When it is broken into two parts then actual velocity of any particle is resultant of two velocities v1 = r1ω 1 and v2 = r2ω 2 Here, r1 is the perpendicular distance of centre of mass from z-axis, ω 1 is the angular speed of rotation of centre of mass from z-axis, r2 is the distance of particle from centre of mass and ω 2 is the angular speed of rotation of the disc about the axis passing through centre of mass. (v) Net v will be horizontal, if v1 and v2 both are horizontal. Since, v1 is already horizontal, because centre of mass is rotating about a vertical z-axis. To make v2 also horizontal, the second axis should also be vertical. Hence, the correct answer is (A). 7.

1 1 2 I ( 2ω ) = kx12 2 2 1 ( 2I ) ( ω )2 = 1 kx22 2 2 From equations (1) and (2), we have x1 = 2 x2 Hence, the correct answer is (C).

8.

9.

− ΔK =

{

}

1 1 1 ⎛4 ⎞ 2 2 I ( 2ω ) + ( 2I ) ( ω ) − ( 3 I ) ⎜ ω ⎟ ⎝3 ⎠ 2 2 2

2

1 2 Iω 3 Hence, the correct answer is (B). ⇒

− ΔK =

Integer/Numerical Answer Type Questions 1.

Initially, 40 N1 = 50 N 2

For First Move:

For translational equilibrium, we have

μk N 3 = μs N 4

…(1)

For rotational equilibrium, we have xN 3 = 40 N 4 ⇒

…(1)

…(2)

x = 32 cm

For Second Move: …(2)

For translational equilibrium, we have

μs N 5 = μk N6

4 ω′ = ω 3

Applying the concept that angular impulse equals change in angular momentum. For any of the disc, we have ⎛4 ⎞ τ t = I ( 2ω ) − I ⎜ ω ⎟ ⎝3 ⎠

…(1)

For rotational equilibrium, we have 32 ( N 5 ) = X R N 6

( I + 2I ) ω ′ = I ( 2ω ) + 2I ( ω )

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 252

Loss of kinetic energy is − ΔK = K i − K f ⇒

Let ω ′ be the common velocity. Then applying conservation of angular momentum, we get

⇒

2 Iω 3 2 Iω τ= 3t

τt =

⇒ 2.

…(2)

X R = 25.6 cm

In case of pure rolling, mechanical energy remains constant since work done by friction is zero. Further in case of a disc, 1 2 Translational Kinetic Energy KT 2 mv = = Rotational Kinetic Energy K R 1 Iω 2 2

09-Feb-21 6:44:20 PM

Hints and Explanations

⇒

⇒

H.253

KT mv 2 2 = = 2 KR ⎛ 1 1 2⎞⎛ v ⎞ ⎜⎝ mR ⎟⎠ ⎜⎝ ⎟⎠ R 2 KT =

2 2 ( Total KE ) = K 3 3

So, total kinetic energy is 3 3⎛ 1 ⎞ 3 KT = ⎜ mv 2 ⎟ = mv 2 ⎠ 4 2 2⎝ 2 Since, decrease in potential energy equals increase in kinetic energy, so K=

3 m v 2f − vi2 4

vf =

4 gh + vi2 3

ω=

)

⇒ 5.

3.

2

ω = 4 rads −1 I1ω 1 = I 2ω 2

4 2 × 10 × 30 + ( 3 ) = 3

4 2 × 10 × 27 + ( v2 ) 3

v2 = 7 ms −1

Since angular impulse equals change in angular momentum, so

⇒

1 ⎞ ⎛ MR2 ⎟ ⎜ ⎛ I1 ⎞ 2 ω 2 = ⎜ ⎟ ω1 = ⎜ ⎟ ω1 1 2 2 ⎝ I2 ⎠ ⎜⎝ MR + 2 ( mr ) ⎟⎠ 2

⇒

2 ⎛ ⎞ 50 ( 0.4 ) ( 10 ) ω2 = ⎜ 2 2⎟ ⎝ 50 ( 0.4 ) + 8 × ( 6.25 ) × ( 0.2 ) ⎠

⇒

ω 2 = 8 rads −1

∫ τ dt = Iω t

⇒

45 × 10 −2 ( 0.5 )

4 gh + vi2 should be same in both cases. 3

value of v f =

⇒

( 4 ) ( 5 × 10 −2 ) ( 9 )( 0.25 )

By conservation of angular momentum, we have

Given that the final velocity in both cases is same, so,

⇒

CHAPTER 3

⇒

(

mgh =

Substituting the values, we get

∫ τ dt = ∫ ω=

3 F sin 30°Rdt

0

I

I

6.

Moment of inertia of this lamina about the point O is

−

IO = ⇒ 4.

ω=

3 ( 0.5 )( 0.5 )( 0.5 ) ( 1 ) 2

1.5 ( 0.5 ) 2

= 2 rads −1

Applying conservation of angular momentum, we get MR ω=0 2 4 mvr ω= MR2 2mvr −

⇒

2

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 253

( 4 m ) ( 2 R )2 2

3 − mR2 2

⇒

3⎞ ⎛ IO = mR2 ⎜ 8 − ⎟ ⎝ 2⎠

⇒

IO =

13 mR2 2

Similarly, MI about point P is IP =

⎛ mR2 ⎞ 3 ( 4 m ) ( 2 R )2 − ⎜ + md 2 ⎟ ⎝ 2 ⎠ 2

09-Feb-21 6:44:27 PM

H.254

JEE Advanced Physics: Mechanics – II

where d 2 = R2 + ( 2R )

7.

⇒

I P = 24 mR2 −

⇒

IP =

2

11 mR2 2

37 mR2 2

37 IP 37 = 2 = ≈3 ⇒ IO 13 13 2 Moment of inertia of system about diagonal of square is ⎛2 ⎞ ⎛2 ⎞ I = 2 ⎜ MR2 ⎟ + 2 ⎜ MR2 + Md 2 ⎟ ⎝5 ⎠ ⎝5 ⎠ ⇒

I=

8 MR2 + 2md 2 5

⇒

2 ⎤ ⎡8 ⎛ 5⎞ I = ⎢ ( 0.5 ) ⎜ + 2 ( 0.5 ) ( 4 )( 2 ) ⎥ 10 −4 ⎟ ⎝ 2 ⎠ ⎢⎣ 5 ⎥⎦

⇒

⎛5 ⎞ I = ⎜ + 8 ⎟ × 10 −4 ⎝5 ⎠

⇒ ⇒

I = 9 × 10 −4 N=9

Mechanics II_Chapter 3_Part 9 Hints and Explanation_3.indd 254

8.

There is no slipping between ring and ground, so f 2 is not maximum. However, there is slipping between ring and stick, so, f1 is maximum.

Moment of inertia of the ring about its centre of mass is 2

I = mR2 = ( 2 ) ( 0.5 ) = 0.5 kgm 2 Since, N1 − f 2 = ma ⇒

N1 − f 2 = ( 2 ) ( 0.3 ) = 0.6 N

Also, a = Rα =

Rτ R ( f 2 − f1 ) R R = = I I

⇒

( 0.5 )2 ( f 2 − f1 ) 0.3 = ( 0.5 )

⇒

f 2 − f1 = 0.6 N

…(1) 2

( f 2 − f1 ) I

…(2)

Since stick applies a force of 2 N on the ring, so we have 2

N12 + f12 = ( 2 ) = 4

…(3)

⎛ P⎞ Further, f1 = μ N1 = ⎜ ⎟ N1 ⎝ 10 ⎠

…(4)

From equations (1), (2), (3) and (4), we get P ≈ 3.6

09-Feb-21 6:44:33 PM

CHAPTER 4: GRAVITATION AND SATELLITES

1.

⇒ U = mV = − (b) When x

Escape velocity from the surface of moon is

⇒

2GMm Rm

ve =

Substituting the values, we have 2 × 6.67 × 10 −11 × 7.4 × 10 22 1.74 × 106

ve = ⇒ 2.

(c)

a , a2 + x 2 ≈ x 2

GMm x {the potential energy of two point masses}

−1 2 dU d = −GMm ⋅ ( a 2 + x 2 ) dx dx

Fx = −

ve = 2.4 × 10 3 ms −1 or 2.4 kms −1

GMm ⋅ x

( x 2 + a2 )3 2

Both the planet and the sun revolve around their centre of mass with same angular velocity (say ω ) r = r1 + r2

…(1)

Gm1m2 m1r1ω 2 = m2 r2ω 2 = r2

…(2)

m1

r2

r1

(d) When x

m2

and Fx = −

cm

Solving these two equations, we get

(e) At x = 0

⎛ m2 ⎞ r1 = r ⎜ ⎝ m1 + m2 ⎟⎠

and ω 2 =

4.

G ( m1 + m2 )

r3 Now, total energy of the system is E = P.E. + K.E. Gm1m2 1 1 ⇒ E=− + m1r12ω 2 + m2 r22ω 2 r 2 2 Substituting the values of r1 , r2 and ω 2 , we get Gm1m2 2r (a) Gravitational potential at point P due to the ring. E=−

M

r

O

x

a

V=−

a , a2 + x 2 ≈ x 2 GMm x2 {force between two point masses} {centre of the ring}

Fx = 0 As the particle is attracted equally from all the four sides.

⎛ m1 ⎞ r2 = r ⎜ ⎝ m1 + m2 ⎟⎠

3.

a2 + x 2

U=−

Fx = − ⇒

GMm

CHAPTER 4

Test Your Concepts-I (Based on Acceleration due to Gravity, Gravitational Field and Applications)

GM GM =− r a2 + x 2

Mechanics II_Chapter 4_Hints and Explanation.indd 255

P m

Net force on M due to the pairs 2M and 2M, 4M and 4M, 5M and 5M, 7M and 7M, M and M is zero. So, the only left out mass is 3M and net force on M is due to 3M, given by F=

3GM 2 , along +x axis d2 GmA mB Given that F = r2 F Since acceleration of A is aA = mA F Gm = 2B ⇒ a = aA = mA r ⇒

5.

G ( 3M ) M d2

F=

GmA mB , then r4 F Gm a′A = = 4B mA r a a′A = 2 r

…(1)

Now if, F =

⇒

{Q of ( 1 ) }

09-Feb-21 6:33:53 PM

H.256

6.

7.

JEE Advanced Physics: Mechanics – II

Since, E =

F m0

4N 20 × 10 −3 kg

⇒

E=

⇒

E = 200 Nkg −1 , along +x direction.

r + dr

Let m1 be the mass of the core and m2 the mass of outer shell, then g A = gB B

To calculate Mtotal, we consider an infinitesimal shell element of inner radius r and outer radius r + dr. If dm be the mass of this infinitesimal element, then dm = ( 4π r 2 dr ) ρ

A R

2R

⇒ ⇒ ⇒ ⇒ 8.

Gm1 G ( m1 + m2 ) = ( 2 R )2 R2

⇒ ⇒

4 m1 = ( m1 + m2 )

4 ⎛4 ⎞ ⎛4 ⎛4 ⎞ ⎞ 3 4 ⎜ π R3 ρ1 ⎟ = ⎜ π R3 ⎟ ρ1 + ⎜ π ( 2R ) − π R3 ⎟ ρ2 ⎝3 ⎠ ⎝3 ⎝3 ⎠ ⎠ 3 4 ρ1 = ρ1 + 7 ρ2

ρ1 7 = ρ2 3

⇒ ⇒ ⇒

L = πR

L π Gravitational field at the centre of a semicircular wire of radius R is ⇒

ρ0 R r ⎛ ρ R⎞ dm = ⎜ 0 ⎟ ( 4π r 2 dr ) ⎝ r ⎠

But, ρ = ⇒

⇒

R=

dm = ( 4πρ0 R ) ( rdr )

⇒

⇒

E=

2π GM L2

Since F = mE ⇒ 9.

2π GMm F= L2

∫

0

4πρ0 R3 Mtotal = = 2πρ0 R3 2 GM E( r ) = 2 r G 2πρ0 R3 π RGρ0 E= = 2 4R2

(

)

GMx

E=

(R

2

+ x2 )

32

M E P R

{

F Q E= m

GMtotal for r ≥ R r2 where Mtotal is the total mass of the sphere.

