An Introduction to Planar Dynamics [4 ed.] 9789814601610

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LIST OF SYMBOLS 1.

Kinematics

0-xyz

Fixed reference system with origin 0

with frame ( i, }, k)

A-x'y'z'

Point A -attached translating reference

A-x"y"z"

Body-attached moving reference system with reference point A on the body f

with frame ( i, }, k) with rotating frame

orf

e e1 , e11

n

j,

s r,

e

and a P, Q, G, .. . OJ

P'

( f"' ]"' k")

Unit vector Tangent and normal unit vectors for path coordinates Tangent and normal unit vectors Radial and transverse unit vectors in polar coordinates Path coordinate

with local frame

Polar coordinates

with local frame ( e, ' ee )

cel' e") Angular velocity and angular acceleration Particles or geometric points Entraining point , which is the point on the moving frame, and coincides with particle P at the instant Position vectors Position vector of point A relative to point B Velocity vector (of point A , B)

w=B ' a=iJ

G for mass center Vp• , Q p • are entrained velocity, acceleration

-

r AB

=

scalar components are , Vy , Vz .

Relative velocity of A observed from the translating frame attached at B (velocity of A relative to B) Acceleration vector (of point A, B)

d

Relative acceleration of A observed from the translating frame attached at B (accel. of A relative to B) Relative velocity of P observed from the moving frame attached at a body /(velocity of P relative to body./) Relative accel. of P observed from the moving frame attached at a body f(accel. of P relative to body./) Corio lis acceleration of particles P based on a rotating reference frame Curvature and radius of curvature Absolute time differentiation

dt 0

Relative time differentiation based on moving framef

V P/f

aPir

K ,

ot

VI

p

-

- rBA

-

-

V Al B = -V B/ A

scalar components are ax, ay, az .. . QAI B

= -Q IJIA

Vx

List of Symbols

2. Kinetics Mass Moment of inertia of a slab about mass center G Moment of inertia of a slab about point C Radius of gyration of a rigid slab about its mass center, or point A

L F M G, M c Ma , Mc fiG,

He

T

Kinetic energy Work done by force from position 1 to 2.

u(con ) 1->2

Work done by conservative forces

u2

Work done by non-conservative forces Potentials of conservative forces (gravitational potential, elastic potential) stiffness of a spring, elastic constant of a spring Total energy Linear impulse of force from t 1 to t 2

- I

mkh,

IA=

Signed magnitudes of M e, M 0 Angular momentum about G or point C Signed magnitudes of fiG, fi c

MG e

=

Linear momentum Force vector Torques, Moment of forces about point G or C

H(j, He

t'

lc

E=T+ V

Angular impulse of moments about a point from t 1 to !2 Inertial force acting at G Inertial moment (torque) about G Restitution coefficient Static friction coefficient Kinetic friction coefficient

VII

Contents CHAPTERl

Kinematics Of Particles

1.1

Reference Systems for Motion

1

1.2

Position, Velocity, and Acceleration 1.2.1 Displacement Vector, Relative Position Vector 1.2.2 Velocity- Time Derivative of Position Vectors 1.2.3 Acceleration - Time Derivative of Velocity Vectors

4

11

1.3

Simple Plane Motion of a Particle 1.3 .1 Rectilinear Motion 1.3.2 Circular Motion

13 13 19

1.4

Planar Curvilinear Motion of Particles 1.4.1 Curvilinear Motion Described in the Cartesian Coordinate System 1.4.2 Curvilinear Motion Described in the Path Coordinate System

28 28 33

1.4.3

Curvilinear Motion Described in the Polar Coordinate System

4

9

38

1.5

Motion Relative to a Point-Attached Translating Frame

47

1.6

Constrained Motion of Connected Particles

58

1.7

Summary

65

1.8

Problem-Solving Guidelines

68

Problems

71

CHAPTER2

KINEMATICS OF RIGID BODIES

2.1

Introduction to Planar Kinematics of Rigid Bodies

79

2.2

Simple Planar Motion of a Rigid Body 2.2.1 Translation 2.2 .2 Rotation

80

General Planar Motion of a Rigid Body 2.3 .1 Combination of Translation and Rotation 2.3.2 Instantaneous Center of Velocity

84 84

Motion Relative to Body-Attached Rotating Reference 2.4.1 Principle of Angular Velocity/ Acceleration Combination

96

2.3

2.4

2.4.2 2.4 .3

Principle ,o f Velocity Combination Principle of Acceleration Combination

80 80

92 96

98 105

2.5

Summary

115

2.6

Problem-Solving Guidelines

117

Problems

119

ix

Contents

CHAPTER3 3.1

Newton's Second Law and Inertial Reference 3 .1.1 Principle of Linear Momentum 3 .1.2 Application of Newton's Second Law for Particles

127 127 128

3.2

Principle of Angular Momentum 3 .2.1 Application of the Principle of Angular Momentum to a Particle in Circular Motion 3.2.2 Application of the Principle of Angular Momentum to a Particle in Planar Curvilinear Motion

137

3.3

Principle of Work and Kinetic Energy of a Particle

143

3.4

Principle of Linear/Angular Impulse and Momentum 3 .4.1 Principle of Linear Impulse and Momentum 3.4.2 Principle of Angular Impulse and Momentum

147 148 150

3.5

Conservative Forces and Conservation of Total Energy 3.5 .1 Potential of Conservative Force Field 3.5.2 Gravitational Force 3.5.3 Linear Spring Elastic Force 3.5.4 Conservative System of Constrained Particles with Ideal Constraints

153 153 153 155 157

3.6

Conservation of Linear/Angular Momentum 3 .6.1 Conservation of Linear Momentum of a Particle 3.6.2 Conservation of Linear Momentum of a System ofParticles 3.6.3 Method of Conservation of Angular Momentum ofMotion of a Particle Under a Central Force

161

Collision 3.7.1 Direct Central Collision 3.7.2 Oblique Central Collision 3.7.3 Collision with Constraints

166

3.8

D' Alembert Principle- A New Look at Newton's Second Law

175

3.9

Summary

179

A b

c

Kinetics Of Particles

1

F c:

c

j

r 3.7

139 141

160 162 164 166 169 172

3.10 Problem-Solving Guidelines

184

Problems

196

CHAPTER4 4.1 4.2

4.3

X

KINETICS OF RIGID BODIES

Principle of Linear Momentum about the Mass Center -The Governing Equation for Translation

205

Principle of Angular Momentum about the Mass Center -The Governing Equation for Rotation 4.2.1 The Governing Equation_ of Relative Rotation about the Mass Center 4.2.2 Principle of Angular Momentum about the Mass Center 4.2.3 Moment of Inertia of a Rigid Slab

209 209 212 215

Application of Newton's Second Law to General Planar Motion of a Rigid Slab

218

Contents

4.4

Principle of Angular Momentum of a Rigid Slab about a Fixed Point

226

4.5

Principle of Work and Kinetic Energy for a Rigid Body 4.5 .1 Kinetic Energy of Rigid Bodies 4.5.2 Principle of Work and Kinetic Energy for a Rigid Body or a System of Rigid Bodies 4.5.3 Conversation ofTotal Energy for Systems of Rigid Bodies with Ideal Constraints

228 228

4.6

Principle of Impulse and Momentum for a Rigid Body

238

4. 7

Application of D' Alembert Principle to General Planar Motion of a Rigid Slab

246

4.8

Summary

249

4.9

Problem-Solving Guidelines

255

Problems

231 235

268

APPENDICES A B

c D E F G

H I

Index

Vector Preliminaries Physical Explanation of Corio lis Acceleration -The Coupling of Relative and Entrained Motion Definition of Various Moments in Planar Dynamics Proof of Equations 4.28, 4.29 and 4.49 Principle of Angular Momentum for a Rigid Body about a Moving Point C Work of a Couple Motion of a Particle in a Non-Inertial Frame -Entrained Inertia Force and Coriolis Inertia Force Equation Maps of Kinematics and Kinetics Moment of Inertia of Some Uniform Slabs

283 285 288 291 293 295 297 299 303

304

XI

CHAPTER 1 KINEMATICS OF PARTICLES

Kinematics of particles studies the geometric properties of the motion of points: position, velocity, and acceleration. A point can represent a particle, which has mass but negligible size and shape, or a geometric point in a rigid body, in which the distance between any two points is always constant.

z

1.1 Reference Systems for Motion All motion is relative. Without a reference system, it is impossible to say whether a point is moving or stationary, or how a rigid body is oriented in the space. A spatial reference system as shown in Fig. 1-la includes:

(i)

an origin 0

(ii) a fram e, i.e. a set of mutually orthogonal unit vectors

(i, }, k ), which follows the right hand rule

X

(a) Basic fixed reference system

(iii) Cartesian (rectangular) coordinates (x, y, z) The reference system can be represented by short, 0-xyz with the default frame (f, },

O-i}k lxyz,

or for

k ). z

A reference system attached to the earth 's surface is considered to be a basic fixed reference system. Any motion observed/measured based on this fixed reference system is called absolute motion.

z

Any motion observed/measured relative to a well-defined moving reference system other than the basic fixed one is called relative motion. There are two types of moving reference systems.

(1) Point-attached Translating Reference System A translating system attached to a particle or a point of a body as shown in Fig. 1-1 b is called a point-attached translating reference system. This system moves together with the particle/point A, which is its origin, keeping the orientation of its frame the same as the basic fixed frame ( i, reference system can be represented by A-x 'y 'z' with the default frame (

i, }, k ).

}, k ). This translating

A-i}k /x 'y 'z ',

---- y j X

(b) Point-attached translating reference system

or for short,

Fig. 1-1

Chapter 1

Kinematics of Particles

z

(2) Body-attached Moving Reference System

z __ y''

A moving system attached to a rigid body with reference point A of the body chosen as its origin is called a body-attached moving reference system . This system can be in translation and/or rotation according to the motion of the rigid body. As shown in Fig. 1-lc, the movmg reference system can be represented by A-i" }" k" /x"y"z" , or for short, A-x"y"z" with the default frame

J

([" ,}" .k" ). In general, the motion of a point can be observed/measured in either a fixed reference system or a moving reference system. The absolute and relative motion of the point is different in path, velocity, and acceleration. The principles governing the relationship of absolute and relative motion will be the main topic in Chapters 1 and2.

c l

X

(c) Body-attached moving reference system

F

Fig. 1-1

Example 1-1 Relative motion observed from moving frames

J t

c

A truck with a boom is moving to the right, while the boom AB is rotating about the pin A and the length BC is being decreased. (i) If a basic fixed reference 0-xy is defined as shown, what is the motion path of point A? (ii) If a point-attached reference frames A-x y ' is attached at A, what is the motion path of point B observed from the translating frame? (iii) If a body-attached reference frame A-x 'y " is attached at boom AB, what is the motion of point C observed from A-x 'y "?

x"

Point A-attached

y"

Basic fixed reference frame

2

c

Body AB-attached reference frame

1.1 Reference Systems for Motion

Solution: (i)

The absolute motion of point A observed from the basic fixed reference system 0-xy is a horizontal straight-line motion. However, the absolute motion paths of B and C cannot be easily determined as they are general curvilinear lines when observed from the fixed frame.

(ii) If we observe the motion of B from a point A-attached translating reference system A-x'y' the relative "motion of point B is observed to be simply a circular path with radius AB. In short, the motion of point B relative to point A is a circular path. It should be emphasized that the meaning of relative to point A is indeed relative to a point-A attached translating reference system. As the observer is moving together with the reference system, point A appears stationary and the boom AB is rotating about the fixed point A.

(iii) If we observe the relative motion of point C from the point A-attached translating reference system A-x 'y ', the path is not circlar but curvilinear as the distance CB is being decreased while the boom AB rotates. However, if we attach a reference system A-x "y " on body AB, the relative motion path of point C observed from this is simply a straight line. In short, the motion of point C relative to body AB is a straight line. As the observer is rotating together with the boom AB, the boom appears stationary. That is why to the observer the point Cis moving in a straight line.

Discussion: In this example, different reference frames are defined to view the motion of different points such that their relative motion observed from the chosen moving frame becomes very simple. That is why we use different moving frames. How do we define a moving frame? A 3D frame is a frame space on which a coordinate system can be defined and used to measure the position of any point in that space. Any rigid body can be chosen as the core of a frame space. For example, planet Earth is modeled as a rigid body in this universe. Using earth as the core, a 3D frame space is spanned. It is an infinite space that includes the finite domain of the earth as its core and the infinite extension of it. The frame moves together with the rigid core, the earth. The motion of the stars we observe from earth are based on this frame . Let us return from the universe to earth. We define this earth-attached frame as our basic fixed reference frame . Any other rigid body moving relative to the earth can serve as the core for defining a new moving frame. In Example 1-1, the rigid boom is treated as the core on which a body-attached frame A-x ''y" is defined and used to observe the motion of point C. Why do we introduce a point-attached translating frame and what is the difference between this and a bodyattached frame? A point or any object modeled as a particle is sizeless. A sizeless particle cannot serve as the core of a frame because the orientation of the frame is "out of control." To control the orientation of the frame attached at a particle, we constrain it such that its orientation remains unchanged. For simplicity we will define all point-attached translating frames that have the same orientation as the basic reference frame . Note that in some cases, a point-attached translating frame may coincide with a body-attached frame if the body is in translation. For example, the frame A-x 'y' can also be treated as a truck-attached frame because the truck is in translation.

3

Chapter 1

Kinematics of Particles

1.2 Position, Velocity, and Acceleration 1.2.1 Displacement Vector, Relative Position Vector . As shown in Fig. 1-2a, in the basic fixed reference system 0-.xyz, points PJ(xb z1) and PJ(xb Y2> z 2) are two geometric points in the coordinate space. The displacement vector from P 1 to P2 is defmed as:

z A b

and the distance between points P 1 and P 2 is defmed as:

0 )-------- y

c T X

(a)

F

If the position of P 1 is given, the displacement vector P2 determines the position of P 2 . Therefore, the displacement vector can be said to be the position vector of point P 2 with respect to (relative to) point P 1• With this interpretation, we now define the

J

relative position vector by

',

of the displacement vector

. Bear in mind that in the notation

which is an alternative notation

P 1 serves as the reference point for P 2 , the point concerned,

c

and hence, P 2 should be written first. Sirnilatly, the position vectors of points P 1 and P2 relative to the origin 0 as shown in Fig. 1-2b are defined as:

=xJ +y}+zJ.

0

rP20 = OP2 = x/ + y) +zJ. X

(1.3a) (L3b)

(b) Fig. 1-2

The pos1t10n vectors relative to the fixed origin are called absolute position vectors. Since the fixed origin 0 is always the default, the position VeCtOrS rpp and r p2o are Simply denoted by

rP, and Yp2 , respectively. According to the law of vector addition (Fig.l-2b ), we have the very useful relation as follows:

(1.4a) (1.4b) tIn some textbooks, be explained in Section 1.5.

4

is used. We prefer to use

because here

P, represents a point instead of a reference frame. More will·

1.2 Position, Velocity, and Acceleration

These say that the position vector of P 2 is equal to the position vector of P 1 plus the position vector of P 2 relative to P 1 • Let us now study the motion of a particle P, which traces a curvilinear trajectory in the space with time (Fig. 1-3). At any instant t, the location of the particle can be determined by one position vector as a function of time:

z

(1.5) Consider two particles P and Q which are moving independently in the space shown in Fig. 1-4. The motion of particle Q is defined by position vector function:

X

Fig. 1-3

(1.6) Similar to Equations 1.4a and 1.4b, we can write: (1.7a)

z

(1.7b) These say that at any instant t, the position vector of particle Q is equal to the position vector of particle P plus the position vector of particle Q relative to particle P. Note that in Equations 1.7a and 1.7b the position vector of the particles vary with time. This relation will serve as the fundamental relation in the study of the relative motion of particles (Section 1.5). X

Note also that

rPQ

- - rQ- = -rQP = rp

y

(1.8)

Fig.l-4

which shows that the relative position vector takes on the opposite direction if the reference point is reversed from one to the other.

5

Chapter 1

Kinematics of Particles

Example 1-2 Position vectors and relative position vectors A mechanism of a four-bar linkage (with OA = CB =rand OC = AB =f) moves from position OABC to position OA 'B 'C as shown. Based on the basic reference frame 0-xy, determine the position vectors of points A, B, C, A' and B ', and the relative position vectors rA'A and rB'B.

y

y

l //.,..

I I

I

I

/

/

B

----

/

I

I I

0

I I

I I

....

\\

',

/

!A'

B'

/

'..................

--------

_..,"""//

Fig. b

Fig. a

Solution: Based on the frame 0-xy as shown, the position vectors of A, C and A ' (Fig. b) are expressed as:

(1)

(2)

(3)

Since OABC and OA 'B 'Care both parallelograms, we have:

-----rBC = rA ' rB'C = rA' ' rBA = rB'A' = rC

(4)

Therefore, we have: (5)

rB'

= rA' + rB'A' = rA' + rc

rB'B

6

= rA'A = rA'- rA

=

(0.866r + /)f- (0.5r)]

= 0.366r f -1.366r

l

= 1.4l4rL -75°

(6) (7)

1.2 Position, Velocity, and Acceleration

Discussion: Translation and rotation of a line In this example, rod AB is said to be in translation in a circular track as line AB keeps parallel to a fixed line OC. Any point P on rod AB is also traveling in a circular path as shown, which has an equal radius center P 1 as shown in Fig. a. On the other hand, the rods OA and CB are said to be in rotation. Any point Q on rod OA is traveling in a circular path as shown in Fig. a, which has a different radius but the same center

0.

Example 1-3 Position vectors and relative position vectors A disk of radius r is rolling without slipping on a horizontal surface. Based on the basic fixed frame 0-xy, point G, the center of the disk, is initially on the y axis, and point P on the periphery is at the origin 0. Determine: (i) the position vector of the center G of the disk (ii) the position vector of point P when the disk has rotated clockwise through angle ¢, which is measured clockwise from the vertical line as shown in Fig a.

y /,.

I

I

'

I I

:

'

... "-

/

G

I

··......

·....

I I I \ \

' ............. p

d

c

X

Fig. a

Solution: When the disk rolls without slipping on the horizontal surface, point G, the center of the circle, is moving in a horizontal straight line. The path of motion of point P, shown as a dotted line is curvilinear. Let G 1 and P 1 be the position points of center of the disk G and point P, respectively, when the disk has rotated clockwise through angle ¢. The no-slip condition requires that the horizontal displacement d equals the circular arc, that is,

.----..

d

=

= r¢

(1)

where ¢is in radian.

7

Chapter 1

Kinematics of Particles

Therefore, we have: (i) the position vector of G1 : (2) (ii) the position vector of P 1:

rp, = rc + rP.c I

I

I I

=ry+r¢t +(-rsin¢i-rcos¢}) = r(¢- sin ¢)i + r(l- cos¢)}

(3)

The trajectory of point P thus obtained in Fig. a is called a cycloidal curve.

Discussion: If a disk is rolling without slipping on the floor, why do we have Equation I? Consider that the floor is covered with wet blue paint. If a disk is placed on the floor vertically, only one point, that which is in contact with the floor surface will turn blue. When the disk rolls, the points on the periphery of the disk will come into contact with points on the floor one by one and turn blue in sequence. Because of the nonslipping, the length of the arc that turns blue must equal the length of the portion on the floor that has been touched by the disk.

