•
ENGINEERING MECHANICS
A N INTRODUCTION TO
•
DYNAMICS
Fourth Edition
•
ENGINEERING MECHANICS
•
A N INTRODUCTION TO
•
DYNAMICS
DAVID J . M C G I L L AND WILTON W . KING
Professors Emeritus, Georgia Institute of Technology
Copyright © 2003 by David J. McGill and Wilton W. King
All rights reserved. No part of this book may be reproduced, stored in a retrieval
system, or transmitted, in any form or by any means —electronic, mechanical,
photocopying, recording, or otherwise—without prior written permission of the
authors.
For more information, contact:
Tichenor
Publishing
Division of T.I.S., Inc.
Tichenor Publishing
5005 North State Road 37 Business
Bloomington, IN 47404-1626
To
OUR WIVES, C A R O L Y N AND
KAY
CONTENTS
PREFACE
1
KINEMATICS OF MATERIAL POINTS OR PARTICLES
1.1
1.2
1.4
1.5
1.6
1.7
2
3
3.3
3.4
3.5
55
Introduction, 56
Newton's Laws and Euler's First Law, 56
Motions of Particles and of Mass Centers of Bodies, 62
Work and Kinetic Energy for Particles, 87
Momentum Form of Euler's First Law, 101
Euler's Second Law (The Moment Equation), 117
Summary, 125
Review Questions, 128
KINEMATICS OF PLANE MOTION OF A RIGID BODY
3.1
3.2
1
Introduction, 2
Reference Frames and Vector Derivatives, 3
Position, Velocity, and Acceleration, 6
Kinematics of a Point in Rectilinear Motion, 8
Rectangular Cartesian Coordinates, 24
Cylindrical Coordinates, 31
Tangential and Normal Components, 43
Summary, 53
Review Questions, 54
KINETICS OF PARTICLES AND OF MASS CENTERS
OF BODIES
2.1
2.2
2.3
2.4
2.5
2.6
xi
129
Introduction, 130
Velocity and Angular Velocity Relationship for Two Points of the Same
Rigid Body, 134
Translation, 147
Instantaneous Center of Zero Velocity, 149
Acceleration and Angular Acceleration Relationship for Two Points of
the Same Rigid Body, 163
vii
Page viii
3.6
3.7
3.8
Rolling, 170
Relationship Between the Velocities of a Point with Respect to Two
Different Frames of Reference, 198
Relationship Between the Accelerations of a Point with Respect to
Two Different Frames of Reference, 207
Summary, 215
Review Questions, 216
KINETICS OF A RIGID B O D Y IN PLANE M O T I O N /
DEVELOPMENT AND SOLUTION OF THE DIFFERENTIAL
EQUATIONS GOVERNING THE M O T I O N
4.1
4.2
4.5
4.6
4.7
5
6
Introduction, 218
Rigid Bodies in Translation, 219
Moment of Momentum (Angular Momentum), 227
Moments and Products of Inertia/The Parallel-Axis Theorems, 229
The Mass-Center Form of the Moment Equation of Motion, 246
Other Useful Forms of the Moment Equation, 272
Rotation of Unbalanced Bodies, 291
Summary, 301
Review Questions, 303
SPECIAL INTEGRALS OF THE EQUATIONS OF PLANE
M O T I O N OF RIGID BODIES: W O R K - E N E R G Y AND
IMPULSE-MOMENTUM METHODS
5.1
5.2
5.3
6.8
6.9
304
Introduction, 305
The Principle(s) of Work and Kinetic Energy, 305
The Principles of Impulse and Momentum, 343
Summary, 375
Review Questions, 377
KINEMATICS OF A RIGID B O D Y IN
THREE-DIMENSIONAL M O T I O N
6.3
6.4
6.5
217
379
Introduction, 380
Relation Between Derivatives/The Angular Velocity Vector, 380
Properties of Angular Velocity, 384
The Angular Acceleration Vector, 398
Velocity and Acceleration in Moving Frames of Reference, 401
The Earth as a Moving Frame, 410
Velocity and Acceleration Equations for Two Points of the Same Rigid
Body, 414
Describing the Orientation of a Rigid Body, 428
Rotation Matrices, 434
Summary, 441
Review Questions, 442
iPage
7
KINETICS OF A RIGID BODY IN GENERAL MOTION
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
8
Introduction, 445
Moment of Momentum {Angular Momentum) in Three Dimensions, 446
Transformations of Inertia Properties, 448
Principal Axes and Principal Moments of Inertia, 455
The Moment Equation Governing Rotational Motion, 472
Gyroscopes, 492
Impulse and Momentum, 499
Work and Kinetic Energy, 504
Summary, 513
Review Questions, 514
SPECIAL TOPICS
8.1
8.2
8.3
8.4
444
516
Introduction, 517
Introduction to Vibrations, 517
Euler's Law for a Control Volume, 535
Central Force Motion, 543
Review Questions, 554
APPENDICES CONTENTS
557
Appendix A
UNITS, 558
Appendix B
EXAMPLES OF NUMERICAL ANALYSIS/THE NEWTONRAPHSON METHOD, 564
Appendix C
MOMENTS OF INERTIA OF MASSES, 567
Appendix D
ANSWERS TO ODD-NUMBERED PROBLEMS, 573
INDEX, 583
SUPPLEMENTARY PROBLEMS
589
x
PREFACE
An Introduction to Dynamics is the second of two volumes covering basic
topics of mechanics. The first two-thirds of the book contains most of the
topics traditionally taught in a first course in dynamics at most colleges of
engineering.
In the writing of this text we have followed one basic guideline — to
write the book the same way we teach the course. To this end, we have
written many explanatory footnotes and included frequent questions
interspersed throughout the chapters. These questions are the same kind
as the ones we ask in class; to make the most of them, treat them as
serious homework as you read, and look at the answers only after you
have your own answer in mind. The questions are intended to encourage
thinking about tricky points and to emphasize the basic principles of the
subject.
In addition to the text questions, a set of approximately one dozen
review questions and answers are included at the end of each chapter.
These true-false questions are designed for both classroom discussion
and for student review. Homework problems of varying degrees of diffi­
culty appear at the end of every major section. There are over 1,100 of
these exercises, and the answers to the odd-numbered ones constitute
Appendix D in the back of the book.
There are a number of reasons (besides carelessness) why it may be
difficult to get the correct answer to a homework problem on the first try.
The problem may require an unusual amount of thinking and insight; it
may contain tedious calculations; or it may challenge the student's ad­
vanced mathematics skills. We have placed an asterisk beside especially
difficult problems falling into one or more of these categories.
Some examples and problems are presented in SI (Systeme Interna­
tional) metric units, whereas others use traditional United States engi­
neering system units. Whereas the United States is slowly and painfully
converting to SI units, our consulting activities make it clear that much
engineering work is still being performed using traditional units. Most
United States engineers still tend to think in pounds instead of newtons
and in feet instead of meters. We believe students will become much
Page xi
Page xii
better engineers, scientists, and scholars if they are thoroughly familiar
with both systems.
Dynamics is a subject rich in its varied applications; therefore, it is
important that the student develop a feel for realistically modeling an
engineering situation. Consequently, we have included a large number
of actual engineering problems among the examples and exercises. Being
aware of the assumptions and accompanying limitations of the model
and the solution method is a valuable skill that can only be developed by
sweating over many problems outside the classroom. Only in this way
can a student develop the insight and creativity that must be brought to
bear on engineering problems.
Kinematics of the particle, or of a material point of a body, is covered
in Chapter 1. The associated kinetics of particles and mass centers of
bodies follows logically in Chapter 2. Here it will be seen that we have not
dwelt at length upon the "point-mass" model of a body. Since the engi­
neering student will be dealing with bodies of finite dimensions, we
believe that it is important to present equations of motion valid for such
bodies as quickly as possible. Thus Euler's laws have been introduced
relatively early; this provides for a compact presentation of general prin­
ciples without, in our opinion and experience, any loss of understanding
on the student's part. This is not meant to deprecate the point-mass
model, which surely plays an important role in classical physics and can
be utilized in a number of engineering problems. As we shall see in
Chapter 2, however, these problems may be attacked directly through
the equation of motion of the mass center of a body without detracting
from the view that the body has finite dimensions. Trajectory problems,
sometimes placed with particle kinematics, will be found in this kinetics
chapter also, since a law of motion is essential to their formulation.
The rigid body in plane motion is treated in detail in the center of the
book — the kinematics in Chapter 3 and the kinetics in Chapters 4 and 5.
In Chapter 3 the topic of rolling is discussed only after both the velocity
and acceleration equations relating two points of arigidbody have been
covered. Further, we treat the equations of velocity and acceleration of a
point moving relative to two frames of reference ("moving frames") in
Chapters 3 (plane kinematics) and 6 (three-dimensional, or general, kine­
matics), after the student has been properly introduced to the angular
velocity vector in these chapters.
Chapter 4 approaches plane kinetics from the equations of motion,
written with the aid of a free-body diagram of the body being studied —
that is, a sketch of the body depicting all the external forces and couples
but excluding any vectors expressing acceleration. Thus the free-body
diagram means the same thing in dynamics as it does in statics, facilitat­
ing the student's transition to the more difficult subject. Moments and
products of inertia are covered right where they appear in the develop­
ment of kinetics. This presentation gives students an appreciation of
these concepts, as well as a sense of history, as they encounter them along
the same paths that were traveled by the old masters.
Chapter 5 is dedicated to solving plane kinetics problems of rigid
Page xiii
bodies with certain special yet general solutions (or integrals) of the
equations of Chapters 2 and 4. These are known as the principles of
work and kinetic energy, impulse and momentum, and angular
impulse and angular momentum.
Chapters 6 and 7 deal comprehensively with the kinematics and
kinetics, respectively, of rigid bodies in three dimensions. There is no
natural linear extension from plane to general motion, and the culprit
is the angular velocity vector which depends in a much more com­
plicated way than
on the angles used to orient the body in three
dimensions. We have found that if students understand the angular
velocity vector they will have little trouble with the general motion
of rigid bodies. Thus we begin Chapter 6 with a study of and its
properties. In three dimensions, the definition of angular velocity is
motivationally developed through the relationship between deriva­
tives of a vector in two different frames of reference. While this point
of view is often associated with more advanced texts, we have found
that college students at the junior level are quite capable of appreciat­
ing and exploiting the power of this approach. In particular, it allows
the student to attack, in an orderly way, intimidating problems such
as motions of gear systems and those of universal joints connecting
noncollinear shafts.
Chapter 8 is an introduction to three special topics in the area of
dynamics: vibrations, mass redistribution problems, and central force
motion.
We have received a number of helpful suggestions from those
who taught from earlier editions of the text, and we are especially
grateful to Lawrence Malvern of the University of Florida, Don Carlson
of the University of Illinois and to our colleagues at Georgia Institute of
Technology, in particular Erian Armanios, Olivier Bauchau, Don
Berghaus, Al Ferri, Janet Hampikian, Satya Hanagud, Dewey Hodges,
Larry Jacobs, Steve Johnson, Manohar Kamat, George Kardomateas,
Harvey Lipkin, Chris Lynch, Richard Neu, John Papastavridis, Mimi
Philobos, Jianmin Qu, Nader Sadegh, Marilyn Smith, Virgil Smith, Jeff
Streator, Ray Vito, Wan-lee Yin, and Minami Yoda.
We thank Meghan Root for her cheerful assistance with prior edi­
tions. And for their many useful suggestions, we are grateful to our
third-edition reviewers, who were:
William Bickford
Arizona State University
Donald E. Carlson
University of Illinois at Urbana
Vincent WoSang Lee
University of Southern California
Joseph Longuski
Purdue University
Robert L. Collins
University of Louisville
Robert G. Oakberg
Montana State University
John Dickerson
University of South Carolina
Joseph E. Parnarelli
University of Nebraska
Page xiv
John F. Ely
Northern Carolina State University
Laurence Jacobs
Georgia Institute of Technology
Mario P. Rivera
Union College
Wallace S. Venable
West Virginia University
Seymour Lampert
University of Southern California
Carl Vilmann
Michigan Technological University
We are grateful to the following professors, who each responded
to a questionnaire we personally sent out in 1991: Don Carlson,
University of Illinois; Patrick MacDonald and John Ely, North Carolina
State University; Vincent Lee, University of Southern California; Charles
Krousgrill, Purdue University; Samuel Sutcliffe, Tufts University; Larry
Malvern and Martin Eisenberg, University of Florida; John Dickerson,
University of South Carolina; Bill Bickford, Arizona State University;
James Wilson, Duke University; Mario Rivera, Union College; and Larry
Jacobs, Georgia Institute of Technology. Their comments were also
invaluable.
Last but not least, we are indebted to Kevin Theile, Manager of
Tichenor Publishing, for seeing this new edition through to comple­
tion.
David J. McGill
Wilton W. King
•ENGINEERING MECHANICS
•
•
A N INTRODUCTION TO
DYNAMICS
1
KINEMATICS OF MATERIAL
POINTS OR PARTICLES
1.1
1.2
1.3
Introduction
Reference Frames and Vector Derivatives
Position, Velocity, and Acceleration
1.4
Kinematics of a Point in Rectilinear Motion
The v-t Diagram
1.5
1.6
1.7
Rectangular Cartesian Coordinates
Cylindrical Coordinates
Tangential and Normal Components
SUMMARY
REVIEW QUESTIONS
Page 1
Page 2
1.1
Introduction
Dynamics is the general n a m e given to t h e study of the motions of bodies
a n d the forces that accompany or cause those motions. The b r a n c h of the
subject that deals only with considerations of space a n d time is called
kinematics. T h e b r a n c h that deals w i t h the relationships b e t w e e n forces
a n d motions is called kinetics, b u t since the force-motion relationships
involve kinematic considerations, it is necessary to study kinematics first.
In this chapter w e present some fundamentals of the kinematics of a
material point or, equivalently, a n infinitesimal element of material. We
shall use the term particle for such a n element, b u t w e shall also use this
term in a broader sense to d e n o t e a piece of material sufficiently small
that the locations of its different material points n e e d not b e distin­
guished. The vagueness of this definition correctly suggests that, for some
purposes, a truck or a space vehicle or even a planet might b e modeled
adequately as a particle.
The key elements of the kinematics of a point (or particle) are its
position, velocity a n d acceleration. Velocity is rate-of-change of position,
a n d acceleration is rate-of-change of velocity. It is the acceleration in
N e w t o n ' s Second Law a n d the position-velocity-acceleration relation­
ships that allow u s to d e d u c e t w o things: the forces that m u s t act for a
particle to achieve a certain motion; or the evolution of a particle's posi­
tion u n d e r t h e action of a set of prescribed forces.
Position, velocity, a n d acceleration vectors are d e n n e d in Section 1.3.
These definitions are i n d e p e n d e n t of the choice of a n y particular coordi­
n a t e system. However, solution of a practical problem almost always
involves the use of some specific coordinate system, the most c o m m o n
being rectangular (Cartesian), cylindrical a n d spherical. The rectangular
a n d the cylindrical coordinate systems are treated in great detail in this
chapter (Sections 1.4-1.6) because w e judge t h e m to b e of greatest prac­
tical use, particularly for problems of motion confined to a plane. These
developments are sufficient for establishing the procedures to follow
should the reader later find it desirable to develop counterpart relation­
ships using some other system.
If w e focus our attention o n the p a t h being traversed by a point (or
particle), w e find that t h e velocity of the point is tangent to the p a t h a n d
that the acceleration h a s parts n o r m a l a n d tangent to the p a t h w i t h
special significances. These characteristics are developed a n d exploited
in Section 1.7.
We n o w begin our study of particle kinematics with a preliminary
section devoted to the calculus of vectors that d e p e n d u p o n scalars. In
particular, w e n e e d to u n d e r s t a n d h o w to take derivatives of vectors with
respect to time a n d to acknowledge the crucial role of the frame of
reference.
Page 3
1.2
Reference Frames and Vector Derivatives
In the next section a n d t h r o u g h o u t the book, w e are going to b e differen­
tiating vectors; the derivative of the position vector of a point will b e its
velocity, for example. T h u s in this preliminary section it seems wise to
examine the concept of the derivative of a vector A, w h i c h is a function of
time t. The definition of dA/dt, w h i c h is also c o m m o n l y written as A, is
deceptively simple:
Figure 1.1
This definition closely parallels the definition of the derivative of a scalar,
such as dy/dx, as found in a n y calculus text. But w h a t w e m u s t realize
about a vector is that it can change with time to two w a y s — i n direction as
well as in m a g n i t u d e . This m e a n s that A is intrinsically tied to the frame
of reference in w h i c h the derivative is taken.
To illustrate this idea, consider the t w o points P a n d Q o n the surface
of the p h o n o g r a p h record in Figure 1.1. The record rests o n a turntable
that revolves in the indicated direction at, say,
rpm.
Suppose w e call R the vector that is the directed line segment from P
to Q, a n d inquire about the rate at w h i c h R changes with time as the
turntable rotates. Even t h o u g h w e perceive the record a n d turntable to
b e h a v e as a rigid b o d y so that the distance b e t w e e n P a n d Q (that is, the
m a g n i t u d e of R) is constant, most of us w o u l d judge dR / dt to be nonzero,
owing to the varying direction of R. This conclusion follows from our
automatically having a d o p t e d the building (or earth) as our frame of
reference. If w e w e r e to ride o n the turntable a n d blind ourselves to the
surroundings, however, our perception w o u l d b e that R is a constant
vector a n d consequently h a s a vanishing derivative. Thus R, relative to
the turntable, is a constant vector a n d , relative to the building, is a
constant-magnitude b u t varying-direction vector. It is therefore seen that
dR/dt cannot b e evaluated except b y specific association w i t h a f r a m e of
reference, w h i c h is n o t h i n g more n o r less t h a n a rigid body. We shall
discuss the frame of reference concept further in the next section a n d
again in C h a p t e r 3.
We shall b e needing several vector derivative relationships that h a v e
analogs in the calculus of scalars; these relationships follow directly from
the definition (Equation 1.1). If a is a scalar a n d A a n d B are vector
functions of t, t h e n
page 4
The first a n d second of these equations allow us to be m o r e specific
about t h e m a n n e r in w h i c h differentiation is linked to a frame of refer­
ence. Suppose that
are mutually perpendicular unit vectors* a n d
A j , A , A are t h e corresponding scalar c o m p o n e n t s of a vector A so that
2
3
If is t h e frame of reference** a n d w e denote t h e derivative of A
relative to b y
, then
N o w if w e
chooseooseto
constant there a n d
h a v e fixed directions in
they are each
which is t h e most straightforward w a y to express t h e derivative of a
vector a n d its intrinsic association with a frame of reference. We n o w give
o n e example of the use of Equation (1.8) and, assuming the reader to be
familiar with Equations (1.1) to (1.5), t h e n m o v e o n to Section 1.3 a n d the
task of describing t h e motion of a point (particle) P.
EXAMPLE 1.1
If the distance from P to Q on the
rpm record in Figure El. 1 a is 3 in. and if A is
the vector from P to Q, find
where the frame is the cabinet of the stereo in
which the axes (x, y, z) are embedded. It is also given that the line PQ is in the
indicated position (parallel to y) when f = 0.
Figure El.la
Solution
At a later time t (in seconds), the vector A is seen in Figure El. l b to make an angle
with y of
* Note that we could use any set of base vectors (that is, linearly independent reference
vectors) here, in which case A,, A , A are not necessarily orthogonal components of A.
Equation (1.6) simply illustrates the most common choice of scalars and base vectors.
When this is the case, the magnitude of A, written | A | or sometimes A, is
**Throughout the book, frames (rigid bodies) are denoted by capital script letters. These
are intended simply to be the capital cursive letters we use in writing; thus we write the
names of bodies and print the names of points. We do this because, as we shall see in
Chapter 3, points and rigid bodies have very different motion properties.
Note that the derivatives of the scalar components of A, such as dA,/d(, need not be
"tagged" since they are the same in any frame.
2
Figure E1.1b
3
Page 5
The vector A, expressed in terms of the unit vectors i and j in the respective
directions of x and y, then has the following form:
Noting that the unit vectors do not change in direction in 3, we obtain, using
Equation (1.8),
We see from this result, for example, that:
1.
2.
3.
4.
s in the x direction.
is in the —y direction.
is in the — x direction.
is in the y direction.
In all four cases, and at all intermediate angles as well, the derivative of A in
is seen to be that of the cross product:
The bracketed vector represents what will come to be called the angular velocity of
the record
in the reference frame (stereo cabinet) In later chapters we shall
see that it is precisely this cross product that must be added to
to obtain
Here, of course, A is constant relative to the turntable 5 0 that its derivative in
(that is,
vanishes.
PROBLEMS
•
Section 1.2
In Problems 1.1-1.8,
are mutually perpendicular
unit vectors having directions fixed in the frame of refer­
ence. In each case t is time measured in seconds ("s" in SI
units). Determine at t = 3 s the rate of change (with re­
spect to time) of vector L.
1.9-1.16 If the vectors enumerated in Problems 1.1-1.8
represent various forces F, find the integral of each force
over the time interval from t = 2 through 5 sec. Let the
metric units become newtons and the U.S. units become
pounds.
Page 6
1.3
Position, Velocity, and Acceleration
In this short b u t important section, w e present the definitions of the
position, velocity, a n d acceleration vectors of a material point P as it
moves relative to a frame of reference
It is important to mention that
while a frame of reference is usually identified b y the material constitut­
ing the reference body (for example, the earth, the m o o n , or the b o d y of a
truck), the frame is actually composed of all those material points plus the
points generated by a rigid extension of the body to all of space. Thus, for
example, w e refer to a point on the centerline of a straight pipe as a point
in (or of) the pipe.
W e n o w consider a point P as it m o v e s along a p a t h as s h o w n in
Figure 1.2. The p a t h is the locus of points of
that P occupies as time
passes. If we select a point O of to be our reference point (or origin), t h e n
the depicted vector from O to P is called a p o s i t i o n vector for P in a n d is
written r .
The first a n d second derivatives (with respect to time) of the position
vector are respectively called the velocity (v ) a n d acceleration (a ) of
point P in
O P
P
Figure 1.2
Position vector for P in
P
(The magnitude of v is called the s p e e d of P.)
P
The derivatives in Equations (1.9) a n d (1.10) are calculated in frame
the only frame u n d e r consideration here. Later, however, w e shall
sometimes find it necessary to specify the frame in which derivatives,
velocities, a n d accelerations are to b e computed; w e shall t h e n tag the
derivatives as in Equation (1.8) a n d write
W h e n e v e r there is just one frame involved, w e shall omit the on both
sides a n d write an equation such as (1.11) in the form of (1.9).
T h r o u g h o u t the text w e h a v e inserted questions for the reader to
think about. (The answer is always on the same page as the question.)
H e r e is the first question:
Question 1.1 Do the velocity and acceleration of a point P depend
upon: (a) the choice of reference frame? (b) the origin selected for the
position vector?
Answer 1.1 (a) Yes; w e could simply define a frame in w h i c h P is fixed, and it w o u l d
then have v = 0 = a . (b) N o ; letting O' be a second origin in and differentiating the
relationship r
= i , + r . in 3 s h o w s that: v (with origin O) = v (with origin O').
The derivative of r - in is, of course, zero! See Problem 1.17.
P
P
o p
00
QO
0
P
P
P
Page 7
At this point it is reasonable to wonder w h y w e have not chosen to
introduce time derivatives of the position vector higher than the second.
The reason is that the relationships between forces and motions do not
involve those higher derivatives. As w e shall see later w h e n w e study
kinetics, if w e know the accelerations of the particles making up a body,
the force-motion laws will yield the external forces; conversely, for rigid
bodies, if w e know the external forces, w e can calculate the accelerations
and then, by integrating twice, the position vectors. The force-motion
laws turn out to be valid only in certain frames of reference; for that
reason writers sometimes refer to motion relative to such a frame as
absolute motion. We have not used the word absolute here because w e
wish to emphasize that kinematics inherently expresses relationships of
geometry and time, independent of any laws linking forces and motions.
Thus in kinematics all frames of reference are of the same importance.
Finally, w e note that positions (or locations) of points are normally
established through the use of a coordinate system. The ways in which
positions, velocities, and accelerations are expressed in two of the most
common systems, rectangular and cylindrical, are presented in the next
three sections.
PROBLEMS
•
Section 1.3
1.17 Show that the velocity (and therefore the accelera­
tion also) of a point P in a frame does not depend on the
choice of the origin. Hint: Differentiate the following re­
lationship in (see Figure P 1 . 1 7 ) :
Figure PI.17
ence; v is the velocity of a point P moving in the frame; t
is time measured in seconds. Determine at t = 2 s the
acceleration of the point for the velocity given.
P
1.23-1.27 The displacement of a point over a time inter­
val f ] to t is denned to be the difference of the position
vectors—that is, r(f ) ~ r(f i). For the cases enumerated in
Problems 1 . 1 8 - 1 . 2 2 , find the displacement and the mag­
nitude of the displacement over the interval t = 4 s to
t = 6 s.
2
2
In Problems 1 . 1 8 - 1 . 2 2 ,
are mutually perpendicular
unit vectors having directions fixed in the frame of refer­
Page 8
1.4
Kinematics of a Point in Rectilinear Motion
In this section w e study problems in w h i c h point P m o v e s along a straight
line in t h e reference frame
this situation is called rectilinear motion,
a n d the position of P m a y b e expressed with a single coordinate x mea­
sured a l o n e t h e fixed line o n w h i c h P m o v e s (see Figure 1.3).
A position vector for P is simply
in w h i c h the unit vector is parallel to t h e line as s h o w n in Figure 1.3 a n d
hence does not change in either m a g n i t u d e or direction in Therefore P
has t h e following very simple velocity a n d acceleration expressions:
Figure 1.3
In rectilinear motion, there are three interesting cases w o r t h y of
special note:
1.
Acceleration is a k n o w n function of time, /(t).
2. Acceleration is a known function of velocity, g(v),where
3. Acceleration is a known function of position, h(x).
In each case, w e can go far with general integrations. W e shall consider
each case in turn a n d give a n example.
in w h i c h Q a n d C are to b e determined b y t h e initial conditions on
velocity a n d position, respectively, once t h e proble m (and t h u s /(f)) is
stated a n d t h e mdefinite integrals are performed. Alternatively, w e might
k n o w t h e values of x at t w o times, instead of o n e position a n d one
velocity. In a n y case, w e n e e d t w o constants.
2
EXAMPLE 1.2
The acceleration of a point P in rectilinear motion is given by the equatior
m / s , with initial conditions i(0) = 2 m / s and x(0) = — 7 m. Find x(t).
2
Solution
We note that this is the problem of a point moving with a quadratically varying
acceleration magnitude and with the initial conditions being the position and
velocity of P at t = 0 as shown in Figure El.2. Integrating as above,
And integrating once more,
Figure El.2
Page 9
The constants are found from the initial conditions to be C = 2 m / s and C
= — 7 m, as follows:
x
2
Thus the motion of the point P is given by the integrated function of time:
Suppose v(t) can be inverted to give t(v); then
If the integral can be found as a function pip), then w e may be able to
and
solve the equation p(v) = t + C for the velocity:
3
v = q(t)
If so, then
so that
EXAMPLE 1,3
x = / q(t) dt + C
4
(1.16)
This
procedure
should become
clearer
the following
example.
Suppose
that the acceleration
of a point
P inwith
one-dimensional
motion
is propor­
tional to velocity according to
with the same initial conditions as in
the previous example. Solve for the motion x(t).
Solution
Page 10
so that, integrating,* we get
Since v = 2 when t = 0, then C = ( - I n 2 ) / 2 and
3
When acceleration is a function of position,
Therefore
combine
and
to obtain the useful relation
w e may
But x = — 7 m when t = 0 s gives C = — 6 m, and so we obtain our solution:
4
Then, if a function r\x) exists such that
we obtain, from
Equation (1.17), the following:
and integrating with respect to time,
EXAMPLE 1 . 4
Thus the square of the speed is
Let x = h(x) = —4x m / s . Find v (x) if the initial conditions are the same as in
Examples 1.2 and 1.3.
Equation (1.19) will be called an energy integral in Chapters 2 and 5.
2
2
* This problem could also be solved b y first integrating the linear differential equation
v + 2v = 0, observing that A e ~ is the general solution.
2t
Page 11
Solution
We are dealing with the equation
Actually we know that the solution to this equation, by the theory of differen­
tial equations, is x = A sin 2t + B cos 2t—which, with x(0) = — 7 m and i(0)
= 2 m / s , becomes x = sin It — 7 cos 2t meters. But let us obtain the desired re­
sult by using the procedure described above, which applies even when h(x) is
not linear. Here h(x) = — Ax, so with r{x) = — 2x , Equation (1.19) gives
2
or
where C has been found by using v = 2 m / s and x = — 7 m at t = 0.
5
The v-t Diagram
In problems of rectilinear kinematics in w h i c h t h e acceleration is a k n o w n
function of time (Case 1), w e sometimes use w h a t is called t h e v-t dia­
gram. W e shall give just o n e example of its use because it is a m e t h o d
s o m e w h a t limited in application. (We discuss this shortcoming at t h e e n d
of the example.)
EXAMPLE 1.5
A point P moves on a line, starting from rest at the origin with constant accelera­
tion of 0.8 m / s to the right. After 10 s, the acceleration of P is suddenly reversed
to 0.2 m / s to the left. Determine the total time elapsed when P is again passing
through the origin.
2
2
Solution
If we graph the velocity versus time, the acceleration (dv/ dt) will of course be the
slope of the curve at every point. The v-t diagram for this problem is shown in
Figure E1.5a.
Figure El.5a
Page 12
We note not only that
but also that
Hence the change in the position x between any two times is nothing more
than the area beneath the v-t diagram between those points. Thus four points, or
times, are important in the diagram for this problem:
t j = starting time (in this case zero)
t = time when acceleration changes (given to be 10 s)
t = time when velocity has been reduced to zero (deceleration causes P to stop
before moving in opposite direction)
t = required total time elapsed before point P is again at origin
The velocity at time f is seen to be 0.8 m / s X 10 s = 8 m / s . To find the
time interval t — t , we use the similar triangles shown in Figure El.5b.
2
3
4
Figure El.5b
2
2
3
2
The total distance traveled before the point (momentarily) stops is thus
This is the distance traveled by the point in the positive direction (to the right).
The point will be back at x = 0 when the absolute value of the negative area
beneath the t axis (the distance traveled back to the left) equals the 200 m traveled
to the right (represented by the area above the axis):
which can be rewritten as
The only root of this equation larger than 50 s is f = 94.7 s, and this is the answer
to the problem.
4
A n alternative approach to the preceding v-t diagram solution is as
follows. Integrating the acceleration during the interval
with x during this interval called x ,
1
Integrating again (over the same interval), w e get
Page 13
Thus at t = 10 s, b y substitution,
Next, after the deceleration starts, using x in this interval,
2
a n d since
Therefore
Integrating again, we get
A n d with x = 40 m w h e n t = 10 s, t h e n C = — 50 m:
2
4
2
x = - 0 . l t + lOf - 50 m
W h e n x = 0, w e can solve for t h e time; the equation is the s a m e as in t h e
v-t diagram solution:
2
2
Of t h e roots, t = 5.28 a n d 94.7 s, only the latter is valid since 5.28 s
occurs prior to the change of acceleration expressions.
Even t h o u g h b o t h approaches yield t h e correct a n s w e r of 94.7 s in
the preceding example, w e m u s t r e c o m m e n d t h e latter a p p r o a c h of inte­
grating t h e accelerations a n d matching velocities a n d positions b e t w e e n
intervals. The reason is that w h e n w e are faced with nonconstant acceler­
ations, the v-t diagram a p p r o a c h requires us to find areas u n d e r curves,
the formulas for w h i c h are not ordinarily memorized.
It is interesting, in using t h e equations, to start a n e w time measure­
m e n t t at t h e beginning of t h e second interval:
4
2
Integrating again, w e get
T h e n x = 0 yields t h e equation
2
which h a s the positive root t = 84.7—which, a d d e d to the 10-s duration
EXAMPLE
1.6
of the first interval,
gives again 94.7 s of total time elapsed. It is slightly
easier
to
calculate
t
h
e integration
with
this a p p rato aacconstant
h of "start­
A point B starts from rest
at the originconstants
at t = 0 and
accelerates
rate
ing
e r " t h a n motion.
to use t hAfter
e s a m6es,t tthe
h r oacceleration
u g h o u t . T h echanges
only price
w etimepay
k m /time
s ino vrectilinear
to the
for this convenience
is that
m uthe
s t opposite
a d d the direction,
times at where
t h e end.
dependent
function
m /ws e in
t = 0 when
2
2
2
2
Page 14
t= 6 s. If the point stops at t = 26 s (from the starting time) and reverses direc­
tion, find the acceleration k during the first interval and the distance traveled by B
before it reverses direction. Then find the total time elapsed before B passes back
through the origin.
Solution
We begin the solution by deterrruning the motion (*i(f)) during the first time
interval; we integrate the acceleration to obtain the velocity and then again to get
the position:
At t! = 6 s, the acceleration changes to a negative value and point B "decel­
erates." At the beginning of this second interval, the speed and position of B are
given by the "ending" values during the first interval. These values are and x
at fj = 6:
t
Note that we start time t at the beginning of the second interval, during which
we have
2
where we note that the minus sign is needed to express the deceleration. Integrat­
ing, we get
where c was computed by using the initial condition that x = 18k meters when
t = 0.
Now we use the fact that x is zero at time t = 26 — 6 = 20 s; this strategy
will allow us to determine k:
4
2
2
2
2
3
0 = -0.002(20 ) + 6fc
k = 2.67 m / s
2
Substituting k into the x expression at f = 20 s gives us the position of B at the
"turnaround":
2
*2STOP
2
4
= -0.0005(20 ) + 6(2.67)(20) + 18(2.67)
= 288 m
Finally, to obtain the time f
2END
when B is passing back through the origin we set
Page 15
Rewriting, we get
On a calculator, the only positive root to this equation* is found in a matter of
minutes to be (to three significant figures):
The total time is f
2END
plus the duration of the first interval, or 38.7 s.
Before w e leave this example w e wish to note that during the first
time interval, while the acceleration is constant,
where
Letting v = x, w e note further that
and eliminating t w e obtain
This expression gives us the magnitude of the velocity in terms of dis­
placement. Most students have used this relationship in high school or
perhaps elementary college physics. There is sometimes a tendency,
however, to forget the conditions under which it is valid; it holds only for
rectilinear motion with constant acceleration. Thus it could not be used
EXAMPLE
1.7 interval of the preceding example, nor could the equa­
during
the second
tions
for in
x and
v from which
it was
derived.a common point (the origin in
Two cars
a demolition
derby are
approaching
Figure El.7), each at 55 mph in a straight line as indicated. Car C does not speed
up or slow down; the driver of C applies the brakes. Find the smallest rate of
deceleration of C that will allow C to precede it through the intersection, if:
x
2
2
x
a. d = 200 ft
b. d = 1 0 0 ft
2
2
* Descartes' rule of signs tells us that the maximum number of positive real roots to
Equation ( 1 ) is o n e (the number of changes in sign o n the left-hand side). A n d there will
be exactly o n e because the left side is negative at
and positive for large values
of
Page 16
Figure El.7
Solution
Placing the origin at the point of intersection of the two cars' paths, we have the
following for
Using the initial condition that x = — 300 ft when t = 0, we get
t
The back of car C will be at the origin (point of possible collision) when x = 0:
x
Now let us study the motion of car C . We use the coordinate q as shown for
this car. Calling the unknown deceleration K, we obtain
2
so that
and
But C = 0, since q = 0 at t = 0.
Next we see that at 3.72 sec the position of
3
Finally, car
just passes the rear of
if q is d at this time:
(f = -6.92K + 300ft
2
is
2
Page 17
Hence:
i . If d = 200 ft then K = 14.5 ft/sec .
I . If d = 100 ft, then K - 28.9 ft/sec .
2
2
2
2
Note also that if d = 300 ft, then K = 0; this is because no deceleration is needed
for the same distances at the same speeds. Further, if d > 300, then K is negative,
meaning that car would have to accelerate to arrive at the intersection at the
same time as car
2
2
Sometimes there are special conditions in a problem that require
ingenuity in expressing the kinematics. If there is an inextensible rope,
string, cable, or cord present, for example, w e may have to express the
constancy of length mathematically. This is the case in the following
example.
EXAMPLE 1.8
Solution
The length L of the rope that passes around both small pulleys is a constant. This
is a constraint equation that must be used in the solution. The procedure is as
follows (see Figure El.8):
Differentiating and noting that L, n, r , r , and K are constants, we get
c
Figure El.8
a
The velocities of C and B are equal since both points move on the same path with
a constant length separating them. Hence
(Note that C moves upward since j is downward!)
Therefore
EXAMPLE 1.9
Figure El.9
The ends A and B of the rigid bar in Figure El .9 are to move along the horizontal
and vertical guides as shown. End A moves to the right at a constant speed of
8 m / s . Find the velocity and acceleration of B at the instant when A is 3 m from
the comer.
Page 18
Solution
In terms of the parameters and unit vectors shown in Figure El.9,
Using the fact that the distance from A to B is a constant 5 m,
2
2
x + y = 25
so that
or
Thus when x = 3 m, y = 4 m and
or
so that
Differentiating again,
Our last example illustrates a different kind of constraint, that of a
Therefore
the instant
of interest
point
on aatbody
mamtaining
contact with a surface (or line) on another
moving body.
or
and
EXAMPLE 1 . 1 0
The curve AB on block B (see Figure E1.10a) is a parabola whose vertex is at A. Its
equation is x = (64/3)y. The block B is pushed to the left with a constant
velocity of 10 ft/sec. The rod slides on the parabola so that the plate p is forced
upward. Find the acceleration of the plate.
2
Solution
We first note that plate p and rod IS together constitute a single body, each of
whose points has one-dimensional (y) motion. The velocities and accelerations of
Page 19
Figure El. 10a
all these points are therefore the same. We shall then focus on point D, the lowest
point of which is in contact with
Defining the ground to be the reference frame we establish its origin at O
as shown in Figure El. 10b.
2
But because D always rests on the parabolic surface of B, y = (3/64)x so that
To get the acceleration of D, we first find its velocity:
To obtain x, we differentiate i
Figure E1.10b
OA
from Equation (1):
since it is given that all points of the body Bhetve the constant velocity 10 ft/sec
to the left.
Substitution of
into Equation (3) then gives
and we see that the velocity of D depends upon x. Differentiating v will give us
the acceleration of D:
D
Equation (6) gives the acceleration of all the points of the plate. Note that the
acceleration of D is a constant.
Question 1.2 Would a be a constant if instead of being quadratically
shaped, the inclined surface were (a) flat or (b) cubic?
D
Answer 1.2
If the surface is flat, then a vanishes. If it is cubic, then a is linear in x.
D
D
Page 20
•
PROBLEMS
Section 1.4
1.28 A slider block moves rectilinearly in a slot (see
Figure PI.28) with an acceleration given by
1.35 x = - 1 1 2 5 m
Time interval: 0 < t < 20 s (See Figure P1.35.)
0
Figure PI.28
Figure PI.35
Find the motion x(t) of the slider block if at t = 0:
a. It is passing through the origin, and
b. It has velocity
1.36 x = 10 m
Time interval: 0 < t £ 30 s (See Figure P1.36.)
0
1.29 Suppose an airplane touches down smoothly on a
runway at 60 mph. If it then decelerates to a stop at the
constant deceleration rate of 10 ft/sec , find the required
length of runway.
2
1.30 A train is traveling at 60 k m / h r . If its brakes give
the train a constant deceleration of 0.5 m / s , find the
distance from the station where the brakes should be ap­
plied so that the train will come to a stop at the station,
How long will it take the train to stop?
2
1.31 A point P starts from rest and accelerates uniformly
(meaning x = constant) to a speed of 88 ft/sec after trav­
eling 120 ft. Find the acceleration of P.
1.32 If in the preceding problem a braking deceleration
of 2 ft/sec is experienced beginning when P is at 120 ft,
determine the time and distance required for stopping.
2
Figure PI.36
1.37 x = 1 0 m
Time interval: 2 s ( < 5 s (See Figure PI.37.)
0
1.33 A car is traveling at 55 mph on a straight road. The
driver applies her brakes for 6 sec, producing a constant
deceleration of 5 ft/sec , and then immediately acceler­
ates at 2 ft/sec . How long does it take for the car to
return to its original velocity?
2
2
1.34 In the preceding problem, suppose the acceleration
following the braking is not constant but is instead given
byx = 0.6r ft/sec . Now how long does it take to return to
55 mph?
2
In Problems 1.35-1.37, the graph depicts the velocity of a
point P in rectilinear motion. Draw curves showing the
position x(t) and acceleration a{t) of P if the point is at the
indicated position x at t = 0.
0
Figure PI.37
1.38 Ahotrod enthusiast accelerates his dragster along a
straight drag strip at a constant rate of acceleration from
zero to 120 mph. Then he immediately decelerates at a
Page 21
constant rate to a stop. He finds that he has traveled a
total distance of \ mi from start to stop. How much time
passes from the instant he starts to the time he stops?
Hint: Sketch a v-t diagram.
1.39 Ben Johnson set a world record of 9.83 seconds in
the 100-meter dash on August 31, 1987. He had also set
the record for the 60-meter dash of 6.40 seconds that
same year. Assuming that in each race Johnson acceler­
ated uniformly up to a certain speed v and then held that
same maximum speed to the end of both races, find
(a) the time t to reach v„; (b) the value of v ; and (c) the
distance traveled while accelerating.
0
0
0
1.40 A train travels from one city to another which is
134 miles away. It accelerates from rest to a maximum
speed of 100 mph in 4 min, averaging 65 mph during
this time interval. It maintains maximum velocity un­
til just before arrival, when it decelerates to rest at
an average speed during the deceleration of 40 mph. If
the total travel time was 110 min, find the deceleration
interval.
1.41 A point Q in rectilinear motion passes through the
origin at t = 0, and from then until 5 seconds have
passed, the acceleration of Q is 6 ft/sec to the right.
Beginning at t = 5 seconds, the acceleration of Q is
12t ft/sec to the left. If after 2 more seconds point Q is
13 feet to the right of the origin, what was the velocity of
Q at t = 0?
2
2
1.42 A point begins at rest at x = 0 and experiences con­
stant acceleration to the right for 10 s. It then continues at
constant velocity for 8 more seconds. In the third phase of
its motion, it decelerates at 5 m / s and is observed to be
passing again through the origin when the total time of
travel equals 28 s. Determine the acceleration in the first
10 s.
2
1.43 An automobile passes a point P at a speed of
80 mph. At P it begins to decelerate at a rate proportional
to time. If after 5 sec the car has slowed to 50 mph, what
distance has it traveled?
b. Determine the distance traveled by the particle
over the same time interval.
1.46 A p o i n t P m o v e s o n a line. The acceleration of P is
gi ven by
The velocity
of P at t = 0 is — 60i m / s , with the point at x = 7 m at
that time. Find the distance traveled by P in the time
interval t = 0 to t = 13 s.
P
• 1.47 The position of a point P on a line is given by
the equation
The point starts moving at
f = 0. Find the total distance traveled by P when it passes
through the origin (counting the start as the first pass) for
the third time.
1.48 A particle moving on a straight line is subject to an
acceleration directly proportional to its distance from a
fixed point P on the line and directed toward P. Initially
the particle is 5 ft to the left of P and moving to the right
with a velocity of 24 ft/sec. If the particle momentarily
comes to rest 10 ft to the right of P, find its velocity as it
passes through P.
1.49 A particle moving on the x axis has an acceleration
always directed to the origin. The magnitude of the accel­
eration is nine times the distance from the origin. When
the particle is 6 m to the left of the origin, it has a velocity
of 3 m / s to the right. Find the time for the particle to get
from this position to the origin.
1.50 A point P has an acceleration that is positiondependent according to the equation
Determine the velocity of P as a function of its position x if
P is at 0.3 m with
1.51 Suppose initial conditions are the same as in the
preceding problem but
Find as a function of
time.
1.52 The velocity of a particle moving along a horizontal
path is proportional to its distance from a fixed point
on the path. When t = 0, the particle is 1 ft to the right
of the fixed point. When v = 20 ft/sec to the right,
a = 5 ft/sec to the right. Determine the position of
the particle when t = 4 sec. (See Figure PI.52.)
2
1.44 Work the preceding problem, but suppose the de­
celeration is proportional to the square of time. The other
information is the same.
1.45 A particle has a linearly varying rectilinear accel­
eration of
Two observations of
the nartirlp's motion are made: Its velocity at t = 1 s is
and its position at t = 2 s is given by
meters.
a. Find the displacement of the particle at t = 5 s
relative to where it was at f = 0.
Figure PI.52
* Asterisks identify the more difficult problems.
Page 22
1.53 A speeder zooms past a parked police car at a
constant speed of 70 mph (Figure PI.53). Then, 3 sec
later, the policewoman starts accelerating from rest at
10 ft/sec until her velocity is 85 mph. How long does
it take her to overtake the speeding car if it neither
slows down nor speeds up?
2
the driver of the car reacts by slamming on her brakes,
giving her car a deceleration a . Find the minimum value
of a for which the car will not collide with the truck. Hint:
Enforce
for all time t before the vehicles are
stopped.
c
c
1.57 Point B of block B has a constant acceleration of
10 m / s upward. At the instant shown in Figure PI .57, it
is 30 m below the level of point A of
At this time, v
and v are zero. Determine the velocities of A and B as
they pass each other.
2
A
a
1.58 The accelerations of the translating blocks
are
respectively. (See Fig­
ure P1.58.)
Figure P1.53
1.54 In the preceding problem, suppose the speeder sees
the policewoman 10 sec after she begins to move, and
decelerates at 3 ft/sec . How long does it take the police­
woman to pass the car if she is actually chasing a faster
speeder ahead of it?
2
1.55 Two cars start from rest at the same location and at
the same instant and race along a straight track. Car
accelerates at 6.6 ft/sec to a speed of 90 mph and then
runs at a constant speed. Car accelerates at 4.4 ft/sec to
a speed of 96 mph and then runs at a constant speed.
2
2
Figure P1.57
a. Which car will win the 3-mi race, and by what
distance?
b. What will be the maximum lead of
3ver ?
c. How far will the cars have traveled when
passes '
* 1.56 A car is 40 ft behind a truck; both are moving at
55 mph. (See Figure P1.56.) Suddenly the truck driver
slams on his brakes after seeing an obstruction in the road
ahead, and he decelerates at 10 ft/sec . Then, 2 sec later,
2
Figure P1.56
Figure PI.58
Page 23
Figure PI.59
Figure PI.60
Figure P1.61
The entire system is at rest at the given instant. Find how
long it will take for block to hit the ground. (Do not
assume that pulleys and
remain at the same level!)
• 1 . 5 9 Block has v = 10 m / s to the right at f = Oanda
constant acceleration of 2 m / s to the left. Find the dis­
tance traveled by block ^during the interval t = 0 to 8 s.
(See Figure P1.59.)
Figure P1.62
A
2
1.60 A man and his daughter have figured out an inge­
nious way to hoist 8000 lb of shingles onto their roof,
several bundles at a time. They have rigged a pulley onto
a frame around the chimney (Figure PI.60) and will use
the car to raise the weights. When the bumper of the car is
at x = 0 (neglect d), the pallet of shingles is on the ground
with no slack in the rope. While the car is traveling to the
left at a constant speed of v = 2 mph, find the velocity
and acceleration of the shingles as a function of x. Do this
by using the triangle to the left of the figure to express y as
a function of x; then differentiate the result.
A
* 1.61 The cord shown in Figure PI .61, attached to the
wall at D, passes around a small pulley fixed to at B; it
then passes around another small pulley and ends at
point A of body. The cord is 44 m long, and the system
is being held at rest in the given position. Suddenly point
B is forced to move to the right with constant acceleration
a = 2 m / s . Determine the velocity of A just before it
reaches the pulley.
2
B
1.62 The ends of the rigid bar in Figure PI.62 move
while maintaining contact with floor and wall. End A
moves toward the wall at the constant rate of 2 ft/sec.
What is the acceleration of B at the instant when A is 6 feet
from the wall?
Page 24
1.63 The velocity of point A in Figure PI.63 is a constant
2 m / s to the right. Find the velocity of B w h e n * = 10 m.
Small wheels
Figure PI.63
1.64 The collars in Figure PI .64 are attached at Q and C
to the rod by ball and socket joints. Point C has a velocity
of •
m / s and no acceleration at the instant shown.
Find the velocity and acceleration of Q at this instant.
2
2
Figure PI.64
1.65 The wedge-shaped cam in Figure PI.65 is moving
to the left with constant acceleration a . Find the acceler­
ation of the follower
0
1.66 In Example 1.10, let the equation of the incline
be given by x = (512/3)y. If the motion starts when
x = y = 0, find the acceleration of the plate
when
y = 2 ft.
3
1.5
Figure P1.65
Rectangular Cartesian Coordinates
In this section w e merely a d d the y a n d / o r z c o m p o n e n t s of position to
the rectilinear (x) c o m p o n e n t studied in the preceding section. This step
allows the point P to m o v e o n a curve in t w o - or three-dimensional space
instead of being constrained to m o v e m e n t along a straight line in the
reference frame
Suppose that P is in a state of general (three-dimensional) motion in
frame
W e m a y study this motion b y e m b e d d i n g a set of orthogonal
axes in as s h o w n in Figure 1.4 on the next page. The position vector of
point P m a y t h e n be expressed as
in w h i c h (x, y, z) are rectangular Cartesian coordinates of P measured
along the e m b e d d e d axes a n d
are unit vectors respectively paral­
lel to these axes (Figure 1.4). Using t h e basic definitions (Equations 1.9
a n d 1.10), w e m a y differentiate i
a n d obtain expressions for velocity
a n d acceleration in rectangular Cartesian coordinates:
O P
Page 25
Figure 1.4
Rectangular Cartesian coordinates of P.
We shall n o w consider examples in w h i c h points m o v e in t w o a n d three
dimensions.
EXAMPLE 1 . 1 1
The position vector of a point P is given as
Find the velocity and acceleration of P at t = 1 sec.
Solution
Differentiating the position vector, we obtain the velocity vector of P:
Another derivative yields the acceleration of P:
Therefore, at t = 1 sec, the velocity and acceleration of P are
Note that the speed (magnitude of velocity) of.
the magnitude of the acceleration at
example in Section 1.7.
/sec
and
We shall return to this
We see from t h e previous example that if t h e position vector of P is
k n o w n as a function of time, it is a very simple matter to obtain t h e
velocity a n d acceleration of the point. In the following example w e are
given t h e acceleration of P a n d asked for its position. Since this problem
requires integration instead of differentiation, initial conditions enter the
picture. These conditions allow us to compute t h e constants of integra­
tion, just as they did for rectilinear motion in t h e preceding section.
Page 26
EXAMPLE 1 . 1 2
A point Q has the acceleration vector
Att = 0, the point Q is located at (x, y, z) = (1, 3, —5) m and has a velocity vector
of
When ( = 3 s , find the speed of Q and its distance from
the starting point.
Solution
Integrating, we get
in which c is a vector constant. Using the initial condition for velocity at t = 0, we
obtain
so that
Therefore
Integrating again, we get
where c' is another vector constant, evaluated below from the initial condition for
the position of Q at t = 0:
so that
and thus
Substituting the required time, t = 3 s, into the expressions for
give the answers:
Thus the speed of Q is given by
Continuing, we have
will
Page 27
The distance d between Q and its starting point is therefore given by
EXAMPLE 1 . 1 3
4f>
m X
The point P in Figure E1.13 travels on the parabola (with focal distance
m)
at the constant speed of 0.2 m / s . Determine the acceleration of P: (a) as a func­
tion of x and (b) at x = 2 m.
Solution
x (ml
We may obtain the velocity components by differentiating:
Figure El.13
Thus
Similarly the acceleration of P is
Since | v |, or v , is constant, we have
P
P
We also see from (2) that we need x; differentiating (3), we get
Substituting
or
When
x = 2 m,
(3) and (4) into (2), we get
Page 28
In closing this section, w e remark that the simple forms of Equations
(1.24) a n d (1.25) are d u e to t h e fact that t h e unit vectors
remain
constant in b o t h m a g n i t u d e a n d direction w h e n t h e axes are fixed in the
frame of reference. For planar applications (Chapter 3), w e shall set t h e z
c o m p o n e n t of velocity identically to zero, obtaining t h e following for a
point in p l a n e motion (moving only in a p l a n e parallel to t h e xy plane):
PROBLEMS
•
Section 1.5
1.67 The moving pin P of a rotating crankhasalocation
defined by
Find the velocity of P when
and 2 s.
1.68 A bar of length 2L moves with its ends in contact
with the guides shown in Figure P1.68. Find the velocity
and acceleration of point C in terms of and its deriva­
tives.
Figure P1.69
Figure PI.71
Figure P1.68
1.72 Point P is constrained to move in the two slots
shown: one cut in the body
the other cut in the refer­
ence frame The constant acceleration of. is given to be
4 c m / s to the left. If point P reaches the bottom of the
slot (in. ) 2 sec after the instant shown in Figure PI.72,
2
1.69 A point P moves on a circle in the direction shown
in Figure PI.69. Express r in (x, y, z) coordinates and
differentiate to obtain v and a . (Angle is in radians.)
0 P
P
P
1.70 Repeat the preceding problem. In this case, how­
ever, the angle . increases quadrationlly, instead of lin­
early,
with
timePaccording
1.71 A
point
starts at to
the origin rad.
and moves along
the parabola shown in Figure PI.71 with a constant
x-component of velocity,
sec. Find the veloc­
ity and acceleration of P at the point (x, y) = (1, 1).
Figure PI.72
Page 29
when
is at rest,
a. Find the velocity Vp(t) and acceleration a (t) of P.
b. Find the position, velocity, and acceleration of P
when t = 4 s.
c. Eliminate the time t from the x and y expres­
sions and obtain the equation of the path of P.
P
a. Through what height h did the marble move?
b. What distance did the marble travel?
1.73 The pin P shown in Figure PI.73 moves in a para­
bolic slot cut in the reference frame and is guided by the
vertical slot in body
For body
m locates
the centerline of its slot.
a. Find the acceleration of P at t = 5 s.
b. Find the time(s) when the x and y components
of a are equal.
1.76 The motion of a particle P is given by x = C cosh kt
and y = C sinh kt, where C and k are constants. Find the
equation of the path of P by eliminating time t.
1.77 In the preceding problem, find the speed of P as a
functionofthedistana
from the origin to P.
P
1.74 A pin P moves in a slot that is cut in the shape of a
hyperbolic sine as shown in Figure PI.74. It is guided
along by the vertical slotted body
all the points of
which have velocity 0.08 m / s to the right. Find v and a
when x = 0.2 m.
P
P
1.78 Describe precisely the path of a particle's motion if
its xy coordinates are given by (2.5f + 7, 6t + 9) meters
when t is in seconds.
2
2
1.79 A particle P moves in the xy plane. The motion of P
is given by
x = 30f + 6 ft
y = 20r - 7 ft
Find the equation of the path of P in the form y = f{x).
1.80 Repeat Problem 1.79 if:
x = 5t m
3
y = -250t m
1.81 Repeat Problem 1.79 if:
x = 2 + 3 sin t ft
Figure P1.73
y = 4 cos t ft
1.82 A cycloid is the curve traced out by a point (such as
P) on the rim of a rolling wheel. In terms of the parameter
(the angle in Figure PI.82), the equations of the cycloid
are:
Noting that
Figure P1.74
1.75 A point P travels on a path and has the following
coordinates as functions of time t (in seconds):
Figure P1.82
changes with time, find the speed of P at
radians, in terms of a and
Page 30
In Problems 1.83-1.86 (see Figures P1.83-P1.86), a
point P travels on the curve with a constant x component
of velocity, x = 3 in./sec. Each starts on the curve at
x = 1 when t = 0. Find the velocity vector of P when
f = 10 sec in each case.
1.83 Logarithmic curve
1.84 Exponential curve
1.85 First-quadrant branch of rectangular hyperbola
1.86 First-quadrant branch of semicubical parabola
1.87-1.90 Find the respective acceleration vectors at
f = 10 s of the points whose motions are described in
Problems 1.83-1.86.
1.91 Two points P and Q have position vectors in a ref­
erence frame that are given by r = 50fl meters and
r = 40i — 20fj meters. Find the minimum distance
between P and Q and the time at which this occurs.
O P
O Q
1.92 Describe the path of a point P that has the following
rectangular Cartesian coordinates as functions of time:
x = a cos cot, y = a sin cot, and z = bt, where a, b, and co
are constants. Identify the meanings of the three con­
stants.
1.93 For the following values of the constants, find the
velocity of P at t = 5 s in the preceding problem: a = 2 m,
b = 0.5 m / s , and co = 1.2 r a d / s .
1.94 The acceleration of a point is given by
At t = 0. the initial conditions are that
m / s and
meters. Find the position vector of P at
f = 3 s, and determine how far P then is from its position
at t = 0.
1.95 A point moves on a path, with a position vector as a
function of time given by
in
units of meters when t is in seconds. Find:
Figure PI.83
a. The speed of the point at t = 0.
b. Its acceleration at t =
c. The component of the velocity vector, at t = 0,
which is parallel to the line tin the xy plane
given by
shown in Figure P1.95.
Figure P1.84
Figure PI.95
• 1.96 A car travels on a section of highway that approxi­
mates the cosine curve in Figure P1.96. If the driver
Figure PI.85
Figure PI.86
Figure P1.96
Page 31
maintains a constant speed of 55 mph, determine his x
and y components of velocity when x = 2500 ft.
1.98 Determine the minimum magnitude of acceleration
of the car in Problem 1.96. Where on the curve is this
acceleration experienced?
• 1 . 9 7 A car travels along the highway of the preceding
problem with a constant x-component of velocity of
54.9 mph. Over what sections of the highway does the
driver exceed the speed limit of 55 mph?
1.6
1.99 Find the maximum magnitude of acceleration of
the car in Problem 1.97. Where does it occur on the curve?
Cylindrical Coordinates
If a point P is moving in such a w a y that its projection into the xy plane is
more easily described with polar (r a n d i coordinates t h a n with x a n d y,
t h e n w e m a y use cylindrical coordinates to a d v a n t a g e . These coordi­
nates are nothing more t h a n the polar coordinates r a n d together with
a n "axial" coordinate z. T h u s r a n d locate the projection point of P in a
plane, while z gives the distance of P from the plane.
Embedding the same set of rectangular axes (x, y, z) in the reference
frame as w e did in the preceding section, we n o w s h o w the coordinates
r and
as well (Figure 1.5). N o t e that
is the projection of P into the
plane xy. From Figure 1.5 w e see that the unit vectors
are d r a w n
in the xy plane a n d that:
Figure 1.5
Cylindrical coordinates of P.
1.
The direction of
2.
is normal to
is that of
in the direction of increasing
It will b e helpful later in the section to note carefully at this point that
zhange (in direction) with changes in but not with r or z. Thus
if the point P m o v e s along either a radial line or a vertical line, the two
unit vectors remain the same. But if P moves in such a way as to alter
then the directions of
and
will vary.
The rectangular a n d cylindrical coordinates (both having z in com­
m o n ) are related t h r o u g h
which can b e differentiated to produce, by Equations (1.24) a n d (1.25),
formulas for velocity a n d acceleration in terms of the cylindrical coordi­
nates (and their derivatives) a n d the unit vectors
and
It is usually
more desirable, however, to express the velocity a n d acceleration in
terms of the set of unit vectors
which are naturally associated
with cylindrical coordinates. Thus it is useful to express a position vector
T p as
0
Question 1.3
Why is there no
term in Equation (1.30)?
Answer 1.3 From Figure 1.5, w e see that
is perpendicular to r . Note, however, that
implicit in the writing and use of Equation (1.30) is the polar angle
o p
Page 32
Differentiating Equation (1.30), w e obtain the velocity of P:
To evaluate
Figure 1.6
w e note from Figure 1.6 that
Hence
and thus the velocity in cylindrical coordinates takes the form
Differentiating again, w e get
Using Equations (1.32), w e find that
Thus the acceleration expression in cylindrical coordinates is
In the special case for which the motion is in a plane defined by
z = constant, w e have
In this case w e need only the polar
coordinates r and
and the directions of
are said to be radial
and transverse, respectively.
Question 1.4 If a point P moves with
W h y then isn't I v I = r?
Before turning to the examples in this section, w e return briefly to
calculation of the derivatives of the unit vectors
We note that
each derivative turns out to be perpendicular to the vector being differ­
entiated. To understand w h y this is the case, w e consider an alternative
derivation
of one
the thing,
formula
. The But
timeI vdependence
is due
to the
Answer
1.4 For
r canfoi
be negative.
| ^ | r | either, of
because
the magni­
tude ofdependence
the derivative of
of the
a vector
is not equalon
to the
absolutedepends
value .of the
derivativethus
of
time
coordinate
which
explicitly;
the magnitude of the vector. Differentiating t produces a term (r8e ) in v in addition
w
e
can
write
tore,.
P
P
OP
e
P
Page 3 3
Let us study the derivative
By definition,
With the aid of Figure 1.7 w e can see that:
1.
The direction of
approaches zero.
2.
The magnitude of
proaches unity as
Figure 1.7
approaches that of
which ap­
approaches zero.
Change in e, as 6 changes.
Thus
and w e obtain
in agreement with Equa­
tion (1.33). The reader may wish to sketch a similar geometric proof of
Equation (1.36).
This mutual orthogonality of a vector and its derivative, incidentally,
is not just restricted to unit vectors. It is in fact a property of all vectors of
constant magnitude. We can show that this is the case by noting that if A
is such a vector, then
and thus
or
Hence,
or
vector
shall
some
make
examples
and
provided
the
use derivative
of
ofthat
frequently
velocity
neither
are
and
mutually
throughout
the
acceleration
vector
perpendicular.
nor
thein
its
book.
cylindrical
derivative
We
This
n ocoordinates.
vanishes,
w
is aproceed
result the
wtoe
Page 34
EXAMPLE 1 . 1 4
The pin P in Figure El. 14a moves outward with respect to a horizontal circular
disk, and its radial coordinate r is given as a function of time by r = 3 f / 2 meters.
The disk 2> turns with the time-dependent angle 6 = 4 t / 3 rad. Find the velocity
and acceleration of P at f = 1 s.
2
2
Solution
Figure E 1.14a
From Equation (1.34) we have
Velocity v
P
Figure El. 14b
Thus
dp
= 21.4 m / s
2
and we note that the speed o f P a t f = l s i s 5 m / s .
SinceContinuing,
at t = 1 wefrom
haveEquation
r = 3 / 2 (1.37)
m andwe8 get
= 4/3 rad, we show the preceding
results pictorially in Figures El. 14b and c.
ThusNote that there is a time, f =
when the
parts of the
radial component of a cancel each u m u , making this component zero at that
instant of time. The reader is urged to compute and sketch v and a at another
time, say t = 2 s.
P
Acceleration a
P
P
Figure El. 14c
P
EXAMPLE 1 . 1 5
In the preceding example, discard the given
Suppose instead that
= constant = 0.3 r a d / s and that the pin slides not only in the slot of disk
(see Figure El. 15a), but also in the spiral slot cut in the reference frame and de­
fined by
meters, with n radians. Find the velocity and acceleration of
the pin when
rad.
Figure El. 15a
Solution
From
Therefore
Page 35
Now for the acceleration:
We shall leam in the next section that the velocity is always tangent to the
path of the point. Thus the angle cf> between the path and the —x axis can be
found from the velocity components, as shown in Figure El.15b, as follows:
Figure El. 15b
In our next examples, there is motion in the z direction as well as the
radial (r) a n d transverse (6) of t h e previous examples.
EXAMPLE 1 . 1 6
A point Q moves on a helix as shown in Figure El. 16a. The pitch, p, of the helix is
0.2 m, and the point travels at constant speed 20 m / s . Find the velocity of Q in
terms of its cylindrical components.
Solution
The meaning of the pitch of a helix is the (constant) advance of Q in the z direction
for each revolution in 8. Therefore
so that
Figure El. 16a
or, for this problem,
Noting that r = 0 since Q travels on a cylinder (with r therefore constant),
Equation (1.34) then gives the following for the point's velocity:
or, using (3),
The speed of Q is constant at 20 m / s ; thus
Page 36
From Equation (3), we then get
Hence the velocity vector of
is (substituting (9) and (8) into (5))
Note that | v | = 20.0 m / s , as it must be. Note also that a larger pitch will spread
out the helix (see Figure El. 16b). The equations of this example then show that
the
component will become larger in comparison to the
component for
larger p.
c
Small p
Larger p
Figure El. 16b
EXAMPLE 1 . 1 7
Find the acceleration of
in the preceding example.
Solution
From Equation (1.37) we get
Because r is constant on the cylinder, this equation reduces to
Furthermore, since | v | is constant, Equations (8) and (3) of the previous
example show that
are constants. Therefore there is only one nonvanishing acceleration component here:
Q
Figure E1.17
Note that even though point
never has a radial component of velocity (see
Figure El. 17), it has only a radial component of acceleration!
EXAMPLE 1 . 1 8
Find the velocity and acceleration vectors of point in Example 1.16 if, instead of
the speed of being constant, we have its vertical position given as the function
of time:
3
z = 0.08f m
Solution
Referring to Example 1.16 (see Figure El.18), we find
Figure El. 18
Therefore
Page 37
This time, however, the velocity is seen to depend on the time; for example,
at t = 10 s,
For the acceleration, we note that
is still zero, so that
This time, all three terms are nonzero. We have
Thus
In t h e final example of this section, w e consider t h e case in which, in
addition to t h e changing a n d z of t h e preceding three examples, the
radius varies.
EXAMPLE 1 . 1 9
A point P moves on a spiraling path that winds around the paraboloid of revolu­
tion shown in Figure El.19. The focal distance
m, and the point P advances
4.0 m vertically with each revolution. If the speed of P is 0.7 m / s , a constant,
determine the vertical component of the velocity vector of P as a function of r.
Solution
2
From z = r , we obtain
Figure E1.19
And from the pitch relationship
we get
Therefore the speed of P may be expressed as
Thus the answer is
Page 38
Let u s extend t h e preceding example slightly. W e can see that varies
with t h e radius r (distance from t h e z axis to P) a n d that it is zero initially
a n d approaches zero again for large r. Its m a x i m u m m a y b e determined
from calculus:
or
This yields the equation a n d result:
at w h i c h
Note from Equation (2) in the example that at this value of z,
FromQuestion
Equation
w einspection
see that at
t h e little
s a m eortime
1.5(1),By
(with
no writing), what is the
maximum magnitude of the radial component of v ?
P
Answer 1.5 W h e n r = 0, w e have z = 0 =
Thus r, the radial component of v , is
maximum there at the value 0.7, w h i c h is the constant speed. (Note that decreases
continuously toward zero from there.)
P
PROBLEMS
•
Section 1.6
a n d therefore, w h e n z is m a x i m u m , the speed is
1.100 The airplane in Figure PI.100 travels at constant
speed at a constant altitude. The radar tracks the plane
and computes the distance D, the angle and the rate of
change of
at all times. In terms of
and D, find the
speed of the airplane.
1.101 A ball bearing is moving radially outward in a slot­
ted horizontal disk that is rotating about the vertical z axis.
At the instant shown in Figure P1.101, the ball bearing is
3 in. from the center of the disk. It is travehng radially
outward at a velocity of 4 in. /sec relative to the disk
as it should be, since it does not change with time.
Page 39
1.103 A particle moves on a curve called the "Lemniscate
of Bernoulli,'' defined by
It moves along
the branch shown in Figure PI. 103 with a"ows, and
passes through point P at f = 0. The angle increases
with time according to
rad, with t measured
in seconds. At the point P, find the velocity and accelera­
tion of the particle.
Figure PI .100
Figure PI. 103
*
A point P moves on the "Spiral of Archimedes" at
constant speed 2 m / s . (See Figure PI.104.) The equation
of the spiral is
Find the acceleration of P when
Figure PI .101
and has a radial acceleration with respect to the disk of
5 in./sec outward. What would
and
have to be
at the instant shown for the ball bearing to have a total
acceleration of zero?
2
1.102 The disk shown in Figure PI. 102 is horizontal
and turns so that
about the vertical. Forces cause
a marble to move in a slot such that its radial distance
from the center equals kt . Note that c and k are constants.
Figure PI. 104
2
a. Find the acceleration of the marble.
b. At what time does the radial acceleration vanish?
Figure P1.105
Figure P1.102
1.105 The cardioid in Figure PI. 105 has the equation
Point P travels around this curve, in the
direction indicated, in such a way that
In terms of K and the length a, find the velocity of P at the
four points where its path intersects the coordinate axes.
Express the result in terms of radial and transverse com­
ponents, and then convert to rectangular components by
expressing
in terms of
at each position.
Page 40
1.106 In the preceding problem, find the acceleration of P
at the same four points. Again, do the problem first in
components and then convert the results to
components.
• 1.110 The point Pin Figure PI.110 moves on the limacon
defined in polar coordinates by
A point P starts at the origin and moves along the
parabola shown in Figure P1.10 7 with a constant x-component of velocity,
Using the following ap­
proach, find the radial and transverse components of the
velocity and acceleration of P at the point (x, y) = (1, 1):
Find v and a in rectangular components (see Problem
1.71); then resolve these vectors along
to obtain
their radial and transverse components.
If the polar angle is quadratic in time according to
rad, find the velocity of P when it is at its highest
point.
P
P
1.111 In the preceding problem, determine the accelera­
tion of P at (a) the same highest point and (b]
rad.
• 1.112 A point P moves on the figure eight in the indicated
direction (Figure PI.112) at constant speed 2 m / s . Find
the acceleration vector of P the next time its velocity is
horizontal.
Figure P1.107
1.108 Solve the preceding problem by a different ap­
proach: Recall the polar coordinate relations
and
and differentiate to obtain t, f, 6, and
for entry into Equations (1.34) and (1.37).
1.109 The four-leaf rose in Figure PI.109 has the Equa­
tion r = 3 sin
A particle P starts at the origin and
travels on the indicated path with
rad/sec
= constant. When P is at the highest point in the first
quadrant, find:
a. the speed of P
b. the acceleration of P
Figure P1.112
' 1.113 An insect is asleep on i
rpm record, 6 in. from
the spindle. When the record is turned on, the insect
wakes up and dizzily heads toward the center, in a
straight line relative to the disk, at 1 in./sec (Fig­
ure PI.113). If the bug can withstand a maximum accel­
eration magnitude of 100 in./sec , does it make it to the
spindle (a) if it starts after the record is up to speed? (b) if
it starts as soon as the record is turned on? Assume
that the turntable accelerates linearly (with time) up to
speed in one revolution, and that
until
in./sec.
2
Figure P1.109
Figure PI. 113
Figure P1.110
1.114 David throws a rock at Goliath with a sling. He
whirls it around one revolution plus 135° more and re­
leases it there, as shown in Figure PI.114, at 50 ft/sec.
Page 41
1.116 Two people moving at 2 ft/sec to the right are
using a rope to drag the box along the ground at the
lower level (Figure P1.116). Determine the speed of as a
function of the angle between the rope and the vertical.
•1.117 The rigid rod in Figure PI. 117 moves so that its
ends, A and B, remain in contact with the surfaces. If, at
the instant shown, the velocity of A is 0.5 ft/sec to the
right, find the velocity of B.
1.118 In Problem 1.117, find the acceleration of B at the
instant in question if the acceleration of A is 2 ft/sec to
the left at that time.
2
1.119 An ant travels up the banister of a spiral staircase
(Figure PI. 119) according to
Find the position and velocity of the ant when t = 30 s.
Figure PI .114
As he whirls the sling, the speed of the rock increases
linearly with the time f; that is, 0 = kt, where A: is a con­
stant. Find the acceleration of the rock just prior to re­
lease.
Figure P1.117
In Problem 1.60 show that the velocity of the shin­
gles may also be obtained by simply taking the compo­
nent of the velocity of the bumper attachment point A
along the rope. Using the cylindrical coordinate expression
for velocity, explain why this procedure works.
Figure PI .116
Figure P1.119
Page 42
Figure PI .121
Figure PI.123
• 1.123 The mountain shown in Figure PI.123 is in the
shape of the paraboloid of revolution H — z = kr , where
H = height = 5000 ft, r is the radius at z, and k is a con­
stant. The base radius is also 5000 ft. A car travels up the
mountain on a spiraling path. Each time around, the car's
altitude is 1000 ft higher. The car travels at the constant
speed of 50 mph. Find the largest and smallest absolute
values of the radial component of velocity on the journey,
and tell where the car is at these two times.
2
Figure PI. 122
1.120 Find the acceleration of the ant (again at t = 30 s) in
the preceding problem.
1.121 A point P starts at t = 0 at the origin and proceeds
along a path on the paraboloid of revolution shown in
Figure PI. 121. The path is described (with time as a pa­
rameter) by
• 1.124 In the preceding problem, find the locations of the
car
for which the following velocity components
are equal.
a. Radial
and transverse
b. Radial and vertical
c. Transverse and vertical
1.125 Show that the velocity of a point P in spherical
coordinates
is given by
Find the position, velocity, and acceleration vectors of the
point when it reaches the top edge of the paraboloid.
(H, R, k-i, and k are constants.)
2
1.122 A bead B slides down and around a cylindrical sur­
face on a helical wire (Figure PI .122). The vertical drop of
the bead as changes by In is called the pitch p of the
helix; R is the radius of the helix.
a. Noting that (and therefore also z) is a function
of time, write the equations for r , v , and a
in cylindrical coordinates.
b. For the values R = 0.3 m, p = 0.2 m, and
rad/s, find and sketch the velocity
and acceleration vectors of B when f = 10 s.
OB
B
B
Figure PI.125
Page 43
See Figure PI .125. Hint: As intermediate steps, obtain the
results
Then differentiate the simple position vector
1.126 Show by differentiating v in the preceding prob­
lem that the corresponding expression for the accelera­
tion in spherical coordinates is
P
Reference frame
Figure PI. 127
• 1.127 The velocity of a point P moving in a plane is the
resultant of one part,
, along the radius from a fixed
point O to the point P, and another part,
which is
always parallel to a fixed line. (See Figure PI. 127.) Prove
that the acceleration of P may be written as
where
1.7
where r is the length of the radius vector from O to P and
is the angle it makes with the fixed direction.
Tangential and Normal Components
In this section w e examine yet a n o t h e r m e a n s of expressing the velocity
a n d acceleration of a point P. Instead of focusing o n a specific coordinate
system, this time w e shall study the w a y in w h i c h the motion of P is
related to its path. Consequently, the c o m p o n e n t s of velocity a n d accel­
eration that result are sometimes called intrinsic or natural.
The p a t h of point P, as m e n t i o n e d in Section 1.3, is the locus of points
of the reference frame
successively occupied b y P as it moves. We
begin, then, b y defining some reference point o n the p a t h . From this
arclength origin w e t h e n m e a s u r e the arclength s along the p a t h to
the point P. Clearly, the arclength coordinate d e p e n d s on the time;
that is, s = s(t).
In Figure 1.8 w e see a position vector, r , for point P. This vector w a s
seen in preceding sections to define the location of P, a n d t h u s it m a y b e
considered a function of the arclength s:
0 P
Forming the velocity of P differentiation (the definition is the same,
regardless of h o w w e choose to represent the vectors), w e get
Figure 1.8 Arclength measurement of point
P on its path.
Page 44
and, b y the chain rule,
Figure 1.9 s h o w s the quantities
Figure 1.9
Changes in \
0P
Figure 1.10
as s changes.
Shrinking As toward zero.
We suggest that the reader sketch a n arc o n a large sheet of p a p e r a n d
t h e n use a straightedge to d r a w the triangle OPP' (Figure 1.10). T h e n the
limit in Equation (1.40) can be taken by, let us say, dividing As in half each
time. After just a few m o r e divisions of As o n t h e large sheet, it will
become clear that as As approaches zero — that is, as P' backs u p t o w a r d
P — t w o interesting things h a p p e n :
1.
Ar
2.
The m a g n i t u d e of
OP
becomes tangent to the p a t h of P at arclength s.
approaches
These t w o results, taken together, prove that dt /ds is always a unit
vector that is tangent to the p a t h a n d pointing in the direction of increas­
ing s. It is for these reasons that this vector is called e , the unit tangent.
Equation (1.40) m a y t h e n b e rewritten as
OP
(
From Equation (1.41) w e see that the velocity vector of point P is
always tangent to its path. The absolute value | s | of the scalar part —
which is the same as the m a g n i t u d e | v | of the velocity vector v — is
called the speed of P in 3, as w e m e n t i o n e d in Section 1.3.
P
P
45
Next w e shall differentiate again in order to obtain the acceleration of
P. Using Equation (1.41), w e get
Since is a unit vector,
is perpendicular to and hence perpen­
dicular, or normal, to the path. Equation (1.42) shows an important
separation of the acceleration into two parts, one tangent and the other
normal to the path of P. The component tangent to the path,
is (for
i the rate of change of the velocity magnitude, or speed, of P. The
component normal to the path reflects the rate of change of the direction
of the velocity vector.
Further examination o f i n Equation (1.42) is facilitated if w e
first restrict our attention to tne case of a two-dimensional (plane) curve.
To that end, let be the inclination of a tangent to the plane curve as
shown in Figure 1.11. We can visualize that as s increases, turns in such
a way that
points toward the inside of the curve — that is, in the
direction of
shown in the figure. We can obtain this result analytically
if w e write
Noting from Figure 1.11 that
Figure 1.11
curve.
Tangent and normal to a plane
which is a unit vector normal to the curve. If dd/ds is positive, as is
w e may differentiate
to obtain
illustrated
in Figure 1.11,
then
is negative (curve
concave downward), then
points toward the outside of the curve,
as the reader may wish to confirm with a sketch, and
again points
toward the inside of the curve. Thus, in either case
where
is understood to point toward the inside of the curve.
From studies in calculus the reader probably recognizes
as
the curvature of a plane curve. The reciprocal of the curvature is the
radius of curvature p. The radius of curvature is the radius of the circle
that provides the best local approximation to an infinitesimal segment of
the curve. Equation (1.43) may thus be written
Page 46
In three dimensions t h e situation is m o r e difficult to visualize. We
cannot use t h e preceding d e v e l o p m e n t because cannot b e expressed as
a function of a single angle such as
Consequently, in the general case
w e adopt a definition of curvature that, in t w o dimensions, reduces to
w h a t w e h a v e just established. That is, w e simply define the curvature
11p to b e t h e m a g n i t u d e of t h e vector
T h e n the unit vector e„ as
defined b y
is called t h e principal unit normal to t h e curve. U p o n substituting into
Equation (1.42), w e t h e n obtain
A n alternative form in w h i c h t h e arclength p a r a m e t e r s is not explic­
itly involved follows if w e choose t h e m e a s u r e m e n t of s so that at the
instant of interest
and
This expression m o r e vividly depicts t h e natural decomposition of accel­
eration into parts related to rate of change of m a g n i t u d e of velocity a n d
rate of change of direction of velocity. W e n o w consider some examples
of the use of tangential a n d n o r m a l c o m p o n e n t s .
EXAMPLE 1 . 2 0
2
A car starts at rest at A and increases its speed around the track at 6 ft/sec ,
traveling counterclockwise (see Figure El.20). Determine the position and the
time at which the car's acceleration magnitude reaches 20 ft/sec .
2
Solution
Figure El.20
The accelerationmagnitudeofQis
2
When a = 20 ft/sec , we obtain the equation
Q
Page 47
At t = 10.3 sec, s = 318 ft, which represents 318/(2jrr) = 0.253 of a revolution,
or 91.1° counterclockwise from the x axis.
EXAMPLE 1 . 2 1
Verify the results of Example 1.13, at x = 2 m, by using e, and e„ components.
Solution
We are given that s = | v | = 0.2 m / s . Since v is tangent to the path of P, we can
calculate e :
P
P
t
x
2
2
x
y = -^ = y f o r / = i
(see Figure E1.21a)
dy
Figure E 1.2la
tan 6 =
= x= 2
(at the given point)
1
0 = tan- (2) = 63.4°
e, = cos 8i + sin 6)
(see Figure El.21b)
= 0.448i + 0.894J
The radius of curvature comes from calculus:
Figure El.21b
EXAMPLE 1 . 2 2
In Example 1.11 find the following for point P at t = 1 sec: tangential and normal
components of acceleration, radius of curvature, and the principal unit normal.
Solution
We obtained
r
3
O P
2
= 2fi + f j + 3f k ft
2
v = 2i + 3f j + 6fk ft/sec
P
a = 6tj + 6k ft/sec
P
2
Page 48
If we write the velocity v as a magnitude times a unit vector, we can determine
and for P:
p
where we note that
since we are choosing the direction of increasing s to be
that of the velocity.
Let us find the tangential and normal components of the acceleration of P at
f = 1 sec:
HgiraEI.22
Now that we have 8„ we can use it to split the acceleration
(at f = 1) into its tangential and normal components (see Figure E1.22). The
tangential component of a (that is, the component parallel to e ) is seen from the
figure to be the dot product of
with
P
(
Next we obtain the normal acceleration component by vectorially subtract­
ing the component
from the total acceleration a . That is, since
P
we obtain
And since
we obtain the radius of curvature:
The unit vector • follows from
as
It isa instructive
function oftotime:
make a direct calculation of
since we here know
Page 49
Thus
which is, of course, the result we have already obtained by investigating the
components of the acceleration vector.
Question 1.6 How would you find the position vector from the origin
O to the center of curvature at t = 1 sec?
In closing this section, w e remark that tangential a n d n o r m a l c o m p o ­
nents of velocity a n d acceleration will b e very useful to us later w h e n w e
h a p p e n to k n o w t h e p a t h of a point (the center C of a w h e e l rolling o n a
curved track, for instance). We can t h e n use Equations (1.41) a n d (1.47) to
express v a n d a .
P
P
Answer 1.6 If w e call the center of curvature C, then
everything evaluated at the time of interest (in this case f = 1 sec).
PROBLEMS
•
with
Section 1.7
1.128 Particle P moves on a circle (Figure P1.12 8) with an
arclength given as a function of time as shown. Find the
time(s) and the angle(s) when the tangential and normal
acceleration components are equal.
1.130 In Problem 1.67 determine the expression for
Integrate, for a motion beginning at t = 0 at (x, y)
= (20, 0) m, and obtain s(f). Evaluate the arclength at
t = 2 s and show that the result, as it should be, is
the circumference of the circle on which P travels.
3
1.131 A point P moves on a path with s = ct where
c = constant = 1 ft/sec . At t = 2 sec, the magnitude
of the acceleration is 15 ft/sec . At that time, find the rad­
ius of curvature of the path of P.
3
2
Figure PI. 128
1.129 In Problem 1.78 find the arclength s as a function of
time.
1.132 A point D moves along a curve in space with a
speed given by
where f is measured from
zero when D is at the arclength origin s = 0. If at a certain
time t' the acceleration magnitude of D is 12 m / s and the
radius of curvature is 3 m, determine f'.
2
Page 50
1.133 At a certain instant the velocity and acceleration of
a point are as shown in Figure PI.133. At this instant find
1.138 Find the radius of curvature of the "Witch of Agnesi" curve at x = 0. (See Figure PI.138.)
a.
b. the radius of curvature of the path
Figure PI. 138
1.139 A point P moves from left to right along the curve
defined in the preceding problem with a constant x com­
ponent
of velocity. Find the acceleration of P when it
reaches the point (x, y) = (0, 2a).
Figure PI. 133
1.134 At a certain instant, the velocity and acceleration of
a point are
1.140 In Problem 1.105, at the same four points express
v in terms of tangential and normal components.
P
1.141 In Problem 1.105, for the position
express
a in terms of tangential and normal components, and
find the radius of curvature of the path of P at that point.
P
At this instant find (a)
(b) the radius of curvature of
the path, (c) the principal unit normal.
1.135 At an instant the velocity and acceleration of a
point are
1.142 A point P starts at the origin and moves along
the parabola shown in Figure PI.142 with a constant
x-component of velocity,
ft/sec. Find the tangen­
tial and normal components of the velocity and accelera­
tion of P at the point (x, y) = (1, 1).
At this instant find:
a.
b. the radius of curvature of the path.
1.136 At a certain instant, the velocity and acceleration of
a point are
Figure PI. 142
* 1.143 In Problem 1.104, find the center of curvature of
the path of P when
Find:
a.
b.
1.137 In Problem 1.103, find the radius of curvature of
the path of P at the instant given. Note that
and
1.144 At a particular instant a point has a velocity
At
this instant find: (a) the principal unit normal, (b) the cur­
vature of the path, and (c) the time rate of change of the
point's speed.
1.145 A point P has position vector
meters. Find the vector from the origin to the center of
Page 51
curvature of the path of P at t = 1 s. Find
same instant.
at the
1.146 The position vector of a point is given as a function
of time by
Find the tangential and
normal components of acceleration at t = 1 sec and de­
termine the radius of curvature at that time.
1.147 A particle P has the x, y, and z coordinates
(3f, 0, 4 In f) meters as functions of time. What is the vec­
tor from the origin to the center of curvature of the path of
P at t = 1 s?
1.148 Show by expressing the velocity and acceleration in
tangential and normal components that
so that
1.149 There is another formula for the radius of curvature
p from the calculus; this one is in terms of a parameter
such as time t, and for a plane curve:
Derive this from the result of the preceding problem
and use it to find the radius of curvature at 8 sec if
2
2
2
1.151 A particle moves on the curve (x — a) + y = a ,
where a is a constant distance in meters. The first and
second time derivatives of the arclength s are related by
in which the constant K has the value 1 second/meter.
The distance s is measured counterclockwise on the curve
from the point (2a, 0) meters. When t = 0, the speed of
the particle is = 1 meter /second and s = a. Find the
normal and tangential components of the acceleration at
time t = 0. Show these components on a diagram.
1.152 Use Example 1.21 to show that the center of curva­
ture does not have to be on the y axis for a curve symmet­
ric abouty. Hint: Use/ = 1 and* = 2, find/?, and compare
with the distance from (2, 1) to the y axis along the normal
to the tangent at this point.
* 1.153 In Problem 1.96 find the tangential and normal
components of the car's acceleration when x = 2500 ft.
Check your result by also computing x and there and
showing that
• 1.154 A particle P starts from rest at the origin and moves
along the parabola shown in Figure PI. 154. Its speed is
given by
where s is in meters per second when
s is in meters. Determine the velocity of P when its x
coordinate is 5 m. Also determine the elaosed time. Hint:
Substitute y' and use a table of integrals to get s(x).
1.150 Find the difference between the velocities (and also
the accelerations) of cars A and B in Figure PI. 150 if, at
the instant shown,
Figure PI. 154
1.155 There is a third unit vector associated with the mo­
tion of a point on its path. It is called the binormal and
forms an orthogonal moving trihedral with
defined by
a. Differentiate
with respect to s. Then,
using
prove that
= 0 and therefore that
is parallel to
b. Using part (a), let „
. (T is called the
tnrsinn of the path or curve.) Then differentiate
with respect to s and prove that
Figure PI. 150
The three equations marked with daggers give the deriva-
Page 52
rives of the three unit vectors associated with a space
curve and are called the Serret-Frenet formulas.
1.156 The derivative of acceleration is called the jerk and
is studied in the dynamics of vehicle impact and in the
kinematics of mechanisms involving cams and followers.
Show that the jerk of a point has the following form in
terms of its intrinsic components:
" 1.157 The following "pursuit" problem is very difficult,
yet it illustrates exceptionally well the idea that the veloc­
ity vector is tangent to the path. Thus we include it along
with a set of steps for the courageous student who wishes
to "pursue" it. A dog begins at the point (x, y) = (D, O)
and runs toward his master at constant speed 2 V . (See
Figure PI.157.) The dog's velocity direction is always
toward his master, who starts at the same time at the
origin and moves along the positive y direction at speed
V . Find the man's position when his dog overtakes him,
and deterrnine how much time has elapsed. Hints: The
man's y coordinate is y (which of course is V*„f). Show
that:
0
2.
3- V = dy /dt.
4. From dividing and rearranging steps 2 and 3, we get
0
M
5. From step 1, we have
6. From steps 4 and 5, we have
7. L e t t i n g f r o m step 6 we get
= dx/2x.
8. By integrating step 7 with a table of integrals, we get
9. From step 8, we get
10. From step 9, we get
11. From step 10, squaring both sides and solving for p,
0
M
12.
when x = D (initial condition).
13. From steps 11 and 12 we have Q = D, so that
14. Integrating step 13, we get
1.
where (x, y ) represent the dog's
D
coordinates at any time.
15. y = 0 when x = D (initial condition).
16. From steps 14 and 15 we have C = 2 D / 3 , so
that
D
2
17.
y =V t.
18. Finally, write the conditions relating to y , y ,
and x when the dog overtakes his master, and
wrap it up!
M
0
M
Figure PI. 157
D
Page 53
COMPUTER PROBLEM
•
Chapter 1
" 1.158 A particle moves in the xy plane according to the
equation
where k is a constant, and has the con­
stant speed v . The particle passes through the origin with
(See Figure PI.158.)
0
a. Show that
b. With the trigonometric substitution 6 = tan </>,
and then consulting integral tables, integrate
the equation and obtain:
Figure PI. 158
c. For the case v /k=
1, use the computer to plot
versus time until 0 has increased from 0 to 2n
radians.
0
S U M M A R Y
•
Chapter 1
In this chapter w e h a v e studied t h e position, velocity a n d acceleration of
a point (or particle). With O being a point fixed in the frame of reference
a n d P denoting the m o v i n g point, t h e n r is a position vector for P a n d
w e defined
O P
Velocity:
Acceleration:
With rectangular coordinates a n d associated unit vectors, a n d with 0
chosen as the origin of t h e coordinate system,
In similar fashion for cylindrical coordinates,
With a p a t h - l e n g t h parameter s(f), describing t h e motion of P o n a given
curve (path) a n d w i t h
and
being unit tangent a n d principal unit
normal, respectively,
Page 54
where p is the radius of curvature of the path at the point occupied by P at
time t. Sometimes it is convenient to choose to measure s(f) so that
in an interval of time of interest. In this case, w e have expressions that
don't involve s explicitly:
REVIEW
QUESTIONS
•
Chapter 1
True or False?
1. The velocity v of a point P is always tangent to its path.
P
2. v depends on the reference frame chosen to express the position
of P.
P
3. v depends on the origin chosen in the reference frame.
P
4. The magnitude and direction in space of v depend on the choice of
coordinates used to locate the point relative to the reference frame.
P
5. a always has a nonvanishing component normal to the path.
6. For any point
P
7. A point can have f = 0 but still have a nonvanishing radial compo­
nent of acceleration.
B. If a ball on a string is being whirled around in a horizontal circle at
constant speed, the center of the ball has zero acceleration.
9. Studying the kinematics of a particle results in the same equations as
studying the kinematics of a point.
10. In our study of the kinematics of a point, the following terms have
not appeared in any of the equations: mass, force, moments, gravity,
momentum, moment of momentum, inertia, or Newtonian (inertial)
frames.
11. The acceleration vector of P, at the indicated point on the path shown
in the figure, can lie in any of the four quadrants.
12. A particle moving in a plane, with constant values of
all times have zero acceleration.
Answflrs: 1. T 2. T 3. F 4. F 5. F 6. T 7. T 8. F 9. T 10. T 11. F 12. F
will at
2
•
KINETICS OF PARTICLES AND
OF M A S S CENTERS OF BODIES
2.1
2.2
Introduction
N e w t o n ' s L a w s and Euler's First L a w
Motion of the Mass Center
2.3
Motions o f Particles and o f M a s s Centers o f Bodies
2.4
W o r k and Kinetic Energy for Particles
The Free-Body Diagram
Work and Kinetic Energy for a Particle
Work Done by a Constant Force
Work Done by a Central Force
Work Done by a Linear Spring
Work Done by Gravity
Conservative Forces
Conservation of Energy
Work and Kinetic Energy for a System of Particles
M o m e n t u m Form of Euler's First L a w
Impulse and Momentum; Conservation of Momentum
Impact
Coefficient of Restitution
2.6
Euler's Second L a w (The M o m e n t Equation)
Moment of Momentum
Momentum Forms of Euler's Second Law
Conservation of Moment of Momentum
SUMMARY
REVIEW
QUESTIONS
Page 55
Page 56
2.1
Introduction
In this chapter w e begin to consider t h e m a n n e r in w h i c h the m o t i o n o f a
b o d y is related to external m e c h a n i c a l actions (forces a n d couples). O u r
kinematics notions o f space a n d time m u s t n o w b e a u g m e n t e d b y those o f
m a s s a n d force, w h i c h , like space a n d time, are primitives o f the subject
o f m e c h a n i c s . W e simply h a v e to agree in a d v a n c e that s o m e m e a s u r e s of
quantity of matter (mass) a n d mechanical action (force) are basic ingre­
dients in a n y attempt to analyze t h e m o t i o n o f a b o d y . W e a s s u m e that
the reader h a s a working k n o w l e d g e , p r o b a b l y from a study o f statics, of
the characteristics o f forces a n d m o m e n t s a n d their vector descriptions.
W e use t h e term body to d e n o t e s o m e material o f fixed identity; w e could
think o f a specific set o f a t o m s , a l t h o u g h t h e m o d e l w e shall e m p l o y is
b a s e d u p o n viewing material on a spatial scale such that m a s s is per­
ceived to b e distributed continuously. A b o d y n e e d n o t b e rigid or e v e n a
solid, but, since o u r subject is classical d y n a m i c s ( n o relativistic effects), a
b o d y necessarily h a s c o n s t a n t m a s s .