Mechanics II_Chapter 4_Hints and Explanation.indd 256

∫

10. Field strength at the axis at distance x from the centre of the ring is,

}

x

Q

dx

∞

Consider an infinitesimal mass element PQ of length dx, mass dm at a distance x from centre of ring. Then dm = λ dx Force on this element will be, dF = Edm =

Since r ( = 2R ) lies outside the sphere, so E=

R

Mtotal = dm = 4πρ0 R rdr

2Gλ E= R M where λ = L ⎛ M⎞ 2G ⎜ ⎟ ⎝ L⎠ E= ⎛ L⎞ ⎜⎝ ⎟⎠ π

r

R

∞

⇒

∞

GMλ x

( R 2 + x 2 )3 2 GMλ xdx

∫ ∫ (R

F = dF = 0

0

dx

2

+ x2 )

32

09-Feb-21 6:34:00 PM

Hints and Explanations

⇒ ⇒ ⇒

∫

dx 2

R +x

1

=−

2

2

R +x

⎛ 1 F = −GMλ ⎜ 2 2 ⎝ R +x

(

∞⎞ 0

⎟ ⎠

Er =

⎛ 1 1⎞ F = −GMλ ⎜ − ⎟ ⎝ ∞ R⎠ GMλ F= R g e = g − Rω 2

60% of his weight at the pole means 60 3 ge = g= g 100 5 3 ⇒ g = g − Rω 2 5 2 ⇒ g = Rω 2 5 2g ⇒ ω= 5R

ω=

2 × 9.8 5 × 6.4 × 106

⇒

ω=

19.6 32 × 106

⇒

ω = 7.8 × 10 −4 rads −1

GM ⎛ r 3 − R13 ⎞ ⎜ ⎟ r 2 ⎝ R23 − R13 ⎠

0 ⎡ ⎢ ⎛ GM r 3 − R13 ⎞ So, Er = ⎢⎢ 2 ⎜ 3 ⎟ r ⎝ R2 − R13 ⎠ ⎢ ⎢ GM ⎢ r2 ⎣

11. At the equator, we have

⇒

r < R1

for

for R1 < r < R2 r > R2

for

13. Reduced by 36% means the value is 64% the original 64 g 100 gR2 Since g h = ( R + h )2 2 64 ⎛ R ⎞ ⇒ =⎜ ⎟ 100 ⎝ R + h ⎠ ⇒

gh =

⇒

8 R = R + h 10

⇒

10 R = 8 R + 8 h

⇒

h=

R = 1600 km 4

Test Your Concepts-II (Based on Gravitational Potential, Potential Energy and Applications)

12. For r < R1 , we have Er = 0 GM For r > R2 , we have, Er = 2 r

1.

M

First of all let us calculate the total number of interactions between these eight particles. Since, total number of interactions ( N ) between n particles is N = nC 2 =

M′ R1 r

)

⎡4 ⎤ G ⎢ π r 3 − R13 ρ ⎥ 3 ⎣ ⎦ ⇒ Er = r2 From (1), we get

2

⇒

R2

N=

n( n − 1) 2

8( 8 − 1) = 28 2 8

Inside i.e., for R1 < r < R2 , we have

(

(

)

Mechanics II_Chapter 4_Hints and Explanation.indd 257

6 4 1

)

4 π r 3 − R13 ρ 3 M where ρ = 4 π R23 − R13 3

7

5

GM ′ Er = 2 r where M ′ is mass of shell from R1 to r . So, M′ =

CHAPTER 4

Since

H.257

…(1)

3 2

Out of these 28 interactions, 12 interactions are due to masses at separation a , 12 interactions are due to masses at separation 2a (the face diagonal) and 4 interactions are due to masses at separation 3a (the body diagonal). So,

09-Feb-21 6:34:05 PM

H.258

JEE Advanced Physics: Mechanics – II ⎛ Gm2 ⎞ ⎛ Gm2 ⎞ ⎛ Gm2 ⎞ U = −12 ⎜ − 12 ⎜ − 4⎜ ⎟ ⎝ a ⎟⎠ ⎝ 2a ⎠ ⎝ 3 a ⎟⎠

⇒ 2.

U=−

Gm ⎛ 12 4 ⎞ 12 + + ⎟ a ⎜⎝ 2 3⎠

r

Consider an infinitesimal element of the rod of length dx at a distance x from m0 . If dm be the mass of the infinitesimal element, then

O

Initial potential energy at O due to this element and m is

M dm = dx L

∫

U i = dU i = − dx

M

m0

⇒

dm

x

4.

∫

GMm0 L

a+ L

∫

dx x

U = dU = −

⇒

GMm0 ⎛ a+ L⎞ U=− log e ⎜ ⎝ a ⎟⎠ L

a

6.

Let gravitational field be zero at a point lying at distance x from M. Then,

⇒

d−x = x

m M

⇒

d −1= x

m M

⇒

⎛ x=⎜ ⎝

⇒

(d − x) = ⎜

Outside the sphere, we have GMm r as if the sphere were a point mass concentrated at its centre. Even if the sphere is replaced by a thin shell, the gravitational potential energy of the particle-shell system will remain the same, because for the shell to the entire mass is concentrated at the centre, so U=−

5.

3 GMm 2 R

GM Gm = 2 ( d − x )2 x

GMm R GMm So, the binding energy is U = R i.e., this much energy is required to displace the particle from the centre of the ring to infinity.

GMm r

0

⎛ 3 GMm ⎞ force is ⎜ − ⎝ 2 R ⎟⎠

U=−

U=−

∫

So, the work performed in the process by gravitational

⎛M ⎞ dx ⎟ Gm0 ⎜ ⎝ L ⎠ Gm0 dm dU = − =− x x ⇒

Ui = −

∫

R

G m dm 3GMm =− r dr r R3

⎛ 3 GMm ⎞ 3 GMm So, Wexternal = U f − U i = 0 − ⎜ − ⎟= ⎝ 2 R ⎠ 2 R

The gravitational potential energy of the infinitesimal mass dm and the point mass m0 is

3.

dr

R

2

⎞ M ⎟d M+ m⎠ ⎛ ⎝

Since, VP = −

…(1)

⎞ m ⎟d M+ m⎠

Gm GM − d−x x

…(2) …(3)

Substituting (1) and (2) in (3), we get VP = − 7.

G ( m + M )2 d

⎛ Potential due to ⎞ ⎛ Potential due to ⎞ VP = ⎜ ⎟⎠ + ⎜⎝ particlle at P ⎟⎠ shell at P ⎝

Consider an elementary hemispherical shell of radius r and thickness dr. If dm is mass of this shell, then dm =

⇒

dm =

M 1 4π r 2 dr 1⎛ 4 3⎞ 2 ⎜ π R ⎟⎠ 2⎝ 3 3M 2 r dr R3

Mechanics II_Chapter 4_Hints and Explanation.indd 258

P

R 2

09-Feb-21 6:34:10 PM

Hints and Explanations Since point P lies inside the shell, so the potential inside the shell is constant and equals the value at the GM surface i.e., − . R

(b) Again consider an infinitesimal element of length dx, mass dm at a distance x from P. Then dm = λ dx

GM GM − ( R/2 ) R

⇒

VP = −

⇒

3M VP = − R

dx L

M

dW = −

0

a (a + L)

If dV be the gravitational potential due to this element at the point P , then

GM 2 = self energy 2R

dV = −

(b) See theory. 9.

⇒ dV = −

(a) Consider a mass element of length dx , mass dm at a distance x from the centre of rod. Then

(

r

dx

Gdm Gdm dV = − =− r a2 + x 2

1.

r r Since, dV = − E ⋅ d l ( 2 , 1, 0 ) A ⇒

⇒

dx

∫

2

a +x

−l

0

dx a2 + x 2

dx 2.

a2 + x 2

(

= log e x + a 2 + x 2

(

⎡ V = −2Gλ ⎢⎣ log e x + x 2 + a 2

Mechanics II_Chapter 4_Hints and Explanation.indd 259

( xdx − 2 ydy + zdz )

)

)

⎤ 0⎥ ⎦ l

( 2 , 1, 0 ) ⎤ ⎡⎛ 2 x z2 ⎞ 2 ⎢ ⎥ =− ⎜ −y + ⎟ ⎢⎝ 2 ⎥ 2⎠ ( 0, 2, 4 ) ⎦ ⎣

⇒

VAB

⇒

VAB = 3 Jkg −1

0

∫

∫

( 0, 2, 4 )

∫ f ( x ) dx = 2∫ f ( x ) dx , iff f ( x ) is even

⇒ V = −2Gλ

VA − VB = −

2

l

l

0, 2, 4

( 2 , 1, 0 )

l

−l

∫ dV = −( ∫ ) ( xiˆ − 2yjˆ + zkˆ ) ⋅ ( dx iˆ + dy ˆj + dz kˆ ) B

a2 + x 2

l

)

Test Your Concepts-III (Based on Relation between Gravitational Field and Potential)

Gλ dx

∫

a+ L a−L

⎛ a+ L⎞ ⇒ V = −Gλ log e ⎜ ⎝ a − L ⎟⎠

The potential at the point P due to this infinitesimal element is

⇒ V = dV = −Gλ

dx x

⇒ V = −Gλ l og e x

x

∫

∫

a−L

a

Also,

G ( λ dx ) x

⇒ V = −Gλ P

Since

Gdm x

a+ L

dm = λ dx

⇒ dV = −

(a − L)

L

Gm R

⎛ Gm ⎞ ⇒ dW = V ( dm ) = − ⎜ dm ⎝ R ⎟⎠

∫

x P

(a) Since, V ( m ) = −

⇒ W=

⎛ l + l 2 + a2 ⎞ V = −2Gλ log e ⎜ ⎟⎠ ⎝ a

CHAPTER 4

8.

⇒

H.259

ur r Since, dV = − E ⋅ d l ⇒ ⇒

dV = − a ( ydx + axdy ) + b ( zdy + ydz ) dV = − ⎡⎣ ad ( xy ) + bd ( yz ) ⎤⎦

Integrating we get, V = − ( axy + byz ) + C

09-Feb-21 6:34:17 PM

H.260

3.

JEE Advanced Physics: Mechanics – II

r r (a) dV = − E ⋅ d l A

( 0, 0, 0 )

B

1, 1, 1 )

∫ dV = −( ∫

( yiˆ + xjˆ ) ⋅ ( dx iˆ + dy ˆj + dz kˆ )

(b) Since V = axy ur ur So, E = −∇V = − ayiˆ − axjˆ 6.

r r ⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞ Since E = −∇V = − ⎜ i+ j+ k ∂y ∂z ⎟⎠ ⎝ ∂x r ⇒ E = ( −5 + 6 xy ) iˆ + ( 3 x 2 − 2z 2 ) ˆj − 4 yz kˆ

( 0, 0, 0 )

∫

⇒ VA − VB = −

( ydx + xdy )

( 1, 1, 1 )

( 0, 0, 0 )

⇒ VAB = −

∫ (

{ as y dx + xdy = d ( xy ) }

d ( xy )

1, 1, 1 )

(b)

r ⇒iˆ E ⇒

( 0, 0, 0 ) ⎤ = 1 Jkg −1 ⇒ VAB = − ⎡ ( xy ) ⎢⎣ ( 1, 1, 1 ) ⎥⎦ r r dV = − E ⋅ d l

⇒

A

( 0, 0, 0 )

B

( 1, 1, 1 )

∫ dV = − ∫

( 0, 0, 0 )

⇒ VA − VB = −

∫ (

( 3x 2 ydx + x 3 dy )

⇒ VA − VB = −

∫ (

(

3

d x y

4.

( )

8.

⇒

∂V = y3 ∂z r E = − ( 6 xy ) iˆ − 3 x 2 + y 2 z ˆj − y 3 kˆ

(

(

)

(

Mechanics II_Chapter 4_Hints and Explanation.indd 260

)

(

)

(

)

)

Given, V = a x 2 + y 2 + bz 2 ur ur Since, E = −∇V r ⇒ E = − 2 axiˆ + 2 ayjˆ + 2bzkˆ

)

ur Hence E = 2 a 2 x 2 + y 2 + b 2 z 2

(

)

r ⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞ E = −⎜ i+ j+ k ∂y ∂z ⎟⎠ ⎝ ∂x where,

∂V ∂ = ( 2x + 3 y − z ) = 2 ∂x ∂x

∂V ∂ = ( 2x + 3 y − z ) = 3 ∂y ∂y

)

(a) Given, V = a x 2 − y 2 ur ur So, E = −∇V = −2 a xiˆ − yjˆ

)

(

9.

∂V = 3x 2 + 3 y 2 z ∂y

5.

(

( 0, 0, 0 ) ⎤ ⎡ = 1 Jkg −1 = − ⎢ x3 y ( 1, 1, 1 ) ⎥⎦ ⎣

∂V = 6 xy ∂x

∂V = 20 ∂x

(

r ⎛ ∂V ˆ ∂V ˆ ∂V ⎞ +j +k Since E = − ⎜ iˆ ∂y ∂z ⎟⎠ ⎝ ∂x where

r ⎛ ∂V ˆ ∂V ⎞ Since, Eg = − ⎜ iˆ +j ∂y ⎟⎠ ⎝ ∂x

∂V = 20 ∂y r ⇒ Eg = −20 iˆ + ˆj r r iˆ F = mEg = − ( 0.5 )( 20 ) iˆ + ˆj Since, r ⇒ F = −10 iˆ + ˆj r ⇒ F = 10 2 N

)

Since in both the cases, the line integral of the field is an exact differential and hence both the fields are conservative in nature.

= −5iˆ − 5 ˆj

and

1, 1, 1 )

⇒ VAB

0, −2 )

E = Ex2 + Ey2 + Ez2

where

1, 1, 1 )

( 0, 0, 0 )

( 1,

2 2 E = ( −5 ) + ( −5 ) + 0 2 = 5 2 Nkg −1

7.

r r E ⋅ dl

V = 5x − 3 x 2 y + 2 yz 2

⇒

∂V ∂ = ( 2x + 3 y − z ) = −1 ∂z ∂z r E = −2iˆ − 3 ˆj + kˆ

09-Feb-21 6:34:23 PM

Test Your Concepts-IV (Based on Conservation Laws, Escape Velocity and Applications) 1.

