Vector Mechanics: A Systematic Approach [9 ed.] 9780998498683

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Vector Mechanics:

A Systematic Approach

Ninth Edition

Kinematics

Li xagru?;d Acceleration A

:

Dynamics/Kinetics

i ]

I

!

Mass and

!

Inertia

'

!

)

.

Moments

J

|

I Linear and

:

Velocity

[

Angular A

;

Equations of Motion

}

!

|

|

Position and Orientation

:

y

: : i

Solutions

| | ! |

Alan P. Bowling

Forces and

ii

Copyright © 2023 Alan P. Bowling or All rights reserved. No part of this publication may be reproduced, distributed,_ or transmitted in any form prior the ut witho by any means, including photocopying, recording, or other electronic or mechal.n cal! meth thods, 1 written permission of the publisher, except in the case of brief quotations embodied in critical reviews and certain

other noncommercial uses permitted by copyright law. Printed in the United States of America Aqualan Press, LLC Mansfield, Texas

www aqualan.com/aqualanpress.html aqualanpresyGaqualan.com Esgth Edition

ISBN-13: 975-0-9984966-8-3

Preface The first part of this book stems from my notes for the first dynamics course that sophomorelevel students take at the University of Texas at Arlington.

The second

part comes

from my

notes for graduate students’ first course in multibody dynamics. The main difference between this book and the most currently available ones is its focus on systema tically developing the equations of motion. It begins with a detailed discussion of topics related to kinematics, which

leads to a discussion of dynamics/kinetics.

Kinematics focuses on the movement of systems

without considering the forces that cause that motion. Dynamics includes these forces and the mass properties that resist them. A critical skill that students learn in dynamics is to combine physical laws with mathematical techniques to develop a mathematical model of a dynamic system. They then use the model to predict the system’s behavior under the influence of forces and moments. Students must become adept in using these tools, which usually have a mathematical encoding. I do not oversimplify mathematics because this hinders students’ understanding and inhibits the development of intuition about the behavior of dynamic systems. Some of this intuition stems from insights they

gain through mathematical modeling. However, mathematics is not a natural language, so I include natural language discussions

of mathematical developments, along with illustrations, to communicate more effectively with beginners. I introduce mathematical concepts gradually within the context of a discussion of modeling a physical system; all concepts are introduced using a physical example. We have to dive into advanced mathematics eventually, but I hope the gradual introduction makes comprehending the more advanced topics easier. Students often stumble on the details of mathematical developments and calculations, so I try not to skip intermediate logical and computational ste.:'ps. Omitting these details can be confusing, and students may spend an excessive amount of time recreating those steps. . Sophomore-level

Part I of this book.

United

students should

have prior exposure to most

of the mathematics

u§ed

in

This assumption is based on the curriculum in many universities in the

States of America.

Part I of this book focuses on planar motion

using concepts

geometry, algebra, trigonometry, calculus, and a small amount of vector algebra.

from

Part II uses

senior or early graduate-level mathematics, mainly calculus and linear algebre.x. I a.s.sume that students obtain greater exposure to mathematics as they advance through their currlculu'm.

The primary purpose of this book is to ensure that students have a clear understanding of the equations of motion, including why you need them, how you derive them, and how you use them to predict the movement of systems under the influence of forces. The systema.!nc approach championed herein addresses dynamic systems modeling in a nearly linear, .stepw1se manner that students can rely on to solve problems. The steps have a logical progression fr?m one to the next. However, they do not necessarily follow a linear progression becaus Figure 3.22: Normal and Tangential Directions.

fi =

—sinf Xn

+ cos(0) Yn

(3.92)

We need the time derivative of t. We can use the chain rule from calculus to obtain the following:

(@ 4tdt 319)_ dtdids df ds dt

p

5didf

where Equation (3.93b) defines the curvature

where

(b)

of a plane curve.

%=;

1

(3.93)

The need for a third coor-

dinate, 6, underscores the ambiguous nature of the normal and tangential coordinates, s and p.

Note that,

dt ¥

=

-

sinf Xy

5

+ cos(d) Yy

-= nn

dt

§ o

-— = 2o n & (5e3)

—

3.94 (3.94)

Substituting Equation (3.94) into (3.91) yields .

Va

=58t

-~

Notice the separation of acceleration into normal

+

$2

7

n

(3.95)

and tangential components.

The

two

com-

ponents imply that a point’s acceleration is not generally tangent to the path; recall that the velocity is always tangent to the path, as was discussed in Section 3.1. The velocity-dependent acceleration in the normal direction is the centripetal acceleration for a point moving along a curved path. This acceleration moves the point toward the concave side of the path and keeps the point moving on a curved path.

We can find $ and t from the point’s velocity, so if we know the point’s acceleration, we can

find p and § as,

§= Va-t .

N

s? - HVA—'s't” .2

.

~

(3.96)

]2

Velocity and Acceleration

Since we know the path, we can also find the distance along the path. An example of the use of normal and tangential coordinates is given in Section 3.15.5.

3.12.2

Velocity-Dependent Accelerations

In this section, we examine the centripetal acceleration acting on a body moving on a curved path, such as the string pendulum in Figure 3.23. The sphere, modeled as a particle, follows a circular path. The osculating circle can only touch this circular path at one point, the current position of the sphere, so it is slightly smaller, infinitesimally smaller, than the circular path. Because of this infinitesimally small difference, the center of the osculating circle is very close, infinitesimally close, to point N. Also, the radius of the osculating circle is infinitesimally close to the radius of the circular path. Thus, in the following discussion, we will assign point N as the center of the osculating circle with the constant radius, p=r, shown in Figure 3.23. The normal and tangential directions are n = -Xsandt= Y, in Figure 3.23. Differentiating the velocity of point A yields, .

Va

=

=

a

dz

dré o

T

d? (T xA) (3.23) dt (r o YA)

d¥a

TV

.~

_d

o

d?

_

d?Pna

dVa

.

=iV

4

d/

=

(3.97)

-

E(—sm(G) Xn + cos(6) ) 0

=

=5 T réY A+T

=

r6Ya

.|

-

dsin(8) o o XN

. @( - sxn(e)/dt

+ ré(—é

cos(f) Xn

- 6 sin(8) ?N)

6|

0

dcos(8) & p YN

+

=r8Ya

—

+ cos(6) 16X,

Y e (3.98)

We can find the same result as in Equation (3.98) in fewer steps using the transport theorem:

Xa a Y2 Yo Z Za (; g g

N dVa _ A AdVa VA= "o wan @ T WxVasTOYat

T

=r6

o

o

YA -2 X, (3.99)

Equation (3.98) reveals the centripetal acceleration term —r§? in the X, direction. We can see this by examining Equation (3.97), which shows that $ = r§. Therefore,

—f8

=

——Xa

= -ré?X,

(3.100)

which is the centripetal acceleration in Equation (3.98) for the pendulum in Figure 3.23. This acceleration pulls the sphere toward the center of its circular path, point N. Figure 3.23: Pendulum A related quantity, centrifugal acceleration, acts outCentripetal Acceleration. ward from the center of the osculating circle when a body moves along a curved path. It acts in the direction opposite of the centripetal acceleration. We can examine the centrifugal acceleration by considering the circular disk in Figure 3.24. It includes a channel within which a sma.ll sphere can move. As the disk rotates about its center, point N, with an angular velocityN =6 ZB, you expecl the sphere to be thrown outward from the center. If we examine the acceleratlon of point B, we

should be able to find this centrifugal acceleration:

.

Vs

=

dVp

.

_ d?Pyp dtz

2

EE(T xB)

|

Velocity and Acceleration

83

—

d

(347)

dTiB

dt(

@

N

T

S

B

S

_

d

/. o

.~

xXs | = E(TXB

PN

N B /. o

+ 19 YB) o

-

B

=

fi3+1‘0'?3+r§?3—r92i3+7‘9.?

=

(ré + 2fé) Ye + (1" - réz) Xs

A7 Xs + 0Y¥8) dt

(3~47.—3.101)

+

X (T’ Xp

Two

(3.101)

r8 YB)

+

3

(3.102)

Compare the accelerations in Equations (3.99) and ( 3.102) for the pendulum traveling on a circular path in Figure 3.23 and the sphere travel in g on a non-circular path in Figure 3.24, respectively. The speed and acceleration of the sphere along its path are found using Equation (3.101), .

3(3.:8_9)

]

X

1/7.2 +T202

.

—

.0'

§

=

L

(3-91)

2'é

e

(3103)

72 4 7262

Using the information in Equation (3.103), we can find the normal direction as

~T

=

(3.89)

X

B+

r

§Y:

B

o

/72 4 242

5

=Zgxt

o~

—rXg + 7V 6 Xp +

=

T Y8

(3.104)

V72 + 1242

Expressing the acceleration of point B in terms of normal and tangential coordinates yields

B

=

(3.95)

7+

72 4+ r26?