In S e c t i o n 2 . 2 w e u s e N e w t o n ' s laws for a particle a n d for interacting
particles to deduce that t h e s u m o f t h e external forces o n a b o d y o f a n y
size is e q u a l t o t h e s u m o f t h e m a ' s o f t h e b o d y , alternatively expressed as
the total m a s s multiplied b y t h e acceleration o f t h e m a s s center. This
result is usually called Euler's first law. Applications o f this are developed
in S e c t i o n 2 . 3 along w i t h a review o f t h e critically important c o n c e p t o f
the free-body diagram.
T h e Principle o f W o r k a n d Kinetic E n e r g y for a particle is developed
in S e c t i o n 2 . 4 , w h e r e i n are found expressions for t h e w o r k d o n e b y
several special types o f forces. T h e c o n c e p t o f a conservative force is
introduced, a n d t h e condition for w h i c h W o r k a n d Kinetic E n e r g y b e ­
c o m e s C o n s e r v a t i o n o f M e c h a n i c a l E n e r g y is established. Finally, the
implications o f W o r k a n d Kinetic E n e r g y for a s y s t e m o f particles are
explored.
In S e c t i o n 2 . 5 the i m p u l s e - m o m e n t u m form o f Euler's first l a w is
developed, a n d conditions for conservation o f m o m e n t u m are d e m o n ­
strated. Applications are m a d e to p r o b l e m s o f impact.
Euler's s e c o n d l a w is t h e subject o f S e c t i o n 2 . 6 . W e return to N e w ­
t o n ' s laws so as to derive this important result w h i c h states that t h e s u m
of t h e m o m e n t s o f the external forces o n a b o d y (or system o f particles)
equals t h e s u m o f the m o m e n t s o f t h e b o d y ' s ma's. M o m e n t u m forms o f
this are developed, primarily for later applications in C h a p t e r s 4 , 5 , a n d 7.
2.2
Newton's Laws and Euler's First Law
T h e usual starting point for relating t h e external forces o n a b o d y to its
m o t i o n is N e w t o n ' s laws. T h e s e were proposed in t h e E n g l i s h m a n Isaac
N e w t o n ' s f a m o u s w o r k t h e Principia, p u b l i s h e d in 1 6 8 7 , a n d are c o m ­
m o n l y expressed today a s :
Page 57
1.
If t h e resultant force F o n a particle is zero, t h e n the particle h a s
c o n s t a n t velocity.
2.
I f F # 0, t h e n F is proportional to t h e time derivative o f the particle's
m o m e n t u m mv (product o f m a s s a n d velocity).
3.
T h e interaction o f t w o particles is t h r o u g h a pair o f self-equilibrating
forces. T h a t is, t h e y h a v e t h e s a m e m a g n i t u d e , opposite directions,
a n d a c o m m o n line o f action.
Clearly the first l a w m a y b e regarded a special c a s e o f t h e s e c o n d , a n d o n e
must add a n assumption about t h e frame o f reference, since a point m a y
h a v e its velocity c o n s t a n t in o n e frame o f reference a n d varying in an­
other. F r a m e s of reference in w h i c h t h e s e l a w s are valid a r e variously
called N e w t o n i a n , G a l i l e a n , or i n e r t i a l . F u r t h e r m o r e , t h e c o n s t a n t o f
proportionality in t h e s e c o n d l a w c a n b e m a d e unity b y appropriate
choices o f units so that t h e l a w b e c o m e s
w h e r e a is the acceleration o f t h e particle.
A s w e m e n t i o n e d briefly in C h a p t e r 1, a particle is a piece o f material
sufficiently small that w e n e e d n o t m a k e distinctions a m o n g its material
points w i t h respect to locations (or to velocities or accelerations). W e also
n o t e d that this definition allows, for some purposes, a truck or a s p a c e
vehicle or e v e n a p l a n e t to b e a d e q u a t e l y m o d e l e d as a particle. In the
Principia, N e w t o n u s e d h e a v e n l y b o d i e s as t h e particles in his e x a m p l e s
a n d treated t h e m as m o v i n g points s u b j e c t o n l y to universal gravitation
a n d their o w n inertia.* N e w t o n did n o t e x t e n d his w o r k to p r o b l e m s for
w h i c h it is n e c e s s a r y to a c c o u n t for t h e actual sizes o f t h e b o d i e s a n d h o w
their m a s s e s are distributed. It w a s to b e over 5 0 years before t h e S w i s s
m a t h e m a t i c i a n L e o n h a r d E u l e r p r e s e n t e d the first o f t h e t w o principles
that h a v e c o m e to b e called E u l e r ' s l a w s .
For a b o d y c o m p o s e d o f a set o f N particles, w e m a y d e d u c e Euler's
laws from N e w t o n ' s laws. A s suggested b y Figure 2 . 1 , w e s e p a r a t e the
forces acting on the I particle into t w o groups: there are the N - l forces
exerted by o t h e r particles o f the system,
b e i n g that exerted by the
particle; then there is F , the net force exerted o n t h e I particle b y things
external to t h e system. Applying N e w t o n ' s s e c o n d law to t h e I particle
w h o s e m a s s is m, a n d w h o s e acceleration is a,,
TH
T H
s
T H
Figure 2.1
in w h i c h w e u n d e r s t a n d
* See C. Truesdell, Essays in the History of Mechanics (Berlin: Springer-Verlag, 1968).
Page 58
N o w w e s u m the N s u c h equations to obtain
But N e w t o n ' s third l a w tells us that
so that
Figure 2.2
T h u s w e c o n c l u d e that
w h i c h is the particle-system form o f Euler's first l a w a n d states that the
s u m o f the external forces o n the s y s t e m equals the s u m o f the ma's o f the
particles m a k i n g u p the system.
For a b o d y w h o s e m a s s is continuously distributed, as depicted in
Motion of the M a s s Center
Figure 2 . 2 , the counterpart to equation (2.3) is
W e close this section b y developing the relationship b e t w e e n the external
w
h e r e acting
dm* is ao ndifferential
m em
n toot ifomn aos sf ,its
a is
a nthis
d
forces
a b o d y a n edl ethe
mits
a s sacceleration,
center. T o do
wise
the
sum
of
the
external
forces
acting
on
the
body.
first construct position vectors for the particles o f a system as s h o w n in
(Fixed in inertial frame)
th
Figure 2 . 3 a . T h u s the acceleration o f the i particle m a y b e written
Figure 2.3a
Applying this to E q u a t i o n ( 2 . 3 ) ,
T h e location o f the m a s s center, C, o f a system o f particles is defined b y
* dm = pdV where p is mass density and dV is an infinitesimal element of volume.
When using rectangular coordinates x, y, z, then dV = dx dy dz.
Page 59
w h e r e m is t h e m a s s
o f t h e s y s t e m . T h u s E u l e r ' s first l a w b e ­
comes
or
Question 2.1
(Fixed in
inertial frame)
Figure 2.3b
W h a t h a p p e n e d to t h e
and
terms in going from
For a c o n t i n u o u s b o d y (Figure 2 . 3 b ) , the counterparts to E q u a ­
tions ( 2 . 5 - 8 ) are
and
resulting again in E q u a t i o n ( 2 . 8 ) . T h u s w e see that t h e resultant external
force o n t h e b o d y is t h e product o f t h e c o n s t a n t m a s s m o f t h e b o d y a n d
t h e acceleration a o f its m a s s center. H e n c e t h e m o t i o n o f t h e m a s s center
of a b o d y is g o v e r n e d b y a n equation identical in form to N e w t o n ' s
s e c o n d l a w for a particle. It is v e r y i m p o r t a n t to realize that for a rigid
b o d y t h e m a s s c e n t e r C coincides at e v e r y instant w i t h a specific material
point o f t h e b o d y or o f its rigid e x t e n s i o n (for e x a m p l e , t h e c e n t e r o f a
h o l l o w s p h e r e ) . T h i s is n o t t h e case for a d e f o r m a b l e b o d y .
c
S o m e t i m e s it is useful to subdivide a b o d y into t w o parts, say o f
m a s s e s m a n d m with m a s s - c e n t e r locations C a n d C . Recalling a
property o f m a s s centers,
1
2
1
2
so that after differentiating twice w i t h respect to time,
Answer 2.1
They are zero since our definition of a body requires that its mass be constant.
Page 60
W e see that Equation ( 2 . 8 ) c a n also b e written
T h e principal purpose o f this section h a s b e e n the derivation of
Equation ( 2 . 8 ) , a n d the n e x t section is devoted w h o l l y to applications o f
it. But the natural o c c u r r e n c e o f the m a s s center b e t w e e n Equations
(2.6) a n d (2.8) motivates a brief review o f the calculation of its location.
T h e e x a m p l e a n d p r o b l e m s that follow are designed to provide that
review in instances for w h i c h the b o d y comprises several parts, the m a s s
centers o f w h i c h are k n o w n .
EXAMPLE 2 . 1
A uniform prismatic rod of density p and length 2L is deformed in such a way that
the right half is uniformly compressed to length L/2 with no change in crosssectional area A. (See Figure E2.1.) The left half of the rod is not altered. Letting
the x axis be the locus of cross-sectional centroids, find the coordinates of the
mass center in the deformed configuration.
Solution
In the first configuration the center-of-mass coordinates are (L, 0, 0); that is, the
center of mass is at the interface of the two segments. In the second configuration,
however,
Figure E2.1
Thus the mass center no longer lies in the interface. This example illustrates that
the mass center of a deformable body does not in general coincide with the same
material point in the body at different times.
PROBLEMS
•
Section 2.2
2.1
Show that the mass center C of a body is unique.
Hint: Consider the two mass centers C and C , respec­
tively:
1
2
and relate R to R . (See Figure P2.1.) Using this relation,
1
2
Figure P2.1
Page 61
show that r = r
the same point!
o1
c 2
, which means that C and C are
1
2
2.2
Find the mass center of the composite body shown
in Figure P2.2. Note that the three parts are composed of
different materials.
2.6
Consider a body that is a composite of a uniform
sphere and a uniform cylinder, each of density p. Find the
mass center of the body. (See Figure P2.6.)
2.7
Find the mass center of the body in Figure P2.7
which is a hemisphere glued to a solid cylinder of the
same density, if L = 2R.
2.8
In the preceding problem, for what ratio of L to R is
the mass center in the interface between the sphere and
the cylinder?
Hollow wood cylinder:
21 slugs
2.9
In Figure P2.9 find the height H of the cone of
uniform density (in terms of R) so that the mass center of
the cone plus hemisphere is at the interface of the two
shapes (i.e., z = 0 ) .
Steel bar, 1 s l u g
- Aluminum sphere.
11 slugs
Figure P2.2
Figure P2.6
2.3
Find the center of mass of the body composed of
two uniform slender bars and a uniform sphere in Fig­
ure 2.3.
2.4
Find the center of mass of the bent bar, each leg of
which is parallel to a coordinate axis and as uniform den­
sity and mass m. (See Figure P2.4.)
2.5
Repeat Problem 2.4 if the four legs have uniform,
but different, densities, so that the masses of
and
are, respectively, m, 2m, 3m, and 4m.
Figure P2.7
2.10 A thin wire is bent into the shape of an isosceles
triangle (Figure P2.10). Find the mass center of the object,
and show that it is at the same point as the mass center of
a triangular plate of equal dimensions only if the triangle
is equilateral. (Area of cross section = A and mass
density = p, both constant.)
Figure P2.3
Figure P2.4
Figure P2.9
Figure P2.10
Page 62
2.3
Motions of Particles and of Mass Centers of Bodies
A l t h o u g h the m a s s center o f a b o d y does n o t always coincide w i t h a
specific material point o f the b o d y , the m a s s center is n o n e t h e l e s s clearly
an important point reflecting the distribution o f the b o d y ' s m a s s . Fur­
thermore, there are a n u m b e r o f situations in w h i c h our objectives are
satisfied if w e can determine the m o t i o n o f a n y material or characteristic
point o f the b o d y . Clearly this is the case w h e n w e attempt to describe the
orbits in w h i c h t h e planets m o v e a r o u n d the sun. Closer to h o m e , a
football c o a c h is overjoyed i f h e finds a punter w h o can consistently kick
t h e ball 6 0 yards in the air, regardless o f w h e t h e r the ball gets there e n d
over end, spiraling, or floating like a " k n u c k l e b a l l . " In such cases the
material point u p o n w h i c h w e focus our attention is unimportant. H o w ­
ever, there is a strong computational advantage in focusing on the m a s s
center: It is that the m o t i o n o f that point is directly related to the external
forces acting o n the body.
W e are m o r e likely to think o f the football as particle-like w h e n
exhibiting the knuckleball b e h a v i o r t h a n w h e n it is rapidly spinning.
N o n e t h e l e s s , the mass center's m o t i o n in e a c h case is g o v e r n e d b y Euler's
first law, although those m o t i o n s might b e quite different b e c a u s e o f the
different sets o f external force i n d u c e d b y the differing interactions o f the
ball with the air.
If the external forces acting o n the b o d y are k n o w n functions o f time,
the m a s s center's m o t i o n can b e calculated from Euler's first law:
or, alternatively,
w h e r e r is a position vector for the m a s s center. It is easily seen that two
integrations o f ( 2 . 1 0 ) with respect to time yield r ( f ) provided that initial
values o f TQC a n d
are k n o w n .
o c
o c
T h e Free-Body Diagram
O n l y o n e thing remains to b e d o n e prior to studying several e x a m p l e s
that m a k e use o f Euler's first l a w to analyze the m o t i o n s o f the mass
centers o f bodies. It is to r e v i e w the c o n c e p t o f the free-body diagram
w h i c h the reader should already h a v e m a s t e r e d in the study o f statics.
W i t h o u t the ability to identify the external forces ( a n d later the m o m e n t s
also), the student will n o t b e able to write a correct set o f equations o f
motion.
A free-body diagram is a sketch o f a b o d y in w h i c h all the external
forces a n d couples acting u p o n it are carefully d r a w n with respect to
location, direction, a n d magnitude. T h e s e forces might result from
pushes or pulls, as the b o y a n d girl are exerting on the crate and rope in
Figure 2 . 4 a . O r the forces m i g h t result from gravity, such as the weight of
Page 63
Figure 2.4a
G r a v i t y a c t s on
each elemental
particle
Resultant gravity
force on crate
and rope
Figure 2.4b
Friction force
distribution
N o r m a l force d i s t r i b u t i o n
Resultant of
friction forces
beneath crate
Resultant of normal
forces beneath cr.ite
Figure 2.4c
the crate in Figure 2 . 4 b . ( N o t e that t h e forces n e e d n o t touch t h e b o d y to
be i n c l u d e d in the free-body diagram; a n o t h e r s u c h e x a m p l e is electro­
m a g n e t i c forces.) O r the forces m i g h t result from supports, such as the
floor b e n e a t h the crate in Figure 2 . 4 c . I f the c r a t e / r o p e b o d y is acted u p o n
simultaneously b y all t h e forces in t h e s e figures, its c o m p l e t e free-body
diagram is as s h o w n in Figure 2 . 5 .
It is i m p o r t a n t to r e c o g n i z e that t h e free-body diagram:
Figure 2.5
1.
Clearly identifies the b o d y w h o s e m o t i o n is to b e analyzed.
2.
P r o v i d e s a catalog o f all the external
3.
A l l o w s us to express, in a c o m p a c t w a y , w h a t w e k n o w or c a n easily
c o n c l u d e a b o u t the lines o f action o f k n o w n a n d u n k n o w n forces. For
e x a m p l e , w e k n o w that t h e pressure (distributed n o r m a l force) e x ­
erted b y t h e floor o n t h e b o t t o m o f t h e b o x h a s a resultant that is a
force with a vertical line o f action. T h e s y m b o l N a l o n g with the
arrow is a c o d e for c o m m u n i c a t i n g the fact that w e h a v e decided to
express that u n k n o w n (vector) force as
T h e fact that w e do not
k n o w t h e location o f t h e line o f action o f that force is displayed b y the
p r e s e n c e o f the u n k n o w n l e n g t h d.
forces ( a n d couples) on t h e b o d y .
In d y n a m i c s , as in statics, the o n l y characteristics o f a force that are
manifest in the equations o f m o t i o n are the vector describing the force
a n d the location o f its line o f action; that is, w e m u s t s u m u p all the
Page 64
Figure 2.6
external forces, a n d w e m u s t also s u m their m o m e n t s about s o m e point.
C o n s e q u e n t l y , everything w e n e e d t o k n o w about the external forces is
displayed o n the free-body diagram, a n d w e m a y readily c h e c k our work
by glancing b a c k a n d forth b e t w e e n our diagram a n d the equations w e
are writing.
W h e n w e focus individually o n t w o or m o r e interacting bodies, the
free-body diagrams provide an e c o n o m i c a l w a y to satisfy—and s h o w
that w e h a v e satisfied—the principle o f action a n d reaction. T h e freeb o d y diagram o f the girl in our e x a m p l e is s h o w n in Figure 2 . 6 . S i n c e w e
h a v e already established b y Figure 2.5 that the force exerted b y the girl o n
the rope will b e
then, b y the action-reaction principle, the force
exerted b y the rope o n the girl must b e
as s h o w n in Figure 2.6. In
other words, consistent forces of interaction are expressed through the
single scalar
a n d the arrow code.
EXAMPLE 2 . 2
Ignoring air resistance, find the trajectory of a golf ball hit off a tee at speed v and
angle with the horizontal.
0
Solution
It is convenient here to set up a rectangular coordinate system as shown in Fig­
ure E2.2 and let time t = 0 be the instant at which the ball leaves the club. With x,
y, and z as the coordinates of the mass center of the ball and since the only
external force on the ball is its weight,
we have from Equation (2.10):
Figure E2.2
Thus, collecting the coefficients of
we obtain
Integrating, we get
Because of the way we have aligned the x and z axes,
Therefore
x(0)
center
Integrating
Our
=location
of
y(0)
the
=again,
ball
of
z(0)
the
is=
we
given
origin
0,get
so that
by
of the
C coordinate
= C = C =system
0 and atthe
thetrajectory
"launch"ofsite
theyields
mass
4
5
6
65
which describes a parabola in the xy plane—that is, in the vertical plane defined
by the launch point and the direction of the launch velocity.
Letting the time of maximum elevation be t , we find that
yields
1
so that t = (v /g)
x
0
sin 6 and the maximum elevation is
If t is the time the ball strikes the fairway (assumed level), then
2
which is, not surprisingly, twice the time (t ) to reach maximum elevation. The
length of the drive is
2
which, with v fixed, is maximized by
That is, for a given launch speed
we get maximum range when the launch angle is 45 °.
The results of this analysis apply to the unpowered flight of any projectile as
long as the path is sufficiently limited that the gravitational force is constant
(magnitude and direction) and we can ignore the medium (air) through which the
body moves. Interaction with the air is responsible not only for the drag (retard­
ing of motion) on a golf ball but also for the fact that its path is usually not planar
(slice or hook!). On one of the Apollo moon landings in the early 1970s, astronaut
Alan Shepard drove a golf ball a "country mile" on the moon because of the
absence of air resistance and, more important, because the gravitational accelera­
tion at the moon's surface is only about one-sixth that at the surface of the earth.
0
EXAMPLE 2 . 3
Figure E2.3a
If the 20-kg block shown in Figure E2.3a is released from rest, find its Speed after
it has descended a distance d = 5 m down the plane. The angle
and the
(Coulomb) coefficients of friction are
Page 66
Solution
In the statement of the problem we are using some loose but common terminol­
ogy in referring to the speed of the block. In fact we may only speak of the speed
of a point, but here we are tacitly assuming that the block is rigid and translating so
that every point in the block has the same velocity and the same acceleration. In
contrast to the preceding example, note that here we do not know all the external
forces on the body before we carry out the analysis, because the surface touching
the block constrains its motion. That constraint is acknowledged by expressing
the velocity of (the mass center of) the block by xi and its acceleration by
Referring to the free-body diagram shown in Figure E2.3b,
Figure E2.3b
or
First we must determine if in fact the block will move. For equilibrium, x = 0 and f
is limited by
Hence
and
or
Figure E2.3c
Thus the block moves (and, as it does, is acted on by N up the plane as shown in
Figure E2.3c). We note that tan
is sometimes called the angle of friction. Here
But
tan
is 16.7°, and this of course is the angle for which tan
it means
that any angle
16.7° (like our 60°) will result in sliding, or a loss of equilib­
rium.
Having checked the statics and briefly reviewed friction, we now solve the
equation of motion for
or
Thus
Page 67
and C = 0 since
Hence
1
if t = 0 is the instant at which the block is released.
and C = 0 if we choose the measurement of x so that x(0) = 0.
If we let t be the time at which x = d, then
2
1
For
we get
from which t = 1.17 s.
Since the velocity is given by
absolute value) of
and
1
the speed at t is merely the magnitude (or
1
Finally we should note that the plausibility of our numerical results can be
verified from the fact that, owing to the steep angle and moderate coefficient of
friction, they should be of the same orders of magnitude as those arising from a
free vertical drop (acceleration g) for which
and
EXAMPLE 2 . 4
A ball of mass m (see Figure E2.4) is released from rest with the cord taut and
Find the tension in the cord during the ensuing motion.
Cord
Solution
In this problem we make two basic assumptions:
Figure E2.4
1.
The cord is inextensible.
2.
The cord is attached to the ball at its mass center (or equivalently the ball is
small enough to be treated as a particle). Either way the point whose motion
is to be described has a path that is a circle. Thus the problem is similar to
Example 2.3 in that the path of the mass center is known in advance (a circle
here and a straight line there) and consequently among the external forces
are unknowns caused by constraints (the tension in the cord here and the
surface reaction in the preceding problem).
Page 68
Using polar coordinates (Section 1.6), we may express the acceleration as
Since the polar coordinate r is the constant I here, referring to the free-body
diagram in Figure E2.4 we have
so that
and
The first of these component equations (Equation (1)) yields the tension T if
we know
the second (2) is the differential equation that we must integrate to
obtain
. In Example 2.3 the counterpart of Equation ,
which
of course was easily integrated.
Here not only do we have a nontrivial differential equation in that is a
function of
but we have the substantial complication that Equation (2) is
nonlinear because cos is a nonlinear function of However, a partial integra­
tion of Equation (2) can be accomplished; to this end we write the equation in the
standard form
and then multiply by
to obtain
which we recognize to be
or
Equation (3) is called an energy integral of Equation (2) and is closely related to
the "work and kinetic energy" principle that is introduced in the next section.
For the problem at hand the constant C, may be obtained from the fact that
when
then
thus
or
Thus from Equation (3) we get
69
which we may substitute in Equation ( 1 ) to obtain
or
Even though we have not obtained the time dependence of the tension,* the
energy integral has enabled us to find the way in which the tension depends on
the position of the ball. As we would anticipate intuitively, the maximum tension
occurs when
,
at which time T = [3(1) — 1] mg = 2 mg.
EXAMPLE 2 . 5
P(planet)
A planet P of mass m moves in a circular orbit around a star of mass M, far away
from any other gravitational or other forces (see Figure E2.5). If the planet com­
pletes one orbit in T units of time, find the orbit radius, using the fact that in a
circular orbit the speed is constant.
Solution
Writing the component of the equation of motion for the planet P in the radial
direction.
Figure E2.5
The only external force on P is gravity, in the
direction (towards the star).
Letting G be the universal gravitational constant, we substitute and obtain:
where from Equation (1.37), the radial acceleration component is
r = R = constant, we have, with
= orbital rate,
But
Since
so that
or
* This would require solving the differential equation (4) for
Equation (5).
and substituting into
Page 70
EXAMPLE 2 . 6
A car accelerates from rest, increasing its speed at the constant rate of K = 6
ft / sec . (See Figure E2.6a.) It travels on a circular path starting at point A. Find
the time and the position of the car when it first leaves the surface due to
excessive speed.
2
Sun
Solution
Before the car (treated as a particle) leaves the surface, the free-body diagram is as
shown in Figure E2.6b. We shall work this problem in general (without substitut­
ing numbers until the end). The purpose is to illustrate the concept of nondimensional parameters. The equation of motion in the tangential
i direction is
Figure E2.6a
Equation (1) shows that the friction exerted on the tires by the road is the
external force which moves the car up the path. Note that after it passes the top of
the circular hill, we have sin
and then the gravity force adds to the friction in
accelerating the car on the way down.
The following equation of motion is the one that will help us in this problem;
it equates
and m a in the normal direction.
c
Figure E2.6b
Question 2.2 Are the components of
tions or just in coordinate directions?
and m a equal in all direc­
c
where
We note that the car will lose contact with the road when N becomes zero.
(The ground cannot pull down on the car for further increases of t, which would
require N < 0!) Therefore, at the point of leaving the ground,
Question 2.3
tion (3)?
What is the meaning of the fact that m cancels in Equa­
Now 6 is related to s according to
Answer 2.2
The components of the two sides of a vector equation are equal in any direction.
Answer 2.3
It means the answer does not depend on the mass of the car.
Page 71
And from
we get another expression for s:
Hence from equations (4) and (5) we get
Substituting for
from (6) into (3), we have
or
Equation
(7) allows us to solve for the dimensionless parameter q =
(Kt /2R),
once we have selected a value of the car's dimensionless acceleration K/g. In this
problem, for example,
2
The following table shows how (with a calculator)* we can quickly arrive at the
value of q that solves Equation (8):
q
0.373q
01
0.5
07 8 5 4
(at the top)
1.0
1.3
1.6
1.7
1.69
1.68
0.7742
0.9595
0.0373
0.1865
1
0.2930
0.9771
08705
0.6862
0.6101
0.6180
0 6258
0 3730
0.4849
0 5968
0.6341
0.6304
0.6266
Thus at
the car leaves the circular track due to excessive
speed. Therefore for K = 6 ft / s e c and R = 1000 ft,
2
* See Appendix B for a numerical solution to this problem using the Newton-Raphson
method.
Page 72
The angle at loss of contact is given by Equation (6):
EXAMPLE 2 . 7
In the system shown in Figure E2.7a, each of the blocks weighs 10 lb and the
pulleys are very much lighter. Find the accelerations of the blocks, assuming the
belt (or rope) to be inextensible and of negligible mass.
Solution
If there is negligible friction in the bearings of the pulley, and the pulley is much
lighter than other elements of the system, then the belt tension won't change
from one side of the pulley to the other. So, referring to Figure E2.7b,
T =T
1
2
For the block in Figure E2.7c:
Figure E2.7a
And for the block and pulley in Figure E2.7d:
We also have a kinematic constraint relationship between y and y because
the belt is inextensible. For this problem it is that
(see Example 1.8) and
consequently
x
Figure E2.7b
2
Figure E2.7c
Solving Equations (1), (2), and (3) simultaneously,
2
10
Figure E2.7d
so the left block accelerates upward at 6.44 ft / sec and the right block accelerates
downward at 12.9 f t / s e c .
2
Page 73
EXAMPLE 2 . 8
Find the accelerations of the blocks shown in Figure E2.8a when released from
rest. Then repeat the problem with the friction coefficients reversed.
Figure E2.8a
Solution
We know from statics that if the two blocks move as a unit, their motion will occur
when
that is, when
which is the case here. But before our solution is complete we must determine
whether either block moves without the other. We consider the free-body dia­
grams of each translating block (see Figure E2.8b) and write the equations of
motion:
We mention that the sum of Equations (1) and (3) gives the "x equation" of the
overall system; the sum of (2) and (4) yields the "y equation" (C is the mass center
of the combined blocks):
Figure E2.8b
Note that f and N , disappear in (5) and (6), as they become internal forces on the
combined system.
1
Page 74
Equation (6) tells us that N = 1870 N, regardless of which motion takes
place. The equation for the x motion (Equation 5) shows again that if f
< mg(sin 25°), then one or both of the blocks must slide:
2
2
and so f cannot be zero. Assuming first that the blocks both move, then f is at its
maximum:
c
2
If they move together as one body, then Equation (5) gives us
Substituting this acceleration into Equation (1), we can check to see if body
additionally slides relative to
/ j = - 1 0 0 ( 0 . 5 8 6 ) + 415 = 356 N
But the maximum value that f can have is given by
x
Hence block slides on and the blocks do not move together; our assumption
was incorrect. We then substitute
into Equation (1) and proceed:
This is then the acceleration of the top block. Substituting f into Equation (3)
gives
x
For no morion of the bottom block, f
clearly needs to be at least 723 N. Since it
is in fact 748 N, the bottom block does not move for this combination of parame­
ters, and
If the friction coefficients are now swapped, nothing changes until we begin
to analyze the six equations. We have
2max
(as before)
Again, then,
cannot be zero. Assuming again that the blocks both move, f is its
maximum and Equation (5) gives
2
Substituting this acceleration into Equation (1), we get
This time we have more friction than we need in order to prevent
on .Thus both
and
are 1.48 m / s .
2
from slipping
Page 75
PROBLEMS
Section 2.3
Figure P2.12
In Problem 1.114 what is the acceleration of the
rock just after release?
2.12 A cannonball is fired as shown in Figure P 2 . 1 2 .
Neglecting air resistance, find the angle a that will result
in the cannonball landing in the box.
2.13 A baseball slugger connects with a pitch 4 ft above
the ground. The ball heads toward the 10-ft-high centerfield fence, 455 ft away. The ball leaves the bat with a
velocity of 125 f t / s e c and a slope of 3 vertical to 4 hori­
zontal. Neglecting air resistance, determine whether the
ball hits the fence (if it does, how high above the ground?)
or whether it is a home run (if it is, by how much does it
clear the fence?).
2,14 From a high vantage point in Yankee Stadium, a
baseball fan observes a high-flying foul ball. Traveling
Figure P2.15
Figure P2.16
vertically upward, the ball passes the level of the observer
1.5 sec after leaving the bat, and it passes this level again
on its away down 4 sec after leaving the bat. Disregarding
air friction, find the maximum height reached by the
baseball and determine the ball's initial velocity as it
leaves the bat (which is 3 ft above the ground at impact).
2.15 A soccer ball (Figure P2.15) is kicked toward the
goal from 60 ft. It strikes the top of the goal at the highest
point of its trajectory. Find the velocity and angle
at
which the ball was kicked, and determine the time of
traveL..
2.16 The motorcycle in Figure P2.16 is to be driven by a
stunt man. Find the minimum takeoff velocity at A for
which the motorcycle can clear the gap, and determine
the corresponding angle 6 for which the landing will be
tangent to the road at B and hence smooth.
Page 76
Figure P2.17
2.17
A baseball pitcher releases a 90-mph fastball 5 ft off
the ground (Figure P2.17). If in the absence of gravity the
ball would arrive at home plate 4 ft off the ground, find
the drop in the actual path caused by gravity. Neglect air
resistance.
2.18
In the preceding problem, find the radius of curva­
ture of the path of the baseball's center at the instant it
arrives at the plate.
2.19 In the preceding problem, the batter hits a pop-up
that leaves the bat at a 4 5 ° angle with the ground. The
shortstop loses the ball in the sun and it lands on second
base,
ft from home plate. What was the velocity of
the baseball when it left the bat?
2.20 The pilot of an airplane flying at 300 km / h r wishes
to release a package of mail at the right position so that it
hits spot A. (See Figure P2.20.) What angle 9 should his
line of sight to the target make at the instant of release?
2.21
A darts player releases a dart at the position indi­
cated in Figure P.2.21 with the initial velocity vector mak­
ing a 10° angle with the horizontal. What must the dart's
initial speed be if it scores a bull's-eye?
Figure P2.20
Figure P2.21
2.22 In the preceding problem, suppose the initial speed
of the dart is 20 f t / s e c . What must the angle a be if a
bull's-eye is scored?
Page 77
Figure P2.23
2.23 Find the angle , firing velocity v and time t of
intercept so that the ballistic missile shown in Fig­
ure P2.23 will intercept the bomber when x = d. The
bomber, at x = D when the missile is launched, travels
horizontally at constant speed v and altitude H. What
has been neglected in your solution?
u
f
0
2.24 The garden hose shown in Figure P2.24 expels
water at 13 m / s from a height of 1 m. Determine the
maximum height H and horizontal distance D reached by
the water.
* 2.25 In the preceding problem, use calculus to find the
angle that will give maximum range D to the water.
* 2.26 Find the range R for a projectile fired onto the in­
clined plane shown in Figure P2.26. Determine the maxi­
mum value of R for a given muzzle velocity u. (Angle
a = constant.)
Figure P2.26
2.28 A child drops a rock into a well and hears it splash
into the water at the bottom exactly 2 sec later. (See Fig­
ure P2.28.) If she is at a location where the speed of sound
is v = 1100 ft / s e c , determine the depth of the well with
and without considering v . Compare the two results.
s
2.27 If a baseball player can throw a ball 90 m on the fly
on earth, how far can he throw it on the moon where the
gravitational acceleration is about one-sixth that on
earth? Neglect the height of the player and the air resist­
ance on earth.
Figure P2.24
s
Figure P2.28
Page 7 8
2.29 At liftoff the space shuttle is powered upward by
two solid rocket boosters of 12.9 X 1 0 N each and by the
three Orbiter main liquid-rocket engines with thrusts
of 1.67 X 1 0 N each. At liftoff, the total weight of the
shuttle (orbiter, tanks, payload, boosters) is about
19.8 X 1 0 N. Determine the acceleration experienced by
the crew members at liftoff. (This differs from the initial
acceleration on earlier manned flights; demonstrate this
by comparing with the Apollo moon rocket, which
weighed 6.26 X 1 0 lb at liftoff and was powered by five
engines each with a thrust of 1.5 X 1 0 lb.) Neglect the
change in mass between ignition and liftoff.
6
6
2.36 The 200-lb block is at rest on the floor ( = 0.2)
before the 50-lb force is applied as shown in Figure P2.36.
What is the acceleration of the block immediately after
application of the force? Assume the block is wide enough
that it cannot tip over.
6
6
6
2.30 What is the apparent weight, as perceived through
pressure on the feet, of a 200-lb passenger in an elevator
accelerating at the rate of 10 f t / s e c upward (a) or down­
ward (b)?
Figure P2.36
2.37
Repeat Problem 2.36 with
2
2.31 When a man stands on a scale at one of the poles of
the earth, the scale indicates weight W. Assuming the
earth to be spherical (4000-mile radius) and assuming the
earth to be an inertial frame, what will the scale read
when the man stands on it at the equator?
2.32 Assuming the earth's orbit around the sun to be
circular and supposing that a frame containing the earth's
center and poles and the center of the sun is inertial,
repeat Problem 2.31. Neglect the earth's tilt.
2.33 In an emergency the driver of an automobile ap­
plies the brakes and locks all four wheels. Find the time
and distance required to bring the car to rest in terms of
the coefficient of sliding friction the initial speed v, and
the gravitational acceleration g.
Figure P2.38
2.39 The two blocks in Figure P2.39 are at rest before the
100-Newton force is applied. If friction between and the
floor is negligible and if = 0 . 4 between and find the
magnitude and direction of the subsequent friction force
exerted on
by
2.34 A box is placed in the rear of a pickup truck. Find
the maximum acceleration of the truck for which the
block does not slide on the truck bed. The coefficient of
friction between the box and truck bed is .
2.35 The truck in Figure P2.35 is traveling at 45 mph.
Find the minimum stopping distance such that the
250-Ib crate will not slide. Assume the crate cannot tip
over.
Figure P2.39
2.40 Find the largest force P for which
will not slide on
in Figure P2.40
Figure P2.40
Figure P2.35
2.41 Work the preceding problem if P is applied to
instead of
Page 79
2.42 The blocks in Figure P2.42 are in contact as they
slide down the inclined plane. The masses of the blocks
are
kg and
kg, and the friction coeffi­
cients between the blocks and the plane are 0.5 for
and
0.1 for Determine the force between the blocks and find
their common acceleration.
Figure P2.47
Figure P2.42
Figure P2.48
2.43 In the preceding problem, let be the coefficient of
friction between
and the plane. Using the two motion
equations of the blocks, find the range of values of for
which the blocks will separate when released from rest.
2.44 If all surfaces are smooth for the setup of blocks and
planes in Figure P2.44, find the force P that will give block
an acceleration of 4 f t / s e c up the incline.
2
2.48 In Figure P2.48 the masses of
ire 1 0 , 6 0 ,
and 50 kg, respectively. The coefficient of friction be­
tween and the plane is
, and the pulleys have
negligible mass and friction. Find the tensions in each
cord, and the acceleration of B, upon release from rest.
2.49 If the system in Figure P2.49 is released from rest,
how long does it take the 5-lb block to drop 2 ft? Neglect
friction in the light pulley and assume the cord connecting
the blocks to be inextensible.
2.50 The coefficient of friction
is the same be­
tween
as it is between and the plane. (See Fig­
ure P2.50.) Find the tension in the cord at the instant the
system is released from rest. Neglect friction in the light
pulley.
Figure P2.44
2.45 Work the preceding problem if the planes are still
smooth but the friction coefficient between
and
is
2.46 Work the preceding problem if the coefficient of
friction is 0.3 for all contacting surfaces.
* 2.47 Generalizing Example 2.8, let the blocks, friction
coefficients, and angle of the plane be as shown in Fig­
ure P2.47. Show that:
a. If tan
b. If
Figure P2.49
motion will occur, and if so:
the blocks move together
c. If
, then
slides on
. In this case, the
lower block does not move if
d. If tan
, then the lower block will not
move. In this case, the upper block slides on it
if and only if tan
Figure P2.50
Page 80
2.51
The system in Figure P2.51 is released from rest.
a. How far does block A move in 2 sec?
b. How would the solution be changed if the coef­
ficient of friction between the floor and A were
2.57
In Figure P2.57 the masses of blocks
are
50, 20, and 30 kg, respectively. Find the accelerations of
each if the table is removed. Which block will hit the floor
first? How long will it take?
Light pulley
Smooth
Block A: 10 kg
Block B: 20 kg
Pulleys: massless
Figure P2.57
Figure P2.51
2.52
A child notices that sometimes the ball m does not
slide down the inclined surface of toy when she pushes
it along the floor. (See Figure P2.52.) What is the mini­
mum acceleration
of to prevent this motion? As­
sume all surfaces are smooth.
2.58
Body
in Figure P2.58 weighs 223 N and body
weighs 133 N. Neglect the weight of the rigid member
connecting
and
The coefficient of friction is 0.3 be­
tween all surfaces. Determine the accelerations of and
just after the cord is cut.
Solid
sphere m
Figure P2.52
Figure P2.58
2.53 In the preceding problem, suppose the acceleration
of is 2a . What is the normal force between the vertical
surface of and the ball? The ball's weight is 0.06 lb.
min
2.54 Let the mass of in Problem 1.57 be 20 kg. What
then must be the mass of
to produce the prescribed
motion? Neglect the masses of the pulleys.
• 2 . 5 5 Find the tension in the cord in Problem 1.61 at the
onset of the motion if the mass of
is 10 kg.
2.56 For the cam-follower system of Problem 1.65 find
the force that must be applied to the cam to produce the
motion. Let the masses of cam and follower be m and m
and neglect friction.
1
2
Figure P2.59
2.59 A particle P moves along a curved surface S as
shown in Figure P 2.5 9. Show that P will remain in contact
with S provided that, at all times,
Page 81
2.60 Find the condition for retention of contact if P
moves along the outside of a surface defined by the same
curve as in the preceding problem. (See Figure P2.60.)
and
is the radius of the circle on which the ball
moves, find the conical speed in terms of
and the ac­
celeration of gravity.
2.63 For an object at rest on the earth's surface, we can
write mg = (GMm) / R , so that the unwieldly constant
GM may be replaced by gR , which for the earth is ap­
proximately 32.2[3960(5280)] ft / sec . Use this, plus the
result of Example 2.5, to solve for the distance above the
earth of a satellite in a circular, 90-minute orbit. (Let the
satellite replace the planet, and the earth replace the star,
in the example.)
2
2
2
Figure P2.60
2.61 A ball of mass m on a string is swung at constant
speed v in a horizontal circle of radius R by a child. (See
Figure P2.61.)
0
a. What holds up the ball?
b. What is the tension in the string?
c. If the child increases the speed of the ball, what
provides the force in the forward direction
needed to produce the ? Explain.
3
2
2.64 Communications satellites are placed in geo­
synchronous orbit, an orbit in which the satellites are
always located in the same position in the sky (Fig­
ure P2.64).
a. Give an argument why this orbit must lie in the
equatorial plane. Why must it be circular?
b. If the satellites are to remain in orbit without
expending energy, find the important ratio of
the orbit radius r to the earth's radius t . Hint:
Use Newton's law of universal gravitation
3
e
together with the law of motion in the radial
direction, and note that if the satellite were sit­
ting on the earth's surface, the force would be
Figure P2.61
2
so that the product Gm may be rewritten as gr ,
as in Problem 2.63. Use r = 3960 mi.
e
e
t
Figure P2.64
Figure P2.62
2.62 There is a speed, called the conical speed, at which
a ball on a string, in the absence of all friction, moves
on a specific horizontal circle (with the string sweeping
out a conical surface) with no radial or vertical component
of velocity (Figure P2.62). If is the length of the string
2.65 Using the result of the preceding problem, show
that a minimum of three satellites in geosynchronous
orbit are required for continuous communications cover­
age over the whole earth except for small regions near the
poles.
Page 8 2
remain against the wall at the same level. Use the equa­
tion
to explain the phenomenon.
Noting that each person is "in equilibrium vertically,"
solve for the minimum
to prevent people from slip­
ping if R = 2 m and the expected friction coefficient be­
tween the rough wall and the clothing is
2.68 In preparation for Problem 2.69, for the ellipse
shown in Figure P2.68, the equation is
Show that the radius of curvature p of the ellipse, as a
function of x, is
Coefficient of friction
Hint: Recall from calculus that if y = y(x), then
Figure P2.66
2.66 In terms of the parameters
and g defined in
Figure P2.66, find the rninimum speed for which the mo­
torcycle will not slip down the inside wall of the cylinder.
2.67 In the "spindle top" ride in an amusement park,
people stand against a cylindrical wall and the cylinder
is then spun up to a certain angular velocity
. (See
Figure P2.67.) The floor is then lowered, but the people
Figure P2.68
2.69 In a certain amusement park, the tallest loop in a
somersaulting ride (Figure P2.69) is 100 ft high and
shaped approximately like an ellipse with a width of
95 ft. The ride advertises "five times the earth's pull at
over 50 mph." Use the result of the preceding exercise to
compute the radius of curvature at the bottom of the
loop. Assuming that the normal force resultant is
5 mg, determine whether or not the maximum speed is
over 50 mph. Treat the cars as a single particle.
Figure P2.67
2.70 If bar shown in Figure P2.70 were raised slowly,
block
would start to slide at the angle
which was seen in statics to be one way of determining
the friction coefficient
Suppose now that the bar is
suddenly rotated, starting from the position
at
constant angular velocity
For
and
= O.lg, compute the angle at which
slips downward
on
and compare the result with t a n
- 1
Page 83
Figure P2.72
Figure P2.69
Figure P2.73
Figure P2.70
2.71 In the preceding problem, let remain at 0.5 but
consider increasing the parameter
. At what value
of this parameter will
slide outward on ? At what
angle will this occur?
2.72 A horizontal wheel is rotating about its fixed axis
at a rate of 10 rad/sec, and this angular speed is increas­
ing at the given time at
r a d / s e c . (See Figure 2.72.)
At this same instant, a bead is sliding inward relative to
the spoke on which it moves at 5 ft/sec; this speed is
slowing down at this time at 2 f t / s e c . If the bead weighs
0.02 lb and is 1 ft from the center in the given configura­
tion, find the external force exerted on the bead. Is it
possible that this force can be exerted solely by the spoke
and not in part by other external sources?
2
2.74 A bead slides down a smooth circular hoop that, at
a certain instant, has
r a d / s e c and
rad/sec
in the direction shown in Figure P2.74. The angular speed
of line OP at this time is
r a d / s e c and
Find the value of
and the force exerted on the bead
by the hoop at the given instant, if the mass of the bead
is 0.1 kg and the radius of the hoop is 20 cm. Hint: Use
spherical coordinates.
2
2
2.73 A ball bearing is moving radially outward in a
slotted horizontal disk that is rotating about the vertical
z axis. At the instant shown in Figure P2.73, the ball bear­
ing is 3 in. from the center of the disk. It is traveling
radially outward at a velocity of 4 in. / sec relative to the
disk. If
r a d / s e c and is constant, find and the force
exerted on the ball by the disk at this instant. Assume no
friction and take the weight of the ball to be 0.05 lb.
Hoop
P (mass = 1 kg]
Figure P2.74
Page 84
•
The four light rods are pinned at the origin and at
each mass in such a way that as these seven bodies are
spun up about the vertical, the masses m move outward
and the mass M slides smoothly up along the vertical rod
Oy. There is a relationship between
g, m, and M
such that at the particular spin-speed i , the bodies be­
have as one rigid body (meaning
remains constant).
Find the relationship. Hint: Use separate free-body dia­
grams of m and M, and write equations of motion for
each. The unknowns are F (force in each top rod) and F
(force in each bottom rod),
and . There will be three
useful equations. See Figure P2.75.
T
2.78 A wintertime fisherman of mass 70 kg is in trou­
ble—he is being reeled in by Jaws on a lake of frozen ice.
At the instant shown in Figure P2.78, the man has a ve­
locity component, perpendicular to the radius r, of
= 0.3 m / s at an instant when r = R, = 5 m. If Jaws pulls
in the line with a force of 100 N, find the value of
when the radius is R = 1 m. Hint:
2
B
Figure P2.75
0
Figure P2.78
Figure P2.76
2.76 A particle P of mass m moves on a smooth, hori­
zontal table and is attached to a light, inextensible cord
that is being pulled downward by a force F(t) as shown in
Figure P2.76. Show that the differential equations of mo­
tion of P are
Then show that Equation (2) implies that
2.77 In the preceding problem, let the particle be at
r = r„ at t = 0, and let the part of the cord beneath the
table be descending at constant speed v . If the transverse
component of velocity of P is
find the
tension in the cord as a function of time t.
2.79 In the preceding problem, show that the differen­
tial equation of the man's radial motion is
to integrate this, and if
when r = 5 m, show that the radial component
of the man's velocity when r = 1 m is 3.04 m / s .
2.80 Particle P of mass m travels in a circle of radius a on
the smooth table shown in Figure P2.80. Particle P is
connected by an inextensible string to the stationary par­
ticle of mass M. Find the period of one revolution of P.
• 2.81 A weight of 100 lb hangs freely from a light rope
(Figure P2.81). It is pulled up by a force that is 150 lb at
t = 0 but diminishes uniformly in magnitude at 1 lb per
foot pulled up. Find the time required to pull the weight
up to the platform from rest, and determine its velocity
upon reaching the top.
c
• 2.82 Rework Problem 2.81, but this time assume that the
force increases by 1 lb per foot pulled up.
Page 85
Figure P2.8D
Figure P2.83
PLATFORM
Figure P 2 . 8 4
2.85 Using the result of the preceding problem and ex­
pressing v as dx/dt, solve for x(f) if x = 0 when f = 0.
2.86 The identical plastic scottie dogs shown in Fig­
ure P2.86 are glued onto magnets and attract each other
with a force F = K / (2x:) , where K is a constant related
to the strength of the magnets. Find the speeds at
which the dogs collide if the magnets are initially sepa­
rated by the distance S.
2
Figure P2.81
2.83 The acceleration of gravity varies with distance z
above the earth's surface as
where g is the acceleration of gravity on the surface and R
is the earth's radius. Find the minimum firing velocity i>j
that a projectile must have in order to escape the earth if
fired straight up (Figure P2.83). Hint: Not to return to
earth requires the condition that
as z gets large for
the minimum possible v .
t
2.84 The mass m shown in Figure P2.84 is given an ini­
tial velocity of v in the x direction. It moves in a medium
that resists its motion with force proportional to its veloc­
ity, with proportionality constant K. By solving for v(x),
determine how far the mass travels before stopping. Then
solve in a different manner for v(t) if v = v„ when f = 0.
0
Initially
Start from
rest
Smooth plane
Figure P2.86
2.87 A ball is dropped from the top of a tall building. The
motion is resisted by the air, which exerts a drag force
given by Dv ; D is a constant and v is the speed of the ball.
Find the terminal speed (the limiting speed of fall) if there
is no limit on the drop height. What is the drop height for
which the ball will strike the ground at 95 percent of the
terminal speed?
2
Page 86
in 12 sec after exiting a stationary blimp. Assuming
velocity-squared air resistance, solve the differential
equation of motion
and determine the constant k.
• 2.91 In the preceding problem, suppose the parachutist
opens his chute at a height of 1000 ft. If the value of k then
becomes 0.63 lb / ( f t / s e c ) , find the velocity at which the
parachutist strikes the ground, if
ft/sec.
2
2.92 The drag car of mass m shown in Figure P2.92,
traveling at speed v , is to be initially slowed primarily by
the deployment of a parachute. The parachute exerts a
force F proportional to the square of the velocity of the
car,
Neglecting friction and the inertia of the
wheels, determine the distance traveled by the car before
its velocity is 40 percent of v . If the car and driver weigh
1000 lb and C = 0.182 l b - s e c / f t , find the distance in
feet.
0
Figure P2.88
d
0
2
2
Figure P2.92
2.93
In the preceding problem, suppose the drag car's
speed at parachute release is 237 mph. Find the time it
takes to reach 40 percent speed.
Figure P2.90
• 2.88 Over a certain range of velocities, the effect of air
resistance on a projectile is proportional to the square of
the object's speed. If the object can be regarded as a parti­
cle, the drag force is expressible as
, in which
p is the density of the air, A is the projected area of the
object onto a plane normal to the velocity vector, and C
is a coefficient that depends on the object's shape. If
pAC = 0.0004 l b - s e c / f t for the 76-lb cannonball of
Figure P2.88, find the maximum height it reaches.
Compare your result with the answer neglecting air
resistance.
D
2
*
A 50-lb shell is fired from the cannon shown in
Figure P2.94. The pressure of the expanding gases is in­
versely proportional to the volume behind the shell. Ini­
tially this pressure is 10 tons per square inch; just before
exit, it is one-tenth this value. Find the exit velocity of the
shell.
2
D
• 2 . 8 9 In the preceding problem, find the velocity of the
cannonball just before it hits the ground; again compare
with the case of no air resistance.
• 2.90 A 160-lb parachutist in the "free-fall spread-stable
position" (Figure P2.90) reaches a velocity of 174 f t / s e c
Figure P2.94
Page 87
2.95 The block of mass m shown in Figure P2.95 is
brought slowly down to the point of contact with the end
of the spring, and then (at t = 0) the block is released.
Write the differential equation governing the subsequent
motion, clearly defining your choice of displacement pa­
rameter. What are the initial conditions? Find the maxi­
mum force induced in the spring and the first time at
which it occurs.
2.98 Continue the preceding exercise and show that the
particle's position is given by the equations
further that both x and y approach asymptotes as
(The limiting value of is known as the terminal
velocity of P, after which the weight is balanced by the
viscous resistance so that the acceleration goes to zero.)
SVinw
FIGURE P2.95
2.96 Show that for a particle P moving in a viscous me­
dium in which the air resistance is proportional to velocity
(Figure P2.96), the differential equations of motion are
* 2.99 A particle moves on the inside of a fixed, smooth
vertical hoop of radius a. It is projected from the lowest
point A with velocity
Show that it will leave the
hoop at a height 3a/2 above A and meet the hoop again
at A.
* 2.100 The two particles in Figure P2.100 are at rest on a
smooth horizontal table and connected by an inextensible
string that passes through a small, smooth ring fixed
to the table. The lighter particle (mass m) is then pro­
jected at right angles to the string with velocity v .
Prove that the other particle will strike the ring with
velocity
Hint: Use polar coordinates
0
and note that
is constant for each particle.
Figure P2.96
Figure P2.100
2.97 In the preceding problem, show by integration that
the components of velocity of P are given by
2.4
Work and Kinetic Energy for Particles
In E x a m p l e 2 . 4 w e w e r e a b l e to get useful i n f o r m a t i o n from a n energy
integral o f t h e governing differential equation. T h e s a m e result m a y b e
o b t a i n e d in general b y a n integration o f
F o r m i n g t h e dot product o f e a c h side with t h e velocity v
center, w e h a v e
c
of the mass
Page 88
Integrating,* w e get
W o r k and Kinetic Energy for a Particle
For a particle,
• v dt is called the w o r k d o n e o n the particle b y
or, for
a particle,
the
resultant
o f external forces. W e n o t e that if there are N forces acting
on the particle, t h e n the resultant
is given b y F ! + F + ... + F a n d
2
N
E a c h term o f this equation represents the rate o f w o r k o f o n e o f the forces.
T h u s the left side o f E q u a t i o n ( 2 . 1 2 ) m a y b e read as the s u m o f the works
of the individual forces acting on the particle. T h e s e statements are all
consistent with the presentation to c o m e in C h a p t e r 5 in w h i c h w e define
the rate of w o r k done b y a force F to b e F • v, w h e r e v is t h e velocity o f the
point o f the b o d y at w h i c h the force is applied. T h e left side o f Equation
( 2 . 1 1 ) m a y t h e n b e interpreted as the w o r k that would b e d o n e b y the
external forces were e a c h to h a v e a line o f action t h r o u g h the m a s s center.
For a particle,
is called the kinetic energy, usually written
T. T h u s for the particle, Equation ( 2 . 1 2 ) is the w o r k a n d kinetic energy
principle:
W o r k d o n e o n the particle = C h a n g e in the particle's kinetic energy
* Sometimes (t , t ), referring to "initial" and "final," are used instead of (t
f
f
u
t ).
2
Thus the appropriate unit of work and of energy in SI is the joule (J), the joule being
i N • m; in U.S. units the ft-lb is the unit of work and energy. The N • m and lb-ft are
usually reserved for the moment of a force. Note, that work, energy, and moment of
force all have the same dimension.
Page 89
or
For a body, the kinetic energy is defined to b e the s u m o f the kinetic
energies o f the particles constituting the b o d y . I f all the points in a b o d y
h a v e the s a m e velocity ( w h i c h is t h e n v ) , t h e n
is the total
kinetic energy o f In general, h o w e v e r , the b o d y is turning or deforming
(or b o t h ) a n d this is n o t the case; the b o d y t h e n h a s additional
kinetic
energy due to its c h a n g e s in orientation (that is, due to its angular m o t i o n )
or due to the deformation. For a b o d y
w e shall also see in C h a p t e r 5
that, in general, the left side o f Equation ( 2 . 1 1 ) does n o t constitute the
total w o r k d o n e on b y the external forces a n d couples. This is b e c a u s e ,
for a b o d y , the forces do n o t h a v e to b e concurrent as t h e y are for a
particle. Equation ( 2 . 1 3 ) still turns out to b e true for a rigid b o d y , h o w ­
ever, with the two sides o f Equation ( 2 . 1 1 ) representing parts o f W
and
Finally, with n o restrictions o n the size o f the body, the energy
integral (Equation 2 . 1 1 ) states that the work that w o u l d b e d o n e if the
external forces acted at the m a s s center equals the c h a n g e in w h a t w o u l d
b e the kinetic energy if every point in the b o d y h a d the velocity o f the
m a s s center. W e could call this result the " m a s s center w o r k a n d kinetic
e n e r g y principle."
c
Work Done by a Constant Force
Before attempting to apply the w o r k - e n e r g y principle to a specific p r o b ­
lem, it is helpful to determine the w o r k d o n e b y t w o classes o f forces.
First, suppose F is a constant force a n d suppose w e let r b e a position
vector for the particle. T h e n
which states that the work done is the dot product of the force with the
displacement o f the particle. W e recall that this dot product c a n b e e x ­
pressed as the product o f the force m a g n i t u d e a n d the c o m p o n e n t of
displacement in t h e direction of the force or as t h e product o f t h e dis­
p l a c e m e n t magnitude a n d the c o m p o n e n t o f force in the direction o f the
displacement.
Work Done by a Central Force
T h e s e c o n d case to w h i c h w e give special attention is that o f a central
force. S u c h a force is defined to h a v e a line o f action always passing
through the s a m e fixed p o i n t in t h e frame o f r e f e r e n c e a n d a magnitude
Page 90
that d e p e n d s o n l y u p o n the distance r o f the particle from that fixed
point, as s h o w n in Figure 2.7.
T h e velocity o f the particle m a y b e expressed as
Figure 2.7
since b y Equation ( 1 . 3 8 ) w e k n o w that
is
If
is a function o f r so that
T h u s the w o r k d o n e by
then the w o r k m a y b e written as
W o r k Done b y a Linear Spring
A special central force is that exerted b y a spring o n a particle w h e n the
other e n d o f the spring is fixed. In the case o f a linear spring o f instanta­
n e o u s l e n g t h r, w e n o t e that f=k(r
— L„), w h e r e L is its natural, or
unstretched, length a n d k is called the spring m o d u l u s or stiffness. In this
case,
or, m o r e simply,
where
is the
spring stretch. T h u s b y equation ( 2 . 1 6 ) ,
0
Question 2.4 What assumption about the mass of the spring is to be
understood in the force-stretch relationship?
Answer 2.4 (k) . (stretch) gives the equal-in-magnitude but opposite-in-direction force
acting at the ends of a spring in equilibrium. If particles in the spring are accelerating,
as is generally the case in dynamics problems, there is no simple force-stretch law. If the
spring is very light, however, so that its mass may be neglected (compared to the masses
of other bodies in the problem),
and the forces on the spring are instan­
taneously related just as if the spring were in equilibrium. The hidden assumption is
that the mass of the spring may be neglected.
Page 91
Work Done by Gravity
A s e c o n d special case o f a central force is the gravitational force exerted
on a b o d y b y the earth. B y N e w t o n ' s l a w o f universal gravitation,
w h e r e r is the radius o f the earth, m is the m a s s o f the attracted b o d y , and
e
g is the gravitational strength (or acceleration) at t h e surface o f the earth.*
By Equation (2.15) we have
W e n o t e t h a t for this case the function
is given b y
If the m o t i o n is sufficiently n e a r t h e surface o f the earth,
a n d so
a c o n s t a n t . In this case,
= (weight) X (decrease in altitude o f m a s s center o f b o d y )
(2.20)
a n d the
function b e c o m e s (if z is positive u p w a r d )
c
* The force of gravity in fact results in infinitely many differential forces, each tugging
on one of the body's particles. For nearly all applications on the planet earth, these
forces may be thought of as equivalent to a single force through the mass center of the
body. For applications in astronomy or in space vehicle dynamics, however, the gravity
moment
that accompanies
Conservative
Forces the force at the mass center becomes important. In Skylab,
for example, three huge control-moment gyros were present to "take out" the angular
momentum
builtwup
moment ofthe
only
the
In e a c h case
e by
h a av egravity
considered,
w ao rfew
k hlb-ft.
a s dAnd
epen
d egravity
d o n l ymoment
o n the
exerted on the earth by the sun and moon's gravitation causes the earth's axis to precess
initial a n d final positions o f the p o i n t w h e r e the force is applied. S u c h a
in the heavens once every 25,800 yr. The gravity moment vanishes if the body is a uni­
forcesphere
w h o s(which
e w o rthe
k isearth
i n dise pnot,
e n dbeing
e n t bulged
o f the atp the
a t hequator
traveled
y the varying
point on
form
and bhaving
density). A further discussion of this luni-solar precession is presented in Chapter 7.
Page 92
w h i c h it acts is called conservative. Furthermore, the w o r k m a y b e ex­
pressed as the c h a n g e in a scalar function o f position; w e s a w this to b e
the case for the central force, and w e m a y m a k e the s a m e s t a t e m e n t for
the constant force b y d e n n i n g
to b e —F • r.
Question 2.5
Why the minus sign in front of F • r?
Conservation of Energy
If all forces acting are conservative a n d if
is the s u m o f all their
functions, t h e n the w o r k a n d kinetic e n e r g y equation ( 2 . 1 3 ) b e c o m e s
or
or
W e call the potential energy a n d
the total (mechanical) energy.
T h u s Equation ( 2 . 2 2 ) is a s t a t e m e n t o f conservation o f m e c h a n i c a l energy
w h e n all the forces are conservative ( p a t h - i n d e p e n d e n t ) .
Question 2.6 How would Equation (2.22) read if instead of we had
chosen to construct the scalar function
so that the work done by a
force is the increase in its ?
In closing it is important to realize that n o t all forces are conservative.
A n e x a m p l e is the force o f friction acting o n a b l o c k sliding on a fixed
surface. T h a t force does negative w o r k regardless o f the direction o f the
motion, a n d a potential function
c a n n o t b e found for it.
North
Answer 2.5 It is needed so that the work equals the decrease in
that is
Answer 2.6
we use the simply so that we may say that mechanical
energy is the sum of its two parts.
EXAMPLE 2 . 9
In an accident reconstruction, the following facts are known:
1.
2.
Identical cars 1 and 2, respectively headed west and south as indicated in
Figure E2.9, collided at point A in an intersection.
With locked brakes indicated by skid marks, the cars skidded to the final
positions B a n d B shown in the figure.
1
Figure E 2 . 9
2
Assuming the cars are particles, determine their velocities immediately following
the collision if the friction coefficient between the tires and road is 0.5.
Page 9 3
Solution
After separation, each car is brought to rest by the friction force acting on its tires.
For Car 1, we have
in which T = 0 since the cars end up at rest, and v is the speed of Car 1 just after
the cars separate. Thus
f
t
Similarly for Car 2,
In the next section, we will return to this example and use the principle of impulse
and momentum to approximate the speeds of the cars before the impact.
EXAMPLE 2 . 1 0
We repeat Example 2.4 (see Figure E2.10): For a ball of mass m released from rest
with the cord taut and
we wish to find the tension in the cord as a
function of
Solution
If, as before, we write the force-acceleration component equation in the radial
direction, we have
Now we apply Equation (2.13) by letting t be the initial time at which
and letting t be the time at which we are applying (1). We note that
t
,
2
Figure E2.10
and the work done by the cord tension T is zero since that force is always
perpendicular to the velocity of (the center of mass of) the ball. By Equation (2.20)
the work of the weight is
Thus Equation (2.13) yields
or
Substituting in Equation (1) above, we get
which is precisely the result obtained previously.
Page 94
EXAMPLE 2 . 1 1
The block shown in Figure E2.11a slides on an inclined surface for which the
coefficient of friction is
Find the maximum force induced in the spring if
the motion begins under the conditions shown.
Solution
We assume that the block can be treated as rigid; thus the end of the spring, once
it contacts the block, will undergo the same displacements as the mass center (or
for that matter any other point) of the block. To apply the mass-center work and
kinetic energy principle we let t be the initial time shown above and let t be the
time of maximum compression of the spring. To catalog the external forces that
do work, we consider a free-body diagram at some arbitrary instant between
and t . (See Figure E2.11b.) Since the mass center has a path parallel to the
inclined plane, there is no component of acceleration perpendicular to it and
1
2
2
or
so that the friction force is 0.3(20) = 6 lb.
Denoting the left side of Equation (2.11) by Work (fj, t ) we have
2
Figure E2.1 l a
where
and the various works are
1.
For N, work t , t ) = 0 since the force is perpendicular to v at each instant.
2.
For friction, work
3.
For the weight, work
4.
For the spring,
1
2
c
Work (t , t ) = work(t , contact) + work(contact, t )
1
2
Figure E2.1 lb
Thus,
or
From the quadratic formula,
1
2
Page 95
from which only the positive root is meaningful:
The corresponding force is 100(1.64) = 164 lb.
EXAMPLE 2 . 1 2
In the preceding example, find the next position at which the block comes to rest.
Solution
At time t the spring force (164 lb) exceeds the sum of the component of weight
along the plane (15 lb) and the maximum frictional resistance (6 lb), so we know
that the block is not in equilibrium and must then begin to move back up the
plane, with the friction force now acting down the plane as shown in Fig­
ure E2.12. Suppose we let r be the time at which the block next comes to rest and
let d represent the corresponding compression of the spring. Then, since
we have
2
3
For the spring, the work is
is
for the weight, the work is
for the friction force, the work
Thus
or
Figure E2.12
The negative sign here tells us that the spring must be stretched 1.22 in. when the
block again comes to rest. If, as intended here, the spring does not become
permanently attached to the block on first contact (that is, contact is maintained
only in compression), our analysis only tells us that contact is broken before the
block comes to rest. We therefore need to modify the expression for the work
done by the spring, which we now see should have been
It is
convenient to let e be the distance from the end of the spring to the block
(measured up the plane). Then
It is instructive to obtain this result by using the work and kinetic energy principle
over the interval t to f , for which the net work done by the spring is zero. Noting
then that the mass center of the block drops
in. vertically and that the
distance traveled by the block on the plane is
in., we have
1
3
Page 9 6
which is the same result we obtained before except for the third significant
figure — a consequence of rounding off at an intermediate step.
EXAMPLE 2 . 1 3
A particle P of mass m rests atop a smooth spherical surface. (See Figure E2.13a.)
A slight nudge starts it sliding downward in a vertical plane. Find the angle 6 at
which the particle leaves the surface.
L
Solution
Mechanical energy is conserved here because (1) there is no friction, (2) the
normal force does not work since it is always normal to the velocity of P, and (3)
the only other force is gravity. (See Figure E2.13b.) Therefore, using Equations
(2.21) and (2.22),
Figure E2.13a
+ r direction
Equation (1) contains two unknowns; to eliminate the velocity v , we use the
equation of motion in the radial direction:
2
Figure E2.13b
But N has just become zero when P is at the point of leaving. Therefore
Equating the right sides of Equations (1) and (3), we get
Page 97
Work and Kinetic Energy for a System of Particles
As h a s b e e n m e n t i o n e d b e f o r e , the kinetic e n e r g y o f a s y s t e m o f particles
is given b y
or
N o w if w e let W, b e t h e w o r k o f all t h e forces acting o n t h e i'th particle,
and, s u m m i n g o v e r all t h e particles,
or
for the system. T h e work, W, h o w e v e r , is the n e t w o r k o f all t h e forces
external and internal
can
that act on t h e particles. S o m e t i m e s E q u a t i o n ( 2 . 2 3 )
b e u s e d effectively b e c a u s e t h e n e t w o r k o f internal forces c a n b e
evaluated. For e x a m p l e , i f t w o m o v i n g particles are j o i n e d b y a linear
spring, n o simple formula c a n b e written for t h e w o r k o f t h e spring force
on one o f t h e particles. H o w e v e r , t h e n e t w o r k o f t h e e q u a l a n d opposite
spring forces on t h e t w o particles is given b y E q u a t i o n ( 2 . 1 7 ) (see P r o b ­
lem
2 . 1 2 8 ) . Particles that are rigidly c o n n e c t e d interact t h r o u g h forces
for w h i c h n o n e t w o r k is d o n e . T h u s , w e shall find in C h a p t e r 5 that
w h e n w e use
for a rigid b o d y , t h e w o r k o n l y involves external
forces.
PROBLEMS
•
Section 2.4
2.101 A truck body weighing 4000 lb is carried by four
light wheels that roll on the sloping surface. (See Fig­
ure P2.101 ) The truck has a velocity of 5 f t / s e c in the
position shown. Determine the modulus of the spring if
the truck is brought to rest by compressing the spring 6 in.
Note: Light wheels with good bearings imply negligible
friction.
2.102 The block shown in Figure P2.102 weighs 100 lb
and the spring's modulus is 10 lb/ft. The spring is unstretched when the block is released from rest. Find the
minimum coefficient of friction such that the block will
not start back up the plane after it stops.
Figure P2.101
Figure P2.102
Page 98
2.103 The block shown in Figure P2.103 is released from
rest. What is its velocity when it first hits the spring?
2.104 How far does the block in the preceding problem
rebound back up the plane after compressing the spring?
2.105 The 6-lb block shown in Figure P2.105 is released
from rest when it just contacts the end of the unstretched
spring. For the subsequent motion, find: (a) the maximum
force in the spring; (b) the maximum speed of the block.
2.106 Block
weighs 16.1 lb and translates along a
smooth horizontal plane with a speed of 36 ft / sec. (See
Figure P2.106.) The coefficient of friction between
and
the inclined surface is
and the spring constant is
100 l b / f t . Determine the distance that
moves up the
incline before coming to rest.
2.107 The weight shown in Figure P2.107 is prevented
from sliding down the inclined plane by a cable. An engi­
neer wishes to lower the weight to the dashed position by
inserting a spring and then cutting the cable. Find the
modulus of a spring that will accomplish this task without
allowing the block to move back up the incline after it
stops. Hint: You are free to specify the initial stretch — try
zero!
Figure P2.107
2.108 At the instant shown in Figure P2.108, the block is
traveling to the left at 7 m / s and the spring is un­
stretched. Find the velocity of the block when it has
moved 4 m to the left.
2.109 Repeat the preceding problem with
will need
You
where a and c are constants.
2.110 The Bernoulli brothers posed and then solved the
"brachistochrone problem." (See Figure P2.110.) The
problem was to determine on which single-valued, con­
tinuous, smooth path a particle would arrive at B in mini­
mum time under uniform gravity, after beginning at rest
at a higher point A. Their solution, beyond the scope of
this book, was that this path of "quickest descent" is a
cycloid. Show that, regardless of the path, the speed on
arrival is as if the particle had been dropped freely
through the same height H.
Figure P2.103
Figure P2.1D5
Figure P2.108
Smooth
Figure P2.10G
Figure P2.110
Page 99
2.111 A small box (see Figure P2.111) slides from rest
down a rough inclined plane from A to B and then falls
onto the loading dock. The coefficient of sliding friction
between the box and plane is
Find the distance D
to the point C where the box strikes the dock.
2.112 A particle is released at rest at A and slides on the
smooth parabolic surface to B, where it flies off. (See
Figure P2.112.) Find the total horizontal distance D
that it travels before hitting the ground at C.
Figure P2.111
2.113 A particle of mass m slides down a frictionless chute
and enters a circular loop of diameter d. (See Fig­
ure P2.113.) Find the minimum starting height h in order
that the particle will make a complete circuit of the loop
and exit normally (without having lost contact with the
loop).
2.114 An 80-lb child rides a 10-lb wagon down an incline
(Figure P2.114). Neglecting all frictional losses, find the
"weight" of the child at A as indicated by a scale upon
which she is sitting.
Figure P2.112
A skier descends the smooth slope, which may be
approximated by the parabola
(see Fig­
ure P2.115). If she starts from rest at "A" and has a mass
of 52 kg, determine the normal force she exerts on the
ground the instant she arrives at "B", and her acceleration
there. Note: treat the skier as a particle, and neglect fric­
tion.
Figure P2.113
Starts at rest here
Figure P2.115
Figure P2.114
Page 100
2.126 The system in Figure P2.126 consists of the 12-lb
body
the light pulley
the 8-lb "rider"
and the
10-lb body
Everything is released from rest in the
given position. Body
then falls through a hole in
bracket
which stops body
Find how far
descends
from its original position.
2.116 Find the speed sought in Problem 2.81 using work
and kinetic energy.
2.117 Use work and kinetic energy to solve Problem 2.83.
2.118 Use work and kinetic energy to solve Problem 2.94.
2.119 Use work and kinetic energy to solve Problem 2.86.
2.127 At the instant shown in Figure P2.127, block is
30 m below the level of block
At this time, v and v
are zero. Determine the velocities of
as they pass
each other.
have masses of 15 kg and 5 kg, re­
spectively. The pulleys are light.
2.12 0 Show that the equation
can
be obtained by conservation of mechanical energy
in Example 2.10. Why can this princi­
ple not be used in Examples 2.11 and 2.12?
A
Check the solutions to Problems 2.78 and 2.79 by
using the principle of work and kinetic energy.
2.122 Show
B
that
for central gravitational force
as distinct from the uniform gravin the text, the potential is given by
where G is the universal gravitational constant and M and
m are the masses of the two attracting bodies. Note that in
view of Equation (2.19), one simply needs to show that
2.123 Using the result of the preceding problem, calculate
the work done by the earth's gravity on a satellite be­
tween the times of launch and insertion into a geo­
synchronous orbit with radius 6.61 times the radius of
earth. (See Problem 2.64.)
Figure P2.126
2.124 For Problem 2.49, use
to find the speeds of
the blocks when the 5-lb block has droppped 2 ft.
Figure P2.127
2.125 Block
in Figure P2.125 is moving downward at
5 ft / sec at a certain time when the spring is compressed
6 in. The coefficient of friction between block and the
plane is 0.2, the pulley is light, and the weights of
are 161 and 193 lb, respectively.
a. Find the distance that
falls from its initial po­
sition before coming to zero speed.
b. Determine whether or not body
move back upward.
Figure P2.125
will start to
2.128 Suppose the ends of a spring are attached to "parti­
cles" of mass m and m . Show that the sum of the works
of the spring forces on the particles is given by Equation
(2.17).
x
•
2
The blocks in Figure P2.129 are released from rest.
Determine where they are when they stop permanently.
What is the spring force then? Hint: Write the workenergy equation for each block, add the two equations,
and use the result of Problem 2.125. Also, think about the
motion of the mass center.
Figure P2.129
Page 101
• 2.130 Show that if the surface in Example 2.13 is the
parabola shown in Figure P2.130, the particle will never
leave the surface. Hint: Show that
and use this in our equation:
Figure P2.130
Earth
• 2.131 Find the least velocity with which a particle could
be projected from the moon and reach the earth. (See
Figure P2.131.) For this problem assume that the centers
together
withand earth are both fixed in an inertial frame.
of
the moon
Moon
Figure P2.131
2.5
Momentum Form of Euler's First Law
T h e momentum o f a particle is defined to b e t h e product o f its m a s s a n d
its velocity. F o r a s y s t e m o f particles (Figure 2 . 8 a ) t h e m o m e n t u m is
defined to b e t h e s u m o f t h e m o m e n t a o f t h e particles in t h e s y s t e m . T h u s ,
if w e d e n o t e t h e m o m e n t u m o f a s y s t e m (or b o d y ) b y L,* t h e n
or, for a b o d y o f c o n t i n u o u s l y distributed m a s s (Figure 2 . 8 b ) ,
(Fixed in inertial frame)
Figure 2.8a
t h
If R is a position v e c t o r for the 1 particle, t h e n
i
and Equation (2.24) becomes
or
(Fixed in
inertial frame)
Figure 2.8b
* Momentum is sometimes called linear momentum.
Page 102
Recalling [see Equation ( 2 . 7 ) a n d Figure 2 . 3 a ] that
then
or
L = mv
c
(2.26)
w h e r e v is the velocity o f the m a s s center o f the s y s t e m or body.
T h e c o n n e c t i o n b e t w e e n external forces a n d m o m e n t u m n o w can b e
m a d e easily b y differentiating Equation ( 2 . 2 6 ) to obtain
c
But
so that
w h i c h is the m o m e n t u m form o f Euler's first law.
Question 2.7
Should we expect Equation (2.27) to be valid for a sys­
tem for which the mass is changing with time, such as a rocket with its
varying-mass contents?
Impulse and M o m e n t u m ; Conservation of M o m e n t u m
A straightforward integration o f the first l a w o f m o t i o n (Equation 2 . 2 7 )
yields
Answer 2.7 No, at several places in the development we have needed to require that
the mass be constant.
* The impulse is sometimes called the linear impulse.
Page 103
a specific time interval. If, during s o m e time interval, t h e s u m o f the
external forces vanishes, t h e n
a n d h e n c e the m o m e n t u m is a c o n ­
stant, or is conserved, during that interval.
S i n c e Equation ( 2 . 2 8 ) is a vector equation (unlike the scalar w o r k a n d
kinetic e n e r g y equation), w e m a y u s e a n y or all o f its c o m p o n e n t e q u a ­
tions. For e x a m p l e :
and similarly for y a n d z. W e n o t e that w e m a y h a v e a planar situation, for
example, in w h i c h
over an interval. I f this is t h e case,
m o m e n t u m is c o n s e r v e d in the x direction but not in the y direction.*
Impact
S o m e t i m e s it is possible, b y conservation o f m o m e n t u m , to obtain limited
quantitative information a b o u t the m o t i o n s o f colliding bodies. A s a rule
this can b e d o n e w h e n the b o d i e s interact for a relatively brief interval
— before a n d after w h i c h it is r e a s o n a b l e to treat their m o t i o n s as rigid.
W h i l e the analysis is b e s t discussed w i t h examples, w e m a k e the obser­
vation h e r e that generally it m a k e s little sense to treat the bodies as rigid
during the collision. I f w e wish to describe the m o t i o n that e n s u e s w h e n a
bullet is fired into a w o o d e n b l o c k , for e x a m p l e , t h e b l o c k clearly cannot
be regarded as rigid during the penetration process. O n the other h a n d , it
m a y b e quite plausible to a s s u m e that rigid m o t i o n o f the b l o c k a n d
e m b e d d e d bullet occurs subsequent to p e r m a n e n t reorientation o f m a t e ­
rial.
A k e y feature in the analysis o f collision (or impact) p r o b l e m s is t h e
fact that the m o m e n t u m o f a b o d y m a d e u p o f t w o parts is the s u m o f the
m o m e n t a o f the individual parts. T h i s feature follows directly from the
definition o f a b o d y ' s m o m e n t u m as the integral
w h e r e the subscripts (1 a n d 2 ) identify the t w o constituent parts o f the
body.
EXAMPLE 2 . 1 4
A wooden block of mass m is at rest on a smooth horizontal surface when it is
struck by a bullet of mass m traveling at a speed v as shown in Figure E 2 . 1 4 a .
After the bullet becomes embedded in the block, the block slides to the right at
speed V. Find the relationship between v and V.
x
2
Smooth
Figure E2.14a
* Ballistics problems are of this type if air resistance is neglected.
Page 104
Solution
Let t be the time at which the bullet first contacts the block and let f be the time
after which the bullet/block composite behaves as a rigid body in translation.
For t < t < t , a complex process of deformation and redistribution of mass is
ocairring within the block. If we isolate the block/bullet system during this
interval (see Figure E2.14b), Equation (2.28) yields
1
2
x
2
Figure E2.14b
But
and
since v is the speed of the mass center of the bullet and the block has no momen­
tum at t . Therefore, equating the coefficients, we get
1
or
We note that, in the absence of an external force with a horizontal component,
the horizontal component of momentum is conserved.
While we cannot calculate the reaction N during the collision, we can calcu­
late its impulse:
or
Similarly, the impulse of the force F exerted on the bullet by the block can be
calculated if we apply Equation (2.28) to the bullet:
The reader should note that with a high-speed collision occurring in a short
period of time, the impulses can be accurately estimated by neglecting the
impulses of the weights of the bodies. In this example we would have
Other examples of impact problems are treated in Chapter 5.
Page 105
EXAMPLE 2 . 1 5
A block is at rest on a smooth horizontal surface before being struck by an
identical block sliding at speed v. (See Figure E2.15a.) Find the velocities of the
two blocks after the collision assuming (1) that they stick together or (2) that the
system experiences no loss in kinetic energy.
Figure E2.15a
Solution
Let v and v be the speeds of the mass centers of the left and right blocks at the
end of the collision; that is,
is the velocity of the left block. The free-body
diagram of the system of two blocks during the collision (see Figure E2.15b)
shows that there is no external force with a horizontal component. Thus the
horizontal component (the only component not zero here) of momentum is
conserved and
L
Figure E2.15b
R
or
If the blocks remain attached after the collision is completed and they are
behaving as rigid bodies, we have
so that
Question 2.8
What would be the common velocity if the right block
had 100 times as much mass as the left block? If it had 10,000 times as
much mass?
If, however, the blocks do not stick together, the conservation of momentum
statement alone is not adequate to deterrnine their subsequent velocities. What
we need is some measure of their tendency to bounce off each other — or, to put
it another way, a measure of how much energy is expended in permanent defor­
mations or vibrations (or both) of the blocks. The parameter used to describe
these effects (the coefficient of restitution) is discussed in the text that follows this
example. At this point we simply note that when the blocks stick together the
kinetic energy of the system is less after the collision than before. That loss is
which is to say that one-half of the mechanical energy was dissipated in the
collision in this case.
The other extreme case is that in which no mechanical energy is expended
during the collision process. In this case
Answer 2.8 v / 1 0 1 ; v / 1 0 , 0 0 1 .
Page 106
But since v = v — v from Equation (2), we have
R
L
or
Therefore either v = 0 and u = v, or v = v and v = 0. The latter case must be
rejected as physically meaningless since it would require the left block to pass
through the stationary right block.
An extension of this result to the case of three blocks is shown in
Figures E2.15c and E2.15d:
L
Rest
R
Rest
L
R
Rest
Rest
After
Before
Figure E2.15d
Figure E2.15c
If we let the spacing between the blocks initially at rest approach zero and add
more of them, then we have the mechanism for a popular adult toy (see
Figures E2.15e,f):
After
Before
Figure E2.15e
Figure E2.15f
Coefficient of Restitution
In t h e p r e c e d i n g e x a m p l e w e n o t e d t h e n e e d for s o m e m e a s u r e o f the
capacity o f colliding b o d i e s t o r e b o u n d off e a c h other. T h e introduction
of a p a r a m e t e r called t h e coefficient o f restitution w h i c h provides this
information is m o s t easily a c c o m p l i s h e d t h r o u g h a s i m p l e e x a m p l e .
S u p p o s e that, as depicted in Figure 2 . 9 , t w o disks are sliding a l o n g a
s m o o t h floor. T h e p a t h s are t h e s a m e straight line a n d disk
to o v e r t a k e a n d c o n t a c t disk
is just a b o u t
at time t . T h e centers o f m a s s o f t h e disks
1
will a p p r o a c h e a c h o t h e r until, at time t , t h e y h a v e t h e c o m m o n velocity
2
v . T h e n t h e y will r e c e d e from e a c h o t h e r until at time t c o n t a c t is
c
3
b r o k e n . T h e e q u a l a n d opposite forces o f interaction F(t) are s h o w n on
t h e disks in Figure 2 . 9 . A p p l y i n g t h e i m p u l s e - m o m e n t u m principle dur-
Page 107
ing the intervals o f approach and t h e n separation o f the centers o f m a s s
of the disks,
and
Defining the coefficient of restitution, e, b y
Figure 2 . 9
w e obtain, after using the i m p u l s e - m o m e n t u m equations above,
Eliminating v , there results
c
w h i c h is s e e n to b e t h e quotient o f the "relative velocity o f separation"
a n d the "relative velocity o f a p p r o a c h . " T h e coefficient o f restitution is
inherently nonnegative, a n d the case e = 0 yields v = v , w h i c h m e a n s
that t h e disks stick together. In E x a m p l e 2 . 1 5 , for the case o f n o energy
dissipation, w e h a d
Af
Bf
Exercise P r o b l e m 2 . 1 4 7 provides an outline o f p r o o f that, u n d e r the
conditions o f our discussion here,
T h e impact just described is called central, b e c a u s e the line o f action
of the equal a n d opposite forces o f interaction is the line joining the m a s s
centers o f the bodies. It is also called direct b e c a u s e the preimpact veloci­
ties are parallel to that line of action. Generalization to the case o f indi­
rect, but still central, impact is easily accomplished, assuming the disks
are s m o o t h a n d t h a t t h e time o f contact is so small that there are n o
significant c h a n g e s in their positions during the collision. T h e n , the v e ­
locity c o m p o n e n t s perpendicular to t h e line o f action o f t h e impulsive
force (called the line o f impact) are u n c h a n g e d b y the collision. T h e
Page 108
e q u a t i o n s a b o v e n o w refer t o v e l o c i t y c o m p o n e n t s a l o n g t h e l i n e o f i m ­
pact. T h u s , t h e coefficient o f restitution i s t h e r a t i o o f r e l a t i v e v e l o c i t y
c o m p o n e n t s a l o n g t h e line o f i m p a c t .
Experiments* indicate that the coefficient of restitution depends
upon just about everything involved in an impact: materials, geometry,
and initial velocities. Therefore, numerical values must be used with care.
Nonetheless, the fact that the coefficient must have a value between zero
and unity is valuable information in bounding the behavior of colliding
bodies. The use of the coefficient of restitution for other than central
impact is discussed in Chapter 5.
EXAMPLE 2 . 1 6
Two identical hockey pucks collide, coming into contact in the positions shown.
Their velocities before the collision are also shown in Figure E2.16a.
Line of impact
Figure E2.16a
If the coefficient of restitution is 0.8, find the velocities of the pucks after the
collision. Then find the impulse of the force of interaction.
Solution
Neglecting friction and assuming insignificant deformation, the forces of interac­
tion will act along the line of impact shown in Figure E2.16a. It is convenient to
choose and parallel and perpendicular to this line as shown. Let m be the mass
of each puck and let v and
be the final and initial velocities of puck. and
similarly for puck
Af
Thus
where v is the unknown component along the line of impact. Also
x
and
* See W. Goldsmith, Impact (London: Edward Arnold Publishers, Ltd., 1960).
Page 109
Since there are no external forces on the system of two pucks, momentum is
conserved:
The component equation for the directions perpendicular to the line of impact is
automatically satisfied, and for the -direction
or
By the definition of the coefficient of restitution
or
Solving (1) and (2),
so that
and
Figure E2.16b
Thus the paths of the pucks are as shown in Figure E2.16b.
To compute the impulse of the force of interaction we apply the impulsemomentum principle to puck
noting that a hockey puck weighs about
6 ounces. Thus with m = ( 6 / 1 6 ) / 3 2 . 2 = 0.0116 slug,
EXAMPLE 2 . 1 7
In Example 2.9 we found the velocities of the identical Cars 1 and 2, just after
they collided in the given position, to be 40.1 and 25.4 ft / s e c , respectively, as
shown in Figure E2.17. Now, using the principle of impulse and momentum,
find the velocities of the cars prior to impact. Remember that Cars 1 and 2 were
heading west and south, respectively. Assume the collision is instantaneous.
Solution
If the impact occurs over a vanishingly small time
then the impulse from the
road (due to the friction force on the tires) during the impact is negligible, so that
the linear momentum of the system of two cars may be assumed to be conserved
Page 110
North
Figure E2.17
during
In expressing this conservation, we shall use "i" for initial (before
impact) and "f " for final (after impact):
or
and
Using the results of Example 2.9 and the angles in the figure above, we obtain the
following post-collision velocity components:
Using Equations (1,2), we find
We remark that the energy lost during the collision may now be calculated:
Note that the work done by the road friction was (see Example 2.9) equal to
0.5(32.2)(50 + 20)m = 1130m, and that this energy change plus the 260m lost in
the collision (to deformation, sound, vibration, etc.) gives the total original kinetic
energy, 1390 m.
Page 111
PROBLEMS
•
Section 2.5
2.132 Figure P2.132 presents data pertaining to a system
of two particles. At the instant shown find the:
a. Position of the mass center
b. Kinetic energy of the system
c. Linear momentum of the system
d. Velocity of the mass center
e. Acceleration of the mass center
2.133 The astronaut in Figure P2.133 is finding it difficult
to stop his forward momentum while jogging on the
moon. Using a friction coefficient of
and a gravi­
tational acceleration one-sixth that of earth's, illustrate
the difficulty of stopping a forward momentum of mv
= (5 slugs)(12 f t / s e c ) . Specifically, use the principle of
impulse and momentum to find the time it takes to stop
on earth versus on the moon.
2.134 A horizontal force F(f) is applied for 0.2 sec to a cue
ball (weighing 0.55 lb) by a cue stick; the form of the force
is as shown in Figure P2.134. If the velocity of the center
of the ball is 8 ft / sec after contact with the stick is broken,
find the peak magnitude F of the force. Neglect friction.
Force F is measured in pounds.
0
Fiqure P2.132
2.135 The 50-lb box shown in Figure P2.135 is at rest
before the force F(f) = 5 + 2f pounds is applied at t = 0.
Assume the box to be wide enough not to tip over and
suppose the coefficient of friction between box and floor
to be 0.2. Find the velocity of (the mass center of) the box
at f = 10 sec.
2.136 Repeat the preceding problem for the case in which
the force F(f) has a vertical component as shown in
Figure P2.136.
2.137 A force P applied to
at t = 0 varies with time
according to
lb, where t is in seconds.
(See Figure P2.137.) How long will it take for to begin
sliding? What will be its velocity at t = 30 sec?
Figure P2.133
Figure P2.135
Figure P2.136
Figure P2.134
Figure P2.137
Page 112
2.138 An unattached 2.2-lb roofing shingle slides down­
ward and strikes a gutter. (See Figure P2.138.) The angle
at which the shingle would be just on the verge of slipping
is 20°. Determine the impulse imparted to the shingle by
the gutter if there is no rebound. If the interval of impact is
0.1 sec, find the average force imparted to the gutter by
the shingle.
2.139 Two railroad cars are coupled by a collision occur­
ring just after the instant shown in Figure P2.139. Ne­
glecting the impulse caused by friction from the tracks,
determine the final velocity of the two cars as they move
together.
2.140 In the preceding problem, find the average impul­
sive force between the cars if the coupling requires 0.6 sec
of contact.
2.141 In a rail yard a freight car moving at speed v strikes
two identical cars at rest. (See Figure P2.141.) Neglecting
any resistance to rolling, find the common velocity of the
three-car system after the coupling has been completed
and any associated vibrations have died out.
I ft/sec
At rest
20 tons
Figure P2.141
2.142 A man of mass m and a boat of mass M are at rest as
shown in Figure P2.142. If the man walks to the front of
the boat, show that his distance from the pier is then
is the ratio of the masses
of man and boat. Explain the answer in the limiting cases
in which m
Neglect the resistance of
the water to the boat's motion.
'2.143 Two men each of mass m stand on the ends of a
flatcar of mass M. The car is free to move on frictionless
level tracks. All is at rest initially. One man runs to the
right end of the car and jumps off horizontally, parallel to
the tracks with a velocity U relative to the car. Then the
other man runs to the left end of the car and jumps off
horizontally, parallel to the tracks also with a velocity U
relative to the car. Find the final velocity of the car and
indicate clearly the direction of its motion.
* 2.144 In Figure P2.144 the man of mass m stands at end A
of a 20-ft plank of mass 3m that is held at rest on the
smooth inclined plane by the cord. The man cuts the cord
and runs down to end B of the plank. When he gets there,
end B is in the same position on the plane as it was origi­
nally. Find the time it takes the man to run down from A
toB.
Figure P2.138
Figure P2.139
Figure P2.142
30 tons
Page 113
Horizontal
Line of centers
Smooth
Goal line
Figure P2.149
Figure P2.144
2.145 A ball is dropped from a height Hand bounces. (See
Figure P2.145.) If the coefficient of restitution is e, find
the height to which the ball rises after the second bounce.
2.146 Two identical elastic balls
and
move toward
each other. Find the approach velocity ratio
that
will result in coming to rest following the collision. The
coefficient of restitution is e. (See Figure P2.146.)
2.147 Use the two equations
and
( = coefficient of restitution)
to prove that the loss in kinetic energy as the bodies
collide (Figure P2.147) is
and
2.148 Use the result of the preceding problem to show
that for a head-on collision at equal speeds v and equal
masses m,
so that if e = 0 then all of the initial T is lost and if e = 1
then none of T is lost. Is this true for differing speeds and
masses?
2.149 In soccer, a goal is scored only when the entire
ball is over the entirety of the 4-in.-wide goal line. (See
Figure P2.149.) Neglecting friction between ball and
post, determine the maximum coefficient of restitution
for which a goal will be scored before the ball hits the
ground. The velocity of the ball's center C makes an angle
with the horizontal of 15°. Neglect the deviation caused
by gravity on the trajectory between post and ground.
2.1 SO Repeat Example 2.16 for the line of impact shown
in Figure P2.150.
Deduce from this result that
Figure P2.146
Figure P2.145
Figure P2.147
Figure P2.150
Page 1 1 4
2.151 Let disk
in Problem 2.150 weigh 8 oz (and
weigh 6 oz as before), and then repeat the problem.
2.152 Repeat Example 2.16 for the case where the line of
impact is parallel to the before-collision velocity of
2.155 Using the angle a that will land the cannonball of
Problem 2.12 in the cart, find the maximum deflection of
the spring. (See Figure P2.155.)
2.153 Let disk
in Problem 2.152 weigh 9 oz (and
weigh 6 oz as before), and then repeat the problem.
2.156 Find the total time after firing for the cannonball
and box to either stop or strike the wall, whichever comes
first. (See Figure P2.156.)
2.154 A 10-kg block swings down as shown in Fig­
ure P2.154 and strikes an identical block. Assume that the
6 m rope breaks during impact and the blocks stick to­
gether after colliding. How long will it be before they
come to rest? How far will they have traveled?
2.157 A cannonball is fired as shown in Figure P2.157
with an initial speed of 1600 f t / s e c at 60°. Just after the
cannon fires, it begins to recoil, and strikes a plate at­
tached to a spring. Find the maximum spring deflection if
the plane is smooth and the spring modulus is 5 0 0 lb / ft.
Figure P2.157
Figure P2.154
Figure P2.155
Figure P2.156
Page 115
bullet is fired with a speed of 1800 f t / s e c
into a 10-lb block. (See Figure P2.158.) If the coefficient
of friction between block and plane is 0.3, find, neglecting
the impulse of friction during the collision:
a. The distance through which the block will slide
b. The percentage of the bullet's loss of initial ki­
netic energy caused by sliding friction, and the
percentage caused by the collision
c. How long it takes block and bullet to come to
rest after the impact.
2.159 Weight W falls from rest through a distance H; it
lands on another weight W , which was in equilibrium
atop a spring of modulus k. (See Figure P2.159.) If the
coefficient of restitution is zero, find the spring compres­
sion when the weights are at their lowest point.
t
2
2.160 Block
in Figure P2.160 weighs 16.1 lb and is
traveling to the right on the smooth plane at 50 f t / s e c .
Block
weighs 8.05 lb and is in equilibrium with the
spring barely preventing it from sliding down the rough
section of plane. Body
impacts
the coefficient of
restitution
Find the maximum spring deflection.