The constraint of rolling without slipping occurs in many situations, for example, cable-pulley systems (Fig. b). The relation d = r¢ holds true for this case as the pulley rolls without slipping along the cable.

d Fig. b

8

1.2 Position, Velocity, and Acceleration

1.2.2 Velocity - Time Derivative of Position Vectors

Assume that at the instant t, the instantaneous position of particle P is defined by the position vector: (Fig. 1.5a)

r = r (t) = x(t)i + y(t)] + z(t)k

In the time interval (from t to t point P to point P' as given by:

(1.9)

the particle moves from /

/o

X

r'= r(t +

= x(t +

+ y(t +At)]+ z(t + At)k

(1.10)

(a) Average velocity

The displacement vector from P to P' is:

= r'-r = &i +

+ &k

(1.11)

& = x(t +

x(t) = y(t +At)- y(t) & = z(t + z(t)

where

Then, the average velocity of the particle traveling from P to P' in the time interval Lit is defined as: vavg

r'-r

&

&

= - - = - = - l +- j + - k

(1.12)

, as shown in The unit is mls and the direction follows that of Fig. l-5a. When the time interval becomes smaller and approaches zero, point P' approaches P. Thus the instantaneous velocity at that instant is given by the vector: v(t)

&

&

= Lim-= Lim-i + Lim-j + Lim-k D.t-->0

dr

=- =

dt

D.t-->0

D.t-->0

D.t->0

0 i(t)i + y(t)j + i(t)k

(1.13)

This is the tangent to the path at point P, as shown in Fig. l-5b. The magnitude of velocity is usually called speed. From the above, we can draw two conclusions: (1)

X

(b) Instantaneous velocity Fig. 1-5

The time derivative (time rate of change) of the position vector function (t) gives the velocity vector function

r

v(t), which is always tangent to the trajectory of the particle.

9

Chapter 1

Kinematics of Particles

(2)

The time derivative of a position vector expressed in terms of the basic fixed Cartesian coordinate system can be obtained by simply differentiating the scalar components with respect to time (Equation 1.13). As the unit vectors (f, }, k) in Equation 1.9 are constant vectors, their time derivative is zero.

Example 1-4 Velocity of a point/particle If the disk in Example 1-3 rotates at a constant rate c:v, i.e. ¢ = OJt, express the velocities of points G and P as functions of the variable ¢.

y

Solution: From Example 1-3, the position vectors of moving points G and Pat any time can be expressed in the Cartesian coordinate system as:

rc=r} + r¢ i

(with ¢ = oJt)

(1)

rp = r(¢- sin¢)i + r(l- cos¢) J

(with ¢ = oJt)

(2)

By definition, the velocity of point G is:

d d { ". A. Vc = - rc=-I(J + r.,., z dt dt

=

(3)

rmz

which shows that point G has a constant speed rc:v to the right. The velocity of point Pis:

vP = .!!_ rp =.!!_ h¢- sin ¢)i + r(l- cos¢) J] dt

=

dt

rm(l- cos¢)i + rmsin¢}

It can be seen that when¢= 0, Vp

10

= 0, and when¢= 7t,

(4)

Vp

= 2rm i.

1.2 Position, Velocity, and Acceleration

1.2.3 Acceleration - Time Derivative of Velocity Vectors Unlike the position vector r(t), the velocity vector v(t) does not start from the origin 0, but from the instantaneous position of the particle. The velocity vectors at instant t and ( t + f..t ) attached toP and P ', respectively, are:

v = v(t) = x(t)i + y(t)} + z(t)k v'= v(t + M)

= x(t + f..t)i + y(t + f..t)} + z(t + M)k

(1.14) (1.15) X

as shown in Fig. l-6a.

(a)

In order to study the change of velocity in magnitude as well as in direction during the time interval !!..! , without a change of the vectors, we bring v and v' to the origin Ov of a newly defined coordinate space Ov- xyi as shown in Fig. 1-6b. This mathematical velocity space shares the same frame ( i, }, k ) as 0-xyz. Comparing Fig. 1-6b with Fig. 1-Sa, the velocity vectors in the velocity coordinate space Ov- xyi appear similar to the position vectors, and the tip of the varying velocity vector traces out a path, which is called hodograph. The change of the velocity vector from instant t to ( t + f..t ) is now the displacement vector 1'1.v from Q to Q'. Similarly, we define the average acceleration in the time interval as: -

aavg

f..v

=/'-,.(

a

dv

= - = xi + YJ + zk dt

(1.16) (b)

and the instantaneous acceleration

dv

a(t) = - = x(t)i + Y(t)J + z(t)k dt

(1.17)

z

This is tangent to the hodograph at point Q. Now we return to the space coordinate system 0-xyz. The instantaneous acceleration vector at instant t is attached to the instantaneous position P of the particle (Fig. 1-6c). Note that the velocity vector is always tangential to the trajectory of the particle, however, the acceleration vector in general is not (why?).

0

Remark: Time Derivative and Velocity

From the above discussion we can see that acceleration is indeed the "velocity" of velocity, and the time derivative of velocity gives acceleration. The time derivative of some varying vector represents the "velocity of it. " We can always bring the vector to the origin of

X

(c) Fig. 1-6

11

Chapter 1

Kinematics of Particles

its own mathematical coordinate space and treat it as a "position" vector in the space. In this way, the "trajectory" of the tip of the position vector can also be seen as a hodograph. Thus, the velocity of the tip of the vector traveling along the hodograph gives the time derivative of the vector. We shall use this method to determine the time derivative of rotating unit vectors in the following sections.

Example 1-5

Acceleration of a point/particle

If the disk in Example 1-3 rotates at a constant rate P as functions of the variable ¢.

m,, i.e. ¢= wt, express the accelerations of points G and

y

·.. ·-•.........

···...

___a_____c________

X

7U"

Solution: From Example 1-4, the velocity vectors of moving points G and P at any time can be expressed in the Cartesian coordinate system as:

vG

=

rm'i

vP = rm(l- cos ¢)i + rm sin¢}

(1) (2)

By definition of acceleration, we have: QG

=

0

(3)

ii P =d- [rm(l- cos ¢)i + rm sin¢ j dt 2 = rm (sin¢'i +cos¢])

(4)

It can be seen that the acceleration of P has a constant magnitude rm 2 and always points to the center point G.

12

1.3 Simple Plane Motion of a Particle

1.3 Simple Plane Motion of a Particle Rectilinear and circular motion are two types of simple motion of a particle. General curvilinear plane motion is a combination I integration of simple motion.

1.3.1 Rectilinear Motion X

When a particle Pis moving along the x-axis, (Fig. 1-7), the position, velocity and acceleration vector functions are:

r(t)

= x(t)i

v(t) = v(t)i = x(t)i a(t)

v =xi (.X> 0)

(1.18)

= a(t)i = x(t)i

The algebraic scalar functions x, v and a represent the and in direction [ . The respective projections of vectors magnitude of the vectors are defined to be ·the positive values lxl ,

r ,v

a

lvl, and lal

respectively. However, the scalar functions x, v, and a are signed magnitudes; their signs indicate the senses of the vectors. It is convenient to use scalar functions directly in the study of the rectilinear motion of particles.

r

=xz

(x > 0)

l

a=

0 ._

xi

(.X< 0)

The two fundamental equations for rectilinear motion of a particle are the definitions of velocity and acceleration: Ii

dx

v=dt dv a =dt

or

dx

= vdt

(1.19)

Fig. 1-7

dv = adt

or

(1.20)

from which all formulas can be derived for any rectilinear motion problems. A few approaches are briefly reviewed below:

(1)

If x = x(t) is given, the velocity v and acceleration a can be obtained directly from these definitions

(2)

If v = v(t) is given, the acceleration can be obtained by differentiation. To determine the position x, we integrate Equation 1.19:

r

fdx Xo

i.e.

x (t)-

X0

to

=

v(t )dt

f v(t)dt to

(1.21)

Here the initial position x 0 should be prescribed.

13

Chapter 1

Kinematics of Particles

(3)

If a(t) is known, integration should be used to determine v andx:

v(t)- v 0 x(t)- x 0

f a(t)dt = f v(t)dt =

to

(1.22)

lo

where both initial position x 0 and initial velocity v0 should be specified. If a = constant, and t0 =0, we simply have:

(4)

v

= v0 +at

x

1 2 = x 0 + v 0 t + -at

(1.23)

2

(1.24)

If either a= a(x) or a= a(v) is glVen, an additional relation:

vdv = adx

(1.25)

is always used for integration. This equation is simply a combination ofEquations 1.19 and 1.20:

vdv

= v(adt) = a(vdt) = adx

(1.26)

With some necessary information (boundary conditions) given at t 0 and t, the integration:

f vdv 0

or

=

rx a(x)dx' given a= a(x) .k.

[ _v_dv = x - x 0 o a(v)

,

given a= a(v)

. (1.27a)

(1.27b)

can be used to study the motion in the period (t 0 , t ). If a = constant, the integration of Equations 1.27a and 1.27b leads to:

(1.28) Note that only two out of the three equations: Equations 1.23, 1.24 and 1.28 are independent. (Why?)

14

1. 3 Simple Plane Motion of a Particle

Example 1-6 Rectilinear motion with constant acceleration

A ball of mass m, subjected to gravity force only, has an initial velocity v0 >0 upward when it is at the initial position y = h > 0 as shown in the figure. Set up the governing equation for the motion of the ball and determine: (i) the position of the ball when it has reached the highest position (ii) the velocity of the ball wheny = 0.5h (iii) the time wheny = 0.5h

y

t = l]

Y1

h

vo

t

r

t=O

r =yj

0.5h

mg

..

t = t2

0 Solution: The position vector of the ball is simply governing equation for motion of the ball:

r = y}.

Using Newton's Second Law

ft' = -mi} = ma = m(.Y})

F = ma'

we have the

(1)

where g is the gravitational acceleration and ji is the signed magnitude of the acceleration. Hence, we get:

ji=-g

(2)

which is the governing equation for the motion of the particle. (i)

Using Equation 1.25 we have:

vdv = jidy = - gdy Integrating (3) over the corresponding domain v

f0

v0

vdv =

P1 h

E

(v 0, 0) andy

(-g)dy

(3) E

(h, y 1):

(4)

15

Chapter 1

Kinematics of Particles

(5)

yields: 2

1 v0 yl =--+h

So,

2 g

(ii) Integrating (3) over the corresponding domain v v2

I

v0

-

(v 0, v 2 ) andy

E

(h, 0.5h):

f0.5h (-g)dy

vdv=

1 2 -v 2 2

yields:

E

h

1 2 -v0 =- g(0.5h- h) = 0.5gh 2

(6)

(7)

So, There are two possible solutions for the velocity at y solution we want should have a negative sign. Hence,

= 0.5h.

Since we are only interested in t > 0, the

(8)

(iii) Using Equation 1.23 when

t = t2.

we have:

Hence,

Discussion: (i)

In this solution, the scalars y, v, and a are treated as signed magnitudes based on the upward positive direction ofthe coordinate 0-y. For example, the negative sign of the scalar v 2 in Equation 8 implies the downward direction of the vector

v =2

gh +

v; ].

(ii) For solving problems of rectilinear motion of a particle with a constant acceleration, Equation 1.28 can always be used to relate the velocity with the position. The relations are in forms of Equations 5 and 6. Multiplying both sides of (6), for example, by the mass of the particle m gives:

1 2

2

1 2

2

-mv 2 --mv0 =-mg(0.5h-h)=mg(0.5h)

which implies that the difference of kinetic energy fromy = h toy= 0.5h equals to the work done by the gravitational force (mg) through the displacement 0.5h. (See Chapter 3, Section 3.3)

16

1. 3 Simple Plane Motion of a Particle

Example 1-7 Rectilinear oscillation motion- Mass-spring system A block of mass m, which is considered a particle, is attached to a spring of stiffness k, and is allowed to move on a frictionless horizontal surface as shown iri the figure. The origin of coordinate x is chosen such that when x = 0, the spring is at its unstretched length. Set up the governing equation for the motion of the block. If the block is released from rest at x = I when t = 0, determine: (i)

the velocity of the block when it reaches x = 0

(ii) where and when the velocity becomes zero again?

ft = -kxi

-;:5J i r =xi

l I

N t= 0

Solution: The position vector of the block is simply

r =xi. Using Newton's Second Law F = ma, we have the

governing equation for the motion of the block:

-kxi =mit

(1)

Hence, the acceleration is a function of x:

.. k x=--x

(2)

m

which 1s the governing equation for the motion of the particle. (i) Using Equation 1.25 we have:

vdv

= xdx = - !5._ xdx m

(3)

W1th mitial position x 0 = l and initial velocity v 0 = 0 known, integrating (3) over the corresponding domain v E (0, v1) and x

E (/,

0) yields:

(4)

So

17

Chapter 1

Kinematics of Particles

Since in the moment concerned the block is approaching the origin from the right, our solution should be

v, ·(i)

(5)

Integrating (3) over the corresponding domain v

E

(0, v) and x

E (/,

x) yields:

12 -0=-1 k(2 -x 2) -v 2

(6)

2m

So the velocity can be expressed as a function of x:

(7) -I, the velocity becomes zero again.

before it becomes zero again. It is seen that when x

=

In order to determine t2 when the block reaches x

-I, we use Equation 1.19 in the form:

=

dx dt=--

(8)

v(x)

Integrating (8) over the corresponding domain t

E

(0, t2) and x

E (/,

-!)gives

(9)

Discussion: (i)

For problems of rectilinear motion of a particle with acceleration given as a function of x, Equation 1.27a can always be used to relate velocity with the position. In this example, acceleration is a linear function of x. Thus the relation is in a form of Equation 6. Multiplying both sides of (6) by the mass of the block m gtves:

k(2

1 2-0 =-1 -x 2) -mv

2

(10)

2

which implies that the difference of the kinetic energy of the block from x = l to x equals to the work done by the spring force through the displacement (see Chapter 3, Sections 3.3 . and 3.5.3). (ii) The general solution of the second order differential equation (2) can be easily obtained mathematically as:

x =A cos m t + B sin m t, with frequency m Using the initial conditions x 0

=

=

I and v0 = 0, the solution is simply:

x= !cosmt which is a harmonic oscillation of the simple mass-spring system.

18

(11)

(12)

1. 3 Simple Plane Motion of a Particle

13.2 Circular Motion The second type of simple motion of a particle is circular motion, in which the path of the motion is a circle of radius r0 as shown in Fig. 1-8a. A circular motion can also be the motion of a geometric point P on a rigid body, say a disk, rotating about the fixed axis z passing through the center of the disk and perpendicular to its plane as seen in Fig. 1-8b.

y

(1) Angular Coordinate, Angular Velocity, and Angular Acceleration

A fzxed reference coordinate system 0-xyz is attached to the fixed center of the circle as shown in Fig. l-8a and Fig. 1-8b. In the study of the circular motion of a particle P or the rotation of a rigid disk, the polar coordinate system is preferred. The position vector of point P can be expressed as:

r =xi+ y] = r0 coset+ r0 sin tf = r0 (coset+ sin tf)

(1.29)

(a)

Since \rl = r0 is a constant, the angular coordinate () of the concerned point is the only variable needed to describe the position of P. The angle (), measured from the positive x-direction and positive in the counterclockwise direction, is the angular coordinate of the point and is a function oft.

ii

Denoting the radial unit vector:

=

(jf

w=Bk

e, =case i-\- 'O.ll\8 3 we rewrite the position vector as: ( 1.31)

z

y

Hence, the circular motion of the particle, the rotation of the rigid d1sk, and the rotation of the position vector as well as the rotation of are all determined by the function B(t). The angular velocity and angular acceleration of the rotation are defmed as:

r,

e,

d()

(J)=-

dt

dw

a=dt

(1.32)

(b)

i.e. UJ = () and a = iJ, respectively (Fig. l-8a). The dot indicates the differentiation with respect tot.

Fig. 1-8

Based on the right-hand rule, the angular velocity vector and the angular acceleration vector are respectively denoted as:

w=wk=Bk a= ai: = ef

(1.33)

Here, m and a are the signed magnitudes of the vectors. In Fig. 1-8b, iJ and 0 are assumed to be positive.

19

Chapter I

Kinematics of Particles

Note that any problem on circular motion concerning t, (), w, and a can be mathematically treated as a rectilinear motion of a representative point moving on the axis B. As a result, all approaches in solving rectilinear motion problems concerning t, x, v, and a in Section.1.3.1. can be used here.

Example 1-8 Circular motion - Angular coordinate .. g A pendulum as shown is released from rest at B= 0. Knowing that a= B =-cos B (see Example 3-5), derive the angular velocity

(V

=

e in terms of B( 0 ::; e ::;

r

7r ).

X

r

a

0

A

e

p

y

Fig. b

Fig. a

Solution: Based on the coordinate system 0-xy, the angular displacement B is measured from the positive x-direction and is positive in a clockwise direction as shown in Fig. a. Thus the angular velocity wand acceleration a of the circular motion are both signed magnitudes with the same sign convention. The circular motion of the pendulum P in Fig. a can be mathematically treated as a rectilinear motion of the representative point A moving on the coordinate B-axis as shown in Fig. b. Thus, the problem is similar to Example 1-7. The acceleration of point A ( iJ = g cos()) is given as a function of coordinate B. Therefore, we r

can use Equation 1.25 in the form:

(J)d(J)=adB =(! cosB )de

·(1)

Since the initial values of B and w are both zero, we can integrate (1) over the corresponding domains w E (0, w) and B E (0, B) and have:

f! So,

20

(J)

=

sine

cos B dB = ! sine

(2)

1. 3 Simple Plane Motion of a Particle

(2)

Velocity in Circular Motion

By definition, the velocity vector of a point is:

v=

The time derivative of Equation 1.30:

err

(1.34)

dt

e,

can be obtained by differentiating

B(-sinO i +cosO

e

where

8

})

(1.35)

=-sinO l+cosO}

is a transverse unit vector, which is perpendicular to the direction of increasing (}(Fig. 1-8a).

(1.36)

e,

and is in

Substituting Equation 1.35 in Equation 1.34 gives: (1.37) Here,

(1.38)

v,

is the signed magnitude of the velocity which is tangent to the circle, and its sense depends on the sign of v. Using the rule of vector cross product, we can simply write Equations 1.34 and 1.37 as:

_ dr

__

v=-=OJxr

dt

(1.39)

z

Remarks: (!) Equation 1.39 is a general formula for calculating velocity of any point in any rigid slab that is rotating about 0 at angular (Fig. 1-9). velocity

m

(2) Equation 1.39 is a general formula for calculating the time derivative of any vector c of constant magnitude, which is rotating at angular velocity aJ :

d -(c)= a> x (c) dt

(1.40)

X

Fig.l-9

21

Chapter 1

Kinematics of Particles

For example, unit vectors

e,

and

e0

at point Pin Fig.l-8a can

be seen as two vectors of constant magnitude 1, rotating together with the disk, so we can have:

y

(1.41a) (1.4lb)

e,

eo

As shown in Fig.l-10, the unit vectors and are brought to the origin, and their time derivative are seen as the velocities of

the tips of the vectors moving along the unit circle. (See Remark in Section 1.2.3.)

Fig.J-10

Example 1-9 Circular motion - Velocity A wheel is rotating about a fixed point 0 . Two particles E and B are attached on the hub and the outer rim of the wheel, respectively, and initially are at the positions shown. A cable is wrapped around the central hub of the wheel and point A on the cable has a constant velocity v = O.lrnls to the left.

v A'

A

d

Determine: (i) the speed of E and Bat the position shown (ii) the position vector and the velocity vector ofB when t = 2

22

X

1. 3 Simple Plane Motion of a Particle

Solution: As discussed in Example 1-3, the motion constraint between the cable and the hub is called rolling without slipping. Assume point A moves to A' through displacement d. The wheel rotates counterclockwise (ccw) through an angle B, and pointE on the hub moves toE' as shown. We then have the constraint condition:

d

=

EE' = 0.3()

(1)

Differentiating (1) with respect to time gives:

d = v = 0.3 B= 0.3w

(2)

which relates v, the velocity of A, to cu, the angular velocity of rotation of the wheel. Hence, the wheel rotates with a constant angular velocity:

v

0.1

1

0.3

0.3

3

OJ=-=-=-

rad/s (ccw)

(3)

(i) Since particles E and B are both on the rotating wheel, each travels in a circle at the same angular velocity cu but at different radii, 0.3m and 0.6m, respectively. Therefore, the speed of particles E and B are:

(ii)

vE

= 0.30J = 0.1m/s

v8

= 0.60J = 0.2 m/s

The angular displacement of the wheel in t = 2 is: 2

(} = OJt =- rad = 38.2°

3

Thus,

(4)

r8 = 0.6L38.2·m

v

8

= 0.2L(38.2° + 90. ) = 0.2L128.2. m/s

Discussion: Angular displacement ()and angular velocity cu are used to characterize the rotation of the wheel. Any point or any particle on the wheel moves in a circular motion with radius r . However, all the points on the wheel have the same angular velocity ro in the relation v = cur .