(a) Increase in potential energy is ΔU = U f − U i GmM ⎛ GmM ⎞ −⎜− ⎟ R + nR ⎝ R ⎠

⇒

ΔU = −

⇒

ΔU =

⇒

⎛ n ⎞ ⎛ GM ⎞ ΔU = ⎜ mR ⎝ n + 1 ⎟⎠ ⎜⎝ R2 ⎟⎠

Since g = ⇒

GmM ⎛ 1 ⎞ ⎜1− ⎟ R ⎝ 1+ n⎠

⇒

v1 =

2Gm22 ⎛ 1 1 ⎞ − m1 + m2 ⎝⎜ r2 r1 ⎠⎟

⇒

v1 =

2G ( 100 ) ⎛ 1 1 ⎞ − ⎟ ⎜⎝ 30 0.5 1 ⎠

⇒

v1 =

20G 3

⇒

v1 = 2.1 × 10 −5 ms −1

H.261

From (2), we get v2 = 4.2 × 10 −5 ms −1

GM R2

4.

Let v is the minimum velocity, then by Law of Conservation of Energy, we have

⎛ n ⎞ ΔU = ⎜ mgR ⎝ n + 1 ⎟⎠

(b) By Law of Conservation of Energy, ⎛ Gain in GPE ⎞ ⎛ Loss in KE ⎞ ⎜⎝ ⎟⎠ = ⎜⎝ of m of m ⎟⎠

2.

⇒

1 ⎛ GMm ⎞ 2 ⎜⎝ − ⎟⎠ 2 + mv = 0 + 0 L 2

2ngR n+1

⇒

v=2

5.

( U + K )at ∞ = ( U + K )surface 0+0= −

3.

( U + K )C = ( U + K )∞

1 ⎛ n ⎞ ⇒ ⎜ mgR = mv 2 ⎝ n + 1 ⎟⎠ 2 ⇒ v=

GMm 1 + mv 2 R 2 2GM R

⇒

v=

⇒

⎛ GM ⎞ v = 2⎜ 2 ⎟ R ⎝ R ⎠

⇒

v = 2 gR = 11.2 kms −1

CHAPTER 4

Hints and Explanations

GM L

Our first job is to find a point where the resultant field due to both is zero. Let the point P be at a distance x from centre of bigger star. ⇒

G ( 16 M ) GM = ( 10 a − x )2 x2

⇒

x = 8a

{from O1 }

O1

O2

By Law of Conservation of Energy, we have

( U + K )at 1 m = ( U + K )at 0.5 m −

Gm1m2 Gm1m2 1 1 =− + m1v12 + m2v22 r1 r2 2 2

…(1)

Further by Law of Conservation of Linear Momentum, we have m1 ( 0 ) + m2 ( 0 ) = m1v1 + m2 ( − v2 ) ⇒

m1v1 = m2v2

⇒

1 ⎛mv ⎞ ⎛ 1 1⎞ 1 Gm1m2 ⎜ − ⎟ = m1v12 + m2 ⎜ 1 1 ⎟ 2 ⎝ m2 ⎠ ⎝ r2 r1 ⎠ 2

⇒

m ⎞ ⎛ ⎛ 1 1⎞ 1 Gm1m2 ⎜ − ⎟ = m1v12 ⎜ 1 + 1 ⎟ m2 ⎠ ⎝ ⎝ r2 r1 ⎠ 2

Mechanics II_Chapter 4_Hints and Explanation.indd 261

…(2) 2

i.e. once the body reaches P the gravitational pull of attraction due to 16M vanishes and the gravitation pull due to M takes the lead to make m move towards it automatically. i.e. a minimum K.E. or velocity has to be imparted to m from surface of 16M such that it is just able to overcome the gravitational pull of 16M . By Law of Conservation of Energy. ⎛ Total Mechanical ⎞ = ⎛ Total Mechanical ⎞ ⎝ Energy at A ⎠ ⎝ Energy at P ⎠

09-Feb-21 6:34:31 PM

H.262

JEE Advanced Physics: Mechanics – II

Total mechanical energy at A is EA =

1 ⎡ G ( 16 M ) m GMm ⎤ 2 mvmin + ⎢− − 2 2a 8 a ⎥⎦ ⎣

⇒

EA =

GMm ⎛ 1 1⎞ 2 mvmin − ⎜⎝ 16 + ⎟⎠ 2 2a 4

⇒

EA =

1 65GMm 2 mvmin − 2 8a

But ve2 =

8.

Total mechanical energy at B is

6.

⇒

GMm 1 2 ( 45 ) mvmin = 2 8a

⇒

vmin =

E=−

⇒

E=0

Given that ve = 11.2 kms −1 =

where, r is the distance of rocket from sun ⇒

3 5GM 2 a

A r

P

9. v

Further by Law of Conservation of Energy, we have mv02 GMm mv 2 GMm − = − 2 2 r0 r

…(2)

where, M is the mass of the sun. Solving equations (1) and (2) for r using the concept of quadratic equations we get two values of r, one is rmax and another is rmin. So,

( (

r rmax = 0 1 + 1 − K ( 2 − K ) sin 2 ϕ 2−K r and rmin = 0 1 − 1 − K ( 2 − K ) sin 2 ϕ 2−K v02 r02 GM

5 2GMe 2G 3 × 10 Me + Re 2.5 × 10 4 Re

⇒

vs =

2GMe ⎛ 3 × 10 5 ⎞ ⎜⎝ 1 + ⎟ Re 2.5 × 10 4 ⎠

⇒

vs =

2GMe × 13 Re

⇒

vs ≈ 42 kms −1

)

Since the gravitational potential energy between the disc of radius a , mass m1 and a particle of mass m2 placed at axis of disc at a distance l from centre is U axis = −

⇒

U axis

Since ⇒

a l

2Gm1m2 a2

(

a2 + l 2 − l

)

…(1)

1 ⎤ ⎡ a2 ⎞ 2 2Gm1m2l ⎢ ⎛ ⎥ =− ⎜⎝ 1 + 2 ⎟⎠ − 1 ⎥ ⎢ 2 a l ⎣ ⎦

⎛ a2 ⎞ 1, ⎜ 1+ 2 ⎟ ⎝ l ⎠

12

1+

a2 2l 2

2 ⎛ 2Gm1m2l ⎞ ⎛ a ⎞ U axis = − ⎜ ⎜ ⎟ 2 ⎝ ⎠ ⎝ 2l 2 ⎟⎠ a

Gm1m2 l Further, when the particle collides with the disc (at its centre i.e., l = 0 ), then ⇒

U axis = −

2Gm1m2 {Q of ( 1 ) } a Applying Law of Conservation of Energy and using the concept of reduced mass we get U centre = −

E=U+K E=−

) )

(

vs =

( U + K )at A = ( U + K )P

⇒

2GMe 2GMs + Re r

⇒

v0

S

7.

vs =

Since, Ms = 3 × 10 5 Me and r = 2.5 × 10 4 Re

r0

where, K =

2GMe Re

1 GMs m GMe m mvs2 − − =0+0 2 r Re

⇒

At minimum and maximum distances, the velocity r vector ( v ) makes an angle of 90° with radius vector. Then, by Law of Conservation of Angular Momentum, we have …(1) mv0 r0 sin ϕ = mvr where, m is the mass of the planet.

⇒

GMm 1 ⎛ 2GM ⎞ + m⎜ ⎟ 2 ⎝ R ⎠ R

Ki + Ui = K f + U f

GMm ( 1 + 4 ) = − 5GMm 2a 2a

EB = −

⇒

By Law of Conservation of Energy, we have

⎛ GMm G ( 16 M ) m ⎞ EB = ⎜ − − ⎟⎠ ⎝ 2a 8a ⇒

2GM R

GMm 1 + mve2 R 2

Mechanics II_Chapter 4_Hints and Explanation.indd 262

09-Feb-21 6:34:38 PM

Hints and Explanations

( U + K )axis = ( U + K )centre −

Gm1m2 ⎛ Gm1m2 ⎞ 1 2 + 0 = −2 ⎜ ⎟ + μ vr ⎝ l a ⎠ 2 m1m2 m1 + m2

where μ =

10.

2.

⇒

⎛ 2 1 ⎞ 1 ⎛ m1m2 ⎞ 2 Gm1m2 ⎜ − ⎟ = ⎜ v ⎝ a l ⎠ 2 ⎝ m1 + m2 ⎟⎠ r

⇒

⎛ 2 1⎞ vr = 2G ( m1 + m2 ) ⎜ − ⎟ ⎝ a l⎠

ves =

2GM = R

⇒

⎛4 ⎞ 2G ⎜ π R3 ⎟ ρ ⎝3 ⎠ = R

(a) Just before the explosion, the orbital velocity is GM corresponding to which the total R energy associated with the satellite in the orbit is NEGATIVE. When a mass Δm is expelled very rapidly with a speed v , then for the satellite to still remain within the gravitational pull of the planet, the new total energy must still be negative. So v0 =

4Gρ R 3

GM ( m − Δm ) 1 ( m − Δm ) ( v0 + vr )2 − ≤0 2 R 2GM ⇒ ( v0 + vr ) ≤ R

ves ∝ R Surface area of 1 is A = 4π R12 Surface area of 2 is 4 A = 4π R22 ⇒

⇒

⎛ 4 3⎞ ⎛ 4 3⎞ ⎛ 4 3⎞ ⎜⎝ π R3 ⎟⎠ ρ = ⎜⎝ π R1 ⎟⎠ ρ + ⎜⎝ π R2 ⎟⎠ ρ 3 3 3

⇒

R33 = R13 + R23

⇒

R33 = 9R13

⇒

13

R3 = 9

⇒

R3 > R2 > R1

⇒

v3 > v2 > v1

⇒

v3 v = 91 3 and 2 = 2 v1 v1

⇒

v33 v = 9 and 2 = 2 3 v1 v1

⇒

3.

⇒

K A rB 4 R = = =2 K B rA 2R

Also, U = −

GMm r

Mechanics II_Chapter 4_Hints and Explanation.indd 263

GM R

⎛ r3 ⎞ Since T 2 = 4π 2 ⎜ ⎝ GM ⎟⎠

⎛4 ⎞ Since M = ⎜ π R3 ⎟ ρ ⎝3 ⎠

rA = R + R = 2R

1 r

⎧ GM ⎫ ⎬ ⎨Q v0 = R ⎭ ⎩

When the satellite is revolving close to the planet, then r≅R ⎛ R3 ⎞ ⇒ T 2 = 4π 2 ⎜ ⎝ GM ⎟⎠

and rB = R + 3 R = 4 R

K∝

2GM R

(b) Since the term ( m − Δm ) cancels (see equation (1)), so, no modification is needed.

Test Your Concepts-V (Based on Satellites, Kepler’s Laws and Applications)

⇒

GM + vr ≤ R

⇒ vr ≤ ( 2 − 1 )

R1

Since, kinetic energy K =

…(1)

2GM R

⇒ v0 + vr ≤

R2 = 2R1

Given that mass of planet 3 is M3 = M1 + M2

1.

1 r U A rB = =2 U B rA U ∝

CHAPTER 4

⇒

⇒

H.263

GMm 2r

4.

R3

⇒

T 2 = 4π 2

⇒

ρT 2 = constant

⎛4 ⎞ G ⎜ π R3 ⎟ ρ ⎝3 ⎠

For the satellite to revolve in a circular orbit of radius r0 the orbital velocity is given by v0 =

GM r0

…(1)

At maximum or minimum distances, velocity is perpendicular to the radius vector. So, on applying Conservation of Angular Momentum and Mechanical Energy, we get

09-Feb-21 6:34:45 PM

H.264

JEE Advanced Physics: Mechanics – II mv0 r0 cos α = mvr

and −

…(2)

GMm GMm 1 =− + mv 2 2r0 r 2

The centripetal force is provided by gravitational force, so Gm1m2 m1r1ω 2 = m2 r2ω 2 = …(1) d2 Solving these equations, we get

…(3)

Solving these three equations, we get two values of r i.e., ( 1 + sin α ) r0 and ( 1 − sin α ) r0 . Therefore, rmax = ( 1 + sin α ) r0 and rmin = ( 1 − sin α ) r0 5.

Let M be the mass of the sun and r be the distance between the two stars, then m2 2 r1 = r= r m1 + m2 3 (b) r r2

⇒

⎛ 2⎞ G ⎜ ⎟ M2 ⎝ 9⎠ r2

⇒

2 GM 2 ⎛ 2 ⎞⎛ r⎞ = m2 r2ω 2 = ⎜ M ⎟ ⎜ ⎟ ω 2 ⎝ 3 ⎠⎝ 3⎠ 9 r2

⇒

ω2 =

2

K1 ⎛ m1 ⎞ ⎛ m2 ⎞ m =⎜ = 2 ⎟ ⎜ ⎟ K 2 ⎝ m2 ⎠ ⎝ m1 ⎠ m1

(c)

L1 I1ω I1 m2 = = = L2 I 2ω I 2 m1

(d)

L = L1 + L2 = ( I1 + I 2 ) ω

(

)

⇒ L = m1r12 + m2 r22 ω

GM r3

⎡ m1m22 d 2 m2 m12 d 2 ⎤ + ⇒ L=⎢ ⎥ω 2 ( m1 + m2 )2 ⎥⎦ ⎢⎣ ( m1 + m2 )

2

⇒

GM ⎛ 2π ⎞ ⎜⎝ ⎟ = 3 T ⎠ r

⇒

4π 2 r 3 T = GM

⇒ L = μω d 2

2

where μ =

Since time period of earth around sun is 4π 2 R3 GM

(e) 7.

r=R

(a) Let the origin be at m1 and the centre of mass be at a distance r1 from it. Then r1 =

2

⇒ m1r1 = m2 r2

m2 = 2M 3

Gm1m2 Centripetal force on m2 is = r2

6.