2.161 The 16-kg body
and the 32-kg body shown in
Figure P2.161 are connected by a light spring of modulus
12,000 N / m . The unstretched length of the spring is
0.15 m. The blocks are pulled apart on the smooth hori­
zontal plane until the distance between them is 0.3 m and
then released from rest. Determine the velocity of each
block when the distance between them has decreased to
0.22 m. Hint: As in Problem 2.129, form the sum of the
work-energy equations for the two blocks.
2,162 The cart and block in Figure P2.162 are initially at
rest, when the bullet slams into the block at speed
and sticks inside it. The combined body then starts sliding
on the cart.
Find:
a. the speed of the block just after impact;
Figure P2.158
b. the energy lost during impact;
• c. the time when the block leaves the cart.
Figure P2.161
Figure P2.159
No friction in
small light wheels
Figure P2.162
Figure P2.160
Page 116
2.163 A chain of length L and mass per unit length is
held vertically above the platform scale shown in Fig­
ure P2.163 and is released from rest with its lower end
just touching the platform. Assume that the links quickly
come to rest as they stack up on the platform and that they
do not interfere with the links still in free fall above the
platform. Draw a free-body diagram of the entire chain
and express the momentum as a function of the distance
through which the upper end has fallen. Then determine
the force read on the scale in terms of this distance.
at rest for t < 0 and is subjected to the uniform pres­
sure (over the end of area A) shown in Figure P2.165b.
If the disturbance has not reached the right end, that is if
t<L/c, then for t > t0 the particle velocities ui and
accelerations
which vary only with x and t, are as
shown in Figures P2.165c and d, where p is the density of
the bar.
The first part of this problem is to evaluate the
integral
The value that should be obtained is
and, since this
equals the external force on the bar, Equation (2.4) is thus
confirmed for this case. It is important to recognize that
only the interval from x = ct — ct to x = ct contributes to
the value of the integral; that is, only the particles in that
region are accelerating.
o
Figure P2.165i
Figure P2.163
• 2.164 A block of mass ml, which can move on a smooth
horizontal table, is attached to one end of a uniform chain
of mass m per unit length. Initially the block and the chain
are at rest, and the chain is completely coiled on the table.
A constant horizontal force mLf is then applied to the
block so that the chain begins to uncoil. Show that the
length x uncoiled after time t is given by
RgireP2.165b
until the chain is completely uncoiled. If the length of the
chain is very large compared with L, show that the veloc­
ity of the block is approximately equal to ( L / ) at the
moment when the chain is completely uncoiled.
Figure P2.165e
• 2.165 An important problem in the dynamics of deformable solids is that of describing the motion which ensues
when pressure is rapidly applied to the end of a slender,
uniform, elastic bar. A useful approximate theory yields
the one-dimensional wave equation as the governing
equation of motion. This theory predicts that a pressure
applied at one end of the bar creates a disturbance (wave)
that propagates into the bar at a constant speed c. To be
specific, suppose the bar shown in Figure P2.165a is
Fi§ureP2.165d
l / 2
Page 117
The second element of this problem is to evaluate the
momentum
In the second,
For the case at hand there is no reason to express a prefer­
ence for the order of differentiating with respect to time
and integrating over the body. If the pressure were sud­
denly applied at full strength (t = 0), however, there
would be a discontinuity in particle velocity (shock wave)
and a consequent undefined acceleration at the wavefront x = ct. Because of this undefined (or infinite) accel­
eration,
becomes meaningless and no longer pro­
vides L. There is no difficulty involved in evaluating L,
however, since the particle velocities are
and zero for x > ct. Thus
The result will be
0
The second term in the brackets, a constant, is the contri­
bution from integrating over the interval ct where the
particles are accelerating. The time dependence of L ap­
pears through the increasing number of particles having
velocity
As expected, we see that
In effect we have confirmed Euler's law,
in
two forms. In the first,
0
and
2.6
Euler's Second Law (The Moment Equation)
A s e c o n d relationship b e t w e e n t h e external forces o n a particle s y s t e m or
a b o d y is o b t a i n e d if, referring to Figure 2 . 1 0 , w e take the cross p r o d u c t o f
r, with b o t h sides o f E q u a t i o n ( 2 . 1 ) :
T h e first term o n the left is r e c o g n i z e d as t h e m o m e n t a b o u t point P o f t h e
external force F,. T h e cross product r, X fy is t h e m o m e n t w i t h respect to
(
P o f t h e force exerted o n t h e I
T H
particle b y t h e
particle. A s b e f o r e , w e
n o w s u m t h e N e q u a t i o n s typified b y E q u a t i o n ( 2 . 3 1 ) to obtain
(Fixed in
inertial frame)
Figure 2.1 G
T e r m s in t h e d o u b l e s u m o c c u r in pairs, s u c h as
But r X f
2
2I
= r X f
1
of f . M o r e o v e r , f
21
2 1
2 1
s i n c e r a n d r b o t h terminate o n t h e line o f action
= —f
2
1 2
1
so that
a n d similarly for o t h e r s u c h pairs. T h a t is, t h e m o m e n t s o f t h e internal
forces o f interaction s u m to zero. T h u s
Page 118
OR
w h i c h is the particle-system form o f Euler's s e c o n d l a w and states that
the s u m o f the m o m e n t s o f the external forces a b o u t a point equals the
sum o f the m o m e n t s o f the ma's a b o u t that point.
For a b o d y w h o s e m a s s is continuously distributed, the counterpart
to Equation ( 2 . 3 3 ) is
Question 2.9 In Equation 2.33 (or 2.34) must point P be fixed in the
inertial frame of reference?
Equations (2.4) a n d ( 2 . 3 4 ) play the s a m e roles in dynamics as do the
equations o f equilibrium in statics. A n d in fact w e obtain those equations,
from (2.4) a n d ( 2 . 3 4 ) , if w e set to zero the accelera­
tions o f all points o f a body.
M o m e n t of M o m e n t u m
Just as Euler's first l a w can b e expressed in terms o f the time derivative of
m o m e n t u m o f a b o d y , so Euler's s e c o n d l a w can b e expressed in terms of
the time derivative o f a quantity called m o m e n t o f m o m e n t u m , or a n g u ­
lar m o m e n t u m . * T h e m o m e n t o f m o m e n t u m w i t h respect to a point
P is designated H a n d is defined to b e the sum o f the m o m e n t s (with
respect to P ) o f the m o m e n t a o f the individual particles m a k i n g u p the
body. Referring to Figure 2 . 1 1 , w h e r e v,- is the velocity in reference frame
of the 1 particle, w e h a v e
P
Reference frame'7
Figure 2.11
t h
Before proceeding to the d e v e l o p m e n t o f several o f the forms of
Euler's s e c o n d law, w e shall develop a very useful relationship b e t w e e n
m o m e n t s o f m o m e n t u m . Noting from the definition, Equation ( 2 . 3 5 ) ,
and from Figure 2 . 1 1 that
and that
t h e n for a n y point P,
Answer 2.9
No. Nowhere in the development did we need to fix P.
* The term angular momentum stems from the fact that the moment of momentum of a
rigid body is related to the angular velocity of the body.
Pat4119
But from S e c t i o n 2 . 5 ,
is t h e m o m e n t u m L, also expressed as
L =
mv
c
so that
T h u s t h e m o m e n t o f t h e m o m e n t u m o f a b o d y a b o u t any point P (not
necessarily fixed in t h e reference frame) is t h e s u m o f t h e m o m e n t o f
m o m e n t u m about its m a s s center, C, a n d t h e m o m e n t o f its (linear)
m o m e n t u m L about P, w h e r e L is given a " l i n e o f a c t i o n " t h r o u g h C.
N o w w e can return to t h e definition o f m o m e n t o f m o m e n t u m ,
Equation ( 2 . 3 5 ) , a n d apply it for t h e case o f a point, O, fixed in t h e frame
o f reference. T h u s , using definition ( 2 . 3 5 ) for t h e third time,
N o w , differentiating w i t h respect to time in
But b e c a u s e O is fixed in
so that
Therefore
Momentum Forms of Euler's Second Law
N o w t h e fundamental form o f Euler's s e c o n d law, Equation ( 2 . 3 3 ) , tells
us that if the frame
in w h i c h the a,< are calculated is an inertial frame,
then
so that, from the last two equations, w e see that
A n o t h e r similar form o f Euler's s e c o n d l a w can b e d e d u c e d i f w e first
use E q u a t i o n ( 2 . 3 6 ) in t h e case o f a fixed point O:
Differentiating with respect to time in t h e reference frame
Page 120
w h e r e w e h a v e u s e d t h e fact that
But w e again recall that
L = mv
(so that v
c
c
X L = 0)
and therefore
N o w w e k n o w from our study o f equipollent force systems in statics that
the external forces o n the b o d y must produce m o m e n t s a b o u t O a n d C
that are related b y
T h i s l a w o f resultants h a s nothing to do with w h e t h e r or n o t the b o d y is in
equilibrium. A n d since is an inertial frame, t h e n w e also k n o w
a n d thus w e m a y subtract the t w o Equations ( 2 . 4 1 ) a n d ( 2 . 4 2 ) to obtain
N e i t h e r o f the a b o v e equations
is a n y m o r e
basic or special t h a n the other, as e a c h o n e c a n b e derived from the other.
T h e y are therefore equivalent forms. H o w e v e r , t h e equation does not
h o l d for a n y arbitrary point P, i.e., in general
Conservation of M o m e n t of M o m e n t u m
W e n e x t n o t e that — as w a s the case with linear m o m e n t u m — there are
situations in w h i c h a m o m e n t o f m o m e n t u m is conserved. In particular,
if for an interval o f time
t h e n during that interval
and
thus H is constant. For e x a m p l e , let the b o d y o f interest b e a single
spherical planet in its m o t i o n a r o u n d its star. T h e gravitational force
exerted o n the planet b y the star a l w a y s passes through the star's mass
center 0 , so
a n d thus H o f the planet is a constant. This result is
G
D
s h o w n in Section 8.4 to lead to the elliptical orbit o f the earth a r o u n d the
sun.
For an arbitrary point P, there is a form o f Euler's s e c o n d l a w that is o f
particular value in analyzing the m o t i o n s o f rigid bodies, although it
remains valid for non-rigid bodies as well. T o derive it w e again use our
k n o w l e d g e a b o u t force s y s t e m s to write
Page 121
T h u s , using Euler's l a w s (Equations ( 2 . 2 7 a n d ( 2 . 4 3 ) )
or
Question 2.10
Must point P be fixed in the inertial frame of reference,
fore Equation
w h i c(a)
h w
shall use(2.36)
later.to be true? (b) for Equation (2.45) to be true?
Finally, w e r e m i n d t h e reader that all o f the relationships o f this c h a p t e r
pertain o n l y to a specific collection o f material — that is, a s y s t e m (or
b o d y ) o f c o n s t a n t m a s s . H o w e v e r , t h e m o m e n t u m forms o f Euler's l a w s
provide t h e natural starting point for developing relationships appropri­
ate to " v a r i a b l e m a s s " systems s u c h as rockets. I f desired, t h e reader n o w
h a s t h e proper b a c k g r o u n d to study that special topic w h i c h is found in
Section 8.3.
Answer 2.10
(a) No; (b) No. Point P was unrestricted in both derivations.
EXAMPLE 2 . 1 8
Two gymnasts of equal weight (see Figure E2.18a) are hanging in equilibrium at
the ends of a rope passing over a relatively light pulley for which the bearing
friction can be neglected. Then the gymnast on the right begins to climb the rope,
while the gymnast on the left simply holds on. When the right gymnast has raised
himself through height h (relative to the floor), what has been the change in
position of the left gymnast?
Figure E2.18a
Solution
Constructing a free-body diagram (Figure E2.18b) of the pulley-rope-gymnasts
system in which we neglect the weights of the pulley and the rope, we see that
Therefore, H
rest,
Q
is constant during the motion, and since everything starts from
Treating the gymnasts as particles and neglecting the moment of momentum
of the pulley,
or
Figure E2.18b
Page 1 2 2
But
H = 0
o
so that
and the left gymnast, "going along for the ride," rises at the same rate as the right
one. Thus when the right gymnast has pulled himself up height h, the left one has
been pulled up the same height h. Note that if the rope is inextensible, the right
gymnast therefore would have climbed 2h relative to the rope!
EXAMPLE 2 . 1 9
Suppose the "counterweight" gymnast on the left in the preceding example were
to weigh twice that of the climbing gymnast, as suggested by Figure E2.19. What
then would be the relationship between their elevation changes?
Solution
From the free-body diagram
Since
grating,
the moment of momentum is not conserved this time. Inte­
Figure E2.19
or
so that
If we define y and y so that y„ = y = 0 at t = 0, then
R
L
L
Thus we see that it's possible for the lighter gymnast to raise the heavier gymnast
by climbing rapidly enough.
For an inextensible rope, the right gymnast climbs, as before, at a rate of
relative to the rope.
PROBLEMS
•
Section 2.6
2.166 The uniform rigid bar in Figure P2.166 weighs
60 lb and is pinned at A (and fastened by the cable DB)
to the frame
If the frame is given an acceleration
2
a = 32.2 ft / s e c as shown, determine the tension T in
the cable and the force exerted by the pin at A on the bar.
Page 123
Now note that (a) if the Q are a set of forces F then
Equation (2.42) results with R being
; and (b) if the Q
are a set of momenta m v of a group of particles, then
Equation (2.36) results with R being
(or L or mv ).
Thus we may conclude that both equations, (2.36) and
(2.42), are practical examples of the very same law of
resultants!
i
i
i
i
i
c
Figure P2.16G
2.171 The angular momentum about point Q is defined as
2.167 A force F causes the carriage to move with rectilin­
ear horizontal motion defined by a constant acceleration
of 20 f t / s e c (see Figure P2.167). A rigid, slender, homo­
geneous rod of weight 32.2 lb and length 6 ft is welded to
the carriage at B and projects vertically upward. Find, in
magnitude and direction, the bending moment that the
carriage exerts on the rod at B.
2
2.168 A uniform slender bar of density p, cross-sectional
area A, and length L undergoes small-amplitude, free
transverse vibrations according to
sin
where y is the displacement perpendicular to the
axis (x) of the bar. (See Figure P2.168.) Neglecting other
components of displacement (and hence acceleration),
calculate the maximum force generated at one of the sup­
ports during the motion.
Differentiate this expression in the inertial reference
frame
and show by the result and that of the preceding
problem that, in general, (i.e., not just at an isolated in­
stant of time):
only if (a) Q is fixed in
v is parallel to v .
Q
2.172 In Problem 2.132 find: (a) the angular momentum
of the system with respect to the origin; (b) the angular
momentum of the system with respect to the mass center.
2.173 A massless rope hanging over a massless, frictionless pulley supports two monkeys (one of mass M, the
other of mass 2M). The system is released at rest at t = 0,
as shown in Figure P2.173. During the following 2 sec,
monkey B travels down 15 ft of rope to obtain a massless
peanut at end P. Monkey A holds tightly to the rope dur­
ing these 2 sec. Find the displacement of A during the time
interval.
Figure P2.167
Figure P2.168
2.169 Show that in Equation (2.36), point C need not be
the mass center, i.e., if Q is another arbitrary point like P,
then show that H = H + r X L (where L is of course
still
).
P
Q
P Q
2.170 L e t S b e a s e t o f vectors Q , Q , . . . , Q . . . , Q o f
equal dimension. Define the resultant of S as
,
and place each Q at a point P . Define the moment of the
set of vectors about a point A by
1
i
2
i
or (b) Q is the mass center C; or (c)
c
N
i
and show that
Figure P2.173
Page 124
2.174 A starving monkey of mass m spies a bunch of
delicious bananas of the same mass. (See Figure P2.174.)
He climbs at a varying speed relative to the (light) rope.
Determine whether the monkey reaches the bananas be­
fore they sail over the pulley if the pulley's mass is negli­
gible
2.176 Define the angular momentum of a particle about a
fixed axis and state the conditions under which the angu­
lar momentum remains constant. A man (to be regarded
as a particle) stands on a swing. His distance from the
smooth horizontal axis of the swing is L when he crouches
and L — H when he stands. As the swing falls he
crouches; as it rises he stands — the changeover is as­
sumed instantaneous. If the swing falls through an angle
a and then rises through an angle
show that
The relative angular momentum of a body
a point P is defined to be
with respect to
(See Figure P2.177, where the velocity v of dm is the de­
rivative of r in inertial frame ) Note that what makes it
"relative" is that the velocity in the integral is the differ­
ence between v (of dm) and v . Now solve the following
problems.
P
Figure P2.174
Figure P2.177
Figure P2.175
2.177 Show that
= H
always!)
H = H
P
P r e l
+ mrpc X v .
P
(Thus H
c
C r e |
2.175 Two gymnasts A and B, each of weight W, hold onto
the left side of a rope that passes over a light pulley to a
counterweight C of weight 2W. (See Figure P2.175.) Ini­
tially the gymnast A is at depth d below B. He climbs the
rope to join gymnast B. Determine the displacement of
the counterweight C at the end of the climb.
COMPUTER PROBLEMS
•
2.178 Show that
not generally equal to
is
!)
2.179 Show that
2.180 Show that H = H
X cLp.
P
P r e (
if and only if v X v
P
c
=
TpC
Chapter 2
2.181 In the table on the next page are data of the mass
center velocity versus time for a 30-lb crate that was lifted
approximately straight up by two people. The "velocity
1" column represents a taller person than the "velocity 2 "
column. Use the computer to integrate numerically the
velocity from t = 0 to 4.4 sec, thereby obtaining and
comparing the heights to which the crate was lifted by the
two people.
Page 125
SUMMARY
•
Time
Velocity 1
Velocity 2
Time
Velocity 1
Velocity 2
(sec)
(in./sec)
(in./sec)
(sec)
(in./sec)
(in./sec)
00
02
00
0 0
2 3
2 5 . 0 (peak)
23.5
00
0 0
24
04
1.5
0.0
25.5
32 0
0.6
08
3 5
5 5
00
00
2.6
2.7
24.0
22.0
20.5
3 7 5 (peak)
1.0
8.5
2 0
190
16 5
35.0
25.0
1.2
1 4
12.0
165
40
6 5
1 6
1.8
190
21 0
20
2 2
24.5
22 5
2 8
3.0
3.2
14.5
21.5
34
13.0
170
10.0
3.6
140
17 0
21 0
3.8
12 0
10.0
15 0
13 0
8.5
9.5
7.5
6.5
85
8.0
40
4 2
4 4
Chapter 2
In this chapter w e h a v e set out the fundamental relationships b e t w e e n
forces o n a b o d y a n d its motion, a n d w e h a v e illustrated their use for the
solution o f a variety o f p r o b l e m s , m a n y of w h i c h are closely associated
with our everyday experience.
T h e starting point h e r e w a s N e w t o n ' s s e c o n d l a w for a particle,
where
is the s u m o f all the forces acting o n the particle, m is the m a s s
of the particle a n d a is its acceleration relative to an inertial frame o f
reference. Extending to a system o f particles, the 1 h a v i n g m a s s m, a n d
acceleration a ,
t h
i
where
is the s u m o f the external
particular value is
forces on the system. A n o t h e r form of
where
is the m a s s o f the s y s t e m o f particles, or the b o d y c o m ­
prising t h e m , a n d a is the acceleration o f the m a s s center C. In addition, it
is sometimes useful to d e c o m p o s e a b o d y into t w o (or m o r e ) parts with
masses m, a n d m a n d m a s s centers C and C , a n d t h e n w e m a y use:
c
2
1
2
T h e preceding equations, for a b o d y o f finite size, are forms o f w h a t is
often called Euler's first l a w a n d are counterparts in dynamics to the
equilibrium equation,
studied in statics. W e h a v e used t h e m to
solve a variety of problems such as finding accelerations and constraining
forces w h e n s o m e forces a n d p a t h s were prescribed a n d also integrating
to find the m o t i o n o f a particle (or m a s s center o f a b o d y ) w h e n external
forces w e r e prescribed. Central to the problem-solving process w a s the
free-body diagram, the i m p o r t a n c e o f w h i c h c a n n o t b e overstated.
Page 126
T h e Principle o f W o r k a n d Kinetic E n e r g y is very useful in solving
problems in w h i c h the speeds o f a particle at different locations in space
are o f interest. T h e Kinetic Energy, T, o f a particle is defined to b e
T h e principle states that the w o r k d o n e b y all the forces acting over an
interval o f time is equal to the c h a n g e in kinetic energy. Or, in s y m b o l s ,
W =
T -T
2
1
This is a derived result following from integrating a n d assigning the term
" w o r k o f a force o n a p a r t i c l e " to
T w o special forces arise frequently e n o u g h in p r o b l e m s to evaluate
the w o r k a n d express it in symbols:
a. C o n s t a n t force: W = F • (r — r ), or in words: (magnitude o f
force) * (magnitude o f displacement) * (cosine o f angle b e ­
t w e e n force a n d displacement). For weight (force exerted b y
gravity n e a r the earth's surface) this m e a n s (weight) * (decrease
in altitude o f m a s s center).
2
1
b. Force exerted b y a linear spring:
where
is spring stretch a n d k is the spring modulus.
A force w h o s e w o r k does n o t d e p e n d u p o n the p a t h o f the point o f
application is called conservative a n d a potential energy, , is associated
with it so that the w o r k d o n e is the negative o f the c h a n g e in that
potential energy,
C o m b i n i n g this with
assuming all forces to b e conservative,
or
w h i c h m e a n s that in this case, kinetic plus potential e n e r g y is conserved.
A potential e n e r g y for a linear spring is
a n d for weight (with z b e i n g elevation)
For a s y s t e m o f particles,
Page 127
a n d the Principle o f W o r k a n d Kinetic E n e r g y applies so l o n g as o n e
considers the w o r k o f all internal forces as well as the w o r k o f external
forces; that is,
This is o f practical value o n l y in special situations w h e r e it's possible to
readily evaluate the w o r k o f internal forces. O n e e x a m p l e is a rigid b o d y
or system o f rigidly c o n n e c t e d particles, for t h e n W
= 0. A n o t h e r
case is that o f a pair o f particles j o i n e d b y a linear spring, for w h i c h the net
w o r k o f the equal a n d opposite internal forces is
internal
T h e c o n c e p t o f m o m e n t u m is particularly useful in p r o b l e m s o f im­
pact or collision in w h i c h very intense forces o f interaction m a y act for a
very brief interval. M o m e n t u m is defined to b e
L = mv
(particle)
(system o f particles)
from w h i c h , for a b o d y in general,
L = mv
c
Euler's first l a w c a n b e written
w h i c h w h e n integrated yields the i m p u l s e - m o m e n t u m principle
S o if external forces do n o t act during the interval,
L(t ) = L(t ),
2
1
a n d so m o m e n t u m is conserved. This is quite often (approximately) the
case in p r o b l e m s o f collision.
Finally in this chapter w e h a v e d e v e l o p e d the counterpart in dy­
n a m i c s to the s e c o n d equilibrium equation,
, in statics. In dy­
n a m i c s this is, for a s y s t e m o f particles,
when
refers, as it did in statics, to the m o m e n t s o f external forces.
This is often called Euler's s e c o n d l a w . T h e r e are several different forms
in w h i c h this l a w c a n b e expressed, a m o n g t h e m expressions involving
the m o m e n t o f m o m e n t u m , defined as
for a s y s t e m o f particles. A useful relationship is
Hp — H + r
c
p c
X L,
Page 128
but the k e y expressions are the forms that Euler's s e c o n d l a w c a n take,
1M
C
= H
c
and
w h e r e "O" is a point fixed in the inertial frame o f reference.
REVIEW QUESTIONS
•
Chapter 2
True or False?
1. A t a given time, t h e m a s s c e n t e r o f a d e f o r m a b l e b o d y c a n b e s h o w n
to b e a unique point.
2. T h e m o m e n t u m o f a n y b o d y (or s y s t e m o f bodies) in a frame c a n b e
s h o w n to b e equal to the total m a s s times the velocity o f the mass
center in
e v e n i f is n o t an inertial frame.
3. Euler's first l a w
applies to deformable bodies w h e t h e r
solid, liquid, or gaseous, as well as to rigid b o d i e s a n d particles.
4. Neither the laws o f m o t i o n n o r the inertial frame is o f a n y value
without the other.
5. T h e m a s s center o f a b o d y
of
h a s to b e a physical, or material, point
6. T h e w o r k d o n e b y a linear spring d e p e n d s o n the p a t h s traversed by
its endpoints b e t w e e n the initial a n d final positions.
7. T h e w o r k d o n e b y the friction force u p o n a b l o c k sliding o n a fixed
p l a n e d e p e n d s o n the p a t h taken b y the block.
8. T h e w o r k d o n e b y gravity o n a b o d y d e p e n d s o n the lateral as well
as the vertical displacement o f the m a s s center o f
9. S i n c e n o external w o r k w a s d o n e o n the two bodies o f E x a m p l e 2 . 1 4
during the impact, their total kinetic energy is the s a m e after the
collision as it w a s before.
10. F o r all b o d i e s o f c o n s t a n t density, t h e centroid o f v o l u m e a n d the
center o f m a s s coincide.
11. In studying the m o t i o n o f the earth a r o u n d the sun, it is acceptable to
treat the earth as a particle; in studying the daily rotation o f the earth
o n its axis, h o w e v e r , it w o u l d n o t m a k e s e n s e to consider the earth as
a particle.
12. T h e external forces acting on a b o d y
w h i c h together form the
resultant F , must e a c h h a v e a line o f action passing through the m a s s
center o f in order for Euler's first l a w to apply.
r
13. Euler's s e c o n d l a w c a n take the form
m o t i o n o f point P in the inertial frame.
Answan:1.T
regardless o f the
2. T 3. T 4. T 5. F 6. F 7. T B. F 9. F 10. T 11. T
12. F 13. F
3
KINEMATICS OF PLANE
MOTION OF A RIGID BODY
3.1
Introduction
3.2
Velocity and Angular Velocity Relationship for T w o Points of
the S a m e Rigid Body
Development of the Velocity and Angular Velocity Relationship
Important Things to Remember About Equation (3.8)
3.3
Translation
3.4
Instantaneous Center of Zero Velocity
Proof of the Existence of the Instantaneous Center
The Special Case in Which the Normals Do Not Intersect
The Special Case in Which the Normals are Coincident
3.5
Acceleration and Angular Acceleration Relationship for T w o
Points of the S a m e Rigid Body
Development of the Acceleration and Angular Acceleration
Relationship
3.6
Rolling
Rolling of a Wheel on a Fixed Straight Line
Rolling of a Wheel
on a Fixed Plane Curve
Gears
3.7
3.8
Relationship Between the Velocities of a Point w i t h Respect
to T w o Different Frames of Reference
Relationship Between the Derivatives of a Vector in T w o Frames
Velocity Relationship in T w o Frames
Relationship B e t w e e n the Accelerations of a Point w i t h
Respect to T w o Different Frames of Reference
SUMMARY
REVIEW
QUESTIONS
Page 1 2 9
Page 1 3 0
3.1
Introduction
In this chapter our goals are to develop the relationships b e t w e e n veloci­
ties, accelerations, angular velocity, a n d angular acceleration w h e n a
rigid b o d y m o v e s in p l a n e m o t i o n in a reference frame
Before doing
so, h o w e v e r , w e shall first explain precisely w h a t w e m e a n b y such terms
as rigid body, plane motion, rigid extension,
reference plane, a n d several
other c o n c e p t s w e shall b e needing in this chapter a n d those to follow.
A r i g i d b o d y is taken to b e a b o d y in w h i c h the distance b e t w e e n
each a n d every pair o f its points r e m a i n s the s a m e throughout the m o ­
tion.* T h e r e is, o f course, n o such thing as a truly rigid b o d y (since all
bodies d o some deforming); h o w e v e r , the deformations of m a n y bodies
are sufficiently small during their m o t i o n s to allow the bodies to be
treated as t h o u g h they were rigid with good results.
T h e significance o f the rigid-body m o d e l is that velocities o f different
points will b e found to differ b y s o m e t h i n g proportional to the rate at
w h i c h the b o d y turns, w h a t w e shall c o m e to call its angular velocity. A n d
accelerations will b e found to b e related through the angular velocity and
its rate o f c h a n g e w h i c h w e k n o w as the angular acceleration. T h u s a very
small a m o u n t o f information will characterize all the accelerations in the
body. T h e r e are a n u m b e r o f w a y s in w h i c h this is important, b u t fore­
most is the fact that the right-hand side o f our m o m e n t equation in
C h a p t e r 2 will in C h a p t e r 4 b e seen to take on a c o m p a c t form involving
the angular velocity a n d the angular acceleration.
P l a n e m o t i o n is treated in this b o o k as m o t i o n in the xy p l a n e (fixed
in ) or in planes parallel to it. Let a point P b e located originally at
coordinates (x , y ,z ).
To say that P h a s p l a n e m o t i o n simply m e a n s that
it stays in the p l a n e z = z throughout its motion. Extending this defini­
tion, w e s a y that rigid b o d y h a s p l a n e m o t i o n w h e n e v e r all its points
remain in the s a m e planes (parallel to xy) in w h i c h t h e y started.
p
p
p
p
Question 3.1
How few points of a rigid body must be in plane motion
to ensure that they all are?
A third c o n c e p t w e n e e d to u n d e r s t a n d in rigid-body kinematics is
that o f the b o d y e x t e n d e d , also called a r i g i d e x t e n s i o n o f the b o d y . This
idea, briefly m e n t i o n e d in C h a p t e r 1, says that w e s o m e t i m e s n e e d to
i m a g i n e points ( w h i c h are n o t physical or material points o f ) m o v i n g
with as t h o u g h they w e r e in fact a t t a c h e d to it. A n e x a m p l e w o u l d be
the points o n the axis o f a pipe that are in the space inside it b u t o f course
m o v e rigidly with it. W e shall imagine a "rigid e x t e n s i o n " o f the b o d y to
pick up such points w h e n e v e r it is useful to do so. N o t e that any point Q
* We have already encountered this concept in Chapter 1, where it was seen to be syn­
onymous with the concept of frame.
Answer 3.1 Three noncollinear points are needed.
Page 1 3 1
m a y b e considered a point o f any b o d y
extended, provided that Q
moves with
as if it w e r e rigidly a t t a c h e d to it.
With these three concepts in mind, w e are n o w prepared to define the
r e f e r e n c e p l a n e . First w e must realize that w e are faced with the p r o b l e m
of determining w h e r e all the points o f a b o d y are as functions o f time t.
T h e s e points' locations (x(t), y(t)) w o u l d take forever to find if w e h a d to
do so for e a c h o f the infinitely m a n y points o f
Fortunately, for a rigid
b o d y in p l a n e motion, if w e k n o w the locations o f all its points in a n y o n e
p l a n e o f ( w h i c h w e shall call the reference p l a n e ) , t h e n w e automati­
cally k n o w the locations o f all its other points in all other p l a n e s . T h e
reason for this is as follows. For e a c h point B o f that does n o t lie in the
reference plane, there is a " c o m p a n i o n p o i n t " o f in the reference plane
(suggested b y A in Figure 3 . 1 ) that h a s the s a m e (x, y) coordinates at the
b e g i n n i n g o f the motion. It t h e n follows that the (x, y) coordinates o f A
a n d B always m a t c h throughout the m o t i o n o f !
xy = reference plane
Axis of symmetry of
(Rigid body in plane motion
parallel to xy plane)
(Axes x, y, z arc
embedded in
frame of reference )
Figure 3.1
Question 3.2
Why is x = x and y = y as time passes?
A
B
A
B
Body in Figure 3.1 is a c o n e pulley, w h i c h turns about its axis o f s y m ­
metry. N o t e from its varying cross-sectional diameter that a b o d y n e e d
n o t h a v e constant cross section to b e in p l a n e motion.
H a v i n g laid the n e c e s s a r y groundwork, w e n o w let xy b e our refer­
ence p l a n e . F r o m Figure 3.1 w e see that
Answer 3.2 If ever
(or both), either the rigid body or the plane mo­
tion assumptions (or both) will have been violated.
Page 132
a n d w e m a y differentiate this equation to obtain
In these equations w e h a v e used t h e facts that x = x , y = y , a n d
z = constant. E q u a t i o n s ( 3 . 2 ) a n d ( 3 . 3 ) s h o w clearly that if w e c o m ­
pletely describe t h e velocities a n d accelerations in o n e reference plane,
w e t h e n k n o w t h e m for all t h e points o f t h e b o d y . T h i s allows us to focus
on o n e p l a n e o f t h e b o d y throughout this chapter a n d m o s t o f t h e next
two as well.
The reference p l a n e is t h u s a v e r y important c o n c e p t , for it allows us
to study t h e m o t i o n o f a n entire b o d y b y c o n c e r n i n g ourselves o n l y with
t h o s e o f its points that lie in this p l a n e . W e say w e " k n o w the m o t i o n " of
w h e n w e k n o w w h e r e all its points are at all times. W e h a v e already
reduced this task to k n o w i n g t h e locations o f the points in the reference
plane. ( T h e rest o f t h e b o d y " g o e s along for the r i d e . " ) But in fact if w e
k n o w the location o f j u s t two points (say P a n d P ) o f the reference plane,
t h e n w e k n o w t h e w h e r e a b o u t s o f all points o f this p l a n e a n d t h u s o f the
w h o l e body! T h i s is b e c a u s e e a c h point o f t h e reference p l a n e must
maintain t h e s a m e position relative to t h e points P a n d P . T h i s idea is
illustrated in Figure 3 . 2 . N o t e in t h e figure that if P a n d P are correctly
located with respect to t h e reference frame
all o t h e r points o f
are
necessarily in their correct positions.
B
A
B
A
B
1
2
1
1
2
2
Position of E
Reference frame
Figure 3.2
Question 3.3 Is knowledge of the locations of two points sufficient for
us to know the motion of a body in general (three-dimensional) motion?
I n s t e a d o f k n o w i n g t h e locations o f t w o points o f the b o d y — ( x y
of Pi a n d (x , y ) o f P — w e m a y alternatively locate t h e b o d y if w e k n o w
w h e r e just one point, P, is l o c a t e d plus t h e value o f t h e orientation angle 8
(about a n axis t h r o u g h P a n d parallel to z); see Figure 3 . 3 .
v ,
2
2
2
l
Answer 3 . 3 No; the body could rotate around the line joining the two points. In three
dimensions it takes three points, not all on the same line!
1
Page 1 3 3
Line fixed in
Position of
Reference frame
Figure 3.3
Question 3.4 Knowing the locations of two points requires four vari­
ables (x , y ,x ,
y ), whereas one point plus the angle takes but three
x ,y , ). Why do these numbers of variables differ?
x
1
1
2
2
1
T h e foregoing is i n t e n d e d to suggest quite correctly that the velocities of
different points in the reference p l a n e will b e linked together b e c a u s e o f
the rigidity o f the b o d y , a n d similarly for accelerations. In the n e x t section
w e will turn to the d e v e l o p m e n t o f t h e relationship b e t w e e n the veloci­
ties o f points s u c h as P a n d P in Figure 3 . 2 .
t
2
Answer 3.4 The 2 X 2 = 4 coordinates of P and P are not independent. The distance
between the points is a constant, so
1
2
can be used to find any one of x , y , x , y in terms of the other three.
1
PROBLEMS
•
1
2
2
Section 3.1
3.1
Which of the bodies shown in Figure P 3 . 1 ( a - f )
are in plane motion in frame _ ?
(a) A turkey being barbecued by
slowly turning on a rotisserie.
Figure P3.1(a-c)
(See next page for d-f)
(b) A cone rolling on a tabletop.
(c) A spinning coin if the base is
fixed.
Page 1 3 4
(d) A can rolling down an
inclined plane
(e) T h e bevel gear
which meshes
with another bevel gear
(f) T h e (shaded) crosspiece of a
universal joint.
Figure P3.1(d-f)
3.2
Give three examples of plane motion besides those
in the previous problem. Then give three examples of
motion that is not planar.
3.2
Velocity and Angular Velocity Relationship for T w o
Points of the Same Rigid Body
In this section w e derive a v e r y useful relationship b e t w e e n t h e
in
o f a n y t w o points in t h e reference p l a n e o f a rigid b o d y
m o t i o n a n d t h e angular velocity v e c t o r o f
in
Let P a n d
t h e s e t w o points o f
a n d let us e m b e d t h e a x e s (x, y, z) in
frame
as s h o w n in Figure 3 . 4 .
velocities
in p l a n e
Q denote
reference
Figure 3.4
W e are saying that e v e n t h o u g h
m a y m o v e with respect to the
reference frame
t h e xy p l a n e o f a l w a y s c o n t a i n s t h e points o f interest
P and Q of
A g o o d e x a m p l e is f o u n d in t h e classroom; let t h e b o d y
b e a b l a c k b o a r d eraser. Letting t h e b l a c k b o a r d itself b e the reference
frame
(so that x a n d y are fixed i n t h e p l a n e o f t h e b l a c k b o a r d ) , the
eraser u n d e r g o e s p l a n e m o t i o n w h e n e v e r the professor erases t h e b o a r d .
O u r points P a n d Q are a n y t w o points o f the erasing surface o f t h e eraser.
N o t e h o w e a c h point o f t h e eraser r e m a i n s t h e s a m e z distance from the
Page 135
blackboard (where z = 0 ) during the erasing. T h e eraser is n o longer in
p l a n e m o t i o n , h o w e v e r , o n c e it leaves the surface o f the b o a r d a n d its
points m o v e with z c o m p o n e n t s o f velocity.
N o t i c e from Figure 3 . 4 that is a unit vector a l w a y s directed from P
toward Q so that
where r
is the distance PQ (that is, the
magnitude o f the vector r ) . N o t e further that the orientation (angular
rotation) o f the b o d y is described b y the angle , m e a s u r e d b e t w e e n any
line fixed in the reference frame
(we shall h e r e use the x axis) and
a n y line fixed in the b o d y (for t h e m o m e n t , w e shall u s e t h e line s e g m e n t
from P to Q ) .
P Q
P Q
Development of the Velocity and Angular Velocity Relationship
W e are n o w ready to develop the velocity a n d angular velocity relation­
ship for rigid bodies. F r o m Figure 3.4 w e see that
so that w e h a v e , u p o n differentiation in frame
Recognizing the first t w o vectors as the definitions o f the velocities o f P
a n d Q (in
w h e r e O is fixed), w e m a y write
In obtaining Equation ( 3 . 5 ) , all derivatives were taken in , so that, for
example, there is n o n e e d to write
In order to write
as a vector w e can use, w e express
as a
magnitude times a unit vector. W i t h the help o f Figure 3.4 w e get
Differentiating this expression as in Section 1.6, w e h a v e
Therefore w e h a v e derived a useful expression for
Question 3.5
In the preceding development, why is
:
?
Substituting Equation ( 3 . 7 ) into ( 3 . 5 ) yields
* Throughout this book
Answer 3.5
Because
and
is rigid,
constitute a right-handed system so that
is the constant distance between points P and Q.
Page 1 3 6
w h i c h relates the velocities o f the points P a n d Q a n d introduces the
a n g u l a r v e l o c i t y o f in reference frame
T h e G r e e k letter o m e g a is
usually u s e d to denote this vector:
W h e n there is n o confusion a b o u t the b o d y a n d reference frame
involved, w e m a y drop the subscripts a n d write
as simply . Also,
s o m e prefer to write
as
rather t h a n
; w e shall use b o t h forms,
feeling that the latter is a nice reminder that in p l a n e motion, angular
velocity is proportional to the time rate o f c h a n g e o f an angle.*
The magnitude
J o f the angular velocity is called the
a n g u l a r s p e e d o f in frame
N o t e that
itself can b e negative.
N o t e further that neither the angular velocity v e c t o r n o r Equation ( 3 . 8 )
depends o n w h i c h body-fixed line s e g m e n t (such as P Q a b o v e ) is c h o s e n
to m e a s u r e . T h e p r o o f o f the preceding s t a t e m e n t is n o t difficult and
will b e given later as a n exercise.
N o t e that if our angle o f orientation were c h o s e n as s h o w n in Fig­
ure 3 . 5 , t h e n the angular velocity w o u l d b e given by
Figure 3.5
The angular velocity vector is always in the direction given b y the righth a n d rule w h e n w e turn our fingers in the direction o f rotation o f the
body. Referring to b o t h Figures 3.4 a n d 3 . 5 ,
and
and
is directed out o f the page if the b o d y is turning counterclockwise,
and into the page if it's turnine clockwise.
Important T h i n g s to Remember About Equation (3.8)
W e n o w h a v e the result that the angular velocity vector is a property of
the overall b o d y , a n d not a property o f its individual points. T h i s idea
cannot be overemphasized. Remember:
1.
A point h a s position, velocity, a n d acceleration.
2.
A body h a s orientation, angular velocity, a n d angular acceleration.
R e m e m b e r too that a point does n o t h a v e orientation,
and a
finite-sized body does n o t h a v e a u n i q u e r, v, a n d a.**
W e shall e m p h a s i z e t h e s e property differences b e t w e e n points and
bodies in the following w a y : throughout this b o o k , points are d e n o t e d by
capital italic printed letters while bodies are d e n o t e d b y ordinary capital
* We remark that in general (three-dimensional) motion, such a simple relationship
as
between angular velocity and body orientation does not exist.
Angular acceleration (a), the derivative of angular velocity, is discussed in Section 3.5.
** Of course the particle, being treated as small enough that we need not distinguish be­
tween the locations of its points, must be considered as having an r, v, and a — and not
an
or a.
Page 137
cursive letters. H e n c e , for e x a m p l e , P, A, a n d B d e n o t e points, w h i l e
and
d e n o t e bodies. Therefore, w e shall simply print t h e n a m e s o f
points, a n d write t h e n a m e s o f bodies in cursive script.
W e n o w m o v e toward a n u m b e r o f e x a m p l e s o f the use o f our n e w
Equation ( 3 . 8 ) ; in e a c h application o f this equation, t h e following three
rules m u s t b e followed without
exception:
from
other
3. This vector extends
the point (P)
on
(right) side of the equation
the point (Q) on the
(left) side.
this
to
2. This is the angular
1. These two
velocity vector of
points are
on the
rigid
body
same
Figure 3.6
Also helpful in using Equation ( 3 . 8 ) is t h e k i n e m a t i c diagram presented in
Figure 3 . 6 . T h e velocity o f Q, from Equation ( 3 . 8 ) , is the s u m o f the two
vectors in Figure 3 . 6 (see Figure 3 . 7 ) . N o t e that d e p e n d i n g o n t h e relative
sizes o f v a n d
, t h e velocity v could lie o n either side, or e v e n along,
line PQ. N o t e further that the difference b e t w e e n t h e velocities o f Q a n d
P — that is, v — v — is simply
. T h i s m e a n s that t h e o n l y w a y
P
0
0
P
in w h i c h t h e velocities o f two points o f a rigid b o d y
in m o t i o n in frame
can differ is b y t h e
term n o r m a l to the line joining t h e m . W e shall
return to this idea following t h e first three e x a m p l e s o f this section.
Incidentally, s o m e b o o k s describe v — v as " t h e velocity o f point Q
relative to point P." W e m e n t i o n this o n l y b y w a y o f explanation; our
definition o f v in S e c t i o n 1.3 s h o w s that points h a v e velocities relative to
frames, not relative to o t h e r points. If o n e uses the p h r a s e " t h e velocity of
point Q relative to point P," o n e m e a n s the velocity o f Q in a reference
frame in w h i c h P is fixed a n d w h i c h translates relative to .*
Q
P
P
Figure 3.7
In e a c h o f the e x a m p l e s that follow, n o t e t h e i m p o r t a n c e o f selecting
and depicting t h e unit vectors to b e u s e d in t h e solution. Also, in e a c h of
the first three examples, p a y careful attention to t h e w a y t h e velocity o f a
point (say, B) is expressed if t h e t a n g e n t to t h e p a t h o f B is k n o w n ; using
w h a t w e l e a r n e d in S e c t i o n 1.7, v is expressed as a single u n k n o w n scalar
(whose absolute value is t h e s p e e d o f B) times a unit vector along the
k n o w n tangent.
B
EXAMPLE 3.1
A 30-ft ladder is slipping down in a warehouse with the upper contact point T
moving downward on the wall at a speed of 2 ft/sec in the position shown in
Figure E3.1. Find the velocity of point B, which is sliding on the floor.
Figure E3.1
* "Translates" means that the frame moves in
cussed in more detail in Section 3.3.
without rotating. Translation is dis­
Page 138
Solution
We relate v and v by using Equation (3.8):
B
T
Noting that v has no -component and that v has no -component, we write:
B
T
Matching the
coefficients, we have
Matching the
coefficients, we have
Thus the velocity of B is
Note that a direction indicator must be attached to
correctly the angular velocity vector of the ladder:
in order to specify
or, alternatively,
Note that the directions of
make sense. Such visual checks on
solutions should be made whenever possible.
EXAMPLE 3.2
At the instant shown in Figure E3.2, the velocity of point A is 0.2 m / s to the right.
Find the angular velocity of rod
and determine the velocity of its other end
(point B), which is constrained to move in the circular slot.
Solution
We shall use Equation (3.8), featuring the points A and B of the rod:
Noting that the velocity of B has a known direction (tangent to its path), we write
v as an unknown scalar times a unit vector in this direction:
B
The component equations are:
Figure E3.2
Page 139
Solving Equations (1) and (2) gives
and therefore the answers (vectors are what are asked for!) are
or, equivalently,
In t h e n e x t e x a m p l e , two bodies h a v e angular velocity; t h u s w e shall
h a v e to subscript t h e
. W e shall simply denote b y
the angular velocity o f
and
t h e angular velocity o f
EXAMPLE 3.3
The crank arm
shown in Figure E3.3a turns about a horizontal z axis, through
its pinned end O, with an angular velocity of 10 r a d / s e c clockwise at the given
instant. Find the velocity of the piston pin B.
Solution
We apply Equation (3.8) first to relate the velocities of A and O on body
and
then to relate v to v on rod .Note that A is a "linking point" of both and ,
since it belongs to both bodies. On body
B
A
On body
Figure E3.3b
and using the Pythagorean theorem (see Figure E3.3b),
Now point B is constrained to move only horizontally. Therefore,
140
Equating the coefficients:
Equating the coefficients:
Therefore,
Substituting
into (1), we have v = 30.3 in./sec and
in./sec.
B
In all three o f the preceding e x a m p l e s , w e r e - e m p h a s i z e that it is
absolutely essential to correctly incorporate the kinematic constraints
i m p o s e d b y slots, walls, floors, a n d so forth.
It is often helpful in studying the kinematics o f rigid bodies to m a k e
u s e o f the following result, w h i c h is a corollary o f Equation (3.8):
Corollary: If P and Q are two points of a rigid body, their velocity
components along the line joining them must be equal.
Intuitively, w e see that the difference b e t w e e n these c o m p o n e n t s is the
rate o f stretching o f the line PQ, a n d this h a s to vanish. Also, w e h a v e
seen that v a n d v differ o n l y b y the term
w h i c h is clearly
normal to the Line PQ joining the points. M a t h e m a t i c a l l y , w e c a n see this
immediately b y dotting Equation (3.8) with the unit vector parallel to r ,
w h i c h is r / r :
Q
P
P Q
P Q
P Q
component of component of
v along PQ v along PQ
Q
zero
r
T h u s , if w e k n o w the velocity o f o n e point o f the b o d y , w e can find
a n y other without involving the angular velocity b y using Equation
(3.11). F o r instance, in E x a m p l e 3.2, t h e unit v e c t o r r / r
is simply
, a n d dotting this with the equation
A B
from that e x a m p l e gives
or
A B
Page 1 4 1
The algebra is seen to be simpler; w e h a v e w o r k e d with one equation in
o n e u n k n o w n rather than t w o in two.
W e n o w return to t h e vector formulation (Equation 3 . 8 ) for t w o final
examples in this section.
EXAMPLE 3.4
In the linkage shown in Figure E3.4, the velocities of A and C are given to be
at the instant given. Find the velocity of point B at the same instant.
Figure E3.4
Solution
On bodyi
On bar
Equating the two vector expressions for v , we get
B
Solving these two equations,
From Equation (1), it follows that
and the same result follows from (2), as a check.
Page 1 4 2
Question 3.6
If the velocities of A and C were given to be
and
for an interval of time, and not just at the instant shown,
would the solution be any different at (a) the same instant? (b) some
other instant?
Answer 3.6
(a) No. (b) Yes, because the geometry would be different.
EXAMPLE 3.5
The end B of rod travels up the right half of the parabolic incline in Figure E3.5a
at the constant speed of 0.3 m / s . Find the angular velocity of, and the velocity of
point A, which is at the origin at the given instant.
Solution
We shall relate v to v using Equation (3.8):
B
Figure E3.5a
A
Next we use Equation (1.41) to express v :
B
To get the unit tangent for point B, we use Figure E3. 5b, noting that
to the parabola at all times:
Figure E3.5b
is tangent
Therefore, for point B,
And thus
Since point A likewise has a velocity tangent to its path, we may write
and so Equation (1) gives
Collecting the coefficients of
and
j coefficients:
so that
coefficients:
we have
Page 143
Substituting for
and solving,
so that
Applications of Equation ( 3 . 8 ) t o rolling bodies are presented in
Section 3 . 6 after w e h a v e e x a m i n e d that topic in detail.
PROBLEMS
•
Section 3.2
3.3
The angular velocity of the bent bar is indicated in
Figure P3.3. Find the velocity of the endpoint B in this
position.
3.4
The velocities of the two endpoints A and B of a
rigid bar in plane motion are shown in Figure P3.4. Find
t h e velocity of the midpoint of the bar in the given posi­
tion.
3.5
If
ure P3.5.
in./sec,
find
and
. See Fig­
3.6
At a certain instant, the coordinates of two points A
and B of a rigid body
in plane motion are given in
Figure P3.6. Point A has
and the velocity of
B is vertical. Find v and the angular velocity of
g
3.7-3.11 in the following five problems involving a
"four-bar linkage" (the fourth bar in each case is the rigid
ground length between fixed pins!), the angular velocity
of one of the bars is indicated. Find the angular velocities
of the other two bars.
Figure P3.3
.7 (reference framel
Figure P3.4
Figure P3.5
Figure P3.6
Figure P3.7
Page 1 4 4
3.12 The equilateral triangular plate
shown in Fig­
ure P3.12 has three sides of length 0.3 m each. The bar
has an angular velocity
r a d / s counterclockwise
and is pinned to at A Body is also pinned to a block at
B, which moves in the indicated slot. At the given time,
find the angular velocity of .
3.13 Crank arm
shown in Figure P3.13 turns counter­
clockwise at a constant rate of 1 rad / s. Rod is pinned to
at A and to a roller at B that slides in a circular slot.
Determine the velocity of B and the angular velocity of
at the given instant.
3.14 The wheel shown in Figure P3.14 turns and slips in
such a manner that its angular velocity is
while
the velocity of the center C is 0.3 m / s to the left. Deter­
mine the velocity of point A.
Figure P3.8
Figure P3.9
Figure P3.12
Figure P3.13
Figure P3.10
Figure P3.11
Figure P3.14
Page 1 4 5
3.15 For the configuration shown in Figure P3.15, find
the velocity of point P of the disk
3.16 The speed of block in Figure P3.16 has the value
shown. Find the angular velocity of rod
and determine
the velocity of pin A of block , when
3.17 Wheel , (Figure P3.17) turns and slips in such a
way that its angular velocity is
r a d / s while the veloc­
ity of C is 0.4 m / s to the left. Determine the velocity
of point B, which slides on the plane. Bar is pinned to
at D.
3.18 Point A of the rod slides along an inclined plane as
in Figure P3.18, while the other end, B, slides on the hori­
zontal plane. In the indicated position,
rad/
sec. Find the velocity of the midpoint of the rod at this
instant.
3.19 Wheel
in Figure P3.19 has a counterclockwise
angular velocity of 6 r a d / s . What is the velocity of point B
at the instant shown?
3.20 Block
in Figure P3.20, which slides in a vertical
slot, is pinned to bars
and
at A. The other ends of
and
are pinned to blocks that slide in horizontal slots.
Block
translates to the left at constant speed 0.2 m / s .
Find the velocity of B: (a) at the given instant; (b) when C
is at point D, (c) when C is at point E.
3.21 The four links shown in Figure P3.21 each have
length 0.4 m, and two of their angular velocities are indi­
cated. Find the velocity of point C and detennine the
angular velocities of
and
at the indicated instant.
Figure P3.18
Figure P3.15
Figure P3.19
Figure P3.16
Figure P3.20
Figure P3.17
Figure P3.21
Page 146
3.22 In the mechanism shown in Figure P3.22, the
sleeve
is connected to the pivoted bar
by the 15-cm
link . Over a certain range of motion of , the angle
varies according to
rad, starting at t = 0 with
and
horizontal. Find the velocity of pin S and the angu­
lar velocities of
and
when
Time f is mea­
sured in seconds.
• 3 . 2 5 Block has a controlled position in the slot given
by
in. for
sec. (See Fig­
ure P3.25.) The time is t = 0 sec in the indicated
position. Find the angular velocities of the rod and the
wheel at (a) t = 0 sec and (b) t = 5 sec.
Figure P3.22
Figure P3.25
• 3 . 2 3 Find the velocity of point B of the rod if end A has
constant velocity 2 m / s to the right as shown in Fig­
ure P3.23. The rollers are small. Compare the use of
Equation (3.8) with the procedure used to solve Problem
1.63.
3.26 Crank
of the slider-crank mechanism shown in
Figure P3.26 has a constant angular speed
Find the
equation for the angular velocity of the connecting rod
as a function of
Figure P3.26
3.27 In the preceding problem, plot
from
Figure P3.23
3.24 Find the velocity of the guided block at the instant
shown in Figure P3.24.
Figure P3.24
as a function of
3.28 Referring to Section 3.2, show that neither the an­
gular velocity vector nor Equation (3.8) depends on which
body-fixed line segment (such as PQ in the text) is chosen
to measure . Use two other points P' and Q' and their
angle
as suggested in Figure P3.28 for your proof.
• 3 . 2 9 Rod
begins moving at
(see Figure P3.290
and is made to rum at the constant angular rate
rad/s. The cord is attached to the end of
and passes
around a pulley. The other end of the cord is tied to
weight
at point B. Observe that
moves downward
until
, when it reverses direction. Write an equa­
tion that gives the velocity of point B as a function of
for
Hint: Using trigonometry, write y as a
function of and the length L of the cord. Then differ­
entiate.
Page 1 4 7
Before (time t = t,)
Later (time t = t )
2
Figure P3.28
• 3 . 3 0 Repeat the preceding problem by using Equation
(3.8) to obtain v ; then resolve v into two components:
(a) one along PA that equals the magnitude of v and
(b) the other normal to PA, which does not affect B.
(These are sometimes called stretching and swinging com­
ponents, respectively.)
A
A
B
3.3
Figure P3.29
Translation
W h e n a rigid b o d y m o v e s during a certain time interval in such a w a y
that its angular velocity vector r e m a i n s identically zero, t h e n t h e b o d y is
said to b e t r a n s l a t i n g , or to b e in a state o f t r a n s l a t i o n a l m o t i o n during
that interval. F r o m Equation (3.8) w e thus see that for translation
T h a t is, all points o f t h e b o d y h a v e the s a m e velocity vector. B y differen­
tiating Equation ( 3 . 1 2 ) , w e see that the accelerations o f all points o f are
also equal for translation. N o t e that if
only at an instant (that is, at a
single value o f time rather t h a n over an interval), t h e n all points o f t h e
b o d y h a v e equal velocities at that instant but need not have equal
accelera­
tions.
Question 3.7
Why is this the case?
Answer 3.7 The derivative of
is not zero merely because happens to be zero
at one instant of time. To be able to differentiate v = v , this equation must be valid
for all values of t and not just one!
e
P
Page 1 4 8
Translation c a n b e either:
1.
Rectilinear:
2.
Curvilinear:
E a c h point o f
E a c h point o f
m o v e s along a straight line in
m o v e s o n a c u r v e d p a t h in
E x a m p l e s o f translation are s h o w n in Figure 3 . 8 . Part (a) s h o w s an
e x a m p l e o f rectilinear translation: B o d y
is c o n s t r a i n e d to m o v e in a
straight slot. P a r t ( b ) s h o w s a n e x a m p l e o f a i r v i l i n e a r translation:
B o d y is c o n s t r a i n e d b y t h e identical links.
(a)
(h)
Figure 3.8 Examples of translation.
P e r h a p s a n e v e n b e t t e r pair o f e x a m p l e s is t h e b l a c k b o a r d eraser
(Figure 3 . 9 ) , w h i c h w e u s e d earlier to explain p l a n e m o t i o n in S e c t i o n 3.2.
In part (a), t h e professor m o v e s t h e eraser s o that e a c h o f its points stays
on a straight line; it is therefore in a state o f rectilinear translation. In part
(b), t h e professor m o v e s t h e eraser o n a curve; b u t if t h e w o r d eraser is
a l w a y s horizontal during t h e erasing, t h e n
a n d the eraser is in a
state o f curvilinear translation. E v e n t h o u g h e a c h o f its points m o v e s o n a
curve, all t h e velocities ( a n d accelerations) are e q u a l at all times. T h e r e is
o n e n o t a b l e exception to our earlier s t a t e m e n t that "points, n o t bodies,
h a v e velocities a n d a c c e l e r a t i o n s . " In this present c a s e o f translation,
since all t h e points h a v e t h e same v ' s a n d a's, o n e c o u l d loosely refer to
" t h e velocity o f t h e e r a s e r " w i t h o u t ambiguity.
(a)
Figure 3.9
Another example of translation.
(b)
Page 149
T h e r e are n o examples or p r o b l e m s in this section b e c a u s e translation
p r o b l e m s o f rigid bodies require n o n e w theory b e y o n d w h a t w a s devel­
oped in C h a p t e r 1.
S u m m a r i z i n g , w h e n a b o d y is translating (either rectUinearly or curvilinearly), its angular velocity
is identically zero, and all its points
h a v e equal velocities (and accelerations). I f
o n l y at an instant, then
all the points o f t h e b o d y h a v e the s a m e velocity at that instant b u t n e e d
n o t h a v e equal accelerations.
3.4
Instantaneous Center of Zero Velocity
If P is a point in the reference p l a n e h a v i n g zero velocity at s o m e instant,
t h e n t h e velocity field o f is t h e s a m e as if the b o d y w e r e constrained at
that instant to rotate a b o u t an axis through P n o r m a l to the reference
plane. T h i s axis is called t h e i n s t a n t a n e o u s a x i s of rotation, a n d point P
is called t h e i n s t a n t a n e o u s c e n t e r (abbreviated
) o f z e r o velocity* of
T h u s if Q is a n y other point o f , t h e n w e h a v e
a n d since v is t h e n n o r m a l to b o t h of t h e vectors
and
, w e see that
each point moves with its velocity perpendicular
to the Urn- joining it to __.
Q
Figure 3.10 Instantaneous center of a
rolling wheel.
This c o n c e p t is illustrated in Figure 3 . 1 0 for a rolling* w h e e l , in w h i c h
is the contact point.
Proof of the Existence of the Instantaneous Center
W e can s h o w that i f a b o d y
h a s an i n s t a n t a n e o u s center.
Question 3.8
has
at a given instant, then it
Why can there be no point
whenever
is zero?
To d e m o n s t r a t e t h e existence o f
, w e shall u s e Equation ( 3 . 1 3 ) in
conjunction with Figure 3 . 1 1 . A s w e h a v e n o t e d a b o v e , t h e vector v ,
being equal to
for t h e point
having
is n o r m a l t o b o t h
. H e n c e w e h a v e these results:
Q
1.
Figure 3.11
T h e vector
lies in t h e reference p l a n e a n d is n o r m a l to v . It thus
lies along the line in Figure 3 . 1 1 .
Q
* The phrase is admittedly redundant, but it is in common usage. "Instantaneous center
of velocity" would perhaps be more concise, and "center of velocity" even more so.
"Instantaneous center," however, is inadequate because of the possibility of confusion
with points of zero acceleration.
Rolling means no slipping, according to the definition we adopt in this book (see Sec­
tion 3.6).
Answer 3.8 If
Equation (3.8) says that v = v ; that is, all points of have the
same velocity vector. This common velocity vector is then zero only if the body is at rest.
Incidentally, some think of as having an instantaneous center
at infinity when
0
P
Page 1 5 0
2.
T h e point
exists (and is unique) b e c a u s e
be
in order that
is therefore seen to
Question 3.9 Why is
below Q in Figure 3.11 instead of being
the same distance above Ql
W e h a v e thus verified the existence o f the i n s t a n t a n e o u s center (unless
), b e c a u s e w e k n o w h o w to get t o it from a n y arbitrary point Q o f
the b o d y w h e n e v e r the angular velocity
o f a n d the velocity v of
the point Q are k n o w n . W e also n o t e again that the velocity magnitude
of every point o f (in the reference plane!) equals the product o f
and
the distance to the point from
S o m e t i m e s b e c a u s e o f a constraint o n the m o t i o n w e k n o w the loca­
tion o f
at the outset. This is the case for the rolling w h e e l o f Figure 3 . 1 0
in w h i c h the b o t t o m point grips the ground and is held at rest (thereby
becoming
) for the instant o f its contact. I f the radius o f the w h e e l is 1 5 "
a n d the velocity o f its center
t h e n from the a b o v e
discussion, the angular speed o f the w h e e l is
Q
N o t e that w h e n v i e w e d from
the senses o f the velocity direction
o f a n y point a n d the angular velocity direction o f the b o d y must always
agree. For e x a m p l e , these are possible situations:
T h e s e are not:
Answoi 3 . 9 If point I were above
then
would give an incorrect direction
for the velocity v — it would be opposite to the actual direction. The senses of v and
to when viewed from
must agree, as we point out later in this section.
D
0
Page 151
Figure 3.12
Therefore the direction of
o f the w h e e l in Figure 3 . 1 0 is clockwise
in order that the k n o w n velocity direction o f point C
a n d the
angular velocity of the b o d y b e in a g r e e m e n t as the b o d y rotates about
at the instant s h o w n .
E v e n i f the angular velocity o f is not k n o w n , w e can still easily find
the instantaneous center o f zero velocity
if w e k n o w t h e velocities —
or really, just the directions o f the velocities — o f two points A a n d B of
Constructing perpendicular lines to the velocities o f A (at A) a n d o f B
(at B) as s h o w n in Figure 3 . 1 2 , w e immediately recognize
as the
intersection point o f the two lines.
Question 3.10
Why?
F r o m Figure 3 . 1 2 a n d the discussion a b o v e , w e k n o w that:
(Recall that there is only o n e
for the b o d y . )
w h e r e w e are abbreviating
by
,and
by .
We n o w present three e x a m p l e s o f the use o f the a b o v e procedure for
locating
w h e n t w o velocity directions are k n o w n in advance.
Answer 3 . 1 0 The point
is unique. Since there is only one common point on the lines
drawn perpendicular to the velocities (to v at B and to v,, at A), that point is the instan­
taneous center.
B
EXAMPLE 3.6
Ladders commonly carry a warning that for safe placement, the distance B in
Figure E3.6a should be of the length L (i.e., of
). Let us suppose that a
careless painter temporarily set a ladder against a wall in a dangerous position
with B/l = 0.5, and went off to get his paint and brushes. Suppose further that
the ladder began to slip, with the top of the ladder, point P, sliding down the wall
and the bottom, point Q, slipping along the ground as shown. When B is 15 ft,
find the instantaneous center of zero velocity
of the ladder, and discuss the
path of
in space as the ladder falls further.
Solution
When B = 15 ft, the normals to v at P and to v at Q intersect at point
as
shown in Figure E3.6b on the next page.
If we imagine a rigid sheet of very light plastic glued to the ladder as in
Figure E3.6c, then the ladder has been "rigidly extended." Note that only for this
instant, we can think of the extended body as rotating about an imaginary pin at
the intersection point
of the normals to two velocities as shown. Note further
from Figure E3.6c that the velocities of all points of the rigid sheet are perpendic­
ular to lines drawn to them from
. The velocity magnitude of each point is
proportional to the distance from that point to
, with the proportionality
P
Figure E3.6a
Q
Page 1 5 2
Line normal
to v
r
Line normal
to v
Q
S
Figure E3.6b
Figure E3.6c
constant being
of the body at the instant. Hence all triangles like the three
shaded in Figure E3.6c are similar.
As the ladder falls, the location of the point _ changes on the imagined
rigid extension (sheet) as time passes, because the perpendiculars to v and v„
intersect at different points of the sheet, as seen below in Figure E3.6d:
A
Figure E3.6d
Note that as point P (and thus all of the ladder) gets closer and closer to the
ground, point _ gets closer and closer to point Q. Thus even though increases,
becomes
gets smaller and smaller, until, in the limit (as P contacts the ground) Q
(the intersection of the perpendiculars) and V is then zero.
Q
W h e n t h e b o d y h a s a pivot (a point that n e v e r m o v e s throughout
the body's m o t i o n , such as a pin), it is clearly always
; in this case the
motion is called pit re rotation. But otherwise, the point
is not the s a m e
Page 1 5 3
point o f t h r o u g h o u t t h e motion, as w e h a v e already s e e n with t h e wheel,
a n d ladder. In e a c h o f t h e n e x t t w o e x a m p l e s , o n e o f the bodies h a s a,
pivot.
EXAMPLE 3.7
Rework Example 3.3 using instantaneous centers. (See Figure E3.7a) The crank
arm
turns about a horizontal z axis, through its pinned end O, with an angular
velocity of 10 rad/sec clockwise at the given instant. Find the velocity of the
piston pin B.
Solution
Since O is the point
for body
, we have
Figure E3.7a
Next we find the
of , using the fact that it is on lines perpendicular to the
velocities of A and B as shown in Figure 3.7b. If we next find the distance D from
to A, then
will be
By similar triangles:
Then
so
Again by similar triangles:
Figure E3.7b
so that
fore.
i n . / s e c to the right, as we have seen be­
EXAMPLE 3.8
At the instant shown in Figure E3.8a, the angular velocity of bar
is
rad/sec. Find the velocity of pin B connecting bar
to the slider
block, constrained to slide in the slot as shown.
Figure E3.8a
Solution
As seen in Figure E3.8b on the next page, the point
for
is O, since it is
pinned to the reference frame. The velocity of A is perpendicular to the line from
to A (that is, from O to A) and has a direction in agreement with the angular
velocity of
as the body turns about O. Its value is
Page 1 5 4
Next we sketch body
and note the position of
Figure E3.8c. Similar triangles yield the height H of
for , as explained in
above A:
Figure E3.8b
Question 3.11
Why does v have to be "southwest" along the slot and
not "northeast"?
B
is on each of these
lines since they are each
normal to the velocity of
a point of £
2
We may now use
of
to get the angular velocity of
; using vectors this
time,
Substituting, we get
Solving gives
Figure E3.8c
Note that when we write
we are saying that
is counterclockwise in
accordance with the sign convention adopted for the problem in the figure if its
value turns out positive. Thus when its value is now found to be negative, we
know that is turning clockwise at the given instant. Of course, as we have seen,
we do not have to use vectors on such a simple problem; we can use what we
know about the instantaneous center in scalar form to get a quick solution:
where we assign the direction in accordance with the known velocity direction of
A and the position of
Next we use
of
to obtain v :
B
The velocity of B is thus
Note that the arrow in this sketch is just as descriptive of the direction of the
vector velocity of B as is the unit vector —
Answer 3.11 The known velocity direction of A dictates that
is turning clockwise
around
so v has to be "southwest" for this to be the case.
B
Page 155
T h e Special Case in W h i c h the Normals Do Not Intersect
T w o things can go w r o n g with the procedure w e h a v e b e e n following o f
intersecting t h e n o r m a l s to two p o i n t s ' velocity vectors to find . T h e
first of t h e s e is that t h e t w o perpendicular lines m a y b e parallel a n d h e n c e
n o t intersect, as suggested b y Figure 3 . 1 3 b e l o w for t h e points A a n d B o f
bar
N o t e that A a n d B are e a c h at t h e top o f t h e vertical p l a n e circles on
w h i c h t h e y m o v e , a n d since their velocity vectors are tangent to their
paths, e a c h is horizontal at this instant.
Perpendiculars
d o not
intersect!
Figure 3.13
Let us e x a m i n e E q u a t i o n 3.8 for this case:
Since v a n d v h a v e o n l y -components w h i l e
has both
and
then
m u s t b e zero at this instant. T h i s does n o t m e a n t h e b o d y
is translating; that occurs w h e n
is zero all the time. Rather, in this case
t h e b o d y is just stopped for o n e instant in its angular motion, as its ang­
ular velocity is c h a n g i n g from clockwise to counterclockwise (see Fig­
ure 3 . 1 4 ) .
T h e equation a b o v e also s h o w s that at such an instant w h e n
all points o f t h e b o d y h a v e identical velocities. S o in Figure 3 . 1 3 ,
v = v = v
at t h a t instant. C o n v e r s e l y , a n y time t w o points o f a
b o d y h a v e equal velocities in p l a n e motion, t h e b o d y ' s angular velocity
vanishes at that instant.
B
A
A
B
B n y p o j n l ofe
is
just before A and E
reached highest points...
Circular path
of B
Circular path
of A
...but is
just after A and B leave highest
points. At the highest points,
Figure 3.14
Page 1 5 6
T h e Special Case in W h i c h the Normals are Coincident
T h e s e c o n d exceptional case occurs w h e n the perpendiculars to two
velocities are o n e a n d the s a m e line (see Figures 3.15(a,b)):
(a)
(b)
Figure 3.15
In this case, w e c a n find the i n s t a n t a n e o u s center using similar triangles
as s h o w n . T h i s simple procedure works, b e c a u s e , for the case s h o w n in
Figure 3 . 1 5 a ,
so that
If w e s h o u l d get coincident n o r m a l s w i t h the directions o f v a n d v
opposite, as in Figure 3 . 1 5 b , t h e n
I lies between P a n d Q, a n d it m a y
again b e f o u n d b y similar triangles. T h i s time,
P
Q
In the e x a m p l e s w e h a v e p r e s e n t e d in this section, n o t e that use of
the i n s t a n t a n e o u s center m a y b e m a d e w i t h or w i t h o u t vector algebra. Its
a d v a n t a g e is in finding a n d using points o f zero velocity in order to
simplify the resulting m a t h e m a t i c s . Instantaneous centers never have to
be u s e d to effect a solution. S o m e t i m e s they are helpful, b u t at other
times, t h e y m a y b e m o r e trouble to locate t h a n t h e y are worth!
E x a m p l e s o f b o t h the a b o v e special cases occur in our last e x a m p l e in
this section. It involves four different positions o f the s a m e system:
EXAMPLE 3.9
Figure E3.9a shows a rolling wheel of a large vehicle that travels at a constant
speed of
. Find the velocity of the piston when equals: (a) 0°, (b) 90°,
(c) 180°, (d) 270°.
Page 157
Figure E3.9a
Solution
First we solve for the velocities by using several approaches. In each case, the
speed of the wheel's center is the same as the speed of the vehicle: 60 mph or
88 ft/sec. And since the piston translates, all its points have equal velocities
and equal accelerations at every instant.
Case (a): As we shall see in detail in Section 3.6, the instantaneous center of a
wheel rolling on a fixed track is at the point of contact. Since velocities increase
linearly with distance from this point
, we have for the point E of :
If we draw lines at E and P perpendicular to v and v (P is constrained to move
horizontally), they are parallel and thus will not intersect (see Figure E3.9b).
Therefore
at that instant. Thus
E
Figure E3.9b
P
Figure E3.9c
Case (b): This time we shall use vectors; on the wheel (see Figure E3.9c),
Page 1 5 8
or
or
On
now (after noting the trigonometry results in Figure E3.9d):
Equating coefficients of
and then of
we obtain:
Therefore
so that
Figure E3.9d
Figure E3.9e
Case (c): In all four cases,
ure E3.9e),
Again, as in Case (a), body
velocity. Thus
rad/sec. This time, then (see Fig­
has
so that all points of
have the same
Page 1 5 9
Figure E3.9f
Case (d): Using Figure E3.9f, we see that, on body
We shall now use the instantaneous center of
above figure,
Therefore, on body
,
. From the similar triangles in the
,
and
N o t e from the four answers in the a b o v e e x a m p l e that the piston is
moving faster during the parts o f the wheel's revolution in w h i c h v
makes small angles with r o d
; conversely, it m o v e s slower w h e n v is
making a large angle with
. This is b e c a u s e the c o m p o n e n t s of v a n d v
along
must always b e t h e s a m e , as w e s a w earlier.
E
£
E
P
Page 1 6 0
PROBLEMS
•
Section 3.4
3.31 The angular velocity of
in Figure P3.31 is
r a d / s = constant. Trace the five sketches and then
show on parts (a) to (d) the position of
for the rod .In
part (e), using the proper length of , draw the positions
of
at the two times when v = 0 . Rod
has length
0.9 m.
B
3.32 Solve
centers.
3.33
Problem 3.16
by
using
instantaneous
Solve Problem 3.7 by using instantaneous centers.
3.34 Solve Problem 3.8 by using instantaneous centers.
i.n
(b)
(d)
(e)
Figure P3.31
3.35 Solve Problem 3.6 by using instantaneous centers.
3.36 Solve
centers.
Problem 3.24
by
using
instantaneous
3.37 Solve
centers.
Problem 3.11
by
using
instantaneous
3.38 In Figure P3.38 the crank arm
is 4 in. long and
has constant angular velocity
rad/sec. It is
pinned to the triangular plate, , which is also pinned to
the block in the slot at D. Find the velocity of D at the
instant shown.
(c)
Page 161
3.42 Rods and
are pinned at B and move in a verti­
cal plane with the constant angular velocities shown in
Figure P3.42. Locate the instantaneous center of, forthe
given position, and use it to find the velocity of point C.
Then check by calculating v by relating it (on ) to the
velocity of B. Note that sometimes
is more trouble to
locate than it is worth!
c
Figure P3.38
3.39 See Figure P3.39. The angular velocity of rod is a
constant:
rad/s. Determine the angular veloc­
ities of plate ^ and bar
in the indicated position.
3.40 The pin at B (Figure P3.40) has a constant speed of
51 c m / s and moves in a circle in the clockwise direction.
Find the angular velocities of bars
and
in the given
position.
3.41 Solve Example 3.5 by using the instantaneous
center of the rod.
Figure P3.42
3.43 The linkage shown in Figure P3.43 is made up of
rods
and
. Rod
has constant angular velocity
rad/sec. Determine the angular velocities of
and
when the angle is equal to 90° as shown.
Figure P3.43
Figure P3.39
Figure P3.40
3.44 A bar of length 2L moves with its ends in contact
with the planes shown in Figure P3.44. Find the velocity'
and acceleration of point C, in terms of and its deriva­
tives, by writing and then differentiating the position
vector of C. Then check the velocity solution by using the
instantaneous center.
Figure P3.44
Page 162
3.45 The piston rod of the hydraulic cylinder shown in
Figure P3.45 moves outward at the constant speed of
0.13 m / s . Find the angular velocity of
at the instant
shown.
3.48 The roller at B, which moves in the parabolic slot, is
pinned to bar
as shown in Figure P3.48. Bar
is
pinned to at A. The angular velocity of, at this instant
is shown. Find the angular velocity of , at this time.
3.49 Bars
and
(see Figure P3.49) are pinned to­
gether at A. Find the angular velocity of bar , and the
velocity of point B when the bars are next collinear. Hint:
To find this configuration, draw a series of rough sketches
of
and as turns counterclockwise from the position
shown, and you will see and
coming into alignment.
3.50 Repeat the preceding problem at the second instant
of time when the bars are collinear. Follow the same hint,
but this time start just past the first collinear position,
found in Problem 3.49.
3.51
The constant angular velocity of wheel
is
rad/sec. It is in rolling (i.e., "no slip") contact
with
, which means the contacting points have the
same velocity. Find the angular velocity of the bar
at
the instant shown in Figure P3.51.
Figure P3.45
3.46 Using the method of instantaneous centers, find
the velocity of point B in Figure P3.46, which is con­
strained to move in the slot as shown. The angular veloc­
ity of
is
r a d / s at the indicated instant.
3.47 The center of block
in Figure P3.47 travels at a
constant speed of 30 mph to the right. Disk is pinned to
at A and spins at 100 rpm counterclockwise. Find: (a)
the velocity of P; (b) the instantaneous center
of
and (c), using
, the velocities of Q, S, and R.
Figure P3.48
Figure P3.49
Figure P3.4G
Figure P3.47
Figure P3.51
Page 163
3.5
Acceleration and Angular Acceleration Relationship for
T w o Points of the Same Rigid Body
T h e a n g u l a r a c c e l e r a t i o n vector o f a rigid b o d y in plane m o t i o n in a
frame
is defined as the derivative in
o f the angular velocity a n d is
called a :
or
where
is a constant vector in b o t h
and
N o t e that, as with
, we
delete the subscript w h e n there is n o confusion a b o u t the frame o f refer­
e n c e being used.
Development of the Acceleration and Angular Acceleration Relationship
W e n o w develop the relationship b e t w e e n the accelerations o f t w o points
P a n d Q o f a rigid b o d y
Differentiating b o t h sides o f Equation ( 3 . 8 )
yields
N o w using Equation ( 3 . 7 ) , * w e m a y rewrite t h e last term as
or
Question 3.12
zero?
Why is the dot product
in Equation (3.17)
Equations ( 3 . 1 6 ) and ( 3 . 1 8 ) t h e n yield the desired relation b e t w e e n the
accelerations o f P a n d Q:
W e n o t e that the s a m e three rules spelled out in E q u a t i o n ( 3 . 1 0 ) also
h o l d for the use o f Equation ( 3 . 1 9 ) . U n l i k e velocities, h o w e v e r , the accel­
erations o f P a n d Q do n o t generally h a v e equal c o m p o n e n t s along the
line P Q joining t h e m ; t h e s e c o m p o n e n t s differ b y
. Likewise,
the c o m p o n e n t s perpendicular
to PQ differ b y n
fin the s a m e w a y as
the velocity c o m p o n e n t s n o r m a l to PQ differ b y
).
Answer3.12
to
Because r
P C
lies in the (xy) reference plane and is therefore perpendicular
* And the vector identity A X (B X C) = B(A • C) - C(A • B).
Page 164
Fi gure 3,1
If t h e acceleration of point A of a b o d y is & , for e x a m p l e , t h e n t h e
acceleration of a n y point B is the s u m o f the three vectors s h o w n in
Figure 3 . 1 6 . If A is a p i n n e d point, or pivot,* t h e n a h a s two c o m p o n e n t s :
o n e along t h e line from B t o w a r d A a n d t h e other perpendicular to it a n d
t a n g e n t to t h e circle (on w h i c h it necessarily travels w h e n there is a pivot
at A ) . N o t e that the direction o f the tangential c o m p o n e n t depends o n the
direction o f a, but t h e " r a d i a l " c o m p o n e n t is always i n w a r d t o w a r d t h e
pivot. T h i s case is s h o w n in Figure 3 . 1 7 . W e n o w illustrate the use o f
Equation ( 3 . 1 9 ) with several examples.
A
B
Figure
3.17
EXAMPLE 3 . 1 0
In Figure E3.10, let the links have length 1 m and let them each have
r a d / s and
r a d / s at a time when they make an angle of 45°
with the ceiling. Find the acceleration of block
(That is, find the acceleration
of any of its points — they are all the same since is translating.)
J
Solution
All points of have the same v and a as point A. Using Equation (3.19) for the
link OA, we get
Figure E3.10
EXAMPLE 3 . 1 1
In Example 3.3 find the acceleration of the translating piston at the given instant
if
rad/sec and
r a d / s e c . See Figure E 3 . l l .
2
Figure E3.11
* Again, pivot means a point of
but is not limited to, a hinge.
that does not move throughout a motion. It includes,
Page 1 6 5
Question 3.13 What does it mean when a is in the opposite direction
from that of ?
Solution
Relating O and A on body
by Equation (3.19), we have
And relating A and B on body
, we have
In the previous example we found
Noting that the piston translates horizontally, we get
The coefficients of yield
The coefficients of then give our answer:
Let us find the acceleration o f the i n s t a n t a n e o u s center o f zero v e l o c ­
ity o f the rod
in E x a m p l e s 3.7 a n d 3 . 1 1 . Using those e x a m p l e s and
Equation ( 3 . 1 9 ) , w e h a v e
W e see from this result that the i n s t a n t a n e o u s center o f zero velocity
does n o t generally h a v e zero acceleration. U n l e s s the point
is a pivot
(i.e., a p e r m a n e n t l y fixed point, s u c h as a pin c o n n e c t i n g the point to the
reference frame), it s h o u l d never b e a s s u m e d that
is zero.*
Answer 3.13 It means that the angular speed of the body is decreasing.
* Actually, in nontranslational cases there is a point of zero acceleration, but unless it is
a pivot point, or a point of rolling contact at an instant when
, it is more trouble
to find than it is worth. See Problem 3.83.
Page 166
EXAMPLE 3 . 1 2
In Example 3.5, find the angular acceleration of the rod and the acceleration of its
endpoint A. See Figure E3.12a.
Solution
Relating the accelerations of B and A with Equation (3.19),
The acceleration of point B is
Figure E3.12a
where
since = constant = 0.3 m / s . The curvature formula from calcu­
lus gives us the radius of curvature at point B:
Therefore
Therefore,
The unit vector
is seen in Figure E3.12b to be
was 63.4°
(From Example 35]
Therefore
Figure E3.12b
Substituting a into Equation (1) gives
B
where
has also been substituted.
Next, the radius of curvature of the path of A at the instant of interest is
Substituting
and
,
and
from Example 3.5 (along with
at point A), we obtain the vector equation:
Page 1 6 7
Writing the component equations, we have
The
equation gives
And the
equation then yields
Therefore
M a n y m o r e examples of the use of Equation (3.19) will be found in
the next section, after w e h a v e discussed the topic of rolling.
PROBLEMS
•
Section 3.5
3.52 In Figure P3.52 the angular velocity of the bent bar
is 0.2 r a d / s counterclockwise at an instant when its angu­
lar acceleration is 0.3 r a d / s clockwise. Find the accel­
eration of the endpoint B in the indicated position.
2
Figure P3.53
Figure P3.52
3.54 End B of the rod shown in Figure P3.54 has a con­
stant velocity of 10 ft/sec down the plane. For the posi­
tion shown (rod horizontal) detenrtine the velocity and
acceleration of end A of the rod.
3.53 The acceleration of pin B in Figure P3.53 is 9.9 f t /
sec down and to the left, and its velocity is 4 f t / s e c up
and to the right, when
passes the horizontal. At this
instant, find the angular acceleration of
Figure P3.54
2
Page 168
3.55 The velocities and accelerations of the two endpoints A and B of a rigid bar in plane motion are as shown
in Figure P3.55. Find the acceleration of the midpoint of
the bar in the given position.
3.60 IfinProblem 3.11 the bar whose angular velocity is
given to be 2 rad / sec has an angular acceleration of zero
at that instant, find the angular acceleration of the 5-inch
(horizontal) bar.
3.56 In Problem 3.42 find the acceleration of C in the
position given in the figure.
* 3.61 The angular velocity of in Figure P3.61 is a con­
stant 3 rad/sec clockwise. Find the velocity and accelera­
tion of point C in the given configuration, and determine
the acceleration of point C when v = 0.
3.57 At the instant given, the angular velocity and an­
gular acceleration of bar
are
rad/sec and
r a d / s e c . (See Figure P3.57.) Find the angular accel­
erations of
and
at this instant.
c
2
3.58 Bar
rotates with a constant angular velocity
of
rad / sec. Find the angular velocities and angular
accelerations of
and
at the instant shown in Fig­
ure P3.58.
3.59 In the position indicated in Figure P3.59, the slider
block has the indicated velocity and acceleration. Find
the angular acceleration of the wheel at this instant.
Figure P3.61
Figure P3.55
Figure P3.64
3.62 Refer to the preceding problem. At the instant of
time when
, find the acceleration of point C.
Figure P3.57
Figure P3.58
3.63 If in Problem 3.24 the angular velocity of the
5-inch bar is constant throughout an interval which
includes the instant shown, find at that instant the accel­
eration of the guided block.
3.64 At the instant of time shown in Figure P3.64, the
angular velocity and angular acceleration of rod
are
rad/sec and
r a d / s e c . At the same time, find the
angular acceleration of bar
2
Figure P3.59
3.65 In Problem 3.26 find the equation for the angular
acceleration of the connecting rod
a a function of r,
, and .
Page 169
3.66 In the preceding problem, plot
(0 < 6 £ In) for
= 1, 2, and 5.
versus
3.67 Crank
in Figure P3.67 is pinned to rod
; the
other end of
slides on a parabolic incline and is at the
origin in the position shown. The angular velocity of is
r a d / s = constant. Determine the acceleration of A
and the angular acceleration of
at the given instant.
Hint The radius of curvature p of a plane curve y = y(x)
can be calculated from
Use this result in computing the normal component of a
as in Example 3.11.
3.72 InProblem 3.49 find the acceleration of point Band
the angular acceleration of body
at the described in­
stant.
3.73 In Problem 3.5 0 find the acceleration of point B and
the angular acceleration of body
at the described in­
stant.
3.74 In Problem 3.22 find the acceleration of S and the
angular accelerations of
and
at the instant when
= 30°.
3.75 In Problem 3.38 determine the angular accelera­
tion of the plate and the acceleration of pin l? in the
indicated position.
A
3.76 In Problem 3.45 determine the angular accelera­
tions of
and
in the indicated position.
3.77 The motion ofarotating element in a mechanism is
controlled so that the rate of change of angular speed
with angular displacement is a constant K. If the angular
speed is
when both and the time t are zero, determine
and the angular acceleration a as functions of time.
• 3 . 7 8 Rod in Figure P3.78 is pinned to disk at A and
B. Disk
rotates about a fixed axis through O. The rod
makes an angle radians with the line as shown, where
= sin t. The time is given by t in seconds. Deteirmine the
horizontal and vertical components of acceleration of the
midpoint P of segment AB when
rad.
Figure P3.67
3.68 The 10-ft bar in Figure P3.68 is sliding down the
13-ft-radius circle as shown. For the position shown, the
bar has an angular velocity of 2 r a d / s e c and an angular
acceleration of 3 rad/sec , both clockwise. Find the x and
y components of acceleration of point B for this position.
2
3.79 A rod is pinned at A and B to the centers of two
small rollers. (See Figure P3.79.) The speed of A is kept
constant at v even after B encounters the parabolic sur­
face. Find the acceleration of B just after its roller begins to
move on the parabola.
0
3.69 In Problem 3.23, find the acceleration of B when
x = 10 m.
3.70 In Problem 3.31 find the acceleration of the upper
end B of rod
in position (a).
3.71 Find the acceleration of P in Problem 3.58 if
at the same instant the body
has angular acceleration
r a d / s e c instead of zero.
2
Figure P3.78
Figure P3.68
Figure P3.79
Page
170
* 3.80 In the preceding problem, find a just after the left
roller has begun to travel on the parabola.
B
* 3.81 The bent bar shown in Figure P3.81 slides on the
vertical and horizontal surfaces. For the position shown,
A has an acceleration of 4 f t / s e c to the left, while the bar
has an angular velocity of 2 rad/sec clockwise and an
angular acceleration of (X.
2
a. Determine a for the position shown.
b. Find, for the position shown, the angle and
the distance PA such that point P has zero ac­
celeration.
• 3 . 8 2 The right end P of bar is constrained to move to
the right on the sine wave shown in Figure P3.82 at the
constant speed
in./sec. The left end A of is con­
strained to slide along the x-axis. At the instant when
in., find (a)
(b)a .
• 3 . 8 3 Show that for a rigid body in plane motion, as
long as
and a are not both zero there is a point of
having zero acceleration. Hint: Let P be a reference point
with acceleration
See if you can find
a vector
from P to a point T of zero ac­
celeration. That is, solve
P
for x and y.
y — 3 sin x in
Bar shown at
Figure P3.81
in
Figure P3.82
3.6
Rolling
Let
and
b e two rigid bodies in m o t i o n . W e define r o l l i n g to exist
between
and
if during their motion:
1.
A c o n t i n u o u s s e q u e n c e o f points on t h e surface o f
c o m e s into
o n e - t o - o n e contact with a continuous s e q u e n c e o f points o n the
surface o f
2.
At e a c h instant during t h e interval o f t h e motion, t h e contacting
points h a v e t h e s a m e velocity vector.
N o t e that according to this definition there can b e n o slipping or sliding
b e t w e e n t h e surfaces o f
and
if rolling exists. M a n y authors, h o w ­
ever, use t h e p h r a s e "rolling without slipping" to describe t h e m o t i o n
defined h e r e . In their context, "rolling a n d slipping" w o u l d denote turn­
ing without t h e contact points h a v i n g equal velocities; in our c o n ­
text, rolling means n o slipping, so w e shall s a y " t u r n i n g a n d slipping" in
cases o f u n e q u a l contact-point velocities.
In this section w e consider three classes o f p r o b l e m s involving rolling
contact:
1.
Rolling o f a w h e e l o n a fixed straight line
2.
Rolling o f a w h e e l o n a fixed p l a n e curve
3.
Gears
Page 171
Rolling of a W h e e l on a Fixed Straight Line
If the w h e e l
s h o w n in Figure 3 . 1 8 is rolling o n t h e g r o u n d ( , t h e
reference frame in this case), t h e n t h e c o n t i n u o u s (shaded) s e q u e n c e s of
points o f
a n d , are in contact, o n e pair o f points at a time. S i n c e the
points o f
are all at rest, e a c h point P o n t h e rim o f
c o m e s instanta­
n e o u s l y to rest as it contacts a point P o f
(and is gripped for a n instant
b y t h e ground). In this case, t h e velocities o f P , a n d P are e a c h zero,
although t h e y n e e d n o t vanish in general for rolling; all that is required is
t h a t v p , = vp .*
1
2
2
2
Figure 3.18
Rolling wheel, illustrating contacting sequence of points.
Using E q u a t i o n ( 3 . 8 ) a n d n o t i n g that t h e center point C m o v e s only
horizontally a n d that t h e c o n t a c t point is a l w a y s t h e i n s t a n t a n e o u s center
of
,
For t h e rolling w h e e l , therefore, t h e velocity o f its center C a n d t h e b o d y ' s
angular velocity are related quite simply b y
Question 3 . 1 4
H o w would this expression be different w e r e to h a v e
been chosen so that it would increase with counterclockwise
turning of
the body?
T h e relation b e t w e e n the d i s p l a c e m e n t o f C a n d the rotation of
follows from integrating Equation ( 3 . 2 3 ) :
w h e r e t h e integration constant is zero if w e c h o o s e x
c
= 0 when
* If is the flatbed of a truck, for example, itself in morion with respect to a ground
reference frame
then
but E still rolls on
x
Answer 3.14 Then we would have
Page 172
A n o t h e r a p p r o a c h to rolling is to b e g i n with a small d i s p l a c e m e n t
while the base point
grips t h e ground. I f t h e angle o f rotation
p r o d u c e d is
, then
if w e envision t h e b o d y turning about its i n s t a n t a n e o u s center. Dividing
b y a small time i n c r e m e n t
a n d taking t h e limit, w e get
as w a s o b t a i n e d in Equation ( 3 . 2 3 ) .
N e x t w e consider accelerations. F r o m Equation ( 3 . 2 2 ) :
a n d the c e n t e r point C accelerates parallel to the p l a n e , as it m u s t since it
(alone a m o n g all points o f t h e w h e e l ) h a s rectilinear m o t i o n . N o w let us
c o m p u t e t h e acceleration o f t h e i n s t a n t a n e o u s c e n t e r
of the wheel.
Using Equation ( 3 . 1 9 ) , w e obtain
T h u s t h e c o n t a c t point o f a w h e e l rolling o n a flat, fixed p l a n e is
accelerated t o w a r d its c e n t e r with a m a g n i t u d e
. W e see o n c e again
that a point o f zero velocity n e e d not b e a point o f zero acceleration,
although o f course it will b e s u c h if it is p i n n e d to t h e reference frame. T h e
point
in t h e e x a m p l e is at rest instantanously, but it h a s an
acceleration — w h i c h is w h y its velocity c h a n g e s from zero as s o o n as it
moves and a new
takes its place in t h e rolling. W e n o w take u p several
examples,* e a c h o f w h i c h deals w i t h a r o u n d object rolling on a flat
surface.
EXAMPLE 3 . 1 3
At a given instant, the rolling cylinder in Figure E3.13 has
r a d / s and
r a d / s . Find the velocity and acceleration of points N and E.
2
Solution
We shall relate the velocities and accelerations of N and E to those of point C.
Calculating these, we have, because of the rolling (with S being
),
Figure E3.13
* The examples of this section make continued use of Equations (3.8), (3.13), and (3.19),
with the added feature that a rolling body is involved in each problem.
Page 173
and
Therefore
Note the agreement of this result with
Continuing, we get
EXAMPLE 3 . 1 4
In Example 3.9 find the piston acceleration when 6 = 9 0 ° . See Figure E3.14.
Figure E3.14
Solution
In
Case E,
(b) of Example 3.9 we found
rad/sec,
For point
ft/sec,
rad/sec, a n d f t
/ s e c . On b o d y , the
velocity of C is constant
so that a = 0. Also, a = ra so that a = 0.
c
c
t
x
and
iThe ,reader
which may
as we
wish
have
to obtain
seen isthe results
not for
zero.
a. and a by relating them instead to
N
E
Page 174
Therefore
Equating the
coefficients first eliminates the unknown a :
P
Then the coefficients of
yield our answer:
EXAMPLE 3 . 1 5
The wheel in Figure E3.15 rolls to the right on plane At the instant shown in
the figure, has angular velocity
r a d / s . Rod is pinned to
at A,
and the other end B of
slides along a plane Q parallel to p. Determine the
velocity of B and the angular velocity of
at the given instant.