23

Chapter 1

Kinematics ofParticles

(3) Acceleration, Tangential and Normal Components By defmition, the time derivative of the velocity vector v m Equation 1.37 gives the acceleration vector:

(1.42)

where the first term represents the rate of change in the magnitude of the velocity and the second term represents the rate of change in the direction of the velocity. Substituting Equation 1-41b in the second term of Equation 1.42 gives:

y

Referring to Fig. 1-11a, the first term acceleration and is denoted by:

IS

called tangential

dv a I =dt- = a r0

(a)

(1.44)

and the second term pointing to the center of the circle is called normal acceleration and is denoted by:

•+w

a=Bk

z

= Bk

an= an(-e,) an = vw = w 2 ro

Note that although

a

1

vz (1.45)

ro

is always tangent to the circle, its sense

depends on the sign of a = B, that is, it depends on the direction of ii . In Fig. 1-1la and b, a is assumed to be positive. As for

a.,

(b) Fig.l-11

24

since its signed magnitude an is always positive, it implies that the rate of change of direction of the velocity of the point is always pointing toward the center of the circle, that is, in direction, regardless of the sense of rotation. That is why normal acceleration is usually called centripetal acceleration.

e,

1. 3 Simple Plane Motion of a Particle

For intuitive physical interpretation, we denote: (1.46)

as the tangential and normal unit vector, respectively. The acceleration is then expressed as:

- dv vl a =a +a =-e +-e I n dt I ro n A

A

(1.47)

Applying the rule of vector cross product, we can rewrite Equations 1.44 and 1.45 as:

-

- ax - r a,=

a- n

(-) 1=OJ X OJ X r =-OJ r

(1.48)

as shown in Fig. 1-11 b. Therefore, Equation 1.47 can also be expressed using the vector cross-product operation:

(1.49)

which is the general formula for calculating the acceleration of a particle traveling in a circular path, or of a point on a rigid body rotating about a fixed axis.

Example 1-10 Circular motion -Acceleration In Example 1-9, if point A on the cable has a constant acceleration a

= 0.05rnls2 and initial velocity

v0

=

0.1m/s, both to the left, determine the acceleration of point Bat time t = 2s. y

a

A

vo

Fig. a

Fig. b

25

Chapter I

Kinematics of Particles

Solution: Using the constraint condition of rolling without slipping (Example 1-9):

d

=

EE' = 0.3()

(1)

and differentiating it twice with respect to time gives:

d = a = 0.3 iJ = 0.3a

(2)

which relates a, the acceleration of A, to a, the angular acceleration of the rotation of the wheel. Hence, the wheel rotates with a constant angular acceleration: a

0.05

I

0.3

0.3

6

a=-=--=- rad/s2 (ccw)

(3)

Using Equations 1.23 and 1.24 in form:

co = co 0 +at () =

1 2

B0 + w 0 t +-at

2

. I With {)0 = 0 and C0 0 =- rad/s, when t = 2 we have: 3

1

I

2

w = w 0 +at= -+-(2) = - radls(ccw) 3 6 3 I

I

II

B=O+w 0 t+-at 2 =-(2)+--2 2 =Irad=57.3 2 3 26

(4) o

(5)

The tangential and normal components of acceleration of particle B can be calculated respectively as:

= 0.6a =0.1m/s

2

and

a; = 0.6w

2

= 0.267m/s

Therefore, the magnitude of the acceleration of point B when t = 2 is

and

26

2

(6)

1.3 Simple Plane Motion of a Particle

with

as shown in Fig. b. Using:

¢=tan -1

=tan -1 (0.267) = 69.5o 0.1

a8

we can express the acceleration vector of point B as:

Discussion: For simple planar problems such as the above example, we may directly calculate the magnitude of the two 2 components of acceleration by using scalar equations a, = ar and an = rw . However, we should keep in mind that the direction of the tangential acceleration depends on the actual direction of the angular acceleration ii , as shown in Fig. c. The normal acceleration always points to the center of rotation.

a r

Fixed point

Fixed point

a

Fig. c

e

Like and w, the angular acceleration a is used to characterize the rotation of the wheel. All points on the wheel move in a circular motion and have the same angular acceleration a in the relation a, = ar.

v v

It should also be emphasized that for point D on the wheel (Fig. a), we always have D = A , but the acceleration of Dis not equal to the acceleration of A. Only its tangential component equals the acceleration

27

Chapter 1

Kinematics of Particles

1.4

Planar Curvilinear Motion of Particles

1.4.1

Curvilinear Motion Described in the Cartesian Coordinate System - Combination of Rectilinear Motion

In the 2D Cartesian coordinate system, the planar motion of a particle P is described by the position vector:

y fJ b

r = r(t) = x(t)i + y(t)}

c 1

vx

Py 0

Px

It shows that a curvilinear motion can be decomposed into two rectilinear motions: that of a point Px and a point Py moving along the x-axis and the y-axis, respectively. The velocity and acceleration of particle P can also be obtained by vectorial addition of the corresponding vectors of the rectilinear motions, Fig. l-12a:

(a)

c

(1.50)

(1.51) (1.52)

where:

y

vx =x(t) , vx

= constant

vy = y (t) ,

.

+g X

0 (b) Fig. 1-12

28

ax = i(t) for the rectilinear motion on the x-axis ay = ji(t) for the rectilinear motion on they-axis

Thus, any planar curvilinear motion described in Cartesian coordinates can be treated as a combination of two rectilinear motions in the x and y directions, and all the approaches studied in Section 1.3 .1. can be used. Projectile problems are a type of curvilinear motion in which the particle moves with constant velocity in the horizontal direction, and with constant acceleration (g) in the vertical direction. (Fig. 1-12b)

1. 4 Planar Curvilinear Motion of Particles

Example 1-11 Curvilinear motion in the Cartesian coordinate system - a Projectile Consider a ball thrown at speed v = 10 mls in the direction as shown. Its motion is influenced by gravitational force only. Determine: (i) the distance d to where the ball strikes the inclined surface at B (ii) when and where the velocity of the ball is parallel to the inclined surface

y v=lOm/s

Solution: ln the Cartesian coordinate system as shown the initial velocity of the ball is given by:

(1) The curvilinear motion of the projectile can then be treated as a combination of a horizontal motion with constant velocity v0 : (2) and a vertical motion with a constant acceleration -g:

y= y 0 +v0 i-

1

2

gt

2

r;;

5"1!2t-0.5gt2

(3)

(i) Assume that the ball hits the surface at point B, we then have the relation: X8

:y 8 =4:(-3)

(4)

Substituting (2) and (3) into (4) yields:

-3( 5 ..fit) Thus,

t2

=

4 (5 ..fit- 0.5gf)

35-fi = --=2.52s

2g

29

Chapter 1

Kinematics of Particles

r;;

175

x 8 =5 vL. t 2 =-m, and

Therefore, we have

g

d=

5 4

-X8

=22.3

ID

(ii) Assume that the ball reaches point A while its velocity have the relation:

v is parallel to the inclined surface, we then (5)

Since: (6) we have:

- 3(5J2)- 4(5J2- gt) = 0

and:

t=

35 J2 4g

XA

= 5 "\/ L. t =

=

(7)

1.26s

So, the point A is at: r;;

175

-

2g

=

8.92m

r;; 2 175 YA = 5 "\/ L. t- 0.5gt = = 1.11m 16g

Discussion: We can also determine the time and the position when the ball reaches its highest altitude by using the condition vy = 0.

Example 1-12 Curvilinear motion in the Cartesian coordinate system - a Circular motion The motion of a particle P is described in the Cartesian coordinate system by: 7r

x=l5cos(-t)

(a)

{ y=l5sin(; t) 4

(b)

Determine: (i)

the equation for the trajectory of the motion

(ii) the equation for the velocity hodogragh

30

1. 4 Planar Curvilinear Motion of Particles

Solution: (i) Eliminating t from (a) and (b) gives: (1) which is the equation of the circle of radius 15 as shown below.

y ,,.,, .......

---

/

I

/

I I

-15:

X

Fig. a From motion equations (a) and (b), the circular motion can be seen as the combination of two oscillatory motions: Px on the x-axis and Py on they-axis. The position vector 1s:

r

r = 15 cos¢ i + 15 sin¢ J

with

(¢

1(

=

wt = - t) 4

(2)

which rotates counterclockwise at an angular velocity w =7t/4 while the point P travels along the circular path. (ii) Differentiating equations (a) and (b) with respect tot yields:

.

X= -

15n sm( . n t)

4

4

n) = 15n cos(n t+l

4

4

{ y=-cos(-t) . 15n n 15n . (n n) =-sm -t+4 4 4 4 2

(3) (4)

Eliminating t from (3) and (4) gives: (5) which shows that the velocity hodograph is a circle of radius 11.8 (Fig. b).

31

Chapter 1

Kinematics ofParticles

The velocity can be expressed as:

v=11.8cos(¢+

with

+11.8sin(¢+

¢

= (J)(

7r

=-t 4

(6)

Comparing the position vector (2) and the velocity (6), we can see that the two vectors are rotating at the same angular velocity m. However, the phase angle of the velocity vector is rc/2 ahead that of the position vector (Fig. b).

y

/

v

y

--

/

a

Q

/

I

/

Py

I

I I I

¢

I I \

I

0

Px / I

\ \

' ' ... ...

I

'

----

/ /

/

/

X

o.

I

······......

...//

__

.....··•·

Fig. b

Discussion: Differentiating (6) once again gives the acceleration of the particle P:

a= 9.25 cos(¢ + 1r) i + 9.25 sin(¢ + n-)]

with

7r

¢=mt=-t 4

(7)

Comparing acceleration (7) with position vector (2) and the velocity (6), we can see that the three vectors are rotating at the same angular velocity m. However, the phase angle of the acceleration is rc ahead that of the position vector (Fig. b).

32

1.4 Planar Curvilinear Motion of Particles

1.4.2

Curvilinear Motion Described in the Path Coordinate System - Integration of Circular Motion

y

Assume a given plane curve is located in the fixed reference system 0-xy as shown in Fig. 1-13a. If a particle P moves along the curvilinear path, the position of the particle may be determined by a path coordinate s, which is the arc length measured from the chosen origin o on the curve, and with the positive direction prescribed. The motion of the particle can then be described by the scalar position function s = s(t).

+s

v = se,

As s = s(t) here plays a similar role as x = x(t) in rectilinear motion, the speed and the rate of change of speed of the particle in and respectively. Therefore, the scalar curvilinear motion are approach used in rectilinear motion (Section 1.3 .1) can also be applied here for relating t, s, and However, the direction of the velocity varies along the curve, and so does the direction of its acceleration.

s

s, s

0 (a)

s.

(1) Velocity Tangential to the Path

With the position function s(t) known, the time derivative of s(t) is the signed magnitude of the velocity: ds

v=-=s dt

(1.53)

The positive direction of velocity is the tangential unit vector:

e, = cos¢ i +sin¢ J

(1.54)

where ¢is measured by rotating a radial vector from the [ direction counterclockwise to the tangential line with the sense toward the +s direction as shown in Fig. 1-13a and Fig. l-13b.

y +s

v = se,

0 (b)

Thus, the velocity of the particle is:

v = ve, = se, Here, both

Fig. 1-13 (1.55)

s and e, are functions oft.

33

Chapter 1

Kinematics ofParticles

(2)

Curvature and Radius of Curvature

With regards to a smooth curve, how "curved" is it at a point P? To measure the curvature, we introduce an osculating circle, which is the "closest" circle tangential to the curve at P (Fig. l-14a). To determine this circle, consider that a particle moves from P to the neighboring point P' through Lls along the curve, and the angular displacement of the tangent line during this motion is ,1t/J in radian. The circle passing through points P and P' and tangential to the curve at both points has the radius p = s/ ¢1. When P' approaches P, the limitation:

✓ -ILim & p-

f-

date

1-1 d¢ ds I

(1.56)

is the radius of the osculating circle, which is the largest among all the circles tangent to the curve at P. Thus, we define pas the radius of curvature of the curve at P. The center C of the osculating circle is called the center of curvature (Fig. l-14a). The curvature of the curve at Pis defined to be:

(a)

(1.57)

+s

y

Note that the curvature K can be positive, negative, or zero. Its sign represents the direction of curvature and depends on the direction of the normal vector which must point to the center of curvature:

en '

Sign(K)

X

0 (b) Fig. 1-14

=(e

1

X

eJ· k

(1.57a)

en

e,

en

e,

For example, at point A in Fig. l-14b, is reached by rotating 90° in a counterclockwise direction. Then K > 0 , the curve is considered concave at A. At point B, is reached by rotating 90° in a clockwise direction. Then K < 0 , and the curve is considered convex at B. At the inflexion point D, we have K = 0. Like any point on a straight line, at point D p = oo . (3) Acceleration, Tangential and Normal Components Consider a curve being cut into a series of small segments. Each segment can be approximately treated as a small arc of a circle. When the segments become smaller and smaller, we say that a curvilinear motion of a particle is a series of infinitesimal segments of circular motion. So, at any instant, the motion of a particle passing through a point P may be thought of as if it is traveling on the osculating circle (Fig. l-14c). With both the center of curvature C and the radius of curvature p known, the acceleration of the instantaneous "circular motion" can be obtained in the form of Equation 1.47:

34

1.4 Planar Curvilinear Motion of Particles

(1.58)

y

en

where the unit vectors el and are determined by the location of the instantaneous osculating circle at P as shown in Fig. 1-14c. With 1 defined by Equation 1.54, can be expressed as:

e

en

(1.59)

which is perpendicular to the tangent line, and the sense depends on the sign of the curvature (K). For example, in Fig. 1-14c, the curvature at point Pis positive. Thus in Equation 1.59, the positive sign should be adopted.

(c) Fig. 1-14

So far we have successfully extended the study of simple circular motion to the study of general curvilinear motion. In simple circular motion, the center of the circle is fixed and the radius r 0 is a constant. But, in a general curvilinear motion, the position of the instantaneous center of curvature C varies, and so does the radius of curvature p.

Example 1-13 Curvilinear motion in the path coordinate system

=..!. x

2 as shown. Determine the 2 acceleration of the particle when it passes point (2, 2) and the radius of curvature of the path at this point.

A particle P is traveling with a constant speed v along the given path y

2

0 2

35

Chapter 1

Kinematics ofParticles

Solution: Since the motion trajectory of the particle is known, we have at point (2, 2):

tan()=

dyl

dx

=2

and then()= 63.4°

(1)

x=2

So, the tangential and normal unit vectors of the curve at the point are respectively: (2) (3) The position vector of particle P can be expressed in the Cartesian coordinate system as:

r =

xz + YJ = xz + -1 x 2

(4)

J

Differentiating (4) with respect tot gives the velocity vector:

- ar = Xl.-:- + YJ = Xl.-:- + XX]

v= dt

(5)

Defining the path coordinator s as shown, we have: · I· 2 +xx 2 •2 s=v=....;x =v

(6) (7)

Atx = 2: Differentiating (5) with respect to t once again yields the acceleration:

d2-

a =---;. =xi + y} =xi + (.x + xx )J dt 2

(8)

Since the tangential acceleration is zero, we have:

s =at = a. et =[xi+ (x

2

+ xX )J] [cos63.4. i +sin 63.4. J]= 0

(9)

Substituting x = 2, and (7) into (9) gives:

i = -0.08v 2

(10)

Hence, at point (2, 2), we have the acceleration: (11)

36

1.4 Planar Curvilinear Motion of Particles

a= a.e•.

This is indeed normal acceleration 2

Consequently, using a.

.

can calculate the radius of curvature of the path at point (2, 2):

p

(12)

Discussion: (1) Mathematical formulas of Kand p Since the path of motion is given, the curvature K and the radius of curvature p at any point on the curve can be determined by the following mathematical formulas:

(13)

In this example, we may use the above formula to obtain pat point (2, 2) directly. (2)

Use of the relation v =

pB

For instantaneous circular motion, we may use the relation v =

pB

at position P (2, 2). Then, how do we

determine () ? Differentiating both sides of equation tan(} = dy = x gives:

dx

Using .X = v cos(}, we have: .

(} =

3

v cos (}

At point P, (} = 63.4°. Thus,

v 1 p=--:- = - = 11.18 (} cos 3 (}

37

Chapter 1

Kinematics of Particles

1.4.3

Bead sliding

B

In the preceding sections, we analyzed the curvilinear motion of a particle by either decomposing it into two rectilinear motions in the Cartesian coordinate system (Section 1.4.1 ), or treating it as an integration of a series of infinitesimal segments of circular motion in the path coordinate system (Section 1.4.2). The third way to study curvilinear motion is to consider it as a combination of rectilinear motion and circular motion in the polar coordinate system.

I

0

Curvilinear Motion Described in the Polar Coordinate System Combination of Rectilinear and Circular Motion

' Rigid wire rotating " about 0

As shown in Fig. 1-15a, a bead P is sliding along a rigid straight wire OB. At the same time the wire is rotating about a fixed point 0. The bead then traces out a curvilinear trajectory.

(a)

The position vector of the bead in the polar coordinate system (Fig. 1-15b) is denoted by:

y I I

(1.60)

B where

er =cos() i +sin() J

(b) Fig. 1-15

38

Similar to x = x(t) in rectilinear motion, the scalar function rp = rp(t) determines the rectilinear motion of the bead on the wire if the wire is stationary (B= constant). The rotation of the wire is then described by the varying angular coordinate B. If the bead stops sliding ( rp = constant), it goes through a circular motion as the wire rotates. However, if both coordinates rp and ()are varying, the bead may trace out a general curvilinear path.

1. 4 Planar Curvilinear Motion of Particles

(1)

Combination of Velocities (Relative and Entrained Velocities) By definition, the time derivative of the position vector of a

y

particle (Equation 1.60) gives the velocity of the particle:

I

Vp

'- /'" \

.... " ,'

(1.61)

which consists of two components: the first term the radial component and the second term rpiJ

/''

\\

B '

/

rpe, is known as

eo is the transverse

component (Fig. l-15c). The physical meaning of the two components can be further discussed as follows:

(c)

Let P' be the geometric point on the wire that coincides with the particle P at the instant t 0 (Fig. l-15d). During the small time interval L1t ( 1'1t

= t1 -

P' moves to

, so we have:

t0

),

the particle P moves to

but the point I

= P'P' 1 +P' 1 P1

B

(1.62)

As f...t is gets smaller and smaller, the velocity of Pis expressed as:

.

. P' P' I

.

P' I

vP = Lzm--= Lzm--+ Lzm--/1{ /;HO f'1t f>HO f'1t

(1.63)

(d)

Since P' is in a circular motion, its absolute velocity is:

(1.64)

e

y I

which says that the physical meaning of rPiJ 0 is the velocity of y" P '. We shall call P' the entraining point for P, and P' the entrained velocity of particle P . Of course, point P' belongs to the j" wire, and at the instant t 0, rp. = rP (Fig. l-15d).

v

Next, let a body-attached reference system 0- x "y" be attached on the rigid wire OB (Fig. l-15e). For the observer on this rotating reference system, the bead P is in rectilinear motion. The position vector of P can be expressed in 0-x 'y" as: (1.65)

(e) Fig. 1-15

39

Chapter I

Kinematics of Particles

rp

It is important to emphasize that the position vector

rP

Equation 1.65 is the same position vector

in

in Equation 1.60.

Equation 1.65 is only an alternative expression of rp in the moving reference 0-x 'y ". Therefore, we have the following coordinate transformation relations:

x"= rp

f"=e, =cosB i +sinB]

(1.66)

where [" is rotating and is a function of B. However, for the observer on 0-x'y", ["is "stationary." Therefore, the velocity of the bead observed from 0-x"y" is the relative velocity and it is expressed as:

P' p

v P/ 0-x"y" =Lim--' _ I =x"i"= f !JJ->O /).( P

e

(1.67)

r

rpe, is the sliding velocity of the bead, i.e. the relative velocity of the bead observed from 0-x 'y ".