+ 1266 5

i

—

-

p

V72 + 7262

i

(3.105)

The acceleration in the normal direction will be positive and act inward toward the center of curvature, point O in Figure 3.21. The issue is that point O may not coincide with point N as the sphere moves along the path shown in Figure 3.24; the center of curvature can translate

so that p will be positive. Therefore, examining the acceleration in normal and tangential coordinates does not reveal the centrifugal acceleration in the system in _Figure 3.24. Instead, we will investigate the acceleratlons expressed in Frame B, although the XB

may not always correspond to t and &. Examining Equation (3.102), we see the centripetal acceleration for circular motion from Equation (3.98), 762 in the X, direction. Recall that this term represented the pendulum’s acceleration toward the center of the osculating circle, which kept it moving on a circular path. This term acts similarly for the sphere in Figure 3.24, accelerating it back toward point N and keeping it moving on a curved path. However, also notice the 7 term, which represents a different acceleration in the radial direction, Xg. This term represents the centrifugal acceleration we expect to see for the system in Figure 3.24. Further-

=

5

directions

=

Figure 3.24: Centrifugal Acceleration.

more, if the su; of forces %n the radial direc-

tion is equal to zero, recall Equation (3.1):

=

and YB

g

(3.106)

Equation (3.106) will yield a positive radial acceleration that will throw the sphere outward &s.the disk spins, as expected. Notice that in the case of Equation (3.106), this centrifugal acceleration,

84

Velocity and Acceleration

7. ends up being dependent on velocity, but it can depend on other terms. A common use of centrifugal acceleration is in a centrifuge to separate fluids into different components. The acceleration in the Yp direction of Equation (3.102) includes a similar 78 term present in the pendulum example, Equation (3.98). This term represents an acceleration in the direction of the rotation in a right-handed sense caused by the rotational acceleration of the disk. This translational acceleration is similar to the translational velocity caused by the disk’s angular velocity discussed in Section 2.6. However, a new velocity-dependent acceleration appears in Equation (3.102), 27, further accelerating the sphere in the Ya direction. This term is referred to as a Comohs acceleration and generally appears as the multiplication of different generalized speeds; note that the other velocity-dependent accelerations involved the square of a generalized speed. The Coriolis acceleration allows for the possibility that the sphere may accelerate in the direction opposite of the rotation, the —Y 4 direction, which is a general characteristic of the Coriolis acceleration. However, the sphere in Figure 3.24 is constrained to move within the channel, which prevents it from moving in the —Y 4 direction. If the sphere were not confined to the channel and could roll without slipping on the disk’s surface, it would exhibit a translation in the —Y 5 direction indicative of the Coriolis acceleration.

The velocity-dependent accelerations are referred to as fictitious accelerations. When multiplied by a mass, recall Newton's Second Law in Equation (3.1), they are referred to as fictitious or velocity-dependent forces. After this multiplication, the centripetal, centrifugal, and Coriolis accelerations are referred to as centripetal, centrifugal, and Coriolis forces. These fictitious accelerations and forces will be examined further in Section 3.12.3. 3.12.3

Fictitious

Accelerations

and

Forces

Velocity-dependent accelerations appear when we express the position of a point, or orientation of a body, in terms of a non-inertial reference frame, which we typically refer to as a rotating reference frame. Because Newton’s Second Law describes absolute motions, the velocity-dependent accelerations appear as correction factors when we cannot model the system using only the inertial reference frame. Therefore, the Coriolis and centripetal accelerations, in particular, are thought of as fictitious accelerations that appear only because of our choice of coordinate system. However, sometimes we must describe some quantities using a non-inertial reference frame because there is no more straightforward way to model the system. For example, reconsider the string pendulum example in Figure 3.23. We could express the position of the sphere as R R Pna

=

zXN

+

vy YN

(3.107)

so that the Cartesian coordinates ¢ and y are referenced

to the inertial reference frame N

= (XN, Yn)

in Figure 3.23.

However,

Section 2.15 discussed

how

these would

be ambiguous

coordinates when the pendulum string remains taught, causing the sphere to move on a circular path, as was discussed in Section 3.12.2. Thus, the Cartesian coordinates should not be used to describe the pendulum’s position. Instead, the angle between the inertial reference frame and a rotating frame is used as the coordinate, resulting in a velocity-dependent, or fictitious, acceleration. We could also attempt to use Equation (3.107) to describe the sphere’s motion in the channel of Figure 3.24. However, the expression in Equation (3.107) will not cause the sphere to remain within the channel as the disk rotates; Equation (3.107) only specifies that the sphere can move in the Xn-Yn plane. In other words, x and y are independent, and the sphere should be able

to move in a straight line in the XN

direction.

This motion

can only occur when

the disk is

not rotating and the channel aligns with the Xy direction. Equation (3.107) cannot define the sphere’s position if the disk rotates. We have to define some constraint that causes the sphere to remain within the channel under all possible motions of the disk and the sphere. It is difficult to determine this type of constraint. Instead, we use polar coordinates that more closely match the sphere’s motion; recall Section 2.16. However, the angle § measures the angle between an inertial and rotating reference frame, and the r coordinate is measured along a direction changing with tune. These non-inertial references lead to fictitious accelerations and forces.

Velocity and Acceleration

3.13

Implicitly Satisfying

Constraints:

85

Acceleration

If we cannot explicitly enforce the constraints, dependent acceler ations may appear in our model, and we must eliminate them. The second derivative of the constraint equations yields relationships between generalized accelerations. Consider differe ntiating the constraint derivative of the distance constraint in Equation (3.35) using the produc t rule:

d

.

% (0 =716

cos(f)

+ L&

cos(a))

(3.108)

0 = 76 cos(d) — ré2sin(8) + Léacos(e) — L&?sin(a)

where

=

0

=

d8

.

it~

e—

d2%¢

a2

=

—

dé

CTE o

=

e—

d%a

T e =

-——

(3.109)

(3.110)

are generalized accelerations. The second derivatives of the constraints allow us to use the substitution method to eliminate dependent accelerations from the model. This elimination implicitly enforces the constraints.

3.14

Acceleration Propagation for Two Points Fixed on a Rigid Body and Relative Acceleration

An acceleration propagation equation exists for two points fixed on a rigid body. This expression is obtained by differentiating the velocity propagation formula, Equation (3.54), for two points fixed on a rigid body: N

d

a

.

Vs

.

(3=4- ) Va

+

.47

No

(VB

=

x Pap

+

Va

Nw?

+

NoA

x

XPAB)

AdBs

(3'111)

0

/{‘f-

NwA

x Pap

(3.112)

We use the transport theorem because Pap is assumed to be expressed in the body-attached frame. Therefore, vB

=

VA

+

V,-e[

=

VA

+

NGJAXPAB

+

NwA

x (NwA

XPAB)

(3.113)

which defines the relative translational acceleration or relative acceleration. The relative angular acceleration, or the relative acceleration, is found from the derivative of Equation

(3.62)

N

dwret dt

_

(3.82)

G

=

(3.62)

AoB

= NGB

_ NA

(3.114)

Equation (3.114) also states the acceleration propagation formula for angular accelerations.

3.15

Examples

Often the problem statement determines the velocities and accelerations we need to find. Th}ls, we consider the given information and objectives in the problem statement in tl.le following examples to assess which velocities and accelerations are required. Enough information must be provided so that the number of equations we develop will be enough to solve the problem.

86

Velocity and Acceleration

You should also continue to develop symbolic expressions for the velocities and accelerations so that their derivatives will be correct. After the accelerations have been found, using the given information to simplify the problem is reasonable. The following examples also show how to address and write up the velocity and acceleration modules in Figure 1.1. Once again, in the write-up, you record the outputs from the Linear and Angular, Velocity and Acceleration modules. This record-keeping tracks the progression of your thought process in applying the concepts from this module to a particular problem.

3.15.1

The Write-up

The format for the write-up is intended to guide you through applying the concepts presented up to now. In this section, we add the velocity and acceleration modules to the write-up as follows: Problem CSD:

Statement:

Givens:

Section 2.11

Position Orientation:

Velocity:

Sections 2.12 - 2.18

Objectives:

Sections 3.2 and 3.3

Here you only find expressions for the velocities that you need to satisfy the objectives of the problem. Finding these velocities involves differentiating the position vector directly or using the transport theorem (Section 3.7). You may also use velocity propagation (Section 3.8) to find the translational velocities of interest. The rotational /angular velocities are found by inspection. You also may need to differentiate the constraints to implicitly enforce them (Section 3.6) by finding the relationships between the generalized speeds (Section 3.5). Acceleration:

Section 3.12

Here you only find expressions for the accelerations that you need to satisfy the objectives of

the problem. Finding these accelerations involves directly differentiating the translational and rotational velocity vectors or using the transport theorem (Section 3.7). You can also use acceleration propagation (Section 3.14) to find the accelerations of interest. You also may need to take the second derivative of the constraints to implicitly enforce them (Section 3.13) by finding the relationships between the generalized accelerations. Solution:

Show how you satisfy each objective and draw a box around your final answer(s).

Depending on the question, you may omit portions of the write-up if they are unnecessary to achieve your objectives. You may also omit any calculations and information not required to fulfill the objectives. However, you must always fully address the position and orientation portion of the write-up. Examples of how to apply this write-up format in answering some different problems are given in the following sections.

3.15.2

Cannonball

As was discussed in Section 2.12.2, the cannonball is an example of a classic type of problefn referred to as projectile motion. It illustrates the use of the particle representation of a rigid body. This example is a continuation of the one in Section 2.19.2.