Figure E3.15
Solution
We shall use Equation (3.8) together with the results we have derived for rolling.
We first seek the velocity of A; when we have v we shall then relate it to v on
rod
We may find v in either of two ways, each on wheel
A
A
B
Page 1 7 5
We note for interest that when point A is beneath
left — that is, it is going backwards!
Next, on
its velocity is to the
Point B is constrained to move horizontally; therefore
or
so that
EXAMPLE 3 . 1 6
so the
thatpreceding example, at the same instant,
In
r a d / s . Find the accel­
eration of point B and the angular acceleration of rod fi •
2
2
Solution
The reader is encouraged to mentally locate the
point for £ and from it deduce
thatwe
the
andofvA, are
As
diddirections
with the of
velocity
we correct.
can relate a to the acceleration of either
orC:
2
B
A
Page 176
Now we relate a to a on the rod; note that the acceleration of B is con­
strained by the plane to be horizontal:
A
B
so that
Rolling of a W h e e l
on a Fixed Plane Curve
In the s e c o n d class o f rolling p r o b l e m s to b e c o n s i d e r e d in this section, t h e
csoo nthat
t a c t surface is curved. L e t
and
b e the principal unit t a n g e n t a n d
n o r m a l vectors for t h e c e n t e r point C o f the w h e e l . ( S e e Figure 3.19.)
T h e n , since w e are again defining a m o t i o n s u c h that t h e c o n t a c t point
h a s zero velocity, E q u a t i o n ( 3 . 8 ) yields
w h i c h gives us the velocity o f C. Differentiating E q u a t i o n ( 3 . 2 7 ) , w e get
Figure 3.19 Wheel rolling on curved path
concave upward.
in w h i c h w e h a v e u s e d Equations ( 1 . 4 2 ) a n d ( 1 . 4 5 ) , w h e r e p is t h e instan­
t a n e o u s radius o f curvature o f t h e p a t h o n w h i c h C m o v e s . If this p a t h is a
circle, as is often the case, t h e n p = constant = radius o f t h e circle.
T h e acceleration o f
or
is interesting, a n d it follows from Equation ( 3 . 1 9 ) :
Page 1 7 7
Comparing the accelerations of the contact points
of Figures 3.18
and 3.19, we observe from our results (Equations 3.26 and 3.31) that the
point contacting the curved track has a greater acceleration than the one
touching the flat track, due to the r/p term of Equation (3.31). This term
represents the normal component of acceleration of C, which was zero on
the flat track.
It is also interesting to examine the acceleration of point Q at the top
of the wheel in Figure 3.19:
Note that it is possible for the
(normal) component of AQ to be either
away from or toward C (or even zero), depending on whether r > pat
r <p(orr
= p), respectively (seeFigure 3.20).
Figure 3.20
If the track is concave downward as shown in Figure 3.21, similar
results may be obtained for the accelerations of C,
, and Q. From
Equation (3.29), which still holds:
Figure 3.21
After relating
and ac on
And relating Q to either C or
, we obtain:
gives
inward
Notenormal
the
that
in this
this
component
time
case.the radius
of the
r cannot
acceleration
exceed,of Q is_ seen
is always
to beoutward;
always
Page 178
T h e reader m a y find it useful to r e m e m b e r t h e following form that
doesn't d e p e n d u p o n a n y particular c h o i c e o f coordinate system:
If w e k n e w either a or t h e tangential c o m p o n e n t o f a , for e x a m p l e , w e
could get t h e other w i t h o u t c o n c e r n for expressing t h e n o r m a l c o m p o ­
n e n t o f a . W e n o w consider several e x a m p l e s that feature a b o d y rolling
on a curved surface.
c
c
EXAMPLE 3 . 1 7
The cylinder
shown in Figure E3.17a is rolling on the fixed, circular track with
the indicated angular velocity and acceleration when
is at the bottom of the
track. Rod is pinned to the center C of , and its other end, B, slides on track
Find the velocity and acceleration of B.
Solution
Figure E3.17i
As we have seen in Section 3.5, problems like this one have two parts. The
"velocity part" must be solved before the "acceleration part" because the veloci­
ties as well as angular velocities are needed in the expressions for acceleration.
We shall use instantaneous centers to get v ; the steps are:
B
Figure E3.17b
1.
The contact point of
ure E3.17b).
2.
v is determined as in the diagram from
3.
The velocity of B is vertical (tangent to the path of the point).
4.
Thus
of
is at the intersection of the normals to v and v , namely at
point O (see Figure E3.17c).
is its
point, since the body is rolling (see Fig­
c
c
5.
Then
6.
Finally,
B
rad/s
m/s
Next, to find a we shall relate it to the acceleration of C, which is, from Equa­
tion (3.29),
B
Relating a to a , we get, using Figure E3.17c:
c
B
Figure E3.17c
In substituting for a we have used the facts that for B we have
(direction
vector of v ) and
(always toward the center of curvature of the path of
R
B
Page 1 7 9
the point). Substituting
scalar equations in the two unknowns
we then have two
and a :
2
Therefore
W e wish to m a k e a very important point regarding the preceding
example. Y o u m a y h a v e n o t i c e d that the values o f a a n d
could h a v e
b e e n o b t a i n e d m o r e quickly from
2
and
TTizs shortcut is very dangerous because it is not always valid! It is essential to
u n d e r s t a n d w h e n a n d w h y this procedure works. (It is not b e c a u s e O is
the instantaneous center o f zero velocity o f
!) For a counterexample,
consider the results o f E x a m p l e s 3 . 3 , 3 . 7 , a n d 3 . 1 1 :
fiqure 3.22
If w e were to divide
(see Figure 3 . 2 2 ) , w e w o u l d erro­
n e o u s l y obtain 2 6 . 6 r a d / s e c with a direction indicator
. But a
is
r a d / s e c ! T h e a n s w e r to the question o f w h e n the procedure
is legitimate is covered b y the following text question:
2
2
2
Question 3.15 When can we use
to obtain the correct accel­
eration component of B normal to line i B?
Answer 3.15 From Figure 3.13 we can see exactly when the component of a„ normal to
line AB is given by r a It is when a* has no component normal to AB. (This result does
not require that A be the point
, incidentally.)
AB
Page 180
EXAMPLE 3 . 1 8
2
The velocity magnitude of G in Figure E3.18a is v = t ft/sec, and G moves on
the 8-ft circle in a clockwise direction. The position shown is at t = 2 sec. Find the
acceleration of B at this instant.
G
-Circular rolling surface
Figure E3.18a
Solution
We shall relate a to ac on bar
B
First, we determine a . Since
= 2(2) = 4 f t / s e c . Therefore
in this problem, we have
G
2
From Equation (1)
in which the acceleration of B is an unknown magnitude in a known direction as
signified by the unit vector
/ 1 7 down the slot.
Question 3.16 If the direction of a turns out to be up the slot, will the
solution be valid?
B
Answer 3.16 Yes; this will be manifested by a turning out neeative. Note that (neg­
ative a ) •
/ l 7] is the same as (positive a ) •
B
B
B
Page 181
We have two equations
coefficients) in the three unknowns
a ,ct ,
and
but the angular speed
may be found from the instantaneous center of
. From the geometry and similar triangles, we use Figure E3.18b to obtain:
B
(7) of
2
(intersection of normals to two velocity vectors)
Figure E3.18b
Thus
Note that until
of is located, we do not know whether v is up or down
the slot; however, the normal to v is the same line in either case. Once
is
established, v to the right gives
clockwise — and then we know that v is
down the slot. Substituting, we get
B
B
G
B
Eliminating a gives a = —181, so that
2
Figure 3.23
B
The final example in this part of Section 3 . 6 will b e very helpful to us
later in situations such as t h e one s h o w n in Figure 3 . 2 3 . Suppose w e n e e d
to k n o w the location of point C of cylinder
after
h a s rolled on
to
Page 182
t h e l o w e r position.* T h e p r o b l e m is to find t h e angle t u r n e d t h r o u g h b y
for a given rotation
o f t h e line OQ. T h e procedure to b e followed in
solving this p r o b l e m is illustrated b y t h e n e x t e x a m p l e .
EXAMPLE 3 . 1 9
Find the relationship between the angle (locating the line OQ in Figure E3.19)
and the angle of rotation of the rolling cylinder
Solution
Treating C as a point whose path is a known circle, we get
Alternatively, we may also treat C as a point on the cylinder with instantaneous
center at
Figure E3.19
Thus, equating the two expressions for v , we have
c
Integrating, and letting
when
, we get
or
F r o m E x a m p l e 3 . 1 9 , i f w e let R = 2r, t h e n w e see (Figure 3 . 2 4 ) that
. E v e n t h o u g h t h e circumferences o f cylinder a n d track are
and
the curvature forces t h e angular velocities o f the line O C a n d
the cylinder to b e t h e s a m e . I f t h e outer track w e r e straight a n d o f length
, the cylinder w o u l d turn two revolutions in s p a c e instead o f just o n e
in traversing it.
It is seen that t h e line AC o f ( a n d h e n c e itself) revolves o n c e as C
completes its circular p a t h for the c a s e R = 2r. W h e n R > 2r, t h e n
S u c h a c a s e is s h o w n in Figure 3 . 2 5 . I f w e n o w let R = 7r, t h e n E q u a ­
tion ( 3 . 3 6 ) gives
a n d n o w re v o l v e s o n c e in s p a c e for e a c h 6 0 ° of
turn o f line OC.
Figure 3.24
This need will arise, for example, in kinetics problems in which we seek the work
done by gravity on
if its mass center is offset from its geometric center.
Which at first glance might lead one to believe that the cylinder would revolve twice
per revolution of OC.
Page 183
End (6 revs)
Start
I rev
5 revs
\
1 revs
4 revs
3 revs
Figure 3.25
a n d the wheel would turn in space three times as fast a n d as far as line OC
(see Figure 3 . 2 6 ) .
End
(3 revs
completed)
2 revs
completed
Start -
1 rev
completed
Figure 3.26
Gears
The final class of rolling problems is c o n c e r n e d with gears. Gears are used
to transmit power. The teeth of the gears are cut so that they will give
constant speed to the driven gear w h e n the driving gear is itself turning at
constant angular speed.
H o w e v e r , gears violate the rolling condition; there is necessarily
some sliding since the contacting points do not h a v e equal velocities
(except at
) , as c a n be seen in Figure 3 . 2 7 for spur gears; n o n e t h e ­
less, the teeth are cut so that w e m a y correctly treat the gears for dynamic
Page 184
,Pitch circles
Figure 3.27
purposes as if t h e y w e r e t w o cylinders rolling on e a c h other at t h e pitch
circles. Thus w h e n the centers are pinned as in Figure 3 . 2 7 , w e m a y use
the relation
where r , a n d r are the respective pitch radii of the gears
and
. We
m a y also use the derivative of Equation 3 . 3 7 (since it is valid for all t):
2
W e note that the radius ratio is inversely proportional to the ratio of
angular speeds (and directly proportional to the ratio of n u m b e r s of gear
teeth, since the shape a n d spacing of teeth must m a t c h ) . W e n o w consider
several examples.
EXAMPLE 3 . 2 0
Find the angular speed of the front sprocket (rigidly fixed to the pedal crank) of
the bicycle in Figure E3.20a if the man is traveling at 10 mph. There are 26 teeth
on the front sprocket and 9 on the rear sprocket (which turns rigidly with the rear
wheel). The wheel diameters are 26 in.
Solution
Spokes
not
shown
Chain
Figure E 3 . 2 0 a
The velocities of A and B, the two ends of the straight upper length of chain, are
equal. (As shown in Figure E3.20b, A is just leaving the rear sprocket ; B is just
about to enter the front sprocket
.) To prove this, we note that the translating
section AB of chain is behaving as if rigid, so that, calling this "body"
,
1 Speed
clunker
But
= 0, so that
Page 1 8 5
Next we relate the equal velocities of A and B to the respective centers of their
sprockets
and :
Now the velocities of C and C are each equal to the "velocity of the bike,"
meaning the common velocity of all the points on the translating part of the bike,
such as points of the frame and seat. Therefore v and v cancel, leaving
t
Figure E3.20b
2
Ci
C2
This says simply that (see Figure E3.20c):
Now if the speed of the bike is to be 10 mph, we have
Figure E3.20c
or
Thus
(Radii are proportional to number of teeth!)
So the rider must turn the pedal crank at
rev/sec.
EXAMPLE 3.21
Frame is a fixed ring gear with internal teeth (not shown in Figure E3.21a) that
mesh with those of the planetary gear . The teeth of, also mesh with those of
the sun gear , which is pinned at its center point O to frame The crank arm
is pinned at its ends to O and to the center point P of . The arm
has angular
speed
counterclockwise. Find the angular velocity of
in terms of R, r,
and
Figure E3.21a
Solution
We take to be our reference frame, to which all motions are referred. We work
first with the crank , since we know its angular velocity and the velocity of one
of its points ( v = 0). From the sketch of
(Figure E3.21b), we see that we can
write
0
Figure E3.21b
(Note that we align parallel to OP for convenience; we need not always draw it
to the right.)
Next, the points of
and
that are pinned together at P have the same
velocity at all times. Furthermore, the points of and
at D (see Figure E3.21c)
Page 186
are in contact and each has zero velocity* since D is fixed in
. Thus
or
Solving for
gives
We are now in a position to obtain the velocity of point Q of
contact with the tooth of
Figure E3.21 c
Substituting for
Pitch circle
in terms of
, the point in
from Equation (2), we get
Note that Q has twice the speed of P since it is twice as far from the instantaneous
center D of
as is P.
Finally, we come to the body
of interest (see Figure E3.21d). Knowing that
the points Q and Q' (the respective tooth points in contact on
and
) have
equal velocities as they move together tangent to the pitch circle, we obtain
Thus the angular speed of
in
is
Figure E3.21d
and the angular velocity of
in
is
N o t e in t h e previous e x a m p l e that t h e point o f
passing o v e r point
Q h a s velocity
, w h i c h is o f course less t h a n t h e velocity v o f t h e gear
teeth in c o n t a c t at Q. T h i s velocity is
, w h i c h is m o r e
Q
t h a n twice as fast as
W e also r e m a r k that since t h e a n s w e r s
• As we have pointed out, the contact points of gear teeth necessarily slide relative to
each other. The points used in the analysis are actually not tooth points, however, but
imaginary points on the pitch circles of the gears. Furthermore, the radii given in the ex­
amples and problems are the radii of these circles.
Page 187
are c o m p l e t e l y general functions o f time, the angular accelerations are
obtainable i m m e d i a t e l y b y differentiation, w i t h
:
R a t h e r t h a n differentiating, h o w e v e r , w e shall obtain t h e s e t w o results in
t h e following e x a m p l e in a n o t h e r m a n n e r : b y r e p e a t e d u s e o f Equation
( 3 . 1 9 ) . T h e purpose is to gain insight into its u s e in gearing situations
involving several b o d i e s . T h e procedure in t h e n e x t e x a m p l e w o u l d h a v e
to b e followed if t h e previous e x a m p l e h a d b e e n w o r k e d using instanta­
n e o u s values instead o f generally (with s y m b o l s ) .
EXAMPLE 3.22
Find the angular accelerations a , and a of the planetary and sun gears in the
previous example in terms of R, r, and
and a , which are given functions of the
time t.
3
2
Solution
Relating the accelerations of P and 0 on body
E3.22a):
gives (see the bar in Figure
This acceleration is then carried over to the coincident point P on
(again see
Figure E3.22a). Relating D and P on the planetary gear
, we have
Recalling that
the coefficients of
then give
Figure E3.22a
and the
coefficients yield
We now need a , where Q is again the tooth point of
gear
:
Q
in contact with the sun
Page 188
We now go to body
to complete the solution. Relating the tooth point Q' to O
on
gives the components of
. (see Figure E3.22b). The tangential accelera­
tion components of Q and Q' are equal* as the teeth contact and move together:
Thus
Figure E3.22b
N o t e that w e m a y express the n o r m a l acceleration c o m p o n e n t o f Q '
in terms o f
b y using t h e result for
from E x a m p l e 3 . 2 1 :
W e also n o t e that w e could h a v e alternatively o b t a i n e d the accelera­
tions o f D a n d Q as points o n rims o f w h e e l s rolling o n curved tracks by
using Equations ( 3 . 3 1 ) a n d ( 3 . 3 2 ) . Noting that p for P is (R + r), w e
present these partial c h e c k s on our solution:
* This is in fact true even when neither body's center is fixed and the geometry is irregu­
lar. As long as there is rolling, the acceleration components of the contacting points in
the plane tangent to the two bodies are equal in plane motion at all times. See "Contact
Point Accelerations in Rolling Problems," D. J. McGill, Mechanics Research Communica­
tions, 7(3), 1 7 5 - 1 7 9 , 1980.
Page 189
•
PROBLEMS
Section 3.6
3.84 The wheel in Figure P3.84 rolls on the plane with
constant angular velocity
rad/ sec. Find the velocity of
point Q by using the instantaneous center
of zero
velocity. Then check by using Equation (3.8) to relate v
to v .
c
P
3.87 Figure P3.87(a) shows the manner in which a train
wheel rests on the track. If the train travels at a constant
speed of 80 mph and does not slip on the track, determine
the velocities of points A, B, D, and £ on the vertical line
through the center C in Figure P3.87(b). Which point is
traveling backward? Why?
Figure P3.84
W
3.85 In the preceding problem, suppose that the plane
on which the wheel rolls is not fixed to the reference
frame but instead translates on it (this time the reference
frame is ) at constant velocity 3 f t / s e c to the left. (See
Figure P3.85.)
a. Find the instantaneous center of zero velocity
b. Find v
Q
again.
(b)
Figure P3.87
Figure P3.85
3.86 The wheel in Figure P3.86 rolls on the bar. If at a
certain instant the bar has a velocity of 2 m / s to the right
and the wheel has counterclockwise angular velocity of
0.5 rad/s, determine the velocity of (a) the center of the
wheel and (b) point P.
Figure P3.86
3.88 In the preceding problem find the velocities of
points F, G, and H.
3.89 Find the velocities of points B and C in Figure P3.89
if the cylinder does not slip on the translating bodies
and
Figure P3.89
Page 1 9 0
Figure P3.92
Figure P3.90
3.90 Two men, a tall one and a short one, travel up
identical inclines, pulling identical spools by means of
ropes wrapped around the hubs. (See Figure P3.90.) The
men travel at the same constant speed
, and the ropes
are wrapped in the opposite directions indicated. If the
spools do not slip on the plane, one of the men will be run
over by his own spool. Prove which one it is, and show
how long it will take, from the instant depicted, for the
spool to roll over him.
3.91 A cylinder of radius r rolls over a circular arc of
constant radius of curvature R (see Figure P3.91). What is
the ratio of the angular speed of the cylinder to
Figure P3.93
eration
(These values are v and a for all points in the
body of the tank and for the centers of its wheels.) Find
the velocities of the five points P , P , P , P , and P if
there is no slipping. The wheels have radius R.
1
2
3
4
5
3.94 The wheel rolls on both
and
(See Figure
P3.94.) The constant angular velocity of the wheel in
frame is shown in the figure. Find:
a. The velocity of the points of
relative to
b. The constant velocity of C in for which the
velocities of T (on ) and B (on ) in are
equal in magnitude and opposite in direction.
3.92 A disk with diameter 1.2 m rolls along the plane as
indicated in Figure P3.92. Its center point C has velocity
Figure P3.94
where t is the time in seconds. Find the velocity of the
point that lies 0.3 m directly below C when ( a ) f = 2 s and
(b) t = 5 s.
3.93 The tank shown in Figure P3.9 3 is translating to the
right, and at a certain instant it has velocity
and accel­
3.95 An inextensible string is wrapped around the cylin­
der in Figure P.3.95, fitting in a small slot near the rim.
The center C is moving down the plane at a constant
speed of 0.1 m / s . Find the velocities of points A, B, D, and
Page 1 9 1
E. Hint: The cylinder is not rolling on the plane, but it is
rolling on
?.
3.98 The cylinder in Figure P3.98 is rolling to the left
with constant center speed
. A stick is pinned to the
cylinder at B, and its other end A slides on the plane. Find
the velocity of A when 0 = 0 ° , 9 0 ° , 180% and 2 7 0 ° .
Figure P3.98
Figure P3.95
3.99 Wheel
in Figure P3.99 has angular velocity 3
rad/sec. Find the angular velocities of wheel
and the
bent bar
3.96 At the instant shown in Figure P3.9 6, point B of the
block (to which rod is pinned) has
ft /sec.
Find the angular velocity of the rolling cylinder.
Figure P3.99
Figure P3.96
3.97 Figure P3.97 shows a circular cam and an oscilla­
ting roller follower consisting of the roller
(which rolls
on ) and the follower bar
. If the cam rums at the
constant angular velocity 0.3 rad/s, find the angular
velocity of the follower bar and of the roller at the given
instant.
Figure P3.97
* 3.100 The cart
in Figure P3.100 travels from left to
right, with its rear wheels
rolling at constant angular
velocity 0.2 rad/sec. The front wheels
are rolling up
the parabolic surface shown. The wheels have radius
0.4 m, and are pinned to the cart. Find the angular veloc­
ity of cart
at the given instant. Hint:
Figure P3.100
Page 1 9 2
3.101 The constant angular velocities of the ring gear
and the spider arm
shown in Figure P3.101 are 2
r a d / s and 10 rad/s, respectively. Determine the angu­
lar velocity of gear
and the velocity of the point of
having maximum speed in the given position. The centers
of
and
are pinned to the reference frame
3.105 The shaded arcs on
and
(Figure P3.105(a)) are
always equal if the two bodies are in rolling contact; how­
ever, the converse is not necessarily true. Just because the
contacting arclengths are equal does not mean that
rolls
on
. For the wheel
on the plane
shown in Figure
P3.105(b), give constant values
and
for which the
arclengths of contact are equal but the velocities of the
contact points are not. Hint: Look at the shaded arcs on
and
in Figure P3.105(b).
Figure P3.101
3.102 The rod which is pinned to cylinder , translates
upward in the y-direction at the constant speed 4 ft/sec
(see Figure P3.102). The rod
is pinned to the reference
frame at O and rests against the rim of as shown in the
figure. There is rolling contact between
and . Find the
angular velocities of
and
at the instant shown.
3.103 If the given velocities of P and Q in Problem 3.89
are constant, find the accelerations of C and B.
3.104 If, in Problem 3.89, the respective accelerations of P
and Q are
, find the angular ac­
celeration of the cylinder.
Figure P3.102
Figure P3.105
3.106 If in Problem 3.99 the angular acceleration of
at the given instant, find and
that time.
is
at
3.107 In Problem 3.93 find the accelerations of the same
five points P , P , P , P , and P . (See Figure P3.107.)
1
2
3
4
5
3.108 During startup of the two friction wheels (see Fig­
ure P3.108), the angular velocity of is
rad/
sec. Assuming rolling contact, compute the acceleration
of the point T, which is at the top of
when f = 3 sec.
Figure P3.107
Page 1 9 3
3.112 In Problem 3.88 find the accelerations of points F,
G, and H.
3.113 A moment applied to gear
in Figure P3.113 re­
sults in a constant angular acceleration
rad/
sec . The other gear,
, is fixed in the reference frame.
Determine:
2
a. The time required for C to return to its starting
point after one revolution around
from rest
Figure P3.108
3.109 The wheel in Figure P3.109 rolls, its center having a
constant velocity of 10 ft/sec to the right. Find
at
the instant shown.
b. The number of revolutions turned through in
space by
during the complete revolution.
3.114 The center of the rolling wheel in Figure P3.114
moves to the right at a constant speed of 10 in./sec. The
bar is pinned to the wheel at A, and end B always stays in
contact with the ground. Find the acceleration of B at the
instant shown.
3.115 Calculate the acceleration of the instantaneous
center of rod
in Example 3.15.
Figure P3.109
3.110 A wheel rolls on a 10-cm-radius track. (See Figure
P3.110.) At the instant shown the wheel has an angular
velocity of
r a d / s and an angular acceleration of —
r a d / s . At the instant shown, find:
2
a. The velocity and acceleration of C
b. The velocity and acceleration of A
c.
Figure P3.114
d. The center of curvature of the path of point T.
3.111 In Problem 3.87 find the accelerations of points A,
B, C, D, and E.
Figure P3.110
3.116 Gear
and crank
have angular speeds
and
angular acceleration magnitudes
at the instant shown
in Figure P3.116 in the indicated directions. Find the an­
gular velocities and angular accelerations of gears
and
at the same time, if
is pinned to ,
and
Figure P3.116
Page 1 9 4
• 3.117 The center C of the small cylinder
in Figure
P3.117 has a speed of 0 . l t m / s as it moves clockwise on
a circle. Body
rolls on the large cylinder
. In the
position given in the figure, t = 10 s. Find the accelera­
tion of point B of the stick that is in contact with
at the
given instant.
2
3.120 Point A of the slider block has, at the instant
shown in Figure P3.120,
and a„
=
. Find the angular acceleration of bar
3.121 Two 5-in.-radius wheels roll on a plane surface.
(See Figure P3.121.)A13-in. bar is pinned to the wheels
at A and B as shown. If C has a constant velocity of 20
ft/sec to the right, find, for the position shown, the accel­
eration of A.
Figure P3.121
Figure P3.117
3.118 In Problem 3.89 let
and a
=
at the instant shown. Again assuming no
slipping, find the accelerations of B and C.
Q
3.119 The two identical cylinders
and
(Fig­
ure P3.119) are connected by bar
(which is pinned to
their centers), and they roll on the surface as shown. If the
angular velocity of
is
, find the
angular accelerations of both
and
at the given in­
stant.
3.122 See Figure P3.122. The velocity of the pin in block
is
and its acceleration is
in the
given position. Find at this instant the angular velocity
and angular acceleration of the cylinder if there is suffi­
cient friction to prevent it from slipping.
3.123 A wheel rolls along a curved surface. In the position
shown in Figure P3.123, its angular velocity and angular
acceleration are
and
. De­
termine at this instant the angular acceleration of bar
and the acceleration of pin B of the slider block.
Figure P3.119
Figure P3.122
Figure P3.120
Figure P3.123
Page 1 9 5
3.124 In Example 3.14 find the piston acceleration when
, 180°, and 2 7 0 ° .
3.125 The center point C of gear
in Figure P3.125
moves in a horizontal plane at constant speed . The ring
gear is fixed in the reference frame, and the constant
angular velocity of
is clockwise. Find the acceleration
of point Q of
Figure P3.127
Figure P3.125
3.126 If the ball in a ball bearing assembly (Figure P3.126)
neither slips on the shaft nor on the fixed housing, find
the velocity and acceleration of the center of the ball in
terms of the angular velocity and angular acceleration of
the shaft
3.128 The angular velocity of crank
in Figure P3.128 is
a constant 3 r a d / s . In the given position, find the ve­
locity of the center C of wheel
and also determine the
angular acceleration of
, which rolls on the circular
track.
3.129 Cylinders
and
in Figure P3.129 have a radius
of 10 in. each and roll on the respective planes. Bar has
length 48 in. and is pinned to the centers of the cylinders.
The center G of
has velocity
. If
the time at the instant shown is t = 5 sec, find and
at
that instant.
Figure P3.126
3.127 The ball in Figure P3.127 rolls on the fixed surface
and at the instant shown has angular velocity
r a d / s and angular acceleration
. At this
instant find:
Figure P3.128
a. The velocities of A and B
b. The accelerations of A and B
c.
Figure P3.129
Page 196
3.130 Bar in Figure P3.130 is 25 cm long and is pinned to
the rolling cylinder at B. The other end of is pinned to
the roller at A as shown. The center of the cylinder has
= 11.2 c m / s and a = 16.8 c m / s down the plane at
the given instant; at this time line
is vertical and
is horizontal, and BC is parallel to the plane beneath it.
Find the acceleration of point A and the angular accelera­
tion of body at the given instant.
2
c
• 3 . 1 3 4 Gears
and
in Figure P3.134 have 25 and 50
teeth, respectively. Rod is 2 ft long, and the radius of
is 1 ft. Determine the acceleration of point A when t = 0 if
x = 0.2 sin
ft (positive to the left) with
= 90° at
t = 0.
B
Figure P3.134
• 3.135 At the instant shown in Figure P3.135, bar
has
rad/sec and
r a d / s e c ; and the
gear
has
=
rad/sec and
=
rad/sec .
At this instant, determine the accelerations of each of the
two gear tooth contacting points.
2
2
Figure P3.130
3.131 Figure P3.131 shows a 10-ft-radius disk that rolls
on a plane surface. It has, at the instant shown, an angular
velocity of 2 rad/sec and an angular acceleration of 3
rad/sec , both counterclockwise. Find a point on the disk
or the disk extended that has zero acceleration at this
instant.
2
3.132 Find the acceleration of point B, the pin connecting
rod to the blockin Figure P3.132, at t = 1 sec. Therodis
horizontal at f = 0, and the velocity of C is v = 2t
ft/sec.
2
c
• 3.136 The outer gear
in Figure P3.136 is stationary.
Crank
turns at the constant angular velocity of 10 r a d /
sec counterclockwise and is pinned at its ends to the
centers of the sun gear
(at S) and the planetary gear
(at P). Find the accelerations of the points of and
that
are in contact with each other, if the radii of and
are,
respectively, 3 in. and 10 in.
• 3.137 Point O is pinned to the reference frame
(See
Figure P3.137.) The pitch radii of gears
and
are each
0.2 m. The angular velocities of
and
are 2 rad/s,
clockwise for
and counterclockwise for
, and both
constant. Find the maximum acceleration magnitude ex­
perienced by any point of
Figure P3.135
Figure P3.131
Figure P3.132
3.133 Referring to Problem 3.121, if at the instant shown
v = 20
i n . / s e c and a = 5
i n . / s e c , find a at this
time.
2
c
c
A
Figure P3.136
Page 1 9 7
c. Observe that P is at its highest point when
and that
there. In this configura­
tion, show that the following two expressions
for the acceleration of P agree:
• 3.140 Show that for a rigid body in the plane motion, as
long as
there is a circle of points P of
whose
accelerations pass through any point C of
Hint: Write
and dot both sides with the
vector
Assume that
and see if you can
exhibit r . For the rolling wheel, show that the points are
as shown in Figure P3.140.
CP
Figure P3.137
• 3.141 Show that for the rolling uniform cylinder in Figure
P3.141 there is a point of zero acceleration at the indi­
cated position J if and are in the given directions and
are not both zero. You should find that the coordinates of
J are
Figure P3.138
• 3.138 The wheel in Figure P3.138 rolls on the plane. Find
the radius of curvature and the center of curvature of the
path of point T at the given time in terms of r.
• 3.139 A cycloid is the curve traced out by a point on the
rim of a rolling wheel. The equations for the rectangular
coordinates of a point on the cycloid, in terms of the
parameter (the angle shown in Figure P3.139), are
Figure P3.139
where a is the wheel's radius. Recall from calculus that
the curvature of a plane curve is
Points accelerating
through C
Figure P3.140
where p is the radius of curvature.
a. Use the chain rule
and show that, for the cycloid,
b. Explain what the minus sign in the expression
for
means.
Figure P3.141
Page 198
3.7
Relationship Between the Velocities of a Point with
Respect to T w o Different Frames of Reference
W h i l e Equation ( 3 . 8 ) gives us the relationship b e t w e e n the velocities in
of t w o points o f the same rigid b o d y , w e often n e e d a n o t h e r equation
relating the velocities o f the same point relative to t w o different frames or
bodies. This relationship ( t o g e t h e r w i t h a c o m p a n i o n equation for accel­
erations to b e developed in the n e x t section) will b e essential in solving
kinematics p r o b l e m s involving bodies m o v i n g in special w a y s relative to
others (such as a pin o f o n e b o d y sliding in a slot o f another).
Relationship Between the Derivatives of a Vector in T w o Frames
T o develop this equation, w e m u s t first find the relationship b e t w e e n the
derivatives o f an arbitrary vector A in t w o frames a n d
T o do this, w e
begin b y e m b e d d i n g axes X a n d Y in
a n d x a n d y in B, as suggested b y
the h a t c h m a r k s in Figure 3 . 2 8 . Further, w e let
and
b e pairs of
unit vectors, always respectively parallel to (X, Y ) a n d to (x, y).
W e n o t e that if
again locates the x axis relative to X as s h o w n ,
then
and
Figure 3.28
Differentiating in frame
obtain
a n d noting that
and
are constants there, w e
N o w let the arbitrary vector A b e written in frame (meaning that A
is expressed in terms o f its c o m p o n e n t s t h e r e — i . e . , in terms o f unit
vectors fixed in direction in ):
Differentiating this vector in
w e get
W e n o w n o t e that the first t w o terms o n the right side o f Equation ( 3 . 4 3 )
add u p to the derivative o f vector A in
b e c a u s e a n d do n o t c h a n g e in
magnitude or direction w i t h time there. T h u s
Substituting the derivatives o f
( 3 . 4 2 ) yields
and
in
from Equations ( 3 . 4 1 ) and
Page 1 9 9
or
(3.44)
T h e angular velocity
h a s reappeared, a n d Equation ( 3 . 4 4 ) s h o w s
us that this vector h a s a m o r e general purpose t h a n m e r e l y relating
velocities in kinematics. It is in fact the link that allows us to relate the
derivatives of a n y vector in two different frames. (This s a m e result is in
fact true in general three-dimensional motion, w i t h three-dimensional
vectors a n d a m o r e general expression for angular velocity substituted, as
will b e seen in C h a p t e r 6.)
Velocity Relationship in T w o Frames
W e n o w use Equation ( 3 . 4 4 ) to relate the velocities o f a point P in t w o
frames a n d
F r o m Figure 3 . 2 9 , the position vectors of P in t h e s e two
frames are related b y
(3.45)
Differentiating this equation in
we have
(3.46)
Figure 3.29
T h e first t w o vectors in Equation ( 3 . 4 6 ) are the velocities o f P a n d
b y definition:
in
(3.47)
Question 3.17 (a) Why is the last vector in Equation (3.47) not the
velocity of P in
(b) Why is it not the velocity of P in
T o replace
tion ( 3 . 4 4 ) , with
b y a vector that w e can operate with, w e use E q u a ­
b e c o m i n g the vector A :
(3.48)
Therefore, recognizing that
into ( 3 . 4 7 ) , w e obtain
is
a n d substituting Equation ( 3 . 4 8 )
(3.49)
A n o t h e r w a y o f expressing Equation ( 3 . 4 9 ) is to think o f frame as a
" m o v i n g f r a m e " with respect to a " f i x e d " reference frame
T h e n the
velocities of P can b e written as simply v w h e n the reference is (thus
and as v w h e n the reference is the " m o v i n g f r a m e "
(thus
P
r e l
Answer 3.17 (a) It is not
because the origin of the position vector is not fixed in
(b) And it is not
because the derivative is not taken in
Page 200
H e n c e w e c a n write E q u a t i o n ( 3 . 4 9 ) in abbreviated notation
as
(3.50)
where
the position of P in t h e m o v i n g frame, a n d
is the
angular velocity o f relative to frame
T h e reader m a y find this form of
Equation ( 3 . 4 9 ) easier to use w h e n there is just o n e " m o v i n g f r a m e . "
N o w let us d e n o t e b y
the point o f (or e x t e n d e d ) that is coin­
cident with P. T h e n
a n d t h e last two terms o f Equation ( 3 . 4 9 )
or ( 3 . 5 0 ) are s e e n ( b y E q u a t i o n 3 . 8 ) to b e the velocity of
in
(3.51)
In words, Equation ( 3 . 5 1 ) says that at a n y time, t h e velocity o f P in
is
the s u m o f the velocity o f P in
plus the velocity in
o f the point o f
coincident with P.*
As a preliminary e x a m p l e , consider Figure 3 . 3 0 , in w h i c h pin Q is
m o v i n g to the right. Let P b e the center o f the o t h e r pin, w h i c h is attached
to frame
T h e n w e m a y write
Figure 3.30
N o w the center o f the
m o t i o n in is necessarily along a straight line
within the slot o f
T h e r e f o r e the velocity o f
(the point o f
extended
coincident with P ) is seen to be also parallel to the slot a n d in a direction
opposite to that o f
. W e n o w consider several detailed e x a m ­
ples of t h e use o f Equation ( 3 . 4 9 ) .
EXAMPLE 3 . 2 3
A yellowjacket walks radially outward at a constant 2 i n . / s e c in a straight line
relative to a record turning at
(See Figure E3.23.) Find the velocity of the
yellowjacket in frame
which is the cabinet on which the stereo rests.
Solution
We shall treat the yellowjacket as a point Y, and we note that the unit vectors in
Figure E3.23 are fixed in our "moving frame"
Also, the "moving"
and
"fixed" (O) origins are coincident. Then we have, using Equation (3.49):
Note how the second term grows linearly with the radius.
Figure E3.23
* This latter term is sometimes called the vehicle velocity of P.
Page 2 0 1
In t h e n e x t e x a m p l e , there is m o r e t h a n o n e " m o v i n g f r a m e . " T o
avoid three levels o f subscripts, w e shall n a m e t h e bodies
rather t h a n the usual
Thus
becomes
and
a n d so on.
EXAMPLE 3 . 2 4
Collar in Figure E3.24 is pinned to rod at P and is free to slide along rod
The angular velocity of is 0.2 r a d / s at the instant shown. Find the angular
velocity of at this time, and determine the velocity of P relative to
Solution
We relate the velocities of P in
and in
Figure E3.24
Thus the pin is moving outward on
Question 3.18
Will
which is turning clockwise.
always have the same
as does
N o t e that in t h e preceding e x a m p l e , w e did n o t n e e d to k n o w the
angular velocity o f N o n e t h e l e s s , it is important for t h e reader to realize
that
. T h e reason is that a n d can o n l y translate relative to each
other; t h u s lines fixed in e a c h will turn at the s a m e time rates in
This
observation will s o m e t i m e s be n e e d e d (as in P r o b l e m s 3 . 1 4 7 a n d 3 . 1 4 9 ) .
EXAMPLE 3 . 2 5
Block translates in a horizontal slot (see Figure E3.25a) and is pushed along by a
bar that turns at angular velocity
rad/sec about the pin at point O.
Find
the velocity of the contact point of
when
=60°.
Figure E3.25a
Answer 3.18
Definitely not! This happened because of the geometry at the given instant.
Page 202
Solution
Let the ground be the reference frame
and note that T is the point of
coincident with Q at the given instant (the point we have been calling in the
theory). Using Equation (3.51), we obtain
Therefore
(1)
Note that the velocity of Q relative to bar
has an unknown magnitude
but a known direction (along ). Now we also know the direction of
so that
(2)
Equating the x components of Equation (2), we get
(3)
And equating the y components:
(4)
Equation (4) gives
Figure E3.25b
= 13.3 ft/sec, from which Equation (3) then yields
The correct triangle relating the velocities of Q and T is shown in Figure E3.25b.
As a check,
and
ft/sec.
Question 3 . 1 9
Can the preceding example also be worked by using
Answer 3.19 Yes, provided we recognize that the direction of
is along the axis of
the rod (see Problem 3.146). That is,
must be tangent to the surface at which and
touch. It is important to realize, however, that while the path of Q in is a straight
line, the path of T in is not. Thus
would not be in the direction of the axis of
EXAMPLE 3 . 2 6
Figure E3.26a
Disk of Figure E3.26a, with its attached pin P, has limited angular motion. After
a 45° clockwise rotation from the original position (see Figure E3.26b), disk has
angular velocity
rad/sec. At this time, find of the slotted triangular
body and determine the velocity of pin P relative to
Page 2 0 3
Figure E3.26b
Solution
Calling the ground frame
we relate the velocities of pin P in
and in
(1)
where
is the point of coincident with P.
We may find
by relating it on body
to
(which vanishes):
0
To
find
we need the orientation of the slot. At 4 5
, the configuration
is as shown in Figure E3.26c. Note that the slot center S moves on a circle about 0
and that a tangent to this circle at S must pass through P at all times, since P must
stay within the slot. From the diagram at the left we get, from geometry and
trigonometry,
Figure E3.26c
Using
to form
in Equation (1), we have:
in which we have used the fact that we know the direction but not the magnitude
of
(It moves in the slot at the angle calculated earlier.) Further, relating the
velocities of points
and O on
we have
Page 204
or
Substituting, and equating the coefficients of
and of
we get
In w o r k i n g out the following p r o b l e m s , the student is urged to begin
Solving
these carefully
equations about
gives the =selection
1.61, so that:
by
thinking
o f a point w h o s e velocities in
t w o frames are to b e related with E q u a t i o n ( 3 . 4 9 ) .
PROBLEMS
Section 3.7
3.142 Boat in Figure P3.142 departs from A and is sup­
posed to arrive at point B some 100 ft downstream and on
the other side of a river with a current of 5 ft/sec. If can
move at 10 ft/sec relative to the water, and if it travels on
a straight line from A toward B, how long will it take?
3.144 Cylinder in Figure P3.144 rolls on a circular sur­
face. When it is at the lowest point of the circle, its angular
velocity and acceleration are
=0.2
r a d / s and
= 0.02 r a d / s . Rod is pinned to at
and is also
pinned to a block at P that slides in the slot of
The
constant angular velocity of
is 0.3
rad/s. Find the
velocity of P in and the angular velocity of at the given
instant.
2
Figure P3.142
3.143 Bar
in Figure P3.143 is turning clockwise with
angular speed 0.25 rad/s, pushing bar as it goes. Find
at the given instant.
Figure P3.144
3.145 Rod is pinned to the ceiling at A and slides on
wedge
at B. (See Figure P3.145.) The wedge moves to
the right with constant velocity of 5 ft/sec. Find the an­
gular velocity of the rod in the position shown.
Figure P3.143
3.146 Referring to Example 3.25 for the meanings of the
symbols, show that
Page 205
3.149 Rod in Figure P3.149 has angular velocity 5
rad / s. It is pinned to another rod which passes through
a slot in as shown. At the given instant, find the angular
velocity of body
and the velocity of any point of
relative to
Hint: All points of translate in
—what
does this mean about the angular velocities of and ?
3.150 A mechanism consists of crank
pinned to
rocker pinned to , and a small body that is pinned to
and slides in the slot of
(See Figure P3.150.) The
length of is
, where D is the distance between O
and
. If has constant angular velocity
over a
range of its motion, find
when: (a)
Figure P3.145
3.147 Bar slides through a collar in body (see Figure
P3.147) and is pinned at P to a second bar Both and
are pinned to the reference frame as shown, and rotates
with limited motion at constant angular velocity
= 1 r a d / s counterclockwise. Find the angular velocity of
when point P is at the top of the circle on which it
travels.
3.151 Collars
and
in Figure P3.151 are pinned to­
gether at C, and they slide on rods and
respectively.
Rod
has a constant angular velocity
for
45°. Find the velocity of point C relative to
as a
function of D, , and in this range of angles.
3.148 Plank slides on the floor at A and on block at
Block
moves to the right with a constant velocity of
6 ft/sec while end A moves to the left with a constant
velocity of 4 ft/sec. For the position shown in Figure
P3.148, find the angular velocity of the plank.
Figure P3.149
Figure P3.150
Figure P3.147
Figure P3.148
Figure P3.151
Page 206
3.152 Figure P3.152 shows a circular cam
and a flatface translating follower
If
rotates with constant
angular velocity
find the maximum velocity in ref­
erence frame of any point of , in terms of
and the
offset distance
3.155 Rods and in Figure P3.155 are pinned at O and
to a reference frame
Rod
is also pinned to the
slotted body at B. The upper end of is pinned at P to a
roller that moves freely in the slot of The angular veloc­
ities of rod and link are constants:
Determine the velocity of P in
of at the given instant.
and the angular velocity
Figure P3.152
3.153 Figure P3.153 illustrates a "Geneva mechanism,"
in which disk
is driven with a constant counterclock­
wise angular speed and produces an intermittent (starting
and stopping, waiting, then repeating) rotational motion
of the slotted disk Pin P is fixed to
and drives disk
by pressing on the surfaces of the slots. Show with the use
of Equation (3.49) that disk will have zero angular speed
in the two positions shown, a varying angular speed in
between these positions, and zero angular speed while P
is returning to the = 135 ° position. Note that the oper­
ation of the mechanism requires that the distance be­
tween O and
be
Figure P3.155
* 3.156 In Figure P3.156 collar isfixedtoarm
andslides
along rod
Arm
is pinned to a second collar at A;
this collar slides on rod At the given instant,
=0.2
rad/s,
= 0 . 1 rad/s, and the velocity of all points of
relative to rod is 0.3 m / s outward (along OC). Give
the angular velocity of by inspection, and find the ve­
locity of the points of relative to
Figure P3.156
Figure P3.153
• 3 . 1 5 4 In the preceding problem let R = 0.1 m and
= 5 r a d / s = constant. Find the angular velocity of the
slotted body at the instant when
=160°.
3.157 Show that the rigid-body velocity equation (3.8)
can be derived from Equation (3.49). Hint: Fix P to and
the equation will relate the velocities in of the two points
P and
of
Page 2 0 7
3.8
Relationship Between the Accelerations of a Point with
Respect to T w o Different Frames of Reference
W e shall n o w derive the c o m p a n i o n equation to ( 3 . 4 9 ) , this o n e relating
the accelerations o f P in
and
Differentiating Equation ( 3 . 4 9 ) in
w e get
(3.52)
Using Equation ( 3 . 4 4 ) to " m o v e the derivative" in the first a n d last terms
o n the right side o f Equation ( 3 . 5 2 ) gives
(3.53)
Recognizing that
a n d that
terms, from Equation ( 3 . 5 3 ) w e obtain
a n d rearranging
(3.54)
in w h i c h
as w e h a v e already seen in Section
3.5.
T h e parenthesized term in Equation ( 3 . 5 4 ) is seen from Equation
( 3 . 1 9 ) to b e
w h e r e , as before,
is the point o f (or extended) that
is coincident with P. T h e r e f o r e w e h a v e the following for our result:
(3.55)
In words: T h e acceleration o f P in equals its acceleration in
plus the
acceleration in
o f the point o f coincident with P, plus the C o r i o l i s
acceleration,
T h e Coriolis acceleration is seen to provide an
u n e x p e c t e d b u t essential difference b e t w e e n the forms o f Equations
(3.51) and (3.55).
Question 3.20 If we differentiate Equation (3.51) instead of (3.49), we
might (erroneously!) obtain
and we come up (incorrectly) short by half on the Coriolis term. What is
wrong with this approach?
In the s a m e procedure w e u s e d for velocities, w e can simplify the
notation o f Equation ( 3 . 5 4 ) if there is b u t o n e " m o v i n g f r a m e "
in­
volved, w h i c h is in m o t i o n relative to the reference frame
(3.56)
Answer 3.20 The error is that the derivative
is not equal to
because
notes a succession of points of which are at each instant coincident with P.
de­
Page 2 0 8
In this equation, a a n d
are t h e respective accelerations o f P in a n d in
T h e vectors
and
are the angular velocity a n d angular acceleration
of in (equal to
and
)and
(the position vector o f P in t h e
P
" m o v i n g f r a m e " ) . W e n o w consider several e x a m p l e s s h o w i n g t h e use o f
Equations ( 3 . 5 4 ) a n d ( 3 . 5 5 ) .
EXAMPLE 3.27
Find the acceleration in frame
of the yellowjacket of Example 3.23.
Solution
Using Equation 3.54, we obtain:
The
in this example is the Coriolis acceleration. Note that the yellow­
jacket has two nonzero acceleration components, even though both and are
zero in this example.
EXAMPLE 3 . 2 8
If in Example 3.24 we have the additional data that
given time, find
and
(See Figure E3.28.)
at the
Solution
Relating the accelerations of P in
and
we have
in which the answers to Example 3.24 are used in the Coriolis term. Filling in the
variables, we get
Figure E3.28
Page 2 0 9
Thus point P is slowing down as it moves outward along
down as it rotates clockwise.
and
is slowing
T h e reader should note in the preceding e x a m p l e that
and
are
identical. T h e sleeve forces the two bodies a n d to translate relative to
e a c h other, so that, as w a s pointed out at the e n d of E x a m p l e 3 . 2 3 ,
By differentiating this equation, w e see that the angular accel­
erations are also always equal.
EXAMPLE 3 . 2 9
In Example 3.25, suppose that at the given instant ( = 6 0 ° ) all the data are the
same and in addition
= 3 0 rad/sec . Find the acceleration of block (see
Figure E3.29).
2
Solution
Note that if we again use Q as our point (of the block ) that is moving relative to
two bodies ( and ), we know the direction of
(It is along since point Q
moves on a straight line in )
Equation (3.55) thus gives
Figure E3.29
Noting that point
is the point
we obtain
This result yields the following scalar component equations:
Thus the acceleration of the translating block is:
Page 2 1 0
EXAMPLE 3 . 3 0
2
If in Example 3.26 we add
= 10
r a d / s e c to the data, find the angular
acceleration of and the acceleration of P relative to
(See Figures E3.30a,b.)
Figure E3.30a
Figure E3.30b
Solution
Again we apply Equation (3.55):
and again we equate the coefficients, and then the
scalar component equations:
Solving these equations results in
coefficients, to obtain the
2
= 5 . 1 2 r a d / s e c , so that:
Page 2 1 1
EXAMPLE 3 . 3 1
Pin P in Figure E.3.31 is attached to cart and slides in the smooth slot cut in
wheel
The wheel rolls on the rough plane
The cart's position is given by
= 0.3t , with
in meters when t is in seconds. Find
and
at the given
instant (which is at t = 3 s), and determine the acceleration of P in the slot at this
time.
2
Figure E3.31
Solution
2
Since
= 0.3t , we have
and
acceleration vectors of all points of the cart
and
Relating the velocities of P in frames
these are the velocity and
in particular of P. At t = 3, we have
and
we obtain
Relating the accelerations using Equation (3.54), we get
Note that we have related
(the point of -extended coincident with P) and the
center of (call it C) to get
Solving, we obtain
Page 2 1 2
Thus the acceleration of P in the slot (which is its acceleration in ) is
m / s ; it is upward because we assumed it to be in the positive y direction
and got a negative answer. Note that although P is momentarily stopped
in the slot (
= 0 ) , it has to be "getting ready" to move outward since is
translating to the right and is rolling that way. This is what is indicated by
= 1.8 m / s .
1.8
2
2
Section 3.8
PROBLEMS
3.158 A bug
is crawling outward at a uniform speed
relative to the rotating arm of 3 f t / sec. In the position
shown in Figure P3.158, for the arm
= 2 rad/sec and
= 4 rad/sec , both counterclockwise. What is the ac­
celeration of the bug? Indicate the direction in a sketch.
2
3.160 Bar in Figure P3.160 has angular velocity 0.25
r a d / s and angular acceleration 0.15
r a d / s at the
given instant. Find the angular acceleration of at this
time. (See Problem 3.143.)
2
figure P3.1S8
Figure P3.160
3.159 The mechanism shown in Figure P3.159 is used to
raise and lower hammer
The 26-cm crank
turns
clockwise at the constant rate of 30 rpm. It is pinned to
block
which slides in a slot in
If at t = 0 point A is
directly above
find the velocity and acceleration of
as a function of time. (The block and hammer are slightly
offset from so they do not interfere with the pin at O.)
3.161 In Problem 3.144 find the acceleration of P in
the angular acceleration of
and
3.162 In Problem 3.155 find, at the same instant of time,
the acceleration of P in and the angular acceleration of
3.163 In Figure P3.163,
and
If the bar stays in contact with both the step and circular
Figure P3.159
Figure P3.163
Page 2 1 3
trough, find its angular acceleration. Hint: Treat point Q
(fixed to the step) as the "moving point," and note that Q
moves on a straight line relative to the bar.
3.164 In Problem 3.151 determine the acceleration of C
relative to as a function of D,
and
• 3.167 Extending Problem 3.149, suppose that the angular
velocity of rod
5
rad/s, is constant in time. (See
Figure P3.167.) Find, at the given instant, the angular
acceleration of
and the acceleration of any point of
relative to
3.165 Rods
and
(see Figure P3.165) pass smoothly
through the short collars, which can turn relative to each
other by virtue of the ball-and-socket connection. If bars
and turn with constant angular velocities 0.4
rad/
sec and 0.2 rad/sec, respectively, find the velocity and
acceleration of the ball-and-socket connection with re­
spect to
and to in the indicated position.
Figure P3.167
3.168 Referring to Example 3.29 for the meanings of the
symbols, show that
Figure P3.165
* 3.169 A circular turntable
(see Figure P3.169) rotates
about a vertical axis through O (normal to the plane of the
paper) with changing at a constant rate
A block
rests in a groove cut in the turntable. If the cable is reeled
in at a constant velocity relative to find expressions
for the radial and transverse components of the block's
acceleration. Check your answers using the expression
for acceleration in cylindrical coordinates.
* 3.166 Wheel in Figure P3.166 has a constant clockwise
angular velocity of 2 rad/sec. It is connected by link to
block
End B of rod slides in a vertical slot in block
For the position shown, find the angular velocity and
angular acceleration of rod if block translates.
Figure P3.169
3.170 Show that the rigid-body acceleration equation
(3.19) can be derived from Equation (3.54). Hint: If you fix
P to the equation will then relate the accelerations in
of the two points P and
of
• 3.171 In Problem 3.153 let R = 0.1 m and
= 5 rad/s
as in Problem 3.154. This time, again at = 160°, find: (a)
the angular acceleration of (b) the acceleration of P in
Figure P3.166
Page 2 1 4
• 3.172 The 26-ft rod in Figure P3.172 slides on a plane
surface at A and on the fixed half-cylinder at Q as shown
below. If end A is moved at a constant velocity of 4 ft/sec
to the right along the plane, determine the vertical com­
ponent of the acceleration of B for the position shown.
• 3.173 Pin P in Figure P3.173 moves along a curved path
and is controlled by the motions of the slotted links and
At the instant shown, each point of has a velocity of
5 ft/sec and an acceleration of 20 f t / s e c , both to the
right, while each point of has a velocity of 3 f t / s e c and
an acceleration of 30 f t / s e c in the direction shown in the
figure. Find the radius of curvature of the path of P in this
position.
2
Figure P3.173
2
(on the rod extended)
• 3.174 Considering the instantaneous center
of the bar
in Figure P3.174 as a point of
extended, find the
acceleration of _ at the instant shown, if v =
in./
A
(on the rod extended)
sec = constant. Also find the acceleration of the point
of
passing over the pin, and note that it is not along
the slot.
Figure P3.174
Figure P3.172
COMPUTER PROBLEMS
•
Chapter 3
Crank
in Figure P3.175 is driven at a constant
angular speed
clockwise. Show that the speed of the
piston is maximum when satisfies the equation
Figure P3.175
Solve with a computer for the first root of this equation
when the lengths of and
are 8 cm and 20 cm, respec­
tively. Note that the answer is independent of the value of
You may wish to read Appendix B and make use of the
Newton-Raphson method described there.
•
Crank
in Figure P3.176 rotates at constant angu­
lar velocity
. Use a computer to generate data for a plot
of the following two quantities as functions of for the
case in which D = 21:
a. The angle
b. The ratio
that locates slider
of the angular speeds of
and
Figure P3.176
Page 2 1 5
SUMMARY
Chapter 3
This chapter h a s been devoted to presentation of t h e velocity a n d accel­
eration relationships that pertain to a rigid b o d y in plane motion.
If A a n d B are t w o points of the b o d y lying in the s a m e plane of
motion, then their velocities are linked through the angular velocity
( is perpendicular to the plane of motion) by
If a point instantaneously has zero velocity, call it
, then:
w h i c h s h o w s that the speed of a n y point is the product of the angular
speed a n d the distance between the point a n d the instantaneous center.
For accelerations, w h e r e
is the angular acceleration, w e h a v e
W h e n o n e b o d y rolls on another, the points of contact h a v e the s a m e
velocity. Rolling of a wheel on a fixed surface m e a n s that the point on the
wheel in contact with the surface is the instantaneous center of velocity,
a n d with C denoting the center of the wheel,
and
+ (a part n o r m a l to the p a t h of C )
Finally, w e h a v e investigated the relationships between velocities
a n d then accelerations of a single point P relative to t w o reference frames.
The results c a n be expressed c o m p a c t l y if w e think of the underlying
basic frame of reference as "fixed" a n d a second body as moving. Then
with
v
P
= velocity of P relative to the fixed frame,
v = velocity of P relative to the m o v i n g body,
rel
= angular velocity of the moving body,
a n d similarly for accelerations a n d with
being a point fixed in the
moving body, w e have:
and
The last term in t h e s e c o n d equation is called t h e Coriolis acceleration.
Page 2 1 6
REVIEW QUESTIONS
Chapter 3
True or False?
In these questions, P a n d Q are points in the reference plane of a rigid
body in plane motion.
1. For a rigid body in plane motion with angular velocity
depends
on a certain choice of points w h o s e velocities are to be related.
2. For a rigid body in plane motion, there is always an instantaneous
center
of zero velocity located a finite distance from the body.
3. a
4.
is not necessarily zero.
if the angular velocity of
is not zero.
5. For a rigid b o d y in rectilinear translation, the velocities of all points
of
are equal, a n d so are the accelerations.
6. For a rigid body in curvilinear translation, the velocities of all points
of
are equal, but the accelerations are not.
7. A t a given instant, a n y point can be considered to lie on a rigid
extension of any rigid body.
8. T h e smallest n u m b e r of scalar parameters required to locate a rigid
body in plane motion is four.
9. F o r a rigid body
in plane motion,
10. A point P can h a v e an angular velocity.
11. T h e c o m p o n e n t s of v a n d v
P
Q
along the line P Q are not always equal.
12. T h e c o m p o n e n t s of a a n d a along the line PQ are not always equal.
p
Q
Answers: 1.F 2 . F 3 . T 4. T 5. T 6. F 7. T 8. F 9 . T 10. F 11. F 12. T
4
KINETICS OF A RIGID BODY
IN PLANE MOTION/
DEVELOPMENT AND SOLUTION
OF THE DIFFERENTIAL
EQUATIONS GOVERNING
THE MOTION
Introduction
4.1
4.2
Rigid Bodies in Translation
4.3
Moment of M o m e n t u m (Angular Momentum)
Inertia Properties
4.4
Moments and Products of I n e r t i a / T h e Parallel-Axis T h e o r e m s
Examples of Moments of Inertia
The Parallel-Axis Theorem for Moments of Inertia
The Radius of Gyration
Products of Inertia
Transfer Theorem for Products of Inertia
4.5
T h e Mass-Center Form of the Moment Equation of Motion
Development of the Equations of Plane Motion
Helpful Steps to Follow in Generating and Solving the Equations
of Motion
4.6
Other Useful Forms of the Moment Equation
Moment Equation in Terms of a
Moment Equation in Terms of a
Moment Equation for Fixed-Axis Rotation (The "Pivot" Equation)
c
p
Rotation of Unbalanced Bodies
SUMMARY
REVIEW Q U E S T I O N S
Page 217
Page 218
4.1
Introduction
In this chapter w e apply Euler's laws to the plane motions of rigid bodies.
Motion of the mass center of any body, rigid or not, is governed by Euler's
first law as discussed in Chapter 2. The rotational motion of a rigid body is
governed by Euler's second law. We saw in Chapter 2 that this law can be
expressed in terms of a moment of momentum for any body, rigid or not.
However, the moment of momentum for a rigid body can be expressed
in a particularly compact way that involves moments and products of in­
ertia of the body and its angular velocity; because of this, the term an­
gular momentum is used synonymously with moment of momentum.
There is, however, one type of plane motion of a rigid body ßthat can
be immediately studied, prior to the introduction of angular momentum.
This class of motions, called translation, is characterized by the angular
velocity of B being always zero. The translation problems treated in the
next section (4.2) differ from the particle/mass-center motion problems
of Chapter 2 in that a moment equation is required for their solution in
addition to the mass-center equation
= ma .
In Section 4.3, w h e n w e develop the expression for the angular
momentum of a rigid body, the moments and products of inertia sud­
denly appear — in the same way the mass center did, back in Chapter 2.
Thus w e spend some time in Section 4.4 studying these inertia properties
before moving on.
In Sections 4.5 and 4.6 w e deduce several especially useful forms of
Euler's second law in terms of the inertia properties. After each form of
the equation, w e illustrate its use with a set of examples. Some of these
examples might be termed "snapshot" problems; in these w e investigate
the relationships between external forces on a body and its accelerations
at a single instant. These problems are rather natural extensions of those
the student has encountered in statics — that is, w e know the geometri­
cal configuration and seek information about forces on the body. Other
problems might be called "movie" problems; in one class of these the
geometry is of sufficient simplicity that Euler's laws produce differential
equations which w e can readily integrate so as to predict the motion of
the body during some interval of time.
c
In Section 4.7, w e take up the very special problem of rotation of an
unbalanced body about a fixed axis. Here w e establish the criteria for the
technologically important problem of balancing.
In the next chapter, w e will continue our study of plane-motion
kinetics of rigid bodies by investigating the use of three special solutions
(which can be obtained in general) to the differential equations of mo­
tion. These special integrals are known as the principles of work and
kinetic energy; linear impulse and momentum; and angular impulse and
angular momentum.
Finally, w e mention to the reader that it is possible to obtain all the
results of this chapter on plane motion of rigid bodies from the general
three-dimensional results developed in Chapter 7. It is not necessary to
Page 219
travel this complex route in order to l e a m p l a n e motion, however, a n d in
this chapter w e take a simpler p a t h . It is w o r t h noting that while the
planar case covers a restricted class of motions, it does in fact contain a
large n u m b e r of problems with important engineering applications.
4.2
Rigid Bodies in Translation
In Chapter 2 we presented the equation
(2.34)
w h i c h is valid b o t h for any b o d y a n d any point P. In particular, if the
b o d y is translating, t h e n b y definition all its points h a v e the same accel­
eration — including its m a s s center — a n d if w e label t h a t c o m m o n ac­
celeration a, t h e n it m a y be factored, leaving:
(using t h e definition
of t h e mass center)
(4.1a)
a n d therefore
(4.1b)
In this section, w e shall apply this simple equation to several examples of
translating rigid bodies in plane motion. The reader should note, h o w ­
ever, that t h e t w o forms of Equation (4.1) apply w h e t h e r the motion is
plane or not. It also d o e s n ' t even require the b o d y to be "physically
rigid," although if it is translating, it is necessarily b e h a v i n g like a rigid
b o d y during that motion.
Question 4.1 If point P is arbitrary in Equation (4.1a), where does the
inertial frame come into the equation?
Before w a d i n g into t h e examples, w e w i s h to n o t e w h a t is n e w here.
In Chapter 2, we were concerned with the mass center motions of bodies,
w h e t h e r they were rigid or not. In those sections there w a s n o n e e d to
s u m m o m e n t s , a n d that is w h a t will distinguish this chapter from those
preceding sections. N o w t h e simplest problems b y far in w h i c h a m o m e n t
equation is sometimes n e e d e d are those involving translation. Transla­
tion is simple because it can b e studied prior to the introduction of
angular m o m e n t u m forms for rigid bodies with their accompanying in­
ertia properties a n d angular velocities.
We n o w examine three examples of translation, a n d in each, the
reader is urged to note t w o things: (1) h o w the problem could not be
Answer 4.1 T h e acceleration a is t h e s e c o n d derivative, t a k e n in a n inertial frame, of t h e
position vector from an ortgin in t h a t frame to a n y p o i n t of t h e b o d y .
Page 220
solved without the use of Equation (4.1); and (2) h o w for translation, the
moments generally do not sum to zero. They do sum to zero at the mass
center, and also at points lying on the line of a drawn through C (for then
r is parallel to a)*, but not otherwise. (Equations for translation at con­
stant velocity are trivial and identical to the equilibrium equations; these
were studied in statics and are not considered in this book.)
pc
EXAMPLE 4 . 1
m,
P
L
Find the angle 6 for which the bar in Figure E4.1a will translate to the right at the
given constant acceleration "a." Then find the force P required to produce this
motion.
Solution
We sum moments at the contact point A, using the FBD in Figure E4.1b and
Equation (4.1):
0
Figure E4.1a
Therefore 0 in terms of a is given by:
mg
P
o
µN
To determine P, we write the mass-center equations:
A
N
Figure E4.1b
(1)
and
Substituting this value of N into Equation (1) gives
P —µmg= ma
or
P =m(µg+ a)
Note that part of P balances the friction and the rest, the unbalanced force in the
x-direction, produces the "ma."
a n d of course at times w h e n t h e e q u a l accelerations of all p o i n t s , a, are zero.
Page 221
2 ft
1ft
EXAMPLE 4 . 2
A 300-lb cabinet is to be transported on a truck as shown in Figure E4.2a.
Assuming sufficient friction so that the cabinet will not slide on the truck, for
forward acceleration find
4 ft
4 ft
a. the forces exerted by the truck bed on the cabinet for 5 ft/sec accel­
eration;
b. the maximum acceleration for which the cabinet will not tip over.
2
Solution
a. The acceleration is
ure E4.2b) and putting
Figure E4.2a
so using the free body diagram (Fig­
= ma into component form,
1 ft
4 ft
300 lb
To locate the line of action of N, that is, to find the distance d, we
need the moment equation:
4F - (1 - d)N = 0
4(46.6) - (1 - d)(300) = 0
Figure E4.2b
1 - d = 0.621
d = 0.379 ft
We might note at this point that the minimum coefficient of friction
for which this motion is possible is
b. Assuming we still have adequate friction to prevent slip, but
treating the acceleration magnitude, a, as an unknown, we can
retrace our steps as in part (a) to obtain
and
4F - (1 - d)N = 0
When the cabinet is on the verge of tipping, the line of action of N is
at the left comer, d = 0, so for that condition:
4F - (1 - 0)(300) = 0
F = 75 lb
Page 222
and then from
we find
The reader should note that this tendency to tip over "backwards" is a
phenomen uniquely of dynamics; there's nothing quite like it in stat­
ics. In addition we sometimes tend to think of friction in oversimpli­
fied terms, as perhaps "always opposing motion"; of course it is pre­
cisely the friction that here provides the motive force to cause the
cabinet to accelerate.
In the preceding example w e saw h o w the state of translation could
be jeopardized by the tendency of a body to "rock." In the next w e
additionally explore a tendency to slip.
Bar
60
0
EXAMPLE 4 . 3
The coefficient of friction at both ends of the uniform slender bar in Figure E4.3a
is 0.5. Find the maximum forward acceleration that the truck may have without
the bar moving relative to the truck.
Figure E4.3a
Solution
One possibility is that the upper end separates from the truck body. The freebody diagram in Figure E4.3b shows the situation when the end is barely about to
break away. Because the bar is in translation, we may write:
60
0
Figure E4.3h
But this much friction cannot be generated because µ= 0.5 < 0.577. Therefore
motion of the rod relative to the truck will not be initiated by this mechanism.
The other possibility, as shown in the second free-body diagram, Fig­
ure E4.3c, is that the bar is on the verge of slipping where it contacts the truck;
this must occur at the two surfaces simultaneously. The equations of motion are
(with a =
(1)
Figure E4.3c
(2)
Page 223
(3)
With µ= 0.5, Equation (3) yields
N, = 16.7N
2
and from (2)
N = 0.0581 mg
2
so that
N = 16.7N = 0.971 mg
1
2
Equation (1) then gives
0.5(0.971 mg) - 0.0581 mg =
= 0.427 g
2
2
2
2
which is 13.8 ft/sec (for g = 32.2 ft/sec ) or 4.19 m / s (for g = 9.81 m / s ) .
The reader is encouraged to rework the previous example using
Equation (4.1a) to sum moments about the bottom point of the translat­
ing bar.
PROBLEMS
•
Section 4 . 2
4.1 For what force P is it possible for the uniform
slender bar (Figure P4.1) to translate across the smooth
floor in the position shown? The bar has mass m and
length
1ft
40 lb
1 ft
1 ft
Figure P4.2
Figure P4.1
4.2 A 100-lb cabinet, rolling on small wheels, is sub­
jected to a 40-lb force as shown in Figure P4.2. Neglecting
friction, find (a) the acceleration of the cabinet; (b) the
reactions of the floor on the wheels.
4.3 Repeat the preceding problem for the case where
the 40-lb force is applied 1 ft above C,
4.4 Find the value of F for which one of the wheels of
the door in Figure P4.4 lifts out of its track. Which one?
Assume negligible friction.
Mass m
Figure P 4 . 4
Page 224
4.5 The force P causes the uniform rectangular box of
weight W in Figure P4.5 to slide. Find the range of values
of H for which the box will not tip about either the front or
rear lower corner as it slides on the smooth floor, if P = W.
4.9 The cords in Figure P4.9 have a tensile strength of
12 N. Cart has a mass of 35 kg exclusive of the 10-kg
and 1.2-m vertical rod which is pinned to it at A. Find
the maximum value of P that can be exerted without
breaking either cord if: (a) P acts to the right as shown; (b)
P acts to the left. Neglect friction, and assume negligible
tension in each cord when the cart is at rest.
4.10 A force F, alternating in direction, causes the car­
riage to move with rectilinear horizontal motion defined
by the equation x = 2 sin
ft, where x is the displace­
ment in feet and t is the time in seconds. (See Fig­
ure P4.10.) A rigid, slender, homogeneous rod of weight
32.2 lb and length 6 ft is welded to the carriage at B and
projects vertically upward. Find, in magnitude and di­
rection, the bending moment that the carriage exerts on
the rod at B when
2b
Figure P4.5
4.6 Repeat the preceding problem for a coefficient of
sliding friction of 0.2.
1.1 m
0.9 m
Cord
4.7 A 400-lb cabinet is to be transported on a truck as
shown in Figure P4.7. Assuming sufficient friction so that
the cabinet will not slide on the truck, what is the maxi­
mum forward acceleration for which the cabinet will not
tip over?
1 ft
2 ft
Small wheels
Figure P4.9
4 ft
4 ft
Figure P4.10
4.11 A child notices that sometimes the ball does not
roll down the inclined surface of toy when she pushes it
along the floor. (See Figure P4.11.) What is the minimum
acceleration
of to prevent this rolling?
Figure P4.7
4.8 The uniform bar in Figure P4.8 weighs 60 lb and is
pinned at A (and fastened by the cable DB) to the frame
If the frame is given an acceleration a = 32.2 ft/sec as
shown, determine the tension T in the cable and the force
exerted by the pin at A on the bar.
Solid
sphere 111
2
a = 32.2 ft/sec
2
Figure P4.11
45°
6 ft
Cable
4.12 In the preceding problem, suppose the acceleration
of
is 2 a . What is the normal force between the
smooth vertical surface of
and the ball? The ball's
weight is 0.06 lb.
m i n
Figure P4.8
Page 225
4.13 A can that may be considered a uniform solid
cylinder (see Figure P4.13) is pushed along a surface
by a moving arm
If it is observed that translates to the
right with
= g/10, what must be the minimum coeffi­
cient of friction between and ? The coefficient of fric­
tion between and
2ft
4 ft
Figure P4.16
Figure P4.13
4.14 The force Pis applied to cart and increases slowly
from zero, always acting to the right. (See Figure P4.14.)
Point C is the mass center of B. At what value of P will the
bodies and no longer move as one?
2 ft-
4.17 The block of mass m in Figure P4.17 is resting on
the cart of mass M. Force P is applied to the cart, starting
it in motion to the right. The wheels are small and frictionless, and the coefficient of friction between and is
Find the largest value of P for which and will
move together, considering all cases.
|B(3 lb)
1.8 ft
µ= 0.3
0.6 ft
6 (2 lb)
Figure P4.17
Figure P4.14
4.15 A nonuniform block rests on a flatcar as shown in
Figure P4.15. If the coefficient of friction between car and
block is 0.40, for what range of accelerations of the car
will the block neither tip nor slide?
4.18 The 25-lb triangular plate is smoothly pinned at
vertex A to a small, light wheel (see Figure P4.18). Find
the value of force P so that the plate, in theory, will trans­
late along the incline. Also find the acceleration.
2 ft
0.5
1
1 ft
1 ft
ft
ft
4 ft
2 ft
2 ft
Figure P4.15
4.16 The truck in Figure P4.16 is traveling at 45 mph.
Find the minimum stopping distance such that the 250-lb
crate will neither slide nor tip over.
Figure P4.18
Page 226
4.19 Repeat the preceding problem if the angle of the
plane is changed from
4.20 The monorail car in Figure P4.20 is driven through
its front wheel and moves forward from left to right. If
the coefficient of friction between wheels and track is
= 0.55, determine the maximum acceleration possible
for the car.
8 m-
Figure P4.23
Sm
2 5m
Figure P4.20
4.24 A uniform rod. of length L and weight W is con­
nected to smooth hinges at E and D by the light members
B and
each of length L. In the position shown in Fig­
ure P4.24, has an angular velocity of rad/sec clock­
wise. Find the forces in members ßand
and deter­
mine the acceleration of center C of rod
in terms of
the given variables.
4.21 A dragster is all set for the annual neighborhood
race. (See Figure P4.21.)
a. In terms of the dimensions b, H, and d and the
coefficient of friction
find the maximum pos­
sible acceleration of the car. Neglect the rota­
tional inertia of the wheels.
b. How would you adjust the four parameters b,
H, d, and to further increase the driver's ac­
celeration?
Figure P4.24
4.25 Find the range of accelerations of that F can pro­
duce without moving in any manner relative to
(See
Figure P4.25.) Note carefully the position of the mass
center of
Figure P4.21
1 ft
1 ft
4.22 Rework the preceding problem for a car with (a)
front-wheel drive and (b) four-wheel drive.
4.23 In an emergency the driver of an automobile ap­
plies his brakes; the front brakes fail and the rear wheels
are locked. Find the time and distance required to bring
the car to rest. Neglect the masses of the wheels, and
express the results in terms of the coefficient of sliding
friction the initial speed v, the gravitational acceleration
and the dimensions shown in Figure P4.23.
3 ft
8
15
Figure P 4 . 2 5
2 ft
Page 227
• 4.26 A slender homogeneous rod weighing 64.4 lb and
20 ft long is supported as shown in Figure P4.26. Bars
and are of negligible mass and have frictionless pins at
each end. The system is released from rest with
a. Derive expressions for the angular velocity and
acceleration of bars
and
as functions of
b. Derive expressions for the axial force in bars
and
as a function of
Figure P4.26
4.3
Moment of Momentum (Angular Momentum)
We recall from C h a p t e r 2 that the m o m e n t of m o m e n t u m of a n y b o d y
with respect to a point P (not necessarily fixed in either the b o d y or in the
reference frame) w a s defined by Equation (2.35):
w h e r e in this section R is the vector from P to the element of m a s s dm
of the body, a n d v is the velocity of dm in the reference frame
(See Fig­
ure 4.1.) W e note that H d e p e n d s o n the location of P as well as the distri­
butions of m a s s a n d velocities in the body; it is seen to b e the s u m of
the m o m e n t s of m o m e n t a of all the m a s s elements of
P
body B
reference frame
Figure 4.1
We shall n o w restrict the general b o d y above to b e rigid, place it in
plane motion, a n d recall from Chapter 3 that the kinematics of a rigid
body in plane motion can be described very simply. If w e k n o w the
velocity of just one point a n d the angular velocity of the body, t h e n we
k n o w the velocity of every point of — quite a bargain. Because of this
simplicity, w e shall see that compact a n d yet completely general expres­
sions can b e written for the m o m e n t of m o m e n t u m (or angular m o m e n ­
tum, as it is often called for rigid bodies). We shall further restrict the
generic point P in the foregoing to b e a point of
The rectangular axes (x, y, z) h a v e origin at P as s h o w n in Figure 4.2,
a n d the xy-plane is the reference plane, or plane of motion, as described
in Sections 3.1, 2.
Page 228
reference
plane
Figure 4.2
Recall also from Section 3.1 that each point of t h e rigid b o d y h a s a
" c o m p a n i o n p o i n t " in t h e reference p l a n e w h i c h always h a s t h e same
a n d as t h e point, a n d h e n c e also h a s t h e same velocity a n d acceleration.
For t h e point at "dm" in Figure 4.2, t h e c o m p a n i o n point is
Thus, using
Equation (3.8),
Substituting this expression for v a n d the coordinates of dm into R,
Hp becomes:
or
The integral R dm in t h e first term above is equal to mr b y the defini­
tion of t h e m a s s center. Making this substitution, a n d also carrying out
the cross products in t h e second term, gives t h e m o m e n t of m o m e n t u m
vector in terms of v ,
a n d certain m a s s distribution integrals:
PC
P
Inertia Properties
We call the integrals in this equation inertia properties. Specifically:
m o m e n t of inertia of mass of
about z axis t h r o u g h P
(4.2a)
product of inertia of m a s s of
with respect to x a n d z axes
through
(4.2b)
*If t h e p r o d u c t s of inertia a r e defined w i t h t h e m i n u s sign as a b o v e , t h e n a n d o n l y t h e n
will t h e inertia properties t r a n s f o r m as a t e n s o r — a topic b e y o n d t h e scope of this book,
however.
Page 229
product of inertia of mass of
with respect to y a n d z axes
through P
(4.2c)
Thus,
(4.3)
We are n o w at t h e point in our d e v e l o p m e n t w h e r e t h e inertia properties,
like the m a s s center in Chapter 2, h a v e arisen naturally. W e shall s p e n d
t h e next section studying the m o m e n t s a n d products of inertia; readers
already familiar with inertia properties m a y wish to skip Section 4.4.
Before leaving Equation (4.3), w e note for future reference that its
first term, r
mv , vanishes if P is t h e m a s s center or h a s zero veloc­
ity* In b o t h these cases, w h i c h will prove valuable to us, H takes the
form
pc
P
P
(4.4)
T h u s in these cases t h e m o m e n t of m o m e n t u m can b e expressed in terms
of the angular velocity of (hence its other n a m e : angular momentum),
along w i t h three m e a s u r e s of its mass distribution.
Question 4.2 In Equation (4.3), does the mass center have to lie in the
reference plane with P? How about in Equation (4.4) when P has zero
velocity?
4.4
Moments and Products of Inertia /
The Parallel-Axis Theorems
Examples of M o m e n t s of Inertia
From the definition of m o m e n t of inertia,
w e see that
is a m e a s u r e of " h o w m u c h m a s s is located h o w far" from
t h e 2 axis t h r o u g h P. In cylindrical coordinates w e h a v e
dm, a n d
thus
measures t h e s u m total of mass times distance squared over the
body's volume. The quantify
is t h u s seen to b e always positive. We
n o w c o m p u t e the mass-center m o m e n t s of inertia of a n u m b e r of com­
m o n shapes. In Examples 4 . 4 - 4 . 1 2 , w e are seeking
* O r if r
ps
Answer 4.2
is parallel to v , a case w e n e e d n o t consider here.
p
No. No.
Page 230
EXAMPLE 4 . 4
Homogeneous solid cylinder if 2 is its axis (see Figure E4.4).
Solution
Noting that dm— p dV, where p = mass density,
Figure E4.4
EXAMPLE 4 . 5
Homogeneous solid cylinder if 2 is an axis normal to the axis of the cylinder (see
Figure E4.5a).
Solution
Figure E4.5a
It may be confusing at first to know which moment of inertia to use for a
cylinder in a plane kinetics problem; the answer is that it is always the value
associated with the axis normal to the xy plane of the motion. If the problem is a
rolling cylinder,
If we have a cylinder turning around a diametral
axis (see Figure E4.5b), then
Figure E4.5b
Page 2 3 1
EXAMPLE 4 . 6
Two special cases of Example 4.5: Slender rods and disks (see Figures E4.6a, b).
Solution
1.
If the body is "pencil-like" — that is, L
the moment of inertia for a
lateral axis through C is approximated as mL /12. This is also a correct result
even if the cross section is not circular but has a maximum dimension within
the cross section much less than L. Such a body is called a slender bar or rod.
If the body is a disk, however, we have R
L and the moment of inertia is
approximately m R / 4 .
2
2.
2
For a cylinder with the dimensions of a pencil, for example, with
and L = 7 in., we see that (see Figure E4.6a)
in.
Figure E4.6a
2
where the second term is less than 0.1 of 1 percent of the retained mL /12 term.
For a typical coin, on the other hand, with
in. and L
we obtain (see
Figure E4.6b).
Figure E4.6b
2
This time it is the mL /12 term that is negligible; it is less than 0.15 of 1 percent of
the m R / 4 term. We emphasize, however, that with respect to the axis of any
sobd homogeneous cylinder (disk, rod, or anything in between), the moment of
inertia is mR /2.
2
2
EXAMPLE 4 . 7
A uniform rectangular solid (see Figure E4.7).
Solution
Figure E4.7
Page 232
This integration yields
EXAMPLE 4 . 8
Special case of Example 4.7: A rectangular plate.
Solution
If the rectangular solid is a plate — that is, it has one edge much smaller than the
other two dimensions — then, referring to Figure E4.8a-c, we have:
Figure E4.8a
Figure E4.Bb
Figure E4.8c
Again it depends on how the body's plane morion is set u p as to which axis is z
(normal to the plane of the motion) and hence which formula to use.
EXAMPLE 4 . 9
Solid, homogeneous, right circular cone about its axis (see Figure E4.9).
Solution
Here we encounter a variable limit,
same z axis, then (see Figure E4.9):
Figure E4.9
From similar triangles,
Noting that C and O are on the
Page 233
which gives the varying radius in terms of z. Then
EXAMPLE 4 . 1 0
Hollow, homogeneous cylinder about its axis (see Figure E4.10).
Solution
Figure E4.10
The same result can be obtained by substracting the moment of inertia of the
"hole" (H) from that of the "whole" (W). The basis for this procedure is that we
may integrate over more than the required region provided we subtract away the
integral over the part that is not to be included:
Note that if the wall thickness is small, we have a cylindrical shell (or a hoop if the
length is small) for which
and
(It is obvious that if all the mass is the same distance R from the axis z we should
indeed get mR .)
2
Page 234
EXAMPLE 4 . 1 1
A uniform solid sphere about any diameter.
Solution
Also:
and
Adding:
= 3 / since they are all
equal by symmetry
Thus (see Figure E4.ll for the spherical coordinates):
dV i n s p h e r i c a l c o o r d i n a t e s
Figure E4.11
A less tricky way to do the sphere is to use spherical coordinates directly; the
integral is
2
which yields the same result of (2/5)mR , as the reader may show by carrying out
the integration.
EXAMPLE 4 . 1 2
An example in which the density is not constant. Sometimes a body's density
varies; if it does, it must stay inside the integral when we calculate the inertia
properties. An example is the earth; we now know that the density of the solid
central core of the earth is about four times that of the outermost part of its crust
and, moreover, that this central density is nearly twice that of steel!
Page 2 3 5
Let us imagine a sphere with the same mass and radius as in the preceding
example but with a density that varies linearly and is twice as high at r = 0 as at
r = R. We shall find I about any diameter. (See Figure E4.12(a).)
Density variations
Solidn
onhomogeneous
(a)
spheres
(b)
Figure E4.12
Solution
The mass of the body is
Letting the density at R be p , then
1
Substituting and integrating with the same limits as before, and then equating the
new mass to the old, gives
Thus to have the same mass as the uniform sphere, the density varies from
to
going outward from the center. Then, integrating to find I,
which gives
(slightly less than
)
Alternatively, if the density varies linearly but is only half as much
at
the center as at R, then the results, if m and R are again the same as in the uniform
case, are (see Figure E4.12(b)):
(slightly more than
)
T h e Parallel-Axis T h e o r e m for Moments of Inertia
In many applications the body consists of a number of different smaller
bodies of familiar shapes. In such cases, there is fortunately n o need to
integrate in order to find the inertia of each part with respect to a common
axis of interest, thanks to what is called the parallel-axis theorem or
transfer theorem. If w e know the moment of inertia about an axis
Page 236
through the mass center C of any body
w e can then easily find it about
any axis parallel to C by a simple calculation. The theorem states that the
moment of inertia of the mass of about any line is the moment of inertia
about a parallel line through C plus the mass of times the square of the
distance between the two axes:
To prove this theorem, w e let (x, y, z) and (x , y , z ) be rectangular
cartesian coordinate axes through P and C with the corresponding axes
respectively parallel, as s h o w n in Figure 4.3. Then w e have, by defini­
tion,
1
l
1
Figure 4.3
Since it is seen that
substitution gives
(4.5)
or
(4.6)
in w h i c h
C a n d P.
the square of t h e distance b e t w e e n z axes t h r o u g h
Question 4.3
Why are the last two integrals in Equation (4.5) zero?
Equation (4.6) is the parallel-axis t h e o r e m for m o m e n t s of inertia. But
Answer 4.3
is equal to
and
dm
A n y integral s u c h as
dm, w h e r e x is m e a s u r e d from a n origin at, say,
this is t h e definition of t h e m a s s center. So if t h e origin is C, t h e n = 0
= 0.
Q
Page 237
note: We can only transfer from the mass center C and not from any other
point A about which we may happen to know
EXAMPLE 4 . 1 3
For the uniform slender rod shown in Figure E4.13 find
the moment of
inertia of the mass of with respect to a lateral axis through one end. (This
exercise will be useful in pendulumlike applications in which a rod is pinned at
one end.)
(mass m)
Solution
We next consider an example of the buildup of the moment of inertia
for a composite body.
EXAMPLE 4 . 1 4
Find
for the body shown in Figure E4.14. The mass densities each =
= constant, so that the respective masses are:
Sphere
Half-cylinder
Slender bar
Side view
Top view
Figure E4.14
Page 238
and
Solution
First we observe that
since the inertia integral may be
carried out over the bodies separately, so long as we cover all the elemental
masses of the total body. Filling in the separate integrals, we get
in which, for the half-cylinder,
Note that we cannot correctly transfer the inertia of the semicylinder from
it must be done from the mass center C . So first we go "through the back door"
to find (since we know the moment of inertia already with respect to
and
only then may we transfer to O.
3
EXAMPLE 4 . 1 5
A closed, empty wooden box is 5 ft X 3 ft X 2 ft and weights 124 lb (see Fig­
ure E4.15).
a. Find its moment of inertia about an axis through C parallel to the 2-ft
dimension.
b. If the box is then filled with homogeneous material weighing 240 lb
(excluding the box), how much does the moment of inertia about the
axis of part (a) increase?
3 ft
5 ft
2 ft
Figure E4.1S
Solution
a. The masses of the various sides of the box are proportional to their
areas (thickness and density assumed constant):
Page 239
total
area
Therefore, taking the contributions from the three pairs of sides,
= 15.9 slug-ft
contents
2
2
(or lb-ft-sec )
slug-ft
2
Even though the box weighs only about half as much as the contents, the position
of its mass makes its moment of inertia over three-fourths that of the contents.
The total moment of inertia is 37.0 slug-ft .
2
T h e Radius of Gyration
There is a distance called the radius of gyration that is often used in
connection with moments of inertia. The radius of gyration of the mass of
a body about a line z (through a point P) is called
, or just k if the axis is
understood to be z, and is defined by the equation
P
(4.7)
If one insists on a physical interpretation of k , it may be thought of as the
distance from P, in any direction perpendicular to z, at which a point
mass, with the same mass as the body, would have the same resulting
moment of inertia that the body itself has about axis z. For example, a
solid homogeneous cylinder has a radius of gyration with respect to
its axis of
since
The usefulness of k is seen
P
c
here, since regardless of the mass of a cylinder (and hence of its density)
k will be the same for all homogeneous cylinders of equal radii.
Note further that (using the parallel-axis theorem)
c
Thus
(4.8)
Page 240
a n d w e see from Equation (4.8) that the radius of gyration, like the
m o m e n t of inertia itself, is a m i n i m u m at C.
Products of Inertia
We n o w turn to t h e other t w o measures of mass distribution that h a v e
arisen in our study of p l a n e motion of a rigid b o d y — n a m e l y
and
taken here to b e with respect to axes (x, y, z) t h r o u g h a n y point P.*
O u r first step is to gain insight into t h e m e a n i n g of products of inertia
as w e s h o w that they in fact vanish for t w o large classes of c o m m o n l y
occurring symmetry. These classes are defined b y the two conditions
(with p constant in both): (1) z is a n axis of symmetry a n d (2) xy is a plane
of symmetry. Let us examine w h y t h e t w o products of inertia
and
are zero in these cases. We recall that their definitions are
(4.9)
Class 1: z Is an Axis of Symmetry. For each dV at (x, y, z) there is a cor­
responding dV at (—x, —y, z). T h u s t h e contributions of these t w o ele­
m e n t s cancel in b o t h t h e
and
integrals. Since each point of
has
a "canceling p o i n t " reflected t h r o u g h t h e z axis,
and
are each
zero for this class of bodies. (See Figure 4.4.)
Class 2: xylsa Plane of Symmetry. In this case each differential v o l u m e dV
at (x, y, z) necessarily h a s a mirror image at (x, y, — z). T h u s the contribu­
tions of these t w o elements cancel in b o t h integrals, a n d taken over the
whole of w e see again that
and
are each zero. (See Figure 4.5.)
Figure 4.4
Figure 4.5
* I n general (three-dimensional) motion, t h e r e arise six distinct inertia properties: three
m o m e n t s of inertia a n d t h r e e p r o d u c t s of inertia.
Page 2 4 1
Just because a b o d y does n o t belong to either of these t w o classes
does not m e a n it cannot h a v e zero products of inertia. However, these are
simply c o m m o n cases w o r t h y of note.
Question 4.4 Think of a rigid body for which both products of inertia
are zero, but which does not fall into either of the two classes.
Transfer T h e o r e m for Products of Inertia
There is a transfer t h e o r e m for products of inertia, just as there is o n e for
m o m e n t s of inertia. To derive it, w e write from Figure 4.3:
The last t w o terms v a n i s h b y virtue of t h e definition of t h e mass center
(for example,
dm = m times t h e z distance from C to C, w h i c h is zero).
Thus
(4.10a)
W e note that t h e factor of m in Equation (4.10a) is alternatively the
product of t h e x a n d z coordinates of P in a n axis system w i t h origin at C.
Similarly, w e h a v e
(4.10b)
Answer 4.4 Both p r o d u c t s of inertia are z e r o for t h e b o d y s h o w n in t h e d i a g r a m at t h e
left. T h r e e cylindrical b a r s of a n y l e n g t h s lying a l o n g t h e (x, y, z) axes are joined at t h e
origin t o form a rigid b o d y . N e i t h e r t h e z axis n o r t h e xy p l a n e is o n e of s y m m e t r y , yet
and
a r e zero.
EXAMPLE 4 . 1 6
Find
and
for the body shown in Figure E4.16; it is composed of eight
identical uniform slender rods, each of mass m and length
2
3
4
8
1
5
6
Figure E4.16
7
Page 242
Solution
We have
since xz is a plane of symmetry. Recall that when this happens,
the two products of inertia containing (as a subscript) the coordinate normal to
the plane are zero.* With superscripts identifying the various rods, we then have
the following for the other product of inertia:
Note that by symmetry each rod has zero about axes through its own center of
mass parallel to x and z. Therefore the eight terms listed above will consist only of
transfer terms in this problem.
Furthermore, since is zero for rods 4 and 5, and since is zero for rods 1 and
8, only four rods contribute to the overall
Note that the "unbalanced" masses lie in the second and fourth quadrants in
this example; hence the sign of
is positive since its definition carries a minus
sign outside the integral. We shall return to this example later in the chapter and
examine the reactions caused by the nonzero value of
when the body is spun
up in bearings about the z axis.
PROBLEMS
•
Section 4.4
4.27 An ellipsoid of revolution is formed by rotating the
ellipse about the x axis as in Figure P4.27. Find the mo­
ment of inertia of this solid body of density 15 slug/ft
about the x axis.
3
4.28 In Figure P4.28, the area bounded by the x and y
axes and the parabola y = 1 — x is rotated about the x
axis to form a solid of revolution. The density is p
= 1000 • (1 — x) k g / m . Find the moment of inertia of
the solid mass about the x axis.
2
8
3
1 ft
1
1
2 ft
1
Figure P4.27
Figure P4.28
* Thus
is also z e r o in this e x a m p l e , b u t
m o t i o n equations, as w e h a v e seen.
d o e s n o t a p p e a r a n y w h e r e in t h e p l a n e -
Page 243
4.29 The slender rod in Figure P4.29 has a mass density
given by
4.32 Use the parallel-axis theorems and the results of the
preceding problem to find, for that plate, the moments
and products of inertia at the center of mass.
4.33 For a uniform thin plate with xy axes (and origin O)
in the midplane, show that
in which
and
are constants. The rod has length L.
Find its moment of inertia about the line defined by x = 0,
y-L/2.
4.30 Find
for the semielliptical prism shown in Fig­
ure P4.30 (density
length normal to plane of paper
= L)•
Confirm this statement with the results of Problem 4.31
for the case when that plate is thin.
4.34 Find
for a uniform thin plate in the form of a
pie-shaped circular sector as shown in Figure P4.34.
4.31 The midplane of a uniform triangular plate is
shown in Figure P4.31. Find, by integration:
Figure P4.34
What would be good approximations were the plate thin?
4.35 Find
for the plate in the preceding problem.
4.36 The surface area of a solid of revolution is formed
by rotating the curve y = x (for 0 x 2 m) about
the x axis. (See Figure P4.36.) The density of the material
varies according to the equation p = 20x, where is in
k g / m when x is in meters. Find
and tell why your
answer is also
2
3
4.37 In the preceding problem,
find
and
4.38 See Figure P4.38. (a) Show that the moment of in­
ertia
for a solid homogeneous cone about a lateral
axis through the base is
= (m/20)(3R + 2H ). (b)
Using the transfer theorem, find the expression for
2
Figure P4.29
2a
Figure P4.30
Figure P4.36
Density = p, thickness = t
Figure P4.31
* Asterisks identify t h e m o r e difficult problems.