This implies that

Thus, we can rewrite Equation 1.61

+ as:

VP/0-x"y"

Absolute velocity

+

VP'

(1.68)

Relative velocity + Entrained velocity

This is the Principle of Velocity Combination and we will study it in Section 2.4.2.

v

Pt O-x"y"

is the relative velocity of particle P observed from the

moving reference 0-x "y ". As a body-attached moving reference system is always attached to a rigid body (Section 1.1), the lengthy notation 0-x 'y" in Equation 1.68 can be replaced by the label of the body, e.g.J, for compactness:

y

(1.69) X

Notice that any label that may appear after the slash '/' always represents a reference frame or a body to which the reference frame is attached.

Fig. 1-15 (cont'd)

40

Equation 1.69 can be illustrated by a vector loop (Fig. 1-15£).

1.4 Planar Curvilinear Motion of Particles

Remark: Absolute and Relative Time Differentiations The absolute velocity of point P is a time differentiation of position vector rp based on the basic reference frame 0-xy. The relative velocity of P is also a time differentiation of the position vector rp . However, the position vector rp should be expressed using: (i) x "y" coordinates (ii) unit vectors t", }" as rp = x" ["

In the time differentiation based on the moving frame 0-x 'y ", unit vectors t" and }" are treated as stationary. Let the relative time differentiation for the particular moving frame 0 -x 'y" be denoted by:

Using this operator, the relative velocity Equation 1.67 can be derived as follows:

-

VP/ 0- x"y"

=

(8) -

rp

bt

=(!__) 8t

0 -x"y "

I

(x"t")= x"t"

(1.70)

For compactness, we can simply write:

(1.71)

Using Equations 1.61, 1.71 and 1.64 in Equation 1.69, we have:

(1.72)

where iiJ is the angular velocity of the moving frame f Note that Equation 1. 72 is generally valid for any vector

function R (t). To prove it, let R be expressed in the moving frame:

R= R x"

+ R y"

(1.73)

41

Chapter I

Kinematics of Particles

According to the rule of vector differentiation:

d -

dt R

.d (

= dt

A

A

0

)

A

= Rx . i"+Rx ..

Rx .. i"+RY. j"

ii"

0

A

dt + Ry.. }"+RY.

cfJ"

dt (1.74)

Since

t"

and }" are rotating at angular velocity -:-11

dl

-

-=OJ

dt

a;_f'.n

A

Xi"

_

A

-=OJ X

dt

w, we have:

j"

(1.75)

Using Equation 1.75 in Equation 1.74

t.e.

dt

8t

wx(R)

(1.76)

This is the general formula relating absolute and relative time differentiation. We shall use it in the following sections: Combination of Acceleration (Relative, Entrained and Coriolis Acceleration) The time differentiation of the velocity vector of a particle (Equation 1.61) gives the acceleration ofthe particle:

eo y

\

Zip

+

I

aop

I I

\_,

B

e, a'p X

(a)

Fig.l-16

42

(1.77)

This consists of two components: the first term, which is the radial component, and the second term, which is the transverse component (Fig. l-16a). To understand the physical meaning of these terms, we may rearrange them into three components: (1.78)

Obviously, the first term represents the sliding acceleration of the bead, i.e. the relative acceleration of P observed in 0-x"y".

1. 4 Planar Curvilinear Motion of Particles

The relative acceleration can be defmed as the relative time derivative of the relative velocity PI1 (Fig. 1-16b): .

v

(1.79)

The second term represents the absolute acceleration of point P' in the circular motion due to the rotation of the rigid wire (refer to Equation 1.43), which is defined as the entrained acceleration iiP' (Fig. 1-16b):

x" y"

(1.80) I

The third term is the most subtle. It is called Coriolis acceleration. Using the rule of vector cross product, we designate:

(b) (with

w= Bk)

( 1.81)

as the Corio lis acceleration (Fig. 1-16b). Clearly, the existence of the Corio lis acceleration is due to the coupling of the rotation of the moving coordinate system cv and the relative velocity Plf .

v

Now we can rewrite vector Equation 1.78: aP

as aP Absolute

+

(rpe,)

(- rpB

+ Relative +

2

e, + rpe e

aP.

a Pit

Entrained

8 )

+ (2rpBe 8 ) -c (1.82) + aP + Corio! is

This is known as the Principle of Acceleration Combination, which will be further discussed in Section 2.4.3. Vector Equation 1.82 can be illustrated by a vector loop diagram (Fig. 1-16c). Note that Coriolis acceleration the relative velocity when either cv = 0 or

vP!f.

a; is always perpendicular to

(c)

This term disappears from Equation 1.82

v =0 .

Fig. 1-16 (cont'd)

Plf

t If we take time differentiation of

v

P'

in Equation 1.64, we will find that

.!!_ vp· * ar dt

as in Equation 1.80. Why? Skip this question in

your first reading.

43

Chapter 1

Kinematics of Particles

Example 1-14 Rectilinear Motion described in the polar coordinate system A collar that slides along a horizontal rod has a pin P, which is guided to move in the slot of arm OB as shown.

.

The arm OB oscillates at an angular position B = tr + tr sin 1.5t, where ()is in radian and t is in seconds. For

2

6

t = 5, determine: (i)

the position vector of

r

(ii) the radial and transverse components of velocity of P and verify the direction of the velocity (iii) the radial and transverse components of acceleration of P

Solution: When t = 5, we have: B

B=

;r

2

+

;r

6

sinl.5t=2.062rad= 118.14°

e. =-1r (1.5) cos 1.5t = 0.2722 radls

(1)

(2)

6

500mm

e.. = - 1r- (1.5)

2

6

(i)

sin 1.5t = -1.1051 radls

2

(3)

Using the constraint condition for P:

rsinB = 500 gtves

500

(4)

r=--

sinB

Thus the position vector of point Pis a function of (): -

500

A

(5)

r=--e

sine

where

r

e, = cosei + sin0

When t = 5 (Fig. a), we have ; = 567.02e,

(6)

0 Fig. a

where

e, =cos118.14•i +sinl18.14·] = -0.47161 + 0.8818]

44

(7)

1.4 Planar Curvilinear Motion of Particles

(ii)

Approach 1: Polar coordinate system The time differentiation of position vector (2) yields:

- ar 500cosB ()• 500 ()• v = - = re +r e0 = - - - -2,...---- e, +-- e0 dt r sin () sin() •A

()•

A

A

A

(8)

Using (1) and (2), the radial comp.o nent of the velocity at t = 5 is:

vr =

500cos() sin 2 ()

iJ =

82 _54 mm/s

(9)

and the transverse component of the velocity at t = 5 is:

V0

5000 sin()

= - - = 154.37 mm/s

(10)

Both components are signed magnitudes and the velocity of P at t = 5 is:

(11) =

82.54 (cos 118.14 ° i +sin 118.14 ° }) + 154.37 (-sin 118.14 ° i +cos 118.14 ° })

= -175.05 i mmls which directs to minus x direction (Fig. b).

y

v,e, v

\/

\

p

voeo r

118.14° X

0 Fig. b

45

I'

Chapter I

Kinematics ofParticles

Approach 2: Path coordinate system Since the path of motion of P is known, we use path coordinate s as shown. Thus:

y

s=500cot()

(12)

s =-500-()-

(13)

0

v

sin2 ()

s = _500 Bsin 2 ()- 02 sin2()

500mm

r

sin 4

(14)

()

Att = 5,

°·

0

2722 s = -500 sin 2 118.14° = -175.03 rnm/s

Fig. c

v = -175.03t vr = v. er = 82.54 rnm/s V 0 = V · e0 = 154.37 rnm/s

Therefore, and

(iii) At t = 5, substituting (1)-(3) into (14):

y

s = 659.65 mm/s Hence, 500mm

a

2

= si = 659.65i

So,

ar=a·er= 0

(15)

ao

-31l.llrnm/s

= a.eo = -581.68 rnm/s

Fig. d

Discussion: For calculating acceleration in (iii), we can also use Equation 1.77 in polar coordinates, however this requires more work. In this example, since the path of the motion is given, it is preferable to use both the path coordinate and polar coordinate systems.

46

1.5 Motion Relative to a Point-Attached Translating Frame

1.5 Motion Relative to a Point-Attached Translating Frame Consider two particles A and B moving independently in the space 0-xyz (Fig. 1-17). Their motion can be described by their position vectors:

YA =xi+ yA}

z

= X 8 f + y 8 } +z 8 k

Y8

--

(1.83)

rB

and their respective velocities and accelerations are:

vA =xi +:YA}+iAk GA =xi +yA}+zAk

vB =xi +:YB}+iBk G8 =xi+ y 8 } +z 8 k

rA (1.84)

B

---

YBA A',

0

\

y

X

As discussed in Section 1.2.1, the relative position vector of particle B relative to A is:

Fig. 1-17

(1.85a) or

(1.85b)

Taking the time differentiation of (1.85b) twice gives: (1.86) and

(1.87)

Since

rBA

is the position vector of particle B relative to particle 2

A, d/dt(r8 A) and d /dr (r8 A) are the velocity and the acceleration of particle B relative to particle A respectively. However, it is Important to note that unlike relative position, a relative motion cannot be defmed without referring to a well-defined moving reference system. Particle A itself cannot serve as such a reference system unless a point-attached translating frame A- i}k /x 'y 'z ' is attached to it (Fig. 1-18a). (I) Velociy Once such a frame is assumed, r 8 A can be alternatively expressed in coordinates of A-x 'y 'z' as:

YBA =X'Bf+y'B}+z'Bk

(1.88)

47

Chapter I

Kinematics ofParticles

The relative velocity of particle B observed from A-xy'z' can be defined by the relative time derivative of YBA : (1.89)

Yl-- - - - y'

Equating Equation 1.85a to Equation 1.88 gives the coordinate transformation relations between 0-xyz and A-x y 'z ':

x'

0

Using A to represent the lengthy notation A-x y 'z ', the relative velocity can be simply expressed as v Bl A •

X

(a)

x'B

= XB

-XA

y'B

= YB- YA

z'B

= ZB

-ZA

(1.90)

Substituting Equation 1.90 into Equation 1.89 leads to:

-

-

-

(1.91)

VB/A =VB-VA

which states that the relative velocity of particle B seen from particle A-attached translating frame is simply the difference of the velocities of the two particles.

(b) Fig. 1-18

z

Equation 1.91 is frequently written as: (1.92)

''

which states that the velocity of particle B equals the sum of velocity of particle A and the relative velocity BIA (Fig. 18b). It needs to be emphasized again that in the subscript B/A, B represents particle B, the slash "/" reads "relative to", and A represents the particle A-attached translating coordinate system A-x )l'z ', not just a particle A.

v

z

0 X

(a)

Interchanging A with B, we may also have:

or

(1.93)

where in the subscript AlB, A represents particle A, and B represents the particle B-attached translating coordinate system B-x y 'z' (Fig. 19a, b).

(b) Fig. 1-19

48

Obviously, we have:

-

-

V AlB =-VB/A

(1.94)

1.5 Motion Relative to a Point-Attached Translating Frame

(2) Acceleration

Following the same logic in the preceding study of relative velocity, we define the relative acceleration of particle B relative to the translating coordinate system A-x'y'z' to be the relative time derivative of BIA :

aB

v

(j

A

aBIA =-(VB/A)= X'B i

lit

A

\

aBIA

\

' ... ... ... z

A

+ ji'B j + z'B k

z

(1.95)

---

Substituting Equation 1.90 into Equation 1.95, we can also have:

y

,

(1.96a) y

or

0

(1.96b) X

Equation 1.96a states that the relative acceleration of particle B seen from particle A-attached translating frame is simply the difference of the accelerations of the two particles (Fig. l-20a, b).

(a)

Similar to Equation 1.94, we have:

- = -aBIA -

(1.97)

a AlB

Remark: Equations 1.91 and 1.96a reveal that:

8 lit

c- )

8 lit

c- )

c- ) dt

- /A =- rBA =d- rBA VB

(1.98a)

Fig. 1-20

c- )

- IA = - VBA =d- VBA aB dt

(b)

(1.98b)

This implies an important conclusion: For translating frames and only for translating frames, the relative

differentiation of any differentiation of it:

vector

R

equals

to

the

absolute

(1.98c) This is indeed Equation 1.76 when angular velocity iiJ of the moving frame vanishes.

49

Chapter 1

Kinematics of Particles

Example 1-15 Motion of two points moving independently Particles A and Bare initially at the positions shown. Both have an initial speed 50 rnls in the vertical plane. If initial velocity is at 30° measured from the x-direction as shown, at what angle () must particle B be launched initially to guarantees the collision of the two particles? How long does it take before they collide? y

A

X

Fig. a

Solution: Let the initial velocities of A and B be expressed as:

VAO= 50L30° = 50( COS 30 of +sin 30o l) vB 0 = 50 L () =50(cos8i+sin8})

(1)

Then, the velocities of A and B can be expressed as functions of time t:

v A= 50cos30 oi

+ (50 sin30 o- gt) }

+ (50 sine- gt)

J

(2)

Examining the velocity of B relative to A:

-

-

-

(3)

VB/A =VB -VA = 50 [ccos()- cos30o) i +(sin()- sin30°)}] =

we find that it is a constant vector. It means that if

v

80 -

v

Ao

vBIA is in the direction towards A initially, it will keep

going towards A . Therefore the necessary conditions of the collision of the two balls can be given mathematically as:

(4)

50

1.5 Motion Relative to a Point-Attached Translating Frame

Using (3) in the first condition in (4) gives: cos f)- cos 30°

cos 150°

sinfJ-sin30o

sinl50°

(5)

from which we can solve the unknown initial angle (O 1 are both known:

iiJ f = 3k

(13)

we have Consequently: VB/f

= ii>AB!f xrBC

as shown in Fig. c.

B Fig. c

104

(12)

=-150 L-30° mm/s

2.4 Motion Relative to Body-Attached Rotating Reference

2.4.3

Principle of Acceleration Combination

As shown in Fig. 2-9, the motion of a rigid body f is described by VA and aA, (i.e. the Velocity and acceleration Of the reference point A), and OJ1 , and a 1 (i.e. the angular velocity and angular acceleration of the body). The velocity P and acceleration P of a particle P are also prescribed.

v

a

Similar to Equations 2.14 and 2.15, for the relative acceleration of particle P observed from the translating reference A-x'y', we have:

a c- )= dtd c-VPIA )

-

(2.24)

aPIA =at VPIA

and

(2.25)

(Refer to Section 1.5, Chapter 1).

To study the acceleration of particle P relative to the rigid body f, we may directly use the defmition of relative time differentiation (refer to Remarks in Section 1.4.3, Chapter 1):

c- )

- =8- VP/f QP/f

5t

'--------------------- x 0 Fig.2-9

(2.26)

Using the general relation in Equation 1.76:

c- )=8- c- )+OJ/ - -

-d VP/f dt

5t

VP/f

XV.Pif

we have:

(2.27)

Substituting

v PI[

= v PIA

-

i!J I

X rPA

into Equation 2.27 leads to:

(2.28) where:

-aI = a PIA -a I

=aPIA

X rPA

-wI

X rPA -

i!J I

x(vP/f

+i!JI

XV PI[ -

i!J I

X rPA)

X

(iJ I

X rpJ

(2.29)

105

Chapter 2

Kinematics of Rigid Bodies

Thus, we have: aP/f =aPIA -[af xrPA +mf x(mf xrPA)+2mf XVP/f]

(2.30)

Furthermore, substituting Equation 2.25 into 2.30 yields: aP/f =ap -aA -[af XYPA +mf x(mf XYpA)+2mf XVP/f] (2.31)

which can be rearranged as: ap =aP/f +[aA +af XYPA +mf x(mf XYPA)] +2mf XVP!f (2.32)

(Fig. 2-10)

'------'--- - - - - - - - x

0

Fig.2-10

With help of Equation 2.8, the physical meaning of the expression in the bracket of Equation 2.32 is identified as the absolute acceleration of the entraining point P ': (2.33)

which is named as the entrained acceleration for point P due to the motion of the reference frame f The

additional

term

2m1 x v Plf

Is

named

Coriolis

acceleration and denoted as: (2.34)

which is due to the coupling of entrained angular velocity f»J and relative velocity v Plf . Fig. 2-11

Thus, Equation 2.31 can be expressed as: (2.35a)

or

+

Absolute accel.

Relative accel. Fig. 2-11

106

+

Entrained accel.

a; J Coriolis accel.

(2.35b)

2.4 Motion Relative to Body-Attached Rotating Reference

which is known as the Principle ofAcceleration Combination:

The absolute acceleration of a point P is equal to the sum of the relative acceleration observed from the body-attached moving frame, the entrained acceleration, which is the acceleration of its coincident point P' on the body or on the extension of the body, and the Corio lis acceleration.

Remarks: The Principle ofAcceleration Combination (Equation 2.35) is the most general equation for relative acceleration of a point P observed from any moving frame f We can see that Equation 2.25 is a particular case of Equation 2.35 when the reference rigid body f is in translation (£0- = 0 and Of = 0) . In this case, any entraining point P' on the body has the same acceleration as that of point A: aP' = aA ' and the Coriolis acceleration vanishes.

Therefore, the equation iiPIA

a

Plf

= ap

-

a a; P' -

Fig. 2-12

can be reduced tO

=iip -aA as shown in Fig. 2-12. Obviously, the point A-

attached translating frame now coincides with a body-attached moving frame f, which happens to be in translation.

Example 2-9 Relative acceleration of a point observed from a body in fixed-point rotation Two slender bars AB and CD maintain contact with each other when bar CD rotates. In the position shown (Fig. a), bar CD has a clockwise angular velocity of 2 rad/s and a counterclockwise angular acceleration of0.2 rad/s2 • Determine the angular acceleration ofbar AB in this position.

B

200mm

400mm

Fig. a

107

Chapter 2

Kinematics of Rigid Bodies

Solution: This example is a continuation of Example 2-7, where we had obtained:

m1

=

(1)

0.933 rad/s (cw)

(2)

with

and

ar

To determine the angular acceleration of bar AB, we first study the acceleration of point D on bar CD and that of its entraining point D' on bar AB. Since both points are in circular motion, their acceleration can be expressed in terms of tangential and normal components as shown in Fig. b. The acceleration of point Dis known to be:

(3) with and

a;

2

= acDCD =

86.19 mm/s

=

1723.76 mm/s

=

2

The acceleration of point D' can be written as:

(4) with

= m}AD' = 402.06 mm/s2

and

n B

/t

\

B

A 2 rad/s

Fig. c

Fig. b

To relate ii D to ii D', we must apply the Principle ofAcceleration Combination

(5) or

108

-

a Dlf

= aD -

-

-c

aD' - aD

(6)

2.4 Motion Relative to Body-Attached Rotating Reference

Using (1) and (2), the Coriolis acceleration can be calculated as: (7) Here, the direction of

a_g

is determined by the right hand rule in cross product of the two vectors,

or by simply rotating vector

v

Dlf

90° in iiJ 1 direction (Fig. c and Fig. d). VD/f

(J)I

)

2 {

{2 A2-

Principle of Work and Energy

0 Newton's Second Law

u1 .... 2

F =rna= m dV dt

=T2

-IJ

is a differential governing equation of motion, which relates

force to the motion of particles at any instant. As linear momentum Z and angular momentum fi A about a fixed point A are defined as state functions of motion, Newton's Second Law appears in forms 2 and 3, which are also differential governing equations of motion valid at any instant. Integrating differential equations 2 and 3 over a time domain leads to two auxiliary principles 4 and 5, respectively. They are no longer differential equations but algebraic equations, relating linear/angular impulses to the changes in the corresponding vector state functions in the concerned time interval. Similarly, integrating differential equation 1 over a path in the spatial domain leads to another auxiliary Principle 6, which is an algebraic equation relating work to the change in kinetic energy during the motion. The kinetic energy is defined as a scalar state function of motion.