Velocity and Acceleration

87

Problem Statement: Find an expression for the magnitude of the absolute velocity of the cannonball in Figure 3.25. Assume that the cannonball will move in a plane. Do not model the cannon and model the cannonball as a particle.

Simplifications: The system boundary includes only the cannonball. Assume that the cannonball will move only in a plane and that we will model it as a particle. Rigid Bodies: ball.

Figure 3.25: Cannon and g Cannonball.

The problem involves a single particle, the cannon-

Inertial Reference Frame and Point: (XN, YN) is the inertial reference frame. Other Points and Frames: Location

e =*

Point N is the inertial reference point, and

N =

Point A is the body-attached point.

Descriptions: Pva

=1

(2.9)

iN

+

vy ?N

(3.115)

Coordinates: Two coordinates, z and y, appear in the location descriptions.

Constraints: Examining the cannonball, we see that z and y can be set independently, so no constraints are needed.

Figure 3.26: Cannonball CSD.

0 constraints Givens:

=

(2.78)

Degrees of Freedom (DOF's): 2 coordinates —

2 DOFs

none.

Objectives: Find ||V4|. Velocity: We find the velocity of point A by differentiating its position vector. rule from calculus yields: Va

= (3.8)

dPna — dt dz

(311)

d = -~ = —|zXN+1yY N) y (3.115) dt (I N

=

dt XN

The product

+

zgdt

+

N

ygdt (3. 13)

N

y

Solution: The magnitude of the absolute velocity is,

IVall = v 3.15.3

+ 42

(3.117)

Mass-Spring-Damper

The mass-spring-damper system in Figure 3.27 is a classic system used to analyze vibrational motion. This example is a continuation of the one in Section 2.19.3.

88

Velocity and Acceleration

Problem Statement: Find the magnitude of the velocity of point A in Figure 3.27, assuming that the block is moving to the right at a speed of 7m/s. Flexible bodies are not modeled, so

/ 7

the spring and damper are omitted.

Rigid Bodies: The problem involves one rigid body,

fW

the block.

Other Points and Frames: body-attached frame.

Yn)

Im

—

At

v =7mis

. N

Z—C{E—N ‘i CTTTI777777I77777777777 Figure 3.27: Mass-Spring-Damper

Inertial Reference Frame and Point: Point N is the inertial reference point, and N = (Xn, inertial reference frame.

'

—

w-

Simplifications:

is the

System.

Point A is the body-attached point, and A = (f(A,

?A)

is the

Location Descriptions:

(“6){ P NA» NR} AR} + +g geom. 'Y

P

*

N A

~ Xa R

(2.19a)

A

w

;

Coordinates: 1 -~

1

rr 7T I

Constraints:

T

T can

Objectives:

7m/a

YN

Zyn

Looking through the location descrip-

be

set

Examining the mechanism, we see that

freely;

thus,

constraints

are

unneces-

Degrees of Freedom (DOF's): 1 coordinate

=

~

sary.

CSD.

VA

Xn

tions, we see one coordinate, z.

Figure 3.28: Mass-Spring-Damper

Givens:

=

( 3.118 )

(3.119) A

rrrr 7P

=

AA

A Xy

e

=

~

=

%% e

= z+w) ) XA Xa + Y A NA (2.9)(

iN.

h=

0.7m,

—

and

0 constra.ints( = w

=

1 DOF

1m.

Find || V,]|.

Velocity: We find the velocity of point A by differentiating its position vector:

Vv

A

=

(3.8)

311

LNa Prua dt

dz + w dt

(. us) dt((

+w)

Xa + (z+w)

xA

+

h Y)

YA

_

d(z + w) Xa

+

+ h%3 Ty * XA

dh¥a dt

(3.120)

Note that the direction vectors that define frame A do not appear to rotate when observed from frame N.

Velocity and Acceleration

Solution:

From the given information, we obtain the following: Va

=

Tm/s

XN

(3.1=20):i:

XA

—_—

T

=

Tm/s

(3.121)

Therefore,

HIVall = Va9m?e® = 7,,,/.% R

3.15.4

et ety e

(3.122)

B

Pendulum

The pendulum in Figure 3.29 is a classic example of rotational motion and the simple rotation. This problem also illustrates the use of the transport theore m. This example is a continuation of the one in Section 2.19.4. Problem Statement: Find the magnitude of the absolute acceleration of the tip of the pendulum in Figure 3.29 when § = 30°, 6=

and

7rad/a,

§ =

5rad/az.

Figure 3.29: Pendulum.

Simplifications: The system boundary contains only the pendulum and excludes the ground. @We will define a subsystem that includes the pendulum and the pin/axle and bearing at its base that allows the pendulum to ro-

tate.

Rigid Bodies: The problem involves one rigid body, the pendulum. Inertial Reference

Frame

and

tial reference point, and N = (X,

Point:

Point N is the iner-

Yn) is the inertial reference

frame.

Figure 3.30: Pendulum CSD.

and Frames: Points Other attached point, and A = (Xa, frame.

Point A is the bodyYa) is the body-attached

Location Descriptions:

_

N

A (2.26) {Pna, AR} + geom b . N

Pna R

YN

R

8

,

0

Np

Xa

N = Yo (2193) |

Zy,

Ry,

=

=

=

= L X,

(3.123)

NA (29) R

~

cos(f) Xy +sin(f) YN

~ ~ -—sin() Xn +cos(d) Yn ~

In

(3.124)

Note that because we are addressing a planar motion problem, we could represent the rotation using

¢

an orientation vector:

NeA

=

(2.17)

R

97,

(3.125)

90

Velocity and Acceleration

Coordinates:

Looking through the location descriptions, we see one coordinate, 6.

Constraints: unnecessary.

Examining the mechanism, we see that 6 can be set freely; thus, constraints are

Degrees of Freedom (DOFs): 1 coordinate Givens:

§ = 30°, L = 2¢,

6=

—

Trad/s and

O constraints 6=

=

1 DOF

(2.78)

Srad/s?.

Objectives: Find | Val|. Velocity: We can find the absolute velocity of point A by differentiating its position vector. Because the position vector is expressed in frame A, we will use the transport theorem to find the derivative. First, we need to find the angular velocity by inspection: N A

= 328

0Z

12 (3.126)

A

Because this is a planar motion problem, we could also find the angular velocity by differentiating the orientation vector:

N

A

e

=

=

—

z

(3.125) dt (0

dt

Y

d ( &

dé 5

=

—

A) @iy dt

0

At

0

dZ4

-

-

.

(3.127)

2,

(3.13)

Using the transport theorem, we obtain, N V

=

A

(3.8)

A

dPNA

dPNA

=

dt

N

—

(347

dt

+

d

-~

L

At At

=

Wi

x Py

L

0

0

(oo

A

A

AA

.

o~

—~

xA

YA

ZA

L

0

0

0

0

+0ZaxLXa=|

A

LOY,

4

(3.128)

Acceleration: We differentiate velocity to find acceleration. Since the velocity is expressed in frame A, we will use the transport theorem again. N

™ (3.82)

=.

19/

=

A

dV,

Tdt dlif

0

o (347

. -

dVa

Tdt

T

N

A

A

= owhx Va (3.128)

.

7

o

7£0YA+L0YA+L(§§Z+ t

LOY, - L X,

dt

.

o~

dLO Y,

.~

dt

.

Xa

o

0

.

.

Ya

o0

Lé

.~

ZoxL6Y T 0Zax LY Z, .

@ 0

(3.129)

Because the generalized coordinate is referenced to a non-inertial frame, a velocity-d ependent

acceleration appears in Equation (3.129). It is a fictitious acceleration in the X, direction classified as centripetal acceleration. The X4 direction is normal to the circular path of point A.

Solution: The magnitude of the acceleration is,

IVall = L2642 + L2 = Ly/é> + ¢4

(3.130)

91

Velocity and Accelerat:on

||vA|| 2ft¢Ead/,z)2 7rad/.9)4 = 98 51;1/., 3.15.5

(3.131)

Cylinder

Here we formally write up the example of a cylinder sliding on a table, Figure 3.31, that has been used throughout this book. This example illustrates the use of normal and tangential coordinates in addition to velocity and acceleration propagation. The CSD for the cylinder is given in Figure 3.32. This example is a continuation of the one in Section 2.19.5.

@

Problem Statement: Find the magnitude of the absolute velocity and acceleration of point B on the cylinder in Flgure 3.32 where 2 = 3m/s, y=

6=

dm/s, &

=

3m/s?,

§

=

dm/s?,

O =

60°,

r

=

1cm,

6 =

3rad/s,

“T T oA "

and

Figure

Srad/s?. Also express the velocity and acceleration of point B in terms of

3.31:

Cylin de;-.

)

normal and tangential directions; give the values of the tangential speed and the radius of curvature, 5 and p, at the given instant. Use velocity and acceleration propagation to solve this problem. Point B is attached to the lower face of the cylinder, which is in contact with the table. Simplifications: The system boundary the cylinder; it excludes the ground.

includes only

Rigid Bodies: The problem involves one rigid body, the cylinder. Inertial Reference Frame

and

Point N is t-he

Point:

inertial reference point, and N = (Xn, Yn) is the inertial reference frame.

.

:

Figure 3.32: Cylinder CSD.