Figure P4.38
2
244
•4.39 In the previous problem, find 1° for the body in the
figure if the part above z = H/2 is cut away to form a
truncated cone.
4.40 The body shown in Figure P4.40 is composed of a
slender uniform bar (m = 4 slugs) and a uniform sphere
(m = 5 slugs). Find Ig for the body, where z is normal to
the figure.
y*.
Find Ig for the semicircular ring Bin Figure P4.48.
Hint: If the dashed portion were present, Ig would be
(2m)R ; by symmetry, our semicircular ring contributes
half of this, so that for B we have
2
2
Ig = mR
Now use the transfer theorem to complete the solution
without integration.
A rod ^of length 1 m is welded on its ends to a disk
2> and a sphere S. (See Figure P4.49.) The uniform bodies
have masses m = 10 kg, m = 5 kg, and m = 15 kg.
The radii of 2) and J ' a r e 0.3 m and 0.1 m, respectively.
Find 7^.
e
Figure P4.40
Figure P4.41
3
s
4.50 The three bodies shown in Figure P4.50, welded
together to form a single body B, have masses of 64 (rec­
tangular plate), 56 (rod), and 48 (disk), all in kilograms.
Find the moment of inertia of with respect to the z axis.
Q
4.41 Two bars, each weighing 5 lb per foot, are welded
together as shown in Figure P4.41. (a) Locate the center of
mass of the body, (b) Find Ig.
y
t
4.42 Find 1 ^ in Problem 4.41.
4.43 Use the result of the preceding problem, together
with the transfer (parallel axis) theorem, to find Ig.
Find Iyy in Problem 4.41.
4.45 Use the result of the preceding problem, together
with the transfer theorem, to determine Igj.
4.46 Find the moment of inertia of the mass of B about
axis z if 7£ = 40 kg • m . (See Figure P4.46.)
2
B
4.47 Find the moment of inertia of a uniform hemi­
spherical solid about the lateral axis x through its mass
center. (See Figure P4.47.)
c
X
Figure P4.49
'-•—0.8 m — - \
Figure P4.47
Figure P4.50
Page 245
4.51 In Figure P4.51,
is a solid sphere, is a solid
cylinder, and
and
are slender rods. The center lines
of
and
pass through the mass centers of and
respectively. Find
for the system of four bodies.
0.3
4.54 The bent slender rod in Figure P4.54 is located with,
the axes of the rods parallel to the x- and z-axes, as shown.
Find the value of
m
10 kg
2 4 kg
4.5 kg
2.4 m
0.3 m
12 k g
2.6 m
(3 kg)
0.7 m
0.4 m
Figure P4.51
(4 k g )
4.52 Find (which equals
for a thin plate (density
thickness t) in the shape of a quarter-circle. (See Fig­
ure P4.52.)
4.0
m
Figure P4.54
4.55 The cylinder in Figure P4.55 has a mass of 6 kg
and a radius of 0.4 m. Show that the moment of inertia
about an axis z normal to the page is 0.48 kg • m , and
that the corresponding radius of gyration is
m.
Figure P4.52
2
c
4.53 Find the moments of inertia of the pendulum about
axes x, y, and z. (See Figure P4.53.) Axes and are in the
plane of the pendulum, is a slender rod, and
is a
semicircular disk, each of constant density.
4.56 In the preceding problem, show that it is possible to
drill a hole through that is below the geometric center
and satisfy all of the following:
a. the remaining mass is 5.5 kg;
b. the distance between
and the new mass
center C (see Figure P4.56) is 0.02 m.
Find the radius r of the hole, the center distance d, and the
new value of
(1 kg)
1 m
0.2 m
Figure P4.53
04
m
[2 kg)
Figure P4.55
Figure P4.56
Page 246
4.57 The antenna
in Figure P4.57 has a moment of
inertia about
of
and the counterweights have a
collective moment of inertia about of . Points C and G
are the respective mass centers of the antenna and its
counterweights. The purpose of the counterweights is to
place the combined mass center at O to reduce stresses.
Thus MD = md, where we neglect the mass of the con­
necting frame for this problem. Compute the values of M
and D that will minimize the total moment of inertia I
(Iof about 0 plus 7 of. about O). Use
where
is a constant.
4.61 Show in the following three ways that the moment
of inertia of a uniform, thin spherical shell, about any line
through its mass center, is (2/3) mr . (See Figure P4.61.)
Which of the three approaches would work if the object
were hollow but not thin — that is,
2
a. Use spherical coordinates:
a
where r is the average radius,
b. Use the idea:
Counterweights
(mass M)
Let R increase by and compute the change,
by using differential calculus with
Note that this change is the moment
of inertia of the shell!
Mass m
Figure P4.57
2R
4.58 Find the product of inertia
lem 4.48.
4.59 Find
for the welded body
for the ring of Prob­
of Problem 4.50.
4.60 Determine
in Problem 4.49. The xy plane con­
tains the centers of
and
4.5
Figure P4.61
The Mass-Center Form of the Moment Equation
of Motion
Development of the Equations of Plane Motion
In Chapter 2 w e developed several different forms of Euler's second law.
In this section w e will continue to study t h e mass-center form, Equa­
tion (2.43):
(2.43)
w h e r e C is the mass center of a n arbitrary b o d y
If w e n o w restrict the
body
to b e rigid a n d t h e reference frame to b e a n inertial frame in
Equation (4.4), then w e m a y substitute from that equation for H into
Equation (2.43) above a n d obtain:
c
(4.11)
Page 247
At this stage w e m u s t m a k e a decision with regard to h o w the a n d
axes of Equation (4.11), w h i c h h a v e their origin at C, will b e allowed to
change relative to t h e inertial frame of reference w h e r e the derivative in
Equation (4.11) is to be taken, a n d to which
of the b o d y is referred.
Note that the direction of the z axis h a s already b e e n fixed perpendic­
ular to the reference plane. If w e fix the directions of x a n d y relative
to t h e inertial frame, t h e n a n d (as well as
are constant relative to
that frame b u t
and
are in general time-dependent. This choice is
very difficult to deal with.
Question 4.5
However,
will not change in this case. Why not?
A m u c h m o r e convenient choice is to let the axes
a n d z all b e fixed in
t h e b o d y so that t h e m o m e n t s a n d products of inertia are all constant.
N o w a n d are t i m e - d e p e n d e n t relative to the inertial frame
Figure 4.6
s h o w s the unit vectors a n d (fixed to
expressed in terms of a n d
(which are fixed in t h e inertial reference frame
Figure 4.6
The derivatives of these unit vectors, obtained earlier as Equations (3.41)
a n d (3.42), are:
(4.12a)
Answer 4.5 Let (x, y, z) b e fixed in t h e reference frame
A s t h e b o d y rotates w i t h r e ­
spect to frame
its m a s s is t h e n distributed differently, at different times, w i t h respect
to a n d
But in p l a n e m o t i o n t h e z axis is fixed in direction in both t h e b o d y a n d in
space. T h u s since t h e s q u a r e of t h e distance from t h e z axis is a l w a y s t h e s a m e r re­
gardless of t h e orientation of a n d
w e see t h a t
d o e s n o t c h a n g e as t h e b o d y t u r n s .
2
Page 2 4 8
(4.12b)
Therefore, carrying out the differentiations in Equation (4.11), w e have
or
(4.13)
This expression represents three scalar equations:
(4.14a)
(4.14b)
(4.14c)
Equations (4.14a,b), along with
tell us about the nature of reac­
tions necessary to maintain the plane motion. If
and
are both zero,*
then
and the system of external forces (loads plus
reactions) has a planar resultant, with this plane containing the mass
center. Thus, with a coplanar system of external loads, the resultant
reactions must have a resultant in that same plane. This is the basis for
"two-dimensionalizing" the analysis of problems for which the two
products of inertia above vanish; w e shall first work with symmetrical
bodies for which this is the case. Then later in this chapter w e shall
examine some problems in which
and
are not both zero. In any case,
however, the rotational motion of the body is governed by the simple
kinetics equation
(4.14d)
We see that while force produces acceleration with the "resistance"
being the mass, it is also true that moment produces angular acceleration
with the "resistance" being the body's moment of inertia. Note also that
the moment of the external forces about the z axis through the mass
center is
and w e have
Thus the resultant m o ­
ment about this axis equals the moment of inertia about the axis multi­
plied by the angular acceleration of the body, regardless of whether or
not the products of inertia vanish.
We will n o w restrict our attention in the remainder of this section to a
special class of problems of plane, rigid-body motions. This class is de­
fined by the following pair of conditions:
1.
(usually because the body is either symmetric about
the plane of motion of the mass center or else has an axis of physical
symmetry which remains normal to the reference plane of motion); and
2. The external loads have a planar resultant with the plane contain­
ing the mass center.
When
or
is not zero, t h e r e h a s to b e a n o n z e r o
p r e s e n t to m a i n t a i n
t h e m o t i o n ; t h e s e are u s u a l l y f o r m e d b y lateral forces (such as b e a r i n g reactions) at dif­
ferent positions a l o n g t h e z-axis.
Page 249
These two conditions necessitate external reactions likewise equipollent
to a coplanar (in the same plane as the loads) force system. Euler's laws
are then effectively reduced to the following three scalar equations:
(4.15a)
(4.15b)
(4.15c)
where x and y are coordinates of the mass center in a rectangular
coordinate system fixed in the inertial frame.* Equations (4.15a,b) are of
course the x and y components of the mass center equation of motion
valid for any body, rigid or not. Since there will be n o confusion about the
axis in question, w e shall often write Equation (4.15c) as simply
c
c
or
2M
C
=
Ia
c
(4.16)
Helpful Steps to Follow in Generating and Solving the Equations of Motion
In some instances Equations (4.15a-c) will yield a differential equation(s)
that can be readily integrated so that w e predict the ongoing motion of a
body. More commonly w e shall be dealing with what w e might call
"snapshot" problems where at a specific instant w e calculate forces and
accelerations. These problems are rather natural extensions of statics, all
of the operations in the analysis being algebraic. For problems of either
class the following steps are recommended:
1. Draw a free-body diagram (FBD) of each body in the problem.
2. Define a set of unit vectors, or, equivalently, a coordinate system,
in terms of which unknown forces and accelerations may be expressed.
3. Substitute into the three equations of motion; some may prefer to
do this in vector algebraic form, explicitly displaying previously defined
unit vectors, while others may prefer to initiate the analysis using the
three component equations (4.15a,b and 4.16).
4. Often the number of scalar unknowns will exceed the number of
independent equations (maximum of three), and w e must look for sup­
plementary information. This might mean nothing more than applying
the Coulomb law of friction, but often the supplementary information
will be in the form of a kinematic constraint. For example, for a wheel
having its mass center at the geometric center, rolling would imply (with
appropriate definitions of variables)
If some point A has
its motion constrained, then the restriction o n its acceleration along
with a = a +
can be used to relate acceleration of
c
A
W h i c h c o m p o n e n t e q u a t i o n s a r e used a n d t h e specific forms they take d e p e n d , of
course, o n w h a t t y p e of c o o r d i n a t e system is used to describe t h e m o t i o n of C. A rectan­
gular coordinate s y s t e m is t h e n a t u r a l choice for m o s t of t h e p r o b l e m s t a k e n u p in this
chapter. A polar c o o r d i n a t e s y s t e m is t h e n a t u r a l choice, h o w e v e r , for p r o b l e m s of or­
bital m e c h a n i c s (see Section 8.4).
Page 2 5 0
the mass center and angular acceleration of the body (presuming veloci­
ties, and thus
are known). (Of course, a and a are the kinematical
variables which naturally appear in the equations of motion, as w e have
seen.)
c
5. Solve for unknown forces a n d / o r accelerations. All problems
appearing in this text are "rigid-body dynamically determinate," so this
will always be possible. In the case of "movie" problems, integrate accel­
erations to get velocities and then velocities to get positions as functions
of time.
6. Be sure to check on dimensional consistency of results and, in the
case of numerical results, check to see that units are correct (especially
important in these days of transition from U.S. to SI units). Also, see if
your answer seems to make sense.
We n o w proceed toward a set of examples which will make use of the
equations (4.15a,b and 4.16), which w e have developed in this section.
These examples are designed to illustrate the kinds of problems the
student should learn to solve. The first one is a "movie" problem in which
one resultant force component and the moment are zero, while the other
force component is not.
EXAMPLE 4 . 1 7
A horseshoe pitcher releases a horseshoe with
= 4 . 7 1 rad/sec in the posi­
tion shown in Figure E4.17a. If the horseshoe turns exactly once in plane motion
and scores a ringer, find the initial velocity of the horseshoe's mass center.
CLANK
3 5
ft
4 0 It
Figure E4.17a
A free-body diagram of the horseshoe in flight is shown in Figure E4.17b. We
write and then integrate Equations (4.17a-c):
(the initial
velocity of C
in x-direction)
Figure E4.15b
since x = 0
at t = 0
c
(the initial velocity of C
in y-direction)
since horseshoe is released
at y = 3.5 ft
c
Page 251
Now when the horseshoe lands, 6 = 2n, so
Thus
and
Hence the initial velocity of the horseshoe's mass center is
The next example involves rolling, and is a "movie" (ongoing time)
problem:
EXAMPLE 4 . 1 8
The cylinder (mass m, radius r) is released from rest on the inclined plane shown
in Figure E4.18a. The coefficient of friction between cylinder and plane is µ.
Determine the motion of C, assuming that µis large enough to prevent slipping.
(How large must it be?)
Solution
Figure E4.18a
As in statics, a good first step is to draw a free-body diagram (see Figure E4.18b).
We are to assume that the cylinder rolls. In this case the friction force is an
unknown and has a value satisfying
where
= µN from the study of Coulomb friction in statics. We shall also use a
kinematic equation expressing the rolling. After solving for
we shall then
impose the condition that it be less than µ.N, since we know the cylinder is not
slipping.
We choose x, y, and Øas shown, motivated by the fact that C will move down
the plane and the cylinder will rum counterclockwise. The equations of motion
are
(1)
Figure E4.18b
(2)
(Note that, kinematically, y is constant so that
c
vanishes.)
(3)
We can solve (2) for N, getting N = mg cos There remain two equations in the
three unknowns
andØ.We must therefore supplement our equations of motion with the remaining kinematics result, which comes from the rolling condition:
(4)
Page 252
Solving Equations (1), (3), and (4) gives
Integrating twice, and noting that the integration constants vanish, we get
And since we are told that
that is,
then we have
for the rolling motion to occur.
In the preceding example, things would be quite different if the value
of the friction force that satisfies the equations, namely = (mg sin ) / 3 ,
were larger than
=
= mg cos
So if the solution were to yield
then there would be insufficient friction to permit rolling. We would then
need to abandon* the rolling condition, replacing it by the known maxi­
m u m value of f,i.e.,
but n o w w e know
where N still equals mg cos
And since Equation (3) still holds,
or
Therefore, integrating twice, w e get
* W h e n y o u a s s u m e s o m e t h i n g a n d later arrive at a contradiction of fact, t h e n t h e logi­
cal conclusion is t h a t t h e a s s u m p t i o n w a s invalid.
253
w h e r e t h e integration constants are zero since
tion (1) n o w yields, with
= 0 at t = 0. Equa­
or
Thus
a n d t h e solutions for t h e motion [x^f) a n d (f)] are i n d e e d quite different
w h e n t h e cylinder turns a n d slips t h a n they are w h e n it rolls. If = 0.5
a n d = 3 0 ° , for example, then
a n d the cylinder rolls. But if
= 0.2 a n d
= 60°, t h e n
a n d t h e cylinder slips. N o t e that in general (as one w o u l d expect) the
cylinder rolls for larger a n d smaller
Finally, note that if w e wish to distinguish b e t w e e n static a n d kinetic
coefficients of friction
and
t h e n the rolling assumption w o u l d b e
correct if tan
. But if t a n
w e w o u l d t h e n use
in
t h e r e m a i n d e r of t h e solution, a n d t h e n in the answers for x a n d would
become
Next w e take u p a n o t h e r rolling problem, only this time t h e m a s s
center a n d geometric center are n o t t h e s a m e point. As a result, the
kinematics equations are m o r e difficult. This example is a " s n a p s h o t "
(occurring at one instant of time) problem.
c
EXAMPLE 4 . 1 9
Figure E4.19a
The rigid body in Figure E4.19a consists of a heavy bar of mass m welded to a
light hoop; the radius of the hoop thus equals the length of the bar. Find the
minimum coefficient of friction between the hoop and the ground for which the
body will roll when released from rest in the given position.
Solution
The free-body diagram is shown in Figure E4.19b along with the base vectors
adopted for the problem.
Question 4.6
problem?
Figure E4.19b
downward
Why is this a good choice of base vectors to use in this
Answer4.6Because t h e initial acceleration of t h e m a s s center C will b e to the right a n d
a n d t h e initial a n g u l a r acceleration will b e clockwise
Page 254
Note that the gravity force resultant acts through the center of the bar since we
are neglecting the weight of the hoop.
Next we write the three differential equations of motion, letting a =
+
and a =
c
(1)
(2)
(3)
These equations contain the unknowns f, N,
and a. We draw upon kine­
matics for two more equations. We know that the accleration of the geometric
center of the hoop is
=
so that
or
Therefore, equating the i coefficients and then the coefficients, we find
(4)
(5)
The student may wish to verify that (4) and (5) also result from relating a to
=
true for any round body rolling on a flat, fixed plane. In this problem,
is then zero at release because, until time passes, is still zero.
Solving Equations (1) to (5) for the five unknowns gives the results:
c
To complete the solution, we must get the coefficient of friction µinto the picture.
We know that for any friction force
Therefore, in our problem,
so that
This means that for the body to roll, a friction coefficient of at least = 6 / 1 3 is
required; this is then the desired minimum.
We emphasize that students should always make "eyeball checks" of their
answers — glancing over the results to be sure they make sense physically. In this
problem, for instance, note that:
Page 255
1.
N < mg as expected, for otherwise the mass center could not begin to move
downward as the body rolls.
2.
is positive, and therefore in the correct direction to (since it is the only
force in the x-direction) move the mass center to the right.
and a are all positive and therefore in the expected directions.
3.
The next example features another "snapshot" problem — one
which w e are examining only at one instant. It also involves the interest­
ing constraint of a taut string.
EXAMPLE 4 . 2 0
A uniform rod is supported by two cords as shown in Figure E4.20a. If the
right-hand cord suddenly breaks, determine the initial tension in the left cord AD.
("Initial" means before the rod has had time to move and before it has had time to
generate any velocities.)
64.4
Ib
Solution
Using the free-body diagram in Figure E4.20b, the equations of motion are, with
a =
+
and a =
c
Figure E4.20a
(1)
(2)
30'
(3)
64 4 lb
Figure E4.20b
Unfortunately, Equations (1-3) contain four unknowns (T, , , and a). Thus
we seek an additional equation in these unknowns from kinematics. The point A
is constrained to move (see Figure E4.20c) on a circle of radius I about D. Thus
point A has the tangential and normal components of acceleration shown (see
Section 1.7). Furthermore, v = 0 at the instant of interest (nothing is moving
yet!).
We may relate this a to a :
A
A
c
(4)
Figure E4.20c
Here now is a nice shortcut:
If we dot Equation (4) with a unit vector in the direction
, we will elimi­
nate a (because it is perpendicular to that direction). This is easier than solving
the two
and equations." Such a unit vector is
A
Page 256
so that, doing the dotting,
or
(5)
Equations (1), (2), and (3) yield:
Substituting these three results into Equation (5) results in:
or
Note that before the right-hand string was cut, the tension, from statics, was:
Forces in inextensible strings (ropes, cables, cords) are capable of changing "in­
stantaneously," and indeed we see that this is the case in this problem.
Question 4.7
Can spring forces change instantaneously in this way?
In the preceding example, back-substitution immediately yields
Thus the mass centeT will
start to m o v e off to the
left and d o w n .
Thus the body will
start to turn clockwise.
The acceleration of A follows from Equation (4):
a n d , as a check, t h e direction of a ,
A
Answer 4.7
N o , t h e l e n g t h m u s t c h a n g e a n d t h a t takes time.
, is 3 0 ° .
Page 257
2
The m a g n i t u d e of n is
= 4.97 f t / s e c , which
is
(see Section 1.7) at t h e initial instant. It is interesting to n o t e that
the initial angular acceleration of t h e string DA is
or 4 . 9 7 /
Finally, w e close t h e section with a n example containing t w o bodies
in rolling contact. The plate is simply translating, b u t t h e pipe h a s a more
complicated motion: it rolls o n t h e pipe, b u t not o n t h e inertial frame
(ground).
A
EXAMPLE 4 . 2 1
Mass m
Mass m
Force P is applied to a plate that rests on a smooth surface. (See Figure E4.21a.)
Find the largest force P for which the pipe will not slip on the plate.
Solution
For the pipe (Figure E4.21b), with a =
(1)
(2)
Figure E4.21a
(3)
(Note that
If the thickness (r — r ) is not
given, assume it is small.)
For the plate (Figure E4.21c), we note that only the x equation of motion is of
help;
= 0 gives N = N + Mg = (m + M)g as expected, and dimen­
sions are not given so moments cannot be taken. (The moment equation would
only give us the location of N , anyway.) Therefore
o
i
2
2
(4)
Figure E4.21b
Elirninating f between (1) and (3) gives
(5)
And between (1) and (4) gives
(6)
We note that if m = m + M and C is the mass center of pipe plus plate, then
Equation (6) could have been written immediately from
for the
combined system. Here
= P; the right side follows from two derivatives of
the definition of the mass center
= mx + Mx ).
The kinematics equation is tricky here. It is a rolling condition, but we must
remember that x and x are necessarily measured relative to an inertial frame,
here assumed to be fixed in the ground. Thus it is
that is related to a For
no slip,
which, when differentiated, yields
T
Figure E 4 . 2 1 c
T
c
G
G
c
(7)
Substituting a from (5) into (7) relates the accelerations of the two mass
centers:
(8)
*S o m e t i m e s G is u s e d to d e s i g n a t e a m a s s center.
This difference is just t h e acceleration of C in t h e frame consisting of t h e translating plate.
Page 258
Then (6) and (8) may be combined to give
(9)
And combining (9) and (1) gives us the relationship between P and
(10)
Since
N for no slip, (10) gives:
Any larger P than (m + 2M)
PROBLEMS
•
Section
, will cause the pipe to slip on the plate.
4.5
4.12 A uniform sphere (radius r, mass m) rolls on the
plane in Figure P4.62. If the sphere is released from rest at
f = 0 when x = L, find x(t).
4 63 A symmetric body has mass m and radius R; a
cord is wrapped around it as shown in Figure P4.63,
Compute the downward acceleration of the center C if is
(a) a cylinder; (b) a sphere (with a small slot to accommo­
date the cord); (c) a thin ring. Hint, Work the problem just
once with radius of gyration k ; then substitute the three
values
and 1R far k .
Smooth
c
c
0
60
Fioure P4.64
4.65 Sally Sphere, Carolyn Cylinder, Harry Hoop, and
Wally Wheel each have mass m and radius R. Wally's
spokes and rim are very fight compared to his hub. (See
Figure P4.65.) They are going to have a race by rolling
down a rough plane. Give (a) the order in which they
finish and (b) the times.
Figure P4.62
Figure P4.63
4.64 The cord in Figure P4.64 is wrapped around the
cylinder, which is released from rest on the 60° incline
shown. Find the velocity and position of as a function of
time t.
Sally
Figure P4.64
Carolyn
Harry
Wally
Page 259
4.66 In the preceding problem, Wally and Carolyn are
connected by a bar of negligible mass and released from
rest on the same incline. (See Figure P4.66.) Determine
the force in the bar.
4.69 The uniform sphere (mass m, radius r) in Fig­
ure P4.69 is at rest before P is applied. If µis the coeffi­
cient of friction between sphere and floor,
a. find the maximum P for there to be no slip;
b. for P twice that found in (a), find a and a.
Note that
does not equal
as it would if
there were no slipping, i.e., if the sphere were
rolling.
c
Rod
Figure P4.69
Figure P4.66
4.67 Repeat the preceding problem, but suppose that
Wally and Carolyn switch places.
4.70 The uniform cylinder in Figure P4.70, of mass m
and radius r, is at rest before it is subjected to a couple of
moment M The coefficient of friction between cylinder
and floor is
4.68 The two pulleys
and
and the block in Fig­
ure P4.68 each have mass m and are connected by the
cord.
a. Find the largest value of M for which there is
no slip.
b. For M twice the value found in (a), find a and a.
a. Write a brief paragraph explaining in words
why the system cannot be in equilibrium. Start
with, "If the system were in equilibrium, the
tension in the rope above would equal mg."
Then follow the rope around the pulleys until
you reach a contradiction.
b. Find the acceleration of C .
0
0
0
c
Figure P4.70
1
Cord
Figure P4.68
4.71 Find the ratio of r to R for which the force T in
Figure P4.71 will cause the wheel to roll (no slip) no mat­
ter how small the friction. Treat the wheel as a uniform
cylinder.
Figure P4.71
Page 260
4.72 Force T is given to be small enough, and the friction
coefficient large enough, that both wheels in Fig­
ure P4.72 will roll on the plane.
a. Give arguments why one wheel rolls left and
the other right.
b. Find the ratio of r to R for which the accelera­
tions of C are equal in magnitude.
4.76 The bowling ball in Figure P4.76 is released with
= 22 ft/sec and co = 0 as it contacts the surface of the
alley. Neglecting the effect of the three finger holes, and
using a coefficient of friction of 0.3, find the distance
traveled by the center of the ball before slipping stops.
4.77 The force P = 60 N is applied as shown in Fig­
ure P4.77 to the 10-kg cylinder
originally at rest be­
neath the mass center of the thin, 5-kg rectangular plate
The coefficient of friction between and is 0.5, and
the plane beneath is smooth. Determine: (a) the initial
acceleration of C; (b) the value of x when is slipping on
both surfaces. The length of is 2 m.
4.78 The constant force F is applied to the cylinder,
initially at rest, as shown in the two drawings constituting
Figure P4.78. Show in the following two ways that the
cylinder will slip provided that
0
Figure P4.72
4.73 The uniform sphere in Figure P4.73, of mass m and
radius r, is at rest when it is subjected to a couple of
moment M . If there is no slip, find the acceleration of the
center of the sphere.
0
4.74 A cylinder spinning at angular speed
rad/sec
clockwise is placed on an inclined plane. (See Fig­
ure P4.74.) Show that the cylinder center will begin mov­
ing up the plane if
tan Why does this result have
nothing to do with the size of
?
a. Assume rolling; then obtain the inequality from
after solving for and N.
b. Assume slipping; then integrate x and to
obtain x and 0, and then find the velocity of
the contact point B; if it is to the right (that is,
positive), this is consistent with
to the
left and we have slipping.
c
c
4.75 The cylinder of weight W and radius r shown in
Figure P4.75 has an angular velocity of 100 r a d / s clock­
wise. It is lowered onto the rough incline. If its center C is
observed to remain momentarily at rest, determine the
coefficient of sliding friction. Find how long the center C
remains at rest.
Figure P4.76
Figure P4.73
Figure P4.74
Figure P4.77
100 rad/s
0.1 m
30
0
Figure P4.75
Figure P4.78
Page 261
Cylinder,
Drum
Coefficient of
friction = µ
Bar
4.5 in.
6 in.
3 lb
Figure P4.82
Bar
3 1b
Figure P4.79
4.79 Two drums of radius 4.5 in. are mounted on each
end of a cylinder of radius 6 in. to form a 40-lb rigid body
with radius of gyration
= 5 in. (See Figure P4.79.)
Ropes are wrapped around the drum and tied to a hori­
zontal bar to which a 3-lb force is applied. As rolls from
rest, tell (a) the number of inches of rope wound or un­
wound (tell which) in three seconds and (b) the minimum
friction coefficient needed for the rolling to take place.
4.80 Find the range of possible values of the couple M
for which the cylinder in Figure P4.80 will not slip in
either direction when released from rest on the incline.
The mass is 15 kg; the radius is 0.2 m; and the coefficient
of friction is = 0 . 2 .
3 ft
1 ft
0
4-. 81 An airplane lands on a level strip at 200 mph. (See
Figure P4.81.) Initially, just before the wheels touch the
runway, the wheels are not turning. After they touch the
runway they will skid for some distance and then roll free.
If during this skidding the plane has a constant velocity of
200 mph and the normal force between the wheel and
the runway is 10 times the wheel weight, find the length
of the skid mark. (The coefficient of friction is the radius
of gyration of the wheel is three-fourths of its radius.)
Figure P4.83
4.82 A hula hoop (mass m, radius r) is thrown forward
with backspin;
=
to the right and
=
counter­
clockwise. (See Figure P4.82.)
a. How long and how far does the mass center
move before
stops slipping?
b. Find the relationship between
and
such
that when
stops slipping: (i) it rolls right; (ii)
it rolls left; (iii) it stops.
4.83 The strong, flexible cable shown in Figure P4.83
is wrapped around a light hub attached to the 130-lb cyl­
inder
Find the angular acceleration of upon release
from rest. Note that it is impossible for the wheel to roll
down the plane (meaning without slipping); to do so
the cord would have to break.
4.84 Repeat the previous problem for
Figure P4.8G
Cord
= 0.25.
4.85 In Figure P4.85, find how far down the incline C
travels in 5 s if the 20-kg cylinder is released from rest.
Airplane
v = 200 mph
Wheel
µ= 0.45
R = 0.5 m
o
60
Figure P4.81
Figure P4.85
Page 262
4.86 The 50-lb body in Figure P4.86 may be treated as
a solid cylinder of radius 2 ft. The coefficient of friction
between and the plane is = 0 . 2 , and a force P = 10 lb
is applied vertically to a cord wrapped around the hub.
Find the position of the center C, 10 sec after starting from
rest.
S(32.21b)
21b
1 ft
2 ft
6 4 . 4 lb
3 1b
4.87 Given that the slot (for the cord) in the cylinder in
Figure P4.87 (mass 10 kg) has a negligible effect on I ,
find:
c
a. The largest for which no motion down the
plane will occur
b. The time required for C to move 3 m down the
incline if = 6 0 ° .
100 ft
Figure P4.88
Thin rim (3 lb)
1 lb (each of 8)
Hub (2 lb)
Figure P4.86
cord
Figure P4.89
0.5 m
0.8 m
Figure P4.87
4.88 A light 100-ft cord is wrapped around the 32.2-lb
spool
which is pinned at C to the cart
(see
Figure P4.88). The radius of gyration of with respect
to an axis normal to the figure at C is 1.3 ft. The cart (with­
out
has weight 64.4 lb. The wheels of are small and
light, so that friction beneath them is negligible. The 2and 3-lb forces are applied to the system at rest. If upon
complete unwrapping the cord is to end up between
points P and Q in the lower figure, where should be
originally parked along PQ?
4.89 A child pulls on an old wheel with a force of 5 lb by
means of a rope looped through the hub of the wheel.
(See Figure P4.89.) The friction coefficient between wheel
and ground is = 0 . 2 . Find I for the wheel, and use it to
determine the location of C after 3 sec.
c
4.90 The wheel shown in Figure P4.90 has a mass of
10 kg, a radius of 0.4 m and a radius of gyration with
respect to the z-axis through C of 0.3 m. Determine the
angular acceleration of the wheel and how far the mass
center C moves in 3 seconds if the wheel starts from rest.
40 N
60 N
4m
µ= 0.5
Figure P4.90
4.91 Rework the preceding problem if the friction coef­
ficient is changed to = 0 . 2 .
Page 263
4.92 Two cables are wrapped around the hub of the
10-kg spool shown in Figure P4.92, which has a radius of
gyration of 500 mm with respect to its axis. A constant
40-N force is applied to the upper cable as shown. Find
the mass center location 5 s after starting from rest if: (a)
= 0.2; (b) = 0.5.
(64.4 lb)
(96.6 lb)
10 lb
Cord
1 ft
3 ft
Cable
600 m m
Figure P4.94
40 N
Cable
200 mm
4.95 Assume that enough friction is available to prevent
the cylinder in Figure P4.95 from slipping.
Figure P4.92
4.93 A sphere of radius ft and weight 16.1 lb is pro­
jected onto a horizontal plane (Figure P4.93). Its center
has initial velocity
at t = 0 and the sphere has initial
angular velocity
defined as shown. If the coefficient
of sliding friction between the sphere and the plane is
0.15, plot graphs of distance gone (x ) against time t up to
t = 3 sec for the following cases:
a. Show that
(i)
rolls to the right if
(r/R).
(ii)
rolls to the left if
(r/R).
(iii)
is in equilibrium if
(r/R) (and
will translate if P increases enough to over­
come friction).
b. Find and a if r = 0.2 m, R = 0.4 m,
P = 20 N, mg = 40 N, and 6 = 45°.
c
a.
b.
c.
= 10 ft/sec;
= 1 0 ft/sec;
= 1 0 ft/sec;
Cord
= 100 rad/sec
= 50 rad/sec
= 30 rad/sec
Figure P4.95
A homogeneous spool of weight W rolls on
an inclined plane; a string tension of amount 4W acts
up the plane as shown in Figure P4.96. With I given
approximately by WR /2g, find the acceleration of C.
Assume unlimited friction.
c
2
Figure P4.93
4.94 The cylinder in Figure P4.94 has a thin slot cut in
it which doesn't affect its moment of inertia appreciably.
A cord is wrapped in the slot and connects to the cart
which rests on the plane on small, light wheels. The force
of 10 lb is applied to with the system initially at rest.
Find the length of unwrapped cord after 4 seconds
elapse. Assume enough friction to prevent slip of on the
plane.
30
Figure P4.96
c
Page 264
4.97 Pulley
in Figure P4.97 weighs 100 pounds and
has a radius of gyration about the z-axis through O of
k = 7 in. Pulley weighs 20 lb and has k = 3 in. Find
the angular acceleration of
just after the system is re­
leased from rest. Assume the rope doesn't slip on
but
that there is no friction between and the rope. Is this the
angular acceleration for later times as well?
0
c
8
in.
4
1
in.
4.99 Wheel is made up of the solid disk
rim
and
four spokes Masses and radii are given in Figure P4.99
and the table.
a. Compute
for the wheel.
b. The coefficient of friction between and the
plane is = 0.3. If a cord is wrapped around
the disk and connected to the 50-kg body de­
termine the acceleration of the mass center C
of
Part
M a s s (kg)
20
5 (each)
10
(100 lb)
Rope
(20 lb)
Cord
2 in. radius
0.2 m
50
(50 lb)
kg
Figure P4.99
Figure P4.97
4.98 The 32.2-lb body in Figure P4.98 is a spool hav­
ing a radius of gyration k = 6 in. about its axis. Cords are
wrapped around the peripheries; one is connected to a
ceiling, the others to the 48.3-lb block Find the acceler­
ations of the centers C (of
and B (of
c
The radius of gyration of the 20-kg wheel in
Figure P4.100 with respect to its axis is 0.3 m. Motion
starts from rest. Find the acceleration of the mass center
C, and determine how far C moves in 5 s.
4in
20°
8 in.
0.5 m
I50N
0.15
Figure P4.98
Figure P4.100
Page 2 6 5
4.101 A string is wrapped around the hub of the spool
shown in Figure P4.101. There are four indicated string
directions. For the direction that will result in the largest
displacement of C in 3 s, find this displacement. Assume
sufficient friction to prevent slipping. The spool has a
mass of 12 kg and a radius of gyration about z of 0.6 m.
Each force equals 10 N, and the spool starts from rest.
0.4 m
1m
0.4 m
10 kg;
k
c
= 1 m
0.2 m
c
Cord
Disk, 7 kg
0.5 m
1m
5 Kg
Figure P4.103
Figure P4.101
4.102 Cylinder in Figure P4.102 has a mass of 4 slugs,
and the effect of the hub on its moment of inertia is negli­
gible. It is connected by means of a cord to the 1-slug
block The mass of the pulley is negligible. The coeffi­
cient of friction between and the plane is = 0.5, and
the radii of are given in the figure. If the system is
released from rest, determine the time that will elapse
before hits the ground.
2 ft
4.104 Disks
and each weigh 64.4 lb and are rigidly
attached to the light shaft that joins their centers. (See
Figure P4.104.) A 96.6-lb cylinder has a hole drilled
along its axis, through which passes. A force of 20 lb is
applied horizontally to an inextensible string wrapped
around If friction is negligible between and and if
and roll on the plane, find:
a. The angular acceleration of the cylinder
b. The angular acceleration of the disks
c. The minimum coefficient of friction between
disks and plane for no slipping.
1 ft
Cord
2 ft
1 ft
2 0 ft
Figure P4.104
Figure P4.102
4.103 Find how long it takes for to roll off the plane in
Figure P4.103, assuming sufficient friction to prevent
slipping. The system is released from rest.
4.105 Rework the preceding problem, but this time as­
sume that the string is wrapped so that it comes off the
bottom of
Page 266
4.106 The two wheels are identical 16.1-lb cylinders with
smooth axles at their centers. (See Figure P4.106.) The
carriage weighs 32.2 lb and has its mass center at C. The
cylinders do not slip on the inclined plane. Find the accel­
eration of point Q.
W r a p p e d cord
Figure P4.109
30
4.110 The semicylinder in Figure P4.110(a) is released
from rest, and there is enough friction to prevent slipping
throughout the ensuing motion (Figure P4.110(b)).
0
Figure P4.106
4.107 Cylinder inFigure P4.107 weighs 100 lb; it is roll­
ing on the plane and is pinned at its center C to the 10-lb
rod
If
is initially 10 ft/sec to the left, and if the
coefficient of kinetic friction between the plane and each
body is = 0.4, determine how long it will take the sys­
tem to come to rest.
a. Find
b. Write the three differential equations of motion
of the body (good at any angle
c. Find the two equations relating and
to
and
d. Eliminate N,
and
and obtain the single
differential equation in the variable (t). Note
the complexity of the equation!
Mass m
Figure P4.107
4.108 A uniform half-cylinder of radius r and mass m is
held in the position shown in Figure P4.108 by the string
tied to B. Find the reaction of the floor just after the string
is cut. There is sufficient friction to prevent slipping.
(b)
(a)
Figure P4.110
B
A
Figure P4.108
4.109 In Figure P4.109 the force P is applied to the cord at
t = 0, when the 25-N cylinder is at rest. Find the position
of the mass center when t = 6 s.
4.111 The 15-lb carriage shown in Figure P4.111 is sup­
ported by two uniform rollers each of weight 10 lb and
radius 3 in. The rollers roll on the ground and on the
carriage. Determine the acceleration of the carriage when
the 5-lb force is applied to it.
Figure P4.111
Page 267
4.112 The 128.8-lb homogeneous plank shown in Fig­
ure P4.112 is placed on two homogeneous cylindrical
rollers, each of weight 32.2 lb. The system is released
from rest. Determine the initial acceleration of the plank if
no slipping occurs. Is this the acceleration for later times
as well?
4.116 The pipe in Figure P4.116 has a mass of 500 kg and
rests on the flatbed of the truck. The coefficient of friction
between the pipe and truck bed is = 0 . 4 . The truck
starts from rest with a constant acceleration a .
0
a. How large can a be without the pipe slipping
at any time?
b. For the value of a in part (a), how far has the
truck moved when the pipe rolls off the back?
0
0
10 ft
2 m
30
0
Figure P4.112
Figure P4.116
Figure P4.113
4.113 Body in Figure P4.113 is a rigid plate of mass M,
resting on a number n of cylinders each of mass m and
radius R. Force F is constant and starts the system moving
from the position shown. If there is no slipping at any
surface, find: (a) the acceleration of the plate and (b) its
position x as a function of M, m, F, n, and time t.
4.117 The uniform sphere (mass = 1 slug, radius = 1 ft)
and the slab (mass = 2 slugs) shown in Figure P4.117 are
at rest before the force P = 24 lb is suddenly applied to
the slab. The coefficient of friction is 0.2 between the
sphere and slab and between the slab and horizontal
plane, (a) Does the sphere slip on the slab? (b) What is the
acceleration of the center of the sphere?
c
4.114 A 6-ft gymnast makes a somersault dive into a net
by standing stiff and erect on the edge of a platform and
allowing himself to overbalance. He loses foothold (with­
out having slipped) when the platform's reaction on his
feet becomes zero; he preserves his rigidity during his fall.
Show that he falls flat on his back if the drop from the
platform to the net is about 43 ft.
Figure P4.117
4.118 The homogeneous cylinder
in Figure P4.118
weighs 64.4 lb. The acceleration of the 96.6-lb cart is
10 ft/sec to the right.
2
The homogeneous cylinder in Figure P4.115 is at
rest on the conveyor belt when the latter is started up with
a constant acceleration of 3 ft/sec to the right. If the
cylinder rolls on the belt, find the elapsed time when the
cylinder reaches the end A .
2
10 ft
a. Determine the acceleration of the center C of
the cylinder and the friction force exerted on
by if there is sufficient friction to prevent
slipping.
b. How large does the friction coefficient have to
be for this to occur?
2.5 ft
Figure P4.115
1 ft
Figure P4.118
Page 268
4.119 The homogeneous cylinder . in Figure P4.119
weighs 64.4 lb and rolls on the 96.6-lb truck The mass
of the truck rollers may be neglected. Find the force P
such that C does not move relative to the plane.
4.123 After release from a slightly displaced position, the
rod in Figure P4.123 will remain in contact with the floor
throughout its fall. Describe the path of C and find the
reaction onto the floor just before the rod becomes hori­
zontal.
4.124 The uniform slender bar of mass m and length L is
released from rest in the position shown in Figure P4.124.
Find the force exerted by the smooth floor at this instant.
6 4 . 4 lb
Rod
30
0
Figure P4.119
Smooth plane
Figure P4.123
Figure P4.124
Small wheels
Figure P4.120
4.120 The system shown in Figure P4.120 is initially at
rest. A force P is then applied that varies with time ac­
a. Determine the minimum value of say
cording to P = 7f , where P is in newtons and t in seconds.
, required of
to prevent
A fromcylinder and cart is
If the coefficient
friction end
between
= 0.5, find how much time elapses before the cylinder
starts to slip on the cart.
4.125 A thin rod AB of length and mass m is released
from rest in the position shown in Figure P4.125. Point A
of the rod is in contact with a surface whose coefficient of
friction is
2
In the previous problem, determine how much time
passes (from t = 0) before the cylinder leaves the surface
of the cart. Initially, the center of the cylinder is 2 m from
the right end of the cart.
4.122 A slender homogeneous bar weighing 193 lb has
an angular velocity of 2 rad/sec clockwise and an angu­
lar acceleration of 8 r a d / s e c clockwise when in the posi­
tion shown in Figure P4.122. The wall at B is smooth; the
coefficient of sliding friction at A is 0.10. Find the reac­
tions at A and B on in this position. Hint: The force P can
be found.
slipping upon release.
b. Find the acceleration of the mass center of the
rod immediately after release for
and
for
B
0
A
30
2
Figure P4.125
13m
U6m
Figure P4.126
4
2 ft
Figure P4.122
1 ft
ft
4.126 The 30-kg sphere and 15-kg rod in Figure P4.126
are welded together to form a single rigid body. Deter­
mine the angular acceleration of the body immediately
after the right-hand string is cut.
4.127 If the right-hand string in Figure P4.127 is cut, find
the initial tension in the left string. The slender rod has
mass m and length L.
Page 269
4.131 The uniform slender bar of mass m is released from
rest in the position shown in Figure P4.131. Find a and
the tension in the inextensible cord at the instant after
release.
A
Figure P4.127
4.128 Repeat the preceding problem if the rod is replaced
by a rectangular plate suspended from the two upper
comers. The width (between strings) is B and the height
is H.
4.129 A uniform slender rod, 10 ft long and weigh­
ing 90 lb, is supported by wires attached to its ends. (See
Figure P4.129.) Find the tension in the right wire just
after the left wire is cut. Assume the wires to be inextensible.
4.130 The left end of a slender uniform bar is attached to a
light inextensible cable as shown in Figure P4.130. If the
bar has mass m and length L and is released from rest in
the position shown, find the angular acceleration of the
bar at the instant after release.
30
s
Figure P4.131
4.132 The uniform 20-lb bar is three feet long and has an
angular velocity = 3 rad/sec with v = 0 at the in­
stant shown in Figure P4.132. Neglecting interaction
with the air, what is the angular velocity of the bar after its
center has dropped 10 feet?
c
0
30
Figure P4.132
Figure P4.129
4.133 The disk shown in Figure P4.133 has mass m and
radius r. Show that at the instant the right-hand string is
cut, the tension in the other string changes to mg, so that
the acceleration of the mass center is
0
45
0
30
Figure P4.130
Figure P4.133
30
0
Page 270
4.142 A slender bar weighing 64.4 lb is attached by
massless cables to a fixed pivot A as shown in Fig­
ure P4.142. The system is swinging about A as a pendu­
lum. At = 0 the angular velocity is 2 rad / sec and
cable AD breaks. Find the tension in cable AB just after the
break.
4.134 A uniform rod
is supported by two cords as
shown in Figure P4.134. If the right-hand cord suddenly
breaks, determine the initial tension in the left cord AD.
("Initial" means before the rod has had time to move and
before it has had time to generate any velocities.)
4.135-4.140 The six equilateral triangular plates (Fig­
ures P4.135-4.140) are each supported on their right­
most comer B by a string; each has a different support
condition at the left comer A. At the instant when
the string at B is cut, find a and a in each case. The
length of each side is s.
4.143 A beam of length L and weight W per unit length is
supported by two cables at A and B. (See Figure P4.143.)
If the cable at B should break, find the shear force V and
bending moment M at section xx just after the cable
breaks. Hint: Euler's laws apply to every part of the body.
c
4.144 See Figure P4.144. Assuming that sufficient fric­
tion is present to prevent slipping between
and the
plane, find the angular accelerations of and just after
force P is applied to the bodies at rest. They are connected
by a smooth pin.
4.141 The uniform 10-lb bar in Figure P4.141 is sus­
pended by two inextensible cables. At the instant shown,
when each point in the bar has a velocity of 10 ft/sec,
the right cable breaks. Find the force in the left cable
immediately after the break.
30
0
13 ft
64.4 lb
Figure P4.134
A
A
B
B
B
C
c
c
Figure P4.135
Figure P4.142
A Moment
spring
Spring
String
Figure P4.137
Figure P4.136
A
A
B
B
A
c
C
A
B
B
C
Figure P4.143
Figure P4.138
Figure P4.140
Counterweight
(pulley is light)
Cylinder,
mass = 5 kg
Figure P4.139
2 ft
2 ft
0.3 m
Starts here at t = 0
at rest
4ft
Figure P4.141
Slender bar,
mass = 2 kg
Figure P4.144
Page 271
4.145 Rods
and
each have mass m. (See Fig­
ure P4.145.) Upon release from rest in the horizontal po­
sition indicated, find the reactions at O, and at A,
onto
Determine the angular acceleration of if is fixed in the
inertial frame of reference. Treat the gears as uniform
disks. The plane of the page is horizontal.
Mass M
Pin
A
B
Figure P4.145
60
(mass m)
0
A
D
Figure P4.147
Figure P4.146
4.146 Two uniform bars and are released from rest in
the position shown in Figure P4.146. Each bar is 2 ft long
and weighs 10 lb. Determine the angular acceleration of
each bar and the reactions at A and D immediately after
release. The rollers are light and the pins smooth.
4.147 A constant torque T is applied to the crank arm
of the planetary mechanism shown in Figure P4.147. The
axes of the identical gears and are vertical, and the
ends of the crank are pinned to the centers of and
0
Figure P4.148
4.148 Cylinder
in Figure P4.148 rolls down a wedge
that can slide without friction on a smooth floor. Show
that the acceleration of wedge is a constant given by the
equation
EXTENDED PROBLEMS
4.149 The stick of mass m, shown in Figure P4.149, origi­
nally at rest with = 0, is disturbed slightly and begins to
slide on a smooth wall and floor. Derive the differential
equation of motion of the stick. Integrate the equation
and find the angle 6 at which contact with the wall is lost.
(1)
(2)
(3)
Then note that there are more unknowns than equa­
tions. Use kinematics to relate a to a , and use the
component" of that equation to obtain a fourth equa­
tion, containing
. Then relate a to a and get a fifth
equation, containing . From your five equations, elim­
inate
N , and N ,obtaining the following differen­
tial equation governing
c
B
c
T
T
B
(4)
Figure P4.149
Multiply (4) by 6 and integrate, using
/dr. Use
the initial condition = 0 at = 0 to evaluate the con­
stant of integration. You will now know as a function of
. Then, with N expressed as a function of , set N = 0
in (1) to find the angle where contact is lost. Your answer
should be
T
Hints: First verify the following equations of motion:
T
Page 272
Disks
and
each weigh 64.4 lb and are rigidly
attached to the light shaft
that joins their centers. (See
Figure P4.150.) A 96.6-lb cylinder
has a hole drilled
along its axis, through which
passes. Let
represent
the rigid body comprised of
and
While the body
is held fixed on the plane, the
cylinder is spun up to an angular velocity 8 rad / sec,
and the system is then released. Assume that part of the
reaction between the axle and the wall of the cylindrical
hole in the cylinder is a friction couple proportional to the
difference in angular velocities, with proportionality con­
stant k. The friction couple acting on the axle will cause
to roll to the right; the opposite couple on will slow its
angular speed down. As time passes, the bodies and
will approach the condition of moving as one. Show this,
and find the common, limiting-case "terminal" angular
velocity shared by
and . There is sufficient friction
between
and the ground to prevent slipping there.
4.6
2 ft
1 ft
Figure P4.150
Other Useful Forms of the Moment Equation
For a rigid b o d y in plane motion, there are several other forms of t h e
m o m e n t equation of m o t i o n [besides t h e translation equation (4.1) a n d
w h i c h are w o r t h y of special study. T h ei d e a is that it is often
convenient a n d helpful to s u m the m o m e n t s about a point other t h a n the
m a s s center C. We will study three of these forms o n e - b y - o n e a n d present
examples of each as w e go along.
Moment Equation in T e r m s of a
c
To develop this form, w e begin with Equation (2.45):
(2.45)
w h e r e w e recall that in this form there is no restriction at all on t h e location
of point P, t h e type of b o d y being studied, or t h e type of motion. Thus,
specializing for a rigid b o d y in plane motion a n d using the r i g h t - h a n d
side of Equation (4.13) to replace H for this case,
c
(4.17)
W h e n e v e r the products of inertia in Equation (4.17) vanish, this equation
takes the particularly simple form
(4.18)
N o t e that t h e translation Equation (4.1) results from Equation (4.18) if
a = 0.
Note further that if P a n d C are in t h e same (reference) p l a n e of
motion, t h e n
is perpendicular to t h e p l a n e of motion — that is,
Page 273
it is parallel to k. W h e n this is the case, w e will rewrite Equation (4.18) in
scalar form*:
(4.19)
in w h i c h ( ) m e a n s the coefficient of the unit vector
within the
parenthesis.
W e n o w present t w o examples of t h e use of Equation (4.18). In the
first one, w e eliminate reactions at a pin b y s u m m i n g m o m e n t s there:
z
EXAMPLE 4 . 2 2
A
C
P
R
Figure E4.22a
The uniform rod in Figure E4.22a (length 80 cm, mass 20 kg) is smoothly
pinned to cart (50 kg) at point A. Force P, applied to with the system initially
at rest, causes to translate with the acceleration
. Find the initial
angular acceleration of the rod.
Solution
By using Equation (4.19), we can sum moments about A of the forces on and
avoid having to use
in this case. We obtain, using the free-body in
Figure E4.22b,
(1)
2
We note that I =
kinematics:
A
/12 = 20(0.8) /12 = 1.07 kg • m , and we get a from
c
y
2
c
A
A
x
C
20(9.81) N
so that, substituting into (1),
Figure E4.22b
Thus the rod starts off with
Question 4.8 Why would the solution to the preceding example be
much more complicated using the mass center form of the moment
Equation (4.16)?
Next w e use Equation (4.19) to rework a n example from t h e previous
section.
* When the products of inertia and both vanish, then by Equations 4.14a,b),
. This leads, when P and C are both in the reference plane, to
= 0=
since X M a l w a y s equals
and
.
Answer 4.8 T h e p i n reaction A w o u l d a p p e a r in
. T h u s w e w o u l d n e e d to also
write Equation (4.15a) to eliminate A .
P
x
x
Page 274
EXAMPLE 4 . 2 3
Find the starting angular acceleration for the body of Example 4.19, shown again
in Figure E4.23a.
Solution
We sum moments about point
with the help of Equation (4.19) and the
free-body diagram in Figure E4.23b:
Figure E4.23a
From the kinematics in the earlier example, we know
Substituting these into the above equation after cancelling "m".
Therefore,
Figure E4.23b
as before.
Note that the
equations of motion, (4.15a,b), were not needed in
this example. They would have been in Example 4.19, however, even if all we
had been seeking was a.
Question 4.9
Why?
Moment Equation in T e r m s of a
P
Another form of the moment equation of motion that is sometimes useful
involves the acceleration and inertia properties at some point other than
the mass center. Recalling that if P is a point of a rigid body in plane
motion, then by Equation (4.3),
(4.3)
And for any point P, w e know from Equation (2.36) that:
(2.36)
Equating these two expressions for H
respect to time,
Answer 4.9
P
and then differentiating with
w o u l d h a v e i n c l u d e d m o m e n t s of b o t h
a n d N.
Page 275
(4.20)
By Equations (2.8) a n d (2.43), w e k n o w that m a
Using these results a n d Equation (2.42), w e m a y replace b y
t w o terms on t h e left-hand side of this equation a n d obtain
c
t h e first
(4.21)
On the right side of Equation (4.21) the first term vanishes since
and the third term is of the same form as except that here the
inertia properties are with respect to axes with origin at P. T h u s retracing
t h e steps b e t w e e n Equations (4.11) a n d (4.13), w e obtain
(4.22)
Question 4.10 Why can we say, as was done above, that
and why does that cause the first term on the right side of
Equation (4.16) to vanish?
W h e n the products of inertia
plifies to
and
vanish, Equation (4.22) sim­
(4.23)
As in Section 4.5, w h e n P a n d C are in t h e s a m e (reference) p l a n e t h e a n d
components of this equation vanish. T h u s t h e scalar form of t h e equa­
tion is
(4.24)
w h e r e ( ) again m e a n s t h e coefficient of within t h e parentheses.
This equation is very useful if w e h a p p e n to k n o w the acceleration of
a point other t h a n the mass center, as in the following examples:
z
Answer 4.10 Let and be position vectors for P and C; differentiating
gives The cross product of parallel vectors always vanishes.
EXAMPLE 4 . 2 4
Solve the problem of Example 4.22 by using Equation 4.24.
Solution
Equation (4.24) makes a problem such as this even simpler than did Equa­
tion (4.19):
Page 276
To determine other information in the preceding example, for in­
stance the starting value of force P, we can write the x-component of the
mass center equation of motion for the cart and then for the bar (see
Figure 4.7):
50 k g
20 k g ,
80 c m
Figure 4.7
For
For
where
or
Page 277
Thus
and
EXAMPLE 4 . 2 5
Repeat Examples 4.19,23, using Equation (4.24) this time to find the starting
angular acceleration.
Solution
We again sum moments about point
0 initially
once again. Notice that the cross-product term is simpler this time, but the mo­
ment of inertia at
has to be calculated by the parallel axis theorem. In Prob­
lems 4.153,154, this problem is to be reworked one last time using the geometric
center of the hoop as point P in Equations 4.19 and 4.24, respectively.
In the preceding example, it accidentally happened that
This is not generally true. For example, an instant later co would not be zero
and the cross-product term would not be zero.
Moment Equation for Fixed-Axis Rotation (The "Pivot" Equation)
Consider n o w the case in which the acceleration of a point P of the rigid
body in plane motion is identically zero. Equation (4.22) tells us that for
such a point,
(4.25)
If w e take the dot product of this equation with w e see that whether or
not the products of inertia
and
vanish, w e always have:
(4.26)
Page 278
W h e n t h e point P is fixed in t h e inertial frame as well as in t h e body, w e
will t h e n usually label the point as O for emphasis. In this case, t h e only
w a y t h e b o d y can m o v e is to rotate about t h e z-axis t h r o u g h O, a motion
w h i c h w e call fixed-axis rotation. The point O, w h i c h does n o t m o v e
during t h e motion of interest, is called a pivot, a n d w e abbreviate Equa­
tion (4.26) as
(4.27)
or as
(4.28)
W h e n the products of inertia vanish at a pivot O, as they will for all
problems in this section, t h e n of course
can be further abbreviated
to simply
for t h e n only t h e z-component of
is non-zero. This is
not always the case, as w e shall later see in the final section 4.7 of this
chapter. W h e n it is, h o w e v e r , w e simply write
(4.29)
Question 4.11 Since Equation (4.4) applies for any point P having
zero velocity, why can we not use equations such as (4.29) for the
instantaneous center
of B when
is not a pivot?
Because of t h e importance of fixed-axis rotation in engineering, w e
present m o r e examples of it t h a n w e did for t h e earlier forms in this
section. In t h e first example, w e examine a composite b o d y rotating about
a pivot:
Answer 4.11 A l t h o u g h H m a y a l w a y s b e w r i t t e n as
whenever v
is zero, its derivative is only equal to
w h e n v is identically z e r o — in o t h e r w o r d s ,
0
D
D
zero all the time.
EXAMPLE 4 . 2 6
The rod and sphere in Figure E4.26a are welded together to form a combined
rigid body which is attached to the ground at O by means of a smooth pin. Find
the force exerted by the pin onto the body, upon release of the system from rest.
Figure E4.26a
Page 2 7 9
Solution
Because O is a pivot of the combined body, we use Equation 4.29:
Figure E4.26b
so that, at release, using the FBD in Figure E4.26b,
or
To calculate the pin reaction we shall use 2F = m a . We first locate the mass
center, C, of the body. The distance from O to C is
c
At this instant we have
Thus
So
An alternate approach to the mass center calculation in the preceding
example would not require that w e explicitly locate C, because
or
from which
Page 2 8 0
and
as above.
In our second example, w e feature distinct bodies connected by an
unwinding rope. One body has a pivot and the others don't.
EXAMPLE 4 . 2 7
Rope
A rope is wrapped around the 10 -lb cylinder as indicated in Figure E4.2 7a. The
rope passes through a hole in the 5-lb annular disk and is then tied to the 15-lb
block When the system is let go from rest with the rope just taut, what is the
reaction exerted on by
Solution
Using the free-body diagrams in Figures E4.27b, 4.27c, and 4.27d, we write the
following equations of motion of the respective bodies. For using the "pivot
equation" (4.29):
Figure E4.27a
which gives us the tension T in terms of
(1)
For the block
by itself,
Figure E4.27b
(2)
where R is the force exerted by
onto
Now, on the disk
(3)
Figure E4.27c
At this point we have three equations in the five unknowns
and
One constraint is that the vertical component of a (see Figure E4.27b) is the
same as the acceleration of the points of the straight portion of rope, and these
accelerations are each
0
(4)
Figure E4.27d
Also, the accelerations of and
are equal. Without any rope tension, they
would each be
with this tension, the acceleration of is slowed, guaranteeing
continuing contact of the two bodies. Therefore:
(5)
Page 281
Adding Equations (1) and (2) and using (4),
or
(6)
Equations (5) and (3) give:
(7)
Adding Equations (6) and (7),
so that, by (6),
R = - 15 + 0.621(25.8) = 1.02 lb
2
Note that the acceleration of is less than "g" (32.2 ft/sec ), as it must be,
and that R is less than the weight of
thereby allowing it to fall, but not freely.
In our third example, w e again have two bodies, but this time both
have pivots and one is in fact in equilibrium. The contact between these
bodies involves sliding friction:
EXAMPLE 4 . 2 8
0.2 m
Brake arm
cross section
is 2 cm X 2 cm;
mass = 2 kg
0.6 m
Just after the brake arm in Figure E4.28a contacts the top of the cylinder
the
cylinder is turning at 1000 rpm
The coefficient of kinetic friction between
and is
. Find (a) how long it takes for
to come to rest under the
constant force P = 40 N; and (b) the pin reactions exerted on to at the pin O.
Solution
Since body is in equilibrium, we may find the normal force between it and the
cylinder by statics. The bar's weight is proportioned between its horizontal and
Cylinder
mass = 10 kg;
radius = 0 2 m
vertical parts as shown in Figure E4.28b. Note that equilibrium requires that since slipping is taking plac
Figure E4.28a
0.1 m
0.3 m
0.2 m
40 newtons
newtons
newtons
Figure E4.28b
0.02 m
Page 282
Summing moments about
we have
0.401N + 0.2(0.3N) - 40(0.401) - 1.5(9.81)(0.301) = 0
from which we get
newtons
and
newtons
Question 4.12 Would the solution for the normal force N be any
different if were counterclockwise?
The motion of body is one of pure rotation. Its free-body diagram is shown
in Figure E4.28c. We use for a since we are ultimately interested in equating to
zero.
Figure E4.28c
Integrating, we get
where the initial condition on
Body
stops when
calculate:
allows us to calculate the integration constant C .
at a time that we are now in a position to
1
Note that the mass center O = C of is fixed in the inertial frame, so that the pin
reactions follow from the mass center equations:
newtons
newtons
In the fourth example, we are concerned with internal forces and with the
fact that the equations of motion can be applied to a part of a body,
considered as a body in itself:
Answer 4.12 Yes, for then the friction force would be in the opposite direction and'.
Page 283
EXAMPLE 4 . 2 9
The slender bar, pinned smoothly at O, is released from rest in the position shown
in Figure E4.29a. After a rotation of
its angular speed is
Find at that instant the axial force, shear force, and bending moment in the bar at
point A, which is one-fourth the length of the bar from its free end.
Figure E4.29a
Solution
First, we find the angular acceleration, making use of the FBD in Figure E4.29b:
0
45
Figure E4.29b
Next we expose the desired forces and moment by drawing a free-body diagram
(Figure E4.29c) of the lower fourth of the bar, and writing the equations of
motion for just that body ( C is its mass center):
so that
Figure E4.29c
where we note that and are the same for this "sub-body" as they were for the
whole bar, and that O is a pivot of the sub-body extended. The other mass center
equation is:
Finally, the "moment equation of motion," written for the sub-body this time, is
The final three results are summarized pictorially on the cut section in Fig­
ure E4.29d.
Figure E4.29d
Page 8 4
We n o w examine
ot problem in which the mass center of the body is
offset from the fixed axis of rotation:
EXAMPLE 4 . 3 0
How must the applied couple C in Figure E4.30a vary with time in order to turn
the unbalanced (but round) wheel at constant angular velocity
? What is
the initial angular acceleration if the couple is absent and the wheel is released in
the position shown in the figure? The moment of inertia of the mass of with
respect to its axis of rotation is , and the mass center of is located at G.
Mass m
Mass M
Solution
Free-body diagrams of
body we have:
and
are shown in Figure E4.30b. Since
(1)
Mass center = G
Figure E4.30a
for
And for the translating block
we may write*
(2)
The point Q of located where the rope leaves the rim has the same velocity
and tangential acceleration component
as does the rope itself at that
point. Since the rope is assumed inextensible, this acceleration has the same
magnitude as
Therefore our kinematics gives us the following additional
equation for the translating block:
(3)
Figure E4.30b
Substituting (3) into (2) gives
And substituting T into the moment equation (1) for
Since
constant, we have
and
then yields
that
and the required couple varies harmonically.
If there is no couple C and the system is released from rest in the position
shown, then the equations are still valid and, with couple C = 0,
The initial angular acceleration of
which is positive
• N o t e t h a t since
(with
) is thus seen to be
if M > 2m.
translates, t h e acceleration of its m a s s center is
Page 285
Question 4.13
What happens if M < 2m? If M = 2m?
In t h e preceding example, it is interesting to write the equations of
m o t i o n in t h e following form for t h e wheel:
(4)
(5)
(6)
Equations (4) a n d (5) are useful if t h e pin reactions are desired,* but
Equation (6) is n o w h e r e near as h a n d y to use as
w h i c h we
have u s e d earlier in the example since the b o d y h a s a pin. The student
m a y wish to eliminate O , O a n d T from (6) b y using (4), (5), a n d the
previous equation for
a n d to s h o w that t h e s a m e result
is obtained (after a good deal m o r e work t h a n in the example) for
(Kinematics m u s t also b e u s e d to relate
and
to
and )
In our last " p i v o t - e q u a t i o n " example, w e take u p a m u c h longer
problem, involving t w o bodies neither of w h i c h is translating. O n l y one
of t h e bodies h a s a pivot, a n d so w e shall h a v e to use b o t h the pivot
equation (4.29) and the m a s s center form of the m o m e n t equation (4.16)
before finally getting t h e problem solved:
x
y
Answer 4 . 1 3 If M = 2 m , t h e n ( w h e n t h e couple is n o t present) T = mg a n d
solutions to t h e p r o b l e m a n d t h e r e is no morion. If M < 2 m , t h e block m o v e s
a n d is negative
are
downward
0.5 m
EXAMPLE 4 . 3 1
0.5 m
The two uniform, slender rods and in Figure E4.31a, each of mass 2 kg, are
pinned together at P, and then is suspended from a pin at O. (This arrangement
is called a double pendulum.) The counterclockwise couple C , having moment
150 N • m, is applied to beginning at t
. Find the angular accelerations of
and of
upon application of the couple, and the force exerted on
at P.
0
Figure E4.31a
Solution
The equations of motion for and
Figures E4.31b and 4.31c, are:
using the respective free-body diagrams in
(1)
(2)
(3)
Figure E4.31 h
* For instance, t h e p i n s m u s t b e designed strong e n o u g h to take t h e forces c a u s e d b y the
accelerations.
Page 286
(Note in Equation (3) that O is a pivot of B .)
1
(4)
(5)
2(9.81) N
150N • m
(6)
Figure E4.31 c
Thus far we have six equations in the ten unknowns 0 ,O ,P ,P
, and
. Kinematics gives the four additional equations we will need:
I
y
x
yl
(7)
(8)
Also,
But
Thus
(9)
(10)
Solving these equations gives:
As a point of interest, the values of the other four variables (besides
are: O = - 1 2 9 N, O = 39.2 N,
m / s , and
m/s .
2
x
2
y
Finally, w e remark that any of the alternative moment equations
discussed in the preceding two sections may be used. However, the
student is cautioned to realize that, just as in statics, once the vector
"force equation" and "moment equation" have been written, n o n e w
(independent) information will arise from summing moments at a differ­
ent point.
Page 287
PROBLEMS
•
SECTION 4 . 6
4.1 I I The rod is pinned to the light roller, which moves in
the smooth slot, and with the system at rest as shown in
Figure P4.151, the 6-N force is applied. Find the angular
acceleration of the rod at the given instant, by using
Equation (4.19) together with
Then check
your solution using only
4.152 If in the preceding problem we replace the 6-N
force by a force F which produces a constant acceleration
of pin P of 0.5 m / s , again find the initial value of a, this
time using only Equation (4.24). Explain why the answers
are the same. (Hint: Solve for F at the given instant.)
2
4.153 Solve the problem of Examples 4.19, 23, 25 using
Equation (4.19), with the point P (about which moments
are taken) being the geometric center of the hoop.
4.1&4 Solve the problem of Examples 4.19, 23, 25 using
Equation (4.24), with the point P again (see the preceding
problem) being the geometric center of the hoop.
4.155 The cylinder shown in Figure P4.155 is made of
two halves of different densities. The left half is steel,
with mass density
slug/ft ; the right half is
wood with
slug/ft . Recalling that the mass
center of each half is located 4r/3n from the geometric
center find the acceleration of when the cylinder is
released from rest. Assume enough friction to prevent
slipping. Hint: Use Equation (4.17) with as point P.
0.5 m
3
48 kg
3
Figure P4.151
Radius r - 1 ft
30
4.156 (a) Use Equation (4.24) to categorize the restrictions
on point P for which we may correctly write
Show that there are only three cases, and that the mass
center form is one of them, while the "fixed-axis-ofrotation form" is but a special case of one of the other two.
(b) Note that the instantaneous center of zero velocity
is not a point P for which, in general,
(c) Fi­
nally, determine in which of the problems in Fig­
ure P4.156 (a-e) it is true that
0
Figure P4.155
Geometric center of circle
Mass center
Cutout
No slip
(rolling)
N o slip
(rolling)
Figure P4,156(b)
Figure P4,156(a)
N o slip
(rolling)
Slipping
Figure P4,156(c)
Figure P4,156(d)
Figure P4,156(e)
Page 288
4.157 Let E be a rigid body in plane motion, in con­
stant contact with a surface
Let Q be the point of in
contact with (Q can be different points of at differ­
ent times). Use the result of Problem 4.156 to prove
that if the following four conditions hold, then
at all times: (1) is fixed in an inertial frame;
(2) is rolling on
(3) is round; and (4) the mass
center of is at its geometric center.
4.158 In the preceding problem, suppose that conditions
(1) and (2) hold, and that at a certain instant,
of is
zero. Using the result of Problem 4.156, show that, at
that instant,
regardless of whether condi­
tions (3) or (4) hold.
4.159 The thin-walled hollow sphere of Figure P4.159
(moment of inertia about any diameter
has
average radius r = 0.5 m and mass 50 kg. It is pinned at
the bottom of the cart. The force F applied to the cart at
rest produces a constant acceleration of all points of the
cart of
a. Find the maximum value of a if the sphere is to
translate and the breaking strength of each of
the cords is 100 N.
b. Suppose a is twice the result of (a), so that one
cord breaks at t = 0. Find the angular accelera­
tion of the sphere at the instant it has turned
through 90°, using Equation (4.24). Does the
answer to (b) depend on the value of a?
Cord 1
Cord 2
30 cm
Figure P4.161
Figure P4.160
4.161 The cylinder in Figure P4.161 has mass 10 kg. The
9-N force is applied to a string wrapped around a thin slot
near the surface of the cylinder. Find the angular acceler­
ation of the cylinder, assuming enough friction to prevent
slip, using (a) Equation (4.19); (b) Equation (4.24). Ob­
serve how both right-hand-sides add up to mr a even
though the individual terms are different.
2
4.162 In Problem 4.152 at a later time, let 6 be the angle
between the vertical and the rod (see Figure P4.162).
Using Equation (4.24), find the angular acceleration of
the rod as a function of
Figure P4.162
Figure P4.159
4.163 The crank arm OP is turned by the couple M at
constant angular velocity
In the position
shown in Figure P4.163, determine the reaction of the
smooth plane onto the 20-kg slender bar PQ. Hints: First
solve for a and for a of PQ using kinematics, then use
Equation (4.24).
0
4.160 The rectangular door of a railroad car has mass m
(Figure P4.160); it is of uniform width
and has its
hinges on the side of the doorway closest to the engine.
Initially the door makes an angle with the train, which
begins to move forward from rest at constant acceleration
a . Find the initial resultant horizontal reaction compo­
nent that the hinges exert on the door.
0
P
4.164 A body weighing 805 lb with radius of gyration
0.8 ft about its z axis (see Figure P4.164) is pinned at its
mass center. A clockwise couple of magnitude e' lb-ft is
applied to starting at t = 0. Find the angle through
which has turned during the interval
c
Page 289
0.5 m
0.3 m
Figure P4.164
Figure P4.166
1.2 m
12 ft
Figure P4.167
4.168 Figure P4.168 shows a scene from Edgar Allen
Poe's "The Pit and the Pendulum." Find the reaction of
the pin onto the bar if the pendulum is instantaneously at
rest in a horizontal position.
Semicircular disk 5 l b
Figure P4.163
Figure P4.165
Smooth pin
4.165 The slender homogeneous rod in Figure P4.165 is
12 ft long and weighs 5 lb; it is connected by a rusty hinge
to a support at A . Because of friction in the hinge, the
hinge exerts a couple of 9 lb-ft on the rod when it rotates.
If the rod is released from rest with
, find: (a) the
angular acceleration of the rod when
, 60" and
90°; (b) the angle at which the angular acceleration of
the rod is zero.
4.166 A wagon wheel spinning counterclockwise is
placed in a comer and contacts the wall and floor. (See
Figure P4.166.)
a. Show with a free-body diagram that the wheel
cannot climb the wall.
b. Show with a free-body diagram that the wheel
cannot move to the right along the floor either.
c. Therefore the wheel stays in the comer. Treat it
as a ring of mass m and radius R, with friction
coefficient at both surfaces of contact. Deter­
mine how long it takes for the wheel to stop,
and find how many radians it has turned
through since first contacting the surfaces.
4.167 The cylinder in Figure P4.167 has a mass of 30 kg
and rotates about an axis normal to the clevis at O. At the
instant shown,
and
rad/s .
Find the force P that acts on the cylinder, and determine
the reactions exerted by the pin onto the clevis at O, all at
the given instant.
Figure P4.168
4.169 The uniform slender bar of mass m is released from
rest in the position shown in Figure P4.169. Find the an­
gular acceleration when the bar has turned through 45 °
2
Figure P4.169
Page 290
4.170 The uniform slender bar in Figure P4.170, of mass
m and length is released from rest at
zero plus a tiny
increment. Find the magnitude of the bearing reaction
when
4.173 Find the angular acceleration of cylinder in Fig­
ure P4.173. The rope passes over it without slipping and
ties to and as shown.
4.174 Body is a pulley made of the cylinders and
which are butted together and rigidly attached. (See Fig­
ure P4.174.)The combined body is smoothly pinned to
the ground through its axis of symmetry (which passes
through its mass center). Ropes wrapped around and
are tied to bodies and respectively. If the system is
released from rest, what will be the angular acceleration
of
40
cm
20 cm
Figure P 4 . 1 7 0
50 N
4.171 Body is a slender bar bent into the shape of a
quarter-circle (Figure P4.171). Find the tensions in strings
OA and OB when the system is released from rest.
20 N
100 N
(mass m)
4.175 The 32.2-Ib particle P rests on the 128.8-lb plank as
shown in Figure P4.175. If the cord at B suddenly breaks,
find the initial acceleration of the particle, and the force
exerted on it by the bar.
Figure P 4 . 1 7 1
4.172 Two weights W and W in Figure P4.172 are con­
nected by an inextensible cord that passes over a pulley.
The pulley weighs W and its mass is concentrated at the
rim of radius R. Show that if the system is released and the
cord does not slip on the pulley, the acceleration magni­
tude of W and W , is
1
1
45 N
Figure P 4 . 1 7 4
2
2
32.21b
128.8 1b
Cord
2 ft
6 ft
Figure P 4 . 1 7 5
4.176 In Example 4.29, note that pivot point O is also a
point of the "subbody" used in that example. Thus for
that body,
Separately compute
and
for the subbody and show that their values are the
same.
(20 kg)
4.177 The uniform slender bar of weight W and length L
in Figure P4.177 is released from rest at
and pivots
on its square end about comer O.
10 kg
Figure P 4 . 1 7 2
Figure P 4 . 1 7 3
30 kg
a. If the bar is observed to slip at
coefficient of limiting static friction
find the
Page 2 9 1
1
ft
4 in.
R
C
=
3 in
9 in
Figure P4.177
4 in
b. If the end of the bar is notched so that it cannot
slip, find the angle at which contact between
bar and comer ceases. Hint: Write the moment
equation of motion about the pivot O, multiply
it by and integrate, obtaining as a function
of Use this relation together with the compo­
nent equation of
in the
direction.
4.178 The chain drive in Figure P4.178 may be consid­
ered as two disks with equal density and thickness. The
larger sprocket has a mass of 2 kg and a radius of 0.2 m. If
the couple is applied starting from rest at t = 0, find the
angular speed of the smaller sprocket at t = 10 s. Hint:
What does a dentist look at?
Figure P4.179
4.180 The slender, homogeneous rod in Figure P4.180 is
supported by a cord at and a horizontal pin at B. The
cord is cut. Determine, at that instant, the location of pin B
that will result in the maximum initial angular accelera­
tion of the rod.
Figure P4.180
2 kg
4.181 The uniform rod of mass m is released from rest in
the horizontal position indicated in Figure P4.181. Con­
sider the force exerted by the smooth pin.
Figure P4.178
4.179 Cylinder in Figure P4.179 with four cutouts is
rotating at 200 rpm initially. A uniform 100-lb cylinder
is placed in the position shown, and the friction produces
a braking moment that will stop The friction coefficient
is
and before the four holes were drilled the uni­
form body weighed 200 lb. For whichever rotation di­
rection of results in a quicker stop, find the stopping
time.
4.7
a. How does the magnitude of the force vary with
the angle through which it has turned?
b. What is the maximum value of this magnitude?
Figure P4.181
Rotation of Unbalanced Bodies
W h e n a rigid b o d y is m o u n t e d in bearings a n d m a d e to rotate b y m e a n s of
a m o m e n t about the axis of the bearings, it is said to be balanced (for
rotation about that axis) if the external reactions exerted b y the bearings
Page 292
onto t h e b o d y are only w h a t are required to support t h e weight of t h e
body. The bearing reactions accompanying imbalance result in vibration
a n d w e a r of rotating machinery a n d are the reason, for example, for
balancing automobile tires.
There are t w o distinct causes for a rotating b o d y to b e out of balance.
The first is if t h e m a s s center is located (a distance "d") off t h e axis of
rotation. T h e n as t h e b o d y turns, there will b e forces at the bearings
producing a n d equaling m a . Clearly, these forces will be constantly
changing in direction (relative to t h e inertial frame) if not also in m a g n i ­
tude.
c
Question 4.14
What would cause them to change in magnitude?
Moving t h e mass center onto the axis of rotation b y addition or deletion
of mass is called static balancing. It carries that n a m e because only t h e n
will the b o d y remain in equilibrium w h e n t u r n e d to a n y position a n d
released, this being true regardless of the orientation of the axis.
The second cause of imbalance is n o n z e r o products of inertia
and/or
w h e r e z is the axis of rotation a n d P is a point on that axis. In
the s a m e w a y that a n off-axis mass center causes bearing forces w h i c h
produce m a , these products of inertia likewise cause bearing forces to
exist; for a system w h i c h h a s b e e n statically balanced, they produce t h e
"bearingmoments"(seeEquations4.14a b)
and
.Andsimilar
to r e m o v i n g t h e offset "d," w e can also a d d or delete material to force t h e
values of
and
to be zero. W h e n this is done, in addition to h a v i n g
ensured that C lies o n t h e axis, the b o d y is t h e n said to b e dynamically
(and of course also statically) balanced. We shall develop t h e equations to
accomplish this in w h a t follows.
c
/
Bearings in the
inertial frame
Figure 4.8
Answer 4.14
C h a n g e s in rotational speed.
Page 293
We show body in Figure 4.8, set in ball bearings at D and E. T is
an externally applied torque about z, the axis of rotation. Let us say that
T is the driving torque less any frictional resistance moments from the
air or the bearings. Finally, note that the and axes are also fixed in
and
are the coordinates of C in this system.
For an unbalanced body rotating about a horizontal axis, the bearing
reactions required to support the weight of the body (when it is not
rotating) may be simply added to the dynamic reactions that would be
generated were there n o gravity. Therefore, for the sake of simplicity w e
shall ignore the effects of gravity in our discussion here.
Figure 4.9
N o w consider the free-body diagram, Figure 4.9, and let the bearingreaction components be referred to the body-fixed axes (x, y, z).
Then
yields the component equations
(4.24a)
(4.24b)
(4.24c)
Using Euler's second law in the form
(4.20)
we obtain the following component equations:
(4.25a)
(4.25b)
(4.25c)
We observe, then, that if w e know
and the geometric and
inertia properties of the body, w e can solve Equations (4.24a,b) and
(4.25a,b) for the bearing reactions D , D , E , and E . We n o w illustrate
such a calculation with an example before taking up the issue of h o w to
balance a body.
x
y
x
y
Page 294
EXAMPLE 4 . 3 2
The body in Figure E4.32 has mass m = 2 slugs, and its mass center is off-axis by
the amount d = 1/64 in. in the x-z plane so that
in. and
. Its
products of inertia are
slug-ft . If the body is spun up to a
constant angular speed of 3000 rpm, what then are the dynamic reactions at
bearings D and E?
2
Solution
By the parallel-axis theorem
Figure E4.32
Also,
Thus by Equations (4.24) and (4.25) we obtain
and
Let u s n o w assume that w e h a v e a rotating b o d y
a n d a not both
zero) m o u n t e d in bearings. For such a body, w e can s h o w that the bear­
ing reactions vanish if a n d only if
and
The "if" proof is simple, for if
Equations (4.25 a,b)
yield E = 0 = E , a n d substituting these zero values along with
into Equations (4.24 a,b), w e find D = 0 = D .
For t h e " o n l y if" part of t h e proof, if t h e bearing reactions are all zero,
Kramer's rule applied to Equations (4.24 a,b) gives
a n d to
Equations (4.25 a,b) yields
W h e n these t w o products of
inertia are zero at a point D, t h e n z is called a principal axis of inertia at D;
this concept is discussed in considerable detail in Chapter 7.
x
y
x
y
In s u m m a r y , then, w e can say that t h e bearing reactions vanish, a n d
h e n c e t h e b o d y is balanced, if a n d only if t h e axis of rotation is a principal
axis of inertia containing t h e m a s s center of t h e body.
N o w let u s see w h a t can b e d o n e about correcting imbalance. S u p ­
pose values of
and
of a b o d y are k n o w n , w h e r e P, lying on
t h e axis of rotation, is t h e origin of t h e coordinates. W e can, for example,
determine t h e coordinates (x , y ) a n d (x , y ) a n d masses (m a n d m ) of
A
A
B
B
A
B
Page 295
a pair of weights which, w h e n placed in two "correction planes" A (at
z = z ) and B (at z = z ), will ensure that the shaft is dynamically bal­
anced. All w e have to do is (a) force the mass center C* of the combined
system (m plus m and m ) to lie on the axis of the shaft, and (b) force the
products of inertia of the combined system to vanish:
A
B
A
B
x-coordinate of
(4.26a)
y-coordinate of
(4.26b)
(4.26c)
(4.26d)
Note that w e assume that the "balance weights" are small enough to
be treated as particles.
These four equations (4.26) may be solved for the four quantities
and . Thus there is some freedom to select two of
the six quantities
and
provided there is n o other
condition linking them; for an example of such a constraint, the weights
might have to be placed o n a circle of given radius (such as w h e n tires are
balanced and weights are clamped to a rim). In this case w e would
and
and n o w there are six equations in six unknowns. Let us illustrate the use
of these equations in the following example.
EXAMPLE 4 . 3 3
In Example 4.32, suppose that we are to balance the body by adding weights in
two correction planes midway between C and the two bearings. Furthermore, the
weights are each to be placed on a circle of radius
ft. Find the masses and
coordinates of the weights.
Solution
We had m = 2 slugs,
in.,
and
choose P to have the same axial position as C, then
Solving these we get
2
slug-ft . If we
ft, and
Page 2 9 6
Similarly,
from which we obtain
Squaring and adding,
slug
So the weight of B is
coordinates,
or 0.932 oz. For the
and
(These add vectorially to
as a check.)
Plane A
0 . 0 9 6 1 ft
Plane B
490 ft
0 . 1 9 1 ft
0 . 4 6 2 ft
Figure E4.33
Now, for the mass and coordinates of m in plane A, we have, by again
squaring and adding,
A
Page 297
Thus the weight of A is
coordinates are
or 1.85 02. The
Again checking,
The above results are all shown in Figure E.4.33.
In our last example, w e shall add mass in the form of two rods to balance
the body in Example 4.16.
EXAMPLE 4 . 3 4
For the body of Example 4.16, find the length L of the pair of rods, each of
mass 3 m, that will dynamically balance the shaft when attached to it as shown in
Figure E4.34.
Figure E4.34
Solution
In the previous example, was computed to be
Note that the mass center
of the original and modified systems is at P, so the system is already statically
balanced. Thus since
(all the mass is still in the xz-plane), all we need for
dynamic balance is:
Page 298
PROBLEMS
Section 4.7
4.182 Explain why the uniform plate in Figure P4.182 is
4.183 A light rod of length with a concentrated end
mass M, is welded to a vertical shaft turning at constant
(See Figure P4.183.) Find the force and moment exerted
by the rod onto the shaft. Include the effect of gravity.
a. Find the dynamic bearing
reactions
at A and B
dynamically
balanced.
in terms of the system parameters shown in the
figure.
b. If
find the radius of a hole (in terms
of r) at Q that will eliminate these bearing reac­
tions.
4.186 Two thin disks are mounted on a shaft, each mid­
way between the center and one of the bearings, as indi­
cated in Figure P4.186. The disks are each mounted off
center by the amount
in. as shown. Determine
the x and y locations of two small 4-oz magnetic weights
(one for each disk), which when stuck to the disks will
balance the shaft. Neglect the thicknesses of the disks,
and treat the weights as particles.
Figure P4.182
Figure P4.183
4.187 In Figure P4.187,
is the axle of a bicycle,
mounted in bearings 2d apart. The cranks are rigidly
connected to the axle and also to the pedal shafts
and
. If the rigid body consisting of axle
the cranks and
the pedal shafts is turning freely about axis z at constant
angular speed
find the forces exerted on the bearings
in the given configuration.
c
4.1 B4 The shaft in Figure P4.184 rums at constant angu­
lar velocity 10 rad/sec. If the bars are light compared
with the two weights, determine the bending moment
exerted on (length
by at the point where they are
welded together. Sketch the way the shaft will deform in
reality under the action of this couple. Ignore gravity.
Each:
W = 5 lb
R = 6 in.
Figure P4.186
Figure P4.184
Figure P4.185
4.185 The circular disk in Figure P4.185 is offset by the
amount 6 from the shaft to which it is attached.
Figure P4.187
;
Page 299
4.188 A solid cylinder (of mass m, radius r, and length
and a light rod are welded together at angle as
shown in Figure P4.188. The rigid assembly is spun up to
rad/sec and then maintained at that speed. Find the
dynamic bearing reactions after
Show that, if a body mounted on a shaft is statically
balanced, and if and are zero for any point D on the
shaft, it follows that and are zero for any other point
Q on the shaft.
4.192 The shaft in Figure P4.192 supports the eccentri­
cally located weights W (0.1 lb) and W (0.2 lb)as shown.
It is desired to add a 0.3-lb weight in plane A and a 0.4-lb
weight in plane B to balance the shaft dynamically. De­
termine the x and y coordinates of the added weights.
1
Bearing
Bearing
45
60
2
0
3 ft
3 ft
3 ft
0
Figure P4.188
Plane B
4.190 The S-shaped shaft in Figure P4.190(a) is made of
two half-rings, each of radius R and mass m / 2 . Find the
dynamic bearing reactions for the instant given. Hint: For
a half-ring, the mass center is located as shown in Fig­
ure P4.190(b).
Plane A
4.189 Repeat the preceding problem if the lower half
(shaded) of the cylinder is missing. (The mass is now
m/2.)
Figure P4.192
4.193 Rotor in Figure P4.193 has a mass of 2 slugs,
and its mass center C is at a 5-in. offset from its shaft
as shown
in.,
in., and
in. in the coor­
dinate system fixed in the shaft at the point B).
The products of inertia of with respect to the center
of mass axes
are
lb-in.-sec and
the two correction planes
and
balance the rotor.
That is, determine the x and y coordinates of each of the
two added masses by ensuring that the mass center of
the final system is on the shaft and that the products of
inertia vanish.
2
(a)
(b)
Figure P4.190
Figure P4.193
Page 300
3 in.
0.01 slug
5 in.
4 in.
0.02 slug
Plane B
2
Plane of m
1
Plane of m
Plane A
View C-C
Figure P4.194
4.194 Balance the shaft in Figure P4.194 by adding a
mass of 0.003 slug in plane A and a mass of 0.004 slug in
plane B.
2 ft
4.195 Two plates, each weighing 32.21b, are welded
to a light shaft as shown in Figure P4.195. A torque T of
10 lb-ft is applied about the z axis until the assembly is
turning at angular speed
then T is removed. If the
bearings can hold a force perpendicular to the shaft of
no more than 320 lb, find the maximum value that
can be without failure. Note that xz is the plane
of the plates and (x, y, z) are fixed to the assembly.
2 ft
1ft
4ft
2 ft
2 ft
4 ft
2ft
Figure P4.195
COMPUTER PROBLEM
•
Chapter 4
4.196 A cylinder of mass m and radius R is rolling to the
left and encounters a pothole of length s, as shown in
Figure P4.196(a). The angular velocity when the mass
center C is directly above O is
We are interested in
the condition(s) for which there will be no slip at O while
the cylinder pivots prior to striking the comer at A.
a. Show that for no slip at O, the equations of mo­
tion are (see Figure P4.196(b)):
where
tuted.
b. Multiply Equation (3) by
taining
have been substi­
and integrate, ob­
Solve these equations for and N, and show that
the no-slip condition
requires that
sin
Note that for very low
this is easily satisfied if
is not too small and s (and therefore is not too
large. But, for example, if
then the cyl-
Page 301
Figure P4.196(a)
Figure P4.196(c)
Figure P4.196(b)
inder will slip regardless of the value of because
then the inequality cannot hold. Note too that if
the cylinder is not slipping just prior to impact, it
has not slipped at all.
c. Next, use Figure P4.196(c) to compute the angle
for which the cylinder will strike the left
comer A of the depression, and show that no
slip will have occurred at any time during the
EXTRA CREDIT PROJECT PROBLEM
•
pivoting if
Finally, use the computer to create data for plots of
the minimum required for no slip at O versus
for
three values of s/2R: 0.1, 0.2, 0.5. Draw the three curves
on the same graph.
Chapter 4
4.197 Construct a round object with a challenging
to
calculate (as an example, see Figure P4.197(a)). On an
inclined plane (see Figure P4.197(b)), roll your object
down a 5-ft, 15° grade and with a stop-watch measure
the descent time. Do this twice and average the times.
Then explain the experiment, calculate the expected time,
compare with the actual time, and give possible reasons
for the difference in a brief report. (Note: It is fun to do all
the students' experiments in the same session.)
5 ft
15°
Figure P4.197(b)
Figure P4.197(a)
SUMMARY
•
Chapter 4
In this chapter w e h a v e developed compact forms for the r i g h t - h a n d
sides of m o m e n t equations for plane motion of a rigid body. The most
general forms w e h a v e studied are, for a n arbitrary point P,
Page 302
and
where moments and products of inertia are defined by
and for which there are the very useful parallel-axis theorems
and
and
Except for the topic of balancing of rotating bodies (Section 4 . 7 ) w e have
restricted our attention to situations in which the products of inertia
vanish, usually because of the body having an xy plane of symmetry. In
those cases,
and
Important special cases are
a. Translation (in which every point has the same acceleration, a,
and of course a = 0) for which:
and so
b. Summing moments at the mass center C:
or, more simply,
c. P is a pivot (body rotates about a fixed axis), so that a = 0:
P
or, more simply,
It is more important to realize that while w e have a number of
options as to form for the moment equation, one moment equation plus
the force equation,
are all w e can bring to bear independently
for a given (free) body. That is, the situation is the same as in statics: we
may sum moments wherever w e like, but the two vector equations —
one force and one moment — give us all the independent relationships
involving external forces on the body. Many practical problems are
solved by augmenting these equations with kinematic constraint condi­
tions that can be invoked to generate relationships between a and
c
Page 303
A body rotating about a fixed axis is said to be statically balanced if
the mass center is located on the axis. It is said to be dynamically balanced
(no bearing reactions induced by the rotation) if in addition the products
of inertia associated with the rotation axis all vanish. Industrial equip­
ment and automobile tires are modified by the addition of "balance
weights" so as to ensure these conditions.
REVIEW QUESTIONS •
Chapter 4
True or False?
These questions all refer to rigid bodies in plane motion.
1. Euler's second law enables us to study the rotational motion of rigid
bodies.
2. The moment of inertia is always positive, whereas the products of
inertia can have either sign.
3. The formula
gives the exact value of the moment of inertia of
a slender rod about a lateral axis through its mass center.
4. Euler's second law,
is valid only in an inertial frame
(meaning that the position vectors and velocities inherent in H , the
origin O, and the time derivative are all taken in an inertial frame).
0
5. In
the quantity d is the distance between the points P
and C. (C is in the reference plane, whereas P is any point of the
body.)
6. Euler's second law,
applies to deformable bodies, liq­
uids, and gases, as well as to rigid bodies.
7. If represents the instantaneous center of zero velocity, then
in general.
8. Products of inertia are not found in the equations of plane motion.
9.
is just as general as
inertial frame.
where O is fixed in an
10. In translation problems, the moments of external forces and couples
taken about any point add to zero.
11. Suppose you buy a n e w set of automobile tires and a dynamic bal­
ance is performed on each wheel by adding weights in two planes
(inner and outer rims). The products of inertia
and
have thus
been eliminated, which otherwise would have, caused bearing reac­
tions and vibration.
12.
applies to deformable as well as to rigid bodies, as long as
they are in plane motion.
13. For two bodies and
for each will be
the sum of the equations
for the combined body.
written
14. If the bodies of Question (13) are turning relative to each other, it
makes n o sense to talk about a combined
equation.