K1 ⎛ m1 ⎞ ⎛ r1 ⎞ = K 2 ⎜⎝ m2 ⎟⎠ ⎝⎜ r2 ⎟⎠

Since, m1r1ω 2 = m2 r2ω 2

m1 = M 3

⇒

d3 2π = 2π G ( m1 + m2 ) ω

1 2 K1 2 I1ω I m r2 = = 1 = 1 12 K 2 1 I ω 2 I 2 m2 r2 2 2 ⇒

CM

T2 =

d3

⇒ T=

m1 r r= and r2 = m1 + m2 3 r1

G ( m1 + m2 )

ω=

m1 ( 0 ) + m2 r m1 + m2

m1

T1 ⎛ r1 ⎞ = T2 ⎜⎝ r2 ⎟⎠

r2

Since, ω =

d

Mechanics II_Chapter 4_Hints and Explanation.indd 264

⇒

32

⎛r ⎞ T2 = T1 ⎜ 2 ⎟ ⎝ r1 ⎠

m2

CM r1

1 1 ( Ι1 + Ι 2 )ω 2 = μω 2d2 2 2

For two satellites revolving around same planet, we have

⇒

⎛ m2 ⎞ ⇒ r1 = ⎜ r and ⎝ m1 + m2 ⎟⎠

K=

m1m2 is the reduced mass m1 + m2

ω1 =

32

⎛ 2 × 10 4 ⎞ = 28 ⎜ ⎝ 10 4 ⎟⎠

32

= 56 2 hour

2π T 2π 2π radhr −1 and ω 2 = radhr −1 28 56 2

09-Feb-21 6:34:54 PM

Hints and Explanations

vA

B

vBA = ω 2 r2 + ω 1r1

⇒

2π ⎤ ⎡ 2π vBA = ⎢ × 2 × 10 4 + × 10 4 ⎥ kmhr −1 28 ⎣ 56 2 ⎦

⇒

vBA = ( 0.1587 + 0.2242 ) × 10 kmhr

⇒

vBA = 3829 kmhr −1

4

…(2)

r′

…(3)

GMm 1 GMm 1 mvp2 − = mv′p2 − 2 r 2 r′

…(4)

G ( M )( M ) GM 2 ⎛1 ⎞ + 2 ⎜ Iω 2 ⎟ = − + Iω 2 ⎝2 ⎠ 2R 2R GM 2 ( GM 2 ⎛ GM ⎞ + MR2 ) ⎜ =− ⎟ 3 ⎝ 4R ⎠ 2R 4R

GM 2 4R Hence, this much amount of energy will be required to separate the two stars to infinity.

⇒ ⇒

5r Hence, the maximum and minimum distances are 3 and r respectively.

r 2 = a2e 2 + a2 − a2e 2 r=a m

vc

b

r

a

M ae

Total energy of a satellite in an elliptical orbit is, E=−

GMm 2a

⇒

EC =

GMm GMm 1 mvc2 − =− r 2 2a

⇒

GMm 1 GMm mvc2 − =− 2 a 2a

Solving the above equations (1), (2), (3) and (4), we get

Mechanics II_Chapter 4_Hints and Explanation.indd 265

2π 4R3 R3 = 2π = 4π ω GM GM

But b 2 = a 2 ( 1 − e 2 )

mvp r = mv′p r ′

5r and r ′ = r 3

E=−

GM 4 R3

r 2 = a2e 2 + b 2

Since, vp lies between orbital velocity and escape velocity, path of the particle would be an ellipse with r being the minimum distance. Let r ′ be the maximum distance and v′p be the velocity at that moment. Then applying the Law of Conservation of Angular Momentum and Conservation of Mechanical Energy, we get

r′ =

M

10. Since, we observe that

Earth v′p

and

R CM

So, binding energy is E =

5 v 4 0 r

R

F = MR

⇒ E=−

5 v0 = 1.25v0 4

GM 2 4 R2

GM 2 = MRω 2 4R2

⇒ T= (c)

=

F = MRω 2

⇒ ω=

−1

GM , where M r is the mass of earth …(1) Absolute velocity of particle would be:

Particle

(b)

⇒

Orbital speed of the satellite is v0 =

vp =

( 2R )

2

M

⇒

vp = v + v0 =

G ( M )( M )

vB

Positions corresponding to maximum separation are shown in figure, so we have r r vBA = vB − vA = vB + vA {for shown position}

8.

(a) F =

CHAPTER 4

9.

A

H.265

{r = a}

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JEE Advanced Physics: Mechanics – II

Single Correct Choice Type Questions

GM a

⇒

vc =

⇒

g vc = R a

1.

{ as GM = gR } 2

dx x

11. Both the stars rotate about their centre of mass. R1

m2

m1

CM

L2 + x2 y dF θ θ M dFsinθ L dFsinθ P θ θ dF

R2

r1

Let the mass M be placed symmetrically.

r2 r

∞

⇒

For the position of CM let Origin be at m1 and CM be at distance r1 from origin. Then r1 =

m2 r m1r and r2 = m1 + m2 m1 + m2

Gm1m2 r2 m2 r Since, r1 = m1 + m2 Also, m1r1ω 2 =

Qω=

2π T

ω2 =

⇒

⎡ G ( m1 + m2 ) ⎤ r=⎢ ⎥ ω2 ⎣ ⎦

…(1)

Gm1m2 Gm1m2 1 =− + μvr2 r ( R1 + R2 ) 2

⇒

vr2 =

2Gm1m2 ⎛ 1 1⎞ − ⎛ m1m2 ⎞ ⎜⎝ R1 + R2 r ⎟⎠ ⎜⎝ m + m ⎟⎠ 1 2

⇒

1 1⎞ ⎛ − ⎟ vr2 = 2G ( m1 + m2 ) ⎜ + R R r⎠ ⎝ 1 2

Substituting the value of r from equation (1), we get 13 ⎡ ⎧⎪ ⎫⎪ ⎤ 1 4π 2 ⎥ −⎨ vr = 2G ( m1 + m2 ) ⎢ ⎬ ⎢⎣ R1 + R2 ⎪⎩ G ( m1 + m2 ) T 2 ⎪⎭ ⎥⎦

Mechanics II_Chapter 4_Hints and Explanation.indd 266

x + L2

dx 2

+ L2 )

3/2

GMm r2

GM = r 2ω 2

Now, g =

…(2)

m1m2 m1 + m2 and vr is the relative velocity between the two stars From equation (2), we get 2Gm1m2 ⎛ 1 1⎞ ⎜⎝ R + R − r ⎟⎠ μ 1 2

x +L

L 2

Let M be the mass of the planet and m the mass of satellite. Then

⇒

GM R2

r 3ω 2 R2 Hence, the correct answer is (C).

where, μ is the reduced mass given by μ =

vr2 =

∫ (x

2

Fnet =

mrω 2 =

Applying Law of Conservation of Mechanical Energy, we get −

Fnet = GMλ L

2

GMλ L (2) L2 2GMλ ⇒ Fnet = L Hence, the correct answer is (C). ⇒

2.

13

GM ( λ dx )

−∞

−∞

}

r3

∫ d F sinθ = ∫ ∞

⇒

{

Fnet =

∞

−∞

G ( m1 + m2 )

⇒

x

⇒

g=

3.

( U + K )A = ( U + K )O ⇒ ⇒

GMm

−

2

r +r

2

+0=

1 GMm mv 2 − 2 r

2

1 ⎞ v GM ⎛ = ⎜⎝ 1 − ⎟ 2 r 2⎠

09-Feb-21 6:35:09 PM

Hints and Explanations

⇒

7.

2GM ⎛ 1 ⎞ ⎜1− ⎟ r ⎝ 2⎠

v=

H.267

Hence, the correct answer is (D).

F=

∫ h

M ⎞ h+ L dx ⎟ m ⎜⎝ ⎠ GMm L = x −2 dx L x2

∫

Since F =

h

⇒ ⇒

GMm ⎛ 1 F=− − ⎜ L ⎝ h+L

⇒

F=

h

⎟ ⎠

⎛4 ⎞ m = ⎜ π R3 ⎟ ρ ⎝3 ⎠

1⎞ ⎟ h⎠

⎛4 ⎞ G ⎜ π R3 ρ ⎟ ⎝3 ⎠ F= 2 4R 4 F = Gπ 2 ρ 2 R 4 9

⇒

GMm h( h + L )

⇒

Hence, the correct answer is (C). 5.

d = 3 R on its axis is 8.

R M d = 3R

GM R2 ⎛4 ⎞ Also, M = ⎜ π R3 ⎟ ρ ⎝3 ⎠ Since, g =

⇒ R

GMd

(R

+d

)

2 32

=

Hence, the correct answer is (D). 6.

3GM R2

2

By Law of Conservation of Energy, we get

( U + K )surface = ( U + K )centre

9.

GMm 3 GMm and U centre = − R 2 R GMm 1 3 GMm 1 2 − + m( 0 ) = − + mv 2 R 2 2 R 2

U surface = −

⇒

GMm ⎛ 3 GMm ⎞ 1 mv 2 = − −⎜− ⎝ 2 R ⎟⎠ R 2

⇒

1 1 GMm mv 2 = 2 2 R

GM v = e R 2 Hence, the correct answer is (D). ⇒

v=

Mechanics II_Chapter 4_Hints and Explanation.indd 267

⇒

g ∝ M1 3 ρ 2 3

⇒

g planet = g ( 8 )

23

13

( 8 )2 3 = 8 g

Hence, the correct answer is (C).

Now, for a solid sphere, we have

⇒

13

⎛ 4πρ ⎞ g = GM ⎜ ⎝ 3 M ⎟⎠

3GM 8R2

Force on sphere is F = ( 8 M ) E =

⎛ 3M ⎞ R=⎜ ⎝ 4πρ ⎟⎠

…(1)

Substituting in (1) we get,

8M

2

2

⇒ F ∝ R4 Hence, the correct answer is (C).

Gravitational field due to the ring at a distance

E=

( 2 R )2

If ρ be the density of each sphere, then

h+ L ⎞

GMm ⎛ x −2 + 1 F= ⎜ L ⎝ −2 + 1

Gm2

CHAPTER 4

4.

h+ L G ⎛

10.

1 and does not depend on the medium. Hence, r2 F F1′ = 1 4 1 Fc ∝ 2 Kr and when air is sucked, force will slightly increase in vacuum, so F F2′ > 2 4 Hence, the correct answer is (C). Fg ∝

2h ⎞ ⎛ g⎜ 1− ⎟ ⎝ R⎠

d⎞ ⎛ g⎜ 1− ⎟ ⎝ R⎠

⇒ d = 2h Hence, the correct answer is (B).

09-Feb-21 6:35:17 PM

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JEE Advanced Physics: Mechanics – II

11. Imagine an inverted hemispherical shell to be placed on this shell so as to complete the spherical shell inside which net gravitational field is zero. Net field can be zero only when field at P is directed along c . Hence, the correct answer is (C).

15.

At equator ϕ = 0° ⇒

⇒ ⇒

−

⇒

v 2 GM GM = − 2 8R R

g 1 = rads −1 R 800 Hence, the correct answer is (C).

ω=

GMm 1 GMm + mv 2 = − R 2 8R ⇒

16.

v 2 7 ⎛ GM ⎞ = ⎜ ⎟ 2 8⎝ R ⎠

7GM 4R Hence, the correct answer is (D). ⇒

⇒

{Q h

R} 17.

2h ⎞ ⎛ Wh = W ⎜ 1 − ⎟ ⎝ R⎠

{Q W = mg }

Given that points P and Q located inside and outside the planet have same acceleration due to gravity equal g to . Thus 4 gR2 g d⎞ ⎛ g⎜ 1− ⎟ = = ⎝ R ⎠ ( R + h )2 4 1 3R R = and 4 R+h 2 ⇒ h=R Maximum separation between P and Q will be when these are diametrically opposite. So, maximum separation is rmax = h + R + ( R − d ) d=

3 R ⎞ 9R ⎛ rmax = R + R + ⎜ R − ⎟= ⎝ 4 ⎠ 4 Hence, the correct answer is (B). ⇒

Mechanics II_Chapter 4_Hints and Explanation.indd 268

4 3 ⎞ πR ρ ⎟ ⎠ 4 3 = π GρR 3 R2

GM r R3

but M =

dE GM = 3 dr R

4 3 πR ρ 3

M 4πρ = ( ρ = density) 3 R3 4π Gρ = constant ⇒ Slope = 3 Hence, the correct answer is (A). ⇒

14.

⇒

E=

i.e., Slope,

Rate of change of weight with height 2mg 2W dWh = 0− = dh R R Hence, the correct answer is (B).

⎛ G⎜ ⎝ GM g= 2 = R ⇒ g ∝ ρR

R is increased by a factor of 2 i.e., to keep the value of g to be the same, the value of ρ has to be changed by 1 a factor of . 2 Hence, the correct answer is (C).

v=

2h ⎞ ⎛ 13. Since, g h = g ⎜ 1 − ⎟ ⎝ R⎠

g ′ = g − Rω 2

For g ′ to be zero

12. ( U + K )surface = ( U + K )R + 7 R GMm 1 GMm − + mv 2 = − +0 R 2 R + 7R

g ′ = g − Rω 2 cos 2 ϕ

18.

r ⎛ ∂U ˆ ∂U ˆ ⎞ F = −⎜ i+ j ∂y ⎟⎠ ⎝ ∂x r ⇒ F = − aiˆ − bjˆ r ⇒ F = a2 + b 2 r a2 + b 2 F ⇒ Acc = = m m Hence, the correct answer is (C).