Points

Other

and

Frames:

Point

A

is the

body-

attached point, and A = (Xa, Ya) is the body-attached

frame.

Location Descriptions:

-

N

=

Pra (29310)

La (2.26) {Pna, AR} + geom.

N

NR A

=

(2.19a)

—zXn-yY TN

T YN

Xa

=

cos(d) Xn + sin(8) Y

Ys

=

—sin(6) Xn +cos(d) Yn

-

(3.132)

-

(3.133) Note that because we are addressing a planar motion problem, we could represent the rotation (3.134) — 67y Noh i ientation vector: N an orientation using Coordinates: 6.

Looking through the location descriptions, we see three coordinates, z, y, and

92

Velocity and Acceleration

Constraints: Examining the system, we see that all three coordinates can be set independently of the others; thus, constraints are unnecessary.

Degrees of Freedom Givens: é =

1

=

3m/-97

(DOFs): y

=

3 coordinates

4M/61

=

3"1/527

y

=

—

4""/32)

0 constraints (2?8) 3 DOFs 6

=

6001

T

=

lcm;

0

=

3rad/3a

and

5rad/a’.

Objectives: Find ||\"B||, 5, p, and express the velocity and acceleration of point B in terms of normal and tangential directions.

Velocity: We can use several methods to find the velocity of point B, but we will use velocity propagation based on the velocity of point A. To do this, we need the angular velocity, which we find by inspection: -

NoA

=

(3.28)

0Zp

= 62y

(3.135)

Because this is a planar motion problem, we could also find the angular velocity by differentiating the orientation vector: N

v

A

dNeA

_

(327)

d (, a —(07Z

=

dt

do -

=

(3128 dt (

— 17

N) @1y

dt

N

+

0

d%

6

=

dt

@

(3.13)

o

.

Zn

(3.136)

The velocity of point A is found by differentiating its position vector 0 v

=

A (3.8)

dPna e— dt

dr &

=

——

(3.11,3.132)

—_

dt Xn

g% -z

dt

0

dy 5

—_——

gAfi

-

dt YN

-y

=

dt

(3.13)

-

A 2 Xy

-

—_

—§ Yn

.

(3.137)

We can find the velocity of point B using velocity propagation as (3—54)

=

=

Va

+

NWAXPAB

=

—i‘iN

Xa

Yo

T

0

0

-tXn-39¥y+

—

y?N

Za

éiAXTiA

R

R

—g¥YN

| =-tXn

6 0

0

+

+

.

1Y,

—#Xn - 5 ¥y + ré(~sin(6) K + cos(8) n) (—:i: —-ré sin(a)) XN

+

(—g +ré cos(fl)) Yn

(3.138)

Acceleration: We find the acceleration of point B using acceleration propaga tion based on the

acceleration of point A. The angular and linear acceleration of point A is: 0 N w- A

'

=

(3.82)

dt

=

—

(3.11,3.135)

dt

Z"

+

b

dt

0 V

=

A (382)

dV, 2YA

=

dt

(3.11,3137)

di _ux

~

s

dt XN

X

z;dt

_

=

(3.13)

o)

Va

+ NoA x Pap

3.

N

(3.139)

0 dvu ~ __.‘li

o

Y

-

dt Yn-y

dt

(3.13)

We can use acceleration propagation to find the velocity of point B:

Vs

§Z

+ NuA x (Nw” x Pap)

_x

~

£Xn

o

~

- §¥n

(3.140)

Velocity and Acceleration

-

S

.

=

.

EXN-§Y¥n+

Ya

Za

r

0

0

0

0

4

[+NAx[

—fi‘iN—ij?N+Té?A+0'2AXT0.?A

=

“EXN-§YN+rVa+|

A

R

R

“EXN-§Yn+

iV,

ré (— sin(9) Xy + cos(f) ?N)

Vg

Xa

=

= Ve

93

Ya

Za

0

rd

o

0

- 2%,

o

— (:r + rfsin(6) + ré? cos(G)) XN

+

Ya

2.“

r

0

0

0

0

¢

4

(3.141)

- r6? (cos(B) XN + sin(8) ?N)

)

=

Xa

Xa

-

XN

(ré cos(d) — 6% sin(6) — y) Y

- i Yn

(3.142)

(3.143)

Notice that the velocity-dependent accelerati ons appear in Vg but not in Va. They occur because the position of point A is described in terms of coordinates associated with the inerti al reference frame. The position vector for point B includes a generalized coordinate, 8, refere nced to the non-inertial frame A. As expected, the veloci ty-dependent (fictitious) accelerations only involve §. Because they involve a squared generalized speed, they can be classified as centripetal or centrifugal accelerations depending on the value of 6. Solution:

IVs] = \/ (—:i:—nésin(a))2 + (—g+récos(0))2

(3.144)

— rfsin(d)

— 1'0'25in(9))2

(3.145)

IVall = \/(—3m/s—0.01m-3rad/ssin(60°))2 + (=4m/s + 0.01m - 3rad/s cos(60°))*

(3.146)

IVe|

=

\/(—:c

— ré2 cos(G))2 +

(—i] + rfcos(d)

Substituting in the given information yields,

[Pt

Sl

ot

gl

! |Ve] = 4.997m/s

(3.147) oy)

IVs|

B|

(3m/s* — 0.01m - 5rad/s?sin(60°) — 0.01m - 3%rad?/s? cos(60°))°

=

Y

2

+

(3.148) )

(4m/s? + 0.01m - Srad/s? cos(60°) ~ 0.01m - 3%rad?/s? sin(60°)) i:‘"\;B” = 4.905m/s=f‘:

(3.149)

The velocity of point B can be expressed in terms of normal and tangential coordinates as,

5= IR

1

T




Note that because we are addressing a planar motio n problem, we could represent the rotations using orient vectors ation

;

21 Coordinates:

A

(21 & 2B

(3.160)

Two coordinates appear in the location descriptions, 6 and a.

Constraints: Examining the mechanism, we can see that if § changes, then o must also change because the gear teeth are engaged. We can use an arc length constrai nt to find the relationship between these coordinates, 3660

=

teetha

6

=

—teethsa

(2.52)

=

—10 «

(3.161)

The constraint in Equation (3.161) is linear in the generalized coordinates, so we can explicitly enforce it in our model. We will choose 8 as the independent coordinate since we are given the value of & We want to solve for a in terms of 6 to eliminate a from the model: =

@ (2.52)

Degrees of Freedom (DOFs): Givens:

N(;JB

—3rad/s?

10

(

2 coordinates — 1 constraint

=

(2.78)

3.162

)

1 DOF

ZN.

Find NwA.

Objectives: Velocity: expected:

=

36 —-—

The

translational velocities of the mass

centers,

points

A

and

0

Va @8

dPa dt

_ e

0

Vs @8)

_ ;gi

dt

@ydt

B, equal

zero,

as

0

+ L}’}-{=

N

dt

(3.13)

0

(

(3.163)

Because this is a planar motion problem, we can find the angular velocities by differentiating the orientation vectors: ~ 36 -~

NeP ®

=

=

(3.160) o Zs 3.162)

-—07Z 10

B

3.164)

(

. . NgB Since we are explicitly enforcing the constraints, we eliminate « from the model, "©®°. Now we can differentiate the orientation vectors to find the angular velocity vectors,

na

_-

(3.21)

dVeh dt

-

2d ( ; AA)

(3.160) dt

=

do 5 Z 74+

(3.11) dt

A

0

dZA4 dt

0

=

(3.13)

0Za

(

3.165

)

96

Velocity and Acceleration

NUB

dNeP

=

(3.21)

dt

d(

= (3.164)

do @ Z &8

=

(3.160)

36

=

dt

d_a

(3.11) dt

-

~

ZB

+

36 df

—=|-—01Z = —— — dt( 10 B) (311) 10 dt Zn

0

dZ,

a

.5

=

dt

aZp

36 , 426

-— 0 10

+

0

(3.166)

36 ;5

= ——01Z (313) 10 B

dt

We can also find the angular velocities of the two gears by inspection as

NoA = §Z Nw (3,23)0ZN

(a)

NoB = 42 Nw AT

(b)

( 3.167 )

However, to explicitly enforce the constraint, we need the relationship between the generalized speeds, which can be found by differentiating the constraints, recall Section 3.6:

d 7360

=-10a)

, = -106 . 366

—

. a_—loo

—

(3.168)

The angular velocity Nw® includes a dependent speed, which we eliminate using the derivative of the constraint:

N,,B

_ 36 = 10 .168) (3.167b,3

6Z

) (3.169)

N

which is identical to the expression in Equation (3.166). Acceleration: velocities:

We find the angular accelerations of the two gears by differentiating the angular

NA Y sen =

N.B v

_

(382)

dNwB dt

dé

dNwh

@ 11,3.1670) dt

dt

_

(311,3.169)

%36 dd df -

10 dt Z

+

0

d7

z

—

-

N + 67?(3.13)0 Zn 3636 . dZ

4t

0

0

_

(313)

%36 . ~

10 6 Zn

.