Answors : 1.T 2. T 3. F 4. T 5. F 6. T 7. T 8. F 9. T
10. F 11. T
12. F 13. T
14. T
5
SPECIAL INTEGRALS OF THE
EQUATIONS OF PLANE
MOTION OF RIGID BODIES:
WORK-ENERGY AND IMPULSEMOMENTUM METHODS
5.1
5.2
5.3
Introduction
The Principle(s) of Work and Kinetic Energy
Kinetic Energy of a Rigid Body in Plane Motion
An Alternative Form for Kinetic Energy
Derivation of the Principle W=
Power and Work of Systems
of Forces and Couples
Restriction of W=
to a Rigid Body; A Notable Exception
Computing the Work Done by Various Types of Forces and Moments
Examples Solved by the Principle W=
Two Subcases of the Work and Kinetic Energy Principle
Potential Energy, Conservative Forces, and Conservation of
Mechanical Energy
The Principles of Impulse and Momentum
The Equations of Impulse and Momentum for the Rigid Body in
Plane Motion
Conservation of Momentum
Impact
The Center of Percussion
SUMMARY
R E V I E W QUESTIONS
Page 304
Page 305
5.1
Introduction
Just as in Chapter 4, the framework here is rigid bodies in plane motion.
But we shall focus our attention now on problems which most efficiently
can be attacked by using work-and-kinetic-energy and/or impulse-andmomentum principles. We shall employ these principles, rather
broadly stated in Chapter 2, taking advantage of the simple forms that
kinetic energy and angular momentum take when the body is rigid and
constrained to plane motion.
In Chapter 2 we defined the kinetic energy of a body to be the sum
of the kinetic energies of the particles making up the body; that is,
T=
Because we have found in Chapter 3 that velocities of dif­
ferent points in a rigid body are related through the body's angular
velocity, the reader should not be surprised to find kinetic energy for such
a body to be expressible in terms of the velocity of one point and the
angular velocity. Moreover, in Chapter 2 we observed that, for a body in
general, change in kinetic energy equals work of external and internal
forces. But for a rigid body the net work of internal forces vanishes, so
that the work W in W = T is the work only of external forces. We shall
derive this work-and-kinetic-energy relationship directly from the force
and moment equations (Euler's laws) as studied in Chapter 4, but it is
helpful to recall the discussion of Chapter 2 and note the consistency of
that material with the result we shall develop here.
The relationship between angular impulse and angular momentum
developed in Section 5.3 takes on a quite useful form for a rigid body in
plane motion, owing to the fact, as shown in Chapter 4, that the angular
momentum can be expressed then in terms of inertia properties and
angular velocity. Thus we shall find ourselves in a position to evaluate
sudden changes in rates of turning for colliding bodies and to study
quantitatively the relationship between the spin rate and the arm-trunk
configuration of a skater.
It is very important for the reader to always keep in mind that the
principle of work and kinetic energy and the principles of impulse and
momentum do not stand as principles somehow separate from Newton's
laws or their extensions to bodies of finite size, Euler's laws. Rather, here
it will be seen, as was observed before in Chapter 2, that these relation­
ships, which involve velocities, are really just special first integrals of the
more fundamental second-order expressions relating forces and acceler­
ations. Thus the principles of this chapter allow us to begin our solutions
halfway between accelerations and positions. They therefore involve
velocities but not accelerations.
5.2
T h e Principle(s) of W o r k and Kinetic Energy
Kinetic Energy of a Rigid Body in Plane Motion
There is a principle, derived from the equations of motion, that will help
us to solve for unknowns of interest in kinetics problems. In this section
Page 306
we shall see that this principle arises from first deriving and then differ­
entiating the kinetic energy of the body.
Kinetic energy, which we have examined in Chapter 2, is usually
denoted by the letter T; for any body or system of bodies, it is defined as
the summation of
over all its elements of mass:
(5.1)
In this section we need to specialize Definition (5.1) for a rigid body in
plane motion. To this end we kinematically relate the velocity v of the
differential mass to the velocity of the mass center C. Using the fact that v
is at all times equal to the velocity of its companion point in the reference
plane containing C (see Figure 5.1), we may write
C o m p a n i o n point of
dm in plane of mass
center C
Rigid body
in plane m o t i o n
Figure 5.1
We note that the x and y axes are fixed in
v , that is,
we have
with their origin at C. Forming
2
(5.2)
Thus the kinetic energy becomes, substituting (5.2) into (5.1),
Recognizing the moment of inertia integral in the second term, we obtain
(5.3)
in which
ity of C.
the square of the magnitude of the veloc­
Question 5.1
Why do
dm and
dm both vanish?
We note that the (scalar) kinetic energy has two identifiable parts (not
components!): one relating to the motion of the mass center
* H e n c e f o r t h , w e shall u s e t h e a b b r e v i a t i o n
Answer 5.1
B y t h e definition o f t h e m a s s c e n t e r .
t h r o u g h o u t this c h a p t e r .
Page 307
and the other to the motion of the body relative to C
This
clear division of T even exists in general motion of rigid bodies (that is, in
three dimensions), though there are more terms in
then.
EXAMPLE 5.1
Calculate the kinetic energy of the round rolling body in Figure E5.1, which has
mass m, radius R, and radius of gyration k with respect to the z axis. The mass
center C (see Figure E5.1) lies at the geometric center.
c
c
Solution
Figure E5.1
Note that if is a solid cylinder, then k =
and
c
In this case, two-thirds of the kinetic energy rests in the translation term of T
If
2
is a ring (or hoop), however, I = mR so that k = R and
c
c
and this time half the kinetic energy is in each of the translational and
rotational terms.
EXAMPLE 5.2
Work Example 5.1 for the case when the mass center C is offset by a distance r
from the geometric center of a round rolling body (See Figure E5.2.)
Solution
In order to use our equation for kinetic energy,
Figure E5.2
we must first calculate
Therefore
Page 308
Substituting, we get
Note that if r = 0, the answer agrees as it should with Example 5.1,
An Alternative Form for Kinetic Energy
There is an alternative means of writing the kinetic energy T of a rigid
body in plane motion by making use of the instantaneous center of zero
velocity
(see Figure 5.2):
Thus by using the parallel-axis theorem we obtain
(5.4)
The translational (T ) and rotational
terms composing the scalar T are
thus seen to collapse into the one term
if we choose to work with
instead of C.
v
Figure 5.2
Figure 5.3
As an example, we consider a rolling cylinder again (see Figure 5.3):
Page 309
We have noted that two-thirds of the cylinder's kinetic energy is asso­
ciated with the "translational part" of T and one-third with the "rota­
tional part." If we now use
we get all of T at once:
(as above)
As a second example of the use of Equation (5.4), consider the
slender rod swinging about a pivot at A as shown in Figure 5.4. The
kinetic energy of may be found in either of two ways:
Figure 5.4
Derivation of the Principle W =
Forces and Couples
Power and Work of Systems of
Returning now to the derivation of our principle, we next compute the
rate of change of kinetic energy:
Therefore
(5.5)
Recalling that
and that the z component of
bodies in plane motion, we may write
for rigid
(5.6)
Page 3 1 0
Question 5.2 Since
can contain x and y components (see Equa­
tion 4.13) why may we substitute the total vector
for just the z
component
in Equation 5.5?
Our next goal is to get the individual external forces and couples
acting on the body into the equation. (See Figure 5.5.) Note the abbre­
viations r = r , r = , and so on, of the vectors to the points of
application of F , F , and so on. We assume that the external mechanical
actions on the body arise from a system of forces ( F , F . . .) and couples
with moment vectors (C C , . . .), as shown in Figure 5.5. Further, we
let (V , v , . . .) be the velocities of the material points (P ,P , . . .) on
which the forces act instantaneously.
CP1
1
l
CP2
l2
2
l
l
l
2
2
2
l
2
Figure 5.5
Clearly, then, the resultant of the external forces is
(5.7)
and the moment of the
s and
s about C is
(5.8)
Substituting Equations (5.7) and (5.8) into (5.6) gives
(5.9)
Now since the dot and cross may be interchanged without altering the
value of a scalar triple product,
(5.10)
Answer 5 . 2
Because
is z e r o !
Page 311
But the velocities of P and C are related:
1
so that
(5.11)
The right-hand side of Equation (5.11) is called the power, or rate of
work, of the external system of forces and couples acting on the body.
The power of a force is its dot product with the velocity of the point on
which it acts; the power of a couple is its dot product with the angular
velocity of the body on which it acts:
Rate of work of force F =
1
Rate of work of couple C =
1
F
= power of F
1
C
1
(5.12)
1
= power of C
1
(5.13)
Hence one form of the principle of this section is
(5.14)
Power
or
Integrating, we obtain another principle.*
or
(5.15)
where the integral of the power is called the work W of the external forces
and couples. It is the work done by the s and
s on the body between
the two times t and t . Hence we have a principle that can be stated in
words:
1
2
Work done by external forces
and couples on
Restriction of W =
Change of kinetic
energy of
to a Rigid Body; A Notable Exception
It is essential to recognize that our derivation of the principle of work and
kinetic energy depends crucially on the body being rigid. In fact the work of
external forces on a def ormable body is not in general equal to the change
in its kinetic energy. That is the case even when the "deformable" body is
* Sometimes
is u s e d t o d e n o t e t h e t i m e i n t e r v a l , r a t h e r t h a n
s t a n d f o r "initial" a n d "final" v a l u e s .
t h e subscripts
Page 312
composed of several individually rigid parts. However, there are a num­
ber of special circumstances, usually easy to recognize, for which the
principle is valid for such a system of rigid bodies. To give an example for
which this is true, suppose we have two rigid bodies,
and
making
up the system, and suppose the bodies are connected by a pin (or hinge)
with negligible friction as shown in Figure 5.6.
Figure 5.6
Let F be the force exerted by on at the pin, and consequently — F
is the force exerted by
on
. Furthermore, let
v = common velocity of attachment points in the two bodies
= power (rate of work) of forces acting on
external to system
= power of forces acting on
that are also
that are also external to system
= kinetic energy of
= kinetic energy of
Now if we apply Equations (5.11) and (5.14) to each of the bodies, we
obtain
and
which may be added to yield
Page 313
or
where P is the power of the external forces on the system and T is the
kinetic energy of the system.
With friction in the pin, however, we also would have interactive
couples C and — C, and the sum of their work rates would be
which in general would not vanish.* This net rate of work of friction
couples would be negative, reflecting the fact that the friction will reduce
the kinetic energy of the system. We can expect the principle of work (of
external forces) and kinetic energy to be valid for a system of rigid bodies
whenever the interaction of the bodies leads neither to dissipation of
mechanical energy by friction nor to a storing of energy as in a spring.
When in doubt, follow the procedure we have just been through — that
is, apply Equation (5.14) to each of the bodies, add the equations, and see
whether the rates of work of interactive forces cancel out.
Computing the Work Done by Various Types of Forces and Moments
Before we can put Equation (5.15) to use, it is essential to demonstrate
how to compute the work W done on by a number of common types of
forces and moments:
Type 1: F is constant. In this case, as in Chapter 2,
1
(5.16)
Type 2: F acts on the same point P of
this case,
1
1
throughout its motion
In
(5.17)
where r
= r and i and f denote starting (initial) and ending (final)
times and positions. It is true, of course, that the velocity v , which
combines with f to produce its power, is at each instant the derivative of
some position vector. If the force acts on different material points of at
different times throughout a motion (such as friction from a brake),
however, the path integral F • d i has no real functional utility and
the general F • V dt must be used.
O P 1
1
1
1
1
1
O P 1
1
Type 3: F is due to gravity. This is an example of both Types 1 and 2.
Thus, letting z be positive downward, we get
1
* It w o u l d v a n i s h , o f c o u r s e , if t h e friction w e r e e n o u g h t o p r e v e n t r e l a t i v e r o t a t i o n s o
that
t h e n t h e s y s t e m w o u l d b e h a v e a s a single rigid b o d y !
W h i c h w a s necessarily t h e c a s e in C h a p t e r 2!
** T h e w o r k d o n e b y any c o n s t a n t f o r c e F a l w a y s a c t i n g o n t h e s a m e p o i n t w i t h p o s i t i o n
v e c t o r r is t h u s F •
Page 314
Expressing the differential of the position vector in terms of rectangular
cartesian coordinates, we get
and substituting we obtain a simple result for the work of gravity:
(5.18)
as we observed in Chapter 2. Note that gravity does positive work if the
body moves downward. (Indeed, a good rule of thumb to remember is
that a force does positive work if it "gets to move in the direction it wants
to" — that is, it has a component in the direction of the motion of the
point on which it acts. If it does not, it does negative work during the
motion of that point.)
Type 4: F is the normal force exerted at the point of contact on a rigid
body that is maintaining contact with a fixed surface, whether rolling or
slipping. Note in the lower portion of Figure 5.7 that the normal force F ,
is always perpendicular to the velocity of P. That is,
1
Figure 5.7
Type 5: F is the friction force exerted at the point of contact when a
rigid body rolls on a fixed surface (Figure 5.8). This time, the force F
(which may or may not be zero) does zero work because it always acts on
1
1
a point of zero velocity:
Figure 5.8
Type 6: F is the force in a linear spring connected to the same two
points P and Q of bodies and during an interval of their motions. (See
Figure 5.9.) We denote:
1
k = spring modulus (which when multiplied by the stretch yields
the force in the linear spring)
(inertial
reference
frame)
Figure 5 . 9
Page 315
unstretched length
stretch
if compressed)
unit vector along spring toward body
We first note that the work of spring S on body
is
Using
we may differentiate and obtain
Therefore, substituting for v , we get
P
Since the derivative of a unit vector is perpendicular to the unit vector,
the last integral vanishes and we obtain
Thus
(5.19)
Figure 5.10
If Q is fixed in the inertial reference frame, the work of S on B alone is
given by the right side of (5.19);* if Q moves, however, we can only say
that the total work on both bodies by S is given by
We note from the spring's force-stretch diagram (Figure 5.10) that
the work done by the spring is in fact the negative of the change in energy
E stored in it; namely, in stretching from
E = (area of triangle OCB) - (area of ODA)
Type 7: We now consider the work done by the force in an inextensible cable (or rope, string, cord) connected to two points P and Q of bodies
and
during an interval of their motions (Figure 5.11). The cable
under consideration may pass over one or more light, f rictionless pulleys
Figure 5.11
* A n d its w o r k o n R is o f c o u r s e t h e n z e r o .
Page 316
between the bodies, but it is assumed to remain taut throughout the
motion.
The work done by the cable tension on the system of
plus
is
zero, which we proceed to prove as follows:
We write v and v in terms of their components parallel and per­
pendicular to the cord:
Q
P
Noting that the perpendicular components
and
have zero dot
products with the unit vectors and (see Figure 5.11), we obtain for the
works of the tensions:
and
But by the cable's inextensibility,
and
so that
W
by
cable on both bodies = 0
Type 8: We have a couple C. In this case, the work of the couple in
plane motion is given by
(5.20)
Thus if C is constant, the work of the couple is given by
(5.21)
That is, the work of C is the strength of the couple times the angle through
which the body turns. As with the work of forces, the couple's work is
positive if it "gets to move" in the direction in which it acts (or turns, in
this case).
Examples Solved by the Principle W —
We are now in a position to solve some problems by using the principle of
work and kinetic energy. A number of examples follow. In the first, work
is done only by gravity, and W =
is used to supplement the equations
of motion.
Page 317
EXAMPLE 5 . 3
Find the pin reaction at O when the uniform bar in Figure E5.3a has fallen
through 4 5 from rest.
0
Figure E5.3a
Solution
We first find the angular speed
0
in thefinal(45 ) position by using the principle
Letting T be the kinetic energy in thefinalposition, and noting that the work
done by gravity
we find:
2
(1)
We must now return to the differential equations to obtain equations in the
desired reaction. (Note that W =
alone can only give us the solution to one
scalar unknown!) In the final position, we have:
Expressing this equation in its tangential and normal components with the help
of the free-body diagram (Figure E5.3b),
(2)
But
Figure E 5 . 3 b
and with k defined as
so that, from the
component of Equation (2),
where we have substituted
nent of the reaction is
from Equation (1). Therefore, the normal compo­
(3)
Next, from the
component of Equation (2),
(4)
Also, since point O is a pivot of the rod, we know:
(5)
Page 318
Substituting
reaction:
from Equation (5) into (4) gives the tangential component of the
Thus the pin reaction is
In the second example, work is done only by a spring; however, calculat­
ing the final stretch is tricky. *
EXAMPLE 5 . 4
One end of the linear spring in Figure E5.4a is attached to a thin inextensible cord
that is lightly wrapped around a narrow groove in the wheel (mass = 1 slug,
radius of gyration about center = 1 . 5 ft). If the wheel rolls, and starts from rest
when the spring is stretched 1 ft,findthe velocity of the center of the wheel when
the center has moved 2 ft. The mass center of the wheel coincides with the
geometric center.
Figure E5.4a
Solution
Wefirstnote that the cord is not attached to a specific material point on the wheel.
However, as time passes, the various "wrapping points" on the end of the
straight portion of the cord (such as in Figure E5.4b) have, at every instant the
cord is taut, the same velocity as the coincident point of the wheel at Q. Thus
Equation (5.19) gives the work done on the wheel by the spring.
We have at all times, by kinematics (see the figure),
and
Figure E5.4b
so that
or
Thus the net shortening of the spring when C has moved 2 ft to the right is ft.
Another way to see this is to let C move to the right the amount x
This com­
presses the spring (if it were able to do so!) the same amount, x . Then turn the
wheel clockwise about C through the angle
radians, until the
correct point is on the ground. ("Correct" means the point that would be on the
ground had the wheel rolled normally over to the final position.) The rotation
wraps
of string around the inner radius and "takes back" of
the compression. Thus
=
is the reduction in the original 1 ft of stretch,
leaving ft, as before. Hence
and
c
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We then find, noting that gravity does no work here,
We note that when the center has moved 3 ft, then
and all the
stretch is gone. At this time, the spring would simply drop out of the problem.
The next example is an actual application of W =
from industry.
EXAMPLE 5 . 5
This example involves a practical application in the antenna industry of the work
and kinetic energy principle. The antenna positioner in Figure E5.5 is equipped
with a mechanical stop spring so that if the elevation drive overruns its lower
limit, the antenna motion (a pure rotation about the horizontal elevation rotation
axis) will be arrested before the reflector strikes another part and is damaged.
Counterweights place mass
center of rotating part on
elevation axis
Elevation rotation axis
M e c h a n i c a l stop
spring
Positioner
Reflector
Figure E5.5
The elevation motor has an armature rotational mass moment of inertia of
0.01 lb-ft-sec (or slug-ft ) and drives the reflector through a gear reducer with a
700:1 gear ratio. The combined moment of inertia of the reflector, its counter­
weights, and the supporting structure is 12,000 slug-ft = I .
It is desired to arrest a rotational speed of 30°/sec during a rotation from
contact to full stop of 3 . The radius from the elevation rotation axis to the stop
spring is 1.5 ft. The spring is unstretched at initial contact and may be assumed to
have linear load-deflection behavior. It is further assumed that the motor is
2
2
2
0
0
Page 320
switched off but remains mechanically coupled while the rotation is being
arrested. Find:
a.
b.
c.
d.
The required stiffness of the spring.
The maximum force induced in it.
The rotational position when it sustains its maximum force.
The angular accelerations of the reflector and motor armature at the
position of maximum force. (Are these the maximum accelerations?)
Solution
Since the spring is linear, its greatest force is the spring stiffness times the maxi­
mum deflection. This is also the position for which motion is completely arrested.
At this position the kinetic energy has been brought to zero with the stop spring
storing the energy; the principle W =
gives
Note that point O is
for the rotating body and that gravity does no work
between contact and stop.
Question 5.3
The values of
Why does gravity do no work?
and
needed in the equation are calculated as follows:
total moment of inertia at axis of rotation
(Note that over the very small angle of 3 ° the spring compression is approxi­
mately the arclength
Solving for the spring's stiffness, we get
The maximum spring force =
= 753,000 X 0.0785 = 59,100 lb. The rota­
tional position is 3 ° beyond contact — that is, the position at full stop. The
Answer 5.3
S i n c e t h e c o u n t e r w e i g h t s p l a c e t h e m a s s c e n t e r o n t h e e l e v a t i o n axis, t h e
mass center does not move.
• A s t h e r e a d e r m a y w i s h to p r o v e , m o m e n t s o f inertia reflect t h r o u g h g e a r t r a i n s f r o m
i n p u t t o o u t p u t w i t h t h e g e a r r a t i o s q u a r e d a s a f a c t o r ; also, t h e t o r q u e i n c r e a s e s ( w h i l e
the speed decreases) with the gear ratio as the multiplying factor.
Page 3 2 1
angular acceleration of the reflector is
2
and that of the motor armature is 5.25 X 700* = 3680 rad/sec .
These are the maximum accelerations, since here the force (and torque) are
greatest. In closing, we note that motor torque and friction, omitted in this
The next example illustrates work done by a force acting on different
points of the body as time passes. A shortcut for calculating this work is
presented.
EXAMPLE 5 . 6
This example illustrates the work done by forces and couples belonging to Types
1, 2, and 8 on the preceding pages. The force F (52 lb) is applied to the uniform
cylinder C at rest in Figure E5.6a at the left. (This type of force might be applied by
a cord on a hub, as is suggested by Figure E5.6b.) If force F continues to act with
a. The work done by F during transit to the dashed position
b. The velocity of C and the angular velocity of the cylinder in the
Figure E5.6a
Figure E5.6b
Solution
We shall work part (a) in two ways. First, the definition of the work of F is
where Q is the point of in contact with F at any time. The geometry in Fig­
ure E5.6c gives an angle of 45.1° between F and v , since
Q
Figure E5.6c
* See preceding footnote.
Page 322
Also,
Therefore
and
A second and simpler approach is to note that F at Q may be moved to C,
using the idea of resultants, as in Figure E5.6d. The force at Q is replaced by the
force and couple at C which produce the same effect on therigidbody. Since, at C,
force F always acts on the same point of the body (it didn't at Q!) we may write
W = work of F at Q = (work of F at C) + (work of couple on
= 6.53F = 340 ft-lb
(as before)
Note that the work of a constant couple in plane motion is simply the moment of
FT sin 22.4°
Figure E5.6d
For part (b) we equate the work to the change in the kinetic energy of C:
Note that the gravity, friction, and normal forces do no work in this problem, for
The hardest part of the next example is finding where the mass center
is in the final position!
5.2 / The Principle(s) of Work and Kinetic Energy
323
EXAMPLE 5 . 7
The unstretched length of the spring in Figure E5.7a is ( = 0.3 m. The initial
angular velocity of body J in the top position is <w = 2.5 ? rad/s. There is
enough friction to prevent slipping of JoxxBdX all times. Determine the modulus
of the spring that will cause ^4 to stop in the <p = 90° position.
u
f
Figure E5.7a
Solution
Part of the work W in this problem is done by gravity. To express this work, we
must determine where the mass center C is located when ^4 reaches the final
position. Reviewing the kinematics, we find that the velocity of the geometric
center Q of J (see Figure E5.7b) is expressible in two ways:
0
1.
As a point of J, v = yfe + ROe,.
2.
As a point of 2>, v = y£ + tye,.
Q
0
Q
Thus we see that R0 = lip. Integrating, we get
Rd = lq>
in which the constant of integration is zero if we select 6 = 0 when g> = 0.
Therefore when <p = n/2, we may find the orientation of body J:
8 = angle that body -J turns through in reference frame (angle seen by
stationary observer in body B)
=
t
] ^
=
0.6 71
oT 2X
=
,
3 j I
Page 324
And so the final position of C is to the left of Q (see Figure E5.7c). We can now
write the work of gravity W because we now know the h moved through by C:
g
C is now here
C w a s initially to t h e right of
but
d u r i n g t h e rolling, line
(and e v e r y o t h e r line o n b o d y
turned
r a d c l o c k w i s e in s p a c e
rad
Figure E5.7c
The work done by the linear spring is always given by
where k is our unknown and the initial and final stretches are computed as
follows:
(unstretched length plus initial stretch =
initial length of spring)
and
so that the stretches are
For the kinetic energy side of W
consider body
Figure E5.7d
At
is now here
and no longer
the instantaneous
center of
we need the moments of inertia;firstwe
We shall use the "short form" of T — namely
(always valid when­
ever
in plane motion). Thus we need I and
* Note that when
is a
different point of a body in the initial and final positions, the value of
is
generally different in the two configurations, as is the case in this problem.
Using Figures E5.7d and E5.7e, we find:
2
2
= 0.392 + 80(0.1 + 0.05 )
= 1.39 k g - m
Figure E5.7e
2
* Since
= 0, w e d o n o t h a v e t o c a l c u l a t e
c e d u r e in g e n e r a l .
= 0.392 + 80(0.1 - 0.05)
= 0.592 kg • m
2
2
h e r e , b u t w e d o s o t o illustrate t h e p r o ­
Page 325
Therefore the kinetic energies of
we need are
(since the final angular
speed is to be zero)
For the bar
is the same in any position since
to the reference frame. Therefore, using
is point 0 , which is pinned
we have
Applying the work and kinetic energy principle, we get
This is equivalent to 829 lb/ft of stiffness in the U.S. system of units, since
1 lb/ft is the same stiffness as 14.6 N/m.
Question 5.4 What happens if k is larger than the calculated value?
What happens if it is smaller?
In the next example work and kinetic energy is used to help deter­
mine the point where rolling stops and slipping starts.
Answer 5 . 4 L a r g e r : t h e r o d will n o t r e a c h t h e h o r i z o n t a l position; smaller: t h e s y s t e m
p a s s e s t h r o u g h this p o s i t i o n w i t h o u t s t o p p i n g .
EXAMPLE 5 . 8
Mass m
Figure E5.8a
As we found in Example 5.3, sometimes it is useful to combine the work and
kinetic energy principle with one or more of the differential equations of motion
in order to obtain a desired solution. This example involves such a combination.
The small cylinder
starts from rest at
in the dotted position (see Fig­
ure E5.8a) and begins to roll down the large cylinder. Find the angle
at which
slipping starts, and show that the small cylinder will always slip before it leaves
the surface for a finite coefficient of friction.
Page 326
Solution
Using the free-body diagram in Figure E5.8b, the equations of motion are
(1)
(2)
(3)
Figure E5.8b
Just prior to slipping, the friction force
while is still equal to
still equal to
Therefore the equations can be rewritten as
and v is
c
(la)
(2a)
(3a)
These equations may be supplemented with the work and kinetic energy
equatio
(4)
Equations (1) to (4) may now be treated as four equations in the unknowns
to , and a , where the last three quantities are the angle, angular velocity, and
angular acceleration at slip. Solving them for yields the equation
2
s
Writing
for sin
quadratic for cos 6., gives
and then squaring and solving the resulting
which plots as shown in Figure E5.8c.
Asymptote
Figure E5.8c
The curve in the diagram gives the slipping angle as a function of the friction
coefficient; this is not the angle at which body leaves the surface. We note that if
we were to look for the angle at which the small cylinder leaves the surface of the
large cylinder, assuming no slipping has occurred, we would be trying to solve a
Page 327
problem with no solution if the friction coefficient is finite. The curve clearly
shows that for C to reach the angle cos (4/7), an infinite coefficient of friction is
required. Since the solution to the "leaving without slip" problem is precisely
cos (4 /7), shown below, then regardless of the friction coefficient (so long as it
is finite) will have to slip before it leaves.
Assuming now that the cylinder leaves withou
the (simpler) solution:
_1
_1
Equation (1)
Equation (4)
Eliminating
gives
As we have noted, this solution is valid only for an infinite coefficient of friction
between the cylinders. If were a particle (no rotational kinetic energy) with a
smooth surface, we would obtain (Example 2.13) = cos (2/3) = 48.2°.Note
the differences between these solutions.
_1
T w o Subcases of the Work and Kinetic Energy Principle
There is an important subcase of the principle of work and kinetic energy
that we have already seen in Chapter 2. Using
we obtained
(5.22)
This principle states again that the work done by the external force
resultant, when considered to act on the mass center, equals the change
in the translational part of the kinetic energy:
(5.23)
The integral of Equation (5.6) is
(5.24)
If we subtract (5-23) from (5.24), we obtain yet another result:
or
(5.25)
Page 328
This second subprinciple says that the work done by the external
moments (about C) on the body, as it turns in the inertial frame, is equal to
the change in the rotational part of the kinetic energy. We may use the
"total" W =
principle or either of its two "subparts" (Figure 5.12).
Figure 5.12
Let us now examine an example in which these subparts may be seen
to add to the "total" W =
equation:
EXAMPLE 5 . 9
As an illustration of the two subcases of the principle of work and kinetic energy,
we consider the cylinder of mass m rolling down the inclined plane shown in
Figure E5.9a. If the cylinder is released from rest, find the velocity v of its mass
center as a function of the distance x traveled by C.
c
c
Figure E5.9a
Figure E5.9b
Solution
Referring to the free-body diagram in Figure E5.9b, we see that the normal and
friction forces do no work because, as the cylinder rolls on the incline, they
always act on a point at rest. That is,
and
Applying the principle that W =
only the component of the gravity force W
that acts parallel to the plane does any work:
(1)
Since W always acts on the same point (C) of
and since
(2)
Page 329
The mass center's velocity is therefore
(3)
Now suppose we apply Equation (5.23):
(4)
In this problem the resultant force acting on is
and Equation (4) becomes
(5)
We see that, as expected, the friction force (though it does no net work)
retards the motion of the mass center C while turning the cylinder, as can be seen
(6)
(7)
or
(8)
And the sum of Equations (5) and (8) indeed gives Equation (2): the total W =
equation!
Potential Energy, Conservative Forces, and Conservation of
Mechanical Energy
In Section 2.4 we introduced the concept of potential energy, or the
potential of a force. When the work done by a force on a body is inde­
pendent of the path taken as the body moves from one configuration to
another, the force is said to be conservative and the work is expressible
as the decrease in a scalar function
the potential (energy). Thus as a
body moves from a configuration at time f j to a second configuration at
time t , the work done by an external conservative force is
2
or simply
If all the external forces that do work on a rigid body are conservative and
Page 330
is now the sum of the potentials of those forces, Equation (5.15) yields
or
or
constant
which expresses the conservation of mechanical energy.
From Chapter 2 and earlier in this section we can easily identify two
common conservative forces: (1) the constant force acting always on the
same material point in the body and (2) the force exerted on a body by a
linear spring attached at one end to the body and at the other to a point
fixed in the inertial frame of reference.
In the case of the constant force, a potential is = — F • r, where r is a
position vector for the point of application. When the force is that exerted
by gravity (weight) on a body near the surface of the earth,
mgz
where h is the altitude of the mass center of the body.
For the linear spring, we recall that =
where k is the spring
modulus, or stiffness, and 8 is the stretch. It is important to recognize that
when a spring is attached to, or between, two bodies that are both moving
(relative to the inertial frame), then
is a potential for the two
spring forces taken together (see Equation 5.19). That is, while neither of
the forces acting on the bodies can be judged by itself to be conservative,
the net work done on the two bodies by the two forces is expressible as a
decrease in the potential,
This is helpful in the analysis of
problems in which we have two or more interacting rigid bodies. We have
already noted earlier in this section that the work of the external forces on
a system of rigid bodies is not in general equal to the change in kinetic
energy of the system; this is because there may be net work done on the
rigid bodies by the equal and opposite forces of interaction. Suppose now
that our system is made up of two bodies joined by a spring, and suppose
the spring forces are the only internal ones that produce net work on the
system. We may then write W =
for each rigid body. Upon adding
these equations there results
(Work of forces external to system) + (Work of pair of spring forces)
= (Change in kinetic energy of system)
If the forces external to the system that do work are conservative, we
may add the various potential energies associated with them to that for
the pair of spring forces and conclude that
- constant
That is to say, in this case the mechanical energy of the system is con­
served.
Page 331
An example of a nonconservative force is sliding friction. A potential
cannot be found for friction, since the work it does depends on the path
taken by the body on which it acts. In this case,
must be used,
and it is seen to be more general than the principle of conservation of
mechanical energy.
EXAMPLE 5 . 1 0
Show that the same equation for the spring modulus in Example 5.7 is obtained
by conservation of mechanical energy. (See Figure E5.10.)
R-
0.1 m
D
(Uniform,
slender bar of
length l = 0.6 m
and mass - 35 kg)
B (Fixed cylinder! inertial
frame for the problem)
Figure E5.10
Solution
The potentials for gravity and for the spring are
Therefore, measuring z from O, we have
c
For the spring, using i for the initial and f for the final configuration, we have
Page 332
Thus, adding the potentials
we get
The kinetic energies were T = 4.34 + 0.365 = 4.71 J and T = 0. Therefore
i
f
574 + 0.0242k + 4.71 = 0.0722k + 0
or, rearranging,
574 - 0.0480k = - 4 . 7 1
This is the same final equation that resulted from
ample 5.7.
PROBLEMS
in the earlier Ex­
Section 5.2
5.1
Find the kinetic energy of the system of bodies ,
, and at an instant when the speed of is 5 ft/sec.
(See Figure P5.1.)
5.3 Upon application of the 10-N force F to the cord in
Figure P5.3, the cylinder begins to roll to the right. After C
has moved 5 m, how much work has been done by F?
Figure P5.3
Figure P5.1
5.4 The suspended log shown in Figure P5.4 is to be
used as a battering ram. At what angle should the ruf­
fian release the log from rest so that it strikes the door at
with a velocity of 20 ft/sec?
5.2
See Figure P5.2. (a) Explain why the friction force f
does no work on the rolling cylinder if the plane f is the
reference frame, (b) If, however, f is the top surface of a
moving block (dotted lines) and the reference frame is
now the groundG,doesfthen do work on ? Why or why
not?
Figure P5.4
Figure P5.2
5.5 The 20-kg bar in Figure P5.5 has an angular veloc­
ity of 3 rad/s clockwise in the horizontal configuration
shown. In that position the tensile force in the spring is
Page 333
30 N. After a 90° clockwise rotation the angular velocity
has increased to 4 rad/s. Determine the spring modu­
lus k.
5.9 Bar is smoothly pinned to the support at A and
smoothly pin-jointed to at B. (See Figure P5.9.) End D
slides on a smooth horizontal surface. If D starts from rest
at
, determine the angular velocities of the rods just
Figure P5.5
Figure P5.9
5.6 A uniform 40-lb sphere (radius = 1 ft) is released
from rest in the position shown in Figure P5.6. If the
sphere rolls (no slip), find its maximum angular speed.
5.10 The prehistoric car shown in Figure P5.10 is pow­
ered by the falling rock m, connected to the main wheel (a
cylinder of mass M) by a vine as shown. If the weights of
the frame, pulley, and front wheel are small compared
with Mg, find the velocity v of the car as a function of y if
c
Figure P5.6
5.7
The unbalanced wheel of radius 2 ft and weight
64.4 lb shown in Figure P5.7 has a mass center moment
of inertia of 6 slug-ft . In position 1, with C above O, the
wheel has a clockwise angular velocity of 2 rad/sec. The
wheel then rolls to position 2, where OC is horizontal.
Determine the angular velocity of the wheel in position 2.
2
Figure P5.10
5.11 A truck body weighing 4000 lb is carried by four
solid disk wheels that roll on the sloping surface. (See
Figure P5.ll.) Each wheel weighs 322 lb and is 3 ft in
diameter. The truck has a velocity of 5 ft/sec in the posi­
tion shown. Determine the modulus of the spring if the
truck is brought to rest by compressing the spring 6 in.
Figure P5.7
Figure P5.8
5.8 Determine the spring modulus that will allow the
2-kg bar in Figure P5.8 to arrive at the position
at
zero angular velocity if it passed through the vertical
(where the spring is compressed 0.1 m) at 8 rad/s
Figure P5.11
Page 334
5.12 For the cylinder of Problem 4.70, assuming no slip
and that the cylinder starts from rest, use work and kinetic
energy to find the speed of its center in terms of the dis­
placement of the center.
5.16 To the data of Problem 5.15 we add a constant
counterclockwise couple of moment
acting as
shown in Figure P5.16. Repeat the problem.
Ideally, the following five problems should be worked
sequentially:
5.13 A cylinder with mass 6 kg has a 20-N force applied
to it as shown in Figure P5.13. Find the angular velocity
of the cylinder after it has rolled through 90° from rest.
Figure P5.16
5.17 To the data of Problem 5.16, we add a spring, at­
tached to a cord wrapped around a second slot in the
cylinder near its outer rim as shown in Figure P5.17. The
spring has modulus 6 N/m and is initially stretched
0.2 m. Repeat the problem.
Figure P5.13
5.14 Rework Problem 5.13 if a slot is cut in the cylinder
and a cord is wrapped around the slot, with the 20-N
force now applied to the end of the cord as shown in
Figure P5.14. Neglect the effect of the thin slot on the
moment of inertia of the cylinder.
Figure P5.17
5.18 The 5-lb cylinder in Figure P5.18 rolls on the in­
cline. If the velocity of the mass center C is 5 ft/sec down
the plane in the upper (starting) position, find v in the
bottom position.
C
Figure P5.14
5.15 Suppose in Problem 5.14 we remove some material
from the cylinder so as to offset the mass center C from
the geometric center Q as shown in Figure P5.15. The
removal reduces the mass to 5.5 kg and makes the radius
of gyration with respect to the axis through C normal
to the plane of the figure k = 0.286 m. Repeat the
problem.
k = 2 lb/ft
Unstretched
length = 4 ft 2 in
C
Figure P5.18
Figure P5.15
5.19 The spring in Figure P5.19 has an unstretched
length of 0.8 m and a modulus of 60 N/m. The 20-kg
wheel is released from rest in the upper position. Find its
angular velocity when it passes through the lower
(dashed) position if its radius of gyration is k = 0.2 m.
C
Page 3 3 5
5.22 The wheel in Figure P5.22 has a mass of5slugs and
a radius of gyration for the z axis through C of 0.7 ft. The
spring has modulus 20 lb/ft and natural length 4 ft. The
wheel is released from rest, and it rolls without slipping
on the plane. Find how far down the plane the mass
center C will move.
Figure P5.19
5.20 Bar, in Figure P5.20 is initially at rest in the vertical
position, where the spring is unstretched. The wall and
floor are smooth. Point B is then given a very slight dis­
placement to the right, opening up a small angle
­
a. Draw a free-body diagram of the slightly dis­
placed bar and use it to show that the bar will
start to slide downward if
b. Find the angular velocity of as a function of
for such a spring.
Figure P5.22
5.23 The wheel in Figure P5.23 weighs 200 N and has a
radius of gyration 0.3 m with respect to the z axis. It is
released from rest with the spring stretched
If there is
C
a. up, and
b. down the plane in the subsequent motion.
Stiff rod
Modulus k
Figure P5.20
20 kg (eachI
Figure P5.23
Figure P5.21
5.21 Find the spring modulus k that will result in the
system momentarily stopping at
after being re­
leased from rest at
if the initial stretch in the
spring is zero. (See Figure P5.21.) Hint: Use symmetry!
5.24 Show that if the rolling body in Example 5.8 is a
sphere instead of a cylinder, it will slip at the angle
satisfying the equation
Page 336
5.25 Use
in Problem 4.87(b) tofindthe velocity
of C when it has moved 3 m down the incline.
• 5.26 In Problem 4.134 use the principle of work and
energy to obtain an upper bound on the rod's angular
speed in its subsequent motion after the right-hand string
5.27 A thin disc of mass m and radius a is pinned
smoothly at A to a thin rod of mass m/2 and length 3a
(see Figure P5.27). The rod is then pinned at B. If the body
is held in equilibrium in the configuration shown, then
released from rest, find the velocity of point A as the
system passes through the vertical.
5.30 For the data of Problem 4.113 use
to find
the speed of the plate as a function of the distance x it
has traveled to the right. Use the x = x (t) result to check
your answer; differentiate and eliminate t to produce the
same
result.
is cut.
C
C
C
5.31 Figure P5.31 shows a fire door on the roof of a
building. The door , 4 ft wide, 6 ft long, and 4 in. thick,
is wooden (at 30 lb/ft ) and can rotate about a frictionless hinge at O. A cantilever arm of negligible weight is
its free end. During a fire the link melts and the door
swings open 45°. Find the angular velocity of the door
just before the 150-lb weight hits the roof: (a) with no
snow on the roof; (b) with snow at 1 lb/ft on the roof.
3
2
5.28 Repeat the preceding problem if the pin at A is re­
placed by a weld.
Root
Hinge
Figure P5.31
Figure P5.27
5.29 The 10-lb wheel shown in Figure P5.29 is attached
at its center to a spring of modulus 20 lb/in. The radius of
gyration of the wheel about the center is 2.5 in. The wheel
rolls (no slip) after being released from rest with the
spring stretched 1 in. Find: (a) the maximum magnitude
of force in the spring; (b) the maximum speed of the
center of the wheel during the ensuing motion.
5.32 Block
in Figure P5.32 is moving downward at
5 ft/sec at a certain time when the spring is compressed
1 ft. The coefficient of friction between block and the
plane is 0.2, and the radius of cylinder is 0.5 ft. Weights
a. Find the distance that falls from its initial po­
sition before coming to zero speed.
b. Determine whether or not body will start to
move back upward.
Figure P5.32
Figure P5.29
5.33 The system in Figure P5.33 consists of a cylinder
(100 kg) and (equilateral) triangular plate
(20 kg)
pinned together at the mass center C of the cylinder. The
other two vertices of the plate are connected to springs,
1
* Asterisks identify the more difficult problems.
Page 337
the left one of which (S ) remains vertical in the slot.
(Spring S is shown only in its initial position.) The initial
stretches of the two springs (in the position shown) are
0.2 m for S and 0.04 m for S . The moduli are 40 N/m
for S and 10 N/m for S . If the system is released from
rest in the given position, find the velocity of C when
vertex B reaches its lowest point in the slot. Assume suffi­
cient friction to prevent, from slipping on the plane. The
moment of inertia of an equilateral triangular plate of
side s about its z axis is ms /12.
1
1
1
2
1
2
1
2
C
Figure P5.35
Figure P5.36
Figure P5.37
Figure P5.33
5.34 The mass center C of a rolling 2-kg wheel of radius
R = 15 cm is located 5 cm from its geometric center Q.
(See Figure P5.34.) The spring is attached at C and is not
shown in position 2; its unstretched length is 0.3 m,
and its modulus is 3 N/m. The radius of gyration is
k = 0.09 m. Find the angular speed in position 2 (onequarter turn from position 1).
C
Figure P5.38
5.36 The center of mass of a uniform triangular plate is
two-thirds of the distance from any vertex to the opposite
side. The moment of inertia of an equilateral triangular
plate is ms /12 with respect to the z axis through C. For
the plate shown in Figure P5.36, with mass 30 kg and
side 2 m, find its angular velocity when it reaches the
dotted position where C is beneath O. The spring has
unstretched length 0.5 m and modulus 20 N/m, and the
plate is released from rest.
2
5.37 The bar in Figure P5.37 weighs the same (W) as the
hoop to which it is welded. The combined body is re­
leased from rest on the incline in the position shown. If
there is no slipping, determine the velocity of Q after one
revolution of the hoop.
Figure P5.34
• 5 . 3 5 The three rods shown in Figure P5.35 are pinned
together with one vertex also pinned to the ground. The
length of the bar labeled is given by 2b = 0.4 m, and
the density of the material of all bars is 7850 kg/m . Their
cross-sectional area is 0.002 m . Find the angular velocity
of the combined body after it swings 90° from rest if: (a)
3
2
5.38 Cylinders
and
in Figure P5.38 are released
from rest and turn without slip at the contact point. A cord
is wrapped around an attached hub of each, which has
negligible effect on the moment of inertia. There is
enough friction to prevent the rope from slipping on the
pulley. Find the velocity of the mass center of the pulley
after body has fallen 20 ft.
Page 338
5.39 A 5-lb cylinder is raised from rest by a force
P = 20 lb. (See Figure P5.39.) Find the modulus of the
spring that will cause the cylinder to stop after its center
has been raised 2 ft. Will it then start back down? The
spring is initially unstretched.
Figure P5.41
Figure P5.39
Figure P5.42
5.40 Body in Figure P5.40 rolls to the right along the
plane and has a radius of gyration with respect to its axis
of symmetry of k = 0.5 m. The corresponding radius of
gyration for is 0.12 m. The spring is stretched 0.6 m at
an instant when
Find
after C has trav­
eled 1 m to the right. (C is an externally applied couple
acting on .)
c
0
5.41 The cylinder and the block each weigh 100 lb. They
are connected by a cord and released from rest on the
inclined plane as shown in Figure P5.41. The spring, con­
nected to the center C of the cylinder, is initially stretched
6 in. Find the velocity of the block at the instant the spring
ping.
5.42 The 20-lb wheel in Figure P5.42 has a radius of
gyration of 4 in. with respect to its (z ) axis. A cable
wrapped around its inner radius passes under and over
two small pulleys and is then tied to the 50-lb block
The spring has a modulus of 90 lb/ft and is constrained
to remain horizontal. There is sufficient friction to pre­
vent
from slipping on the plane, (a) If the system is
released from rest, find the angular speed of after the
block then falls 1 ft. (b) Would the answer be different if
block
were replaced by a device that keeps the cable
unstretched,
force constant at 50 becomes
lb? Why or
why not? if there is sufficient friction be­
C
* 5.43 Rod
and disk
in Figure P5.43 have weights
W = 5 lb and W = 6 lb. The rod's length is 8 in., the
disk's radius is 4 in., the mass center offset of the disk is
2 in. from Q, and the radius of gyration of the mass of
with respect to the z axis through C is 3 in. It is desired to
attach a spring between point Q and afixedpoint so that
1
Figure P5.40
2
Figure P5.43
Page 3 3 9
the disk and rod come to a stop (in the dotted position)
after
turns 90° clockwise from rest. The spring has a
modulus of 25.5 lb/ft and an unstretched length of 4 in.;
it is to be unstretched initially. Find the final spring
stretch, and from this result determine where to attach the
fixed end of the spring. (There are two possible points!)
5.44 The bodies in Figure P5.44 have masses m = 0.3
slug, m = 0.5 slug, and m negligible. A spring is at­
tached to A that is stretched 25 in. in the dotted position
when everything is at rest. Find the spring modulus if
when is horizontal.
1
2
3
5.45 A vertical rod is resting in unstable equilibrium
when it begins to fall over. (See Figure P5.45.) End A
slides along a smooth floor. Find the velocity of the mass
center C as a function of L, g, and its height H above the
floor.
5.47 In Problem 4.108 determine the velocity of corner
B of the half-cylinder when the diameter AB becomes
horizontal for the first time.
5.48 Body translates in the slot without friction. (See
Figure P5.48.) Disk
(radius R) is pinned to block
through their mass centers at G. Body and body each
has mass m; body has mass 2m. The system is released
from rest a distance D above the floor. Find: (a) the start­
ing accelerations of and G; (b) the velocity of when it
hits the floor, using
Light pulley
Cord
5.46 Pulley weighs 100 lb and has a centroidal radius
of gyration k = 7 in. (See Figure P5.46.) The disk pulley
weighs 20 lb. Find the velocity of weight (50 lb) after
it falls 2 ft from rest. (Assume that the rope does not slip
on the pulleys.)
C
Figure P5.48
N o slip
5.49 Link weighs 10 lb and may be treated as a uni­
form slender rod (Figure P5.49). The 15-lb wheel is a cir­
cular disk with sufficient friction on the horizontal sur­
face to prevent slipping. The spring is unstretched as
shown. Link is released from rest, and the light block
slides down the smooth slot. Neglecting friction in the
pins, determine: (a) the angular velocity of the link as A
strikes the spring with
horizontal; (b) the maximum
deflection of the spring. (The modulus k of the spring is
10 lb/in.)
Initial position
Figure P5.44
Figure P5.45
Figure P5.46
Figure P5.49
Page 340
5.50 The masses of four bodies are shown in Fig­
ure P5.50. The radius of gyration of wheel with respect
to its axis is k = 0.4 m. Initially there is 0.6 m of slack
in the cord between
and the linear spring. (Modulus
k = 1000 N/m, and the spring is initially unstretched.)
Determine how far downward body will move.
C
a. Find how far to the right the mass center moves
in the ensuing motion, assuming sufficient fric­
tion to prevent slipping.
b. When stops instantaneously at its farthest
right point, what increase in the 50-N force and
what minimum friction coefficient are needed
to keep it there?
5.58 The slender nonuniform bar in Figure P5.58 (the
mass is m and the radius of gyration with respect to the
mass center C is L/2) is supported by two inextensible
wires. If the bar is released from rest with
, find the
tension in each wire as a function of
5.59 The system depicted in Figure P5.59 is released
from rest with 2 ft of initial stretch in the spring. There is
sufficient friction to prevent slipping at all rimes. Deter­
mine whether will leave the horizontal surface during
the subsequent motion. Note that the string S goes slack if
the stretch tries to become negative.
Figure P5.50
Wrapped
cords .
Figure P5.I1
Figure P5.57
5.51 Cylinder, in Figure P5.51 is moving up the plane
with v = 0.3 m/s at an initial instant when the spring is
stretched 0.2 m. If does not slip at any time, determine
how far down the plane the point C will move in the
subsequent motion. Note: The spring, connected to the
cord, cannot be in compression.
C
5.52 Solve Problem 4.64 for v as a function of x using
C
c
Figure P5.58
5.53 In Problem 4.165, find the angular velocity of the
rod when
5.54 Solve Problem 4.65 (a) by
5.55 For each of the wheels in Problem 4.65, solve for v
as a function of x using
. The wheels start from
rest.
C
C
• 5.58 Solve Problem 4.177 with the help of
nore the hint.
. Ig­
5.17 The radius of gyration of the wheel and hub in
Figure P5.57, with respect to its axis of symmetry through
C, is k = 2.5 m. The springs are unstretched at an initial
position of rest, when the 50-N force is applied.
cC
Figure P6.59
Page 341
5.60 The system is released from rest in the position
shown in Figure P5.60. Force P is constant, 60 lb, and the
cord is wrapped around the inner radius of . Note the
mass center of is at C. Find the normal force exerted
onto by the plane (after using
to get ) at the
instant when has rotated
. The spring is initially
unstretched, and there is enough friction to prevent slip­
ping.
Figure P5.60
5.63 The cord connects the slotted cylinder to the cyl­
inder
as shown in Figure P5.63. Assume that neither
body slips after the system is released from rest. The
spring is initially unstretched, and is stiff and guided so it
can take compression. Find the angular velocity of after
its center C has moved 1 m.
5.64 The 12-ft, 32.2-lb homogeneous rod
shown in
Figure P5.64 is free to move on the smooth horizontal
and vertical guides as shown. The modulus of the spring
is 15 lb/ft and the spring is unstretched when in
the position shown. Rod
is released from rest with
and nudged to the right to begin motion, (a)
Determine the angular velocity of the rod when it be­
comes horizontal, (b) What is the angular acceleration of
the rod in this position
?
5.65 The 50-kg wheel in Figure P5.65 is to be treated as a
cylinder of radius R = 0.2 m. If it is rolling to the left with
v = 0.07 m/s at an initial instant when the spring is
unstretched, find: (a) the distance moved by C before v is
instantaneously zero; (b) the minimum coefficient of fric­
tion that will prevent slip.
c
5.61 The two identical links and in Figure P5.61,
each of mass m and length are pinned together at A, and
is pinned to the ground at B. The end C of slides in the
vertical slot. Friction is negligible, and the system is re­
leased from rest. Find the velocity of point C just before
point A reaches its lowest point.
5.62 The body of mass 2m in Figure P5.62 is composed
of two identical uniform slender rods welded together. If
friction in the bearing at O is neglected and the body is
released from rest in the position shown, find the magni­
tude of the force exerted on the rod by the bearing after
the body has rotated through 90°.
c
Cylinder 1 0 0 N
Figure P5.63
120
Figure P5.64
Figure P5.61
Figure P5.62
Figure P5.65
N/m
Page 342
5.66 The cylinder in Figure P5.66 is rolling at
rad/sec in the initial (i) position, where the spring is
unstretched. Other data are:
m = 2 slugs
r = 3 ft
k = 3 lb/ft
5.69 A slender uniform rod of weight W is smoothly
hinged to a fixed support at A and rests on a block at B.
(See Figure P5.69.) The block is suddenly removed. Find:
(a) the initial angular acceleration and components of re­
action at A; (b) the components of reaction at A when the
rod becomes horizontal.
5.70 Two quarter-rings are pinned together at P and re­
leased from rest in the indicated position (Figure P5.70)
on a smooth plane. Find the angular velocities of the rings
when their mass centers are passing through their lowest
points. Hint: By symmetry, point P always has only a
vertical velocity component; this means that no work is
done on either ring by the other, because (again by sym­
metry) the force between the rings has only a horizontal
component normal to the velocity of P. More generally, as
long as the pin is smooth, the work done by two pinned
bodies in motion on each other will be the negative of
each other because the velocities will be equal whereas
the forces will be opposites.
Figure P5.66
Find the final position of C (x ) at which either the cylin­
der has stopped (for an instant) or started to slip, which­
ever comes first. Hint: Try one, check the other!
C
• 5.71 A slender rod is placed on a table as shown in Fig­
ure P5.71. It will begin to pivot about the edge E and, at
some angle , it will begin to slip. Find this angle, which
will depend on the coefficient of friction and on k. Hint:
Use all three equations of motion together with
Eliminate and
, obtaining expressions for f and N.
Setting
then permits a solution for . Solve the
resulting equation when
and k = 0.25.
5.67 The uniform slender rod in Figure P5.67 (mass = 5
slugs, length = 10 ft) is released from rest in the position
shown. Neglecting friction, find the force that the floor
exerts on the lower end of the rod when the upper end is
6 ft above the floor. Hint: First use a free-body diagram
and the equations of motion to deduce the path of the
mass center.
5.68 In Figure P5.68, the ends of the bar are constrained
to vertical and horizontal paths by the smooth rollers in
the slots shown. The bar, originally vertical, is very gently
nudged at its lower end to initiate motion. Find the reac­
tions onto the bar at A and B just before the bar becomes
horizontal.
Figure P5.69
Figure P5.70
Figure P5.67
Figure P5.68
Figure P5.71
Page 343
• 5.72 The uniform equilateral triangular plate ABE in Fig­
ure P5.72 weighs W and is pinned to a fixed point at A and
to a rope at E. The rope passes over a small, frictionless
pulley at D and is then tied to the (equal) weight W which
is constrained to move vertically. If the system is released
in the given position with the angular velocity of ABE
being , find the angular velocity of ABE in the
dashed position (i.e., when side AE becomes horizontal).
Next, show that an identical result is obtained (as it must
• 5.74 The cylinder
in Figure P5.74 has a spring at­
tached and is released from rest. Assume that there is
sufficient friction between and the plane to prevent slip
throughout the motion, and that the slot around which
the cord is wrapped has a negligible effect on the cylin­
der's moment of inertia. Find the velocity of B when the
unwrapped length of rope is completely vertical (that is,
when C has 0.3 m left to travel before it would be directly
above E). Assume also that the weight
moves only
vertically — that is, that the rope does not start to sway.
Figure P5.72
Smooth,lightpulley
•5.73 This problem, continuing and using the results of
Problem 4.150, is to verify that
for that solution.
First, note that only the friction couple does net work on
the system of plus , , and verify:
5.3
Figure P5.74
T h e Principles of Impulse and Momentum
The Equations of Impulse and Momentum for the Rigid Body in Plane Motion
The principle of work and kinetic energy is very helpful when the prob­
lem is posed in terms of positions and velocities. When time, rather than
position, is the main concern, we often draw on a pair of principles
concerned with impulse and momentum vectors. Just like
, these
principles are obtained by general integrations of the equations of mo­
tion, but now the integration is directly with respect to time, without first
dotting the equations with velocity. Thus they leave us with a set of
vector equations instead of a single scalar result.
We have encountered one of the principles in Section 2.5 in our study
of mass center motion: The impulse of the external forces imparted to any
system equals its change of momentum over the same time interval. In
Chapter 2 the system was general, so this principle holds for the rigid
bodies we are now studying. From Equation (2.27),
so that
Page 344
(5.26)
Reviewing, we note that the integral
dt is called the impulse (or
linear impulse) imparted to the system by external forces. The vector
is the change in the system's momentum (or linear momentum) from the
initial to the final time.
We need only the x and y components of Equation (5.26) for the rigid
body in plane motion:
(5.27)
(5.28)
There is also a corresponding principle of angular impulse and momen­
tum. From Equation (2.43),
so that
(5.29)
This equation may be put into a convenient form for rigid bodies in plane
motion by recalling Equation (4.4) for the angular momentum:
For symmetric bodies in which the products of inertia vanish and
this equation becomes
Therefore
or
(5.30)
The integral
is called the angular impulse imparted to the
system by the external forces and couples, and the quantity
is the
change in angular momentum, both taken about C.
A subtle but important point regarding Equation (5.30) must be
understood here. We note from Equation (5.29) that angular impulse
equals the change in angular momentum for any body (deformable as
well as rigid); therefore the use of Equation (5.30) only requires that the
body of interest behave rigidly at the start (t ) and end (t ) of the time
interval (t , t ). At those times the moment of momentum is
even though this simple expression for H may not apply between t, and
t . A good example is an ice skater drawing in her arms to increase angular
speed, as we shall see later in Example 5.14.
i
i
C
f
f
f
i
Page 345
In summary, for the rigid body in plane motion we have the follow­
ing two principles at our disposal:
1. Linear impulse and momentum:
from which we get
(5.31)
(5.32)
where the directions of x, y, and z are fixed in the inertial frame.
2. Angular impulse and momentum:
(5.33)
We note also that if the products of inertia are not zero, we have
(5.34)
and
(5.35)
Oust lion 5.5 Are the coordinate axes associated with Equations (5.34)
and (5.35) the same as those of (5.31) and (5.32)?
In the remainder of this section we treat only examples of symmetric
bodies, for which Equations (5.31) to (5.33) are the impulse and momen­
tum equations. The first example deals with both linear and angular
impulse and momentum for a single body.
Answer S.5 Yes.
Figure E5.1 l a
EXAMPLE 5 . 1 1
A cylinder has a string wrapped around it (see Figure E5.11a) and is released
from rest. Determine the velocity of C as a function of time.
Solution
We choose the sign convention to be as shown in Figure E5.1 lb, since the cylin­
der turns clockwise as C moves downward. Applying the impulse and momen­
Figure E5.11b
tum equations in the y and J directions (note that
gives
means
so that
Page 346
(1)
and
(2)
We note that
it is itself an unknown and should be treated as such. (In this
problem, use of the equations of motion would separately tell us that T = mg/3,
so that the integral is in fact Tt. But sometimes T is time-dependent, in which case
Tt would be incorrect for the value of the integral.)
Eliminating
However,
because the cylinder rolls on the rope, so that
which gives our result:
In the next example, several bodies are involved, making the solution
a little more difficult. Two limiting case checks are discussed at the end.
The cylinder falls with '
EXAMPLE 5 . 1 2
Figure E5.12a
of a g" because of the retarding force of the rope.
The cart shown in Figure E5.12a has mass M exclusive of its four wheels, each of
which is a disk of mass m/2. The front wheels and their axle are rigidly con­
nected, and the same is true for the rear wheels. If the axles are smooth, find the
velocity of G (the cart's mass center) as a function of time. The system starts from
rest. Assume that there is enough friction to prevent the wheels from slipping.
Page 347
Solution
We first consider the free-body diagram (Figure E5.12b) of a wheel pair (either
front or back). Since the front and rear wheels are constrained to have identical
angular velocities at all times, the ffs (front and back) must produce identical
la's; hence the friction force is the same for the rear wheels as for the front. And
since the wheels' mass centers must always have identical velocities, the forces
acting down the Diane on each pair (front and back) must also be equal. These
resultants are
so the reaction A is also the same on each pair
of wheels.
x
Figure E5.12b
Question 5.6
Are A and N also the same for front and rear wheels?
y
From the free-body diagram, we may write the following linear and angular
equations of impulse and momentum:
(1)
(2)
We may also isolate a free-body diagram of the translating cart (Figure E5.12c)
and write its equation of impulse and momentum in the x direction down the
plane:
(3)
Figure E5.12c
Next we note that Cj and C , the mass centers of the front and back wheel pairs,
are also points of the cart; thus
Using this relation, and adding
Equation (3) to twice Equation (1), eliminates the unknown impulse of the reac­
tion A :
2
x
(4)
Similarly, adding Equation (4) to twice Equation (2) results in an equation free of
the unknown friction force:
Answer 5.6
N o t in g e n e r a l . T h e y d e p e n d o n t h e p o s i t i o n o f G r e l a t i v e t o C a n d C .
x
2
* T h i s is t h e e q u a t i o n o f l i n e a r i m p u l s e a n d m o m e n t u m f o r t h e total ( n o n r i g i d ) s y s t e m o f
cart plus w h e e l s .
Page 348
Carrying out the integration and solving for
, we get
Smooth
Figure
I5.12d
Figure E5.12e
We note from this result that if the wheels are very light compared to the
weight of the cart (m
M), then
which is the answer for the
problem of Figure E5.12d. Thus light wheels on smooth axles makes the cart
move as if it were on a smooth plane, as expected.
The reader may wish to examine the other limiting case, that of the cart being
light compared to heavy wheels (M
m). In this case the result, using the
free-body diagram in Figure E5.12e, is
In the next example, the two bodies — one rolling and the other
translating — are connected by an inextensible cord.
EXAMPLE 5 . 1 3
Force P acts on the rolling cylinder C beginning at f = 0 with C at rest. (See
Figure E5.13a.) Force P varies with the time t in seconds according to
(positive to the left as shown)
:
Cylinder C and body t- respectively weigh 100 and 40 N. Find the velocity of G
(the mass center of l<) when f ™ 10 s. Neglect the effect of the hubs in Fig­
ure E5.13b (and the drilled hole to accommodate force P) on the moment of
inertia of C.
0.1m
0o m
P
Cord
Light
pulley
Figure E 5 . H a
Figure E8.13b
Solution
Using the free-body diagrams (Figures E5.13c and E5.13d), we may write the
equations of impulse and momentum. On C, using Figure E5.13c,
Page 349
at
(1)
Also on C:
Figure E5.13c
or
(2)
On B, using Figure E5.13d,
or
Figure E5.13d
(3)
Subtracting Equation (3) from (1), after integrating the sine function, we obtain
(4)
To obtain a second, independent equation that is also free of the integral of
the unknown tension T, we add Equation (1) to twice Equation (2):
(5)
Multiplying Equation (4) by 0.6 and subtracting from Equation (5) gives
(6)
Kinematics now relates
and
the lowest point of C has the same
velocity magnitude as does G because of the inextensibility of the cord (see
Figure E5.13e):
Kinematic conditions:
Figure E5.13e
Substituting these expressions foi
and
into Equation (6) gives
Hence the velocity of the mass center of # a t t = 10 s (when the force changes
direction) is
Page 350
We emphasize that Equations (1), (2), and (3) in the preceding exam­
ple are merely first integrals of the equations of motion studied in Chap­
ter 4.
Conservation of Momentum
As we saw in Section 2.5, if the force in any direction (let us use x, for
example) vanishes over a time interval, then the impulse in that direction
vanishes also:
Since this impulse equals the change in momentum in the x direction, we
have zero change when
and thus the momentum is conserved in
that direction between r- and t :
(
f
or
mx = mx^
(5.36)
C(
We would of course also have conservation of momentum in the y (or
any other) direction in which the force resultant vanished.
Finally, if the z component of 2M,- is zero between t and t , then the
angular impulse vanishes and we have conservation of angular mo­
mentum:
{
f
or
(5.37)
Fnr nlane motion of symmetric bodies (l£ and J £ = 0) we have
there is then no need for the z subscript, and Equation (5.37)
may be rewritten
(5.38)
We now consider a well-known example of conservation of angular
momentum.
EXAMPLE 5 . 1 4
A skater spinning about a point on the ice (see Figure E5.14) draws in her arms
and her angular speed increases.
Figure E5.14
a. Is angular momentum conserved?
b. Is kinetic energy conserved?
c. Account for any gains or losses if either answer is no.
Page 351
Solution
We begin with part (a). Before the skater draws in her arms, we may treat her as a
rigid body and thus
. The same is true after the arms are drawn in, so
that
If we neglect the small friction couple at the skates and the small
drag moments caused by air resistance, then the answer to part (a) is yes because
XMc is then zero. Thus
Therefore
(showing an angular speed increase since 1 > I )
1
2
For part (b) the kinetic energies are
Thus kinetic energy is not conserved.
For part (c) the change in kinetic energy is seen to be positive:
Since there is no work done by the external forces and couples,* it is clear that this
kinetic energy increase is accompanied by an internal energy decrease within the
skater's body as her muscles do (nonexternal) work on her (nonrigid) arms in
drawing them inward. Since total energy is always conserved (first law of ther­
modynamics), the skater has lost internal energy in the process.
The next example is similar to Examples 2.18 and 2.19 except that
now the pulley has mass.
EXAMPLE 5 . 1 5
Two identical twin gymnasts, L and R, of mass m are in equilibrium holding onto
a stationary rope in the position shown in Figure E5.15a. The rope passes over the
pulley B, which has moment of inertia I with respect to the axis through O normal
to the figure. The gymnasts then begin to move on the rope at speeds relative to it
of y upward and y downward. When gymnast R reaches the end of the rope,
he discovers he is in the same spot in space at which he began. How far up or
down (tell which) has L moved (a) relative to the rope? (b) in space?
L rel
Figure E5.15a
R rel
* A s s u m i n g h e r a r m s a r e d r a w n in at t h e s a m e level.
Page 352
Solution
If we select our system to be "everything": L, R, the rope, and B, then
and thus angular momentum about O is conserved. (Note that the gymnasts'
gravity forces' moments about O cancel, and the pin reactions and weight of B
pass through O.) Therefore,
(since all bodies are at rest initially)
(1)
After motion begins, the angular momentum about 0 of Bis simply \ o) \sL, since
O is a pivot of B. For L and R, however, the situation is different and needs some
discussion. The gymnasts, of course, are not rigid bodies, and we shall resort to
Equation (2.38) to write their angular momenta with respect to O. For L, with
velocity v in the inertial frame (the ground), and mass center Q ,
e
0
B
L
We now assume that the gymnast is a particle; this is equivalent to neglecting
, the gymnast's angular momentum about his mass center, in comparison
with the "r X mv" term. This is a good assumption because whatever body
motions are not translatory are caused by parts (arms, mostly) in relative motion
fairly close to the mass center. Thus,
where P is shown in Figure E5.15b. Similarly,
Question 5.7
Why is r
OCl[
X mv = r
R
0 0
X mv„?
Now, if we locate (see Figure E5.15b) the gymnasts' position in space with the
coordinates y and y , then
L
Figure
E5.15b
R
Note that if the gymnasts move in opposite vertical directions on opposite sides of
the pulley, their angular momenta about O will be in the same direction. Note
further that even though the "particles" L and R are in rectilinear motion, they
still have angular momenta about points such as O that do not lie on their lines of
motion.
Substituting the three angular momenta into Equation (1), we find:
(2)
Next, we know from the data that the rope moves counterclockwise around the
pulley. Therefore, calling its speed y^, we have (see Figure E5.15c)
and
Starting p o s t i o n of
end of rope at t = 0
Figure
E5.15c
Answer 5.7
Because r ^ , = r
O Q
+ r
Q C i
, and r
Q C <
X
mv
Cm
= 0 since T
QCK
and v
C | (
a r e parallel
Page 353
Substituting these into Equation (2), and simplifying the result, we obtain the
equation
(3)
Integrating,
ButC = 0 since y . = y = y = 0 at t = 0. Now at thetimewhen R is at the
end of the rope, y = H, and y
H also, so that
1
rope
r e l
R rel
=
R rel
rope
Therefore
and gymnast L moves u p r e l a t i v e
to the rope and
in space from
his original position.
We can check Example 2.18 by going back to Equation (2) of the
preceding example, and letting J —» 0 (note that at this point nothing has
been said about the relative motions):
or
Therefore, regardless of relative motions (with respect to the rope), the
two gymnasts rise in space equal amounts. The reader may wish to show
that for the problem of Example 2.18 but with I > 0, the left gymnast is
pulled up less than the height of climb in space of the right gymnast.
We now consider an example in which (unusually) both angular
momentum and kinetic energy are conserved.
EXAMPLE 5 . 1 6
The 2-kg collar Cin Figures E5.16a,b turns along with the smooth rod I? (see
Figure E5.16a), which is 1 m long, has a mass of 3 kg, and is mounted in bearings
with negligible friction. The angular speed is increased until the cord breaks (its
tensile strength is 60 N), and at that instant the external moment is removed.
Determine the angular velocity of I? and the velocity of C (the mass center of C)
when the collar leaves the rod.
Page 354
Figure E5.16a
Figure E5.16b
Solution
The string provides the force causing the centripetal (inward) acceleration until it
breaks. At that instant, we may solve for the angular velocity of C:
In the accompanying free-body diagram, Figure E5.16c, Nj and N are the verti­
cal and horizontal resultants of the pressures of the inside wall of exerted by
After the rope breaks at time f,, collar C moves outward in addition to
turning with /?; this is because there is no longer any inward force to keep it from
"flying off on a tangent." Between times t, and f (when it leaves R), we have the
following for the system
2
2
Figure E5.16c
1.
Conservation of angular momentum H about z (because the external
forces have no moment about z ).
Conservation of kinetic energy T (since no net work is done on the system).
Note that the normal forces between rod and collar, being equal in magni­
tude but opposite in direction, act on points with equal velocity components
in the direction of either force; hence their net work vanishes.
a
Q
2.
Condition 1 gives
Until the string breaks,
O is a point of both bodies!
Thus
c
Page 355
The component of v perpendicular to the rod /? is thus v = 1.05afy
= 3.87 m/s. We can now obtain the radial component by conservation of T
(condition 2):
c
Ois
for both e
and/Pinitially
c
components of y parallel
and perpendicular to *
c
Thus since the initial kinetic energy was 59.2 J and since
we see that 63 percent of the original energy has gone into the outward motion of
the collar.
Impact
We studied the impact of a pair of particles in Section 2.5. In this section
we shall extend this study to two bodies colliding in plane motion.
The large forces occurring during an impact between two bodies
and <B obviously deform the bodies. Because of vibrations and perma­
nent deformations that are produced, some of the mechanical energy will
be dissipated in the collision. However, it is often possible to treat a body
as rigid before, and then again after, the impact in order to gain informa­
tion of value. In impact problems we assume that:
2
1.
Velocities and angular velocities may change greatly over the short
impact interval
2.
3.
Positions of the bodies do not change appreciably.
Forces (and moments) that do not grow large over the interval At are
neglected (such as gravity and spring forces). Such forces are called
nonimpulsive; the large contact forces are called impulsive. It is the
impulsive forces and moments that produce the sudden changes in
velocities and angular velocities.
In Chapter 2 we introduced the coefficient of restitution as a measure
of the capacity for colliding bodies to rebound off each other. We shall
continue to use this parameter in this section, where now the relative
velocities of separation and approach are of the impacting points of B
and B . Thus rigid-body kinematics will be needed to relate these velocix
2
Page 356
ties to those of the mass centers of the bodies. We emphasize again that
the coefficient of restitution "e" is not the best of physical properties to
measure; it depends upon the materials, geometry, and initial velocities.
But as long as we take "e" with a grain of salt and remain aware of the
limiting values e = 0 (bodies stick together) and e = 1 (no loss of energy),
the definition of e does provide an approximate, much-needed equation
that allows us to solve many problems of impact. We now consider two
forms of the angular impulse and angular momentum equation that are
applicable at the beginning and end of impacts involving the plane mo­
tion of bodies that may be regarded as rigid except during the collision
phase of the motion.
If the body has a pivot O, we recall that
With O fixed in the inertial frame we have
Thus we may replace the C by an O in the angular impulse and momen­
tum equation (5.33) for such pivot cases. The resulting equation about
the axis of rotation is
(5.39)
This formula is of considerable value in impact problems because impul­
sive pivot reactions have no moment about O and thus do not appear in
the equation. Note that if O is C, then Equation (5.39) is the same as our
previous Equation (5.33) written about the mass center.
Another useful equation follows from
(5.40)
In scalar form, for rigid bodies in plane motion this equation is
in which P is an arbitrary point. If we state that P is now a fixed point O of
the inertial frame J, we may integrate this equation, getting
(5.41)
* W e e m p h a s i z e a g a i n t h a t t h e a n g u l a r m o m e n t u m H (or H ) is n o t e a u a l t o its rigid
body form
during t h e i m p a c t , b u t t h e s e s u b s t i t u t i o n s m a y b e m a d e a t t,
b e f o r e t h e collision a n d a t t a f t e r w a r d .
O
f
C
Page 357
QUESTION 8 . 8 Why is the right-hand side not the integral of the right
side of Equation (5.40) if O is moving?
Answer 5.B If 0 is n o t fixed in
then
t h e s e c o n d t e r m is n o t z e r o t h e n !
and
We shall now use these principles to solve a pair of example prob­
lems.
EXAMPLE 5 . 1 7
An arrow of length L traveling with speed strikes a smooth hard wall obliquely
as shown in thefigure.End A does not penetrate but slides downward along the
wall without friction or rebound. Find the angular velocity of the arrow after
impact.
Solution
The only impulsive force acting on the arrow during its impact with the wall is the
normal force N shown in the free-body diagram in Figure E5.17. We note that the
gravity force over the short time interval is nonimpulsive:
This is negligible in magnitude if
is very small, since mg does not grow large
and this is non-negligible since N grows large "impulsively" during the short
interval
with average value
.
In what follows, we shall delete the
subscript and simply denote the impulse of
The impulse and momentum equation is then:
Figure E5.17
or
(1)
and
(2)
where we note that momentum is conserved in the y-direction during impact.
The angular impulse and angular momentum principle yields:
(3)
"No rebound" means the x-component of
is zero; thus:
Page 358
Using
tion is:
,
we find that the x-component of this equa­
(4)
We have four equations in the unknowns
and
.
Solving, we
find
Note the obvious, that a head-on impact
stop with
and, from Equation
brings the arrow to a dead
also.
The next example, and the comments following it, constituted the
solution to an actual engineering problem.
EXAMPLE 5 . 1 8
A 770-ton steel nuclear reactor vessel is being transported down a 6.5 percent
grade using a specially designed suspended hauling platform together with
crawler transporters. (See Figure E5.18a.) Determine the maximum velocity at
which the reactor can be transported without tipping over if it should strike, and
Figure E5.18a
(Courtesy American Rigging Co.)
Solution
The reactor vessel will tip over if there is any kinetic energy left after it pivots
about the front edge at O (see Figure E5.18b) and the mass center reaches its
highest point B, directly above O. Thus we solve for the velocity that will cause C
to reach B; any higher velocity will cause overturning.
Page 359
C = position of mass
center at instant of
contact at O
B = position of mass
center at point of no return
Figure E5.18b
We begin the solution with some preliminary geometric and trigonometric
calculations based on the diagram. The 6.5 percent slope means
.* The turnover height h, is given by the distance OC:
Also
and
The initial height h of C, above the horizontal line through O, is
t
and thus the vertical distance through which C will move in reaching B is
* W e u s e f o u r significant digits in this e x a m p l e .
Page 360
(If we had adhered to three significant digits, the subtraction would have reduced
us to just one good digit.)
There now remain two separate main parts to the solution of this problem.
We first have to consider that mechanical energy is lost during the impact of
with the obstacle at O. Thus we are prevented from using the principle of work
and kinetic energy over the short period of impact. What does apply, however, is
conservation of angular momentum about O. This is because the impulsive forces
(in both the x and y directions!) causing the sudden changes in the mass center
velocity v and in the angular velocity to are acting at O, so that
Therefore
c
or
(1)
We shall use Equation (2.36) to express
this is the best formula for
for translation problems because
in that case. For
however, we have
a nonzero to. Thus we draw on the fact that since the vessel does not bounce at O,
we may consider O a fixed point of both and the inertial frame during and
following the short period of impact. This in turn means that
is simply
after impact. Therefore Equation (1) becomes
(2)
Now since
and
the left side of Equation (2) is simply
(3)
By the parallel-axis (transfer) theorem for moments of inertia we have
The vessel is essentially a thick shell; considering then that
(4)
we see that
(5)
By substituting Equations (3) and (5) into (2), we thus obtain
and the angular velocity after impact is then
(6)
Page 361
We are now ready to proceed to the second part of the solution.
Between the start of pivoting (immediately following impact) and the arrival
at point B, the system is easily analyzed by work and kinetic energy:
To obtain the least possible value of v for no overturning, we set
. The
only work done in this phase of the vessel's motion is by gravity, so that
ci
where is now an initial angular speed for this final stage of the problem and O
is still a pivot point for Thus
(7)
or
There are several important follow-on remarks to be made about the
preceding example. The first is that it can be shown (with a coefficient of
restitution analysis) that more energy is lost with no bounce at O than if
rebounding takes place. This energy loss for the
case just studied is
where and refer to the instants just before and after the impact.
Substituting (for the case of no bounce), we get
For = 65 ft and r = 10 ft, we obtain
Thus 32.2%, or nearly a third of the original mechanical energy, is
lost during impact if the lower front comer of sticks to, and pivots about,
point O. Of great importance here is the fact that we do not know how
much rebounding would actually occur in the physical situation and
hence how much energy would be lost. This means that 9.4 f t / s e c may
Page 362
not be a conservative engineering answer for the safe speed. If the plane is
flat
,
for example, it can be shown that:
1. The speed corresponding to pivoting as in this example is
11.9 ft/sec. (It has farther to pivot so it can be going faster prior to
impact.)
2. At this speed initially, and with a no-energy-lost rebound, the vessel
will easily overturn even though the striking comer backs up.
A conservative safe speed of the vessel in the inclined plane case can
be obtained by assuming that no energy is dissipated during the impact
and that all the vessel's initial kinetic energy goes into tilting it up about
O. This approach gives
In practice, the engineers in this case decided not to exceed 3 ft/sec, in
view of the importance of the work and the danger involved.
The Center of Percussion
We now turn our attention to a new topic. Besides the mass center C (but
of much lesser importance), there is another special point of interest
associated with a rigid body in plane motion, a point that differs from C
in that it depends not only on the mass distribution of but also on the
motion of the body. This point lies along the resultant of the ma vectors of
all the body's mass elements. The point is called the center of percussion,
and it has value in certain applications such as impact testing.
Before getting into the theory behind the center of percussion, we
first illustrate its existence and demonstrate its value by means of an
example. If a youngster hits a baseball with a stick and does not translate
his hands too much, we may model the situation as shown in Figure 5.13
and ask where the ball should hit the stick in order to eliminate the
"sting" (transverse reaction
of the stick onto the boy's hands). If the
boy hits the ball at just the right place, called the "sweet spot," he hits it a
long way while hardly feeling it and is said to have gotten "good wood"
on the ball. Assuming the stick to be rigid at the beginning (f = 0) and end
of the short impact interval, the two principles of impulse and
momentum are used as follows.
The impulse-momentum equation for the stick in the direction is
Figure 5.13
(5.42)
The angular impulse-angular momentum equation is
(5.43)
Page 363
Using
and
for "no sting," setting
by kinematics at
and multiplying Equation (5.42) by d gives
(5.44)
Dividing Equation (5.44) by (5.43) gives
Thus if the ball is struck two-thirds of the way from O to the end of the
stick, the transverse reaction
will be zero.
In this example the point at which the ball is struck
is the
center of percussion of the stick. To show this, at least for the case when
the bat is rigid, we first recall that in Chapter 2 we saw that for any point P
(moving or not, fixed to or not), we can always write
in which R is the position vector from point P to a generic differential
mass element. Therefore, since the integral of the vectors ( R X a dm) over
the body in fact represents the resultant moment about P of the ma
vectors over the mass of this integral vanishes for all points
on the
line along which the resultant of the ma vectors lies and hence
for these points.
Armed with the fact that
for the center of percussion, we
can now derive the general equation for the distance from a pivot O to the
center of percussion
in plane motion (Figure 5.14):
Figure 5.14
Therefore
(5.45)
and we see that, for the stick,
(as before)
Note from Equation (5.45) that the center of percussion is always farther
from the pivot than is the mass center. We make one final remark about
the center of percussion. If we treat the "a dm's" of as a collection of
vectors, its resultant may be expressed (for a rigid body in plane motion)
at the mass center (Figure 5.15), where
Figure 5.15
Page 3 6 4
In a manner identical to reducing a force and couple to its simplest
form, we may reduce this resultant of the ma vectors as shown in Figure
5.16, where the distance D is
.
We note that there is in fact a line
of points along the resultant of the ma vectors, making the location of a
single point
ambiguous. However, the concept of the center of percus­
sion is usually used in conjunction with problems in which the body has a
pivot O (as in the previous example). In these problems
is the welldefined single point at the intersection of lines and OC as in Figure 5.16.
Figure 5.16
EXAMPLE 5 . 1 9
Find the center of percussion
for a pendulum consisting of a rod plus disk,
each of which has equal mass m. (See Figure E5.19.)
Solution
The mass center of is located at a distance
from O given by
(Note that with equal masses C lies halfway between the mass centers of the rod
Figure E5.19
Thus
m and Equation (5.45) then gives us the location of :
Striking the pendulum at
have seen.
eliminates the horizontal pin reaction at O, as we
Page 3 6 5
PROBLEMS
•
Section 5.3
5.75 Drum has a radius of gyration of mass with re­
spect to a horizontal axis through 0 of 1 m and a mass of
800 kg. Body has a mass of 600 kg and a velocity of
20 m/s upward when in the position shown in Figure
P5.75. Find the velocity of
later.
5.77 The sinusoidal force P is applied to the string in
Figure P5.77 for a half-cycle. If the cylinder (initially at
rest) does not slip, find its angular velocity at the end of
the load application
5.78 A massless rope hanging over a frictionless pulley
of mass M supports two monkeys (one of mass , the
other of mass 2M). The system is released at rest at
as shown in Figure P5.78. During the following 2 sec,
monkey B travels down 15 ft of rope to obtain a massless
peanut at end P. Monkey A holds tightly to the rope dur­
ing these 2 sec. Find the displacement of A during the
time interval. Treat the pulley as a uniform cylinder of
radius R.
Figure P5.75
5.76 The hollow drum shown in Figure P5.76 weighs
161 lb and rotates about a fixed horizontal axis through O.
The diameter of the drum is 2.4 ft, and the radius of
gyration of the mass with respect to the axis through O is
0.8 ft. The angular speed changes from 30 rpm to
90 rpm during a certain time interval. Find the time
interval.
Figure P5.76
Weight = W
Figure P5.78
5.79 Force F in Figure P5.79 varies with time according
to
newtons, where f is measured in seconds. If
there is enough friction to prevent slipping of the cylinder
on the plane, find the velocity of C at: (a) t = 3 s; (b)
f = 10 s. The cylinder starts from rest at t = 0.
Figure P5.77
Figure P5.79
Page 366
5.80 The cylinder in Figure P5.80 has mass m = 3 slugs
and radius of gyration
ft with respect to C. There
is sufficient friction to prevent slipping on the plane. A
rope is wrapped around the inner radius, and a tension
T = 40 lb is applied parallel to the plane as shown. Use
impulse / momentum principles to find the velocity of C
after 3 sec if motion starts from rest.
5.83 Acting on the gear is a couple C with a timedependent strength given by C = (6 + 0.8f)N-m, where t
is measured in seconds. (See Figure P5.83.) If the system
is released from rest at t = 0, find the velocity of block
when (a) t = 3 s; (b) t = 10 s. The centroidal radius of
gyration of the gear is 0.25 m.
Figure P5.83
Figure P5.80
Solve the following problems by making use of the im­
pulse and momentum, and/or angular impulse and an­
gular momentum, methods.
Figure P 5 . 8 1
5.84 Problem 4.107
5.85 Problem 4.166(c)
5.81 The 161 -lb round body is rolling up the plane with
at the instant shown in Figure P5.81.
The radius of gyration of the mass of the body with re­
spect to the axis through the mass center C normal to the
page is 0.7 ft. Find the time required for the mass center to
reach its highest point.
5.82 The cylinder , turning at
,
is brought to
rest by applying the 50-1b force to the light brake arm as
shown in Figure P5.82. Friction in the bearings at O pro­
duces a constant resistance torque of 71b-ft,and the coefficient of friction at the contact point A between and
is
(a) Find the stopping time, and (b) find the
number of revolutions turned by during the braking.
5.86 Problem 4.179
5.87 Problem 4.102
5.88 Problem 4.101
5.89 Problem 4.87(b)
5.90 A pipe rolls (from rest) down an incline (Figure
P5.90). Using the equations of motion, find:
a.
b.
at time t
after C moves the distance
.
Then use work and energy to verify the answer to part (b)
and impulse and momentum to verify part (a). Finally,
give the minimum to prevent slipping.
5.91 A body weighing 805 lb with radius of gyration
0.8 ft about its axis (see Figure P5.91) is pinned at its
mass center. A clockwise couple of magnitude 1b-ft is
Figure P5.82
Figure P5.90
Figure P5.91
Page 367
applied to £ starting at t = 0. Find the angular velocity of
B when t = 3 seconds.
5.92 Given that the slot (for the cord) in the cylinder in
Figure P5.92 (mass 10 kg) has a negligible effect on I ,
find the velocity of the mass center C as a function of time,
if
c
* 5 . 9 6 A child pulls on an old wheel with a force of 5 lb by
means of a rope looped through the hub of the wheel.
(See Figure P5.96.) The friction coefficient between wheel
and ground is
Find I for the wheel, and use it to
determine the velocity of C 3 sec after starting from rest.
c
T h i n rim ( 3 lb)
1 lb (each of 8)
Figure P5.96
Figure P5.92
5.97 Two cables are wrapped around the hub of the
10-kg spool shown in Figure P5.97, which has a radius of
gyration of 500 mm with respect to its axis. A constant
40-N force is applied to the upper cable as shown. Find
the velocity of the mass center C 5 sec after starting from
rest if: (a)
(b)
Figure P5.93
The cart B is given an initial velocity v to the right
at t = 0. The rod B is pinned to B at its mass center G, as
shown in Figure P5.98(a). At t = 0, the mass center C of B
is heldfixedat the instant the cart starts off, then immedi­
ately released. At a later time (see Figure P5.98(b)), it is
observed that B has
at an instant when B has
turned 90° clockwise. If M = m, find the velocity of G at
that instant. Use
and an impulse and momentum
principle.
x
{
2
x
2
2
Smooth
2
Cable
600 m m
Figure P5.94
Figure P5.95
40 N
Cable
5.93 A uniform sphere (radius r, mass m) rolls on the
plane in Figure P5.93. If the sphere is released from rest at
t = 0 when x = L, find i(f).
5.94 The cord in Figure P5.94 is wrapped around the
cylinder, which is released from rest on the 60° incline
shown. Find the velocity of C as a function of time f.
* 5.95 The 50-lb body C in Figure P5.95 may be treated as
a solid cylinder of radius 2 ft. The coefficient of friction
between C and the plane is
and a force P = 10 lb
is applied vertically to a cord wrapped around the hub.
Find the velocity of the center C 10 sec after starting from
rest.
200 m m
Cart B ,
1
Mass M
Figure P5.97
Rod
S
2
Mass m
length /
Figure P5.98(a)
Figure P5.98(b)
Page 368
5.99 Two gymnasts at A and B, each of weight W, hold
onto the left side of a rope that passes over a cylindrical
pulley (weight W, radius R) to a counterweight C of
weight 2W. (See Figure P5.99.) Initially the gymnast A is
at depth d below B. He climbs the rope to join gymnast B.
Detenrune the displacement of the counterweight C at the
end of the climb.
5.100 DiskB1and the light shaft in Figure P5.100 rotate
freely at 40 rpm. Disk B (initially not turning) slides down
the shaft and strikes B ; after a brief period of slipping,
they move together. Find the average frictional moment
exerted on B by B if the slipping lasts for 3 sec.
2
5.101 Two disks are spinning in the directions shown in
Figure P5.101. The upper disk is lowered until it contacts
the bottom disk (around the rim). Find how long it takes
for the two disks to reach a common angular velocity, and
determine its value. Finally, determine the energy lost.
Show that if = I and
your solution pre­
dicts that 100 percent of the energy is lost (as it should).
Determine which of the three answers (time,
energy
loss) are the same if the two disks are instantaneously
locked together instead of slipping.
2
1
1
2
5.102 Figure P5.102(a) shows a rough guess at a skater's
mass distribution. Calculate the percentage increase in his
angular speed about the vertical if he draws in his arms as
shown in Figure P5.102(b). Assume that his arms are
wrapped around the 6-in. radius circle of his upper body.
A'JIO lb)
-Sphere: 0 0 8 m r = 4 in
;
Figure P5.99
*,(15
lb)
Stick: 0 0 9 m (each arm)
1.5
h
Cylinder: 0 2 8 m r = 6 in
;
Figure P5.100
Cylinder: 0 13m (each thigh); r = 3 in
Cylinder: 0 0 7 m (each); r = 1 5 in
Each foot a concentrated mass of 0 0 3 m
Coefficient of
friction = n
Figure P5.102(a)
Shoulders
Figure P5.101
Figure P5.102(b)
Page 3 6 9
• 5.103 A starving monkey of mass m spies a bunch of
delicious bananas of the same mass. (See Figure P5.103.)
He climbs at a varying speed relative to the (light) rope.
Determine whether the monkey reaches the bananas be­
fore they sail over the pulley of radius R if:
a. The pulley's mass is negligible
b. The pulley's mass is fm, where f > 0 and the
radius of gyration of the pulley with respect to
its axis is k.
If either answer is yes, give the relationship between d
and H in the figure for which overtaking the bananas is
possible.
• 5.104 A circular disk of mass M rotates without friction
about O. (See Figure P5.104.) A string passed over the
disk (and not slipping on it) carries a mass M at each end.
The system is released at rest as shown with the righthand mass carrying a washer of mass M. As the system
moves, the left-hand weight picks up a washer of mass M
at the same instant the right mass deposits its washer.
Find the velocity of the right-hand weight just after this
exchange of washers.
• 5.105 A bird of mass m, flying horizontally at speed v
perpendicular to a stick, lands on the stick and holds fast
to it. (See Figure P5.105.) The stick (mass M, length i) is
lying on a frozen pond. Find the angular velocity of the
bird and stick as they move together. (Answer in terms of
m, M, l, and v .) Assume that the bird lands on the end of
the stick.
B
0
Figure P5.103
Figure P5.105
* 5.106 A hemispherical block of mass M and radius a
whose surfaces are smooth rests with its plane face in
contact with a smooth horizontal table. A particle of mass
m is placed at the highest point of the block and is slightly
disturbed. Show that as long as the particle remains in
contact with the block, the radius to the particle makes an
angle 6 with the upward vertical where
Figure P5.104
Page 370
Solid cylinder
- Tlate
(m, l = 2r, r)
(mass M)
Light bar
(at impact)
Figure P5.109
(at impact)
Figure P5.107
5.107 The bar in Figure P5.10 7 is welded to the end of the
cylinder, which is traveling downward in translation. The
bar strikes the tables at speed v , and the cylinder begins
to rotate about the bar without rebound.
0
a. Find the angular velocity of the cylinder when
C is at its lowest point.
b. Find the percentage of energy lost during the
impact; that is
Figure P5.111
5.110 Work the preceding problem but assume that the
plate is initially free. If you work both problems, show
that the difference in the energies is I /(2M).
2
5.111 A bullet (see Figure P5.111) of mass m strikes a
square homogeneous block of mass m , where
The bullet is traveling with initial velocity v and be­
comes embedded in the block. After impact, the block is
observed to be pivoting about corner A. What is the maxi­
mum speed
of the bullet such that the block will not
tip all the way over?
1
5.108 A CARE package (Figure P5.108) consists of the
box plus contents described in Example 4.12. At impact
the crate has
ft/sec and is translating. If there
is no rebound, find the angular velocity of the box and the
velocity of its mass center G just after the impact.
5.109 The plate in Figure P5.109, supported by ball joints
at its top corners, is suddenly struck as shown with a force
that produces the impulse I normal to the plate. Find the
kinetic energy produced by the impact.
2
O
5.112 The cylinder B (radius 10 cm, length 40 cm) swings
down from a position of rest where
and strikes the
particle P of mass 5 kg. (See Figure P5.112.) The coeffi-
Taxpayer
Figure P5.108
Page 371
5.115 A homogeneous cube of side a and mass M slides
on a level, frictionless table with velocity v . See Figure
P5.115.) It strikes a small lip on the table at A of negligible
height. Find the velocity of the center of mass just after
impact if the coefficient of restitution is unity. (The centroidal moment of inertia of a cube about an axis parallel
to an edge is Ma /'6.)
0
21) k«
2
5.116 There is only one height H above a pool table at
which a cue ball may be struck by the stick without the
ball slipping for a while after the impact. (See Figure
P5.116.) Find this value of H, in terms of R, for which the
ball immediately rolls.
Figure P5.112
cient of restitution is e = 0.5, and at impact the particle
has
Find the angle through which Bwill
turn about its pivot O after impact.
1
5.113 An equilateral triangular plate of mass 2 slugs and
side 2 ft is released in the upper position from rest (see
Figure P5.113). It swings down and strikes the stationary
cylinder. The coefficient of restitution for the impact is
e = 1/2. Find elapsed time after impact until the cylinder
no longer slips on the plane.
Figure P5.115
5.114 A block slides to the right and strikes a small ob­
struction at a speed of 20 ft/sec. (See Figure P5.114.)
a. If the coefficient of restitution is zero, find the
energy loss caused by the impact.
b. What is the minimum striking velocity required
to overturn the block after collision?
Figure P5.116
5.117 Compute the error in in Example 5.18 that was
incurred in assuming the vessel to be a shell (so that
was m( /12 + mr /2). Use the weight, height, outer
radius, and density to compute the thickness of the vessel;
then calculate a more accurate and compare.
2
2
5.118 A uniform rod of length L is dropped and translates
downward at an angle with the vertical as shown in
Figure P5.118. If end A does not rebound after striking the
ground at speed v , find: (a) the energy lost during the
impact of A with the ground; (b) the speed at which
the other end B then hits the ground.