19. Let R be the radius of earth and g the acceleration due to gravity on earth’s surface. Then the desired ratio (say x ) is h⎞ ⎛ g⎜ 1− ⎟ 2 ⎝ ⎠ ⎛ h⎞⎛ h⎞ R = ⎜1− ⎟ ⎜1+ ⎟ x= ⎝ g R⎠⎝ R⎠ 2 h⎞ ⎛ ⎜⎝ 1 + ⎟⎠ R

09-Feb-21 6:35:27 PM

Hints and Explanations

2h h⎞ ⎛ R , so ⎜ 1 + ⎟ ≅ 1 + ⎝ R⎠ R

⇒

h⎞⎛ 2h ⎞ ⎛ x = ⎜ 1− ⎟ ⎜ 1+ ⎟ ⎝ R⎠⎝ R⎠

⇒

h x ≈ 1+ R

From this expression we see that x increases linearly with h. Hence, the correct answer is (C). 20.

⎛ GM ⎞ ⎛4 ⎞ E ( r ) = ⎜ 3 ⎟ r where M = ⎜ π R3 ⎟ ρ ⎝3 ⎠ ⎝ R ⎠ , Gr ⎛ 4 ⎞ ⎛4 ⎞ ⇒ E ( r ) = 3 ⎜ π R3 ρ ⎟ = ⎜ πρG ⎟ r ⎠ ⎝3 ⎠ R ⎝3 Hence, the correct answer is (A).

21. Energy required to raise a satellite to a height h is ΔE = U f − U i = −

GMm ⎛ GMm ⎞ −⎜− ⎟ R+h ⎝ R ⎠

⇒

1 ⎞ ⎛ 1 ΔE = GMm ⎜ − ⎝ R R + h ⎟⎠

⇒

GMmh ΔE = R( R + h )

Further, if v0 be the orbital velocity in the orbit, then v0 = ⇒

KE =

GM R+h 1 GMm mv02 = ( 2 2 R + h)

and U f = −

Hence, external work done W is given by W = ΔU = U f − U i =

ΔΕ 2 h = KE R Hence, the correct answer is (C). 22.

GMm GM mR = 2 × 12R 12 R

mgR 12 Hence, the correct answer is (C). ⇒

W=

23. Since, F1 = Also, F2 = ⇒

F2 =

GMm 9R 2 GMm G ( M/8 ) m − 9R 2 ( 5R/2 )2

GMm GMm 41 GMm − = 450 R2 9R 2 50 R2

F2 41 = F1 50 Hence, the correct answer is (B). ⇒

24. By the principle of superposition of fields r r r E = E1 + E2 r where,rE is net field at the centre of hole duerto entire mass, E1 is field due to remaining mass and E2 is field due to mass in hole = 0. r r ⎛ GM ⎞ R Since, E1 = E = ⎜ 3 ⎟ r where r = ⎝ R ⎠ 2 ⇒

So,

GMm GMm =− 4R R + 3R

CHAPTER 4

2

Since h

H.269

r GM E= 2R 2

Hence, the correct answer is (C). 25. Initially field due to both is along positive x-axis. Due to the ring, field will first increase and then decrease to zero at centre. While field due to the solid sphere, will continuously increase in positive x-direction. On the other side of the ring field is now towards negative x-axis. Hence, the correct answer is (B). GM R and potential due to shell at its own surface is G ( 3M ) V2 = − R 4GM So, total potential is V = V1 + V2 = − R Hence, the correct answer is (D).

26. Potential due to particle at the surface is V1 = −

Since U = − So, U i = −

GMm R+h

GMm GMm =− 2R + R 3R

Mechanics II_Chapter 4_Hints and Explanation.indd 269

09-Feb-21 6:35:36 PM

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JEE Advanced Physics: Mechanics – II

27.

By Law of Conservation of Angular Momentum, we get mv1r = mv2 r 2

m1

m2

VA = −

v1r1 = 2v2 r2

GMm and the magni2r tude of gravitational potential energy is

Gm1 Gm2 − and R 2R

U= −

U 2 Hence, the correct answer is (B).

32. Total mechanical energy, at earth’s surface is

Gm2 Gm1 − R 2R

E=U+K = −

Since WA→ B = m ( VB − VA ) Gm ( m1 − m2 ) ( 2 − 1 )

WA→ B =

⇒

Since ve =

2R Hence, the correct answer is (B). 28.

⇒

1 1 1 ⎡ 1 ⎤ Eg = G ⎢ 2 + 2 + 2 + 2 + ... ⎥ 2 4 8 ⎣1 ⎦

Eg = G

⇒

⎛ 1 ⎞ ⎜ 1⎟ ⎜⎝ 1 − ⎟⎠ 4

⇒

⎛4 ⎞ Since, M = ⎜ π R3 ⎟ ρ ⎝3 ⎠ T=

3π ρG

i.e., T is independent of R . Hence, the correct answer is (D). 30.

v1 r r2

r1 S

v2

Mechanics II_Chapter 4_Hints and Explanation.indd 270

GMm 1 ⎛ 2GM ⎞ + m⎜ ⎟ =0 R 2 ⎝ R ⎠

{Q r ≅ R }

GmM GmM 1 ⎛ R ⎞ =− + k⎜ ⎟ R 2 ⎝ 2⎠ 2R 2

2

GMm ⎛ 1 ⎞ kR2 ⎜⎝ 2 − ⎟⎠ = R 2 8

12GMm R3 Hence, the correct answer is (D). ⇒

29. For a satellite revolving near the surface of planet, we have

⇒

−

1st term ⎧ ⎫ ⎨ S∞ = ⎬ 1-Common Ratio ⎩ ⎭

R3 GM

2GM R

33. By Law of Conservation of Mechanical Energy, we get

4G ⇒ Eg = 3 Hence, the correct answer is (B).

T ≅ 2π

E=−

GMm 1 + mve2 R 2

Hence, the correct answer is (D).

1 1 1 ⎡ ⎤ Eg = G ⎢ 1 + + + + ... ⎥ 4 16 64 ⎣ ⎦

⇒

GMm GMm = r r

So, K =

⎛ Potential at ⎞ ⎛ Potential at ⎞ + VB = ⎜ ⎝ B due to A ⎟⎠ ⎜⎝ B due to B ⎟⎠ VB = −

⇒

31. Kinetic energy of satellite is K =

Similarly,

⇒

mv1r1 sin 30° = mv2 r2

Hence, the correct answer is (C).

⎛ Potential at ⎞ ⎛ Potential at ⎞ VA = ⎜ + ⎝ A due to A ⎟⎠ ⎜⎝ A due to B ⎟⎠ ⇒

⇒

k=

34. By Law of Conservation of Energy

( U + K )surface = ( U + K )r ⇒

−

GMm 1 GMm + mv 2 = − +0 R 2 r

⇒

−

GMm 1 2 2 GMm + mk ve = − R 2 r

But ve =

2GM R

⇒

−

GM GM 1 2 ⎛ 2GM ⎞ + k ⎜ ⎟⎠ = − ⎝ R 2 R r

⇒

−

1 k2 1 + =− R R r

09-Feb-21 6:35:45 PM

Hints and Explanations 1 1 − k2 = r R

GM ⎛ 2 2 + 1 ⎞ ⎜ ⎟ R ⎝ 4 ⎠ Hence, the correct answer is (D). r r r r 39. dW = F ⋅ d l = m0 ( E ⋅ d l )

R 1 − k2 Hence, the correct answer is (B). ⇒

r=

35. When kinetic energy < E , total energy will be negative, when kinetic energy equals E , total energy is zero and when kinetic energy > E , total energy is positive. Hence, the correct answer is (C). 36. For r ≤ r1

dV Gm1 = 2 dr r

G ( m1 + m2 ) , slope of r G ( m1 + m2 )

V=−

For dV = dr

ˆ + ˆjdy dW = m0 4iˆ + ˆj ⋅ idx

⇒

dW = m0 ( 4 dx + dy )

V -r

graph is

r2

G ( m1 + m2 ) Gm1 to i.e., slope increases. r22 r22 Hence, the correct answer is (D). changes from

c=

2GMstar R 8

⇒

3 × 10 =

⇒

R

⇒

∫ dW = 0 ∫ ( 4dx + dy ) = 0 ∫ d ( 4x + y ) = 0

40. Total energy of a planet of mass m , in an elliptical orbit is GMm E=− { m = mass of planet} 2a By Law of Conservation of Energy, we have KE + PE = E ⇒

1 GMm GMm =− mv 2 − r 2 2a

⇒

⎛ 2 1⎞ v = GM ⎜ − ⎟ ⎝ r a⎠

2 ( 6.67 × 10 −11 ) ( 3 × 2 × 10 30 ) R

3 ve 4 By Law of Conservation of Energy,

41. Since u =

( U + K )surface = ( U + K )r

9 km

38. Gravitational force on each particle due to the other three particles will provide the necessary centripetal force.

GMm 1 −GMm + mu2 = +0 R 2 r

⇒

−

⇒

GMm −GMm 1 ⎛ 9 2 ⎞ + m ⎜ ve ⎟ = − ⎝ ⎠ R 2 16 r

Since ve2 =

2

GM 2

(

2R )

)

Hence, the correct answer is (A).

Hence, the correct answer is (D).

⇒

)(

Given that, W = 0

At the boundary of outer shell, slope of V -r graph

37.

(

⇒ 4x + y = constant ⇒ y + 4 x = 2 , satisfies the condition. Hence, the correct answer is (A).

Gm2 Gm1 − r2 r

Slope of V -r graph

⇒

⇒

For r1 ≤ r ≤ r2 V=−

v=

⇒

Gm1 Gm2 − = constant r1 r2

V=

⇒

CHAPTER 4

⇒

H.271

cos 45° + 2

Mechanics II_Chapter 4_Hints and Explanation.indd 271

GM 2

( 2 R )2

=

Mv 2 R

2GM R

⇒

−

GMm GMm 1 9 ⎛ 2GM ⎞ + m ⎜ ⎟=− R 2 16 ⎝ R ⎠ r

⇒

−

7 GMm GMm =− 16 R r

16 R 7 Hence, the correct answer is (B). ⇒

r=

09-Feb-21 6:35:56 PM

H.272

42.

JEE Advanced Physics: Mechanics – II

⎛ Work ⎞ ⎛ Increase in Gravitational ⎞ ⎜⎝ done ⎟⎠ = ⎜⎝ ⎟⎠ Potential Energy

45. Let σ be the surface density, then M A = σ 4π RA2 , mB = σ 4π RB2

mgh Since, ΔU = h 1+ R mgR mgR ⇒ W1 = = R 2 1+ R mgh and similarly, W2 = h 1+ R Since, W1 = 2W2

Since VA = ⇒

mgR 2mgh = h 2 1+ R R ⇒ h= 3 Hence, the correct answer is (C).

⇒

y

)

The work done by this field is independent of the path followed between any two states i.e. W( 0 , 0 )→( a , a ) = W( 0 , 0 )→( 0 , a )→( a , a ) = W( 0 , 0 )→( a , 0 )→( a , a ) r r Since, dW = mEg ⋅ dr r where dr = $i dx + $j dy

⇒

dW = mk ( y dx + x dy ) ( a, a )

∫

W = dW = mk W = mka

∫ d ( xy ) = mka ( )

2

0, 0

2

{where m is mass of particle}

W( 0 , 0 )→( a , a ) = W( 0 , 0 )→( 0 , a )→( a , a ) = W( 0 , 0 )→( a , 0 )→( a , a ) = mka 2 Hence the field is conservative. Hence, the correct answer is (B).

Mechanics II_Chapter 4_Hints and Explanation.indd 272

…(2)

M A + MB R 2 = 2 MA RA

…(3)

Now

V M RA ⎛ M A + M B ⎞ RA = = ⎟⎠ 2 VA R M A ⎜⎝ MA RA + RB2

From (3), we have

(

⇒

M A RA2 = MB RB2

R2 = RA2 + RB2 x

r r ⎛ ∂V $ ∂V ⎞ Eg = −∇V = − ⎜ $i +j = k yi$ + x $j ∂y ⎟⎠ ⎝ ∂x

⇒

…(1)

From (2) and (3), we get

dA

O dA abt ⇒ = dt 2 Hence, the correct answer is (B).

44.

and

vdt

b

MA MB M +M = = A 2 B 4π RA2 4π RB2 4π R

From equation (1), we get

dt

dA bv ⇒ = dt 2 Since, v = at , so

VA RA 3 = = VB RB 4

RB =

σ=

1 ( vdt )( b )

dA = dt 2

) )

4 RA 3 Since the new shell formed also has same surface mass density. So, we have ⇒

⇒

43.

( (

2 VA M A RB σ 4π RA RB RA = = = VB MB RA σ 4π RB2 RA RB

Given that,

dA 1 base × height = × dt 2 dt

−GM A −GMB and VB = RA RB

…(4)

M A + MB R 2 = 2 MA RA

Substituting in equation (4), we get RA2 + RB2 5 V = = VA RA 3 Hence, the correct answer is (C). 46. Since, T 2 = kr 3 ΔT Δr ⇒ 2 =3 T r ΔT 3 Δr ⇒ = T 2 r Hence, the correct answer is (A). 4 ⎡4 ⎤ 3 47. Net mass, M = ⎢ π ( 2R ) − π R3 ⎥ ρ 3 ⎣3 ⎦ where ρ is the density of the material of sphere. So, M ⎛ 3M ⎞ =⎜ 4 ⎡ ⎝ 28π R3 ⎟⎠ 3 3⎤ π ⎣ ( 2R ) − ( R ) ⎦ 3 Now V = V2 R at − V R at

ρ=

centre

…(1)

centre

09-Feb-21 6:36:05 PM

Hints and Explanations GM1 GM2 and VR = − 2R 2R

⇒ ⇒

M ⎛4 ⎞ 3 and M2 = ρ ⎜ π ⎟ ( R ) = ⎝3 ⎠ 7 Substituting in equation (1), we get 9GM 14 R Hence, the correct answer is (B).