(3.170)

(3.171)

Solution: Because the system has only one DOF, we expect that if given the value of one generalized speed, we should be able to find all others. Also, if we know one generalized acceleration, we should be able to find all the others if the generalized speeds are known. Thus, we expect that we can fulfill our objectives. From the givens, we obtain

: “

N-B

_

am

36 ; 10 62n

= St

—

Iy

10 . b = —3g (—3rad/s?’) = 0.83raa/s® (3.172)

Therefore,

L NoA(3.30)9 zN = osamd/s= sz ST St TR

3.15.7

ST

OVE

T S

3 BRI

T

T

L“J

No-Slip Rolling

Consider the system in Figure 3.35, where a ball rolls without elipping over a circular mound. This example illustrates the velocity form of the no-slip rolling constraint over curved surfaces. Problem

L

(3.173)

Statement:

The center of the circular mound

is point O in Figure 3.35.

A ball begins at the top of

Ball has

radius 0. /=

T

Figure 3.35: No-Slip Rolling.

Velocity and Acceleration

97

the mound and rolls without slipping along the surface to the configuration shown. Find the position of point A, expressed in the inertial reference frame, when the simple rotation between the inertial and body-attached frame has an angle of 30°. Simplifications: The system boundary includes only the ball, which rolls without slipping over the surface of the

-------------T

circular mound.

Ball has radius r

Rigid Bodies: The problem involves one rigid body, the ball.

R l

i

Figure 3.36: No-Slip Rolling.

Other Points and Frames:

isA=

(XA,

YA)

Inertial Reference Frame and Point: Point N is the inertial reference

point, and N = (X,

Yn)

is the iner-

tial reference frame.

Point A is the body-attached point, and the body-attached frame

An extra frame

B = (XB,

YB) is used to track the position of point A. Point

Bi represents the contact point on the ball, and C; represents the contact point on the surface. Location Descriptions: (2 26) {PNA,

s

AR}

+ geom.

" SR (2.1=9 "

Pna

=

9)

(

(R+n)

+T)

¥

B

(

Xp

=

~ cos(a) Xy +sin(a) Yn

?B

=

ios(a) Y-

Zg

=

2y

3.174

)

sin{a) Xn

(3.175) Yo

-v

4

o Coordinates: Velocity:

NoB

=

BR

Xa

=

cos(8) Xp +sin(6) Y

Yo

=

cos(d) Yg —sin(6) XB

Zpn

=

In

N

=

(2.19a)

~

M

N

-

-

(3.176)

Looking through the location descriptions, we see two coordinates, o and 6.

The angular velocities can be found by inspection as:

NA

= 62N

BuA

= 4 Zy

(3.28)

(3.28)

= NB

(3.34)

BuA — (a + é) Zn

(3.177)

We will use the velocity form of the no-slip rolling constraint, so we need the velocity of the ball’s center: Va (3—._—3)

dPna dt

(( +1) ¥) =, ~(R+7)a X5

(. 174) dt

(3.178)

We now use velocity propagation to determine the velocity of the contact point attached to the ball, point B;, VBI

— (337)

Va

+

N,

w A XPABl

98

Velocity and Acceleration

=

—(R+naXs + (d+é) Zp x (—r ?B)

=

~ —-(R+r)aXg+|

Xs 0

Yo 0

0

=

—(R+r)aXp

=

(1'0 -

Zs &+6

-r

0

+ r(d+é) Xp

Rd) X

(3.179)

Because the surface is stationary, we know that the contact points on the surface have a velocity

equal to zero:

= G (3.69) If we now enforce the no-slip velocity constraint, we obtain,

0= Vo Constraints:

=

(3

Vs = (r0 - Ra) Xs

.180

)

(3.181)

Because the ball rolls without slipping, you should be able to convince yourself

that the two angles are not independent. Thus we need one constraint to eliminate the extra coordinate, which we found in Equation (3.181) =

Ra

=

/ Raé dt

=

Raf(t)

(3.181)

/ rédt

m(t)

+

D

(3.182)

where D is a constant of integration. We can solve for this constant by assuming that at the initial time when the ball is at the top of the mound, frames A and B are aligned: 0(t=03)

=

a(t=0;)

=

Therefore

Orad

=

(3.182)

—

D=0

(3183)

Ra

(3.184)

Compare Equation (3.184) with the constraint in Equation (2.116). Degrees of Freedom

(DOFs):

2 coordinates

—

1 constraint (2__;3) 1 DOF

Givens: R = Im, r = 0.1m, and N®* = 30° = Zra4. Objectives:

Find Pyy.

Solution: From the givens we obtain NQA e

= (2.24) ( 0 +01)

5

Zn

=

T 5

-6

Zn

—

6

+

a

T

-6

=

3

.

185)

Using the constraint, we obtain —a+ r

a

=

(3.185)

@

Therefore

6

r

o m

R+r -6

=

e

" -

o}

=

e,V EE

F"‘ 3oy (Im + 0:1m) (cos(2.7272°) Y PR

N

-

0.1m

e

T

1m+01m —6

—————

=

AT

v

(0.0476rad

LKL

- sin(2.7272°) iN) ~-

VLT

A

LI R

N T

=

LTI 2

2.7272°

TV

L

= 1.099Yn TR

N

SR, LTI

3.186 (3.186)

— 0.052iN§_ T

TR

©(3.187)

Velocity and Acceleration

3.15.8

99

Blocks

The system in Figure 3.37 is an exam-

ple of a vector loop constraint and the

calculation

of the

instantaneous

cen-

ter of zero velocity. This example is a continuation of the one in Section 2.19.8.

Problem Statement: At the instant Figure 3.37: Connected Blocks. shown in Figure 3.37, find the velocity ?f the leftmost block if the rightmost block moves up the incline with a speed of 4m/s. At the instant shown, 6 = 30°, 8 = 40°, L = 10e¢m,

£ = 2em, y = 9.14em, and the angle between the

rod and the horizontal ground is 19.17°. Also, find the instant aneous center of velocity for the rod. Assume that neither of the blocks rotates. Simplifications:

The system boundary includes only the blocks, the rod, and any attached

parts; the inclines are attached to the ground, which is omitted.

We will compose three subsys-

tems represented by the main bodies, including all axles, bearings, and any other parts required

to allow the bodies to translate and rotate.

We will only consider motions where the leftmost

block remains in contact with the inclined surface on the left, and the rightmost block remains in contact with the inclined surface on the right. Also, assume that neither block rotates.

Rigid Bodies: The problem involves three rigid bodies, the two blocks, and the connecting rod. Inertial (Xn,

Reference

Frame

and

Point:

Point N is the inertial reference point, and N =

is the inertial reference frame.

YN)

Other Points and Frames: Points A, E, and B are body-attached points on each body, where points A and E occupy the same position. Frames A = (Xa, Ya), E = (Xg, Yg), and B = (iB, ?B) are the body-attached frames. Frames A and B are also inertial reference frames because they do not rotate; frame E does rotate. Recall that we can draw the body-attached frames anywhere we like; the important thing is to show their alignment in relation to the features of the attached body. Location Descriptions:

=

N

A (226) {Pra,

:

AR} +geom.

=

{Png,

La (2.26) {Pxg,

=z¥Y

Pws

Pxa (2.9) TrA

A

pR} +geom.

= y X

N

A

X ¢

AR

Ry

Kn

_

e

=

{Png, SR} + geom.

L (2.26) {Pre,

Png

.

A

. Lg

NP (2.9) vae

[ ou

s

BR

=

NE (2.9)

ER}+9

Y2

(3.189)

.-

XA

=

cos(¢) Xn +sin(d) YN

\:A

=

- sin(¢) Xn + cos(¢) YN

-~

Zy

_

=

(3.188)

o

Iy

~

~

(3.190)

100

Velocity and Acceleration

A

—> XN

Figure 3.38: CSD for Connected Blocks.

.

? B

)

-

N

A

~

= cos(8) Xy +sin(8) ¥ S

_

sR 2100

\:B

~

=

~

- sin(6) X~ + cos(8) Yn

Zn

=

Zs

A

e

£

N

Xp

—

Xp

(3.191)

Xy ~

fox N

Xe

E

R

Xeg

=

cos(a) Xn

Y

=

-—sin(a) Xn

=

Zn

~

=

(2.19a)

E

QN

a

+ sin(a) Y +

cos(a)

N

Y

N

(3.192)

Note that because we are addressing a planar motion problem, we could represent the rotation

of the rod using an orientation vector:

N@E Coordinates:

=

o Zy

(2.17)

(3.193)

Looking through the location descriptions, we see three coordinates, z, y, and

a; 6 and ¢ are constants.

Constraints: Examining the mechenism, we can see that if one of the coordinates changes, r,

for example, the other two coordinates must also change. Thus, for the system to have at least one DOF, we must find two constraints that allow us to eliminate two dependent coordinates.

We can use the vector loop constraint for this purpose: PNN

or

(2=)PNA

.64

+

PAB

+

PBN

0=Pw=z¥

=

0

=

OxN

+

OYN

+LXg - yXs

(3-194)

(3.195)

We can orthogonalize ang separate the vector in Equation (3.195) into scalar equations using

the dot product:

0 =

Pyn- XN

=

(3.195)

Lcos(a)

— zsin(¢)

—

ycos(6)

(3.196)

101

vVelocity and Acceleration

0 =

Pyn- Yy

Lsin(a)

(3.1=95)

+ zcos(¢)

— ysin(6)

(3.197)

These constraint equations are nonlinear and transcendental, so we cannot easily use them to eliminate coordinates from the model. We must implicitly satisfy them, as discussed in Sections

2.14.1 and 3.6.