0
Figure P5.113
Figure P5.114
Figure P5.11B
Page 372
5.119 A uniform bar AB of length L and mass M is mov­
ing on a smooth horizontal plane with
and
. when end B strikes a peg P (see Figure P5.119).
If
/ L and the coefficient of restitution
find
the loss of kinetic energy.
5.120 The 80-lb solid block hits a smooth, rigid wall
(see Figure P5.120) and rebounds with a coefficient of
restitution of e = 0.2. Prior to impact the block had:
rad/secand
ft/sec. Find the
angular velocity of the block immediately following the
impact.
Figure P 5 . 1 2 3
5.121 The rod in Figure P5.121isfreely falling in a verti­
cal plane. At a certain instant it is horizontal with its ends
A and B having the velocities shown. If end A is suddenly
fixed, prove that the rod will start to rise around end A
provided thatv >,< 2v .
1
2
Figure P 5 . 1 2 4
Figure P 5 . 1 1 9
Figure P 5 . 1 2 5
Figure P5.120
Figure P5.121
5.122 In the preceding problem show that the energy loss
in instantaneously stopping point A is independent of v .
2
5.123 The bar in Figure P5.123 swings downward from
the dotted horizontal position and strikes mass m. The bar
has mass M and length The collision takes place with a
coefficient of restitution of zero. If the coefficient of fric­
tion between m and the plane is find the distance
moved by m before stopping. Treat m as a particle.
5.124 A wooden sphere weighing 0.644 lb swings down
from a position where the rod is horizontal, and it impacts
the block. (See Figure P5.124.) The coefficient of restitu­
tion is e = 0.6. The block weighs 3.22 lb and is initially at
rest. Find the position of the block when it comes perma­
nently to rest. (Assume that the sphere is removed from
the problem after impact and that the spring cannot re­
bound past its original unstretched position.)
5.125 A 4-lb sphere is released from rest in the position
shown in Figure P5.125, and two observations are then
made: (1) The sphere comes immediately to rest after the
impact; (2) the 5-lb block slides 3 ft before coming to rest.
Using these observations, find the coefficients of restitu­
tion (between sphere and block) and friction (between
block and floor).
Page 373
5.126 The rod-sphere rigid body in Figure P5.126 is re­
leased from rest in the horizontal position. It swings down
and at its lowest point strikes the box. Find how far the
box slides before coming to rest if the coefficient of resti­
tution is e = 0.5. The data are:
1.
2.
3.
4.
Rod: length = 1 m; mass = 3 kg
Sphere: radius = 0.2 m; mass = 10 kg
Block: b = 0.3 m; H = 0.35 m; mass = 5 kg
Coefficient of friction between block and plane
= 0.3
Assume the sphere hits the block just once.
5.128 Two toothed gear wheels, which may be treated as
uniform disks of radii a and b and masses M and m, re­
spectively, are rotating in the same plane. They are not
quite in contact and have angular velocities w and w
about fixed axes through their centers. Their axes are then
slightly moved so that the wheels engage. Prove that the
loss of energy is
1
2
5.129 Sphere B has mass m and radius r, and it rolls with
mass center velocity
on a horizontal plane. (See Fig­
ure P5.129.) It hits squarely an identical sphere B that is
at rest. The coefficient of friction between a sphere and
the plane is µ, and between spheres it is negligible. The
impact is nearly elastic
1
2
a. Find v and w of each sphere right after impact.
b. Find v of each sphere after it has started rolling
uniformly.
c. Discuss the special case when µ= 0.
C/
f
c
Figure P5.126
5.127 The assembly in Figure P5.127 is turning at co, = 2
rad/sec when the collar is released. All surfaces are
smooth, and the disk is fixed to the bar; the light vertical
shaft ends in bearings and turns freely. The data are:
The collar moves outward and impacts the disk without
rebound. Find: (a) the angular speed of the bar just before
and just after impact; (b) the percentage of energy lost
during impact. The radii of B and B are small compared
to their lengths. Treat B as a particle.
1
Figure P5.129
•5.130 The 30-kg bent bar in Figure P5.130 falls from the
dashed position onto the spinning cylinder, which was
initially turning at 3000 rad/
If the bar does not
bounce (coefficient of restitution is zero), find the stop­
ping time for the cylinder following the impact.
2
3
Figure
P5.127
Figure P5.130
Page 374
* 5.131 In Example 5.18 assume that the lower front strik­
ing comer of the vessel B rebounds back up the plane
with velocity
where e is the coefficient of restitution
Use the equations of impulse and momen­
tum in the x and y directions, and the equation of angular
impulse and angular momentum, to find the two masscenter velocity components and the angular velocity of B
after impact. Compare the results of
for
e = 1 with those at e = 0. Show that no energy is lost
when e = 1; that is, show
5.133 Prove statement 1 near the end of Example 5.18 for
the case when the plane is level
and e = 0. Hint:
Note carefully that the angle of the plane does not affect
Equation (6), so you only need to alter the Ah in Equation
(7) to obtain the new result.
5.134 Prove statement 2 near the end of Example 5.18.
Again the plane is to be level in this problem, but now
Hint: The
component of velocity is constant
after impact, since with
all external forces (mg and
N) are vertical. To find this velocity
use
and the
velocity of the striking comer Q just alter impact (co is the
same as with
in Problem 5.131 with
is
Then use
f
• 5.132 After the impact in the preceding problem, show
that the equations of motion of the vessel are
to show that C reaches the top with energy to spare.
where N is the normal reaction at the comer Q (see Figure
P5.132). Observe that until there is another impact, the
mass center has a constant x component of acceleration.
Use kinematics to prove that
Use Equations (2) and (4) to eliminate N from (3), thus
obtaining a single differential equation in joveming the
rotational motion of B, and note its complexity.
• 5.135 Show that the cylinder in Figure P5.135, following
release from rest, will reach the lower wall. Find the ve­
locity with which C will rebound up the plane following
impact, and determine the amount of energy lost. All data
are shown on the figure.
" 5.136 A sphere rolling with speed v on a horizontal sur­
face strikes an obstacle of height H. What is the largest
value that H can have if the sphere is able to make it over
the obstacle? Consider the coefficient of restitution to be
zero during the sphere's impact with the comer point O.
The answer will be a function of g, r, and v —in fact, H/r
may be solved for as a function of the single nondimensional parameter
(See Figure P5.136.)
c
c
Figure P5.132
Figure P5.135
Figure P5.136
Page 375
" 5.137 Using a density of wood of 673 kg/m , find the
mass center C of the baseball bat in Figure P5.137 and
then determine its moment of inertia with respect to the z
axis, perpendicular to the axis of symmetry of the bat. Use
the parallel-axis theorem to obtain
and find the bat's
center of percussion if it is swinging about a fixed point O.
3
0
Figure P5.138
Figure P5.137
* 5.138 The hammer in Figure P5.138 strikes the sphere
and imparts a horizontal impulse I to it. Determine the
initial angular velocity of the sphere.
COMPUTER PROBLEM
•
5.139 Repeat the preceding problem but suppose that the
sphere and rod are welded together to form one rigid
body.
Chapter 5
* 5.140 The system in Figure P5.140 is released from rest in
the given position. With the help of a computer, generate
data for a plot of the angle turned through by wheel
beforefirststopping, as a function of the mass ratio M/m.
Hint: First show using
that the equation govern­
ing 6 is
Figure P5.140
SUMMARY
•
Chapter5
For a rigid body in plane motion, the -kinetic energy, T, can be expressed
as
or as
With W standing for the net work over a time interval
of all the
external forces on the body, the principle of work and kinetic energy
states
or more compactly
Page 376
The work done by a force, F, is defined as
where F • v is called the power, or rate of work of force F, with v being
the velocity of the material point being acted upon instantaneously by
the force. This definition is necessary to accommodate the possibility that
the force moves around on the body. When the force always acts on the
same material point for which a position vector is r,
The work done by a couple of moment C is
which, with
is
For some special cases:
(a) F is constant:
and for weight mg, with y being elevation,
(b) Spring forces (on two bodies):
(c) Workless force:
if
as with a normal force acting on a sliding body, or if v = 0
at each instant as for the contact point of a rolling body,
(d) Constant couple:
A conservative force does work independent of path and can have
associated with it a potential which we define so that its change is the
negative of the work done by the force. Examples are weight, for which
and a linear spring,
The minus sign is used for the
convenience that follows when all forces acting on a body are conserva-
Page 377
rive, so that
or
which expresses the conservation of mechanical energy.
For a rigid body in plane motion, we found in Chapter 4 that
and similarly for a pivot. Concerning ourselves only with the case when
the products of inertia vanish,
As long as a body is behaving rigidly before and after an interval of time
of interest, the principle of angular impulse and angular momentum from
Chapter 2 gives
even though it may be that
as in the example of the spin­
ning ice skater. The above is, of course, paired with the principle of linear
impulse and momentum,
to effect solution of collision problems.
REVIEW QUESTIONS
•
Chapter 5
True or False?
These questions all refer to rigid bodies in plane motion.
1. If you raise a 2-lb object 3 ft from rest and stop it there, gravity has
done — 6 ft-lb of work and you have done + 6 ft-lb on the object.
2. The work of a constant couple
on body # i s always
where is the angle through which the body turns.
3. The work done by a linear spring is always
where
and are the amounts of initial and final stretch. (If negative, they
represent compression.)
4. There are actually three separate work and kinetic energy principles;
two of the equations add to give the third.
5. The principles of work and kinetic energy, and (linear and angular)
impulse and momentum, result from general integrations of the
equations of motion, and thus they are free of accelerations.
Page 3 7 8
6. Not all forces acting on a body have to do non-zero work on it in
general.
7. The friction force beneath a rolling wheel does work on it if the
surface of contact is curved and fixed.
8. The normal force exerted on a rolling wheel by a surface, whether
fixed or in motion, never does work on the wheel.
9. The principle
bodies.
10. The principle
bodies.
is valid for deformable
is valid for deformable
11. Any problem that can be solved by
can likewise be solved by
using "kinetic + potential energy = constant."
12. The formula
gives all the kinetic energy of the rigid body
in plane motion, assuming the body is not translating.
6
KINEMATICS OF A RIGID BODY
IN THREE-DIMENSIONAL
MOTION
6.1
6.2
Introduction
Relation B e t w e e n D e r i v a t i v e s / T h e A n g u l a r Velocity Vector
6.3
Properties of A n g u l a r V e l o c i t y
6.4
6.5
The Derivative Formula
Uniqueness of the Angular Velocity Vector
The Addition Theorem
Simple Angular Velocity
Summary of Properties of Angular Velocity
T h e A n g u l a r Acceleration V e c t o r
V e l o c i t y and Acceleration i n M o v i n g Frames of Reference
The Velocity Relationship in Moving Frames
The Acceleration Relationship in Moving Frames /
Coriolis Acceleration
6.6
6.7
T h e Earth as a M o v i n g Frame
V e l o c i t y and Acceleration Equations for T w o Points of t h e
Same Rigid Body
Does A n Instantaneous Axis of Rotation Exist in General?
Describing t h e Orientation of a Rigid Body
The Eulerian Angles
6.9
Rotation M a t r i c e s
SUMMARY
REVIEW Q U E S T I O N S
Page 379
Page 380
6.1
Introduction
In this chapter w e study the kinematics of a rigid body in general
motion — w e n o w do for general motion what w e did in Chapter 3
for plane motion. There w e found that at any instant the velocities
of different points are linked together because of the rigidity of the
body, and the connecting link is the angular velocity of the body. We
also found there that it is angular velocity that links the derivatives
of the same vector relative to two different frames of reference. We n o w
wish to remove the plane-motion restriction.
The principal difficulty encountered in the study of general motion
of a rigid body is that the angular velocity
does not always take the 6k
form of plane motion. The fact that can be changing in direction as the
body moves causes it to be difficult to visualize. A n efficient way to deal
with this abstract concept is to start with derivatives of the same vector in
different frames of reference (rigid bodies). Angular velocities will natu­
rally arise out of connecting these derivatives, and properties of relative
angular velocities amongst several bodies, almost self-evident for the
case of plane motion, will surface for the general case.
Application of the derivative/angular-velocity relationship to posi­
tion vectors, in two frames, of a point leads us to the velocities of the point
as observed in those frames, connected, in part, by the relative angular
velocity of the frames. Subsequent mathematical analysis leads to: a
relationship between accelerations of a point in two frames; a relation­
ship between velocities, relative to a frame, of two points fixed in the
same rigid body; and a relationship between accelerations, relative to a
frame, of two such body-fixed points.
We close this chapter with development of methods to describe the
orientation of a rigid body (relative, of course, to some reference frame).
This was easy to do in plane motion — an angle 8 is all that was required.
In the general case w e shall see the need for three angles, called Euler
angles in one popular scheme.
The reflective reader will notice that the sequence of coverages in this
chapter is almost precisely the reverse of its counterpart in Chapter 3,
which is after all just a special case. There w e were able to capitalize on an
ease of visualization not available to us here.
6.2
Relation Between Derivatives/The Angular Velocity Vector
In this section w e consider the relationship between the derivatives of a
vector taken in two different frames. In the process w e shall arrive at a
concise and useful definition of angular velocity. The reader is strongly
encouraged to persevere until this section and the next are fully under­
stood. Even though the angular velocity vector in three dimensions is a
difficult subject at first, it must be comprehended before w e can consider
the kinematics and kinetics of general rigid-body motion. The angular
velocity vector is the key to the subject. It will either make life easier for
Page 381
students of three-dimensional motion (if they work hard at understand­
ing it) or much more difficult (if they do not).
Let Q be an arbitrary vector. We may express Q in terms of its
components (Q , Q , Q ) associated with directions fixed in a frame by
x
y
2
(6.1)
in which the unit vectors (
) are parallel at all times to the respective
axes of a Cartesian coordinate system fixed in . N o w consider another
reference frame , in which w e wish to differentiate vector Q (see Fig­
ure 6.1). As an example, w e may wish to find the velocity of a point in
frame even though the point's location may be defined in (say by the
vector Q). In this case, part of the solution will require that w e be able to
differentiate Q in
even though it is expressed in terms of its compo­
nents in .
Figure 6.1
Vector Q and frames
and
Therefore it is n o w time to learn h o w to relate derivatives of a vector
taken in two different frames. We emphasize at the outset that these
vectors are completely arbitrary — they need not even be related to dy­
namics! Nor does the derivative have to be with respect to time, although
this is the independent variable of interest to us in dynamics and thus the
one w e shall use in the development to follow.
Letting
represent (see Equation 1.8) the derivative of Q with
respect to time in , w e have
(6.2)
Recognizing the first three terms on the right of Equation (6.2) as the
derivative of Q in , w e have
(6.3)
Clearly the last three (parenthesized) terms in Equation (6.3) represent a
vector depending upon both Q and the change of orientation of frame
with respect to . We n o w proceed to obtain a useful and compact
expression for this vector; in the process, the angular velocity vector will
arise.
Since
(6.4)
it follows that
(6.5)
Page 382
so that the three derivatives of the unit vectors in Equation (6.3) are each
perpendicular to the respective unit vectors themselves.*
Question 6.1 Will this be true for any vector of constant magnitude
(not necessarily a unit vector)?
This means that there are three vectors
and
for which
(6.6)
The cross products ensure that
and are each perpendicular to their
derivatives
(
and so on) and the magnitudes of
and give to
and
their correct magnitudes.
In terms of their components in
w e can write
and as
(6.7)
Substituting these component expressions into Equations (6.6) results in
(6.8)
and w e see that
Question 6.2
and
, at this point, remain arbitrary.
Why do they remain arbitrary?
Here w e are seeking to relate the components of the vectors
and
in the hope of finding a w a y to express the last three terms of Equation
(6.3). To this end w e note that, for all time t,
from the first of which, differentiation yields
(6.9)
* This assumes that the unit vectors are n o t constant in frame . If t w o of t h e m are con­
stant in , then all three are and the angular velocity vanishes, if only one is constant
in , w e have a simple special case to be considered later.
Answer 6.1 Sure, as w e h a v e seen in Section 1.6.
Answer 6.2
tions (6.6).
Since
, then a can b e anything and n o t affect the first of Equa­
z
Page 383
Substitution of the first two equations of (6.6) into (6.9) yields
(6.10)
Interchanging the dot and cross in each term (which leaves the scalar
triple product unchanged) results in
(6.11)
so that
(6.12)
Similarly from
and
(6.13)
w e respectively obtain (as the student should verify)
and
(6.14)
The only components not involved in Equations (6.12) and (6.14) are
and , which were arbitrary. If w e n o w select them as follows,
(6.15)
then all three vectors are identical, and we call the resulting common
vector
:
(6.16)
If w e n o w dot the three equations (6.8) respectively with
the three components of
:
and
w e get
(6.17)
Thus the vector
may be expressed as*
(6.18)
We call the vector
defined by Equation (6.18) the angular
velocity of frame w i t h respect to frame
or more briefly, the angu­
lar velocity of i n . It is clear that the angular velocity vector depends
intimately on the w a y frame is changing its orientation with respect to
. In the next section w e examine some special properties of this vector.
We shall see that
is unique, which means that w e lost no generality
w h e n w e let
, and
in our develop­
ment above of angular velocity.
* This is the definition of angular velocity set forth by the dynamicist T. R. Kane. See his
books Dynamics: Theory and Applications ( N e w York: McGraw-Hill, 1985), p. 16 and
Spacecraft Dynamics ( N e w York: McGraw-Hill, 1983), p. 49.
Page 384
6.3
Properties of Angular Velocity
T h e Derivative Formula
We n o w return to Equation (6.3). Substituting from Equations (6.6) for
and
and using Equation (6.16) to replace
and by
we
obtain
(6.19)
We call this the derivative formula, which may be expressed, using
Equation (6.1), as
(6.20)
Equation (6.20) will turn out to be of vital importance in this chapter
and, moreover, to be equally invaluable in our later study of the kinetics
of rigid bodies in general motion in Chapter 7. It permits us easily to
calculate the derivative of a vector in one frame if it is expressed in terms
of base vectors fixed in another; the only price w e have to pay is to add
the cross product
. Thus the first property of
is that it
allows us to relate (by Equation 6.20) the derivatives of any vector in two
different frames. We have already encountered this for the special case of
plane motion in Section 3.7, where Equation (3.44) may be seen to be the
plane motion counterpart of Equation (6.20).
Uniqueness of the A n g u l a r Velocity Vector
There remains the nagging question of whether there might be more than
one vector satisfying Equation (6.20); remember that w e arbitrarily se­
lected the components
and in the preceding section in order to
make
. We n o w proceed to s h o w that the angular
velocity vector is indeed unique. We do so by postulating that two vectors
and
both satisfy Equation (6.20) and then showing that they
are necessarily equal.* We have
(6.21)
(6.22)
so that, subtracting, w e get
(6.23)
Finally, since Q is arbitrary, the parenthesized expression of Equa­
tion (6.23) must vanish, and thus the angular velocity vector has been
shown to be unique as the two 's are one and the same.
Nothing has yet been said about dynamics in this section or in the
preceding one; thus it is clear that angular velocity is a far more general
* Let
be calculated with
and as described in Section 6.2, for example, and let
be computed with another triad of unit vectors fixed in . The question of unique­
ness is whether the resulting
's are the same.
Page 385
vector than one that is simply useful in describing rotational motions of
rigid bodies. We have seen that angular velocity is in fact the vector that
may be used in relating the derivatives in two frames of any arbitrary
vector. Furthermore, even though w e have used time as the independent
variable, these derivatives may be taken with respect to any scalar vari­
able. Finally, w e note from the defining equation (6.18) for
that
angular velocity is a vector relating two frames; thus it is meaningless to
talk about the angular velocity of a point.
N o w let us consider several additional properties of
that will
prove useful in what is to follow. First w e note for emphasis that if two
frames
and maintain a constant orientation (even if they are each in
motion in a third frame ), then
.* The proof is simply to observe
that n o unit vector fixed in direction in can change with time in
if
there is no change in orientation between
and . Thus from Equation
(6.18),
Next w e shall prove that the angular velocity of in is the negative
of the angular velocity of in If w e add Equation (6.20) to the equation
(6.24)
we obtain
(6.25)
Again, since Q is arbitrary, w e have the expected result:
(6.26)
T h e Addition T h e o r e m
We n o w prove the addition theorem, which states that
(6.27)
For the proof, w e know from the first property of
that
(6.28)
(6.29)
(6.30)
Adding Equations (6.29) and (6.30) yields
(6.31)
and subtracting Equation (6.31) from (6.28) gives
(6.32)
* Constant orientation m e a n s that
and m o v e as if they were rigidly attached except
for a possible translation of o n e with respect to the other.
Page 386
Therefore, again since Q is arbitrary,
(6.33)
and the theorem is proved. It may seem intuitively obvious to the reader
that Equation (6.33) is true, but in the next section w e s h o w that such a
relationship does not exist for angular acceleration!
The addition theorem is an extremely powerful result. With it w e are
able to build up the angular velocity, one pair of frames at a time, of a
body turning in complicated ways relative to a reference frame. This
theorem makes it possible for us to avoid using the definition (6.18),
which has served us well but in practice is normally supplanted by the
properties described in this section.
We recall for emphasis that if
and both move in and maintain
constant orientation with respect to each other, then
Thus, by
the addition theorem,
so that, as expected, the angular velocities in of two frames
and
maintaining constant orientation with each other are identical.
Conversely, if two frames and have equal angular velocities in
w e may s h o w that their relative orientation is constant. Using the addi­
tion theorem,
thus with
still fixed in direction in
then from Equations (6.6) and (6.16)
Therefore,
are constant in
so the orientation of in
is
constant.
Thus, for two frames
and , the descriptions "constant orienta­
tion" and "
" are completely equivalent. Note also that the addi­
tion theorem can be extended to any number of frames by repetition of the
following procedure, two frames at a time:
Simple A n g u l a r Velocity
Next w e show that w h e n there exists a unit vector whose time deriva­
tive in each of two frames
and vanishes (that is,
), then
(6.34)
in which
is the angle between a pair of directed line segments
and fixed respectively in
and
each perpendicular to . The
angle is measured in a reference plane containing projections of the
two lines intersecting at point P as s h o w n in Figure 6.2. The sign of
the angle is given by the right-hand rule: If the right thumb is placed in
Page 387
the positive direction at P, then the direction of positive is that of the
right hand's fingers w h e n they curl from into as shown.
Figure 6.2
Simple angular velocity.
The type of rotational motion given by Equation (6.34) is called
simple angular velocity. One case in which Equation (6.34) holds is that
of plane motion; note, however, that there are more general cases of
simple angular velocity in which the body may also have a translational
motion in
parallel to that would prevent the plane motion designa­
tion.
Figure 6.3 Unit vectors drawn in the
reference plane for simple angular velocity.
To prove Equation (6.34), w e make use of Figure 6.3. (The reference
plane is the plane of the paper, and the unit vector that is constant in
and in is k, perpendicular to the paper.) From this figure w e may write
(6.35)
Therefore, differentiating Equation (6.35), w e get
(6.36)
388
and Equation (6.18) yields, upon direct substitution of Equation (6.36),
(6.37)
which is the desired result.
One interesting final property of
is that its derivative is the same
whether computed in
or in
Using Equation (6.20) and letting Q be
itself, w e get
This result is not true for any other nonvanishing vector, unless it hap­
pens to be parallel to
S u m m a r y of Properties of A n g u l a r Velocity
The properties of
1.
that w e have examined are summarized here:
It is a unique vector that satisfies
which is called "The Derivative Formula."
2.
is synonymous with "the orientations of and
do not
change." And if and maintain constant orientation, their angular
velocities in any third frame are equal.
3.
4.
"The Addition Theorem":
which can be further extended to any number of frames.
5.
"Simple Angular Velocity": If k is constant in both
where
and
then
was defined earlier.
6.
In the following example w e use the addition theorem to write an angular
velocity, and then w e express it in three different frames.
EXAMPLE 6 . 1
Figure E6.1 a
Body in Figure E6.la rotates in frame about the vertical at constant angular
speed
; in disk rotates about its pinned axis at constant angular speed
relative to
(The directions of rotation are as shown. Determine the angular
velocity of in
Page 389
Solution
The coordinate axes shown are fixed in E. By the addition theorem,*
(1)
We see from this answer that expressing
in terms of its components in the
intermediate ("between"
and ) frame has yielded a neat, simple result. If
we had chosen instead to write
in terms of its components in then (see
Figure E6.1b) with
fixed in
Figure E6.1b
so that, substituting into Equation (1),
And if we had written
in Figure E6.1c) fixed in
in terms of its components along directions (
then with
we obtain, again using Equation (1),
Figure E6.1c
We see that expressing
in terms of its components in either or gives a
lengthier expression than in moreover, these expressions become even more
complicated if
or
vary with time.
Question 6.3
Why?
The reader should note, however, that even though each of the three above
representations of
appear to be different, they all yield the same vector.
* While the defining equation (6.18) is always available for directly computing the angu­
lar velocity, it is usually easier to build u p the
vector b y using the addition theorem.
Answer 6.3 Because then the angles (arguments of the sines and cosines) are not simply
or
but integrals of
or
with respect to time.
Page 390
In the next example w e illustrate the use of the "Derivative Formula"
(Equation 6.20) three times.
EXAMPLE 6 . 2
Two children are playing in the park on a seesaw mounted on a merry-go-round
as shown in Figure E6.2. The merry-go-round rotates about the ground-fixed
(frame ) vertical at
rad/sec, and at the instant shown the seesaw turns at
rad/sec relative to the merry-go-round The vector from the girl to the
boy is always
(
) being fixed in the seesaw board Find
and
at the given instant.
parallel to
board B
see-saw
board
radial line
fixed in
merry-goround
ground
Figure E6.2
Solution
is constant relative to so
To
find
where
we shall use the derivative formula:
Page 391
so
One way to find
is to use
where
so
Had we not desired to obtain
we might have used
where, by the addition theorem,
so that
as before.
We n o w present an extended practical example of the use of the
properties. In this example three separate bodies are in motion in a
reference frame, and their angular velocities are related by using the
simple angular velocity and addition theorem properties.
EXAMPLE 6 . 3
The Hooke's joint, or universal joint, is a device used to transmit power between
two shafts that are not collinear. Figure E6.3a shows a Hooke's joint in which the
shafts
and
are out of alignment by the angle
Each shaft is mounted in a bearing fixed to the reference frame . The shafts,
whose axes intersect at point A, are rigidly attached to the yokes and . A rigid
cross is the connecting body between the yokes. One leg of the cross (indicated
by the unit vector ) turns in bearings fixed in at D , and E , while the other leg
(unit vector ) turns in bearings fixed in
at D and E . The arms of cross are
identical; they form a right angle with each other, and each is perpendicular to its
respective shaft.
1
2
Figure E6.3a
1
2
Page 392
Figure E6.3b shows that
measures the angular position of
in
If
(considered to be the drive shaft) has angular velocity
and the
resulting angular velocity of is
find the ratio of
to
in terms
of and and plot
versus for
0,20,40,60, and 80°. Letting be
the rotation angle of , further investigate versus for the same five values.
Axis of
Question 6.4 In Figure E6.3b note that
has no
does this not represent a loss of generality?
Axis of
component. Why
Solution
Using the addition theorem, we may relate the angular velocities of the four rigid
bodies
(shaft
plus its yoke ),
(shaft
plus its yoke ), and
(1)
Figure E6.3b
Since
and
both have simple angular velocity in
we may write
(2)
We also know from Figure E6.3a that the cross has a simple angular
velocity in each of and . For example, the only motion that can have with
respect to is a rotation about DJEJ , the line fixed in both bodies. The same is true
for the motion of in . Thus
(3)
in which
and
are the unknown magnitudes of the respective vectors,
Next we must express all of
and
in terms of a common set of
unit vectors. Then we shall be able to obtain three scalar equations from (1) and
hence solve for
in terms of
. From Figure E6. 3b, three of the unit vectors are
obvious:
(4)
To obtain , we note that it is perpendicular to both
and . Crossing
then gives the assigned direction of
(note that
is opposite!);
is not generally a unit vector, however, so to get
we divide this vector
by its magnitude:
into
(5)
Substituting Equations (4) and (5) for the four unit vectors into Equations (2) and
(3), and then substituting the resulting angular velocity expressions into Equa­
tion (1), we get a vector equation that has the following three scalar component
equations:
coefficients:
(6)
Answer 6.4 The (xy) plane of the paper can be chosen to be the plane containing
without loss of generality.
and
Page 393
c
o
e
f
f
i
c
i
e
n
t
s
:
(
7
coefficients:
in which
gives
)
(8)
. Eliminating
between (6) and (8)
(9)
and substitution of (9) into (7) yields
so that
(10)
A plot of this expression (Figure E6.3c) shows the manner in which
changes over a quarter-turn of in space. Note that since cos is squared, the
curves reflect around the vertical line at
for
; between
180° and 360° we again have a mirror image, this time of the curves between 0°
and 180°. Note that for large misalignment angles
shaft
must turn very
rapidly at and near
; in fact, when
the bodies reach a configuration
in which they cannot turn at all. This is called gimbal lock. Note further that a
misalignment of as much as 10° results in an output speed variation ( ) over a
revolution of only about 3 percent.
(degrees)
Figure E6.3c
Next we examine the angles of rotation
and both have simple angular velocity in
so that (from Equation 10):
(of
) and
we have
(of
). Since
and
,
(11)
394
Integrating (11) gives
(12)
The constant of integration is zero if we define
when
. Then
may be plotted as a function of
for the same representative values of (see
Figure E6.3d). If
, then
and the curve is a 45° line. If there is
90
80
70
60
SO
40
30
20
10
0
0
10
20
30
40
^0
60
70
80
90
(degrees)
Figure E6.3d
misalignment
, then note from the curves that for
always turned more than .Then catches up at
180° the angle of
lags behind .
Question 6.5
, shaft
, and from
is
to
Explain why this is so by using Equation (12).
From 180° to 360°, the cycle repeats and everything returns to the same starting
position
at the same time.
Answer 6.5 For in the second quadrant, and with
we see from Equa­
tion (12) that tan is a more negative number than tan (unless
, in which case
). Therefore, and are angles between 90° and 180°, and
Page 395
PROBLEMS
•
Section 6.3
(See also the Project Problem 6.93 at the end of this chapter.)
6.1
Verify, in Example 1.1, that
is indeed
where
. (See Figure P6.1.)
,
6.2 The angular velocities of
and in a reference
frame are, respectively,
rad/sec and
rad/sec.
Find the angular velocity of in expressed in terms of
and (See Figure P6.2.)
A vector v is given as a function of time t by v =
, where
are unit vectors whose di­
rections are fixed in a frame . The angular velocity of M
in frame is
r a d / s . Find the deriv­
ative of v in frame that is,
(a) as a function of t; (b) at
f = 1 s; (c) at t = 2 s.
6.4
In the preceding problem, find
6.5
Note the three frames
and and the vector A
defined in Figure P6.5 expressed in terms of its com­
ponents in
Also,
and
Find
at f = / 4 sec.
6.6
Review Problem 1.155 in which the unit tangent,
normal, and binormal of a curve in space are defined.
Let be a frame moving relative to in such a way that
, and are always fixed in Use the definition (Equa­
tion (6.18)) of angular velocity to find the angular velocity
of in
Note that
6.7 The antenna
in Figure P6.7 is oriented with the
following three rotations:
1.
Azimuth, about y fixed in
at the rate
= 3t rad/sec
Elevation, about Z fixed in a first intermediate frame
, at the constant rate
A polarization rotation about the antenna axis x
(fixed in both the second intermediate frame
and in ) at
= 4t rad/sec
2
2.
4 in.
3.
1
2
If the structure is in the
position at t = 0, find
at the time f = / 2 sec. Use these unit vectors fixed in
direction in
parallel to y, parallel to z and
l7
Figure P6.1
6.8
Show that the output angle of the Hooke's joint
in Example 6.3 can be alternatively obtained by dotting
with
Figure P6.2
Azimuth
bearing
Figure P6.7
Figure P6.5
Page 396
6.9
A device for simulating conditions in space allows
rotations about orthogonal axes as shown in Figure P6.9.
Determine the angular velocity in frame of the capsule
containing the astronaut. Express the result in terms of
unit vectors
fixed
in the beam Note that the
rotation is about an axis fixed in and in but not in
this axis is parallel to y at t = 0, and
is a constant.
6.10 The outer cone in Figure P6.10 has the following
prescribed motion with respect to the fixed inner cone
I.
2.
3.
4.
The line AB (a base radius fixed in ) always lies in
some vertical plane parallel to XY.
Cone slides on ; that is, there is always a line of
contact between O and a point of the base circle of
Point A of revolves around the x axis in a vertical
circle at constant speed
Use the addition theorem to show that the angular veloc­
ity of in is given by
The vertices remain together.
6.11 In the preceding problem find
if the projection
of AB into the YZ plane through A is always aligned with
the radius. (See Figure P6.11.)
6.12 Find
in the preceding problem if the outer cone
rolls on the fixed inner cone. (See Figure P6.12.)
Figure P6.9
A popular method of stabilizing shipboard antennas is by
means of pendulous masses together with the gyroscopic
effect of spinning flywheels. In Figure P6.13 the ship
(frame ) pitches (about x), rolls (about y), and yaws
(about z) in the sea (frame ). Frame
just above a
Hooke's joint, is to form a stable platform on which the
antenna can then be easily positioned in azimuth (angle
A) and elevation (angle E). The frame remains level by a
"depitching" rotation P above the "derolling" rotation R.
The following three problems are based on this system.
The INMARSAT communications satellite system
required that shipboard antenna systems remain opera­
tional up to the following oscillatory limits:
Pitch:
± 1 0 ° in 6 sec
Roll:
± 3 0 ° in 8 sec
Yaw:
± 8 ° in 50 sec
Figure P6.11
Figure P6.10
Figure P6.12
Page 397
*6.13 Assume sine waves for each of these three motions
and assume yaw over roll over pitch — that is, the as­
sumed order of ship rotations is (1) pitch, from frame to
an intermediate frame ; and (2) roll, from to a second
able to extend (and retract) up to 5 in. in 30 sec. The wrist
has two motions: It is able to pivot up to 180 ° about y' in
10 sec and to rotate (about x') up to 350° in 4 sec. Axes
(x', y', z') are fixed in
Finally, the gripper is able to
open (and close) 3.5 in. in 3 sec, but is assumed here to be
a closed circle with a 2.5-inch diameter. Approximate di­
mensions are shown in the figure.
For this problem, assume that all the robot's motions
(except the gripper opening) are occurring simultaneously
about positive axes with their respective average speeds.
Find the angular velocity of the gripper relative to and
Ship
Sea
(not to scale)
Figure P6.13
Figure P6.16(a)
intermediate frame ; and (3) yaw, from
to frame
Write the angular velocity of the ship in the sea (earthfixed frame ), expressed in the ship-fixed axes (x, y, z).
Hint: For example,
will be
/180sin
/ 6 rad.
6.14 Write the angular velocity of
the axes {x, y, z).
in
expressed in
6.15 Write the angular velocity of in using the re­
sults of the preceding two problems together with the
addition theorem.
* 6.16 A robot manufactured by the Heath Company has
the mechanical arm shown in Figure P6.16(a) and ac­
companying photograph. Its shoulder extends from the
head
which can itself rotate 350° about z in 30 sec
relative to the reference frame The arm is able to travel
150° about axis y in 26 sec. (The axes (x, y, z) are fixed in
the head ) The part of the arm to the left of point £ is
* Asterisks identify the more difficult problems,
Figure P6.16(b)
Company.).
(Courtesy of the Heath
398
expressed in terms of unit vectors in
these two conditions hold:
I.
The shoulder rotation angle
at an instant when
2.
The wrist is pivoted 30°:
is — 60 °:
6.4
The Angular Acceleration Vector
In applying what w e have learned about angular velocity to the kine­
matics of rigid bodies, w e also need to understand its derivative. The
angular acceleration of frame relative to frame
is defined to be
(6.38)
(Note from the last property of
in the previous section that the
derivative could equally well be taken in but generally not in any other
frame.)
It is important to note that the addition theorem (Equation 6.27) does
not hold for angular acceleration. Watch:
(6.39)
We see that there is an extra term (the cross product of two angular
velocity vectors) that prevents the simple theorem w e have derived for
from working for
. This term is sometimes called a gyroscopic term;
note that it vanishes for plane motion, in which case w e do have an
addition theorem for the
(which are then of the form
).
In each of the two examples to follow, the reader should notice h o w
the various properties of
— simple angular velocity (Equation (6.34)),
the addition theorem (6.27), and the derivative formula (6.20) — are
used to great advantage.
EXAMPLE 6 . 4
Body in Figure E6.4 rotates in frame about the vertical at constant angular
speed
; in disk rotates about its pinned axis at constant angular speed
(The directions of rotation are as shown.) Determine the angular acceleration of
in
Axes (x, y, z) embedded in
Figure E6.4
Solution
As we saw in Example 6.1,
Page 399
Next, using Equation (6.20) and noting that
is expressed in terms of axes
embedded in
we "move the derivative" using the derivative formula and
obtain*
Note that the same result is obtained by using Equation (6.39) with frames
and replaced by
and
respectively.
Question 6.6
Why was
in the above example?
EXAMPLE 6 . 5
Determine the angular acceleration of the cross relative to frame in Exam­
ple 6.3, for the case
constant. Express the result in terms of unit vectors ,
and
fixed
in
Solution
Using the definition of angular acceleration [Equation (6.38)], the addition
theorem, and the derivative formula,
(1)
Let us first concentrate on the first term on the right side of Equation (1). In
Example 6.3 we had
(2)
where we are using the notation
and c = cos . We also know from Example 6.3 that
x
(3)
Substituting from Equation (3) into (2) and differentiating the result in
noting
is constant there) yields, after simplifying,
(and
(4)
* By "moving the derivative," w e m e a n shifting it from a frame in w h i c h it is i n c o n v e ­
nient to differentiate, to a frame in w h i c h w e desire to differentiate.
Answer 6.6
and
are constant scalars, and and are unit vectors fixed in direc­
tion in
Page 400
The second term in Equation (1) is
which upon simplification is
(5)
The last term in Equation (1) is:
(6)
where we have differentiated Equation (3).
The solution for
is therefore the sum of the vectors in Equations (4), (5),
and (6):
We note that up to this point in Chapter 6 w e have not mentioned
velocities or accelerations. As long as the angular velocities from one
body to another are all simple, w e can do a considerable amount of
angular velocity and angular acceleration computation merely by using
the definitions and properties of
and
•
PROBLEMS
Section 6.4
6.17 We know that one of the properties of the angular
velocity vector is that
. Show that this is not
a property of the angular acceleration vector
6.18 In Problem 6.6, find the angular acceleration of 8
in
6.19 The components of two angular velocity vectors are
shown in the following table as functions of time. The
orthogonal unit vectors
are fixed in direction in
frame
Find the angular acceleration of in
(a) as a
function of time; (b) at t = 0 sec; (c) at t = 0.5 sec. (See
Figure P6.19.)
Figure P6.19
(The unit vectors are
fixed in )
Find the restriction on frame
2,3).
2
4t
sin t
2t
cos t
6
7t
(i = 1,
6.21 The antenna in Figure P6.21 (see Problem 6.7) is
oriented with the following three rotations:
1.
6.20 Let the angular velocity and angular acceleration
vectors
and
be expressed in terms of their com­
ponents in a third frame :
for which
Azimuth, about y fixed in
at the rate
= 3f rad/sec
Elevation, about ZJ fixed in a first intermediate frame
, at the constant rate
= 1 rad/sec
2
2.
Page 401
6.23 In Example 6.2 find the angular acceleration of the
see-saw board in the ground if in addition to the given
data
rad/sec and
= 1.5 rad/sec .
4t rad/sec
3t rad/sec
2
2
6.24 See Figure P6.24. Axes x, y, and z are fixed in body
which rotates in about the z axis with angular veloc­
ity
The arm attached rigidly to
supports a bear­
ing about which turns with angular velocity
relative
to
Finally, body turns about the direction (which
lies along axes of symmetry of both and
with
relative to If
and
are all functions of time,
find the angular acceleration of in at an instant when
makes angles with x and z of 135 ° and 45 °, respectively.
1 rad/sec
Azirmuth
bearing
Figure P6.21
3.
A polarization rotation about the antenna axis x
(fixed in both the second intermediate frame
and
in ) at
rad/sec
2
If the structure is in the
position at t = 0, find
at the time
sec. Use unit vectors fixed in direction
in
Figure P6.24
6.22 In problem 6.9 find the angular acceleration of the
capsule C in Take
and
to be constants.
6.5
Pin P (fixed to )
moves in both
and
Velocity and Acceleration in Moving Frames of Reference
In certain practical situations a point is moving relative to two frames (or
bodies) of interest. For example, a pin P may be sliding in a slot of a body E
that is itself in motion in another frame (see Figure 6.4). In problems
such as these, w e are often interested in the relationship between the
velocities (and also the accelerations) of P in the two frames and
We
studied this problem in plane motion in Sections 3.7 and 3.8, and w e now
wish to expand the treatment to three dimensions.
T h e V e l o c i t y Relationship in M o v i n g Frames
Figure 6.4 Example of a point moving
relative to two frames.
We shall arbitrarily choose as a reference frame for the moving body
but w e emphasize that both are frames and both are bodies; as long as
they are considered rigid, the terms mean the same. Figure 6.5 shows the
general picture.
Figure 6.5 Point P moving with respect
to frames and
Page 402
Letting O and O' be fixed points of
tiate the connecting relation
and
respectively, w e differen­
(6.40)
in
and obtain*
(6.41)
Because O is fixed in
are the velocities in
the first two of the three vectors in Equation (6.41)
of P and O':
(6.42)
The last vector in Equation (6.42) presents a problem. It is not the
velocity of P in
because the point O' is not fixed in
nor is it the
velocity of P in because the derivative is not taken there. To overcome
this dilemma, w e shall rewrite the term by moving the derivative from
to using Equation (6.20):
(6.43)
(6.44)
We are n o w in a position to derive the equation relating the velocities
of two points of the same rigid body from the (therefore more general)
equation (6.44). Temporarily let P be a fixed point of then
and
(6.45)
or
(6.46)
which is the same as the plane motion equation (3.5), except that n o w the
r and v vectors may also have z components and the vector can have x
and y components in addition to z.
Returning to the general case in which P is not necessarily attached to
either or let us denote the point of (or extended) coincident with P
by
. Then Equation (6.44) becomes
(6.47)
In words, w e may restate Equation (6.47) as follows:
Velocity of
Pin
Velocity of
Pin
Velocity in of
the fixed point of
coincident with P
Equation (6.47) has the virtue of compactness. However, it is a less con-
* The superscripts in Equation (6.41) are now necessary to denote the frame in which
Page 403
venient form than is (6.44) for differentiation to produce a corresponding
relationship of accelerations.
Question 6.7
Why?
Another common alternative means of stating Equation (6.44) is to
view body as a moving frame — that is, as a body moving relative to
another frame
(See Figure 6.6.) We may then rewrite Equation (6.44)
as
(6.48)
in which
v = velocity of P in reference frame
P
= velocity of moving origin =
Figure 6.6
= velocity of P in moving frame =
= angular velocity of moving frame =
r = position vector of P in moving frame =
We n o w consider two examples involving the use of the velocity equation
for moving frames (Equation 6.44, 6.47, or 6.48).
EXAMPLE 6 . 6
Find the velocity in
Figure E6.6.)
of point A at the bottom of the disk in Example 6.1. (See
Solution
We select as the moving frame, and Equation (6.44) gives the following (where
is and A is P):
In this case
A in is given by
also
from the previous example. The velocity of
Axes (x, y, z) embedded in
Figure E6.6
Therefore
Answer 6.7
cause
If w e differentiate Equation ( 6 . 4 7 ) in
w e encounter the term
which is not recognizable as a standard kinematic quantity; this is be
denotes a succession of points of w h i c h are at each instant coincident with P.
Page 404
EXAMPLE
6.7
Crank in Figure E6.7 rotates about axis z through point O. Its other end, Q, is
attached to a ball and socket joint as shown. The ball forms the end of rod
which passes through a hole in the ceiling find the velocity of point P of the bar,
which is passing through the hole when
as shown.
Figure E6.7
Solution
We denote by H the point of
at the center of the hole; then:
(1)
in which P is the point of coincident with point H. Knowing that the motion
of H in must be along the axis of we have:
(2)
or
(3)
Collecting the coefficients of
and
respectively, we find
(4)
(5)
(6)
Page 405
These three equations (4 - 6) obviously cannot be solved for unique values of
all four unknowns
. However, an answer for
is obtainable by
subtracting Equation (5) from (4) and then adding
times Equation (6). The
result is
Substituting this result back into Equations (4-6) gives three equations in
and
whose coefficient matrix is singular (has a zero determinant). Thus they
cannot be solved for the angular velocity components. A more physical reason
for this is that the component of
along bar B cannot affect the answer for
because can turn freely in its socket about its axis without altering
. Mathe­
matically this is manifested by this "axial" component of
being parallel to
R and thus canceling out of Equation (2). These components are not needed
for a solution, however, because from Equation (1) we can obtain our desired
result:
QP
An added note is that although we cannot solve for the component of
along with the given information, we are able to calculate the component of
normal to It will be made up of values
and
that enforce the
relationship
That is,
(7)
This equation states that these
components form a vector normal to the line
QP; again, this vector is the only part of
that can affect
After adding primes to
in Equations (4-6), the solution of Equa­
tions (4-7), as the reader may verify, is
And we may now calculate the velocity of P from Equation (1) as before or from
Equation (2) as follows:
(as before)
T h e Acceleration Relationship in M o v i n g F r a m e s / C o r i o l i s Acceleration
After these examples w e are n o w ready to derive the corresponding
relationship between the accelerations of P in two frames and . Differ­
entiating Equation (6.44) gives
(6.49)
Page 406
or, again using Equation (6.20) (once in each of the last two terms), w e get
(6.50)
Rearranging the terms, w e have
(6.51)
(6.52)
The middle three terms on the right side of Equation (6.51) make up the
acceleration of the point
of (or extended) coincident with P. (The
proof is brief: If P is fixed to at point , then the other two terms vanish
since r becomes a constant vector in and what remains is necessarily
.) The term
(which is
, with both derivatives taken in ) is
clearly the acceleration of P in The last term,
, is called the
Coriolis acceleration of P. Note that due to the presence of the Coriolis
acceleration it is not true that the acceleration of P in frame
is its
acceleration in plus the acceleration of the point of with which it is
coincident (as was in fact the case with the velocity of P). This result is
interestingly analogous to the fact that the addition theorem for angular
velocity is not true for angular acceleration.
As w e did for the velocity equation, w e n o w restate Equation (6.52) in
words:
O'P
Acceleration
of P i n
Acceleration
of P i n
Acceleration in of
the fixed point of
coincident with P
Coriolis
acceleration
In the abbreviated notation of Equation (6.48) w e have
(6.53)
in which
a =
=
=
a =
P
rel
acceleration of P in reference frame
acceleration of moving origin =
angular acceleration of moving frame =
acceleration of P in moving frame =
and in which all other terms in Equation (6.53) are defined directly after
Equation (6.48).
EXAMPLE 6.8
Axes (x, y, z) embedded in
Figure E6.8
Compute the acceleration in
Figure E6.8).
of point A in Examples 6.1, 6.4, and 6.6 (see
407
Solution
We shall use Equation (6.51):
is again the moving frame; the reference
frame is and the moving point Pis A:
The various terms on the right side are calculated as follows:
1.
2.
Note that O' is fixed in
3.
Note that
does not change in direction in
in this example.
since
is a constant, and
4.
5.
Thus the answer for the acceleration of point A is
In Section 6.7, w e shall rework the above example using another
approach.
PROBLEMS
•
Section 6.5
6.25 In Example 6.2, the children are each 5 ft from
the fulcrum O. At a later time the girl slides a stone P
along the see-saw board toward the boy. The velocity
of t h e stone relative to the board is — ft/sec at
an instant when (a) the stone is 2 ft from the girl;
(b)
(c)
rad/sec; and (d)
rad/sec.
Find
and
at this instant.
6.26 In Problem 6.25, if at the given instant
rad/sec , and the stone's acceleration relative
to t h e board is 0 . 8 ft/sec , find
and
6.27 The large disk in Figure P6.27 rotates at 10 rad/
sec counterclockwise (looking down on its horizontal sur­
face). A small disk rolls radially outward along a radius
OD of
At the instant shown, the center C of is 4 ft
from the axis of rotation of
and this distance is increas­
ing at the constant rate of 2 ft/sec. Determine the velocity
and acceleration of point E, which is at the top of at the
given instant.
2
2
4
Figure P6.27
ft
1 ft
Page 408
6.28 Shaft in Figure P6.28 turns in the clevis at 2
rad/sec in the direction shown. The wheel simulta­
neously rotates at 3 rad/sec about its axis as indicated.
Both rates are constants. The bug is crawling outward on a
spoke at 0.2 in./sec with an acceleration of 0.1 in./sec
both relative to the spoke. At the instant shown, find:
(a) the angular velocity of the wheel; (b) the velocity of
the bug.
2 rad/sec
(5 in. long)
2
2 in.
Find the angular acceleration of the wheel and the
acceleration of the bug in Problem 6.28.
1 in.
3 rad/sec
6.30 The crane in Figure P6.30 turns about the ver­
tical at
rad/sec = constant, and simultane­
ously its boom is being elevated at the increasing rate
rad/sec. The (x, y, z) axes are fixed to the crane
zero. When the boom makes a 60° angle with the
horizontal, find: (a)
(b)
(c)
(d)
Figure P6.28
6.31 A platform translates past a turntable at 6 mph.
(See Figure P6.31.) People step onto the turntable and
walk straight toward the center where they exit onto
stairs. There is rolling contact between platform and turn­
table. Suppose the people walk at the constant rate of
approximately 3 mph relative to the turntable. If it is de­
sired that they do not experience more than 3 ft/sec of
lateral acceleration, find the required turntable radius.
2
6.32 The truck in Figure P6.32 moves to the left at a
constant speed of 7.07 ft/sec. At the instant shown, the
loading compartment has an angular speed
rad/
sec and an angular acceleration of
rad / sec . The
cylinder shown on the truck bed comes loose and rolls
4 ft
Figure P6.30
2
6 mph
Figure P6.31
Figure P6.32
Page 409
toward the ground at an angular velocity, relative to the
compartment, of 1 rad / sec at this instant; it is speeding
up at a rate of rad/sec , also relative to the compart­
ment. Find the velocity and acceleration of the center of
the cylinder relative to the ground. Use the rotating refer­
ence frame shown.
2
6.33 The bent bar in Figure P6.33 revolves about the
vertical at
rad/sec. The center C of the collar has
velocity and acceleration relative to of
in./sec and
—
respectively, where is in the direction of
the velocity of C in At the given instant, find the veloc­
ity and acceleration of C in the frame in which turns.
6.34 A centrifugal pump turns at 500 rpm in
and
the water particles have respective tangential velocity
and acceleration components relative to the blades of
120 ft/sec and 80 ft/sec outward when they reach
the outermost point of their blades. (See Figure P6.34.)
At the instant before exit determine the velocity and
acceleration vectors of the water particle at P with
respect to the ground
Figure P6.33
2
2 ft
1 ft
Each blade is
parabolic:
6.35 A man walks dizzily outward along a sine wave
fixed to a merry go-round that is 40 ft in diameter and
turns at 10 rpm. (See Figure P6.35.) If the man's speed
relative to the turntable is a constant 2 ft/sec, what is the
magnitude of his acceleration when he is 10 ft from the
center?
6.36 Disk rotates about its axis at the constant angular
speed
r a d / s . T h e round wire is rigidly affixed to
at points A and B as shown in Figure P6.36. A bug
crawls around the wire from A to B; its speed relative to
the wire (initially zero) is always increasing at the con­
stant rate of 0.001 m / s . Find the velocity and accelera­
tion of the bug, relative to the reference frame in which
the disk turns, when it arrives at B.
2
Figure P6.34
Wire
10 ft
Bug
10 rpm
Figure P6.35
Figure P6.36
Page 410
Figure P6.38
Figure P6.37
3 ft
6.37 A bird flies horizontally past a man's head in a
straight line at constant speed v toward the axis of a
turntable on which the man is standing. The axes xy have
origin at the man's feet; the z axis is vertical and y always
points toward the center of the turntable, which has
radius R and angular velocity
Derive x(t) and
y(t) of the bird in terms of R, v ,
, and the time t. (See
Figure P6.37.) Use Equation (6.47), integrate, then check
your results by inspection.
0
0
* 6.38 In Figure P6.38 the axes x and y, and the origin O,
are fixed on the deck of a ship. The ship has an angular
velocity relative to the earth of
where x and y are fore-and-aft and athwartships axes,
respectively; thus
is a rolling component and co a
pitching component of angular velocity. Point T is a target
fixed relative to earth, such as a geosynchronous satellite.
Find the angular rates and
in terms of
and
that are required to track point T.
P
• 6.39 In Problems 6.9 and 6.22, let
rad/sec and
rad/sec in the directions indicated. With the
dimensions given in Figure P6.39, find the maximum
value of
for which the acceleration magnitude of the
6.6
10 ft
2 ft
Figure P6.39
astronaut's head will not exceed 5g at the given instant.
Let
*6.40 In Problem 6.16 find the velocity of point P at the
tip of the gripper at the instant given.
*6.41 Prove that if two bodies are in rolling contact, the
shaded arcs in Figure P6.41, representing the loci of
former contact points, are equal in length. (Note from
Problem 3.105 that the converse is not true.)
Figure P6.41
The Earth as a Moving Frame
In this section, w e shall make use of Equation (6.53) to set up the differ­
ential equation governing the position of the mass center C of a body
that is in morion near the earth. This equation will allow us to measure
Page 411
North
the position of C with respect to a desired site 0 ' at latitude
which is
itself in motion as the earth turns on its axis from west to east. We assume
that for describing certain motions near the earth, a frame with origin at
the earth's mass center O is "sufficiently fixed" to be justifiably called
inertial. The frame moves as does the (assumed rigid) earth, except that
it does not share the earth's daily spin. Thus the site O' has an accelera­
tion
in
directed toward the earth's north-south polar axis.
We set up the moving frame
as shown as Figure 6.7. The frame
is the turning earth, and the (x, y, z) axes are embedded in it at O'
with x pointing east, y north, and z in the direction of local vertical.
The acceleration of the mass center C of a body
moving near the
earth and whose position is desired relative to O', is known from
Equation (6.53) to be
where r is the position vector of C in and v and a are the velocity and
acceleration vectors of C in Further,
and
We n o w
use the mass-center equation of motion from Chapter 2:
rel
rel
Figure 6.7
to obtain
where F represents all external forces on
written separately.
besides gravity, which is
Question 6.8 This is a good place to ask: Why is Equation (6.54)
restricted to bodies in motion near the earth?
We n o w proceed to compute the various terms in Equation (6.54).
First w e note that
Next w e compute the acceleration
of the site (O'). We utilize Equa­
tion (6.53) again, this time with the "moving point" being O' and the
origin in the moving frame being O:
Answer 6.8 It has been assumed that the strength of the gravitational field is constant,
but of course gravity decreases as C m o v e s farther and farther from the earth's surface.
Page 412
This time, however, note that v and a^, are zero; this follows from the
fact that O' has no motion relative to the moving frame (the earth). Thus
the only term surviving is:
rel
Therefore, from Equation (6.54),
Neglecting | r | with respect to | R | and expressing F in terms of its compo­
nents (F , F , F ), w e arrive at a set of differential equations governing the
motion of C:
x
y
z
(6.55)
We complete this brief section with an example illustrating the use of
Equations (6.55).
EXAMPLE 6.9
Due to the earth's rotation, the resultant force it exerts on a particle P at rest on its
surface is not quite directed toward its center of mass. Use Equations (6.55) to find
this deviation, assuming a spherical earth.
Solution
We have
and all zero, so that the equations of morion of P (which
moves!) in the inertial frame are:
Calling
we see (Figure E6.9a) that the earth must push on P at a
small angle (
) with the "geometric vertical" in order for P to remain at rest in
the "moving frame" rigidly attached to the surface of the planet. It is the angle
not the angle that defines "local vertical"; this is because is the angle that a
plumb bob string makes with the axis X in the equatorial plane.
Figure E6.9b shows that (with
)
Figure E6.9a
Page 413
We note that k, in comparison with mg, is quite small:
This means that we can use small angle approximations on the angle :
Figure E6.9b
so that the deviation of the local "plumb-line" vertical from the "geometric
vertical" is
Of course there is no deviation in direction at the poles (where
) or the
equator (where
). The maximum, at 45° latitude, is 0.0017 rad, or about
0.1°.
PROBLEMS
•
Section 6.6
6.42 If it were possible for a train to travel continuously
around the world on a meridional track as shown in Fig­
ure P6.42(a), one side of the track would wear out in time
due to the Coriolis acceleration. Explain which side will
wear out in each of the four numbered quadrants of the
circular path. Figure P6.42(b) shows how the train's
wheels rest on the track.
6.43 Explain in detail how the Coriolis acceleration is
related to the deflection of the air rushing toward a lowpressure area, thereby forming a hurricane. (See Fig­
ure P6.43.) Do the problem for each hemisphere!
Figure P6.42(b)
Equator
Figure P6.42(a)
Figure P6.43
Page 414
6.44 A car travels south along a meridian at a certain
time; its speed relative to the earth is 60 mph, increasing at
the rate of 2 ft/sec . (See Figure P6.44.) Find the acceler­
ation of the car in a frame having origin always at the
center of the earth and z axis along the polar axis of rota­
tion, but not rotating about the axis with the earth.
2
4v
* 6.46 A projectile is fired from a site at latitude with the
initial velocity components (at time t = 0)
. De­
termine the maximum height reached by the projectile,
neglecting air resistance and the
terms in Equations
(6.55).
* 6.47 Refer to Example 6.9. Show that if the component
equations of (6.54) are written in terms of axes (x', y', z')
(see Figure P6.47) instead of (x, y, z), they will have the
same form as do Equations (6.55) without their
terms, provided that
replaces g; replaces
place y and z.
4000 mi
and y' and z', respectively, re­
Figure P6.44
6.45 If in the preceding problem the car is traveling from
west to east at 45 °N latitude instead of along a meridian,
find the car's acceleration in the same frame. (See Fig­
ure P6.45.)
Figure P6.47
* 6.48 Using equations from the preceding problem, find
the location at which a falling rock will strike the earth if
dropped from rest on the z' axis from a height H. Neglect
air resistance and assume the rock strikes the earth when
z' = 0.
45°
Figure P6.45
6.7
Velocity and Acceleration Equations for Two Points of the
Same Rigid Body
We next apply the concepts of position, velocity, acceleration, angular
velocity, and angular acceleration (developed in Chapter 1 and Sec­
tions 6.2 to 6.5) to the kinematics of a rigid body
in general mo­
tion in a frame
The equation relating the velocities in
of two
points of a rigid body to its angular velocity
is a special case of
Equation (6.44). We have seen in the text following that equation
Page 415
Figure 6.8
Points P and O' of a rigid body
in general motion.
that if P joins O' as a fixed point of (see Figure 6.8), the relationship
between the velocities of these two points is given by the equation
(6.56)
From the text following Equation (6.52), w e also know the relationship
between the accelerations of these two points of :
(6.57)
We emphasize that Equations (6.56) and (6.57) follow from the general
equations (6.44) and (6.51) w h e n and if r is a constant in in that case,
Furthermore, the two points P and
O' in Equations (6.56) and (6.57) may be replaced by any pair of points
fixed to
since being fixed to is the only restriction on either of them.
Speaking loosely, with Equations (6.56, 6.57) w e are interested in two
points on one body, whereas with Equations (6.44,6.51) w e were studying
one point in motion relative to two bodies. We n o w consider examples
involving the use of the two rigid-body equations (6.56) and (6.57).
O'P
EXAMPLE 6.10
Rework Examples 6.6 and 6.8 by treating point A as a point of body
as a point moving with a known motion in
Solution
From Equation (6.56), and Figure E6.10, we have
Now recognizing that Q is also a point of
we obtain
(as we obtained in Example 6.6)
Axes (x, y, z) embedded in
Figure E6.10
Next we relate the accelerations of A and Q with Equation (6.57):
instead of
Page 416
The first term on the right side, using Q and O' as points of body
is
Thus
or
which we previously obtained in Example 6.8 by another approach.
In the next example w e will see that sometimes there is an indeter­
minate component of angular velocity.
EXAMPLE 6.11
Collars and
in Figure E6.11a are attached at Q and C to rod by ball and
socket joints. At the instant shown, C is moving away from the origin at speed
Find the velocity of Q at the same instant. Can the angular velocity of
2
2
Solution
The velocity of C is determined by (see Figure E6.11b):
2
14 c m
where
8 cm
so
Figure E6.11a
Now we can calculate the velocities of C and C using Equation (6.56):
1
2
(1)
We note that the component of the angular velocity
along the line C C
cannot be determined from the given information, because any value of it what­
soever will not affect Equation (1). However, dotting this equation with
t
6 cm
8 cm
Figure E6.11b
2
Page 417
so that
and
EXAMPLE 6 . 1 2
The cone in Figures E6.12a,b rolls on the floor i n such a way that the center Q
of the base of the cone travels on a horizontal circle at constant velocity
Let
denote an i n t e r m e d i a t e f r a m e ( " b e t w e e n " c o n e
and the ground
) in
which
and are fixed. T h e u n i t v e c t o r is a l w a y s d i r e c t e d a l o n g OQ, a n d
is normal to the p l a n e of a n d t h e c o n t a c t line, in a d i r e c t i o n parallel to
finally,
Find the angular velocity of
in
Figure E6.12a
Figure E6.12b
Solution
We shall denote
by simply
. Since v = v = 0, then their difference,
must also vanish. This requires
to be parallel at all times to the
line of contact of
with the ground. Thus
A
Next, using Equation (6.56),
Therefore
so that
Thus
O
Page 418
We wish to make some further remarks about the preceding exam­
ple. The addition theorem gives:
(1)
where
is given by
Substituting
w e find
and
into the addition theorem equation (1) above,
Note the check on the direction
since the only way
can move
relative to
is to rotate around OQ.
Consider finally two ways of depicting the components of
From Figure 6.9a, note that the vector sum of the two components of
is parallel to the contact line since
Figure 6.9a
Figure 6.9b illustrates the addition theorem. Note that the "direction
check" again results in
being along the contact line, this time be­
cause
Figure 6.9b
EXAMPLE 6.13
In Example 6.12, find the angular acceleration of
Solution
Differentiating
We had
and
Thus
,
in
Page 419
and we obtain
or
We see in the above example that acceleration equations need not be
used to compute if
is known.
The final example in this section illustrates the workings of a compli­
cated three-dimensional gear train.
EXAMPLE 6.14
Very large alterations in speed along a given direction may be obtained by using
the gear arrangement shown in the diagram. Gears
and all rotate about
the x axis in but has a more complicated motion:
1.
2.
3.
It rolls on the fixed (to ) gear currently contacting it at P.
It rolls on
(The contacting teeth are at A in Figure E6.14a.)
It rolls on (at B in the figure).
4.
It turns with respect to
Considering
about the line
to be the driven gear, find the ratio of
43 cm
30 cm
9 cm
(given)
Figure E6.14a
which is fixed in both
to
and
Page 420
Solution
The velocity of the contact points of
and
is, using body
(1)
in which we use just one subscript on when it is the angular velocity of a body
with respect to
Next we find another expression for v , this time by relating the velocity of
the tooth point of at B to that of the point of that contacts the reference frame
( is fixed to ) a t P .
B
(2)
To obtain r , we use Figure E6.14b and see that
PB
Therefore
36 cm
45 cm
(3)
Also, by the addition theorem,
9 cm
Figure E6.14b
(4)
in which we have used the fact that we know the directions (but not the magni­
tudes yet) of
and
Substituting Equations (3) and (4) into (2) and substituting the result into
Equation (1) gives us
or, simplifying,
(5)
18 cm
60
45 cm
27 cm
18 cm
To get another equation in these variables, we shall use the point T, which
belongs to both and and is shown in Figure E6.14c along with some essential
geometry. First, as a point of
we have
Ml
9 cm
Figure E6.14c
(6)
Page 421
Next, as a point of
(7)
The cross product in Equation (7) is one-half the cross product in Equation (2).
Question 6.9
Why is this so?
Therefore
(8)
Equating the right sides of Equations (6) and (8) gives
or
(9)
Substituting Equation (9) into (5) leads to
(10)
and
(11)
Substituting Equations (10) and (11) into (4) gives us
the last step of the problem. The result is
which we shall need in
(12)
We can now relate the velocities of the contacting points of bodies and
point A; first, using points A and £ of
we get
at
(13)
To get another expression for v , we relate the velocities of the two points A and P
on body
A
(14)
18 cm
60°
To obtain the position vector r , we use the geometry shown in Figure E6.14d.
The distance x, needed in forming r , is equal to (27 — d)/sin 60°:
PA
PA
60°
27cm
30 cm
60°
Figure E6.14d
(15)
9 cm
Answer 6.9 Because r
PT
= r /2.
PB
Page 422
Therefore
(16)
Substituting Equations (12) and (16) into (14), we get
(17)
Equating the two expressions for
in Equations (13) and (17) gives
(18)
It is seen that the angular speed of gear
opposite direction.
is 36 times that of
and in the
Question 6.10 Give an argument why
has to be turning in the
opposite direction from that of (Hint: Use the original figure and focus
your attention on points P and O of )
Does A n Instantaneous A x i s of Rotation Exist in General?
We recall that in Chapter 3 w e were able to s h o w that in plane motion a
rigid body
except w h e n its angular velocity vanishes, always has a
point of zero velocity (the instantaneous center
and hence a line of
points of zero velocity exists which w e may call the instantaneous axis of
rotation). We n o w show that in general (three-dimensional) motion, such
an axis does not always exist. We start with an arbitrary point P with
velocity v , and sketch its velocity along with the angular velocity vector
of in the reference frame
(See Figure 6.10(a).)
Note that in Figure 6.10a there is a plane defined by the vectors v
and
drawn through P, unless the two vectors are parallel. If they are,
then the motion of in is like that of a screwdriver — the body turns
around a line that translates along its axis. The general case ( v not
parallel to ) may also be reduced to a screwdriver motion as follows.
First w e replace v by its components parallel
and perpendicular
P
Plane
P
P
(a)
P
to
(See Figure 6.10b.) Next w e consider a plane parallel to
and separated from by the distance d as shown in Figure 6.10(b). Point
Q is the projection of P into the plane
and w e may write its velocity in
(b)
Figure 6.10
Answer 6.10 Each point of (extended) which lies on line
has zero velocity. Thus
the velocity of B is its distance from
times
and the same is true for point A. The
former is seen to be coming out of the paper, and the latter going into it, both about
Therefore
as determined by the direction of v , is in the negative x direction, oppo­
site to
A
Page 423
of P by Equation (6.46):
Note from Figure 6.10b that
this is the unit vector
directed from P to Q.
The vector triple product is equal to
so that
Therefore if d is chosen equal to
the line will be a "screwdriver
line" and the motion of will be as s h o w n in Figure 6.10(c).
Figure 6.10(c)
All points on have velocities along at the given instant, while those
off the line have the same velocity component parallel to in addition,
they rotate around it. This is the simplest reduction possible for the
motion of
and it is clear that unless
no points can have zero
velocity. Thus in three dimensions w e are no longer assured of having an
instantaneous axis as w e were in dealing with plane motion.
There are, however, special cases in three dimensions in which an
instantaneous axis exists; a good example is a cone rolling on a plane
(Figure 6.11). Note that in this case the entire line of contact is at each
Figure 6.11
Page 424
instant at rest on the plane. Since the angular velocity of the cone is
parallel to this line, all the contact points have
which w e have
shown must be true if the instantaneous axis is to exist in three dimen­
sions.
Section 6.7
PROBLEMS
6.49 A youngster finds an old wagon wheel
and
pushes it around with the center C moving at constant
speed in a horizontal circle. (See Figure P6.49(a).) If the
end O of the axle stays fixed while C returns to its starting
point in T seconds, compute the angular velocity vector of
the wheel
where
is the ground frame (Fig­
ure P6.49(b)). Give the result in terms of b, and T.
6 In.
3 in,
1 in.
4 in.
Figure P6.50
6.51 In the preceding problem,
find
and
6.52 Find the angular acceleration of the wagon wheel
of Problem 6.49.
(a)
6.53 Using the data of Problems 6.25 and 6.26, find the
velocity and acceleration in of the point Q at the end of
board
(beneath the little girl) having position vector
Do this using Equations (6.56, 6.57) of this
section, then check your answers using Equations (6.44,
6.51).
6.54 The angular velocity of a rigid body in motion in
frame is
rad/sec. If possible, locate (from
P) a point Q of with zero velocity when:
(b)
Figure P6.49
6.50 The bevel gears
and
in Figure P6.50 support
the turning shaft
whose angular velocity is given
by
rad/sec, in which
is parallel to the z
direction in both frames and (Gears and are seen
to be part of ) Find
and
In both cases explain why Q exists or not in light of the
discussion about points of zero velocity at the end of the
preceding section.
6.55 Disk
in Figure P6.55 spins relative to the bent
shaft at constant angular speed
rad/sec; rotates in
the reference frame
at the constant rate
rad/sec.
(The directions are indicated in the figures.) Using the
rigid-body equations (6.56 and 6.57),
find
and
for point Q on the periphery of the disk. Express the result
in terms of components along the (x, y , z) axes, which are
fixed in
Page 425
6.59 The bevel gear in Figure P6.59 is fixed to a refer­
ence frame in which the mating gear moves. The axis
OC of turns about the z axis at the constant rate
= 0.2 rad/sec, and the angle OQC is 30°. Find the angu­
lar velocity of in
Ball joint
Figure P6.59
Figure P6.S5
In the preceding problem, find the angular acceler­
ation of in for the same defined motion.
6.56 Rework the preceding problem, this time using the
moving-frame concept of Section 6.5. Let bar be the
frame in motion with respect to and let Q be the point
moving relative to both and
6.57 A wheel of radius r turns on an axle that rotates
with angular velocity
about a vertical axis (z) fixed
relative to ground (Figure P6.57). If the wheel rolls on the
horizontal plane and
is constant, find:
6.61 A differential friction gear can be made with either
bevel gears, as shown at the top of Figure P6.61, or fric­
tion disks, as shown at the bottom. In each case, body
rolls on and and may rum without resistance on the
crank arm Find
and the velocities of points
A and B of
a. The angular velocity and angular acceleration of
the wheel relative to the ground
b. The acceleration, relative to the ground, of the
point on the wheel in contact with the horizon­
tal plane.
Wheel
Axle
Figure P6.57
6.58 Rework Example 6.14, but this time suppose that
the radius of gear is 44 cm instead of 30 cm (and that
its center is at the same point of ). Explain why gears
and are now moving in the same direction.
Tight
fit
Figure P6.61
Page 426
6.62 The uniform, solid, right circular cone in Fig­
ure P6.62 rolls on the horizontal plane Let represent a
frame in which the vertical axis z and the cone's axis
are fixed. (Hence has a simple angular velocity in
about the vertical.) Show that the cone can roll so that
are constants and
Figure P6.65
Figure P6.62
17.5 m
6.63 The center C of the bevel gear in Figure P6.63
rotates in a horizontal circle at a constant speed of 40
m m / s (clockwise when viewed from above). The mating
gear is fixed to the reference frame
shaft
rigidly
attached to is connected to through a ball and socket
joint at O. Find the angular velocity vector of in
7.5 m
5m
Figure P6.69
6.66 In Example 6.14 find the radius of gear for which
it will remain stationary in as
and turn.
30°
6.67 In Example 6.14 label the radius of gear as H and
call the radius of the 28-cm gear R. Show that the rela­
tionship between
and
is given by
50 mm
( T h e c e n t e r of
is fixed i n
)
6.68 If the speed of point C is constant in Example 6.11,
find the acceleration of Q at the instant given.
Figure P6.63
2
6.69 Collars and in Figure P6.69 are attached at Q
and C to rod by ball and socket joints. Point C has a
motion along the x axis given by x = — 0.012t m. Find
the velocity of C as a function of time.
2
2
3
6.64
Find
2
for the gear in Problem 6.63.
1
*6.65
Plate in Figure P6.65 has the following motion:
1. Corner A moves on the x axis.
2. Comer B moves on the y axis with constant veloc­
ity in./sec.
3. Some point of the top edge of (point Q at the
instant shown) is always in contact with the z axis.
Find the angular velocity vector of the plate when x
= 3 in.
A
*6.70 Cone rolls on cone so that its axis of symmetry
(x) moves in a horizontal plane through O, turning about z
at rate
rad/sec. (See Figure P6.70.) Cone is rotating
about (—z) at
rad/sec. Find, with respect to the frame
in which turns, the angular velocity and angular
acceleration of
and the acceleration of point A. (Axes
(x, y, z) are fixed in frame
which turns so that x is
always along the axis of symmetry of
and z is always
vertical. Also,
and
are constants.)
Page 427
To drive shaft
Pin
60°
120°
Pin
(a)
Figure P6.70
30 ft
20 m p h
6 ft
(b)
Figure P6.73
Figure P6.71
6.71 The two shafts
and
are fixed to bevel gears
and as shown in Figure P6.71. (a) Prove that if the ve­
locities of each pair of contacting points are to match
along the line of contact of the gears, then points A , B, and
C must coincide, (b) Let A , B, and C coincide and find the
ratio of to
6.72 (a) In the preceding problem show that if
, the result is still true about A , B, and C coinciding
(b) Find
for this case.
6.73 Depicted in Figure P6.73(a) are the main features of
an automobile differential. The left and right axles, and
are keyed to the bevel gears and . Gear is fixed to
the case and the combination is free to turn in bearings
around line Gear meshes with gear attached to the
car's drive shaft. As the casing turns about the common
axis of and its pins bear against the other two bevel
gears within which are and . (Observe that these
two gears do not turn about their axes at all on a straight
road.)
Suppose the car makes a 30-ft turn at a speed of 20
mph (Figure P6.73(b)). If the tire radius is 14 in., find the
angular velocities of and and use them to compute the
angular velocity of
In the process note that the differ­
ential allows the driven wheels to turn at different angu­
lar speeds.
It is assumed in the following three problems that not all
points of body have zero acceleration.
6.74 Show that for a rigid body in general motion,
there is a point Q of zero acceleration if
and
is not parallel to Hint: Let P be an arbitrary point, and
let
and
noting that there is no loss in generality in these assump­
tions. Set
set
and solve for x, y, and z.
6.75
(a) Following up the previous problem, show that if
and
at a given instant, then at this instant
there is a point Q of zero acceleration if and only if the
accelerations of all points of B are perpendicular to (b)
Investigate the case
and
6.76 Here is another follow-up on Problem 6.74: Show
that at any instant when the two vectors
and
are
parallel, there is a point of with zero acceleration if and
only if the accelerations of all its points are perpendicular
to and
* 6.77 Find the angular acceleration of the plate of Prob­
lem 6.65 at the same instant of time.
Page 428
6.8
Describing the Orientation of a Rigid Body
In the case of plane motion of a rigid body
velocity
if w e know the angular
as a function of time, w e may clearly integrate to find the orientation of
at any time t:
Thus in plane motion w e may completely specify the position of
by
giving the xy coordinates of a point (usually the mass center C is chosen to
be the point) and the orientation angle
T h e Eulerian A n g l e s
A major difference between planar and general motion is that the angles
which yield a body's orientation in space in three-dimensional motion
are not the integrals by simple quadratures of the angular velocity com­
ponents of the body. In fact, finding (in general) the orientation of a body
in closed form, given
and the orientation of
at t = 0, is an
unsolved fundamental problem in rigid-body kinematics. We n o w intro­
duce the Eulerian angles in order to s h o w the difficulty of detennining a
body's orientation in space w h e n the motion is nonplanar.
We begin with the body
oriented so that the body-fixed axes
{x, y, z) initially coincide respectively with axes (X, Y, Z) embedded in
the reference frame
Let
and
be sets of unit vectors re­
spectively parallel to (x, y, z) and (X, Y, Z). Three successive rotations
about specific axes will n o w be described that will orient in
(See
Figure 6.12(a))
In Figure 6.12(b) the first rotation is through the angle about the Z
axis. Let the n e w positions of (x, y, z) after this first rotation be denoted by
(x , y , z ) as shown; these positions are embedded in an intermediate
frame
. Note that axes Z and z are identical and that
has the simple
angular velocity
in Note also from Figure 6.12 that
and
are unit vectors that are respectively and always parallel to x , y , z . Next
(Figure 6.12(c)) a rotation through the angle about axis J/J moves the
body axes into the coordinate directions (x , y ,z )
of a second interme­
diate frame
having unit vectors
. A final rotation, this
time of amount about z (Figure 6.12(d)), turns the body axes into their
final positions in
indicated by (x, y,z).
It is clear that w e may use the addition theorem to express the angular
velocity of in as follows:
1
1
1
1
1
2
2
1
1
2
2
(6.58)
Page 429
(a) Showing all three rotations
(b) First Eulerian angle
rotation
,
about Z
Figure 6.12
(c) Second Eulerian angle
rotation
about y,
(d) Third Eulerian angle
rotation
,
about z
2
Eulerian angles.
in which w e again remark that
is a unit vector in the jth coordinate
direction of frame
If the preceding expression for
is to be functionally useful, its
three terms should all be expressed in the same frame — that is, in terms
of components associated with the directions of base vectors all fixed in
direction in a common frame. N o w in practice, as w e shall see, they are
sometimes expressed by components associated with body-fixed direc­
tions, at other times with space-fixed directions, and at still other
times with directions fixed in intermediate frames such as
or
. For
example, let us write
in body components. This means, looking at
Equation (6.58), that w e must express
and
in terms of
and .
Page 430
We see from Figure 6.12 that
(6.59)
and that
or
(6.60)
where
and so forth. Substituting these expressions
into Equation (6.58) then gives
in body components:
(6.61)
Alternatively, w e may express
in terms of its components
in frame by writing and
in terms of
Again referring to
the figure, w e see that
(6.62)
and
or
(6.63)
so that, substituting into Equation (6.58), w e have
written in
(6.64)
Finally, w e notice that expressing
in the intermediate frames is
easier still. To write it in terms of its components in
, w e note that
(6.65)
and
(6.66)
so that from Equation (6.58) w e get
(6.67)
All these expressions for
appear different, but of course they all
represent the same vector written in different frames. The components
may vary from frame to frame, but the vector is the same. In the first
exercise at the end of this section, the reader will be asked to write
in
yet another frame,
The angles
are known as the Eulerian angles. They represent
one way of orientating a rigid body in space. Unfortunately the Eulerian
angles
do not carry the same symbol from one book to the next;
worse still, the order and even the directions of the rotations vary from
writer to writer. Obviously, then, it is important to choose a set to work
with and then be consistent.
Page 431
reference frame
(axes X, Y, Z)
A physical feel for the Eulerian angles may be gained by considering
a gyroscope spinning in a Cardan suspension as s h o w n in Figure 6.13.
In this system the Eulerian angles
may be used as follows to
pinpoint the orientation of the rotor B in space:
1. First Rotation: With respect to frame
w e rotate the plane of the
outer gimbal G„ (frame in the earlier theory) about axis Z through angle
. Axis X is turned into x and axis Y into y ; frames (bodies) G and are
not s h o w n yet, but they move rigidly with G in this first rotation. (See
Figure 6.14.)
t
1
i
O
2. Second Rotation: Next w e turn the plane of the inner gimbal
about axis y , through angle thereby tilting G with respect to G .
Axis z is thereby rotated into z and axis x into x . Body (not s h o w n in
Figure 6.15) goes along for the ride.
2
G = outer gimbal (axes x,, y z,)
G, = inner gimbal (axes x y z )
= rotor (axes x, y, z)
0
u
2)
2)
2
1
i
2
1
O
2
3. Third Rotation: The third and last rotation turns the rotor about
axis z through angle . (See Figure 6.16.) This allows to spin relative to
G . Axis x is turned into x and axis y into y.
Figure 6.13
2
i
2
2
Through these three Eulerian angle rotations, the body
can be
positioned in any desired orientation in space
. We are n o w able to see
the difficulty of solving for the orientation of a rigid body in general
motion. If w e write
then Equation (6.61) is
equivalent to
(6.68)
Solving for the rates of change of the Eulerian angles gives
(6.69)
Figure 6.14
First rotation.
Figure 6.15
Second rotation.
Figure 6.16
Third rotation.
Page 432
We see from these equations that even if w e knew the
components as
functions of time in closed form,* it would still be a formidable task to
integrate Equations (6.69) analytically to obtain the Eulerian angles and
thus to know the body's orientation in space. This is usually not even
possible, and resort is made to computers that can numerically carry out
integrations with a step-by-step scheme such as Runge-Kutta.
Incidentally, the sin denominators in Equations (6.69) present seri­
ous obstacles in the dynamics of space vehicles; whenever is zero or a
multiple of
the equations develop a singularity. Sophisticated pro­
gramming or, in some cases, completely different mathematical schemes
for orienting the body are required to overcome such difficulties.
We mention that use of the preceding set of Eulerian angles as de­
fined requires that w e maintain the order of rotation. To illustrate the
importance of rotation order, w e remark that if this book is rotated
through two
rotations about the space axes Y and Z, in opposite
orders as suggested by Figure 6.17, it will end up in a different position.
We should point out, however, that there are ways of setting up the axes
and angles which make a body's final orientation independent of order.
For instance, just as in our Eulerian angle development, let Z be fixed in
let z be fixed in
and let y be always perpendicular to both Z and z. But
n o w restrict Z and z to be nonparallel. (Let z, the axis of
lie along X
initially, for example.) In this case, the angles
as defined earlier
may be performed in any of the six possible orders and the body's
resulting orientation in will be the same each time! †
First Y
BUT! . . . first Z
Figure 6.17
..then Z . . .
.
then
Y...
gives this view
. gives this!
Finite rotations.
* The equations governing these three components of angular velocity are the Euler
equations of rigid-body kinetics. They themselves are also nonlinear and unsolvable in
closed form in general three-dimensional motion except for a few special cases. We shall
be studying these equations in Chapter 7.
†Also, subsequent reorientations will likewise be order-independent; see "Successive
Finite Rotations," by T. R. Kane and D. A. Levinson, Journal of Applied Mechanics, Dec.
1978, Vol. 45, pp. 9 4 5 - 9 4 6 .
Page 433
A n alternative set of three rotations has become popular in the litera­
ture in recent years:
1.
Rotate through
2.
Then rotate through
about y .
3.
Then rotate through
about z .
about X.
1
2
This sequence results in angular velocity components in
the body-axis system (x, y, z) — of
— that is, in
(6.70)
Although (6.70) is not made up of classic Eulerian angles, the result is an
equally valid set of relations between the
components and the
angles
of rotation. The order and axes of rotations of the
in this
system are quite simple to remember.
There are alternatives to using the Eulerian angles or the angles in
Equation (6.70) to orient a body in space. One such alternative is to use
the quaternions of Hamilton, which are free of the singular points caused
by zero denominators at certain values of the Eulerian angles. (For this
reason, quaternions were used in the Skylab Orbital Assembly's attitude
control system.) Another approach to the orientation of a body in space is
to determine the direction cosines of a unit vector, fixed in direction in
space
with respect to a set of axes fixed in the body
Let this unit
vector (call it ) have direction cosines (p, q, r) with respect to axes (x, y, z)
fixed in Further, let
be unit vectors, always respectively parallel
to (x, y, z). Then
and, differentiating in
w e obtain
This vector equation has the following scalar component equations:
(6.71)
If the
components are either prescribed or else found from kinetics
equations (to be studied in Chapter 7), then Equations (6.71) may be
solved for the direction cosines, thereby orienting in Equations (6.71)
are known as the Poisson equations. They are alternatives to equations
such as (6.68) and (6.70).
Page 434
PROBLEMS
•
Section 6.8
6.78 Using the Eulerian angles
discussed in this
section, express
in (terms of its components in) the
frame
6.83 Chasle's theorem states: The most general dis­
placement of a rigid body is equivalent to the translation
of some point A followed by a rotation about an axis
through A. Show that this result follows immediately
from the previous problem.
6.79 Show that the magnitudes of
are all the same
as expressed (a) in in Equation (6.61); (b) in in Equa­
tion (6.64); and (c) in
in Equation (6.67).
6.80
6.84 The circular drum of radius R in Figure P6.84 is
pivoted to a support at O, where O is a distance R/2 from
the center C of the drum. A weight
(particle) hangs
from a cord wrapped around the drum. The drum is
slowly rotated
rad clockwise about O. Find the dis­
placement of
Hint: Use the result of the preceding
problem, with C being the point A and with a
rotation following the translation. Add the displacements
of during each part.
Derive Equations (6.70).
6.81 Write
in by using the successive rotations ,
, and that resulted in Equation (6.70) when expressed
in
6.82 Euler's theorem for finite rotations is stated as fol­
lows: The most general rotation of a rigid body with
respect to a point A is equivalent to a rotation about some
axis through A. Prove the theorem. Hint: Let A be consid­
ered fixed in the reference frame in which moves. (See
Figure P6.82.) Let point P be at P prior to the rotation and
at P afterward; assume the same for point Q (Q before,
Q after). Bisect angle P,AP with a plane normal to the
plane of the angle. Do the same for Q , A Q and consider
tho intersection of the two planes.
6.85 In Figure P6.85, the sphere
rolls on the plane,
and its angular velocity in the reference frame in which
(x, y, z) are fixed, is given by Equation (6.64). Noting
that
write the constraint equations (the
"no slip" conditions) relating and
to the Eulerian
angles
and their derivatives.
1
2
1
2
2
2
Intersection
of planes
Drum
Figure P6.82
Figure P6.85
Figure P6.84
6.9
Rotation Matrices
Suppose that a vector Q is written in a frame and that its components
are the elements of a column matrix
. It is possible to develop a set
of 3 X 3 matrices [T ], [T ], and [T ] each of which, w h e n postmultiplied
by
, gives the components of Q in a n e w frame rotated about an
x, y, or z axis, respectively, of
These matrices are handy work-savers.
For example, they reduce the work involved, in going from Equa­
tion (6.58) to (6.61) or (6.64), to a pair of simple matrix multiplications.
We shall develop [T ] and then state the results for [T ] and
[T ], which are derived similarly. Let
in
which
are a triad of unit vectors having fixed directions
x
y
z
2
y
r
Page 435
along axes (x, y, z) of frame
Suppose further that
is a frame
whose orientation may be obtained from that of
by a rotation
through the angle
about z. The rotated axes, which were aligned
with (x, y, z) prior to the rotation, will be denoted (x , y , z ) with
associated unit vectors
We note that in the "new" frame
w e may write
where Q = Q and
since the rotation is about
this axis, common to both frames. To get Q and Q in terms of Q , Q ,
and , w e use (see Figure 6.18):
1
Z1
1
1
Z
x1
y1
x
y
(6.72)
and similarly,
(6.73)
Figure 6.18
Together with Q = Q , the Equations (6.72, 6.73) give the compo­
nents of Q in the rotated frame in terms of its components Q , Q , Q )
back in N o w w e are ready to observe that if the matrix [T ] is defined as
Z1
Z
x
y
z
z
then the same results for the components of Q in the rotated frame
obtained from the matrix product
are
Page 436
The rotation matrices for rotations of
respectively given by [T ] and [T ]:
x
about x, and
about y, are
y
The student may wish to verify one or both of these matrices, as w e
did for [T ]. Note the change in the "sign of sine" in the [T ] matrix.
Moreover, if w e must turn about an axis through a negative angle, we
need only change the signs of both sine terms; this follows from the fact
that
while
We n o w consider ex­
amples of the use of the rotation matrices. We shall use some shorthand
common in the literature of kinematics:
for
for
etc.
z
y
EXAMPLE 6.15
Use rotation matrices to obtain the components of
in body coordinates,
given its representation (Equation 6.64) in the reference or space frame
Solution
We premultiply
expressed in in matrix form, with rotation matrices of
about the 3-axis, then about the new 2-axis, and then about the new and final
3-axis:
components of
This gives the components
of
This yields the components
of
Finally, we obtain the components of
in
Page 437
Comparing the elements of this matrix with the components in Equation (6.61),
we see that rotation matrices indeed furnish us with a rapid means of "convert­
ing" a vector from one frame to another. Note also that the bracket in the second
line above contains
expressed in , previously derived as Equation (6.67).
The bracket in the third line gives the components of
in
The next example illustrates a use that was made of rotation matrices
by one of the authors in the development of earth stations.
EXAMPLE 6.16
Using rotation matrices, compute the angles A (azimuth) and E (elevation)
through which an antenna must respectively turn about (a) the negative of local
vertical and then (b) the new, rotated position of the elevation axis in order to
sight a satellite in geosynchronous orbit (see Problems 2.64,65). Angles A and E
are called look angles, and an antenna that performs azimuth followed by eleva­
tion in this manner is said to have "el over az" positioning. The azimuth angle A is
a function of the local latitude and the relative west longitude of the satellite;
the elevation angle E depends additionally on R /R the ratio of the earth and
orbit radii. (See Figure E6.16a.)
e
Site P
Satellite
Equator
Geosynchronous orbit
Figure E6.16a
Solution
The frame
has origin at the center of the earth as shown; the xy plane
contains the site P and its meridian. The coordinates of the satellite in this frame
are seen to be given by the position vector
First we rotate the frame through the latitude angle about z in order to line up
the new axis x with the local vertical at the site P. We call the resulting rotated
1
Page 438
frame
and obtain the following for the new components of
and so forth):
Next we translate the axes to the site, as shown in Figure E6.16b. The only
component of
that changes is x , and we see by inspection that
1
Note that we must subtract the earth radius R from x to get the proper x
coordinate of the satellite.
The second rotation is the azimuth rotation about—x ; if we call the rotated
frame , the coordinates (x , y , z ) of S in this frame are given by
e
t
2
2
3
3
3
Figure E6.16b
Here we take an important step. We want the z component of r to be zero
because we wish to rotate next in elevation about z and end up with the "boresight" (axis) of the antenna aiming at the satellite. Thus angle A (see Fig­
ure E6.16c) is determined by setting the third element of the preceding matrix to
zero:
3
PS
3
(1)
Figure E6.16c
Finally we rotate through angle E about the z axis:
3
Page 439
Now we come to the condition that will allow us to determine the value of
angle E (see Figure E6.16d): We wish the antenna to aim directly at the satellite.
Since the antenna boresight is now in the — y direction, we wish the elevation
rotation to stop when the x coordinate is zero:
3
3
(2)
Figure E6.16d
If
then (2) becomes
in which the azimuth angle A is given by Equation (1), so that
(3)
There is a single circle in the sky in which geosynchronous satellites
can exist. This circle, which was examined in Problem 2.64, has rapidly
become very crowded, however. As of the summer of 1994, there were
nearly 500 satellites in geosynchronous orbit. Six years earlier, there
were just over 100, and in early 1982, only around 30.
PROBLEMS
•
Section 6.9
6.86 For the United States the eastern and western limits
of usable satellite positions in the geosynchronous arc
(see Problem 2.64) are about 70° and 143°W longitude,
respectively. Find the ranges in azimuth and elevation
that are required if the antenna in Example 6.16 is to
sweep from the eastern to the western limit for a site at: (a)
34°N latitude and 84°W longitude; (b) your home town.
(Select a city in the contiguous United States if you are
from another country.)
6.87 Use Equations (6.70) together with rotation matri­
ces to compute the angular velocity components in spacefixed axes.
6.88 An antenna has three rotational degrees of free­
dom (see Figure P6.88):
1.
2.
3.
Azimuth angle A about local vertical z,
Elevation angle E about an axis originally parallel
to x,
Polarization angle P about the axis of symmetry of
the dish (originally parallel to y).
Use the addition theorem together with rotation matrices
to calculate
in terms of its components in
Figure P6.88
6.89 In the preceding problem calculate
its components in
in terms of
Page 440
• 6 . 9 0 Plot the elevation angle E versus the satellite
angle (see Example 6.16) for the following values of
(on the same graph):
and 80°.
What do the crossings of the
axis of these curves
physically represent?
• 6.91 Calculate the look angles (see Example 6.16)forthe
case in which the elevation rotation is performed prior to
azimuth ("az over el" positioning).
6.92 It takes six orbital parameters to establish the loca­
tion of a planet with respect to a frame fixed in space. (See
Figure P6.92.) To find its orbital path, we first turn
through the angle ft in the ecliptic plane (the plane con­
taining the path of the sun as we see it from earth) to the
ascending node (the intersection of the ecliptic plane with
the planet's path when going north). Next we turn in
inclination through the angle i about x\ to obtain the tilt of
the planet's plane. (Thus earth's inclination is defined as
zero.) Finally, the angle
locates the perihelion of the
planet's orbit. This is the closest point of the orbit to the
sun's center S. Two other quantities give the orbit's shape,
and a sixth one locates the planet in its orbit with respect
to the perihelion. If v = (v , v , v ) is a vector defined in
x
PROJECT PROBLEM
y
•
z
Chapter 6
6.93 After reading Example 6.3, construct a simple
model that illustrates the workings of a universal joint
(see Figure P6.93 for some examples).
Figure P6.93
Vernal
equinox
Ascending
NODE
Figure P6.92
the space frame
use rotation matrices to obtain the
components of v in the frame
located as shown
in the orbital path at P, in terms of
i, and
Page 441
SUMMARY
•
Chapter 6
The key concept of this chapter is angular velocity. If
and are two
rigid bodies (or frames of reference), the angular velocity of relative to
(or of in ),
is the unique vector that connects the derivatives
relative to
and of any vector, say Q, by
The angular velocity describes the rate at which the orientation of one
body changes relative to another. Among the important properties are:
and the addition theorem (a chain rule):
which can be extended by repetition to
and in fact to any number of bodies.