51. Since

V=−

⇒

48. By Law of Conservation of Energy

( U + K )surface = ( U + K )h −

{Q At maximum height velocity is zero } ⇒

GMm 1 GMm GMm + =− 2 R R+h R ⇒ h=R Hence, the correct answer is (B). ⇒

−

T12 r13 = T22 r23 3

1 T12 ⎛ R ⎞ =⎜ ⎟ = 64 T22 ⎝ 4 R ⎠

(Three earth radii from the surface means four radii from the centre).

GMm 1 GMm + mv 2 = − +0 R 2 R+h

GMm 1 GMm − + mgR = − R+h R 2

GMm GMm GMm + = R+h R R+h

⎛ GMm ⎞ GMm 2⎜ = ⎝ R + h ⎟⎠ R ⇒ 2R = R + h ⇒ h=R Hence, the correct answer is (A).

8 ⎛4 ⎞ 3 where, M1 = ρ ⎜ π ⎟ ( 2R ) = M ⎝3 ⎠ 7

⇒

−

{

Qg =

GM R2

}

⇒

T1 1 = T2 8

⇒

T2 = 8T1 = 8 ( 83 ) = 664 minute

CHAPTER 4

V2 R = −

H.273

Hence, the correct answer is (C). 52.

49. Since angular momentum of the particle about origin O, L = mvb = constant, therefore, areal velocity of the particle about origin O is constant. So, it sweeps equal areas in equal intervals of time. Thus, area swept in smaller time interval will also be smaller. Hence

The direction of net gravitational intensity at various points is shown above. Taking gravitational intensity towards right as positive, the graph will be

A1 > A2 if ( t4 − t3 ) < ( t2 − t1 ) Hence, the correct answer is (B). 53.

Hence, the correct answer is (A). 50.

GMm ⎛ GMm ⎞ −⎜− ⎟ R+h ⎝ R ⎠ when taken from surface to h , we have ΔU1 = −

GMm GMm ΔU1 = − + R+h R Now, when taken from h to infinity, we have ⎛ GMm ⎞ ΔU 2 = 0 − ⎜ − ⎝ R + h ⎟⎠ Since ΔU1 = ΔU 2

Mechanics II_Chapter 4_Hints and Explanation.indd 273

Gm 5R After perfectly inelastic collision, velocity of combined mass will become zero. Therefore, if v be the desired speed, then by Law of Conservation of Mechanical Energy, we have v0 =

( U + K )r =5 R = ( U + K )surface GM ( 2m ) 1 ⎛ GmM ⎞ ⎛ GmM ⎞ −⎜ + ( 2m ) v 2 − +0= − ⎝ 5R ⎟⎠ ⎜⎝ 5R ⎟⎠ R 2 ⇒

GM ( 2m ) GM ( 2m ) 1 ( 2m ) v 2 = − + R 2 5R

8GM = 2 2v0 5R Hence, the correct answer is (A). ⇒

v=

09-Feb-21 6:36:13 PM

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JEE Advanced Physics: Mechanics – II

54. Since Areal velocity =

Area Swept Time for one Revolution of Earth about the Sun

( U + K )surface = ( U + K )at h ⇒

L Further Areal velocity = 2M ⇒

GMm 1 GMm 1 + mve2 = − + mv 2 R 2 R+R 2

Since, ve =

Time for one ⎞ ⎛ L ⎞⎛ Area Swept = ⎜ Revolution of Earth ⎝ 2 M ⎟⎠ ⎜⎜ about the Sun ⎟⎟ ⎝ ⎠

1( 4.4 × 1015 ) ( 365 × 24 × 60 × 60 ) 2 ⇒ Area Swept 7 × 10 22 m 2 Hence, the correct answer is (D). Area Swept =

2GM R

⇒

1 GMm GMm GMm mv 2 = − + + 2 R R 2R

⇒

v=

GM R

Let θ be the angle of its velocity with horizontal, then by Law of Conservation of Angular Momentum about centre of earth, we have

55. Force on element dx on rod is GM ( dm ) x2 M where dm = dx 2R

mve R cos ( 45° ) = mvr cos θ

dF =

⇒

−

⎛ GM 2 ⎞ dx dF = ⎜ ⎝ 2R ⎟⎠ x 2

ve R cos ( 45° ) vr

⇒

cos θ =

⇒

2GM 1 R 2 =1 cos θ = 2 ( 2R ) GM R

⇒

θ = 60°

R

Hence, the correct answer is (C). Total force on the rod is

∫

F = dF =

⇒

F=

GM 2R

2

GM 2 ⎛ 1 ⎞ ⎜− ⎟ 2R ⎝ x ⎠

2R

57.

1

∫ x dx 2

R

2R

=− R

1⎞ GM 2 ⎛ 1 − ⎟ ⎜ 2R ⎝ 2R R ⎠

GM 2 4R2 Hence, the correct answer is (A). ⇒

⎧ GMm ⎪⎪ − r U(r) = ⎨ ⎪ − GMm ⎪⎩ R

, r≥R , r r1 and r > r2 . ⎝ ⎠ r If P lies inside one shell and outside the other. Then field at point P due to the shell enclosing it is zero and Gm is 2 2 due to the other shell. r If P lies inside both the shells, then Ep = 0 . Hence, (A), (B) and (C) are correct.

09-Feb-21 6:37:13 PM

Hints and Explanations Similarly, as in SOLUTION to PROBLEM 3, G ( m1 + m2 ) Vp = − for r > R1 , r > R2 r Which satisfies OPTION (A) For the point P lying inside shell 1 and outside shell 2, the total potential at P equals the potential at P due to shell 1 plus potential at P due to shell 2.

8.

R Since sin θ = 2 x

Vp = Vp1 + Vp 2 ⇒ ⇒

Vp =

−Gm1 −Gm2 + r1 r

⇒

m ⎞ ⎛m Vp = −G ⎜ 1 + 2 ⎟ r ⎠ ⎝ r1

Since cos θ =

R 2

5.

Conceptual (A), (B), (C) and (D) are correct.

6.

Inside the inner sphere field is zero but potential is constant. Between two, field is due to inner sphere and potential is due to both but it is constant due to outer shell. Outside the outer shell, field and potential are due to both and it decreases in both the cases. Hence, (A) and (C) are correct. Nature of Path

v = v0

Circular path around the earth.

v < v0

Elliptical path and body returns to earth.

v > v0 but < ve

Elliptical path around the earth and will not escape.

v = ve

Parabolic path and it escapes from the earth.

v > ve

Hyperbolic path and escapes from earth.

Hence, (A), (B), (C) and (D) are correct.

R2 − x2 4 x

gx R2 − 4x 2 R R we have So, curve is parabolic and at x = 2 a=0 Hence, (B) and (C) are correct. ⇒

Which satisfies OPTION (D) Hence, (A), (C) and (D) are correct.

Mechanics II_Chapter 4_Hints and Explanation.indd 283

R 2x

⎛ mgx ⎞ ⎛ R ⎞ N=⎜ ⎝ R ⎟⎠ ⎜⎝ 2x ⎟⎠ mg ⇒ N= 2 Constant and independent of x Tangential force is given by F = ma = mg ′ cos θ ⇒ a = g ′ cos θ

m ⎞ ⎛m V = −G ⎜ 1 + 2 ⎟ for r < r2 r2 ⎠ ⎝ r1

Velocity of Satellite

sin θ =

⇒

Which satisfies OPTION (C) (Since field inside shell 1 is zero, so potential at a point lying inside shell 1 is a constant and that equals the value of the potential at the surface of shell 1 i.e. Gm1 − ). r1 Similarly, for point P lying inside both the shells, we have Gm1 Gm2 − V=− r1 r2

7.

Net force towards centre of earth is mg′, where ⎛ x⎞ g′ = g ⎜ ⎟ ⎝ R⎠ mgx ⇒ mg ′ = R Normal force N = mg ′ sin θ

CHAPTER 4

4.

H.283

9.

a=

V=− ⇒ ⇒

GM R

GM R R2 V = − gR V=−

Hence, (C) and (D) are correct. 10.

E-r and V -r graphs for a spherical shell and a solid sphere are shown here.

r

r For a shell

r

r

For a solid sphere

Hence, (A), (B) and (C) are correct.

09-Feb-21 6:37:18 PM

H.284

11.

JEE Advanced Physics: Mechanics – II

GMm 2a So, E ∝ m E=−

⇒

where, μ = reduced mass =

E ∝ ms

4π 2 ⎛ rA + rP ⎞ T = ⎜ ⎟ GM ⎝ 2 ⎠ 2

3

{

r +r Qr = A P 2

}

π2 ⇒ T = ( rA + rP )3 2GM By Law of Conservation of Angular Momentum mvA rA = mvP rP vA rA = vP rP

Hence, (B), (C) and (D) are correct. 13. Time period in both the cases is R3 ≈ 84.6 min GM However, v1 > v2 , because the difference in potential energy between the extreme position and mean position will be more in the first case. Hence, (B) and (D) are correct. T1 = T2 = 2π

14.

dV dx If Eg = 0 , then V = constant and this constant may also be zero. Hence, (A) and (C) are correct. Eg = −

15. At two positions, when the planet is closest to the sun (perigee) and when it is farthest from the sun (apogee), velocity vector is perpendicular to force vector i.e., work done is zero. In one complete revolution work done is zero. Hence, (A) and (D) are correct. 16. The field inside the shell is zero and so potential inside the shell is constant equal to the value that exists at the GM . surface i.e. − a Hence, (A) and (D) are correct. 17. Hence, (B) and (D) are correct. 18. By Law of Conservation of Mechanical Energy, we get

( U + K )∞ = ( U + K )r ⇒

0+0= −

Gm ( 4 m ) 1 2 + μ vr r 2

Mechanics II_Chapter 4_Hints and Explanation.indd 284

( m )( 4 m ) m + 4m

=

4m and 5

Substituting in equation (1), we get ⇒

10Gm r

vr =

From equation (1), the total kinetic energy is

2

⇒

…(1)

vr = relative velocity of approach

1 a Hence, (A), (B) and (C) are correct. E∝

12.

G ( m )( 4 m ) 1 2 = μν r r 2

K=

G ( m )( 4 m ) r

4Gm2 r Net torque of two equal and opposite forces acting on two objects is zero. Therefore, angular momentum will remain conserved. Initially both the objects were stationary i.e., angular momentum about any point was zero. Hence, angular momentum of both the particles about any point will be zero at all instants. Hence, (A), (B), (C) and (D) are correct. ⇒

K=

19. When total force is zero, geff = 0 V=−

Gm r

For a shell of radius R, Vinside = Vsurface = − shell

GM R

I inside = 0 whereas I outside =

GM r2

So, plot of V vs r is continuous whereas plot of I vs r is discontinuous Hence, (A), (B) and (C) are correct. 20. Orbital radius of first satellite is 2R and that of second satellite is 8R. For a satellite K .E. =

GMm 2r

P.E. = −

GMm r

GMm 2r Hence, (A), (B) and (C) are correct. T .E. = −

21. Conceptual (A), (B) and (C) are correct.

09-Feb-21 6:37:23 PM

Hints and Explanations

23. According to Right Hand Thumb Rule curl the fingers of right hand in the direction of rotation then thumb gives the direction of the areal velocity/angular momentum. Hence, (A) and (C) are correct. 24. At centre, V = − and E = 0

29. Distance from centre of sun and hence the kinetic energy and potential energy keep changing. Hence, (C) and (D) are correct. 30.

⇒

( Work done ) = ( Change in G.P.E. )

3 GM 2 R

Hence, (A) and (D) are correct. Gm1m3 , attractive i.e. towards O . r2 Hence, (A), (B) and (D) are correct.

25. Force on P is

26.

GM v0 = r 2π r 3 2 T= GM

⇒

1⎞ ⎛ 1 W = U f − U i = −GMm ⎜ − ⎝ R + h R ⎟⎠

⇒

W=−

GMm ⎡ ⎛ h⎞ ⎢ ⎜⎝ 1 + ⎟⎠ R ⎣ R

⇒

W=−

GMm ⎡ h ⎤ 1 − − 1⎥ R ⎣⎢ R ⎦

⇒

W≈

⇒

W ≈ mgh for h

⇒

31.

r2

W = U f − Ui =

⇒

W=

T2 =

4π 2 ( R + h )3 GM

Since h Let CM of system be at a distance r1 from m and r2 from 2m . Then, mr1 = 2m ( r − r1 ) r 2r and r2 = r − r1 = 3 3

Since, T22 = ⇒

T22 =

4π 2 r23 Gm

R

GMm 2R

⇒

{

Q g=

GM R2

}

R

T 2 = MIN =

4π 2 R3 GM

2π R T v is MAXIMUM

Also, v = ⇒

GMm = MINIMUM 2R Hence, (A), (B) and (C) are correct. and total energy is E = −

32π 2 r 3 27Gm 3

GMmh R2

1 mgR 2 Hence, (A) and (D) are correct.