152

3.5 Conservative Forces and Conservation ofTotal Energy

3.5

Conservative Forces and Conservation of Total Energy

3.5.1

Potential of Conservative Force Field

Consider a particle moving in space 0-xyz. Suppose that there is a force acting on it at any position. The force can be gravity force, spring force, or some other type of force. We may say that the particle is moving in a force field. The field, e.g. gravity field, is not just a spatial domain but a field characterized by the presence of force acting on the particle. There are special force fields in which the force acting on the particle depends on the position of the particle only. It means that the force IS a function of position coordinates x, y, and z:

F= Fx(x,y,z)t + FY(x,y,z)] + F,(x,y,z)k.

In such a case, there

exists a scalar function V(x, y, z) (its unit is Nm, the same as unit of work), so that:

av

Fy = _

Ox

-

( av

oV F = _ oV oy ' &

av

av

F=- - - i + - - j + - - k Ox Oy

t.e.

oz

(3.45)

(3.46)

Such a force field is so special that it is known as a conservative force field. The scalar function V is defined as its potential energy function or

simply potentia/function; and the force is called a conservative force. Why do we use the word "conservative"? In this section, two force fields, gravitational and linear spring elastic force fields will be used as examples to show that the conservation of total energy holds for particles moving in a conservative force field.

3.5.2

y

Gravitational Force

The gravitational force acting on a particle of mass m is an "action in distance" conservative force (Fig. 3-12a), since there exists a gravitational potential function:

vg = mgy + constant

YI . ) W

w

(3.47) X

0 so that the gravity force (weight) can be derived as:

oy

J

-mg j

(a) Gravitational force is a conservative force

(3.48)

Fig. 3-12

153

Chapter 3

Kinetics ofParticles

The constant in Equation 3.47 is arbitrary, that means potential Vg depends on the choice of datum, where

vg = 0.

Work done by gravitational force from position 1 to position 2 is:

U 1-+ 2

=

! W ·eli= fy' (-mg)dy=mgy -mgy 2 -

y,

I

1

Path A

(3.49) (3.50)

i.e.

y

2

This equation implies that the work U 1-+ 2 is independent of the

path of the motion. As shown in Fig. 3-12b, a particle may move from point 1 to point 2 by either path A or path B. However, with the same end points, the work done by gravitational force is the same. Because of this conservative feature, the Principle of Work and Energy (Equation 3.31) takes a new form: (3.51) which can be rearranged as: X

0

(3.52)

(b) Work done by gravitional force from I to 2 is independent of the

paths.

or

Fig.3-12

1 2 1 2 mgy1 +- mv1 = mgy2 + - mv2 2 2

(3.53)

The sum of the potential energy and the kinetic energy of the particle is defined as total mechanical energy: (3.54) which serves as a state junction for describing the motion state in the sense of mechanical energy ofthe system. It is concluded that for a particle moving in a gravitational field, its total (mechanical) energy remains unchanged (in other words, conserved)

E 1 = E2

or

E = constant

(3 .55)

This is known as the Conservation of Total Energy:

In a conservative force field, a particle subjected to only the conservative force moves in such a manner that the total mechanical energy of the system remains constant all the time.

3.5 Conservative Forces and Conservation of Total Energy

3.5.3

Linear Spring Elastic Force

The elastic force of a linear spring is also a conservative force. For simplicity, we consider a one-dimensional mass-spring system as shown in Fig. 3-13. There exists a quadratic elastic potential function:

F=-kxt

(3.56)

where k is the stiffness of the spring and x is measured from the origin 0, which is chosen to be the position when the spring is unstretched (V, = 0). The spring force can then be derived from v. as:

F

0 1 2 = --(-kx )z = -kxz

zx

(3.57)

2

Q

.__...i...___ XI

X

X2

Spring force is a conservative force

Fig.3-13

Similar to gravity, work done by the elastic force only depends on the position of the end-points of the path:

U

=

1F ·ar = ix2 (-kx)dx 2 -

x,

I

=

1

2

1

2

-2 kxI --kx 2 2

(3.58)

(3 .59)

i.e.

Again, the Principle of Work and Energy (Equation 3.31) takes the fonn: (3.60) (3 .61)

or

Defining the sum of the potential energy and the kinetic energy of the particle as the total mechanical energy of the mass-spring system: I kx2 I . 2 +-mx E = V + T =e

2

2

(3 .62)

we again have the Conservation of Total Energy: E =constant

(3 .63)

for the conservative system.

155

Chapter 3

Kinetics ofParticles

Remarks: ( 1) Any force associated with a potential function in relation to Equation 3.46 is a conservative force. The total energy E is defmed for the system as consisting of both the moving particle and the conservative force field. The system is called a conservative system.

mg 1

y (a) Trend of driving illustrated by the slope of gravitational potential.

(2) A conservative force vector can always be expressed as the negative of the gradient of a scalar potential function (Equation 3.46). The gradient, which implies the trend of driving, can be graphically illustrated by the slope of the potential function. For the gravitational field near the earth, the slope of Vg is constant everywhere as shown in Fig. 3.14a. However, for the spring elastic force, the slope of v. changes from point to point (Fig. 3-14b). The negative sign in Equation 3.46 implies that the force should be pointing in the direction of the reducing potential function. (3) For a system of particles, all the forces acting on the particles can be divided into two categories: conservative forces and nonconservative forces. Let the total potential of the system be defined as the sum of the potential of all conservative forces, e.g. V = Vg + V• . The work done by all the conservative forces from

2

i

1

position 1 to position 2 is equal to the difference in total potential between the two positions:

x

F; = -kxJ F2 = -kx2 i (b) Trend of driving illustrated by the slope of spring elastic potential

Fig. 3-14

u 2

However, when blocks A and Bare considered as a system, the total work done by the tension forces of the inextensible cable vanishes:

U(T on A) 1-->2

+

u2

= 2Tx A

-

Tx B = 2Tx A

-

2Tx A

=0

(11)

That is why we can treat the constraint of the massless and frictionless pulley-cable system as an ideal constraint.

T

t Fig. b

159

Chapter 3

Kinetics ofParticles

(3) If the friction force acting on block A is not negligible (Fig. c) and the kinetic friction coefficient is given as 0.1, what is different?

0

Fig. c

In this case, the sliding contact constraint is no longer an ideal constraint because the friction force is a non-conservative constraint force:

(12)

Now we have to use Equation 3.64c:

u g tan(¢- B) block A will slide on the incline.

____.a

Fig. P3.14

I97

Chapter 3

Kinetics of Particles

3.15 A 2 kg collar P fits loosely on the inclined part of rod ABC, which rotates about the zaxis (Fig. P3.15). The coefficient of static friction between collar P and the rod is 14 = 0.2. If the rod rotates at a constant angular velocity, determine the maximum and minimum angular velocity it can have so that the collar remains in the position 0.6 m from B.

z c

---------- -------

R

B

(J)

0.4m

Fig. P3.15

Fig. P3.16

3.16 Particle P has a mass of 1 kg and is confined to move along the smooth vertical slot due to the rotation of the arm AB (Fig. P3.16). The rod is rotating counterclockwise at a constant angular velocity of w = 2 rad/s. Determine the force of the rod on the particle and the normal force of the slot on the particle in the position shown in Fig. P3.16. 3.17 A 0.5 kg ball P is guided to move along a circular path of r = 0.4 m in the vertical plane using an arm OC, which is pivoted at 0 and rotates at a constant angular velocity iJ = 0.4 rad/s counterclockwise (Fig. P3.17). At = 30° , determine the force of the arm on the ball and the force of the circular surface on the ball. Neglect friction and the size of the ball.

e

B

A

---a

r

Fig. P3.17

Fig. P3.18

3.18 A 2 kg collar C is free to slide along an inclined smooth rod AB, which is welded onto collar A. If the assembly is sliding along the horizontal shaft in the vertical plane, determine (a) the acceleration of the assembly necessary to maintain collar C in a fixed position on rod AB, and (b) the force acting on collar C by rod AB.

198

Problems

3.19 Block A has a mass of 10 kg and cylinder B has a mass of 5 kg (Fig. P3.19). Determine the acceleration of block A (a) if the surface at A is smooth and (b) if the coefficient of kinetic friction between block A and the surface is 14 = 0.1. Neglect the mass of the pulleys and cords.

'

YB

o··-B

Fig. P3.19

+---------· A

Fig. P3.20

0 3.20 Sliders A and B, which are connected by a light rigid rod of length l = 0.5 m, both slide in the slots as shown in vertical plan (Fig. P3 .20). A force F = 60 N is acting on slider A towards the right. Assume that when xA= 0.3 m, the velocity of slider A, vA = 0.8 m/s is to the right. Determine the acceleration of each slider and the force in the light rod at the instant. Neglect the frictions. 3.2

Principle of Angular Momentum

3.21 The moment of linear momentum vector of a particle about a fixed point is an analogue of the moment of a force vector about the point. True or false? 3.22 In Example 3-6, force T is the only unbalanced force acting on ball A. Since the moment ofT about 0 is always zero, the angular acceleration i.J must be zero. True or false? 3.23 A satellite is traveling along an elliptical path about the earth as shown (Fig P3.23). Given that the radius of the earth is 63 70 km, and the altitudes of the nearest point A and farthest point B of the satellite are 439 km and 2384 km, respectively, determine the ratio of the speed of the satellite at A and B.

A

B

tv=O.Itmls

Fig. P3.23 Fig. P3.24

199

Chapter 3

3.3

Kinetics o[Particles

Principle of Work and Kinetic Energy of a Particle

3.25 Kinetic energy, being a scalar function of time, describes the state of motion of a particle. True or False? 3.26 Work is the product of a force vector and a distance. Thus, it is along the direction of the force. True or False? 3.27 Linear momentum, angular momentum, and kinetic energy are three state functions of motion of a particle. Work can also be defined as a state function of motion. True or False? 3.28 Using the polar coordinate system, derive expressions similar to Equations 3.32 and 3.33 for a particle in a circular motion. Hint: Express Note

ar = (rdB)e 8 •

F = Frer + Feee

and

r = rer.

3.29 In Example 3-7, when we calculated the work U 1-.. 2 in equation (3), we did not count the work done by the tension force ofthe cable acting on block A. Why? Try to solve the problem alternatively using Newton's Second Law to block A. 3.30 Two crates of weight shown in Fig. P3.30 rest on a horizontal floor and are connected by a light cord. A horizontal force F = 18 N is applied to crate Band moves them to the right. The coefficient of sliding friction between the crates and floor is Jlk = 0.2. Determine (a) the work done by forces acting on each of the blocks as the force moves them 4 m to the right, (b) the work done by all the forces acting on the system (A + B) during the period, and (c) the speed of blocks as they move 4 m to the right. 30N 15 N

1---lf ",,$

.8 .N

Fig. P3.30

F

Fig. P3. 31

3.31 Two blocks A and B of mass 1 kg and 2 kg, respectively, rest on a horizontal floor (Fig. P3.31). They are connected by a frictionless cable-pulley system as shown. The coefficient of sliding friction for all surfaces is Jlk. A constant force F is applied to block B pulling it to the right. Determine the speeds of blocks A and B in terms ofF, pk, and x 8. Discuss the minimum force needed for moving the blocks. 3.32 Block A of mass 1 kg rests on an inclined surface of a 2 kg block B, which in tum rests on a horizontal floor (Fig. P3.32). They are connected by a frictionless cablepulley system as shown. The coefficient of sliding friction for all surfaces is Jlk = 0.1. A constant force F = 200 N applied to block B pulls it to the right. Determine (a) the normal forces at two contact surfaces, (b) the speeds of blocks A and B when block B movesx 8 = 0.1 m. (See Example 1-19 for the constraint equation).

200

F

Fig. P3. 32

Problems

3.4

Principle of Linear/Angular Impulse and Momentum

3.33 Linear impulse and linear momentum are both state functions of motion. True or False?

3.34 Linear/angular impulse is not an instantaneous value but a value that measures the action of a force over a time interval. True or False? 3.35 Particle P of mass m is in counterclockwise circular motion at a constant speed v on a smooth horizontal surface as shown in Fig. P3.35. Determine the linear impulse applied upon it during the period when P travels (a) from point A to point B, and (b) from point B to A.

p

y

v

1.2 m

,A

B

_....

I /

/

/

0.9m Fig. P3.35

0.9m

Fig. P3.36

3.36 A uniform beam of 25 kN is initially at rest as shown (Fig. P3.36). (a) Determine the average tension in each of the two cables AB and AC, if the beam is accelerated upward to a speed of 2.4 mls in 2 s. (b) Given that the ml!Ximum tension each cable can sustain is 25 kN, determine the shortest possible time when the beam reaches a speed of 5 mls. Neglect the mass of the cable. 3.37 A pendulum of mass m is released from rest at B = 0. Determine the angular impulse about point 0 applied upon it by all the forces during the period when P travels (a) from point A to point B, and (b) from point A---+ B---+ C---+ B (Fig. P3.37).

y:

! G)---------------Q_ : --- ----------------0------ X 8 ;A .j ' \ l I

.

C \

\,

\.,,,,

l r !

--------------Jr----------

,/

,//

p

iB Fig. P3.37

Fig. P3.38

3.38 The circular table in Fig. 3.38 is rotating at a constant angular velocity of OJ = 18 rad/s about the center. Block A of mass m with the string (R = 0.5 m) taut is lowered down onto the table with zero initial velocity. If the coefficient of kinetic friction between the block and the surface is Jlk = 0.3, determine the time before the block stops slipping.

201

Chapter 3

3.5

Kinetics of Particles

Conservative Forces and Conservation of Total Energy

3.39 A conservative force vector can always be expressed as the gradient of a potential function defined in a force field, therefore: (a) the potential function must be a constant. (b) the potential function must have a constant slope against the position variables. (c) the potential function is a scalar function of position variables. (d) the potential function is a vector function of position variables. 3.40 Potential energy of a particle in a conservative force field at a given point is equal to the work done by the conservative force acting on the particle from the given point to the datum, at which the potential is zero, regardless of the path. True or False? 3.41 A block of mass m = 1 kg is attached to two identical springs of negligible mass as shown in (Fig. P3.41). Each spring has a stiffness of k = 2 Nm· 1 and a natural length of 0.6 m. The block is released from rest in the position shown. Determine the maximum speed of the particle in the subsequent motion. Ignore friction and the size of the block.

1.5 m

1.5 m

h 0.35 m

0.65 m

Fig. P3.41

Fig. P3.42

3.42 If the 10 kg cylinder in Fig. P3.42 is released from rest at h = 0, determine the required stiffness k of each spring so that its motion momentarily stops when h = 0.5 m. Each spring has an unstretched length of 1 m. 3.6

Conservation of Linear/Angular Momentum

3.43 A wagon with a mass M can move without friction along horizontal rails. A simple pendulum with a mass m suspended from a long string l is fastened on the wagon. At the position shown in Fig. P3.43, the wagon and the pendulum are both at rest and the pendulum is released at an angle Bfrom the vertical line. What will be the velocity of the wagon when the pendulum string passes through the vertical line? Assume that m .

/r)

)

t

t

Kinetic diagram

Free body diagram

Fig. c

221

Chapter 4

Kinetics of Rigid Bodies

According to the equivalency of the two vector systems, we can establish three independent scalar component equations:

T cos45.

=

cos45.

(5)

T- mgcos45. =-maL cos45"

(6)

4

(7)

Since in (6) and (7) only two unknowns T, a are involved, we can solve the two simultaneous equations for:

T

= 4 -.fi mg 11

and

12g 11L

a=--

Discussion: Alternative scalar equation based on moment equation about any point Instead of using equation (6), which takes the components of vectors in direction to set up the scalar component equation:

t

Tcos45"- mg =

n direction, we may choose }

cos45"- maL 4

(8)

However, it is not a good choice because there are three unknowns (T, and a) in (8) and we have to deal with three simultaneous equations (5), (7),_ and (8) for solving them. To simplify the calculation, we prefer to have one equation with only one unknown if possible. In some cases, it is possible. For this example, to prevent unknowns T and from appearing in the equation, we take the moment about C, which is the intersection of the two vectors as shown in Fig. d.

L T -

L

8

G l

l

}

J

L

ma4

t Free body diagram

Kinetic diagram

Fig. d

222

4.3 Application ofNewton's Second Law to General Planar Motion of a Rigid Slab

Thus, we successfully establish a moment equation about point C:

(1 1)

2 L L L mg-=ma-(-)+laa = - + - mL a

8

4 8

in which there is only one unknown a, and a=

12 32

(9)

12 g . IlL

Based on the equivalency of the vector systems, any direction may be used to set up a scalar component equation from

F = maG, and any point in the plane may be used to set up a moment equation based on

Equation 4.43. However, there are only three independent equations among them for a rigid body in planar motion. Experience is needed to make the right choice to simplify the calculation.

223

Chapter 4

Kinetics of Rigid Bodies

Example 4-5 Application of Newton's Second Law to a Rigid Slab in General Planar Motion A spool (Fig. a.) has a mass of 50 kg and radius of gyration 0.2 m. A cable wrapped around the central hub of the spool is pulled by a force 100 N. Knowing that the spool rolls without slipping on the horizontal floor, determine the angular acceleration of the spool and the friction force.

Solution:

lOON

Kinematic Analysis (Fig. a): Relating a G to a (Example 2-3), we have: aG = 0.45a

aG =0.45a

J

(1)

l

Kinetic Analysis (Fig. b):

c

Kinematics diagram

Draw the free body diagram and kinetic diagram as in Fig. b; there are three scalar unknowns: a, F1 , and N.

Fig. a

Here F1 is assumed to be to the right.

lOON maG =

C' .

0.45ma

C' N

Free body diagram

Kinetics diagram

Fig. b According to the equivalency of the two vector systems, we can establish three independent scalar component equations as follows:

-------

t a!) 224

100 + F 1 = 0.45ma

(2)

mg-N=O

(3)

100(0.3)- F 1 (0.45) = ! Ga

with

IG

=50(0.2 2 )

(4)

4.3 Application ofNewton's Second Law to General Planar Motion of a Rigid Slab

Adding (2) multiplied by 0.45 to (4) gives: 100(0.3 + 0.45) =[50( 0.45 2 ) + 50(0.2 2 ) ]a

a = 6.186 rad/s2

(5)

(6)

So, from (2) we obtain:

(An alternative approach for this problem will be presented in Example 4-14.)

Discussion: (a)

Direction of friction force In the above solution, we assumed correctly the direction (to the right) of the friction force since the solution we obtained for F1 is a positive value. However, our assumption could have been wrong

because the direction of the friction force at the point of rolling without slipping cannot be determined before the governing equations have been solved. For example, if the radius of gyration of the spool is 0.4 m instead of 0.2 m, equation (5) becomes: 100(0.3 + 0.45) = [ 50(0.45 2 ) +50( 0.4 2 )]a

Then the friction force is calculated to be:

a = 4.138 rad/s

F1 =- 6.897N

2

(7)

(8)

The negative sign means that the actual direction of the friction force is to the left. Our assumption has been proven to be wrong. It is interesting to note that the radius of gyration kG plays a crucial role in the direction of the friction force. For this example, we can also determine a critical value for the radius of gyration ( kgr) = 0.367 m), at which the friction force becomes zero. If kG> 0.367 m, then the friction force points to the left.

(b)

Can we use M c =Ica in this problem? Alternatively, based on the equivalency of the two vector systems shown in Fig. b and to avoid F1 in the moment equation, we use the moment about point C to set up the moment equation:

(9) which is exactly the same as equation (5). It can also be written in the form: (10)

Obviously, since point C' belongs to the disk and acceleration of C' is parallel to

rGC, Equation 4.47 is

valid for this special case (Fig 4.11 c).