Degrees of Freedom (DOFs):

3 coordinates — 2 constraints

= 1 DOF (2.78)

Givens: Vg(t = 0s) = 4m/s X, 8 = 30°, ¢ = 50°, L = 10cm, T = 2em, y = 9.14cm, and a = 19.17°, Objectives:

Find V(¢ = 0s) and the rod’s instantaneous center of zero velocity at the given

instant.

Velocity: The two blocks do not rotate, but we can find the angular velocity of the rod by inspection as N,

WF ,E =y= &- aZ I

(3.198) .

Because this is a planar motion problem, we could also find the angular velocity by differentiating the orientation vector:

@

(3.93)

dt

(3.21)

&

d/

dNeE

,E

N,

N)

dt (a

(3.11)

dt

N+

dt

0

dz

da 4

.

(3‘13)0‘

~

N

19

(

)

We find the translational velocities of the two blocks as

Va

— (3.8)

dPna NA dt

%

dr S

— Y (3.11,3.189) dt

0

dt

Atz

P

= Y A (3.13)I

3.200

(

)

0 Vs

_ (3.8)

dPnB dt

dy Xs< — 2 B (3.11,3.189) dt

+

.S

% y

= X 4B (3.13) y

dt

(

3.201

)

The ?A and iB directions do not rotate, so they do not appear to change direction over time to an absolute observer, such as the inertial reference frame. Since we know we have to satisfy the constraint equations implicitly (recall Section 3.6), in Equations (3.196) and (3.197), we differentiate them here to find the relationships between the

generalized speeds as,

%(0

= Lcos(a)

= -L & sin(a) (3.196)

— zsin(@)

— ycos(d))

— Zsin(¢)

(3.202)

— ycos(f)

%(0 = Lsin(a) + zcos(¢) — ysin(6)) = L& cos(a) (3.197) Solution:

+ Ecos(¢p)

— y sin(f)

(3.203)

First, notice that we are given only one velocity, which leads us to the value of one

generalized speed: Ve

=

y Xg

=

4m/s

)A(B

—

Yy

=

4m/s

(3204)

102

Velocity and Acceleration

Since this is a one DOF system, we expect to find all the other generalized speeds if given one, using the constraints. Once we have all of the generalized speeds, we can find any velocity we need, so we expect that we can achieve the objectives. We will use the substitution method to solve the system of equations composed of Equations (3.202) and (3.203) for the other speeds. This system of equations has two unknowns, & and &, and two equations, which are linear in the unknowns, so we have confidence that we can solve it. The main operation in the substitution method is to solve one equation for one unknown and substitute the solution into the remaining equations, thereby eliminating that unknown from the system formed by the remaining equations. Thus, we solve for ¢, _ —isin(@) — ycos(h) (3.202) Lsin(a)

(3.205)

and eliminate it from Equation (3.203) by substitution, _

—L & cos(a)

(3.203)

. sin(¢) cos(a) ’ (1 " cos(¢) s'm(a))

¥ sin(6)

cos(¢) cos(¢) (Z sin(¢) + y cos(8)) cos(a)

+

y sin(0)

cos(¢) sin(a)

_ B

cos(®)

. [ cos() cos(a) sin(8) (cos(¢) sin{a) + cos(¢)>

(3.206)

The system of equations resulting from this elimination consists of the single remaining Equation (3.206), which includes only one unknown, &. If the remaining system had additional equations, we would perform the substitution operation again, solving for another unknown and so on, until we obteined a single equation in a single unknown. The solution for % is

.

cos(¢) sin(a)

sin(6)

(cos(6)cos(er)

_

2 0¥ (Satgrontel * ie)) (et —snigremeer) To evaluate Equation (3.207), we need to find the necessary angles.

(3200

Since the system only has

one DOF, we know we can find all angles if given one. The procedure for doing this is discussed in the example given in Section 2.19.8. The same angles found earlier are used in this problem.

Plugging in all of these angles yields,

b= (cos(30°)cos(19.17°) cos(50°)sin(19.17°)

sin(30°)

cos(50°) sin(19.17°)

~ cos(50°) / \ cos(50°)si—n(19.17°) sin(50°) cos(19.17°)

.

y

(3.208)

£ = —1929 = —1.92-4m/s = —7.67Tm/s

(3.209)

Va = £ YA = —7.6Tm/s Ya

(3.210)

So finally We

can find the instantaneous

center of zero velocity for the rod

using the propagation

formula if we know the velocity of one point on the rod and the angular velocity of the rod:

Va =

(3.66)

N UE

= &Zax (P: Xa

X Paz

(3.211)

+ py YA)

where point Z is the instantaneous center of zero velocity.

.

-76Tmis Yo

=

.

N

& 24 x (p, Xs

.

+ Py YA)

=

Xa

Ya

2Za

Pz

Dy

0

0

0

&

=

-~

pz& YA

—

-

pyé Xa

(3.212)

103

Velocity and Acceleration

The above equation yields two scalar equations —7.6Tm/s

=

pg&

0

=

—pyd

—

—7.67m/s &

Pz =

p, = 0

(3.213)

We can find & as

_

_Zsin(¢)

(3.205)

+ ycos(h)

Lsin(e)

Therefore

Dy

=

= =

=

_Z1.6Tm/ssin(50°)

+ 4m/scos(30°)

0.1msin(19.17°)

_

—7.67m/s

T3ddreay

—0.104m

=

= T3 Adrad/e (3.214)

-10.44cm

(3.215)

So the instantaneous center of zero velocity for the rod is ’Z.

RS

R

il

ghruapaiy f‘(_:.;"_’.:i’.‘ifl;

Pa = z —1044em Xp |

(3.216)

Equation (3.2.16) implies that if we consider point Z to move as though it is rigidly attached to the rod, at this instant, Vz = 0, and the rod is rotating about point Z.

3.16 3.16.1

Exercises Short

Answer

Section 3.2 Translational Velocity

Provide brief answers to the following questions. Assume that all velocities and accelera-

tions mentioned are absolute unless otherwise stated in the problem. S3.1 S3.2

What

is the

tional velocity? What

definition

S$3.10 What is the rationale behind omitting the translational observer notation on the velocity vector when considering the absolute velocity of a point?

of transla-

is the definition of rotational

velocity?

S3.3 What is the purpose of parameterizing our generalized coordinates by time?

S3.11 Name two rules from calculus used in differentiating a position vector. $3.12 Find the velocity of point A given Pna

3.1 The

Observer

S3.5 Why do we need to consider the motion of the observer when taking the derivative of a position vector? S3.6

What

is an absolute observer?

$3.7

What

is the rotational observer?

S3.8

What

is the translational observer?

S3.9 Why is the velocity vector always tangent to the path followed by a moving point?

L XA

+ h Y4

and using the

relationship between the inertial and body-attached frames shown in Figure 2.26. Assume that L and h are constants, and express the result in frame A.

S3.4 What is the function of the variable time parameter in parameterizing the generalized coordinates? Section

=

Section

3.3 Rotational

Velocity

$3.13 Describe two methods for finding the angular velocity vector.

$3.14 If NP S3.15

I’t: NwD

= _& Yp, what is PwN? =

—

?B

and

EwD

=

& Zg, what is NwE? S3.16

If frame E rotates with a constant speed when observed from frame N, NwE

=

0.5rad/s

iN

+

0.3rad/s

?N

104

Velocity and Acceleration

find the orientation vector assuming that the initial angle is equal to Zero.

Section

3.4

The

Function

Designator

“(t)”

S53.17 Why do we omit the “(t)” function designator from most quantities developed in this text?

Section 3.11 Quasi-Velocities in No-Slip Rolling S3.26 What is the definition velocity?

Section

3.12 Acceleration

S3.27

Section 3.5 Generalized Speeds

Section

S$3.18 What is a generalized speed? Section

3.6 Implicitly Satisfying

straints

ConS3.28

53.19 Under what circumstances can you only satisfy constraints implicitly? Section 3.7 The Transport Theorem 53.20 What is the main reason for using the transport theorem?

that L and h are constants,

and express the result in frame A.

Section 3.8 Velocity Propagation Two Points on a Rigid Body

3.9

Relative

Versus

3.12.1 Normal Coordinates Is

the

and

Tangential

translational

acceleration

vector always tangent to the path of

Section 3.12.2 Velocity-Dependent celerations

Ac-

S3.30 What is centrifugal acceleration? Section

53.31

3.12.3 Fictitious and Forces What are and forces?

Accelerations

fictitious

accelerations

Section 3.13 Implicitly Satisfying Con-

S3.23 Point B lies on the same rigid body as point A in problem $3.21 such that Pag =r X4, Use velocity propagation to find the velocity of point B.

Velocity

is the definition of accelera-

a point’s motion?

for

53.22 What is the reason for using the velocity propagation for two points fixed on a rigid body?

Section

What tion?

S3.29 What is centripetal acceleration?