27.

r1 =

⎤ − 1⎥ ⎦

GMm , and R GMm GMm Uf (h = R) = − =− R+R 2R

−GMm 2r GMm K= 2r Hence, (A) and (D) are correct.

⇒

−1

Also, U i = −

U=

r1

GMm R GMm Uf = − R+h

Ui = −

CHAPTER 4

22. According to Kepler’s Laws Areal Speed is constant. Also, by Law of Conservation of Angular Momentum L must be constant. Hence, (B) and (C) are correct.

H.285

32. Since, t = nT −

1

⇒ T2 ∝ r 2 and T2 ∝ m 2 Hence, (A) and (D) are correct. 28. Conceptual (A), (B) and (D) are correct.

Mechanics II_Chapter 4_Hints and Explanation.indd 285

t 40 = n 20

⇒

T=

⇒

T=2s

Now, ΔT = nΔT

09-Feb-21 6:37:31 PM

H.286

⇒ So, ⇒

JEE Advanced Physics: Mechanics – II Distance of any mass from centre is

Δt ΔT = t T

a 3 So, radius of circular path followed is r=

1 ΔT = 40 2

a 3 a Mass is moving in circular path of radius r = 3 Such that gravitational force on the particle due to other two provides the necessary centripetal force.

ΔT = 0.05

r=

l Since, time period, T = 2π g ⇒

1 Δg ΔT =− 2 g T

⇒

−

Δg ΔT =2 g T

So, percentage error in g is

⇒

Δg × 100% g

0.05 ΔT × 100% = −2 × × 100% T 2 ⇒ % Error = 5% Hence, (A) and (C) are correct. % Error = −2

33. The force of attraction between any two point masses is responsible for providing the necessary centripetal force to a mass to revolve in a circle of radius r. Using trigonometry, we get cos 30 = r=

⇒

r=

⇒

mrω 2 =

Gm a

⎛ a ⎞ 2π ⎜ ⎝ 3 ⎟⎠ T= = v

2π a 3 3Gm

Total kinetic energy of the particles is 2 ⎛1 ⎞ 3 Gm K = 3 ⎜ mv 2 ⎟ = ⎝2 ⎠ 2 a

⎛ Gm2 ⎞ U = −3 ⎜ ⎝ a ⎟⎠ ⇒ Total energy E is

l 3

3 ⎛ Gm2 ⎞ E=U+K = − ⎜ ⎟ 2⎝ a ⎠

Gmm l2 l 2 Gm2 m ω = 2 3 l

3 ⎛ Gm2 ⎞ ⎜ ⎟ 2⎝ a ⎠ Hence, (B) and (C) are correct. ⇒ Binding Energy =

3Gm l3

⇒

ω=

⇒

T = 2π

⇒

T ∝ l 2 and

l3 3Gm

1 2

T ∝m Hence, (A) and (D) are correct. 34.

Mechanics II_Chapter 4_Hints and Explanation.indd 286

GMm 2r GMm Potential energy, PE = − r GMm and the total energy, E = − 2r 1 Kinetic energy is always positive and KE ∝ r 1 Potential energy is negative and PE ∝ r

35. Kinetic energy, KE =

3 −

⇒

v=

3Gm2 a

Potential energy U of the system is

l 2 cos 30

⇒

⇒

l2 l = r 2r

⇒

mv 2 = ⎛ a ⎞ ⎜⎝ ⎟ 3⎠

{OPTION (A)} {OPTION (D)}

1 r Also E < PE , so from the graph we observe that A is kinetic energy, B is potential energy and C is total energy of the satellite. Hence, (A), (B) and (D) are correct. Similarly, total energy is also negative and E ∝

09-Feb-21 6:37:38 PM

Hints and Explanations

GM 3GM GM GM and Vcentre = − + =− R 2R R 2R Hence, (C) and (D) are correct.

GM , so R

2Gm = 2 gR R Hence, the correct answer is (A). ⇒

V∞ =

4.

Although no gravitational field is produced inside a symmetric shell, it produces a field at points outside of shell. Hence, the correct answer is (D).

5.

Work will be done only in bringing the unit mass from infinity upto the surface of shell because inside shell there is no gravitational field and in moving inside the shell no work will be done. Hence, the correct answer is (D).

37. Conceptual (A), (B), (C) and (D) are correct. 38. Both the stars will revolve about their centre of mass. So, if the centre of mass be at a distance x from 2m , then x= So, r1 =

2m ( 0 ) + mr r = 3m 3 r 2r and r2 = 3 3

ω and T will be same for both the stars, so K1 =

6.

1 1 I1ω 2 and K 2 = I 2ω 2 2 2

⎛ 2r ⎞ m⎜ ⎟ ⎝ 3 ⎠ K1 I1 = = =2 2 K2 I2 ⎛ r⎞ 2m ⎜ ⎟ ⎝ 3⎠

dW d d d 2 gh = ( mΔg ) = m dh ( Δg ) = m dh ⎛⎜⎝ R ⎞⎟⎠ dh dh dW 2mg = = constant ⇒ dh R Hence, the correct answer is (B).

L1 I1 = =2 L2 I 2

7.

Hence, (A), (B) and (C) are correct.

V is not same everywhere as indicated by Vin . Hence, the correct answer is (C).

dA 1 2 dθ 1 2 = r = r ω dt 2 dt 2

8.

2.

3.

r r dL The torque on a body is given by τ = . In case of dt planet orbiting around Sun no torque is acting on it. So, r dL =0 dt r ⇒ L = constant Hence, the correct answer is (A). Using energy conservation

Mechanics II_Chapter 4_Hints and Explanation.indd 287

GMm 1 mve2 − =0 2 R

The time period of satellite which is very near to earth R = 84 min = 1 hr 24 min g Hence, the correct answer is (A).

2

dA mr ω L = = = constant 2m 2m dt Hence, the correct answer is (A). ⇒

GM ( 2 3R − r 2 ) 2R 3

GM R 3 GM At centre, r = 0, so Vcentre = 2 R 3 ⇒ Vcentre = VS 2 ⇒ Vin > VS

r r dA L = dt 2m r Since L = constant r dA = constant ⇒ dt Also,

Vin =

At surface, r = R, so Vs =

Reasoning Based Questions 1.

⎛ 2h ⎞ Δg = g − g ′ = g ⎜ ⎝ R ⎟⎠

⇒

L1 = I1ω and L2 = I 2ω ⇒

2h ⎞ ⎛ g′ = g ⎜ 1 − ⎟ ⎝ R⎠ ⇒

2

⇒

ve =

CHAPTER 4

36. At all the places potential will increase by

H.287

is given by T = 2π

9.

Work done in conservative field in cyclic process is zero. Hence, the correct answer is (A). GM 10. Acceleration due to gravity is given by g = 2 , so it r does not depend on mass of body on which it is acting. Also, it is not a constant quantity because it changes with the change in value of both M and r (distance between two bodies). Hence, the correct answer is (C).

09-Feb-21 6:37:44 PM

H.288

JEE Advanced Physics: Mechanics – II

Linked Comprehension Type Questions 1.

2 gR 5 Hence, the correct answer is (D). ⇒

According to Kepler’s Law, we have T 2 ∝ r3 ⇒

Tm2 rm3 = Te2 re3

⇒

⎛r ⎞ Tm = ⎜ m ⎟ ⎝ re ⎠

5.

Since Te = 1 yr ⇒ ⇒

⎛ 6 × 1010 ⎞ Tm = ⎜ ⎝ 1.5 × 1011 ⎟⎠ ⎛ 6 × 1010 ⎞ Tm = ⎜ ⎝ 15 × 1010 ⎟⎠

32

( 1 yr )

2.

⇒

( v o )m = ( v o )e

⇒ ⇒

( v o )m = ( v o )e ( v o )m = ( v o )e

6.

GMsun re

7.

re rm 1.5 × 1011 15 = 10 6 6 × 10

8.

5 2

ΔU = U f − U i =

1 GMm 5 R

mgR ⇒ ΔU = 5 Hence, the correct answer is (C). 4.

By Law of Conservation of Energy ΔU =

2π ( 10 4 ) kmh −1 1

v2 − v1 r2 − r1 Hence, the correct answer is (B).

ω=

⎧ GMm ⎪ F = ⎨ r2 ⎪⎩ 0

, r≥R , r R , g = 2 d Hence, the correct answer is (D).

Vsphere at P = −

For d = R , g s =

Now, Vcavity at P

38. Here, the weight of person on the equator is W. If the 3W earth rotates about its axis, then weight is 4 Radius of the earth = 6400 km The acceleration due to gravity at the equator is g e = g − Rω 2 ⇒ ⇒ ⇒

3 g = g − Rω 2 4 g Rω 2 = 4

ω=

g = 4R

2 GM ⎛ 2 ⎛ R ⎞ ⎞ 11GM R − 3 ⎜ ⎟ ⎜ ⎟⎠ = − 3⎝ ⎝ ⎠ 2 8R 2R

⎛ M⎞ G 3 ⎜⎝ 8 ⎟⎠ 3GM =− =− 2 ⎛ R⎞ 8R ⎜⎝ ⎟⎠ 2

GM 11GM ⎛ 3GM ⎞ −⎜− =− ⎝ 8 R ⎟⎠ R 8R Hence, the correct answer is (B). ⇒

VP = −

44. Centripetal force is provided by the gravitational force 1 1 which is proportional to . So, F ∝ r r mv02 1 ⇒ ∝ r r

10 = 0.625 × 10 −3 rads −1 4 × 6400 × 10 3

⇒ ω ≈ 0.63 × 10 −3 rads −1 Hence, the correct answer is (A). 39. The correct answer is (A). 40.

v0 =

⇒

GM = gR R

Since T =

2GM = 2 gR R So, increase in velocity is

45. Potential V ( r ) due to a large planet of radius R is given by

Δv = gR ( 2 − 1 ) Hence, the correct answer is (A). 41. Let the area of the ellipse be A . According to Kepler’s Second Law, areal velocity of a planet around the sun dA is constant, i.e., = constant , so we have dt A A 3A t1 Area of abcsa 2 + 4 = = = 4 =3 A t2 Area of adcsa A − A 2 4 4 ⇒ t1 = 3t2

GM for r > R r −GM V(r) = for r = R R V0 ( r ) = −

r2 ⎞ 3 GM ⎛ ⎜⎝ 1 − ⎟ for r < R 2 R 3R2 ⎠ Hence, the correct answer is (B). Vin = −

46.

F = Fgravitational force on M =

Hence, the correct answer is (C). Net force on the astronaut is zero. Hence, the correct answer is (C). 43.

VP = Vsphere at P − Vcavity at P GM Since, V = − 3 ( 3 R2 − r 2 ) 2R

Mechanics II_Chapter 4_Hints and Explanation.indd 304

2π r v

⇒ T ∝r Hence, the correct answer is (C).

ve =

42. Gravitational pull on the astronaut FG =

v0 = constant

GMm

Mv 2 R

⇒

⎛ GM 2 1 ⎞ GM 2 Mv 2 + F = 2⎜ = ⎟ 2 2 R ⎝ ( 2R ) 2 ⎠ ( 2R )

⇒

GM 2 GM 2 Mv 2 + = R 2R 2 4 R 2

( R + h )2

1 GM (1 + 2 2 ) 2 R Hence, the correct answer is (D). ⇒

v=

09-Feb-21 6:41:43 PM

Hints and Explanations

⇒ Required energy is 5GMm 6R Hence, the correct answer is (A). ⇒

ΔE = E f − Ei =

48. Energy required to make the spaceship reach the free space is GMm R GM Since g = 2 R ΔE =

⇒

Gm G ( 4 m ) = ( r − x )2 x2

⇒

1 ⎛ x ⎞ ⎜⎝ ( ⎟ = r − x)⎠ 4

2

r …(1) 3 r Gravitational potential at a point P i.e. at x = is 3 Gm G ( 4 m ) Gm G ( 4 m ) − V=− =− − ( ) r⎞ x r−x ⎛ r⎞ ⎛ ⎜⎝ ⎟⎠ ⎜⎝ r − ⎟⎠ 3 3 ⇒

x=

3Gm 3G ( 4 m ) Gm − = −9 r 2r r Hence, the correct answer is (D). ⇒

m R

⇒

ΔE = gR2 ×

⇒

ΔE = mgR

⇒

ΔE = 1000 × 10 × 6400 × 10 3 = 64 × 109 J

⇒

ΔE = 6.4 × 1010 J

V=−

50. The acceleration due to gravity at a height h from the g ground is given as . 9

CHAPTER 4

GMm R GMm GMm In orbit, E f = − =− 2 ( 3R ) 6R

47. At surface, Ei = −

H.305

Hence, the correct answer is (C). 49. Let x be the distance of the point P from the mass m where gravitational field is zero.

GM ⎛ GM ⎞ 1 =⎜ 2 ⎟ ⎝ R ⎠9 r2 ⇒

r = 3R

The height above the ground is 2R Hence, the correct answer is (A).

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems

⇒

mv 2 = 2K

1.

⇒

M=

For a particle revolving in a circular orbit of radius r due to the gravitational attraction of inner cloud of mass M , we have GMm mv 2 = r r2 ⇒

M=

v 2 r 2mv 2 r = G 2Gm

1 Since K = mv 2 = constant 2

2Kr Gm

2Kdr Gm Also, we know that ⇒

dM =

…(1)

dM = ρ ( r ) dV ⇒

dM = ρ ( r ) 4π r 2 dr

So, equation (1) becomes,

ρ ( r ) 4π r 2 dr = ⇒

2Kdr Gm

K ρ(r ) = n( r ) = m 2π Gm2 r 2

Hence, the correct answer is (B).