225

Chapter 4

Kinetics of Rigid Bodies

4.4

Principle of Angular Momentum of a Rigid Slab about a Fixed Point

Let us define the angular momentum of a slab in general planar motion about a point C ftxed in the plane to be the sum of the moments of linear momentum about C of all mass elements

(M; =

):

y

i=l

Substituting

"i;e = F;G+ rGe

and

vi= v

G

+

ifj

X

F;G into it, we can

easily show (Fig. 4-12a):

[He

L_______________ X 0

xmVc +leW

]

(4.49)

which is the moment of linear momentum mVG about C plus the Angular momentum about a fixed point C for a slab in planar motion (a)

relative angular momentum about mass center

HG=liiJ

(See Appendix D3). Differentiating both sides of Equation 4.49 with respect to time:

d H- e = r-Ge dt

-

- + I Ga-

X maG

(4.50)

The right-hand side of Equation 4.50 is identical to the right-hand side of Equation 4.43. Hence:

-

d dt

(4.51)

M e =-He

which is known as the Principle of Angular Momentum of a Rigid Body about a Fixed Point.

0 Angular momentum about a fixed point C of a slab in pure rotation

Note that Equation 4 .51 is valid only when Cis ftxed. If C is a moving point, then time differentiation of both sides of Equation 4.49 may not lead to Equation 4.50 and then Equation 4.51 is not true. See Appendix E. Only for special cases when the slab is in pure rotation about ftxed point C, i.e. C and C' are both fixed (Fig. 4-12b ):

(b)

(4.52a)

Fig. 4-12

So

which is Equation 4.47. 226

]

(4.52b)

4. 4 Principle ofAngular Momentum of a Rigid Slab about a Fixed Point

Example 4-6 A rigid slab in pure rotation about a fixed point A uniform slender rod of mass m and length L is pivoted at A in the position as shown (Fig. a). The rod is released from rest. Determine the initial angular acceleration of the rod and the constraint force at A.

1

2

Ua =12.mL ). A

r

L

t

4

)

L

aG =a4

Fig. a

Solution: Kinematic Analysis (Fig. a):

Letting a be the angular acceleration of the rod. We have:

aL

(1)

aG=4

A

Kinetic Analysis (Fig. band c):

Use moment equation about the fixed point A: (2)

I.e.

L

L

4

2

mg-= I Aa = [IG + m(-) ]a 4 4 1 12

mg

L

Free body diagram

Fig. b

L

= [-mL2 + m(-) 2 ]a 4

12g

and

a=--

7L

Next, using the governing equation

aG

F = maG :

3 =-g 7 A

)

(

3 7

mg-FJ =m(-g) Kinetic diagram

Fig. c

227

Chapter 4

Kinetics ofRigid Bodies

4.5 Principle of Work and Kinetic Energy for a Rigid Body We now study the motion of a rigid slab in planar motion from the viewpoint of work and kinetic energy. 4.5.1 Kinetic Energy of Rigid Bodies

y'

The kinetic energy of a rigid slab consisting of N elements tvn, (Fig. 4.13), is defmed as the sum of the kinetic energy of all the elements: (4.53)

Using (4.54)

So,

L_______________ X

0 Fig. 4-13

Since G is the mass cente, the second term in Equation 4.55 vanishes: N

N

N

i= l

i=l

i=l

L:(&niVi/G )= L(Lim;aJ X r,G) = aJ X L:(Lim;F;a)= 0

(4.56)

The first term in Equation 4.43 reads:

N(l-Lim v 2) L . 2 I

1=1

G

=

1 2 -mv 2 G

(4.57)

and the third term can be written as: (4.58)

So, the kinetic energy of the slab can be defined as:

1 2 1 2 T = -mva + -]G{J)

2

2

which consists of two components:

228

(4.59)

4. 5 Principle of Work and Kinetic Energy for a Rigid Body

- - y' vG =wxrGC

(a) The kinetic energy of the body due to the entrained translation together with G, which equals the kinetic energy of "particle G" with total mass m.

(b) The kinetic energy of the body due to the relative rotation about the mass center G.

y

If a slab is rotating about a fixed point C as shown in Fig. 4-14, (or about an instantaneous center of rotation C,) we have vG = i!J X rGC. The kinetic energy of the slab can be expressed as:

1 2 2 1 2 T=-mrGc w +-1Gw

2

2

1(mr

2 GC

=-

2

+1G ) w 2

=

1 Cw 2 -1 2

(4.60)

Fig. 4-14

Example 4-7 Kinetic energy of a rigid body in general planar motion

A uniform disk of mass M = 1 kg and radius r = 0.1 m rolls without slipping on a horizontal surface. The center of the disk 0 is pin-jointed to one end of a uniform rod of mass m = 0.5 kg and length I= 0.3 m. The other end of the rod rests against a smooth vertical wall. At the instant shown in Fig. a, the center of the disk has a velocity v = 1 rn/s and the angle of inclination of the rod is B = 45°. Determine the total kinetic energy of the system at this instant.

Solution: Kinematics Analysis:

Disk:

v = 1 rn/s

wD

v

1

r

0.1

= - =-

=

10 radls

Rod: Using the instant center C,

v

1 0.3cos45"

WR

=- =

VG

= WRrGC = 4.714(0.15) = 0.7071

roc

=

4.714 radls

rn/s

Fig. a.

229

Chapter 4

Kinetics of Rigid Bodies

Total kinetic energy of the system: Total K.E. of system = K.E. of disk + K.E. of rod. Disk: K.E. of disk = Translational K.E. with 0 + Rotational K.E. about 0

=..!..Mv +..!_I a/ =..!..Mv +..!..(..!..Mr )mD 2 2°D2 22 2

2

2

2

2

2

= ..!..xlxl +..!..(..!..xlxo.e)x l0 = 0.75 J

2

2 2

Rod: K.E. of rod = Translational K.E. with G + Rotational K.E. about G 1 2 1 2 1 2 =lmVG +2/G(J)R =lmVG 1 =2

X

0.5

X

1 2) 2 +21(Urn/ (J)R

1 ( -1X 0.5 0.7071 2 +2 12

X

0.3 2 ) 'X 4.714 2 =0.167 J

T= 0.75 + 0.167 = 0.917 J

Total K.E. of system:

Discussion: Calculation of kinetic energy using instantaneous center of velocity Alternatively, when we calculate the kinetic energy of the rod, we may assume that at that instant, the rod is rotating about the instant center C. Therefore: K.E. of rod = Translational K.E. with G + Rotational K.E. about G =

Rotational K.E. about the instantaneous center C 1

2

= -JCOJR =

2

230

-(J G + mrGC 2 1

2

2 [12

2112

(2I J2] OJR

)wR =- -mf + m -

2

=

0.167 J

4.5 Principle of Work and Kinetic Energy for a Rigid Body

4.5.2

Principle of Work and Kinetic Energy for a Rigid Body or a System of Rigid Bodies

The Principle of Work and Kinetic Energy for a particle can be logically extended to that for a rigid body: (4.61)

Here T1 and T2 are calculated by applying Equation 4.59 to the rigid body in state 1 and state 2, respectively:

1 2 1 2 T2 =-mvG2 +-1Gm2

2

2

(4.62)

The work U 1_. 2 is the work done by all the forces including internal and external forces acting on the body. However, for a rigid body there is no net work done by the internal forces. Why? Fig. 4-15

Consider any two particles of the rigid body, i and j, as shown in Fig. 4-15. By Newton's Third Law, the forces of interaction between the particles exist in equal but opposite, and collinear pairs. The total work done by these forces on the particles in any small displacement is zero: (4.63)

d(J

Here diiJ is the displacement of particle i relative to j. Since in the rigid slab particles i and j are always a fixed distance apart, arij must be perpendicular to the line of action of the forces joining the two particles. Therefore, the dot product in Equation 4.63 is zero. Another way of looking at this is that both particles must move the same displacement along the line joining them (to keep a fixed distance apart), and since the forces are in opposite directions, the work done by one is cancelled out by the other. We can now conclude that U 1_. 2 is the work done by all the external forces only, including all applied forces and constraint forces (Fig. 4-16a):

(4.64a)

M.J

y

'---" ' ,...__ [!]

X

0 (a) Fig. 4-16

231

Chapter 4

Kinetics of Rigid Bodies

which is the sum of all the work done by each individual external force and couple on the body during the motion. Note that the work done by a couple (pure torque) is

Jo,

M 1 dB. (See

Appendix F). The total work done may be calculated alternatively by using the equivalent resultant force (Fig. 4-16b):

y

F and resultant torque MG

(4.64b)

(b)

where the first term is the work. done by the resultant force through the path of the mass center G and the second term is that by the resultant torque through the angular displacement of the slab from state 1 to state 2.

Fig. 4-16 If two or more rigid bodies are connected in such a way that the internal forces between the bodies do no work during the motion (e.g. bodies connected by pin joints), we may apply the Principle of Work and Kinetic Energy to the system as a whole. In this case work U 1-+ 2 is the sum of the work done by all the external forces acting on the system, and the kinetic energy of the system T 1 (or T2) is the sum of kinetic energy of all the bodies (Example 4-9).

232

4.5 Principle of Work and Kinetic Energy for a Rigid Body

Example 4-8 Principle of Work and Energy for a rigid body A spool as shown has mass 50 kg and radius of gyration ka = 0.2 m. A cable wrapped around the central hub of the spool is pulled by a force 100 N. The spool is initially at rest and rolls without slipping on the horizontal floor. Determine the angular velocity of the spool after the center of the spool G has moved 2 m to the right.

Solution: This is the same as Example 4-5 except we are now asked to determine the motion state after G has traveled though x=2m. Using (4.51) and the relation Ul->l = (100

0.45

lOON

have:

+ F1)x + [100(0.3)- Fj(0.45)]B

X

"'

j = 100(2) +

0.45

= 333.33 J l

Note that the work done by friction force F1 due to the constraint of rolling without slipping vanishes. Knowing that T1 = 0 and the kinetic energy at x = 2 m is: 1 1 1 1 T2 =-m(0.45lV) 2 +-lalV 2 =-(50)(0.45lU) 2 +-(50 x 0.2 2 )lV 2 2 2 2 2 ..!_(50)(0.45@ f +_!_(50 x 0.2 2 )lV 2 = 333.33 2 2

we have:

lV =7.8 rad/s

Example 4-9 Principle of Work and Energy for rigid bodies

A double pulley consists of two wheels, which are attached to one another and tum as one body. The pulley has a mass of 15 kg and a radius of gyration k 0 =110 mm. A 40 kg block A is originally at rest. A constant force F is applied to the rope, so it gives the block a speed of 0.4 m/s when A moves 0.6 m. During the motion, there is a friction torque A1j of magnitude 50 Nm acting on the axle of the pulley. Determine (i) the constant force F, and (ii) the tension force acting on block A during the motion.

Solution:

1-----.-0

Letting s be the distance moved by the block A, d the distance moved by force F, and e the angle of rotation of the pulley, we have:

s= 0.6 m

vA

75 d = 0.6 ( - ) = 0.225 m 200

= 0.4 rn/s

(} =

d

06 · = 3 rad

0.2

v lV =_A_= 2 rad/s 0.2 233

Chapter 4

Kinetics of Rigid Bodies

(i) Considering the system as a whole, the work done by all the external forces equals the change in the total kinetic energy of the system: 2 1 Fd- (mg )s-M/) = mvA

2

1 2 +21 0m

F(0.225)- (40)(9.81) 0.6- 50(3) = _.!_( 40)0.4 2 +_.!_[ (15)(0.11 2 ) )2 2

2

2

F= 1729N (ii) Considering block A, the work done by the external forces equals the change in its kinetic energy

(T -mg)s =_!_mv3 2

[T -( 40)9.81 )0.6 = _.!_( 40)0.4 2

' 234

2

T=397.7N

4. 5 Principle of Work and Kinetic Energy for a Rigid Body

4.5.3

Conservation of Total Energy for Systems of Rigid Bodies with Ideal Constraints

In Section 3.5.4, we discussed the conservation of total energy for conservative systems of constrained particles. The concepts and the method can be logically extended and applied to conservative systems of constrained rigid bodies. Consider a system of rigid bodies subjected to conservative and non-conservative forces. The Principle of Work and Energy for the system reads:

U

1->2

= u(nan-con) + u(con) = T _ T 1->2 1->2 2 1

(4.65)

If all applied forces acting on the system are conservative, and all the constraints involved in the system are ideal ones, then u(non-con) = O· 1->2

.

Frictionless rotational pin joints

y (4.66a)

c

t.e.

E2

= E1

=constant

(4.66b) Frictionless sliding guide

So, the total energy of the system is conserved.

(a)

For example, in the system shown in Fig. 4-17a, the frictionless rotational pin joints (at A , B, and C) and the frictionless sliding guide are typical ideal constraints. Since the applied forces are gravitational forces, the total energy of the system is conserved. For a rigid body, the potential energy in a gravitational field equals the potential of a point mass at its mass center:

mg

(4.67)

y

where y G is the height of the center of mass above the chosen datum 0. In Fig. 4-17b, a disk rolls without slipping on an inclined surface. The weight mg and the spring force Fs can be considered as conservative applied forces. The normal force N and the friction force F1 are constraint forces due to the constraint of rolling without slipping at point C. It can be easily understood that the normal force N does no work during the motion. That the work done by F1 is zero is not obvious. However, it can also be proven (Example 4.8). Hence, the constraint of rolling without slipping is also a typical ideal constraint. The total energy of this system is conserved.

'I 0 (b)

Fig. 4-17

235

Chapter 4

Kinetics of Rigid Bodies

Example 4-10 Conservation of Total Energy of Conservative System A 10 kg and 0.4 m slender rod AB moves in a vertical plane. It is constrained such that the two pin-connected sliders A and B at its ends move along the grooved slots as shown. The spring, which connects slider B to a fixed point Cas shown, has an original unstretched length 0.2J2 m and stiffness k = 800 N/m. The system is released from rest in the position when B = 45· (Fig. a). Determine the angular velocity of the rod when it passes the horizontal position (Fig. b). Neglect friction and the mass of sliders A and B.

())

0.2.J2 m

Fig. b

Fig. a

Solution: The applied forces acting on the rod, gravity force and spring force, are both conservative forces. The constraints of smooth sliding contact are ideal constraints. So the system is conservative in total energy. Let the gravitational potential be zero when the rod is in the horizontal position. At this position, the potential of the spring force is also zero. Thus, we have:

Total energy at initial position:

El

= Vgl + Vsl + Tl =

10(9.81)(0.2sin45. )+ _!_800(0.4-0.2.J2) 2 +0

2

Total energy at the horizontal position:

E 2 = Vg 2 + Vs2 + T2 Equating E 1 to E 2 gives:

=

0 + 0 + T2

T2 = 19.363 J

When the rod is at the horizontal position (Fig. b), point A is the instant center, so:

T2

236

=_!_! w 2 2

A

=

_!_(_!_10 x 0.4 2 +10 x 0.2 2 )w 2 = 19.363 2 }2

w

=

8.52 rad/s

4. 5 Principle of Work and Kinetic Energy for a Rigid Body

Example 4-11 Conservation of Total Energy for Conservative System

Four identical uniform rods, each of length I= I m and mass m = 0.5 kg, are connected at the frictionless pins A, B, C, and D. A spring of spring constant k = 500 N/m connects pins A and D, and a block E of mass M = 3 kg is supported at pin D . The conservative system is released from rest at = 45°. If the spring is unstretched at this configuration, determine the maximum deflection x of the block E.

e

Solution: This is a system of 5 rigid bodies, 4 of them are identical rods and one block E can be modeled as a particle. The system is conserved in total energy. Notice that point G, the midpoint of AD, is the mass center of the 4 rods. When D moves a distance x o

downward as shown, G moves a d!Stance

X

2 downward. So, it is easier to calculate the potential energy of the gravity forces acting on the four rods together rather than individually.

X

2

c

Since at the initial position and the final position the system has zero velocity,

Letting the gravitational potential of the system be zero when x spring is also zero, thus we have E 1 = 0, and:

X

I

2

2

=

0, at this position the elastic potential of the

2

- 4(0.5)(9.8I)-- 3(9.8I)x + -(500)x = 0

x=O.I57m

237

Chapter 4

Kinetics of Rigid Bodies

4.6 Principle of Impulse and Momentum for a Rigid Body In Section 3.4, we studied the principles of impulse and momentum for particles. The concepts and principles for particles can be logically extended and applied to rigid bodies.

Consider a moving rigid slab that is subjected to a system of external forces/torques (Fig. 4-18a). The total linear impulse and angular impulse acting on the system from t 1 to t 2 are defined as the sum of all the impulses by the external forces and torques: (i)

Total linear impulse: (4.68a)

y where F is the resultant force. (ii) Total angular impulse about mass center G:

L________________ 0

X

c (4.68b)

(a)

where M G is the resultant moment about moving mass center G.

(iii) Total angular impulse about point C fixed in the plane:

=

f

, M-c dt

(4.68c)

/I

where M c is the resultant moment about fixed point C.

(1)

Principle of Linear Impulse and Momentum for a Rigid Body

- dL

Using the Principle of Linear Momentum: F = - m

dt

Equation 4.68a yields: (4.69)

(b) Fig. 4-18

238

(Fig. 4-18b ). The principle states that:

4. 6 Principle ofImpulse and Momentum for a Rigid Body

fidt I

(2)

Principle of Angular Impulse and Momentum about the Mass Center of a Rigid Slab

y

- di! Next, using M G = __G_ in Equation 4.68b gives: dt

J (ang )G (4.70) (Fig. 4-18c ). The principle states that:

X

(c) Fig. 4-18

(3)

Principle of Angular Impulse and Momentum about any Point C Fixed in the Plane . E quatwn . 4.51, M-c U smg

JCang)c

= fz dH c df 1

dHc m . E quatwn . 4. 6 8c gtves: . dt

=--

dt

=ifC2 -ifCl

(4.71)

where

He= rGC X mvG + IGiiJ

(Equation 4.49).

This is the principle of angular impulse and momentum about a fixed point for a rigid body in general planar motion (translation and rotation). For special cases when the fixed point C belongs to the rigid slab, we have

if c

1-(ang)C-

= I c i.J. Thus:

fz dHc dt -I

dt

J - -J C{J)2

C{J)I

(4.72)

239

Chapter 4

Kinetics of Rigid Bodies

This is the principle of angular impulse and momentum about a fixed point for a rigid body in pure rotation (Section 4.4 Fig. 4-12b).

(4)

Principle of Conservation of Momentum for a Rigid Body or a System of Rigid Bodies •

Conservation of Linear Momentum For a rigid body or a system of rigid bodies, the total

linear momentum I is conserved (i.e. constant) in all or certain directions under the following conditions:

(i) If /I-+ 2 = 0, i.e. no resultant external linear impulse acts on the system, then:

; -I2

-'4 -

N

I=

t.e.

Z:m;liG;

=constant vector

(4.73)

i= l

where N is the number of bodies.

(ii) If 11..... 2

•

e= o, i.e. there is no component of resultant

external impulse in a direction

I .

t.e.

e, then:

e = constant scalar

(4.74)

Note that wen eed only consider external forces because internal forces that act between the bodies within a system cancel each other out.

•

Conservation of Angular Momentum

For a rigid slab, the total angular momentum about its mass center G is conserved if no resultant external angular impulse about G acts on a rigid slab. Then: fiG = I Gi}j = constant vector

(4.75)

For a rigid slab or a system o frigid slabs, the total angular momentum about a fixed point C is conserved if =

0, i.e. no resultant external angular impulse

about C acts on the system: N

He =Z:(rG,c xmvGi +IG;m;) =constant vector i =l

where N is the number of bodies.

240

(4.76)

4. 6 Principle of Impulse and Momentum for a Rigid Body

Example 4-12 Application of the Principle of Impulse and Momentum to a rigid slab in general planar motion

A spool (Fig. a) has mass 50 kg and radius of gyration lea= 0.2 m. A cable wrapped around the central hub of the spool is pulled by a force F = 20t N . The spool is initially at rest and rolls without slipping on the horizontal floor. Determine the angular velocity of the spool when t = 5.