S3.21 Use the transport theorem to find the velocityof point A given Pyy = L XA + h Ya and using the relationship between the inertial and bodyattached frames shown in Figure 2.26. Assume

of a quasi-

Absolute

53.24 Given the velocities of two points A

straints Revisited

S3.32 What is the purpose of differentiating the constraint equations twice? Section 3.14 Acceleration Propagation for Two Points Fixed on a Rigid Body and Relative Acceleration S53.33 How do we obtain the acceleration propagation formula for two points fixed on a rigid body?

and B, Vi

Vg

= 3m/a XN+5m/s YN and = 2in/s XN + 2in/s YN, find the

relative velocities Section

3.10

Instantaneous

Zero Velocity 53.25

NV 45 and NVsa. Center

of

What is the instantaneous center of

zero velocity?

3.16.2

Practical

Write up the mat given in velocities and are absolute, problem.

following problems using the forSection 3.15.1. Assume that all accelerations requested or given unless otherwise stated in the

Velocity and

Acceleration

05 105

Sections 3.1 through 3.6

P3.1

velocity of the slider at the configuration shown in Figure 3.41 if the wheel is rotating clockwise at a rate of 4.3rad/s. (an-

[P2.1] At the configuration shown in Fig-

swer: ||V| = 24.15¢m/s.)

ure 3.39, the upper link is rotating in the counterclockwise direction at a rate of 3rad/s observed from the inertial reference frame. The lower link is rotating in the clockwise direction at 2rad/s observed from the frame attached to the upper link. Under these conditions, find

\)\

Statlonary Channel

Ié\ -

' }

the absolute velocity of point B expressed in the inertial reference frame. (answer: HV”

=

44.13cm/s.)

Figure 3.41: P3.4

Figure 3.39:

Double Pendulum.

the right at a constant speed of 2m/s, find

the velocity of the rightmost block. Use the inertial reference point N given in the

at the configuration shown in Figure 3.40 assuming that the tip of the rod sticks to while

In Figure 3.42, the smaller pulleys are of identical size and the larger pulleys are of

identical size. Since we are not interested in the rotation of the pulleys, model them as particles. If the rope is being pulled to

P3.2 [P2.8,P2.23] Find the velocity of Point A the ground

Crank-Slider Mechanism.

figure.

the block slides to the

right with a speed of 2t/s. Assume that the rod always remains in contact with the block and the ground, and that the entire lower surface of the block is always in contact with the ground. Use the transport

gravity

s

theorem to solve this problem.

Figure 3.42: Blocks and Pulleys. Fi

P3.3 The 3.41

3.40:

B crank-slider

consists

.

P3.5

mechanism

of a circular

in Figure

wheel

whc?se

center is stationary. A rod is pinconnected to the wheel, which in turn is

connected to a slider that must remain within the horizontal channel. Find the

o

y

[P2.11,P2.30] The thin rod in Figure 3.43 .

.

:

always remains in contact with the le ft-

most corner of the circular channel where

At the configuration r = R = 0.75m. shown in Figure 3.43 the absolute angular

speed of the rod is 1.6rad/s in the counteri clockwise direction. Find the velocity of

Velocity and Acceleration

point A on the rod at this instant.

swer:

||V = 2.4m/s.)

(an-

the right expressed in the inertial reference frame.

| Figure 3.43 Figure 3.45: Planetary Gears.

P3.6 {P2.11,P2.30,P3.5) Repeat problem P3.5 assuming that the channel is elliptical with r =1m

and

R

= 0.75m.

P3.7 [P2.9,P2.25] The angular speed of body C in Figure 3.44 is 5rad/s in the counterclockwise direction. Find the velocity of point A at the configuration shown in Figure 3.44 by differentiating Pna. Express the answer in the inertial reference frame. (hint: examine the Solution in Section

3.15.8. answer: 4.9m/s.) im

L2l

P3.9 Figure 3.46 shows a circular cam and a follower shaft that slides within a stationary channel. The spring attached to the follower forces it to remain in contact with the cam always. Point A is the center of the circular cam. Find the magnitude of the velocity of the follower at the instant shown, if the cam rotates at an angular speed of 27rad/s in the counterclockwise direction.

2Ly

(answer:

169.42m/s.)

Follower

Figure 3.44: Linkage.

Figure 3.46:

P3.8 [P2.37) The small gears in the planetary gear system in Figure 3.45 have an effective radius of r = 1cm, and the large center gear has an eflective radius of R = 4em. Assume that the center of the center gear is stationary and the outer gear is stationary. If the small gear on the far right side of the planetary gear system rotates counterclockwise at a rate of 3.3rad/s at the instant shown in Figure 3.45, find the angular velocity of the large center gear.

Also

find

the translational

veloc-

ity of the center of the smaller gear on

P3.10

The

ball in Figure

3.47

rolls without

slipping on top of the block. The block is free to slide horizontally on the ground but does not rotate. We are modeling for the condition where the ball stays in contact with the top surface of the block.

If

the block slides to the right with a speed of 4ft/s and the ball, of radius 1y¢, rotates counterclockwise with an angular speed of 0.2rad/s, find the velocity of point A at the configuration shown in Figure 3.47.

Velocity and Acceleration

107

upper two pulleys are identical in size, with a radius of 2cm, and their centers are the same distance from the ceiling. Model the pulleys as particles represented by points at their geometric centers. The pulley at point A has a radius of 3.5¢m. Find the angular velocity of the horizontal rod if the rope is pulled toward the left at a speed of 4cm/s.

Figure 3.47:

P3.11 In Figure 3.48, the smaller pulleys are of identical size and have a radius r = Im, the concentric pulleys are of identical size and the outer radius is R = 2m; the concentric pulleys consist of a small pulley rigidly attached to a larger pulley. We don’t care about the rotation of the smaller pulleys so you can model them as particles. The concentric pulleys are connected by a belt and cannot be modeled as particles. If the rope is being pulled to the right at a constant speed of 2m/s, find the velocity of the rightmost block. Use the inertial reference point N given in the

! e

LT

Yem g

Figure 3.49: Four Bar Pulley.

figure. (answer: ||V|| = 1m/s.)

L

L

P3.13 In slides cular ter of lar.

Figure 3.50, a along the surface block. Point A the cylinder if

The

semicircular

rectangular block of another semicirrepresents the cenit were fully circublock

rolls

along

the ground without slipping. However, as the semicircular block rolls toward the left, the rectangular

Figure 3.48: Blocks and Pulleys. P3.12 In Figure 3.49, assume that point A only moves in the vertical direction. A pulley and rod are both pin-connected to the ground at point E, but the rod and

pulley are not attached to each other. The

block begins to slip

on its surface. At the configuration shown in Figure 3.50, the semicirle has rolled toward the left 15° from when its top surface

is horizontal. At the configuration shown,

point A is moving toward the left w1tt‘1 a

speed of 2.7cm/s, a..nd the block is sliding towa.'rd the left wl?h. a speed of 0.603/3 relative to the semicircular bl(fck. Un ler these conditions, find the velocity of point

B. (answer: ||V = 4em/s.)

108

Velocity and Acceleration

Figure 3.52: Two Balls Rolling.

Figure 3.50:

P3.14 In Figure 3.51, the pulleys are of identical size and the blocks are of identical size. The blocks are connected by ropes and they do not rotate. Model the pulleys as particles. Find the velocity of block A if block C moves toward the right with a speed of 7rt/s.

P3.16 In Figure 3.53, the smaller pulleys are of identical size and their centers are connected to the horizontal bar. Assume that the bar always remains horizontal. A sphere hangs from the larger pulley. All bodies only translate in the vertical direction. Model the pulleys as particles located at their geometric centers. If the sphere moves downward with velocity 27t/s, find the velocity of the bar.

G

Figure 3.51: Blocks and Pulleys.

P3.15 [P2.29] Figure 3.52 shows two balls rolling without slipping on flat ground. Both balls stay in contact with each other and with the ground. The larger ball has a radius of 3in and the smaller ball has a radius of 1.2in. If larger ball rotates clockwise through 3.7revolutions at constant rate of 0.7rad/s, find the velocity of point D at this instant and express it in the inertial reference frame. (answer: nV"

= l.4|'n/l.)

L

R Tl

Figure 3.53: Bar and Pulleys.

P3.17 In Figure 3.54, a 6in X 3.5in rectangular block can rotate and slide along a horizontal surface, but must always remain in contact with the surface. At the instant shown in Figure 3.54, the contact point is stationary and the block is rotating clockwise at a rate of 1.4rad/s.

Find the veloc-

ity of the geometric center of the block at

the instant shown in Figure 3.54. (answer: "V"

= 4.86:’11/3.)

Velocity and Acceleration

38° &

/A S

Figure 3.54: Rocking Block.

~—In— Figure 3.56 Sections 3.7 through 3.11

P3.18 In Figure 3.55, the disk attached to a rope rolls without slipping on an incline. The disk has a radius of 6cm. If the rope is pulled downward in the direction shown in Figure 3.55 at a speed of 1.4cm/s, find the wheel’s angular velocity and the velocity of the wheel’s center. Model the pulleys as particles located at their geometric centers and use the inertial point N given in Figure 3.55. A

P3.20 In Figure 3.57, a peg to a rod such that the slide within a slot in rotates about point N,

is rigidly attached rod can rotate and a disk. The disk which is the iner-

tial reference point. At the instant shown,

w = 3rad/s. The peg is moving away from point N within the slot at a speed

of 2¢m/s, and the absolute angular speed

of the rod is 0.5rad/s in the clockwise direction. The angle given in Figure 3.57 is measured from the centerline of the slot. Find the velocity of point C on the rod at the instant shown in Figure 3.57.