Mechanics II_Chapter 4_Hints and Explanation.indd 305

09-Feb-21 6:41:59 PM

H.306

2.

JEE Advanced Physics: Mechanics – II

Given that ve = 11.2 kms −1 =

R

λ gp ⎛ x2 ⎞ ⇒ F= ⎜ ⎟ R ⎝ 2 ⎠ 4 R/ 5 Substituting the given values, we get F = 108 N Hence, the correct answer is (B).

2GMe Re

By Law of Conservation of Energy, we have Ki + Ui = K f + U f ⇒

1 GMs m GMe m mvs2 − − =0+0 2 r Re

4.

In circular orbit of a satellite, potential energy U is 1 U = −2 × ( kinetic energy ) = −2 × mv 2 = − mv 2 2 Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be +mv 2 . Hence, the correct answer is (B).

5.

For annular disc, gravitational potential at the point P lying on the axis at a distance x from centre is 2GM VP = − 2 R22 + x 2 − R12 + x 2 R2 − R12

where, r is the distance of rocket from sun ⇒

2GMe 2GMs + Re r

vs =

Since, Ms = 3 × 10 5 Me and r = 2.5 × 10 4 Re ⇒

vs =

2GMe 2G 3 × 10 5 Me + Re 2.5 × 10 4 Re

⇒

vs =

2GMe ⎛ 3 × 10 5 ⎞ ⎜⎝ 1 + ⎟ Re 2.5 × 10 4 ⎠

vs =

2GMe × 13 Re

⇒

(

GMplanet 2 Rplanet

=

GMe ( 10 )

( 10 )

3

Re2

2

=

2GM (4 2 − 5) 7R Hence, the correct answer is (A). ⇒

6.

R

∫

4R 5

( λ dx ) g p ⎛⎜

x⎞ ⎝ R ⎟⎠

Mechanics II_Chapter 4_Hints and Explanation.indd 306

WP→∞ =

mv 2 GmM = r r2 ⎛4 ⎞ Here, M = ⎜ π r 3 ⎟ ρ0 ⎝3 ⎠ For r ≤ R ,

…(1)

Substituting in Equation (1), we get v ∝ r i.e. v -r graph is a straight line passing through origin. For r > R, we have ⎛4 ⎞ Gm ⎜ π R3 ⎟ ρ0 ⎝3 ⎠ mv 2 = 2 r r 1 ⇒ v∝ r The corresponding v -r graph will be as shown in option (C). Hence, the correct answer is (C).

gp =

F=

VP = −

Since WP →∞ = m0 ( V∞ − VP ) , where m0 = 1 unit

GMe 10 Re2

ge 10 The value of g inside the planet at a distance x from centre of the planet is ⎛ x⎞ ⎛ x⎞ ginside = gsurface of planet ⎜ ⎟ = g p ⎜ ⎟ ⎝ R⎠ ⎝ R⎠ So, total force acting on wire is ⇒

)

2GM ( 4 2R − 5R ) 7 R2 Also, V∞ = 0 ⇒

R Given, Rplanet = R = earth 10 Mearth Since, density ρ = 4 3 π Rearth 3 Mplanet Also, ρ = 4 3 π Rplanet 3 M Me = ⇒ Mplanet = earth 1000 10 3 Let the acceleration due to gravity at surface of planet and at the surface of earth be g p and g e respectively. Then gp =

(

where R2 = 4 R, R1 = 3 R and x = 4 R

⇒ vs 42 kms −1 Hence, the correct answer is (C). 3.

)

7.

In case of binary star system, angular velocity and hence, the time period of both the stars are equal. Hence, the correct answer is (D).

8.

Time period of a satellite very close to earth’s surface is 84.6 min . Time period increases as the distance of the satellite from the surface of earth increases. So, time period of spy satellite orbiting a few 100 km above

09-Feb-21 6:42:18 PM

Hints and Explanations the earth’s surface should be slightly greater than 84.6 min . Therefore, the most appropriate option is (C) or 2 h Hence, the correct answer is (C).

T2 = 2π Now g h =

⇒

l g

Since T1 = 2π

⇒ ⇒

T2 = T1

ΔU = −

GMm GMm + 2R R

{Q h = R }

ΔU =

GM R2 1 g∝ 2 R Δg ΔR = −2 g R

14. Since, g =

2

( R + h )2

g gh = 4

GMm ⎛ GMm ⎞ −⎜− ⎟ R+h ⎝ R ⎠

mgR GMm 1 ⎛ GM ⎞ = m⎜ 2 ⎟ R = ⎝ ⎠ 2 2 2R R Hence, the correct answer is (A). ⇒

l gh gR

ΔU = −

⇒

{Q h = R }

⇒

g =2 g 4

So, g will increase, if R decreases

Hence, the correct answer is (D). 10. Force on satellite is always towards earth, therefore, acceleration of satellite S is always directed towards centre of the earth due to which net torque of this gravitational force F about centre of earth is zero. Therefore, angular momentum (both in magnitude and direction) of S about centre of earth is constant throughout. Since, the force F is conservative in nature, therefore mechanical energy of satellite remains constant. Speed of S is maximum when it is nearest to earth and minimum when it is farthest. Hence, the correct answer is (A). 11. From Kepler’s Third Law, we have T 2 ∝ r3 32

⇒

T ∝ (r)

⇒

T2 ⎛ r1 ⎞ = T1 ⎜⎝ r2 ⎟⎠

⇒

⎛r ⎞ T2 = T1 ⎜ 2 ⎟ ⎝ r1 ⎠

⇒

T2 ≈ 129 days

32

32

CHAPTER 4

9.

13.

H.307

⎛ 1⎞ = ( 365 ) ⎜ ⎟ ⎝ 2⎠

32

⇒

Δg = −2 ( −1% ) = 2% g

Hence, the correct answer is (C).

Multiple Correct Choice Type Problems 1.

Gravitational field at a distance r due to mass ⎛ 4 ⎞ m ⎜ = π r 3 ρ ⎟ is ⎝ 3 ⎠ ⎛4 ⎞ Gρ ⎜ π r 3 ⎟ ⎝3 ⎠ 4Gρπ r E= = 2 3 r Consider a small element of width dr and area ΔA at a distance r from the centre. Pressure force on this element is due to the gravitational force on dm from m inwards towards the centre. ⇒ ( dP ) ΔA = E ( dm ) ⎛4 ⎞ where dm = ( ΔA ) ( dr ) ρ and m = ⎜ π r 3 ⎟ ρ ⎝3 ⎠ ⇒

⎛4 ⎞ − dPΔA = ⎜ Gπρr ⎟ ( ρΔAdr ) ⎝3 ⎠

⇒

⎛ 4Gρ 2π ⎞ − dP = ⎜ rdr ⎝ 3 ⎟⎠

Hence, the correct answer is (B). 12.

F ∝ R −5 2 This gravitational force of attraction provides necessary centripetal force to planet to revolve about the massive star. ⇒

mRω 2 ∝ R −5 2

⇒

ω 2 ∝ R −7 2

⇒

4π 2 ∝ R −7 2 T2

⇒ T 2 ∝ R7 2 Hence, the correct answer is (B).

Mechanics II_Chapter 4_Hints and Explanation.indd 307

P

∫ 0

⇒

−P =

r

∫ R

4Gρ 2π ( 2 r − R2 ) 3×2

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H.308

JEE Advanced Physics: Mechanics – II 2Gρ 2π ( 2 R − r2 ) 3

⇒

P=

⇒

P = k ( R2 − r 2 ) , where

Mass of R is MR = MP + MQ 2Gρ 2π k= = constant 3

⎛ ⎛ 7 R2 ⎞ 3R 9R 2 ⎞ , P1 = k ⎜ R2 − For, r = ⎟ = k ⎜⎝ ⎟ ⎝ ⎠ 4 16 16 ⎠ ⎛ ⎛ 5R 2 ⎞ 2R 4R2 ⎞ , P2 = k ⎜ R2 − For, r = = k⎜ ⎟ ⎝ ⎝ 9 ⎟⎠ 3 9 ⎠ ⇒

P1 63 = P2 80

⇒

4.

RR3 = RP3 + RQ3

⇒

RR3 = 9RP3

⇒

RR = 91 3 RP

⇒

RR > RQ > RP

⇒

VR > VQ > VP 1 V VR = 91 3 and P = VQ 2 VP

Gravitational field is the acceleration due to gravity. ⎡ GM 2 ⎢ So, g = ⎢ r ⎢ 4 π Gρ r ⎢⎣ 3

⎛ ⎛ 3R2 ⎞ R R2 ⎞ For r = , P5 = k ⎜ R2 − = k⎜ ⎟ ⎝ ⎝ 4 ⎟⎠ 2 4 ⎠

⇒

⇒

Hence, (B) and (D) are correct.

P3 16 = P4 21

For r =

⎛4 3⎞ ⎛ 4 3⎞ ⎛4 3⎞ ⎜⎝ π RR ⎟⎠ ρ = ⎜⎝ π RP ⎟⎠ ρ + ⎜⎝ π RQ ⎟⎠ ρ 3 3 3

Also,

⎛ 16 R2 ⎞ 3R 9 ⎛ ⎞ , P3 = k ⎜ R2 − R2 ⎟ = k ⎜ For, r = ⎝ ⎝ 25 ⎟⎠ 5 25 ⎠ ⎛ ⎛ 21R2 ⎞ 2R 4R2 ⎞ , P4 = k ⎜ R2 − For, r = ⎟ = k ⎜⎝ ⎟ ⎝ ⎠ 5 25 25 ⎠

⇒

⇒

⎛ ⎛ 8R2 ⎞ R R2 ⎞ , P6 = k ⎜ R2 − = k⎜ ⎟ ⎝ ⎝ 9 ⎟⎠ 9 ⎠ 3

r≥R

( Outside )

r R and r2 > R

P5 27 = P6 32

and

Hence, (B) and (C) are correct.

F1 r1 ( Inside ) = F2 r2

where r1 < R and r2 < R Hence, (A) and (B) are correct.

2. 5. Let v is the minimum velocity, then by Law of Conservation of Energy, we have

The spherical cavities can be assumed to be negative masses placed symmetrically about origin O. So gravitational force due to this object at origin is zero.

( U + K )C = ( U + K )∞ ⇒

1 ⎛ GMm ⎞ 2 ⎜⎝ − ⎟ 2 + mv = 0 + 0 L ⎠ 2

GM L Hence, (B) and (D) are correct. ⇒

3.

v=2

ves ∝ R Surface area of P is A = 4π RP2

The dotted circle is lying in the yz plane and is an equipotential surface (as we can see that two cavities are lying symmetrically on each side of the EPS (Equi-Potential Surface) so gravitational potential is same at all the points lying on the circle y 2 + z 2 = a 2 where a 2 = constant .

Surface area of Q is 4 A = 4π RQ2

Hence, (A), (C) and (D) are correct.

ves =

⇒

2GM = R

⎛4 ⎞ 2G ⎜ π R3 ⎟ ρ ⎝3 ⎠ = R

4Gρ R 3

(

)

RQ = 2RP

Mechanics II_Chapter 4_Hints and Explanation.indd 308

09-Feb-21 6:42:52 PM

Hints and Explanations

Reasoning Based Questions

⇒

v e = 2v

1.

⇒

N=2

2.

GMm

( 3 l )2

Integer/Numerical Answer Type Questions 1.

At height h , we have g gh = 2 h⎞ ⎛ ⎜⎝ 1 + ⎟⎠ R

For point mass at distance r = 3l , we have

⇒

Gm2 = ma l2

Gm2 = ma ( 4l ) l2 Equating these two equations, we get GMm 2

…(1)

+

GMm Gm2 GMm Gm2 − 2 = + 2 9l 2 l 16l 2 l

2

⇒

h=R

Applying Law of Conservation of Energy, from A to B, we get

( U + K )A = ( U + K )B GMm 1 GMm + mv 2 = − +0 R 2 R+h Since h = R , so we have

⇒ 3.

⇒

gp ge

=

GM p Rp2

=

Re ρe

Also, ve = 2 gR

⇒

v 2 GM = 2 2R

⇒

⇒

v=

GM R

⇒

3 vP = ve 11

⇒

vp = 3 kms −1

Mechanics II_Chapter 4_Hints and Explanation.indd 309

vp ve

=

4 Gπ Rp ρ p 3

Rρ ρ p

1 GMm GMm mv 2 = − 2 R 2R

2GM R

…(2)

7GMm 2Gm2 = 144 l2 7M m= 288

Since, g p =

−

Since ve =

…(1)

For point mass at distance r = 4l , we have

g Given that g h = 4 Substituting in equation (1) we get, 1 ⎛ R ⎞ =⎜ ⎟ 4 ⎝ R+ h⎠

−

CHAPTER 4

Force acting on astronaut is utilised in providing necessary centripetal force, thus he feels weightlessness, as he is in a state of free fall. Hence, the correct answer is (A).

H.309

⎛ g p ⎞ ρe 6 3 =⎜ = × g e Re ⎝ g e ⎟⎠ ρ p 11 2

g p Rp

09-Feb-21 6:43:02 PM

Mechanics II_Chapter 4_Hints and Explanation.indd 310

09-Feb-21 6:43:02 PM