Solution: Unlike Example 4-5, in this problem the applied pulling force is a function of time and we are only interested in the motion state after a period of 5 seconds. Kinematics Analysis (Fig. a):

Relating v G to

{J),

F=20t

we have: (1)

VG

= 0.45{i)

1

Kinetics Analysis (Fig. b):

First, we draw the free body diagram showing the impulses of all forces during the interval of 5 s. Next, draw the kinetic diagram for state 2 and initial state 1 showing the linear momentum and angular momentum about G.

c

l

Kinematics Diagram

Fig. a

c

c

N(5) Free body diagram (Impulses in the interval)

Linear momentum and angular momentum at state 2

Linear momentum and angular momentum at state 1

Kinetics diagram

Fig. b

241

Chapter 4

Kinetics of Rigid Bodies

Based on the Principle of Linear/Angular Impulse and Momentum, we have the equivalency of the two vector systems in Fig. b. Since we are only interested in determining m2 , to avoid the two unknown impulses by the constraint forces F1 and N, we take moment about C to set up the moment equation:

(0.75)

Knowing

IG

=

J: 20tdt =

I GOJ 2 + [m(0.45) OJ 2 ](0.45)- 0

= 50(0.2 2 )

OJ 2 =

(2)

2

kgm and m =50 kg, we have:

15.46 rad/s

Discussion: Comparing this example with Examples 4-5 and 3-8, we can see that the Principle of Impulse and Momentum is nothing new but an integration over a time interval of the differential equations of the Newton's Second Law (see the Remarks in Section 3.4). So the procedure for problem-solving using the principle is similar to those studied before. In this example, the friction force F1 is also a function of time. To determine it, we have to use the conventional method of Newton's law. Note that the friction force will reach a maximum value Jl.sN at a

critical timet" when the spool starts rolling with slipping. When t > r·, the friction force becomes a constant f-lkN. In this example, we assume that Jl.s is big enough that the critical time r· is longer than 5 seconds, such that during the period, the spool remains rolling without slipping.

242

4. 6 Principle of Impulse and Momentum for a Rigid Body

Example 4-13 Application of the Principle of Angular Impulse and Momentum to a rigid slab in fixed-point rotation A uniform slender rod of mass m = 12 kg and length L = 2 m is suspended at A in the position as shown. The rod is initially at rest when a horizontal force 1000 N is applied at B in a very short duration of 0.03 second. Determine the maximum angle to which the rod swings after the impact.

1

(I G = U mL2 ). Neglect

friction.

Solution: Let w1 be the angular velocity of the rod right after the impact. Applying the Principle of Angular Impulse and Momentum about point A, we have:

(1)

Here:

/ 0-+ 1 =1000(0.03)=30Ns

B

and

Hence:

w1 = 30 (1. 5) 16

=

2.8125 rad/s

(2)

Assume that the rod picks up the angular velocity in 0.03 s without any angular displacement. After the impact, the system is conservative in total energy. Letting h be the maximum vertical displacement of the mass center after impact, we have:

1 h = .!..(16)(2 .8125) 2 = 0.5376 m 2 12(9.81)

(3)

So the maximum angle to which the rod swings after the impact is:

¢ = cos -1

(1-h)

= 62.46°

243

Chapter 4

Kinetics ofRigid Bodies

Example 4-14 Application of the Principle of Conservation of Momentum for rigid bodies

Two identical balls, A and B, of mass 2 kg each, are attached to the ends of a massless rod of length 0.8 m. Both balls are initially at rest. At the instant shown in Fig. a, a ball C of mass 0.5 kg that is moving at a velocity I 0 i mls collides with ball A. If ball C rebounds at a velocity of -5 i mls, what are the velocity of the mass center G and the angular velocity of the rod? Assume that the size of the balls is negligible.

C

I 0 m/s

A

0

E

5 m/s

0.8m

B Fig. a

B Fig. b

Solution: Kinematic Analysis (Fig. b)

After the collision, the rigid body AB has both translational and rotational motion. It has a velocity G at its center of mass G and an angular velocity iii .

v

Kinetic Analysis (Fig. c)

Considering rigid body AB and ball C as a system, we can apply the Principle of Conservation of Linear Momentum and Angular Momentum about G because there are no external forces acting during the collision. The only forces that act are the collision forces between the bodies of the system. These internal forces always act in equal but opposite pairs and therefore cancel each other out.

244

4. 6 Principle of Impulse and Momentum for a Rigid Body

mc vc 0 = 0.5 (10)

c

c

A

---o

--o

A

G

B

B

Initial momentum diagram

Final momentum diagram

Fig. c

Conservation of linear momentum:

0.5 ·10 = -0.5 · 5 +4v0 VG

= 1.875

Conservation of angular momentum about G:

- 0.4 · 0.5 ·1 0 k = 0.4 · 0.5 · 5 k- (2 · 2 · 0.4 2 {i)

v0

)m k

= 4.6875 =

1.875f m/s

iiJ = -4.6875k rad./s

245

Chapter 4

Kinetics ofRigid Bodies

4. 7 Application of D' Alembert Principle to General Planar Motion of a Rigid Slab In Section 4.3, the two governing equations of Newton's Second Law: (4.7la) (4.7lb)

and

were illustrated as the equivalency of two vector systems shown in free body diagram and in the kinetic vector diagram, respectively (Fig. 4.19a and b).

-L_______________

X

0

(b) Kinetic vector diagram

(a) Free body di agram (System of external forces/torques)

y

0

0

(c) Slab is in dynamic equilibrium when imaginary inertial force and torque are added onto the mass center.

Fig. 4-19

246

X

0

G

X

(d) The slab is stationary observed from the bodyattached non-inertial frame

4. 7 Application of D 'Aiembert Principle to General Planar Motion of a Rigid Slab

Following the idea discussed in Section 3.8, let us introduce an imaginary inertial force: (4.78a) and an imaginary inertial torque about G: (4.78b) Adding each side of Equation 4.78a to Equation 4.77a, and Equation 4.78b to Equation 4.77b gives: (4.79a) (4.79b) respectively. The two equations are now in form of dynamic equilibrium. It can be illustrated by Fig. 4-19c and d. As shown in Fig. 4-19c, the force system including all the external forces/torques and the imaginary inertial force/torque is in "equilibrium. " There is NO vector shown in the kinetic diagram (Fig. 4-19d). This means that the body is stationary relative to the non-inertial body-attached reference frame . Similar to what we studied in Section 3.8, we then have the D' Alembert Principle for dynamic equilibrium of a rigid body:

Observed from a moving reference attached to a rigid body, the rigid boliy is relatively stationary. lt is then said to be in dynamic equilibrium. The resultant force and moment of all the forces/torques including the real external ones and the fictitious inertial force and the inertial toque acting at the mass center vanish.

247

Chapter 4

Kinetics of Rigid Bodies

Example 4-15 Application of D'Aiembert Principle to a rigid slab in general planar motion

Solve the Example 4-5 using the D' Alembert Principle.

Solution:

lOON

According to kinematic analysis, we have:

aG =0.45a A

aG =0.45a

j

(1)

l

Following the procedure in Example 4-5, we draw the free body diagram with all the applied force and constrained forces as in Fig. b. Now, on top of it, at the mass center G we add in:

% Kinematics diagram

Fig. a

(1) the inertial force with magnitude maG but in

opposite direction to aG (2) an inertial moment with the same magnitude I Ga but in opposite direction to a as shown.

lOON

In the generalized free body diagram, the system of the forces including the imaginary inertial force and inertial moment is under dynamic equilibrium. The equilibrium conditions of it can be easily obtained as:

----- ·

t:

100 + F1 - 0.45ma= 0

(2)

mg-N=O

(3)

c N

c:): -100(0.75)+0.45ma(0.45)+ JGa =0 From (4), we can directly solve:

a= 6.186 rad/s 2 and from (2):

248

maG= 0.45ma

(4)

Generalized free body diagram with inertial force and moment at the mass center

Fig. b

4.8 Summary

4.8

Summary

In this chapter, three important principles on Kinetics of Rigid Bodies have been covered:

(i) (ii)

Principle of Linear/Angular Momentum (Newton's Second Law) Principle of Work and Energy

(iii) Principle of Linear/Angular Impulse and Momentum These principles are valid for motions measured in any inertial reference frame. (1)

Different Forms of Newton's Second Law • Principle of Linear Momentum of Mass Center G

The linear momentum L = mvG is defined as a vector state function characterizing the motion state of a rigid slab in translation. Newton's Second Law in the form of the Principle of Linear Momentum for a Rigid Body is:

(4.11)

• Principle ofAngular Momentum about Mass Center G The angular momentum ii G = I G(jj is defined as a vector state function characterizing the rotational motion state of a rigid slab relative to mass center G. Newton 's Second Law in the form of the Principle of Angular Momentum for a Rigid Slab about Mass Center G is:

d M a =-Ha =laa I dt I I N'

N

LMJ + IF;GX F; J =l

(4.33)

I

J(x'

2

2

+y' )p(x' , y')dx' dy'= m

i=l

249

Chapter 4

Kinetics of Rigid Bodies

• General Moment Equation about Point C Newton's Second Law in the form of moment about any point Cis:

Me

I

N'

= rGe X maG+ ! Gii

(4.43)

N

Ir:exF;

LMJ + i=l J=l

This is the direct result based on the two basic principles Equations 4.11 and 4.33 . In the special case of a uniform disk rolling without slipping about the contact point C:

(4.47)

• Principle of Angular Momentum about Point C Fixed in an Inertial Frame The angular momentum

H e = rGC

X

mvG + !Gm is

defined as a vector state JUnction characterizing the rotational state of a rigid slab in general planar motion about a point C, which is }txed in a inertial frame. Newton's Second Law in the form of the Principle of Angular Momentum for a Rigid Slab about Point C Fixed in an Inertial Frame is:

(

d M- e =-He

J

(4.51)

In the special case of C belonging to the slab, that is, if the slab is in pure rotation about a fixed point C:

250

4.8 Summary

(2) Principle of Work and Kinetic Energy

Defining kinetic energy T = ..!_ mv 2 as a scalar state function 2 for describing the motion state of a particle, we have the Principle of Work and Kinetic Energy for a rigid body moving from position 1 to position 2:

(4.61) J

I 1 2 1 2 T = -mvG + -!GCtJ

2

2

or

or 2 _

J I

F ·diG+

F MG

Jo, MGdB

1

T=-1 0 m

81

0: Instantaneous centre of rotation or fixed pivot.

: Resultant force :

2

2

Resultant moment about G

• Conservative Forces The work done by a conservative force as a rigid body moves from position 1 to position 2 is equal to the difference in the potential between the two positions:

u2 -

(

J

(4.66a)

.______

which is, of course, independent of the path of the motion . The potential energy of a rigid body in a gravitational field IS :

Vg

=

mgy G + COnStant

(4.67)

The Principle of Work and Kinetic Energy can be written as:

U

1->2

= u(non-con) + u{con) = T _ T. 1->2 1->2 2 I

(4.65)

or u(non-con) 1->2

+ (v.I _ V,2 ) -_ T2 _ T.I

251

Chapter 4

Kinetics ofRigid Bodies

• Conservation of Total Energy The total mechanical energy of a rigid body or a system of rigid bodies is defined as:

E=T+V

with V= V8 +V.

Here Tis the sum of the kinetic energy of all the bodies and V is the sum oft he potentials of all conservative applied forces acting on the system. We then have: (non-con) E U 1-+2 == 2

-

E

1

If U 1

FY =g-aA

(3)

Bar AB: In j direction:

(4) Taking moment about A:

_LMA =]Ga+(rGA xmiiG)·k 0.5(R -m 2 gcos45°) = !Ga + 0

(5)

258

I

4.9 Problem-Solving Guidelines

STEP 3: Solving the Equations We now have five independent equations involving the five unknowns a A' a, aG, R, and FY. So, it is possible to solve them. Substituting (1) and (3) in (4):

3g -

_.B._ + (g - a ) = _2_ . a A

J2

J2J2

A

R

5 2

:::::} 4g--=-a

J2

(6) A

From (5) and (2):

(7)

Substituting (7) in (6):

:::::}

5

aA =-g=8.175

6

Substituting in (2):

:::::}

Answers:

a= 11.56

(i) Acceleration ofblockA = 8.175 m/s 2 (ii) Angular acceleration of bar AB = 11.56 rad/s 2

259

Chapter 4

Kinetics ofRigid Bodies

(2)

Principle of Work and Energy This is often used to solve for the linear/angular velocity of rigid slabs after they have moved a certain linear and angular displacement. It is a scalar method.

STEP 1: Kinematic Analysis This is used to establish the kinematic relationship between the linear/angular velocities of objects within a system of slabs. • Draw a kinematic diagram, defming clearly the coordinate systems and coordinate axes, showing the positive direction of each axis. • Set up any kinematic equations such as position equations for the bodies in terms of defined coordinates, constraint equations, and relative velocity equations. STEP 2: Kinetic Analysis This sets up the work and energy equations of the system. • Define the system to be analyzed. Draw a free body diagram for the system to account for all the non-conservative forces that do work on the system. Check to see if the system is conservative. • Draw an initial position diagram showing the initial velocities and configuration of the system. • Draw a final position diagram showing the final velocities and configuration of the system. • With the help of these diagrams, set up the work and energy equations for the system of slabs:

(UI-+2 )Non-conservative

= (T2 + V2)-

+ v;)

• Positive signs for the forces, moments, and velocities correspond to the positive axes and right-hand rule used in defining the coordinate systems. If force and displacement, or torque and angular displacement, are in opposite directions, then work done by the force or the torque is negative. STEP 3: Solving the Equations It is sometimes helpful to check that the total number of independent kinematic and kinetic equations is equal to the total number of unknowns. If so, the equations may be solved for the unknowns.

Example 4-17 Principle of Work and Energy

In the mechanism shown in Fig. a, the uniform links OA and AB, each of mass m = 1 kg and length l = 0.5 m, move in a vertical plane. Link AB is connected to a slider of mass M = 5 kg, which translates on a smooth horizontal surface. The slider is connected to one end of a spring of stiffness k = 400 N/m. At the instant shown, B= 30° and the spring is unstretched. Given that the system was released from rest

260

4.9 Problem-Solving Guidelines

when (} = 0°, calculate the angular velocity of OA at the instant when (} constraints in the system are ideal ones.

=

30°. Assume all the

A

k

Fig. a

A

Fig. b

Solution: In this problem, we are asked to determine an angular velocity after the system has been moved by forces over certain displacements. This can be solved by applying the Principle of Work and Energy. STEP 1: Kinematic Analysis Kinematic diagram: In the kinematic diagram (Fig. b), the combined center of mass G of the two identical links is located by considering symmetry. It will be easier to consider the kinetic energy of the two links together rather than individually.

Kinematic equations: The position of block B is x 8

= -2/ cosO

(1)

The velocity of block B is given by:

v = _:!_{- 2/ cos B)= 2(0.5)(sinB)B = m sin() dt

(2)

The position of G is:

l sin() YG= - 2

(3)

261

Chapter 4 Kinetics ofRigid Bodies

The velocity components of the combined centre of mass G are given by:

X

v 2

(4)

VG = -

Y

d ( l sin B)

v G = dt - 2 - =

2l cos B . B· = 2l OJ cos B

(5)

STEP 2: Kinetic Analysis System:

The system consists of the two links, the block and the spring. Initial &final position diagrams:

v=O

B

A

w=O

0

Initial position

A

Final position

Kinetic equations of motion: Since gravitational and spring forces are conservative and the ideal constraints do no work, the system is conservative. Conservation of energy equation: (6)

262

4.9 Problem-Solving Guidelines

We define the initial gravitational potential energy of the system to be zero.

At(}= 0,

Total energy £ 1 =Potential energy in spring=

Y

2

= _!._ ( 400)(2/ - 2/ cos 30° 2

y = 3.59 J

(7)

Total energy =Energy oflinks + Energy of slider + Energy of spring

Links: . energy= 2mgyG = 2mg (/sin(}) Potential - = 2.4525 J 2

(8)

Kinetic energy = Translation K.E. + Rotational K.E. =

m[( =

ml'w')

+

0.1302m 2

(9)

Slider:

Y= 0.625m 2

Kinetic energy= _!._ Mv 2 = _!._ · 5 · (m sin 30°

2

2

(10)

Spring:

Potential energy= 0 (unstretched)

(11)

STEP 3: Solving the Equations

Substituting (7) to (11) into (6): 3.59 = 2.4525 + 0.1302m 2 + 0.625m 2

=> m = 1.23 rad/s (cw)

263

Chapter 4

Kinetics of Rigid Bodies

(3)

Principle of Impulse and Momentum

This is often used to solve for the linear/angular velocities of rigid bodies in a system after a certain period of time, or after an impact. It is a vector method. STEP 1: Kinematic Analysis This is used to establish the kinematic relationship between the linear/angular velocities of objects within a system. • Draw a kinematic diagram, defining clearly the coordinate systems and coordinate axes, showing the positive direction of each axis. • Set up any kinematic equations such as constraint equations and relative velocity equations. STEP 2: Kinetic Analysis This sets up the impulse and momentum equations of the system. • Draw impulse and momentum diagrams. Impulse diagram:

This is a free body diagram showing the impulses acting on the system in the period of motion. Momentum diagrams:

These are sketches of the initial and final configurations of the system, showing the momentum vectors. • With the help of these diagrams, set up the impulse and momentum equations. Define clearly which bodies are included in the system. It is important to determine what principles should be used: Principle ofLinear Impulse and Momentum:

Principle ofAngular Impulse and Momentum about G:

I (ang)G

-

H G2

-

H G1 -- I GOJ2

-

I GOJ1

Principle ofAngular Impulse and Momentum about a Fixed Point C: j(ang)C

• Resolve the vector equations in the directions of interest. Check whether momentum is conserved along any direction for the system. Note that impulses of internal forces between bodies of the same system cancel each other out.

264

4.9 Problem-Solving Guidelines

STEP 3: Solving the Equations [t is helpful to check that the total number of independent kinematic and kinetic equations is equal to the total number of unknowns. If so, the equations may be solved for the unknowns.

Example 4-18 Principle of Impulse and Momentum The combined weight of a cyclist and his bicycle is 75 kg. The rear wheel has a mass of 2 kg and a moment of inertia about the axle of 0.25 kgm 2 • The front wheel has a mass of 1 kg and a moment of inertia 2 about the axle of 0.15 kgm • Each wheel has a radius of 0.35 m. The bicycle starts from rest with the chain providing a constant clockwise couple of 10 Nm on the rear wheel. Assuming that the wheels do not slip on the ground and the effect of wind resistance is negligible, determine the speed of the bicycle after 10 s.

lONm

Fig. a

Fig. b

Solution: We can apply the Principle of Impulse and Momentum to solve for the velocity after the system has been moved over a certain period of time.

STEP 1: Kinematic Analysis Kinematic diagram (Fig. b): Kinematic equation:

v

v 0.35

{j)=-=--

r

(1)

STEP 2: Kinetic Analysis Impulse and momentum diagram: Rear wheel (Fig. c):

JR.asdt

Impulse diagram

Initial momentum diagram

Fig. c

Final momentum diagram

265

Chapter 4

Kinetics of Rigid Bodies

Front wheel (Fig. d):

Impulse diagram

Final momentum diagram

Initial momentum diagram

Fig. d Kinetic equations of motion: System: Rear wheel Apply the Principle of Angular Impulse and Momentum about the rear wheel centre G8

:

(2) Note that the moment of impulse ofthe unknown force

RGabout G8 is zero. 8

System: Front wheel Apply the Principle of Angular Impulse and Momentum about the front wheel centre Gc :

::::::>

(3)

Note that the moment of impulse of the unknown force RGc about Gc is zero.

System: Bicycle and cyclist Apply the Principle of Linear Impulse and Momentum to the whole system:

i:

Note that the unknown forces RGc and RG8 are internal forces for the whole system.

266

(4)

4.9 Problem-Solving Guidelines

STEP 3: Solving the Equations We now have three equations (2), (3), and (4), for solving three unknowns, F8 , Fe, and v2 . Combining equations (2), (3), and (4) as (2) + (3) + (4) x r:

10(10) =

0.35

+

0.35

+ 75v2 (0.35)

v2 = 3.65 m/s

267

Chapter 4

Kinetics of Rigid Bodies

Problems 4.1

Principle of Linear Momentum about the Mass Center for a Rigid Body

4.1

When a rigid body is made to move in translation, the center of mass always moves in a straight line. True or false?