Figure 3.55

P3.19 In Figure 3.56, a disk is pin connected to the ground and to an L-shaped link, which is in contact with a wedge attached

to the ground.

Assuming that the L-

shaped link remains in contact with the wedge, find the velocity of point C at the configuration shown in Figure 3.56 if the disk is rotating counterclockwise

with a spedd of 6rad/s. 25.2Tm/s.)

(answer: ||V =

Figure 3.57

P321 (P29 pP2.25P3.7 using velocity point C.

Repeat

problem

P3.7

in Figure

3.58

propagation

P3.22 [P2.10,P2.27] The wheel

starting

at

rolls without slipping and maintains con-

tact with the ground.

The pin-connected

110

Velocity and Acceleration

rod has a width of 10cm and also maintains contact with the ground. Find the velocity of point C in Figure 3.58, using the instantenous center of zero velocity and velocity propagation if the wheel rotates clockwise with an angular speed of 2.3rad/s and an angular acceleration equal

speed of v = 2cm/s, find the angular velocity and the linear velocity of the center of

the barrel.

(Hint: use the instantaneous

center of zero velocity to solve this prob-

to Orad/s®.

§ dt

dPny arnNy _

o

mg + mp + mg

dt

v

my

B VI

Inertia Dyadic and Angular Momentum

In Section 4.1, we found the linear momentum of a rigid body as its total mass multiplied by the velocity of its mass center. However, if the velocity of the mass center is equal to zero, this does not mean that the body has no momentum. The

body may rotate about its mass center such that every point other than the mass center has some linear momentum; recall

the circular saw example in Figure 4.2. The linear momentum

of the other

points

momentum;

can

be described

as the body’s

engular

recall that the position of all of the other points

is included in geom. of the location description. The body’s angular momentum is similar to its linear momentum, the ro-

~

‘i} Ne

tational mass multiplied by the angular velocity. We find this P form by taking the sum of moments for the forces in Newton'’s e Second Law for all of the tiny particles comprising the rigid Figure 4.8: Particle B; Relative to Mass Center A. body. This calculation results in taking the sum of the moment of linear momentum for each tiny particle to obtain the angular momentum of the rigid body. You know from your course on statics that you can take moments about any point you like. We will sum moments about the body’s mass center in Section 4.2.1 and sum moments about arbitrary points in Section 4.2.8.

4.2.1

Inertia Dyadic for a Rigid Body About its Mass

Absolute Rotational Observer\o

EM)..

O-

=

dH

bidnio d 2 (5

Body ™ Mass Center

Figure 4.9: Euler’s Second

Center

The moment of the linear momentum of body B; in Figure 4.8 about the mass center can be defined as,

(XM)g,a

= Pap, x (3 F)g,

(4.3) =

Pap, x

JK

B

dt -

(422)

where (Y M BiA is the sum of moments on particle B; about .

point A. Summing Equation (4.22) over the entire body yields:

Law.

(EM)py = —fl::“

(4.23)

122

Mass and Inertia

where (3"M), , is the sum of moments on body A is the angular momentum of body A about its mass Euler’s Second Law. We will consider this law for physical meaning of the notation in Equation (4.23)

about its mass center, point A, and Hxx center. Equation (4.23) is a statement of arbitrary points later in Section 4.2.8. The is shown in Figure 4.9.

We know that moments can be taken about any point and still yield a valid representation of the rotational motion. Therefore, we consider the sum of moments, the resultant moment, independent of any point. Yet to determine the resultant moment, we must choose a point. The angular momentum,

Haa

in Equation

(4.23), must

be consistent with that choice.

The

rota-

tional mass contained within the angular momentum is considered to have a physical meaning that depends on the chosen point.

Proof: Equation (4.23) The sum of moments of linear momenta in Equation (4.22) can be expressed as:

(EM)s

22)2(2M)B|A §PAB, X (ZF)s, =, D Pas x dI;tB' i=1

(4.24)

The term on the far right side of Equation (4.24) is not useful because it involves an infinite sum. We can find a different expression for this term as follows: =

dPAB.

d

e

b

gd— S

x Kp, + ZPAB, dKs, dt i=1

ildt

(4.25)

Define the angular momentum as the sum of the moment of linear momentum about the mass center, point A, yields o0

Haa “=

.23

o0

ZHB,A

=

) i=1

00

ZPAB'

xXKp,

=

ZPABI

i=1

x mp,

VB,

(4.26)

i=1

Substituting the definition of angular momentum in Equation (4.26) into Equation (4.25) yields,

e

dHan

At

e

g

dI'IAA

TyZHBIA

—

d(PNBl

S~ 4Pxp

—

= 3i=1

ZmBM

E

m

x K,

o

3 Xma Ve an & dt

dPna

=

gPAB‘

+

i=1

- !

(4.27)

dK

—2

(4.28)

X dI:tB'

(4.29)

X

= xmp Ve, + ) Pap x i=1

o0

AA

dKB

oo

- PNA)

g R

o0

dl;?A

x mgp, Vg,

+

ZPAB'

i=1

i=1

Therefore,

dHan

=

@)-go~

dP dHaa th @12) .am

dPxA A P Vama

x ma, (vA +

-

d;“x

N, Vo A

prB,)

o N“’AXZ

= .>< +§PAB

(4.30)

0

Pas,

+ZPAB. '(431)

Mass and Inertia

123

which yields Euler’s Second Law in Equation (4.23) dHaa

dt

(431)§PAB'X

dKp,

dt

(424 (M) a4

u

The angular mome_ ntum , Hpp in E quati ion (4.23), isi equal to the sum of the angular momentum of each particle B; about the mass center of body A, Hp,4: ¢ [o o]

o0

o0

H AA (423) = ;HBIA == ;PAB, xKp, = ZPAB. x mp, Vp, = i= i=1 In 'fmalyzing the summation in Equation (4.26), the goal is to 1sol.ate the angular velocity in the angular momentum, Absolute Rotational revealing a description of the rotational mass.

Observer

The result of Equation (4.26) is

Hyn=

Haa (429) Ina - Nt )

:

]

Bod

(4.32)

*_

(4.26)

Body-Attached

Ing-

Frame

(DO‘)

Mass Center

Figure 4.10: Angular Momentum about the Mass Center.

The physical meaning of the notation in Equation (4.32) is shown in Figure 4.10. The term I, 4 is the rotational mass, which describes the resistance of body A to rotation.

Ina

This term can be expressed as:

= Iz XpXa

- Iy (iA?A +?AiA)

R

o

o

+ Iy Ya¥a

o

o

I (YaZa+2a¥n) + L. ZaZa — L. (XaZa +ZaXa) The expression Ips in Equation (4.33 is referxi d to as :lfe ‘ine‘r‘(zia dyat(iic 01_

Table 4.1: Rules for Dyadic; Vector Algebra.

body A about point A, where I, I, and I,. are moments of inertia and Iifyv I,:, and I,; are products of iner-

Dyadic product is non-commutative -~ -~ =~ = XaZa # ZaXa

t.za. Compared to a single scalar quantity for the translational mass, six scalar

Dot multiplication from right side — —

quantities

are required

rotational mass.

to represent

the

The rotational mass is

directional, quite different from the trans-

lational mass.

Considering the different

directions, the inertia dyadic has nine differcnt terms.

In

Equation

(4.33),

the

rotational

mass is expressed( in .'):L form that uses

(4.33)

XAZA

s

5

XaZa

-Xa

=

=Xa0=0




Ya + Low. Zp - I;.w, Xa

(4.35)

Furthermore, if we consider the cylinder example in Figure 4.8, where the products of inertia are equal to zero, Iy, = I;, = 0, the angular momentum

HAA

(4=32)

IAA

et

ZA

(4=35)

YA

_MJZ

Izzw=

+

which only has a component in the EA direction.

R

0 _

-

0 .

~

becomes,

-

ZA

}4)2

xA

=

Iz:wz

ZA

(4.36)

Therefore, knowing that the cylinder exhibits

planer motion and that products of inertia are equal to zero, its inertia dyadic is,

Ina = L. ZaZa where I, completely describes its rotational mass.

(4.37)

This description occurs frequently in the

examination of planar motion. The moments and products of inertia are defined in Table 4.2, where,

Pyp, (436) zi Xa

+ 4 Ya

+

7 Za

(4.38)

The distances z,, y;, and z; are variable parameters that range over the body’s dimensions.

We

express P ap, in the body-attached frame A so that z;, y;, and z; will be constant for each point B;. These position vectors are included in the geometry, geom., of the location description, recall Section 2.12. If, instead, we express P,

in the inertial reference frame, z;, y;, and z; may be

varisbles that change as body A moves. This alternate choice makes it difficult to obtain the simple. constant expressions in Table 4.2 for the cylinder’s inertias in Figure 4.11. As you may imagine, it is difficult to evaluate the moments and products of inertia using the expressions in Table 4.2. Thus, in the end, we will use integration techniques from calculus rather

Mass and Inertia

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Table 4.2: Moments and Products of Inertia.

A A A Z A ZN

Moments of Inertia

i=1

Ly =