4.2

A 10 kg block is subjected to a constant force of 5 N. If the block starts from rest, what will be the velocity and distance moved after 3 s?

4.3

A I 0 kg thin plate supported by two massless rollers A and B is positioned as shown in Fig. P4.3. If the plate is released from rest, determine the acceleration of the plate in translation and the reaction forces of the surface acting at the rollers. Ignore the friction forces at A and B.

600m

Fig. P4.3

Fig. P4.4

4.4

A 15 kg thin triangular plate ABC is held in position by the wire CF and the two massless links AD and BE as shown in Fig. P4.4. If wire CF is cut suddenly, determine the acceleration of the plate in translation and the tension forces in the two links at the instant the wire is cut.

4.5

A 7 kg thin semicircular plate of radius r = 300 mm is part of a four-bar linkage as shown in Fig. P4.5. Parallel links AB and CD are assumed to be massless. Knowing that in the position shown links AB and CD are rotating clockwise at 200 rpm, determine (a) acceleration of the plate in circular translation, and (b) reaction forces at A and C by the two parallel links. (The mass center G of the uniform semicircular plate is as shown in Fig. P4.5.)

0.7m

c 600mm

Fig. P4.5

268

O.Sm

2m

lm

3m

Fig. P4.6

Problems

4.6

A 3500 kg lorry is carrying a 1500 kg crate in the position shown in Fig. P4.6. The coefficient of static friction f..ls between the crate and the lorry is 0.5. Assume that the crate will not tip over before the slipping occurs. Determine: (a)

the maximum acceleration the lorry can achieve before the crate starts to slip

(b)

the normal reaction forces NA , N8 and friction force F1 at points A and B

Assume that the rear wheel is driven and the friction coefficient at B is big enough, and that the friction at A is neglected. 4.7

A 20 kg rod BC is connected to a massless block A by a wire as shown in Fig. P4.7. Neglecting the mass of the wire, determine the minimum constant acceleration of A needed in order that the rod will remain translating in the angle as shown. Find the corresponding tension in the wire. Massless block A

60mm

c

lOOmm

Fig. P4.7

Fig. P4.9

4.2

Principle of Angular Momentum for a Rigid Body

4.8

The moment of inertia of a rigid body with respect to its center of mass is always the largest possible moment of inertia of the whole body. True or false?

4.9

A hoop of 10 kg is supported on a fulcrum as shown in Fig. P4.9. Its inner radius is 60 mm and the outer radius is 100 mm. Determine the mass moment of inertia of the hoop about point A.

4.10 Fig. P4.1 0 shows a simple pendulum. Determine the moment of inertia of the assembly about point 0.

Cylinder A mass = 6 kg, radius = 40 mm Slender barB mass = 4 kg, length = 400 mm Thin disk C mass = 10-kg, radius = 100 mm

c Fig. P4.10

269

Chapter 4

Kinetics ofRigid Bodies

4.11 A 60 kg double pulley has a radius of gyration of 0.5 m. The outer radius is 0.6 m and the inner radius is 0.3 m.

(a) The 20 kg block A is connected to the pulley as shown in Fig. P4.1la. Determine the acceleration of A when th.e is released from rest. (b) For the system shown in Fig 4.11 b, where block A is 20 kg and block B is 50 kg, determine the velocity of A d the distance moved by B 2 seconds after the system is released from rest.

30 rnrn

30 rnrn

60 rnrn

60 rnrn

(a)

A

q:ig.

Double pulley

A P4.11

(b)

4.12 Solve Question 4.5 if both links AB and CD are not massless but of mass 4 kg each. 4.13 A 9 kg disk of radius 120 rnm is g a 166 rad/s. A 4 kg brake rod AB is mounted to the cylinder as shown in Fig. P4.1 . o ing that the coefficient of kinetic friction f.-ik between the cylinder and the brake r is .4, determine the force F needed to stop the disk in 50 revolutions.

400

15 m/s 250 rnm

100 mm

50rnm

F Fig. P4.13

Fig. P4.14

4.14 A 10 kg wheel is lowered onto a conveyer belt that is moving at a constant speed of 15m/s, as shown in Fig. P4.14. The radius of gyration of the wheel is 0.3 m. The wheel skids on the belt until it gains enough angular velocity to roll on it without slipping. The frictional force of the belt on the wheel is 420 N. Determine the number of revolutions the wheel turns while it is skidding.

270

Problems

4.3

Application of Newton's Second Law to General Planar Motion of a Rigid Body

4.15 A uniform metal cylinder of mass 18 kg is released from rest on a 30° incline (Fig. P4.15). The radius of the cylinder is 50 mm. If the coefficient of static friction is Jl. s = 0.5 and the coefficient of dynamic friction is Jl. k = 0.45, determine the acceleration of the center of cylinder and the distance it moved after 3 seconds. 4.16 A uniform slender pole of mass 45 kg is initially at rest when a force of 100 N is applied as shown in Fig. P4.16 . Assuming that the coefficient of static friction Jl. ·' = 0.3 and the coefficient of dynamic friction Jl. k = 0.2, determine the pole's angular acceleration at the instant the force is applied.

2m 100 N 0.5 m

Fig. P4.15

1m

Fig. P4.16

Fig. P4.17

4.17 A uniform slender bar with a mass of 100 kg is attached to a uniform disk of 150 kg in the manner shown in Fig. P4.17. Determine the angular accelerations of the bar and the disk in the position shown as they are released from rest. Assume that there is no friction between the bar and the wall, and that the disk rolls without slipping. 4.18 A 10 kg unbalanced wheel (Fig. P4.18) has a mass center at G and a radius of gyration lea= 0.1 m. If the wheel is released from rest at the position shown, determine its angular acceleration. Assume no slipping occurs. 150mm

1 kg

Fig. P4.18

Fig. P4.19

271

Chapter 4

Kinetics of Rigid Bodies

4.19 A ring of negligible mass and radius 0.5 m has three point masses attached to it (Fig. P4.19). What is the angular acceleration in the position shown if the ring is released from rest? Assume no slipping occurs. 4.20 A uniform bar of mass 10 kg and length 1.2 m is attached to a 15 kg disk of radius 0.3 m as shown in Fig. P4.20. Assume that the disk rolls without slipping. When the system is released from rest, determine the acceleration of the center of the disk and the angular acceleration of the rod.

c

B

lOON

Fig. P4.20

B

D

3m

200N

Fig. P4.21

4.21 A uniform 250 kg bar is suspended from cords at C and D as shown in Fig. P4.21. If cords A and B are subjected to constant forces of 100 N and 200 N, respectively, determine the acceleration of the center of the bar and the angular acceleration of it at the position shown. Neglect the mass of the pulleys. 4.22 Disk A of radius 0.3 m has a mass of 45 kg and a radius of gyration of 0.2 m. It is connected to a block B of mass 20 kg by a flexible inextensible cord that passes over a massless and frictionless pulley, as shown in Fig. P4.22. Assume that there is no friction between block B and the plane, and that disk A rolls without slipping. Determine the acceleration of the center of the disk and the tension in the cord.

0.3 m

Fig. P4.22

272

4.10 Problems

4.23 As shown in Fig. P4.23, two uniform disks A and B are connected by a cable. Disk A rolls without slipping on a horizontal plane while disk B moves downward rolling without slipping along the cable. The inextensible cable with one end fixed at 0, passing around disk Band a massless and frictionless pulley, is attached to the center of disk A. Determine the accelerations of the centers of disks A and B, and tension forces for different portions of the cable. 5 kg

Massless pulley

Fig. P4.23

4.4

Application of Newton's Second Law for a Rigid Slab Rotating about a Fixed Point

4.24 An unbalanced 20 kg plate has a radius of gyration kG = 0.4 m about its mass center G as shown in Fig. P4.24. Initially the plate rotates about the fixed pivot 0 at a clockwise angular velocity of 8 rad/s in the horizontal plane. Determine the constant moment M needed such that the plate will have a clockwise angular velocity of 20 rad/s in 5 seconds.

0.3 m

G

2m

Fig. P4.24

Fig. P4.25

4.25 The 30 kg thin circular disk shown in Fig. P4.25 is rotating in the vertical plane, and in the position shown, it has an angular velocity of ro = 6 radls clockwise. Determine the angular acceleration of the disk and the constraint force at point 0 at this instant.

273

Chapter 4

Kinetics o[Rigid Bodies

4.26 A pendulum as shown in Fig. P4.26 has a mass of 6 kg with center of mass at G. It has a radius of gyration about the pivot 0 of ko = 0.4 m. If the pendulum is released from rest at () = 0, determine its angular acceleration and the constraint force at the pivot 0 when ()= 60°. Neglect friction.

a,

OJ

Fig. P4.27 Fig. P4.26

4.27 A 20 kg uniform slender bar of length 0.3 m is mounted on a vertical shaft as shown in Fig. P4.27. At this instant, the bar is rotating in the horizontal plane at an angular velocity m = 15 radls, and angular acceleration a= 80 rad/s2. Determine the horizontal constraint force by the bearing. 4.28 OA and BC are two uniform slender bars, each with a mass of 18 kg. The bars are welded at A. If the bars are released from rest in the position shown in Fig. P4.28, determine the total force R supported by the bearing at 0.

1.2 m B

0.6m

A

) m= 4 rad/s

0

0.6m

c Fig. P4.28

Fig. P4.29

4.29 A uniform semicircular disk of mass 18 kg and radius 0.6 m is rotating freely about a horizontal axis through 0. The disk has an angular velocity of 4 rad/s as it passes through the position as shown in Fig. P4.29. Determine the constraint force by the bearing at 0 at that instant.

274

4.10 Problems

4.5

Principle of Work and Kinetic Energy for a Rigid Body

4.30 The system shown in Fig. P4.30 consists of three elements: a 7 kg block A, a 9 kg uniform disk B, and a 12 kg uniform disk C. An inextensible cable connects the block A to disks B and C. If disk C rolls clockwise without slipping at an angular velocity me = 10 rad/s at this instant, determine the total kinetic energy of the system. 12 kg

9 kg

We= 10 rad/s

7 kg

Fig. P4.30

4.31 A bar of length 3 m and mass 10 kg is pivoted at one end and the other end is attached to a spring as shown in Fig. P4.31. The spring has an unstretched length of 1.5 m at B = oo and its spring stiffness k = 30 N/m. In the vertical plane, the bar is subjected to a constant clockwise moment of M = 60 Nm. Determine the total work done by all the forces acting on the bar when it has rotated downward from B= oo to B= 30°.

- -- --,-------

-· Fig. P4.31

Fig. P4.32

4.32 A 5 kg uniform slender rod of length 1.5 m rotates about a vertical axis through one end of it (Fig. P4.32). Determine the constant M required to increase its angular speed from 15 rpm to 40 rpm in 10 revolutions. .-..o.k..CH-i

275

Chapter 4

Kinetics of Rigid Bodies

4.33 A bar AB is oflength 3m and mass 12 kg with mass center at G. As shown in Fig. P4.33, two massless sliders pinned to the bar at G and A are engaged in a vertical and horizontal smooth slots, respectively. Bar AB, which is initially in the vertical

position (i.e. B = 0), is released from rest and falls down. When slider A has moved 1.2 m, it engages with a unstretched spring of a stiffness coefficient k = 20 N/m, and compresses the spring as the bar goes on falling. Determine (a) the angular velocity of the bar when B = 30°, and (b) the velocity of point G when the bar strikes the horizontal surface. Neglect the size of two sliders.

Spring 1.5 m

Fig. P4.33 1.2 m

4.34 A system shown in Fig. P4.34 consists of three elements: a 15 kg block A, a 20 kg uniform cylinder B, and a 30 kg block C. Assume that the bearing of cylinder B is frictionless, and the friction coefficient between the block C and horizontal surface is 14 = 0.2. In the position shown, the spring with stiffness coefficient k = 20 N/m, the ends of which are attached to the wall and block C, is compressed 0.3 m, and block A moves down with a velocity of 25 m/s. What will be the speed v of A after it has dropped I m?

Fig. P4.34

276

4.10 Problems

4.35 A 5 kg slender rod AB of length 0.8 m is supported on a pivot 0 as shown in Fig. P4.35. The other end is pressed against (but not attached to) a spring of constant k = 120 KN/m until the spring is compressed 15 mm, the rod is then in a horizontal position. If the rod is released from this position, determine its angular velocity and the constraint force at pivot 0 as the rod passes through a vertical position.

0.2m

0.6m

0 A

Fig. P4.35

Fig. P4.36

4.36 The assembly of a uniform 12 kg slender rod AB and a 4 kg wheel is released from rest in the position as shown in Fig. P4.36. Assume that the wheel rolls without slipping, and neglect the friction between rod AB and the wall. Determine the angular velocity of the rod when () = 0°. 4.37 In the initial position as shown in Fig. P4.37, the compound pulley is at rest, the spring is unstretched, and a constant force P = 20 N is applied on the cable wrapped around the inner pulley of radius r = 0.2 m. A cable wound around the outer pulley of radius R = 0.4 m is attached to a spring of stiffuess k = 100 Nm- 1. The moment of ine1tia of the compound pulley about its center is 500 Nm2 • Determine the maximum angular velocity of the pulley in the rotational oscillation.

R= 0.4 m

r= 0.2 m

_ ____,_ p

Fig. P4.37

277

Chapter 4

Kinetics ofRigid Bodies

4.38 A 3 kg stepped cylinder of inner radius r = 0.1 m, outer radius R = 0.2 m, and h = 0.06 kg.m 2 is connected to two blocks A and B through cables and pulleys as shown in Fig. P4.38. The mass of block A is 2 kg and that of block B is 1 kg. The system is released from rest, and the cylinder rolls without slipping on the 30° incline. Neglect the friction and the inertia of the two pulleys. Determine the velocities of blocks A and B after 3 seconds and the tension forces in the cables connecting the cylinder to blocks A and B.

Fig. P4.38

4.6

Principle of Impulse and Momentum for a Rigid Body

4.39 A rocket of 20 kg is launched from a moving tank (Fig. P4.39a). The rocket launcher on the tank is designed to exert a horizontal force on the rocket that varies with time (Fig. P4 .39b) and attain a peak force of Fp at t = 0.2 second in the time interval of 0.45 second. At the same time, the rocket itself exerts a constant thrust of 300 N in the direction of the launch. Knowing that the tank is traveling at a constant speed of 5 m/s and that the rocket reaches an absolute speed of 25 m/s in 0.45 s, determine the peak force Fp of the rocket launcher. F catapult

t (b)

(a)

Fig. P4.39

278

4. 10 Problems

4.40 A 15 kg disk is pin-supported at its center as shown in Fig. P4.40. If it is acted upon by a constant moment of 40 Nm and a force of 100 N which is applied to a cord wrapped around its periphery, determine the angular velocity of the disk 2 seconds after starting from rest. Also find the constraint force at the pin.

lOON Fig. P4.40

Fig. P4.41

4.41 Disk A has a mass 5 kg and radius 0.5 m while disk B has a mass 10 kg and radius 0.75 m. The system is at rest when a torque of magnitude 5 Nm is applied to disk A. If no slipping occurs between them, determine the time required for the angular velocity of disk A to reach 500 rpm and the frictional force that gear A exerts on B. 4.42 As shown in Fig. P4.42, a cylinder of radius r, mass m, and moment of inertia I a (about the center G) rolls without slipping from rest down a plane inclined at ()degree with the horizontal. What is the speed of its center t seconds from rest?

Fig. P4.42

Fig. P4.43

4.43 A spool in Fig. P4.43 has a mass 75 kg and a radius of gyration about its center of 0.2 m. At time t = 0, the pulley is rolling without slipping to the left with a velocity of its mass center G of 0.8 m/s. The force P, which is applied to the cable wrapped around the central hub of the spool, is a linear function of timet: P = 7.5! N. Determine the angular velocity m of the pulley 5 seconds after Pis applied.

279

Chapter 4

Kinetics of Rigid Bodies

4.44 A homogeneous sphere of radius 50 mm and 1.5 kg is mounted at A on an L-shaped rod as shown in Fig. P4.44. The L-shaped rod rotates freely about the vertical axis z at a counterclockwise angular velocity of 12 rad/s. The sphere is held in position by a cord, which is suddenly cut. Knowing that the moment of inertia of the rod about the vertical z-axis is 0.45 kg.m2, and the moment of inertia of a sphere about the axis passing its mass center is 0.4mr 2 , determine (a) the angular velocity of the L-shaped rod after the and (b) the energy lost in the plastic collision between sphere has moved from A to the sphere and the stopper at A Assume that there is no friction during the sliding of the sphere.

z

Cord

Stopper

Fig. P4.44

4.45 A stream of water moving at 15 m/s enters a horizontal curved pipe as shown in Fig. P4.45. The pipe has a cross-sectional area of 3000 mm2. Assuming the speed of the water relative to the pipe is constant and it is frictionless between the water and the pipe surface, determine the force of the pipe exerted on the stream of water.

Fig. P4.45

280

4. 10 Problems

4.46 A uniform steel square plate is suspended from 2 cords as shown in Fig. P4.46. The plate has mass 8 kg. Particle P of 0.02 kg travels at an initial speed of 260 m/s and strikes the plate at a point, which is 50 mm below mass center G of the plate. The collision is perfectly plastic. Determine the angular velocity of the plate, the velocity of G, and the velocity of particle P at the instant after the impact. Assume that there is no displacement of the plate just after the collision.

G

. . . . . .SOmm . . . . . . .... . .

!

260 m/s p

20mm

200mm

:

200mm Fig. P4.46

4.7

Application ofD' Alembert Principle to General Planar Motion of a Rigid Slab

4.47 As shown in Fig. P4.47, a rod AB rests against the wall of a lorry at A. The other end of it is attached to the lorry by a frictionless pin at B. The acceleration of the lorry keeps increasing until the contact force between the rod and the wall at point A vanishes. What is this critical acceleration?

A

Fig. P4.47

281

Chapter 4

Kinetics of Rigid Bodies

4.48 The spool in Fig. P4.48 has a mass of 7 kg and a radius of gyration of lea = 0.3 m. Assuming that the cords have negligible mass and that the spool rolls without slipping along the cords, determine the spool's angular acceleration when a constant force 100 N applied to the cord as shown.

lOON

B

Fig. P4.48

Fig. P4.49

4.49 In a horizontal smooth plane, block A is made to translate in i direction at a constant acceleration a. A rod AB attached to block A by a frictionless pin is initially in the position as shown in Fig. P4.49. The mass of the rod is m, its mass center G is at a distance r from A, and the radius of gyration of the rod about A is kA. Derive an expression for the angular velocity m of the rod in terms of Bas it swings from the initial position due to a very small disturbance in B. Assume that the disturbance makes rod rotating in counterclockwise direction. 4.50 A 30 kg uniform rod shown in Fig. P4.50 is initially at rest. Determine the constraint forces at A and B when it is released from the position shown. Assume that there is no friction between the rod and the surfaces.

Fig. P4.50

282

Appendix A

APPENDIX A VECTOR PRELIMINARIES A1 Notation of Vectors Quantities such as density or work, which can be specified by a real value (positive, negative, or zero) are called scalar. Quantities such as velocity or force, whose specification requires a direction as well as a magnitude, are called vectors. A vector in 0-xyz space can be expressed as:

y, j

y

(I)

where

i, },

and

k are unit vectors, and Ax, Ay, and Az are projections ?, ], k directions, respectively. A planar vector A in

of the vector in 0-xy plane (Fig. a) may also be expressed as:

.A= A) +Ay] =A e=AL.e where A

=

IAI

A' and

-

A y

0 Fig. a

e is

i

X,

y, j

the unit vector. The unit vector can always be expressed as:

e= cose l +sine J= L.()

e

I

(2)

+A; is the magnitude of vector

=

A

(3)

where e is measured by rotating from +x direction to the vector direction (positive as rotating from x axis to y axis, i.e. counterclockwise, and negative for clockwise).

0 '---------

A2 Products of Two Vectors (1) Scalar product of two vectors in 0-xy (Fig. b):

X,

i

Fig. b

A· B = l.4lli3l cos(A, B)= I.AliBI cos¢

(0