This rather abstract-seeming concept of angular velocity reduces to
the familiar form of Chapter 3 w h e n w e have "plane-motion" situations
in which a direction is fixed in both of two bodies. So if is the constant
vector (denotes a fixed direction) in both of the bodies and w e have
This equation, along with the addition theorem, gives us an important
tool to deal with systems of bodies that appear complex but where each
body moves relative to a neighbor by rotating about an axis fixed in the
neighbor.
Angular acceleration is the derivative of angular velocity, and the
derivative may be calculated in either of the two bodies (frames) in­
volved.
Angular accelerations cannot be added with the simplicity of an addition
theorem such as (6.27), but instead obey the formula:
Velocities of a point P relative to two frames
where 0' is a point fixed in
and
are related by
Similarly for accelerations, w e have
By letting P be fixed in
(as is O') w e can deduce from the two
previous equations a pair of expressions relating the velocities and then
the accelerations (relative to a frame ) of two points such as P and O',
Page 442
both fixed in :
These are the counterparts in three dimensions of Equations (3.8) and
(3.19).
Finally, w e have discussed a system that is often used to describe the
orientation of a body relative to a frame of reference — Euler angles.
Angular velocity in terms of rates of change of Euler angles has been
developed.
REVIEW QUESTIONS
•
Chapter 6
True or False?
1. The angular velocity of body in frame depends only upon the
changes in orientation of with respect to
2. The addition theorem for angular velocity applies equally well to
angular acceleration.
3. The formula relating the velocities of two points of a rigid body in
plane motion,
applies to three-dimensional
problems provided that
and the v's become threedimensional vectors.
4. The formula relating the accelerations of two points of a rigid body in
plane motion,
applies to threedimensional problems provided that
and the a's become
three-dimensional vectors.
5. The equation
in plane motion extends to three similar linear
equations in general motion for determining the orientation angles.
6. For a point P to have a nonvanishing Coriolis acceleration, there
must be both a relative velocity of P with respect to the "moving
frame" and an angular velocity of the moving frame relative to the
reference frame.
7. If w e premultiply a vector {v} by a rotation matrix [T], the 3 X 1
vector w e get contains the "new" components of v in the rotated
frame.
8. The Eulerian angles are used to orient a body in three-dimensional
space.
9. The Eulerian angles are three rotations
were originally distinct and orthogonal.
and about axes which
10. It is possible, for any moving point P, to choose a moving frame such
that the Coriolis acceleration of P vanishes identically.
11. The angular velocity vector is used to relate the derivatives of a vector
in two frames.
12. If one yoke of a misaligned universal (Hooke's) joint turns at constant
angular speed, so does the other.
443
13. In general motion of a rigid body as long as
zero velocity of or -extended.
there is a point of
14. The order of rotations is important in orienting a body if the Eulerian
angles are used as defined in this chapter (in conjunction with the
Cardan suspension of the gyroscope).
ANSWERS:
1.T
2.F
14. T
3.T
4.F
5. F
6.T
7. T
B.T
9. F
10. T
11. T
12. F
13. F
7
KINETICS OF A RIGID BODY IN
GENERAL MOTION
7.1
7.2
Introduction
Moment of Momentum (Angular Momentum) in Three
Dimensions
7.3
Transformations of Inertia Properties
Transformation of Inertia Properties at a Point
7.4
Principal Axes and Principal Moments of Inertia
Principal Axes at C
Calculation of Principal Moments of Inertia
Calculation of Principal Directions
Principal Axes at Any Point
Orthogonality of Principal Axes
Equal Moments of Inertia
Maximum and Minimum Moments of Inertia
7.5
7.6
7.7
7.8
The Moment Equation Governing Rotational Motion
The Euler Equations
Use of Non-Principal Axes
Use of an Intermediate Frame
Gyroscopes
Steady Precession
Torque-Free Motion
Impulse and Momentum
Work and Kinetic Energy
SUMMARY
REVIEW QUESTIONS
Page 444
Page 445
7.1
Introduction
I n C h a p t e r 2 w e f o u n d t h a t , r e l a t i v e t o a n i n e r t i a l f r a m e of r e f e r e n c e ,
m o t i o n of a b o d y is g o v e r n e d b y
(7.1)
and a m o m e n t equation
(7.2)
or
(7.3)
w i t h O b e i n g fixed in t h e i n e r t i a l f r a m e . T h e s e g e n e r a l e q u a t i o n s w e r e
s p e c i a l i z e d t o plane m o t i o n of a rigid b o d y
in C h a p t e r 4 a n d will n o w
b e u s e d t o s t u d y t h e general m o t i o n of
in three dimensions.
A s w e i n d i c a t e d i n C h a p t e r 2 , t h e first o f t h e t w o v e c t o r e q u a t i o n s
g i v e n a b o v e d e s c r i b e s t h e m a s s c e n t e r m o t i o n of any s y s t e m . * It is a p p l i ­
c a b l e , for e x a m p l e , t o rigid o r d e f o r m a b l e s o l i d s , s y s t e m s of s m a l l m a s s e s ,
liquids, a n d gases. For a b o d y in general (three-dimensional) motion,
E q u a t i o n (7.1) n o w p o s s e s s e s t h r e e n o n t r i v i a l s c a l a r c o m p o n e n t e q u a ­
t i o n s w h o s e s o l u t i o n s a l l o w u s t o l o c a t e t h e m a s s c e n t e r C.
W e n o t e that, as w a s the case w i t h p l a n e motion, the m a s s center can
m o v e i n d e p e n d e n t l y of t h e b o d y ' s c h a n g i n g o r i e n t a t i o n ( p r o v i d e d t h a t
t h e external forces d o n o t t h e m s e l v e s d e p e n d o n t h e b o d y ' s a n g u l a r
m o t i o n , w h i c h is f r e q u e n t l y t h e c a s e ) . W e s a w s u c h a n e x a m p l e i n S e c t i o n
6.6 w h e n w e e x a m i n e d t h e m o t i o n o f (a p a r t i c l e o r ) t h e m a s s c e n t e r of a
body near the rotating earth. W e emphasize that such a simple a n d
n a t u r a l e x t e n s i o n f r o m t w o t o t h r e e d i m e n s i o n s w i l l not o c c u r w i t h t h e
orientation (or a n g u l a r ) m o t i o n of
a s w e shall s e e in Section 7.5. T h e
r e a s o n is t h a t
i n E q u a t i o n (7.2) cannot b e w r i t t e n a s t h e s u m of
t h r e e t e r m s of t h e f o r m
I n t h e r e m a i n d e r of t h e c h a p t e r w e s h a l l first d e v e l o p t h e e x p r e s s i o n
for t h e m o m e n t of m o m e n t u m of a rigid b o d y
in general motion. This
w i l l t h e n l e a d u s i n t o a s t u d y of t h e i n e r t i a p r o p e r t i e s of
Having
p a r t i a l l y e x a m i n e d t h e c o n c e p t of i n e r t i a i n C h a p t e r 4 , w e s h a l l e x t e n d
this s t u d y to include transformations at a point as well as principal m o ­
m e n t s a n d a x e s of i n e r t i a . T h e n a n d o n l y t h e n s h a l l w e b e f u l l y p r e p a r e d
t o d e r i v e t h e E u l e r e q u a t i o n s t h a t g o v e r n t h e r o t a t i o n a l m o t i o n of a rigid
b o d y in g e n e r a l m o t i o n . W e shall also e x a m i n e , a s w e d i d for t h e p l a n e
* Excluding throughout, of course, relativistic effects occurring when velocities are not
small compared to the speed of light.
Page 446
m o t i o n i n C h a p t e r 5 , s o m e s p e c i a l i n t e g r a l s of t h e e q u a t i o n s of m o t i o n ,
w h i c h a r e k n o w n a s t h e p r i n c i p l e s of i m p u l s e a n d m o m e n t u m , a n g u l a r
impulse a n d angular m o m e n t u m , a n d w o r k a n d kinetic energy.
7.2
Moment of Momentum (Angular Momentum) in
Three Dimensions
W e s a w i n C h a p t e r 4 t h a t w h e n it is r e a s o n a b l e t o t r e a t a b o d y a s rigid,
t h e e q u a t i o n s of m o t i o n of
a r e greatly simplified. T h e m a s s center
b e c o m e s fixed i n t h e b o d y , a n d t h e m o m e n t of m o m e n t u m is e x p r e s s i b l e
i n t e r m s of t h e a n g u l a r v e l o c i t y a n d t h e i n e r t i a p r o p e r t i e s of — h e n c e t h e
o t h e r n a m e of m o m e n t of m o m e n t u m : angular m o m e n t u m . W e s h a l l
p r o c e e d n o w t o s t u d y t h e a n g u l a r m o m e n t u m H of a b o u t a p o i n t P i n
general (three-dimensional) motion. W e shall see that t h e equations that
result are m u c h m o r e complicated t h a n their plane m o t i o n counterparts.
P
L e t u s b e g i n b y i n t r o d u c i n g a s y s t e m of r e c t a n g u l a r a x e s (x, y, z)
w h i c h h a v e t h e i r o r i g i n a t P . T h e a n g u l a r v e l o c i t y of i n r e f e r e n c e f r a m e
m a y t h e n b e e x p r e s s e d i n t e r m s of its c o m p o n e n t s a l o n g t h e s e a x e s b y
(7.4)
T h e l o c a t i o n , r e l a t i v e t o P , of a t y p i c a l p o i n t i n t h e b o d y is g i v e n b y
(7.5)
T h e m o m e n t of m o m e n t u m of
tions
r e l a t i v e t o P is n o w d e f i n e d ( s e e S e c ­
2.6 a n d 4.3) t o b e
(7.6)
i n w h i c h v, t h e v e l o c i t y of t h e m a s s e l e m e n t dm, i s n o t t h e d e r i v a t i v e of r
b u t r a t h e r of t h e p o s i t i o n v e c t o r t o t h e e l e m e n t f r o m a p o i n t fixed i n t h e
reference frame
as s h o w n in Figure 7.1.
•Path
of
dm
Figure 7.1
Since
obtain
is a rigid b o d y , w e k n o w f r o m E q u a t i o n ( 6 . 5 6 ) t h a t
a n d w e m a y substitute this expression into E q u a t i o n (7.6) to
Page 447
U s i n g t h e m a s s c e n t e r d e f i n i t i o n , t h e i n t e g r a l i n t h e first t e r m o n t h e
r i g h t - h a n d s i d e is
mr
PC
Therefore
(7.7)
I n t h e c a s e s w h e r e e i t h e r (a) P is c h o s e n t o b e t h e m a s s c e n t e r C, o r ( b )
v = 0, o r (c)
, t h e first t e r m o n t h e r i g h t s i d e of E q u a t i o n (7.7)
v a n i s h e s . For these cases,
P
(7.8)
Substituting
a n d r f r o m E q u a t i o n s (7.4) a n d (7.5) a n d u s i n g t h e i d e n t i t y
(7.9)
w e obtain the result
(7.10)
Question 7.1
W h y may the
c o m p o n e n t s b e b r o u g h t outside t h e
various integrals in Equation (7.10)?
O n c e w e recognize t h e inertia p r o p e r t i e s (see Sections 4.3 a n d 4.4),
t h i s a n g u l a r m o m e n t u m e x p r e s s i o n b e c o m e s , for t h e c a s e w h e n P is C ,
(7.11)
T h e f o r m of t h e e q u a t i o n is i d e n t i c a l if p o i n t P is n o t C, b u t r a t h e r e i t h e r
v = 0 or
; t h e o n l y d i f f e r e n c e is t h a t t h e i n e r t i a p r o p e r t i e s a r e
c a l c u l a t e d w i t h r e s p e c t t o a x e s a t P i n s t e a d of a t t h e m a s s c e n t e r :
P
(7.12)
B o t h of t h e s e f o r m s for t h e a n g u l a r m o m e n t u m v e c t o r ( E q u a t i o n s 7 . 1 1
a n d 7.12) will p r o v e i m p o r t a n t to u s in t h e sections to follow.
Answer 7,1 As we have seen in Chapter 6,
depends only on how a set of unit vec­
tors, locked into change their directions in The angular velocity is a constant with
regard to integration at a particular instant over the body's volume.
Page 4 4 8
PROBLEMS
•
Section 7.2
7.1
Find the angular momentum vector H
wagon wheel of Problem 6.49.
O
of the
7.2
Find the angular momentum vector of the disk
Problem 6.27 about (a) C and (b) O.
7.5
Depicted in Figure P7.5 is a grinder in a grinding
mill that is composed of three main parts:
in
1.
The vertical shaft
lar speed
7.3
Find the angular momentum vector for the bent bar
of Example 4.16 about the mass center, if it turns about
the z axis at angular speed
2.
The slanted shaft
and rums with it
3.
The grinder
of radius r, turning in bearings at C
about
and rolling on the inner surface of
7.4
A thin homogeneous disk of mass M and radius r
rotates with constant angular speed
about the shaft
(Figure P7.4). This shaft is cantilevered from the vertical
shaft and rotates with constant angular speed
about
the axis of
Find the angular momentum of the disk
about point Q, and s h o w the direction of the vector in a
sketch.
which rotates at constant angu­
of length
which is pinned to
As shaft
gets up to speed
body swings outward,
and then the angle remains constant during operation.
Treat the grinder as a disk and find its angular momen­
tum vector H in convenient coordinates. (A suggested set
is shown.)
7.6
In the preceding problem note that point O is a
fixed point of all three bodies
and extended. Com­
pute the angular momentum H of
and verify that
c
O
H = H + r X L.
O
Figure P7.4
C
OC
Figure P7.5
7.3
Transformations of Inertia Properties
S o m e t i m e s w e n e e d t h e m o m e n t s a n d p r o d u c t s of i n e r t i a a t p o i n t s o t h e r
t h a n t h e m a s s c e n t e r C of a rigid b o d y
These properties can be found
w i t h o u t further integration b y using the parallel-axis theorems, w h i c h
w e r e derived in C h a p t e r 4. These are restated below, w h e r e
are
t h e c o o r d i n a t e s of t h e m a s s c e n t e r C r e l a t i v e t o a x e s a t P. F o r t h e m o m e n t s
of i n e r t i a a t P:
(7.13a)
(7.13b)
(7.13c)
Page 449
A n d for t h e p r o d u c t s of i n e r t i a a t P :
(7.14a)
(7.14b)
(7.14c)
EXAMPLE 7 . 1
As a review example, compute the inertia properties at the corner B of the uni­
form rectangular solid of mass m s h o w n in Figure E7.1.
Solution
For the moments of interna w e obtain
Note that the distance between x and x is
B
In the same way,
c
For the products of inertia,
Figure E7.1
Transformation of Inertia Properties at a Point
We n o w consider a second a n d equally important transformation, which
w i l l d e m o n s t r a t e t h a t if w e k n o w t h e m o m e n t s a n d p r o d u c t s of i n e r t i a
a s s o c i a t e d w i t h a s e t of o r t h o g o n a l a x e s t h r o u g h a p o i n t P , w e c a n e a s i l y
c o m p u t e t h e m o m e n t s a n d p r o d u c t s of i n e r t i a a s s o c i a t e d w i t h a n y other
s e t of a x e s h a v i n g t h e s a m e o r i g i n . C o n s i d e r t w o s e t s of a x e s w i t h a
c o m m o n o r i g i n a t P . ( S e e F i g u r e 7.2.) L e t
b e t h e d i r e c t i o n c o s i n e s of
x' r e l a t i v e t o x, y, a n d z, r e s p e c t i v e l y . T h e n t h e r e c t a n g u l a r c o o r d i n a t e x' of
a p o i n t Q i n t h e b o d y is r e l a t e d t o t h e r e c t a n g u l a r c o o r d i n a t e s x, y, z b y
1
unit vector along x axis
Figure 7.2
(7.15)
Page 450
W e s e e k a f o r m u l a for
i n t e r m s of t h e i n e r t i a p r o p e r t i e s w r i t t e n
w i t h r e s p e c t t o t h e (x, y, z) a x e s . T h e d e f i n i t i o n of
is
(7.16)
Since
r , we may
tion (7.16):
PQ
( e a c h is t h e s q u a r e of t h e l e n g t h of
add
a n d s u b t r a c t x'
2
to p r o d u c e this quantity in Equa­
(7.17)
S u b s t i t u t i n g x' f r o m E q u a t i o n ( 7 . 1 5 ) i n t o ( 7 . 1 7 ) g i v e s
E x p a n d i n g t h e trinomial a n d r e a r r a n g i n g , w e get
(7.18)
Since the
a r e t h e d i r e c t i o n c o s i n e s of t h e v e c t o r i n t h e d i r e c t i o n of t h e
x' a x i s , w e k n o w t h a t
U s i n g t h i s r e l a t i o n i n t h e first t h r e e t e r m s of t h e i n t e g r a n d i n E q u a ­
tion (7.18) gives
Rearranging, we have
(7.19)
R e c o g n i z i n g t h e six i n t e g r a l s i n E q u a t i o n ( 7 . 1 9 ) a s t h e i n e r t i a p r o p e r t i e s
a s s o c i a t e d w i t h t h e (x, y, z) d i r e c t i o n s a t P, w e a r r i v e a t o u r g o a l :
(7.20)
T h i s f o r m u l a a l l o w s u s t o c o m p u t e t h e m o m e n t of i n e r t i a of t h e m a s s of
a b o u t a n y l i n e t h r o u g h P if w e k n o w t h e p r o p e r t i e s a t P f o r a n y s e t of
o r t h o g o n a l a x e s . W e n o w i l l u s t r a t e its u s e w i t h a n e x a m p l e .
EXAMPLE 7 . 2
Compute the moment of inertia about the diagonal BA of the rectangular solid of
Example 7.1.
Solution
We define the axis x' to emanate from B, pointing toward A as in Figure E7.2. The
inertia properties at B were computed in the prior example. The direction cosines
Page 4 5 1
of x' are seen by inspection to be
in which
Substituting the
and the inertia properties at B into Equation (7.20) gives
Figure E7.2
I n t h e p r e c e d i n g e x a m p l e w e o b s e r v e t h a t t h e l i n e BA a l s o p a s s e s
t h r o u g h t h e m a s s c e n t e r C. T h e m o m e n t of i n e r t i a a b o u t l i n e BA is of
course the s a m e n o matter w h i c h point o n the line o n e uses to m a k e the
c a l c u l a t i o n . (It w o u l d a c t u a l l y b e e a s i e r t o c o m p u t e a t C i n t h i s c a s e ,
because the
a r e t h e s a m e w h i l e t h e p r o d u c t s of i n e r t i a v a n i s h ! )
A r e s u l t s i m i l a r t o E q u a t i o n ( 7 . 2 0 ) for p r o d u c t s of i n e r t i a w i l l n o w b e
d e r i v e d . L e t n , n , a n d n b e t h e d i r e c t i o n c o s i n e s of a x i s y ' . N o t e t h a t t h e
r e c t a n g u l a r c o o r d i n a t e y' m a y t h e n b e w r i t t e n , i n t h e s a m e w a y a s E q u a ­
tion (7.15), as follows:
x
y
z
(7.21)
By d e f i n i t i o n ,
T h e r e f o r e , e x p a n d i n g a n d r e c o g n i z i n g t h e p r o d u c t of i n e r t i a i n t e g r a l s ,
(7.22)
Since
a n d ( n , n , n ) a r e c o m p o n e n t s of u n i t v e c t o r s a l o n g t h e
m u t u a l l y p e r p e n d i c u l a r a x e s x' a n d y', w e m a y d o t t h e s e v e c t o r s t o g e t h e r
and obtain
x
y
z
T h e i n t e g r a n d in E q u a t i o n (7.22) m a y therefore b e w r i t t e n as
(7.23)
Page 452
Substituting E q u a t i o n (7.23) into (7.22) t h e n yields t h e desired transfor­
m a t i o n e q u a t i o n f o r t h e p r o d u c t s of i n e r t i a :
(7.24)
EXAMPLE 7 . 3
In Examples 7.1 a n d 7.2 let t h e solid b e a cube (a = b = d) a n d let y' b e defined as
follows (see Figure E7.3):
1.
y' is perpendicular to x'.
2.
y' is the s a m e p l a n e as z a n d x'.
Find
Solution
From t h e equations in Example 7.2 w e h a v e
for t h e unit vector along
w e n o w force t h e c o m p o n e n t s of
is, (n , n , n ) — to b e s u c h t h a t conditions 1 a n d 2 are satisfied:
x
y
— that
z
Figure E7.3
(a vector perpendicular to the plane of x'and y')
These t w o equations give n = n a n d n = 2n . W e m u s t also ensure t h a t
a unit vector:
y
x
z
x
is
Substituting t h e c o m p o n e n t s of
a n d t h e inertia properties at B (letting the
general point P in (7.24) b e B in this problem) into Equation (7.24) gives
W e n o t e t h a t i n E x a m p l e 7.3 t h e z e r o r e s u l t is n o t o b v i o u s a t t h i s
p o i n t i n o u r s t u d y . W h i l e it is t r u e t h a t , f o r t h i s c a s e of a = b = d, t h e x ' y '
Page 4 5 3
p l a n e is a p l a n e of s y m m e t r y , t h i s g u a r a n t e e s ( s e e S e c t i o n 4 . 4 ) t h a t
and
are zero but not necessarily
N o t e also that there are t w o
d i r e c t i o n s ( 1 8 0 ° a p a r t ) for y' t h a t b o t h satisfy c o n d i t i o n s 1 a n d 2 i n t h e
preceding example.
Question 7.2
W h e r e in the solution did w e choose o n e of these direc­
tions? (And does it matter?)
W e close this section b y n o t i n g t h a t E q u a t i o n s (7.20) a n d (7.24) are
t h e t r a n s f o r m a t i o n e q u a t i o n s satisfied b y a s y m m e t r i c s e c o n d - o r d e r t e n ­
sor; t h u s t h e inertia p r o p e r t i e s d o i n d e e d f o r m s u c h a tensor. W e also n o t e
f r o m t h e s e t w o e q u a t i o n s t h a t o n l y if t h e p r o d u c t s of i n e r t i a a r e d e f i n e d
w i t h t h e m i n u s s i g n (see E q u a t i o n s 4 . 2 ) d o w e g e t t h e c o r r e c t t e n s o r
transformation equations.
Answer 7.2 When we said that
that is, took the positive square root, we
chose y' to be in the direction making an acute angle with x ; had we chosen
we would have gotten the opposite direction for y'. And had I been nonzero, the sign
of the answer would have been opposite also.
B
x'y'
PROBLEMS
•
Section 7.3
7.7
T h e three h o m o g e n e o u s rods in Figure P7.7 are
w e l d e d together at O to form a rigid b o d y . Find the m a s s
m o m e n t s a n d products of inertia at point Q w i t h respect
to axes there that are parallel to x, y, a n d z.
7.9
In t h e preceding problem, find t h e m o m e n t of iner­
tia about the line OP.
7.10 C o m p u t e t h e m o m e n t of inertia w i t h respect to line
AB for the bent bar in Figure P7.10. The bar lies in a plane
a n d h a s mass 4 m .
Figure P7.10
Figure P7.7
Figure P7.8
7.8
Find the mass m o m e n t s a n d products of inertia of
the b o d y in Figure P 7 . 8 , w i t h respect to a set of axes
t h r o u g h the point P at
, respectively, parallel to
x, y, a n d z. Each of the t w o perpendicular r o d s of the " T "
h a s m a s s m a n d length
Figure P7.11
7.11 Find the product of inertia
for the h o o p
mass m a n d radius R in Figure P 7 . l l . The p l a n e of
misaligned w i t h the xy p l a n e by angle
of
is
.
Page 454
7.12 C o m p u t e t h e m o m e n t of inertia a b o u t line BA in
Example 7.2 b y using Equation (7.20) a t C instead of B.
S h o w that if a = b = d, t h e a n s w e r becomes equal to
(which equals
in this case).
7.16 Find t h e inertia properties at O for t h e b o d y s h o w n
in Figure P7.16, w h i c h is c o m p o s e d of a rod a n d ring t h a t
h a v e equal cross sections a n d densities. T h e r o d is per­
pendicular to t h e p l a n e of t h e ring.
7.13 T h e centroidal m o m e n t s of inertia for t h e solid el­
lipsoid in Figure P 7 . 1 3 are
Figure P7.16
Further, t h e mass of is
pabc, w h e r e is the mass
density. Find t h e m o m e n t of inertia of t h e m a s s of about
the line m a k i n g equal angles w i t h x, y, a n d z.
7.14 S h o w that t h e s u m of a n y t w o of
always exceeds t h e third.
and
7.15 Part of a special-purpose, dual-driven a n t e n n a sys­
t e m consists of an octagonal rotator as s h o w n in Fig­
ure P 7 . 1 5 . Each of t h e eight equal sections is a square steel
tube with t h e indicated dimensions a n d thickness in.
Find t h e m o m e n t of inertia of t h e rotator a b o u t t h e axis of
rotation. (Consider each section to h a v e squared-off e n d s
at the average 18-in. length a n d ignore t h e small overlaps.
Use a density of 15.2 s l u g / f t . )
3
Figure P7.18
* 7.17 Find
for t h e p r o b l e m of Example 7.3 if b = d
= a/2. T h e axes h a v e their origin at B just as in t h e exam­
ple.
* 7.18 For t h e rigid b o d y in Figure P7.18 let t h e inertia
properties at P b e k n o w n
. Further, let there be
m e a s u r e d along t h e s a m e axes t h e quantities
Figure P7.13
where
and
are as defined in Section 7.3. S h o w
that Equation (7.20) t h e n implies
This is t h e equation of an ellipsoid centered at P. Devel­
oped b y C a u c h y in 1827, it is called t h e ellipsoid of inertia.
S h o w that t h e m o m e n t of inertia a b o u t a n y line x'
t h r o u g h P equals t h e reciprocal of t h e square of t h e dis­
tance from P to t h e point w h e r e X' intersects t h e ellipsoid.
4 in.
18 in.
Figure P7.15
* Asterisks identify the more difficult problems.
Page 455
7.19 I n t h e preceding problem, if t h e p r o d u c t s of inertia
vanish, t h e equation of t h e ellipsoid of inertia written in
terms of t h e resulting m o m e n t s of inertia
is
2
2
Round plates
(mass m each)
2
S h o w that n o t all ellipsoids of t h e form AX + BY + cz
= 1 can b e ellipsoids of inertia. HINT: T h e s u m of a n y t w o
m o m e n t s of inertia m u s t always exceed t h e third, as
stated in Problem 7.14.
**7.20 Calculate t h e inertia properties at O of t h e three
circular fan blades connected b y light rods in Fig­
ure P7.20. T h e blades are tilted 30° w i t h respect to t h e
axes OC OC , a n d O C ; t h e s h a d e d halves of each are
b e h i n d t h e plane of the d r a w i n g a n d t h e u n s h a d e d halves
are in front of it.
v
2
Figure P7.20
3
7.4
Principal Axes and Principal Moments of Inertia
I n t h i s s e c t i o n w e d e m o n s t r a t e a p a r t i c u l a r l y u s e f u l w a y of d e s c r i b i n g t h e
i n e r t i a c h a r a c t e r i s t i c s of a r i g i d b o d y
It h a p p e n s t h a t , a t a n y p o i n t P of
it is a l w a y s p o s s i b l e t o f i n d a s e t of r e c t a n g u l a r a x e s s o t h a t t h e p r o d u c t s of
inertia at P w i t h respect t o t h e s e axes all v a n i s h . T h e s e axes a r e called t h e
principal axes of inertia a t P , a n d t h e m o m e n t s of i n e r t i a w i t h r e s p e c t t o
t h e m a r e k n o w n a s t h e principal m o m e n t s of inertia f o r t h e p o i n t .
S p e c i f i c a l l y , a n axis x is a p r i n c i p a l a x i s a t P if
w h e r e is a n y
axis t h r o u g h P t h a t is p e r p e n d i c u l a r t o x. ( S e e F i g u r e 7.3.) W e c a n s h o w
t h a t if
, i n w h i c h (x, y, z) f o r m a t r i a d of r e c t a n g u l a r a x e s a t P ,
t h e n x is a p r i n c i p a l axis a t P . F o r t h e p r o o f , w e s h a l l s h o w t h a t
implies
where
is t h e a r b i t r a r y a x i s t h r o u g h P n o r m a l t o x.
U s i n g E q u a t i o n (7.24) w e m a y write
Figure 7.3
In this equation
a r e t h e d i r e c t i o n c o s i n e s of x ( t h e first
subscript in
) w i t h r e s p e c t t o x, y, a n d z, r e s p e c t i v e l y , w h i l e (n , n , n )
a r e t h e d i r e c t i o n c o s i n e s of
(the second subscript in
), a l s o w i t h
r e s p e c t t o x, y, a n d z. S i n c e
w e h a v e (n , n , n ) = ( 0 , n , n ).
Substituting these
a n d n's, w e get
x
x
y
z
y
y
z
z
But
so that
a n d w e s e e t h a t all it t a k e s t o m a k e a n
a x i s , s u c h a s x, a p r i n c i p a l a x i s a t a p o i n t is t o s h o w t h a t
and
w h e r e (x, y, z) f o r m a n o r t h o g o n a l t r i a d a t P . W e s h a l l n e e d t h i s
r e s u l t i n w h a t is t o f o l l o w .
Page 4 5 6
Principal Axes at C
W e n o w p r o c e e d t o a c o m p u t a t i o n a l p r o c e d u r e for f i n d i n g t h e p r i n c i p a l
a x e s a n d m o m e n t s of i n e r t i a . W e s h a l l d o t h e d e r i v a t i o n w h e n P is C a n d
t h e n l a t e r e x p l a i n h o w it a p p l i e s e q u a l l y w e l l t o all p o i n t s of
F o r t h e m o m e n t l e t (x, y, z) b e c e n t e r e d a t C a n d a t s o m e i n s t a n t l e t x
be parallel to t h e angular velocity
of
in a reference frame
Then
a n d E q u a t i o n ( 7 . 1 1 ) g i v e s , f o r t h e a n g u l a r m o m e n t u m of
a b o u t C, t h e s i m p l i f i e d e x p r e s s i o n
(7.25)
W e n o t e f r o m E q u a t i o n ( 7 . 2 5 ) t h a t H is p a r a l l e l t o
if a n d o n l y if
— t h a t i s , if x is a p r i n c i p a l axis a t C. T h u s w e s e e , a s d i d
L e o n h a r d E u l e r h i m s e l f i n t h e m i d d l e of t h e e i g h t e e n t h c e n t u r y , t h a t t h e
p r i n c i p a l a x e s h a v e t h e p r o p e r t y t h a t w h e n t h e a n g u l a r velocity lies a l o n g
o n e of t h e m , s o w i l l t h e a n g u l a r m o m e n t u m . E u l e r w a s s e e k i n g a n axis
t h r o u g h C f o r w h i c h , w h e n w a s s e t s p i n n i n g a b o u t it, t h e m o t i o n w o u l d
c o n t i n u e a b o u t this axis w i t h o u t a n y n e e d for external m o m e n t s to m a i n ­
t a i n it. N o t e f u r t h e r t h a t w h e n
­­H
is p a r a l l e l t o
"), the propor­
t i o n a l i t y c o n s t a n t is n e c e s s a r i l y t h e m o m e n t of i n e r t i a I a b o u t t h e i r c o m ­
m o n axis.
c
C
N o w w e a r e r e a d y for t h e big s t e p . W e let
be a
u n i t v e c t o r a n d w e s e e k t h e d i r e c t i o n of fi t h a t w i l l e n s u r e its b e i n g a
p r i n c i p a l a x i s . I n o t h e r w o r d s , w e w a n t t o f i n d t h e v a l u e s of t h e d i r e c t i o n
c o s i n e s of
(n , n , n ) s u c h t h a t if
then
Writing
in
c o m p o n e n t form gives
x
y
z
(7.26)
S u b s t i t u t i n g f r o m E q u a t i o n ( 7 . 1 1 ) for H , t h e v e c t o r r e l a t i o n
t h e n gives t h e following t h r e e scalar c o m p o n e n t equations:
c
(7.27)
If w e d i v i d e all t h r e e of E q u a t i o n s ( 7 . 2 7 ) b y
we have
rearrange t h e equations as follows:
and
a n d note that since
then we may
(7.28)
W e n o w h a v e a s e t of t h r e e e q u a t i o n s t h a t a r e a l g e b r a i c , l i n e a r , a n d
h o m o g e n e o u s i n t h e t h r e e v a r i a b l e s n , n , n . S u c h a s y s t e m is k n o w n t o
h a v e a n o n t r i v i a l s o l u t i o n if a n d o n l y if t h e d e t e r m i n a n t of t h e coeffix
y
z
PAGE 4 5 7
c i e n t s o f t h e v a r i a b l e s is z e r o . * I n t h i s c a s e , w e m a y d r o p t h e " n o n t r i v i a l "
a d j e c t i v e b e c a u s e t h e t r i v i a l s o l u t i o n ( n = n = n = 0) fails t o s a t i s f y t h e
side condition
x
y
z
(7.29)
w h i c h m u s t a l w a y s b e t r u e f o r t h e d i r e c t i o n c o s i n e s of a v e c t o r .
Calculation of Principal Moments of Inertia
S e t t i n g t h e d e t e r m i n a n t of t h e coefficients i n ( 7 . 2 8 ) e q u a l t o z e r o w i l l l e a d
first t o t h e s p e c i a l v a l u e s of J for w h i c h t h e t h r e e e q u a t i o n s h a v e a
s o l u t i o n . E a c h s p e c i a l v a l u e J is c a l l e d a n eigenvalue,
or
characteristic
value, a n d w i l l b e a p r i n c i p a l m o m e n t of i n e r t i a ; t h e c o r r e s p o n d i n g
( w i t h c o m p o n e n t s n , n , n ) is c a l l e d t h e eigenvector a s s o c i a t e d w i t h t h i s
e i g e n v a l u e I. T h e u n i t v e c t o r p o i n t s i n t h e d i r e c t i o n of a p r i n c i p a l axis of
i n e r t i a a t C. T h e d e t e r m i n a n t is e q u a t e d t o z e r o b e l o w :
x
y
z
(7.30)
If w e e x p a n d t h i s c h a r a c t e r i s t i c d e t e r m i n a n t , w e c l e a r l y g e t a c u b i c p o l y ­
n o m i a l i n I:
(7.31)
T h e a a r e o f c o u r s e f u n c t i o n s of t h e i n e r t i a p r o p e r t i e s . N o w w e k n o w
f r o m a l g e b r a t h a t if p o l y n o m i a l s w i t h r e a l coefficients h a v e a n y c o m p l e x
roots, they m u s t occur in conjugate pairs. T h u s the polynomial derived
a b o v e h a s a t t h i s p o i n t a t l e a s t o n e r e a l r o o t I (It is p o s i t i v e b y t h e
d e f i n i t i o n of t h e q u a n t i t y " m o m e n t of i n e r t i a " t h a t it r e p r e s e n t s . ) W e
n o w a r e g u a r a n t e e d a t l e a s t o n e p r i n c i p a l m o m e n t of i n e r t i a a n d c o r r e ­
s p o n d i n g p r i n c i p a l a x i s of i n e r t i a .
In order to s h o w that there are t w o others, w e next reorient our
o r t h o g o n a l t r i a d of r e f e r e n c e a x e s s o t h a t o n e of t h e m (x) c o i n c i d e s w i t h
t h e a l r e a d y identified p r i n c i p a l axis; this t h e n a l l o w s u s to write
and
w h e r e y a n d z a r e n o w a n e w p a i r of axes n o r m a l to
o u r n e w (principal) x axis. E q u a t i o n s (7.28) n o w a p p e a r as
i
1
(7.32)
and the determinantal equation becomes:
(7.33)
* Cramer's rule clearly gives n = n = n = 0 as the only solution if the equations are
independent, in which case the determinant D of the coefficients is not zero. If D = 0,
then Cramer's rule yields the indeterminate form 0 / 0 for the n's and there is a chance
for other solutions, as the equations are then dependent.
x
y
z
Page 458
This time t h e resulting cubic b e c o m e s factorable. E x p a n d i n g t h e determi­
nant, w e have
(7.34)
T h e p r i n c i p a l m o m e n t s of i n e r t i a a t C a r e t h e r o o t s of E q u a t i o n ( 7 . 3 4 ) .
T h e first r o o t ( t h e o n e w e a l r e a d y k n e w ) is r e a f f i r m e d b y s e t t i n g t h e first
factor to zero:
T h e o t h e r s will b e s e e n to c o m e from e q u a t i n g t h e s e c o n d factor to zero:
(7.35)
T h i s is, of c o u r s e , a q u a d r a t i c e q u a t i o n i n I. R e c a l l i n g t h a t t h e t w o r o o t s t o
are
w e s e e t h a t w e s h a l l h a v e t w o ( m o r e ) r e a l r o o t s I a n d I if t h e d i s c r i m i ­
n a n t is p o s i t i v e o r z e r o :
2
3
(7.36)
T h e r e f o r e all t h r e e r o o t s of t h e c h a r a c t e r i s t i c c u b i c e q u a t i o n a r e r e a l ( a n d
p o s i t i v e ) , a n d s o w e a l w a y s h a v e t h r e e p r i n c i p a l m o m e n t s of i n e r t i a a t C,
e a c h w i t h its o w n c o r r e s p o n d i n g p r i n c i p a l axis.*
Calculation of Principal Directions
W e m e n t i o n a t t h i s p o i n t t h e p r o c e d u r e for o b t a i n i n g t h e p r i n c i p a l d i r e c ­
tion, g i v e n b y n ,n ,
a n d n , for each of t h e p r i n c i p a l m o m e n t s of i n e r t i a
(I 1 , o r 7 ) . E q u a t i o n s ( 7 . 2 8 ) , b e i n g d e p e n d e n t , m a y n o t b e s o l v e d for t h e
t h r e e c o m p o n e n t s of e a c h
in themselves; however, together with the
identity
a s o l u t i o n m a y b e f o u n d . T h e i d e a is t o s o l v e
x
v
2
y
z
3
for, s a y , n a n d n i n t e r m s of n f r o m t w o of E q u a t i o n s ( 7 . 2 8 ) ; t h e n w e
s u b s t i t u t e i n t o E q u a t i o n ( 7 . 2 9 ) a n d s o l v e for n . E i t h e r s i g n m a y b e u s e d i n
t a k i n g t h e final s q u a r e r o o t , b e c a u s e t h e r e a r e o b v i o u s l y t w o l e g i t i m a t e
s e t s of d i r e c t i o n c o s i n e s . T h e s e t w o s e t s a r e n e g a t i v e s of e a c h o t h e r , a n d
e a c h yields t h e correct p r i n c i p a l axis. In Figure 7.4, e i t h e r or — defines
a p r i n c i p a l a x i s t h r o u g h C. T h e p r i n c i p a l axis is a n u n d i r e c t e d l i n e .
y
2
x
x
I n t h e first of t w o e x a m p l e s t o f o l l o w , w e s h a l l a g a i n s e e ( a s i n t h e
p r e c e d i n g d i s c u s s i o n ) t h a t if a t l e a s t t w o of t h e p r o d u c t s of i n e r t i a v a n i s h ,
Figure 7.4
* This was proved in 1755 for the first time by Segner, a contemporary of Euler. Segner
also showed that the principal axes (for distinct principal moments of inertia) are ortho­
gonal.
Page 4 5 9
t h e n t h e cubic e q u a t i o n (7.31) b e c o m e s factorable. In this case, w e d o n o t
h a v e t o s o l v e it n u m e r i c a l l y .
EXAMPLE 7 . 4
The inertia properties of a right triangular plate (see Figure E7.4a) are
Figure E7.4a
Find t h e principal m o m e n t s of inertia of the plate. T h e n find their associated
principal axes w h e n B = H.
Solution
Equations (7.28), w h i c h lead to t h e principal m o m e n t s a n d axes of inertia, b e ­
come, for t h e plate,
(1)
(2)
(3)
2
The algebra is simplified by dividing by m H / 3 6 a n d defining
(4)
This gives, in terms of t h e n o n d i m e n s i o n a l parameters B a n d I*,
(5)
(6)
(7)
Therefore t h e determinantal equation becomes
(8)
Expanding across t h e third r o w (or d o w n t h e third column), w e get
(9)
Therefore o n e of the brackets m u s t v a n i s h a n d the roots come from
(10)
Page 460
and
(11)
Equation (10) gives, using Equations (4),
(12)
Equation (11) gives, b y t h e quadratic formula,
(13)
Thus
(14)
The t h r e e principal m o m e n t s of inertia of t h e plate are given b y Equations (12)
a n d (14). In t h e case w h e n t h e right triangle is isosceles (b = H), w e h a v e B = 1 so
that, from Equation (13),
or
and
(15)
Also, from Equation (10),
C h a n g i n g back to dimensional inertias b y Equation (4), w e see t h a t for a right
isosceles triangular plate,
(16)
We shall n o w d e t e r m i n e t h e principal axis associated w i t h each of these principal
m o m e n t s of inertia. W e first substitute
(with B = 1) in each of Equations
(5) to (7) a n d get
- n - n = 0
x
-n
y
- n
z
= 0
y
n = 0
z
T h e third of these equations says that t h e principal axis for
plate (xy); t h e other t w o equations b o t h give
n=
x
-n
y
(17)
is in t h e p l a n e of the
(18)
Substituting this result into
(19)
gives
(20)
461
(21)
Thus, b y Equation (18),
(22)
so t h a t either
(23)
or
(24)
T h e lines
defined b y these t w o sets of direction cosines are s h o w n in Fig­
u r e E7.4b a n d Figure E7.4c. It is seen that t h e two preceding results represent the
same line; t h e positive directions are opposite b u t u n i m p o r t a n t . T h e inertia value,
being t h e integral of r dm, is i n d e p e n d e n t of t h e directivity of t h e line.
2
Principal axis
for I,
(a) Line of Eq. (7.12)
(b) Line of Eq. (7.13)
Figure E7.4h
For
Figure E7.4c
Equations (5) to (7) b e c o m e
n -n
x
= 0
y
—n + n = 0
x
y
n = 0
z
(25)
This time n = n w i t h n again equaling zero, so this principal axis m a k e s equal
angles w i t h x a n d y . (See Figure E7.4d a n d Figure E7.4e.) N o t e from t h e p r e x
y
c
2
c
Principal axis
for I
2
Figure E7.4d
Figure E7.4e
Page 462
ceding diagrams t h a t t h e r e is m u c h m o r e inertia a b o u t the principal axis for J,
t h a n there is for I ; t h e m a s s is m o r e closely clustered a b o u t t h a n it is a b o u t
The third principal axis is found from Equations (5) to (7) w h e n , for B = 1,
2
(26)
These equations are
-2n
-n
x
- n = 0
z
y
-
2n = 0
y
[2(2) - 4]n = 0
(27)
z
The first t w o of these equations h a v e the solution n = n = 0. The third leaves n
indeterminate. But from Equation (19) w e h a v e n = 1 or -1. T h u s t h e principal
axis for I is t h e line n o r m a l to the plate at C. This will in fact always b e true: W h e n
a b o d y is a plate (that is, flat w i t h negligible thickness c o m p a r e d to its other
dimensions), t h e m o m e n t of inertia a b o u t t h e axis n o r m a l to the plate at a n y point
is principal for that point. It is e v e n true t h a t it is the s u m of the other t w o a n d
therefore the largest.
x
y
z
z
3
Principal Axes at Any Point
W e n o w w i s h t o r e m a r k t h a t a s e t of p r i n c i p a l a x e s e x i s t s a t every p o i n t of
n o t j u s t a t t h e m a s s c e n t e r C. T o s h o w t h i s w e first r e c a l l E q u a t i o n (7.7),
w h i c h w i t h E q u a t i o n (7.12) g i v e s t h e a n g u l a r m o m e n t u m H a b o u t a n y
p o i n t P of b o d y
p
Whenever the cross-product term in H vanishes, the terms that remain
a r e i d e n t i c a l t o t h o s e of E q u a t i o n (7.11) if P r e p l a c e s C. T h e r e f o r e , for
cases in w h i c h r X
= 0, w e n e e d o n l y r e c a l l o u r a r g u m e n t s m a d e for
C a n d w e shall b e led, t h r o u g h a n identical procedure, to t h e principal
a x e s a n d m o m e n t s of i n e r t i a f o r any p o i n t P of
T h e r e f o r e w e o n l y n e e d t o i m a g i n e t h a t a t s o m e i n s t a n t o u r p o i n t P of
interest h a s either v = 0 or
In either case, r X v = 0 a n d H
t h e n h a s t h e f o r m of E q u a t i o n (7.12). R e t r a c i n g o u r s t e p s f r o m t h a t p o i n t ,
w e a r r i v e a t t h e t h r e e a x e s ( t h r o u g h P t h i s time) f o r w h i c h a l l t h r e e
p r o d u c t s of i n e r t i a a r e z e r o ; t h e s e a r e t h e s a m e t h r e e a x e s t h r o u g h P for
which
whenever
is a l i g n e d w i t h o n e of t h e m a n d v e i t h e r
P
PC
P
PC
P
P
P
v a n i s h e s o r is p a r a l l e l t o r . O f c o u r s e , all r e f e r e n c e s t o t h e m o t i o n
c o n d i t i o n s t h a t l e d t o t h e d e t e r m i n a n t a l e q u a t i o n a r e a g a i n l o s t (as t h e y
w e r e f o r C ) , s o t h a t t h e p r i n c i p a l m o m e n t s a n d a x e s of i n e r t i a d e p e n d
only o n the body's m a s s distribution.
PC
Page 4 6 3
EXAMPLE 7 . 5
Find t h e principal m o m e n t s of inertia at O a n d the directions of their associated
principal axes for the b o d y s h o w n in Figure E7.5a. It is m a d e u p of three rigid,
identical slender rods w e l d e d together at right angles to form a single rigid b o d y
Solution
Using t h e m o m e n t s of inertia s h o w n for o n e r o d in Figure E7.5b, plus t h e paral­
lel-axis (transfer) theorem, the six inertia properties are calculated below. The
reader s h o u l d verify each of t h e entries.
Figure E7.5a
I is
small
Figure E7.5b
Page 4 6 4
For this problem, then, Equations (7.28) m a y b e expressed as
(1)
in w h i c h w e h a v e multiplied t h e three equations b y
b y a n d write t h e d e t e r m i n a n t as
w e m a y replace
Expanding gives t h e characteristic cubic equation:
If a c o m p u t e r or p r o g r a m m a b l e calculator is n o t available,* w e can a l w a y s solve a
cubic b y trial a n d error r a t h e r quickly. N o t i n g t h a t
is 1342 at
a n d is
negative at
for example, o n a calculator w e m a y proceed a n d within a few
m i n u t e s obtain t h e root b e t w e e n these v a l u e s . †T h e p r o c e d u r e is as follows:
3
2
2 6
2.61
2 608
2 607
2.6076
2 6077
2 60769
3.10
-0.93
-0.123
0 2802
0.03835
- 0 00195
0 00208
2 607695
2 607696
0 00006
- 0 00034
(still positive)
(so it is > 2 6 and < 2 6 1 , closer to the latter)
(back up slightly!)
(so it is about two-thirds of the way from 2.607 to 2 608)
(so just a little farther)
(the root is close to this number!)
(should be halfway between the last one and this
one . . .)
(now double-check)
Next w e use synthetic division to obtain t h e r e d u c e d quadratic:
* See Appendix B for a numerical solution to this problem using the Newton-Raphson
method.
†We abandon our three-digit consistency in numerical analyses like this one in order to
illustrate the speed of convergence.
Page 4 6 5
Using t h e quadratic formula, w e obtain
The value of
strongly h i n t s that 22 m i g h t b e a rational root. Synthetic division
s h o w s t h a t it is, a n d refined values from t h e r e d u c e d quadratic are t h e n
Since
icant figures,
our dimensional principal m o m e n t s of inertia are, to six signif­
We next illustrate t h e c o m p u t a t i o n of t h e direction cosines, w h i c h locate for
us t h e principal axes of inertia. We find t h e m from Equations (1), for w h i c h
and
are t h e only special values (eigenvalues) of for w h i c h these equations
h a v e a solution. First w e seek t h e principal axis associated w i t h
The first of Equations (1) becomes
7.392305n - 9n + 3n = 0
x
y
z
Solving for n in terms of n a n d n a n d substituting t h e result into t h e second of
Equations (1) gives
z
x
y
n = 0.732051n
y
x
Thus
n = -0.267950n
z
Substituting these expressions for n a n d n into
y
z
x
yields
T h u s t h e angles t h a t t h e principal axis of m i n i m u m m o m e n t of inertia m a k e s
w i t h x, y, a n d z are, respectively, 3 7 . 9 4 ° , 54.74°, a n d 102.20°. This axis h a s
to be t h e o n e to w h i c h , loosely speaking, t h e m a s s finds itself closest. Examina­
tion of Figure E7.5c at t h e left, together w i t h these angles, s h o w s that this m a k e s
sense.
Next w e m a y follow t h e s a m e p r o c e d u r e for t h e principal axis of m a x i m u m
m o m e n t of inertia
. T h e results are, as the reader
m a y verify,
Figure E7.5c
Page 466
A n d for t h e intermediate m o m e n t of inertia
principal axis is defined by
the
Orthogonality of Principal Axes
N o t e in the preceding example that
a n d that
* These are very good checks o n the solution
s i n c e t h e p r i n c i p a l a x e s , w h e n t h e p r i n c i p a l m o m e n t s of i n e r t i a a r e
distinct
are orthogonal. To p r o v e this in general,
let
a n d l e t x lie a l o n g t h e a x i s of I T h e n f r o m t h e first of
E q u a t i o n s ( 7 . 3 2 ) , w h i c h is
1
(I -I)n
1
x
= 0
w e s e e t h a t if I i s e i t h e r I o r I , t h e n n = 0 ; t h a t i s , t h e c o s i n e of t h e a n g l e
b e t w e e n x a n d t h e c o r r e s p o n d i n g principal axis h a s to b e zero. This
m e a n s t h a t x is p e r p e n d i c u l a r t o t h e o t h e r t w o p r i n c i p a l a x e s . I n t u r n ,
these t w o axes are n o r m a l to each other; this follows from reorienting the
a x e s o n c e m o r e s o t h a t x still lies a l o n g t h e axis of I b u t n o w y lies a l o n g
t h e axis of I . T h i s t i m e
is a l s o z e r o , s o t h e n e w s e c o n d e q u a t i o n
becomes
2
3
x
1
2
(I - I)n
2
y
= 0
T h i s s h o w s t h a t for
t h e v a l u e of n for t h e t h i r d p r i n c i p a l
a x i s v a n i s h e s a n d it is t h e n n o r m a l n o t o n l y t o x ( w h i c h it still i s , s i n c e w e
h a v e o n l y r o t a t e d it a b o u t x) b u t a l s o t o y. T h u s t h e t h r e e p r i n c i p a l a x e s
a r e o r t h o g o n a l if t h e p r i n c i p a l m o m e n t s of i n e r t i a a r e all d i f f e r e n t . W e
n o w t u r n o u r a t t e n t i o n t o w h a t h a p p e n s if t h e y a r e n o t .
y
Equal Moments of Inertia
A c o m m o n l y o c c u r r i n g c a s e is for t w o of t h e p r i n c i p a l m o m e n t s of i n e r t i a
at a p o i n t P to b e e q u a l to each o t h e r b u t different from t h e third. W h e n
t h i s h a p p e n s , w e c a n s h o w t h a t every l i n e t h r o u g h P i n t h e p l a n e o f t h e
t w o a x e s (call t h e m x a n d y) w i t h e q u a l m o m e n t s of i n e r t i a ( s a y I = I ) is
a p r i n c i p a l a x i s h a v i n g t h i s s a m e v a l u e for its m o m e n t of i n e r t i a .
1
2
T o d o t h i s , w e first s h o w t h a t if
t h e n t h e axis a s s o c i a t e d
w i t h I (call it z) is p e r p e n d i c u l a r t o t h o s e (x, y) of I a n d I . T h e t h i r d
e q u a t i o n of ( 7 . 3 2 ) g i v e s
3
1
(I - I)n
3
z
2
= 0
T h u s w h e n I is I o r I , t h e n n = 0. H e n c e t h e a n g l e b e t w e e n z a n d x ( a n d
1
* This could have been
2
z
—equally correct!
Page 4 6 7
b e t w e e n z a n d y ) is a l s o 9 0 ° . It d o e s n o t f o l l o w i n l i k e m a n n e r f r o m t h e
equations, h o w e v e r , t h a t axes x a n d y are perpendicular. To h a n d l e this
c a s e , w e b e g i n b y s h o w i n g t h a t if t h e a x e s of t h e o t h e r t w o p r i n c i p a l
m o m e n t s of i n e r t i a (I = I ) a r e n o t p r e s u m e d t o b e o r t h o g o n a l ( s a y t h e y
a r e a a n d x i n F i g u r e 7.5 w i t h I = I = I ), then all is w e l l b e c a u s e I is
also e q u a l t o I .
To p r o v e this, w e u s e E q u a t i o n (7.20):
1
2
p
qq
p
1
xx
yy
1
x and z are principal!
Figure 7.5
T h u s w e see that
I =I
yy
1
N e x t w e let b e a u n i t v e c t o r i n t h e d i r e c t i o n of a n a r b i t r a r y a x i s x' i n
t h e p l a n e of t h e p e r p e n d i c u l a r a x e s x a n d y. T h e n u s i n g E q u a t i o n ( 7 . 2 0 )
again gives
A n d e q u a t i o n ( 7 . 2 4 ) y i e l d s ( w i t h y' p e r p e n d i c u l a r t o x' a n d l y i n g i n t h e
p l a n e of x', x, a n d y a s i n F i g u r e 7 . 6 :
P
N o t i n g t h a t z is p e r p e n d i c u l a r t o x' a n d t h a t I = 0 s i n c e z is p r i n c i p a l , w e
t h e n h a v e t h e r e s u l t t h a t x' is p r i n c i p a l w i t h t h e s a m e m o m e n t o f i n e r t i a a s
x a n d y, a n d t h i s i s w h a t w e w a n t e d t o p r o v e . N o t e t h e n , for a b o d y w i t h
a n a x i s of s y m m e t r y , f o r e x a m p l e , t h a t a n y a x i s w h i c h p a s s e s t h r o u g h a n d
is n o r m a l t o t h e s y m m e t r y axis of, is a l w a y s p r i n c i p a l ; t h i s is t r u e even if
the axis is not fixed in the body, w h i c h w i l l p r o v e u s e f u l t o u s l a t e r .
x'z
Figure 7.6
F i n a l l y , if all t h r e e p r i n c i p a l a x e s (x, y, z) t h r o u g h P h a v e t h e s a m e
c o r r e s p o n d i n g p r i n c i p a l m o m e n t of i n e r t i a I , t h e n every a x i s t h r o u g h P is
p r i n c i p a l w i t h p r i n c i p a l m o m e n t of i n e r t i a I . L e t b e t h e u n i t v e c t o r
a l o n g a n a r b i t r a r y a x i s x' t h r o u g h P i n t h i s c a s e o f t h r e e e q u a l p r i n c i p a l
m o m e n t s of i n e r t i a . A l s o l e t y' a n d z ' c o m p l e t e a n o r t h o g o n a l t r i a d w i t h
x', w i t h
and
b e i n g u n i t v e c t o r s i n t h e r e s p e c t i v e y' a n d z ' d i r e c t i o n s .
Thus
, (m , m , m ), a n d (n , n , n ) a r e t h e r e s p e c t i v e s e t s of
1
1
x
y
z
x
y
z
d i r e c t i o n c o s i n e s of x', y', a n d z ' w i t h r e s p e c t t o (x, y, z). E q u a t i o n ( 7 . 2 4 )
Page 468
t h e n gives
In the s a m e way, I
p
x'z'
= 0 since
a s w e l l . T h u s t h e a r b i t r a r y a x i s x'
t h r o u g h P i s p r i n c i p a l , a n d E q u a t i o n ( 7 . 2 0 ) s h o w s t h a t its m o m e n t of
inertia is also I :
1
E x a m p l e s of t h e p r e c e d i n g r e s u l t s r e g a r d i n g t w o a n d t h r e e e q u a l p r i n c i ­
p a l m o m e n t s of i n e r t i a a r e g i v e n i n t h e t a b l e o n t h e n e x t p a g e .
Maximum and Minimum Moments of Inertia
A n i m p o r t a n t p r o p e r t y of p r i n c i p a l m o m e n t s of i n e r t i a is t h a t t h e l a r g e s t
a n d s m a l l e s t of t h e s e a r e t h e l a r g e s t a n d s m a l l e s t m o m e n t s of i n e r t i a
a s s o c i a t e d w i t h any a x i s t h r o u g h t h e p o i n t i n q u e s t i o n . T o s h o w t h a t t h i s
is t r u e , l e t
b e t h e p r i n c i p a l m o m e n t s of i n e r t i a a t P a n d l e t t h e
c o r r e s p o n d i n g p r i n c i p a l a x e s b e x, y, a n d z. T h e m o m e n t of i n e r t i a a b o u t
s o m e o t h e r a x i s , x', is g i v e n b y E q u a t i o n ( 7 . 2 0 ) :
or
since
and
Thus
s o t h a t no l i n e t h r o u g h P h a s a s m a l l e r m o m e n t of i n e r t i a
t h a n t h e s m a l l e s t principal
m o m e n t of i n e r t i a . B y a s i m i l a r a r g u m e n t
which demonstrates that
so that n o line t h r o u g h
P h a s a l a r g e r a s s o c i a t e d m o m e n t of i n e r t i a t h a n t h e l a r g e s t
prin­
cipal m o m e n t of i n e r t i a . T h u s t h e l a r g e s t a n d s m a l l e s t m o m e n t s of i n e r t i a
a t a p o i n t P a r e f o u n d a m o n g t h e p r i n c i p a l m o m e n t s of i n e r t i a for P .
It m a y n o w b e s h o w n q u i t e e a s i l y t h a t t h e s m a l l e s t m o m e n t of i n e r t i a a t
t h e m a s s c e n t e r is t h e m i n i m u m I for any l i n e t h r o u g h any point
or
extended.
of
Page 4 6 9
Two Equal I's
Three Equal I's
Solid
cylinder
Solid sphere:
in any direction through C.
All axes through C in the shaded plane have this same
inertia and are principal. Note that I = mR /2 and is
generally not equal to / and l ; if, however,
then
every axis through C is principal with the same principal
m o m e n t of inertia!
C
2
ZZ
xx
yy
Solid cube:
in any direction through C. Therefore if line is a diagonal o1
the cube, then lg = ms /6 even though it would be
formidable t o obtain this result bv intearation. Note that the
direction cosines of are
since it makes equal angles with x, y, and z. Thus Equa­
tion (7.20) gives
2
Question 7.3
Write a one-sentence proof of this statement by using
the preceding results together with the parallel-axis theorem.
Answer 7.3 The moment of inertia about any line through any point P other than C is
larger than the moment of inertia about the line through C parallel to by the transfer
term md , and thus the smallest I at C is the smallest of all.
2
Page 470
Problems
•
Section 7.4
7.21 Find the principal axes a n d associated principal
m o m e n t s of inertia at O for t h e semicircular plate of mass
M a n d radius R s h o w n in Figure P 7 . 2 1 .
7.22 Find the principal axes a n d associated principal
m o m e n t s of inertia for the p l a n a r wire s h o w n in Fig­
ure P7.22, at t h e m a s s center.
7.23 Find the vector from O t o the m a s s center of t h e
b e n t bar in Example 7.5. Observe t h a t it d o e s n o t lie along
any of t h e three principal axes at O.
7.27 S h o w that if a line is a principal axis for t w o of its
points, t h e n it is a principal axis for t h e m a s s center.
7.21 S h o w that t h e three principal axes for a n y point
lying o n a principal axis for C are parallel to t h e principal
axes for C.
7.IS Find the principal m o m e n t s of inertia a n d their as­
sociated axes at O for t h e thin plate (Figure P7.29) in
terms of its density a n d thickness t.
P
P
P
• 7,31 Find t h e m o m e n t s of inertia I , I , a n d I for the
b o d y depicted i n Example 4.16. T h e n find t h e principal
m o m e n t s of inertia a n d corresponding principal axes at P.
xx
yy
zz
7.24 Use the definitions of the m o m e n t s of inertia to
prove that if a b o d y lies essentially in t h e xy p l a n e (that is,
it has very small dimensions normal t o it), t h e n
w h e r e P is any point in the plane, a n d t h a t z is a
principal axis at P .
• 7.31 In Problem 7.16 extend t h e p r o b l e m a n d find t h e
principal m o m e n t s of inertia, a n d their principal axes, at
p o i n t O.
7.25 S h o w t h a t if an axis through the m a s s center C of
body is principal at C, t h e n it is a principal axis for every
point on that axis. Hint: Use t h e transfer theorem for
products of inertia, together w i t h t h e orthogonality of
principal axes. (See Figure P7.25.) Transfer I a n d I to P !
• 7 . 3 2 Calculate t h e principal m o m e n t s of inertia at O, a n d
t h e direction cosines of their respective principal axes, for
b o d y in Figure P 7 . 3 2 . It is m a d e u p of three b e n t bars
w e l d e d together; all legs are either along, or parallel to,
the coordinate axes.
7.26 S h o w that if a principal axis for a point (such as P in
the preceding problem) passes t h r o u g h C, t h e n it is also
principal for C. (Same hint!)
• 7.33 Find the principal m o m e n t s of inertia a n d related
principal axes at t h e origin for t h e b o d y in Figure P 7 . 3 3 .
C
YZ
C
XZ
Figure P7.21
Figure P7.29
Figure P7.32
Figure P7.22
Figure P7.25
Figure P7.33
Page 4 7 1
• 7.34 In Example 7.5 find t h e principal m o m e n t s a n d
axes of inertia at t h e m a s s center C.
•
7.35 Find t h e principal m o m e n t s of inertia a n d their as­
sociated principal axes for t h e plate of Example 7.4 w h e n
b = 2H.
7.36 Find t h e principal m o m e n t s a n d axes of inertia if
the b o d y of Example 7.4 h a s d e p t h L instead of being a
thin plate. (See Figure P7.36.) Hint: T h e xy p l a n e is still
one of symmetry, so I = I = 0 again!
C
C
xz
• 7.39 Four slender bars, each of m a s s m a n d length are
w e l d e d together to form t h e b o d y s h o w n in Figure P7.39.
Find: (a) t h e inertia properties at t h e m a s s center C; (b) t h e
principal axes a n d principal m o m e n t s of inertia at C.
• 7.40 In Figure P7.40, t h e axis of symmetry, y , of t h e
disk is parallel to y; t h e p l a n e of t h e disk is parallel to xz.
Find t h e principal m o m e n t s of inertia of D at t h e origin O,
a n d for t h e smallest one, d e t e r m i n e t h e angles t h a t its
associated principal axis forms w i t h x, y, a n d z.
c
yz
* 7.37 For the h o m o g e n e o u s rectangular solid s h o w n in
Figure P7.37, find t h e smallest of t h e three angles b e ­
t w e e n line AB a n d t h e principal axes of inertia at A.
* 7.38 At t h e origin, find t h e principal m o m e n t s of inertia
a n d associated principal axes for a b o d y consisting of
three square plates w e l d e d along their edges as s h o w n in
Figure P7.38. (The axis of t h e smallest value of I s h o u l d
be t h e o n e t h a t t h e m a s s lies closest to in an overall sense.
Make this r o u g h check o n your solution.) Mass = 3m,
side = a.
•* 7.41 Figure P 7 . 4 1 s h o w s p a r t of a space station being
constructed in orbit. Find t h e principal axes a n d m o m e n t s
of inertia at C . T h e m o d u l e s h a v e 33-ft diameters, b u t
d u e to t h e material within t h e y are n o t hollow. For t h e
p u r p o s e s of this problem, treat each as a uniform hollow
shell w i t h a radius of gyration about its axis of 12 ft.
3
Figure P7.39
Figure P7.36
D (mass 100 kg)
Figure P7.40
Each of 3 modules; 16,000 lb length = 70 ft
;
Figure P7.37
Solar
panels:
6000 ft
and 8000 lb total
in x z plane
2
3
Figure P7.38
Figure P7.41
3
Page 4 7 2
7.5
The Moment Equation Governing Rotational Motion
The Euler Equations
In this section w e derive t h e t h r e e differential e q u a t i o n s g o v e r n i n g the
a n g u l a r m o t i o n of a r i g i d b o d y
T h e i r s o l u t i o n , w h i c h is difficult t o
o b t a i n i n c l o s e d f o r m i n m o s t c a s e s , y i e l d s t h e t h r e e c o m p o n e n t s of t h e
a n g u l a r v e l o c i t y of in a n i n e r t i a l f r a m e
We b e g i n w i t h E q u a t i o n (7.2),
n o t i n g t h a t t h e t i m e d e r i v a t i v e is t a k e n i n
However, the angular m o ­
m e n t u m H was most conveniently expressed "in"* body
in E q u a ­
tion (7.11). T h u s w e shall u s e E q u a t i o n (6.20) to m o v e t h e derivative in
(7.2) f r o m
to
:
c
(7.37)
W e n o w fix t h e a x e s (x, y, z) t o b o d y
so that, relative to
t h e inertia
p r o p e r t i e s a r e c o n s t a n t . U s i n g E q u a t i o n ( 7 . 1 1 ) , t h e first t e r m o n t h e right
s i d e of ( 7 . 3 7 ) is
(7.38)
w h e r e the unit vectors
a n d a r e r e s p e c t i v e l y p a r a l l e l t o x, y, a n d z, a n d
t h e r e f o r e n o w a r e fixed i n d i r e c t i o n i n . T h e s e c o n d t e r m i n E q u a ­
tion ( 7 . 3 7 ) , a f t e r c o m p u t i n g t h e c r o s s p r o d u c t , is
(7.39)
T h e s u m of E q u a t i o n s ( 7 . 3 8 ) a n d ( 7 . 3 9 ) y i e l d s t h e r i g h t s i d e of E q u a ­
t i o n ( 7 . 3 7 ) , w h i c h i n t u r n e q u a l s t h e m o m e n t a b o u t t h e m a s s c e n t e r C of
all t h e e x t e r n a l f o r c e s a n d c o u p l e s a c t i n g o n
It is c l e a r t h a t t h i s e q u a t i o n is e x t r e m e l y l e n g t h y a n d c o m p l i c a t e d . If
w e s e l e c t t h e b o d y - f i x e d a x e s (x, y, z) t o b e t h e principal a x e s t h r o u g h C,
h o w e v e r , t h e n all p r o d u c t - o f - i n e r t i a t e r m s v a n i s h a n d w e o b t a i n
and
so that, substituting i n t o E q u a t i o n (7.37) a n d e q u a t i n g t h e respective
* "Expressing a vector in a frame" simply means the vector is expressed in terms of unit
vectors fixed in that frame.
Page 4 7 3
coefficients of
and
we obtain the Euler equations:
(7.40)
W e n o t e that the Euler equations are nonlinear in t h e
components and
that the plane-motion equation
does n o t extend simply to
general motion.
It is v e r y i m p o r t a n t t o r e a l i z e t h a t , if a b o d y h a s a p i v o t ( p e r m a n e n t l y
fixed p o i n t ) , e q u a t i o n s a n a l o g o u s t o t h e p r e c e d i n g p e r t a i n . A n d i n f a c t
t h e s e a r e m e r e l y w h a t is o b t a i n e d b y s u b s t i t u t i n g O ( t h e p i v o t ) f o r C i n all
e q u a t i o n s f r o m (7.37) t h r o u g h (7.40).
W e c a n u s e E q u a t i o n s (7.40) to m a k e a n i m p o r t a n t o b s e r v a t i o n a b o u t
t h e s p e c i a l c a s e of " t o r q u e - f r e e " m o t i o n , m e a n i n g
. S u p p o s e at
an instant
w h e r e x, y, a n d z a r e p r i n c i p a l a x e s .
Then, with
E q u a t i o n s ( 7 . 4 0 ) tell u s t h a t
t h a t is, if t h e b o d y w e r e i n i t i a l l y t o b e s p u n a b o u t a p r i n c i p a l axis it w o u l d
c o n t i n u e t o s p i n a b o u t t h a t axis a n d a t c o n s t a n t r a t e . C o n v e r s e l y , if w e
s e e k c o n d i t i o n s for w h i c h
w e find f r o m E q u a ­
t i o n s ( 7 . 4 0 ) t h a t t w o of
and
m u s t v a n i s h . T h u s t h e s p i n will
p e r s i s t i n t h e a b s e n c e of e x t e r n a l m o m e n t if a n d o n l y if t h e axis of i n i t i a l
s p i n is a p r i n c i p a l a x i s . T h i s i n v e s t i g a t i o n is w h a t l e d E u l e r t o d i s c o v e r t h e
principal-axis concept in 1750.
Question 7.4
C a n a similar conclusion b e d r a w n for spinning a b o u t
a n axis t h r o u g h a pivot?
Use of Non-Principal Axes
S o m e t i m e s o t h e r f o r m s of m o m e n t e q u a t i o n s a r e m o r e a d v a n t a g e o u s t o
a p p l y i n p a r t i c u l a r p r o b l e m s t h a n E q u a t i o n s ( 7 . 4 0 ) . Firstly, w e m a y f i n d
it c o n v e n i e n t t o u s e r e f e r e n c e a x e s t h a t a r e b o d y - f i x e d b u t n o t p r i n c i p a l .
O f t e n it is l e s s t r o u b l e t o s i m p l y d e a l w i t h n o n z e r o p r o d u c t s of i n e r t i a a n d
a x e s t h a t a r e c o n v e n i e n t t o t h e b o d y (or its a n g u l a r v e l o c i t y ) t h a n t o
c o m p u t e principal directions a n d associated inertia properties. The com­
p o n e n t equations are formed b y combining Equations (7.37)-(7.39).
W h e n s p e c i a l i z e d t o (x, y) p l a n e m o t i o n , w e r e c o v e r t h e f o l l o w i n g e q u a ­
tions d e v e l o p e d in C h a p t e r 4:
(7.41)
Answer 7.4
Yes, provided there is n o net m o m e n t about the pivot.
Page 4 7 4
Use of an Intermediate Frame
S e c o n d l y , w e o f t e n f i n d it c o n v e n i e n t t o e x p r e s s e x t e r n a l m o m e n t s
a n d / o r a n g u l a r m o m e n t u m i n t e r m s of c o m p o n e n t s a s s o c i a t e d w i t h
d i r e c t i o n s fixed n e i t h e r i n t h e b o d y n o r i n t h e i n e r t i a l f r a m e . T h a t is, w e
m a y choose to involve a n intermediate frame, say
and use
(7.42)
or, w h e n t h e r e is a p i v o t O ,
(7.43)
T h i s a p p r o a c h is p a r t i c u l a r l y u s e f u l w h e n , u s u a l l y b e c a u s e of s y m m e ­
t r i e s , m o m e n t s a n d p r o d u c t s of i n e r t i a of t h e b o d y r e m a i n c o n s t a n t r e l a ­
t i v e t o a x e s fixed i n t h e i n t e r m e d i a t e f r a m e .
W e c l o s e t h i s s e c t i o n w i t h five e x a m p l e s , t h e first e m p l o y i n g E u l e r ' s
e q u a t i o n s t o s t u d y t o r q u e - f r e e m o t i o n w h e n t h e i n i t i a l s p i n is n o t a b o u t a
principal axis. I n t h e s e c o n d w e u s e body-fixed axes t h a t a r e n o t principal
i n a p r a c t i c a l p r o b l e m of a s a t e l l i t e d i s h a n t e n n a . T h e final t h r e e e x a m p l e s
i l l u s t r a t e t h e u s e of i n t e r m e d i a t e f r a m e s of r e f e r e n c e .
EXAMPLE 7 . 6
A satellite* is m o v i n g t h r o u g h d e e p space far from t h e influence of atmospheric
d r a g a n d gravity. (See Figure E7.6.) If t h e z axis is o n e of s y m m e t r y a n d if at some
instant called t = 0 w e h a v e
along the body-fixed mass-center
axes,
find
. A s s u m e t h e satellite to b e a rigid b o d y .
Solution
The Euler equations, if I
C
xx
= I
C
yy
= I and I
C
zz
= J, are
(1)
Figure E7.6
(2)
(3)
in w h i c h the m o m e n t c o m p o n e n t s are zero in the absence of external forces a n d
couples. Equation (3) gives
(4)
so t h a t Equations (1) a n d (2) b e c o m e linear a n d are:
(5)
(6)
Differentiating Equation (6) a n d solving for
w e get
(7)
* Such as the Voyager spacecraft, which left our solar system in 1983.
Page 4 7 5
This expression m a y b e substituted into (5) to yield a n equation free of :
or
in w h i c h
Since
. The solution to this equation is harmonic:
at t = 0, w e see t h a t
Finally, Equation (6) gives
or
T h e initial condition for
gives us
so that t h e other t w o c o m p o n e n t s (besides
of
are
EXAMPLE 7 . 7
For reasons of interference with other bodies, an a n t e n n a w a s recently designed
a n d built w i t h a n offset axis as s h o w n in Figure E7.7a a n d Figure E7.7b. The
a n t e n n a is c o m p o s e d of a 12-ft, 1200-lb parabolic reflector« a (ourrteroeteht W,
a reflector s u p p o r t structure S, a n d a positioner. The positioner consists of (1) a
pedestal P t h a t is fixed to the (inertial) reference frame; (2) a n azimuth bearing at
Counterweight
Figure E7.7a
Back View
Figure E7.7b
Side View
(x, y, z) are
fixed in S
Page 4 7 6
O a n d ring gear b y m e a n s of w h i c h t h e a r m A is m a d e to rotate about t h e vertical;
a n d (3) a n elevation torque m o t o r at E t h a t rotates t h e s u p p o r t structure S with
respect to t h e a r m A. Rotations of t h e reflector consist of an a z i m u t h angle
a b o u t Z, a n d an elevation angle
a b o u t x.
a. Find t h e weight W of t h e counterweight W.
b. Find t h e inertia properties at point O of t h e (assumed rigid) b o d y
c o m p o s e d of R +
W+S.
c. Write t h e equations of rotational m o t i o n of
d. D e t e r m i n e t h e n e t m o m e n t s t h a t m u s t be exerted about axes t h r o u g h
O for t h e position s h o w n if first (in units of r a d i a n s a n d seconds)
and then
Solution
Let us m o d e l t h e a n t e n n a as follows. T h e reflector is treated as a thin disk, t h e
counterweight as a point mass, a n d t h e s u p p o r t structure as being rigid b u t light.
Axes (X, Y, Z) are fixed in t h e inertial frame w h i l e (x, y, z) are attached to .
In p a r t (a) t h e p u r p o s e of t h e counterweight is to place t h e m a s s center of o n
its elevation axis. T h u s
1200(4) = W(2)
W = 2400 lb
For part (b) w e generate t h e inertia properties of a b o u t t h e point O. (See the
table o n t h e next page.)
Therefore t h e inertia matrix m a y be written as
slug-ft
2
(1)
For part (c), o n e w a y to proceed is to c o m p u t e t h e principal axes a n d m o ­
m e n t s of inertia from this matrix a n d t h e n use the Euler equations (7.40). This
w o u l d b e a very u n w i s e a p p r o a c h in this case, h o w e v e r . N o t only is it tedious to
locate t h e principal axes, b u t following this w e w o u l d h a v e to break u p the
angular velocity into its c o m p o n e n t s along these directions; w e w o u l d t h e n ob­
tain not-so-useful m o m e n t c o m p o n e n t s a b o u t axes s k e w e d with respect to t h e
rotation axes. It is m u c h simpler to use t h e second of Euler's laws:*
T h e only prices w e will h a v e to p a y are (1) to retain a n d deal with t h e n o n z e r o
product of inertia I a n d (2) to m o v e t h e derivative from frame to . T h e angular
m o m e n t u m of a b o u t O is, for our problem,
O
xy
(2)
• All that is required of O for this equation to be valid is that it be a point of the inertial
frame; in what follows, however, it also needs to be, and is, a pivot of
Page 4 7 7
Reflector
+
Counterweight
O
=
Total
O
* The y axis through O is an axis of symmetry of R; hence I = 0 = I . And the third
product of inertia of R, I , vanishes because zy is a plane of symmetry for body R.
xy
yz
O
xz
T h e angular velocity of in frame is found by the addition t h e o r e m . The
reflector a n d its supporting structure rotate in elevation with a simple angular
velocity
with respect to t h e h o u s i n g A; likewise, A rotates in a z i m u t h with a
simple angular velocity
with respect to t h e pedestal (which is rigidly fixed to
t h e reference frame ). Therefore
(3)
Page 4 7 8
Substituting t h e c o m p o n e n t s into Equation (2) gives t h e angular m o m e n t u m of
i n , expressed in (terms of its c o m p o n e n t s in)
(4)
Next w e use Equation (6.20) to differentiate H
a n d . ):
0
(note t h a t point O is fixed in
(5)
Taking t h e derivative a n d performing t h e cross product, w e find t h a t t h e three
c o m p o n e n t equations are as follows. (Note that t h e inertia properties d o not
c h a n g e in )
(6a)
(6b)
(6c)
As a n indication of t h e increased difficulty of three-dimensional d y n a m i c s p r o b ­
lems, n o t e t h a t all four of t h e n o n v a n i s h i n g inertia properties contribute to each
component of t h e external m o m e n t acting o n at O!
In t h e indicated position,
. Therefore
(7a)
(7b)
(7c)
In part (d), for t h e case specified,
while
Also,
while
. This case physically corresponds to t h e a n t e n n a , at 45 ° elevation,
swinging a r o u n d t h e vertical at 30 ° / s e c a n d s u d d e n l y sensing a n object traveling
t o w a r d zenith; t h e controls activate a m o t o r w h o s e torque produces a n angular
acceleration t h a t will s e n d t h e a n t e n n a u p w a r d in elevation. T h e angular acceler­
ations are large because t h e n e e d is to get t h e r e quickly.
Substituting these values of
and
into Equations (7), along w i t h
t h e inertia values, gives o u r answer:
Page 4 7 9
(8)
These are t h e m o m e n t s exerted b y P o n t o A, excluding t h a t required to balance
the d e a d weight of t h e a n t e n n a . In t h e inertial reference frame, t h e m o m e n t s are
(9)
In the opposite case w h e n t h e a n t e n n a is tracking in elevation, say
r a d / s e c , a n d receives a s u d d e n c o m m a n d resulting in
, t h e m o m e n t c o m p o n e n t s b e c o m e (here
" at
(10)
O n c e again w e see t h e considerable effect of the product of inertia term.
T h e negatives of the X, Y, Z c o m p o n e n t s respectively b e n d , b e n d , a n d twist
the pedestal a n d are considerations in its design; far larger a n d m o r e i m p o r t a n t
m o m e n t s , h o w e v e r , arise from the w i n d blowing against the " d i s h " a n d from
gravity. T h e r e are also forces exerted on A at O d u e to gravity a n d the m a s s center
acceleration.
EXAMPLE 7 . 8
Figure E7.8a
A symmetric wheel D spins at angular speed
a b o u t its axis, which is a line fixed
in b o d y , as well as in D. (See Figure E7.8a.) The m o m e n t s of inertia of D at C are
I = J and I = I = I. Body h a s negligible m a s s a n d rotates at angular speed
C
C
yy
xx
C
zz
Page 480
about t h e x axis at O. If a m o m e n t M is applied to parallel to t h e x axis, find,
at t h e instant s h o w n , t h e rates of c h a n g e of
and
a n d t h e force a n d m o m e n t
exerted o n D at C b y t h e pin. Neglect friction.
0
c
Solution
Let u s d e n o t e t h e inertial reference frame (in w h i c h moves) b y
t h e o r e m for angular velocity t h e n gives, for t h e w h e e l ,
T h e addition
w h e r e t h e axes (x , y , z ) are fixed in a n d their associated unit vectors
are fixed in direction in
N o t e that w e m a y use t h e equation
c
c
c
for t h e angular m o m e n t u m of D because even t h o u g h (x , y , z ) are n o t fixed in
b o d y D, t h e y are nonetheless p e r m a n e n t l y principal. Therefore
c
c
c
T h e second l a w of Euler t h e n yields
Figure E7.8b
w h e r e for convenience w e differentiate H in frame since t h e vector h a s b e e n
written in terms of its c o m p o n e n t s there. Continuing,
c
From Figure E7.8b, a free-body d i a g r a m of D, w e get t h e c o m p o n e n t s of
so t h a t
T h u s w e see t h a t
(1)
(2)
(3)
In addition,
for t h e disk yields
so t h a t
(4)
(5)
(6)
have
T u r n i n g n o w to Figure E7.8c, a free-body d i a g r a m of t h e light b o d y
so that
we
(7)
C o m b i n i n g Equations (1), (5), a n d (7),
Figure E7.8c
(8)
Page 4 8 1
and
(9)
and
(10)
The results (4), (6), (8), (9), a n d (10) are recognizable as w h a t w o u l d h a v e b e e n
obtained h a d t h e disk b e e n frozen in its bearings, in w h i c h case a n d D w o u l d
constitute a single rigid b o d y in plane motion.
Equation (3), h o w e v e r , does n o t follow intuitively from the s t u d y of p l a n e
motion. The term
is sometimes called a gyroscopic m o m e n t , a n d t h e e q u a ­
tion says t h a t a m o m e n t of this m a g n i t u d e m u s t act o n D a b o u t z if t h e given
m o t i o n is to occur. N o t e t h a t in this case t h e b o d y D is not allowed to t u r n about z
as it spins (about y ) . If it were, say b y m e a n s of a bearing b e t w e e n C a n d O, t h e n
the m o m e n t c o m p o n e n t M w o u l d b e c o m e zero, a n d a third
component
(about z ) w o u l d appear. N o t e also from Figure E7.8c t h a t the gyroscopic m o ­
m e n t twists the shaft of This, for example, is a consideration in t h e retraction of
the wheels of some airplanes.
c
c
c
PZ
c
EXAMPLE 7 . 9
The thin disk D t u r n s o n the light arm b y w a y of a s m o o t h bearing t h a t keeps
the axes of D a n d
aligned. The a r m is h i n g e d to a shaft driven at constant
angular speed
by a motor. The system is set u p so t h a t the a r m is horizontal
w h e n the disk contacts t h e g r o u n d as s h o w n in Figure E7.9a. A s s u m i n g t h e disk,
of m a s s m, to roll o n t h e g r o u n d , find t h e forces exerted by the g r o u n d o n the disk,
the forces a n d m o m e n t s exerted b y the a r m on t h e disk, a n d the forces a n d
m o m e n t s exerted b y t h e shaft on the arm.
Solution
We shall begin the analysis b y applying to the disk the equations of motion:
Figure E7.9a
Page 482
and
Because C is o n t h e axis of t h e a r m w h i c h t u r n s at constant rate
w e k n o w that
w h e r e w e are using t h e axes a n d unit vectors s h o w n in Figure E7.9a a n d these are
fixed in the arm
To obtain a n expression for H w e n e e d to d e t e r m i n e
So,
using t h e addition t h e o r e m ,
c
w h e r e t h e only m o t i o n D can h a v e in is to t u r n (in simple angular velocity)
a b o u t their c o m m o n axis.
Because t h e disk rolls, its point E in contact w i t h t h e g r o u n d h a s zero veloc­
ity; using Equation (6.56), w e obtain
or
and
Therefore, since w e a s s u m e t h e disk to b e relatively thin so t h a t to good approxi­
mation
a n d since
N o t e that since
a n d so
we have
is constant,
Page 483
N o w using Figure E7.9b, t h e free-body d i a g r a m of t h e disk, t h e equation
yields
(1)
(2)
(3)
a n d t h e equation
likewise yields
Figure E7.9b
(4)
(5)
(6)
Using Figure E7.9c, t h e free-body d i a g r a m of t h e massless bar, w e h a v e
(7)
Figure E7.9c
We shall delay using
and
w h i c h will ultimately
yield
R , R ,R ,T a n d T , a n d focus o n Equations ( l ) - ( 7 ) from w h i c h w e obtain
x
y
z
y
z
Page 484
which is as far as w e can go because this problem is actually "dynamically"
indeterminate. However, if w e assume friction to be small (just enough to pro­
duce rolling as this system is slowly brought up to speed) so that
, then w e
have M ,B , B and N uniquely determined. With that condition w e may n o w use
the remaining "equilibrium" equations for the light rod to obtain
x
y
z
Let us make a couple of observations: first, note that T = 0 means that the motor
doesn't have to supply any driving torque in order to maintain the constant
— remember w e have smooth bearings; and secondly note that if is large, so
is N:
y
and that extra part of N, over and above the weight mg, is often said to be due to
gyroscopic action.
EXAMPLE 7 . 1 0
It is possible for the homogeneous cone of base radius R in the Figure E7.10a to
roll steadily around on a flat horizontal table
in such a way that its uncon­
strained vertex O remains fixed and the center point Q of its base travels on a
horizontal circle at constant speed. Let this motion be begun by forces which are
then released. Assume that there is sufficient friction between the cone and the
table to prevent slipping. Besides having enough friction, there is yet another
special condition that must be satisfied in order that the motion occur. Find the
friction and normal force resultants, their lines of action, and the special condi­
tion. Refer to Example 6.12 for some related kinematics.
Figure E7.10a
Solution
Since the mass center C moves on a horizontal circle at the constant speed
Figure E7.10b), the friction force / is given by:
(see
Page 4 8 5
or
Figure E7.10b
since
. T h e n o r m a l force N is simply mg since a = 0. There is n o
friction force c o m p o n e n t n o r m a l to t h e p l a n e of t h e paper, since a also vanishes.
T h e directed line of action of f is as s h o w n above, along AO. For t h e line of action
of N, w e u s e t h e m o m e n t equation of motion; n o t e t h a t
is our inertial frame:
cz
cy
w h e r e frame
contains t h e Z axis a n d t h e axis of t h e cone. N o w t h e angular
m o m e n t u m H is, since
are in principal directions at t h e fixed p o i n t O:
o
In t h e preceding equation, t h e
t h e previous examples,
Noting t h a t
-components are those of t h e b o d y , in
and, substituting for
J,
a n d I,
, w e find
a n d from t h e free-body diagram, using N = mg,
. From
Page 4 8 6
or
w h i c h gives t h e line of action of t h e n o r m a l force resultant.
The "special c o n d i t i o n " m e n t i o n e d i n t h e p r o b l e m s t a t e m e n t is t h a t the
n o r m a l force m u s t intersect t h e p l a n e at a point of physical contact w i t h the cone,
i.e.:
Thus
w h i c h w h e n simplified is
Holding
constant, w e
small) is t h a t the n o r m a l
physically h a p p e n , t h e
another, t h a t of tipping
PROBLEMS
•
see that w h a t h a p p e n s if v is too large (or g or R too
force n e e d s to act b e y o n d the point A. Since t h a t cannot
specified m o t i o n will n o t occur b u t will give w a y to
outward.
Q
Section 7.5
7.42 The r o d is rigidly attached to shaft S w h i c h is free
to turn in the t w o bearings as indicated in Figure P7.42.
The y a n d y' axes point into the p a g e at C. S h o w t h a t
the m o m e n t w i t h respect to C t h a t m u s t b e supplied
by the bearings to shaft S to sustain t h e m o t i o n m u s t
h a v e the c o m p o n e n t s :
D o this in t w o w a y s : (1) Use the Euler equations (7.40)
with the principal axes (x, y, z); (2) u s e t h e Equa­
tions (7.41) w i t h the axes (x', y', z') in t h e figure.
Figure P7.42
7.43 Find the reactions exerted b y t h e bearings o n the
shaft to w h i c h a 30-kg thin plate p is w e l d e d . (See Fig­
u r e P7.43.) T h e assembly is turning at t h e constant angu­
lar speed of 30 r a d / s . Work the p r o b l e m b y using
that is, b y expressing H using principal directions in p
(which omits t h e n e e d for c o m p u t i n g t h e n o n z e r o product
of inertia I ).
c
xz
Figure P7.43
Page 4 8 7
Figure P7.45
7.44 Rework a n d check t h e results of t h e preceding
p r o b l e m b y using Equations (7.24) to calculate I ; t h e n
use Equations (7.41) to obtain t h e bearing reactions.
xz
7.45 T h e equilateral triangular plate
w a s to be
m o u n t e d o n t o t h e rotating shaft as s h o w n in Figure P7.45
in t h e solid figure. T h e installation resulted in t h e mis­
alignment angle (see t h e d a s h e d position). T h e plate h a s
m a s s 2 slugs, a n d t h e shaft is light. Determine t h e d y ­
n a m i c bearing reactions at
a n d B caused b y t h e mis­
alignment, in terms of , by: (a) using principal axes; (b)
w i t h o u t using principal axes (i.e., calculate a n d utilize t h e
p r o d u c t s of inertia).
2
7.46 A slender r o d A a n d a small ball each w e i g h 0.3 lb.
(See Figure P7.46.) T h e bodies rotate about t h e vertical
along w i t h t h e slender shaft S a n d are s u p p o r t e d b y a
s m o o t h step or thrust bearing at D a n d b y t h e cord C. Find
t h e tension in the cord if t h e angular speed of the system
is 20 r a d / s e c .
7.47 T h e disk (mass m, radius r) in Figure P7.47 is rigidly
attached to t h e shaft, a n d t h e assembly is s p u n u p to
angular speed
a b o u t t h e z axis. Determine t h e bearing
reactions at A a n d B in terms of m, r, g,
a n d L.
c
7.48 A bicycle w h e e l weighing 5 lb a n d h a v i n g a 14-in.
radius is misaligned b y 1° w i t h t h e vertical. T h e t o p of t h e
w h e e l tilts t o w a r d t h e right w i t h t h e bike m o v i n g for­
w a r d . If t h e bike is driven along a straight p a t h at 15 m p h ,
find t h e m o m e n t
exerted o n t h e w h e e l . Use t h e re­
sult of P r o b l e m 7.11, neglecting t h e spokes a n d h u b .
7.49 In t h e preceding problem, s u p p o s e instead t h a t t h e
bicycle is in a 50-ft-radius t u r n to t h e (a) right a n d (b) left.
Determine t h e n e w values of t h e m o m e n t exerted o n t h e
misaligned w h e e l . Neglect t h e lean angle.
Figure P7.46
Figure P7.47
Page 488
7.50 Writing Euler's l a w s as
and
versed effective force (— m a ) a n d t h e inertia torque
If these quantities are a d d e d to the free-body diagram, t h e
object m a y b e treated as t h o u g h it w e r e in equilibrium.
For a car at t h e instant of o v e r t u r n i n g while traveling at
speed v on a curve of r a d i u s R a n d b a n k angle , s u c h a
d i a g r a m w o u l d a p p e a r as s h o w n in Figure P7.50.
c
observed b y S. C h a n d l e r in 1 8 9 1 . T h e difference is attrib­
results in w h a t is k n o w n as t h e re­
u t e d to t h e nonrigidity of t h e earth. A l t h o u g h e n e r g y dis­
sipation s h o u l d d a m p o u t this " w o b b l e , " in fact it d o e s
not. The ongoing cause of t h e w o b b l e is a n u n s o l v e d
p r o b l e m in g e o d y n a m i c s at this time. See Science News, 24
October 1981.)
7.56 A screwdriver-like m o t i o n b e t w e e n t h e p l a n a r case
a n d general (three-dimensional) m o t i o n is defined as fol­
lows . All points of t h e b o d y h a v e , at a n y time, identical z
c o m p o n e n t s of velocity in a reference frame F. T h e unit
vector of this
c o m p o n e n t is constant in b o t h a n d F,
t h o u g h can vary w i t h time. T h u s t h e a n g u l a r velocity
vector is still expressible as
Derive a m o m e n t
equation for
t h a t is valid for this motion.
7.57 The disk in Figure P 7 . 5 7 is s p i n n i n g a b o u t t h e light
axle, a n d t h e axle is precessing at t h e constant rate of
r a d / s . If t h e axle is o b s e r v e d to r e m a i n horizontal, find
t h e m a g n i t u d e a n d direction of t h e spin of t h e disk.
Figure P7.50
Calculate t h e m o m e n t a b o u t t h e p o i n t Q a n d , n o t i n g
that
for this free-body diagram, c o m p a r e t h e
effects of t h e reversed effective force (or inertia force)
— ma w i t h that of t h e inertia t o r q u e —
o n t h e over­
turning tendency. W h a t factors m a k e t h e car m o r e likely
to turn over?
c
7.51 S u p p o s e that t h e c o m p o n e n t s of t h e m a s s center
velocity v are written in b o d y instead of in a n inertial
frame
U s e t h e p r o p e r t y (6.20) of
to derive the
scalar e q u a t i o n s of m o t i o n of t h e m a s s center from Euler's
first law:
c
7.52 S h o w that if a rigid b o d y u n d e r g o i n g torque-free
m o t i o n in a n inertial frame
h a s t h r e e equal principal
m o m e n t s of inertia at its m a s s center, t h e n its a n g u l a r
velocity is constant in
7.53 In Example 7.8 define t h e (x, y, z) axes at O as prin­
cipal for
w i t h associated respective m o m e n t s of inertia
a n d . Rework t h e p r o b l e m w i t h o u t a s s u m i n g t h a t
h a s negligible mass. T h e t w o sets of axes are respectively
parallel prior to t h e application of M .
Figure P7.57
7.58 A single-engine aircraft h a s a four-bladed propeller
weighing 128 lb w i t h a radius of gyration a b o u t its center
of m a s s of 3 ft. It rotates counterclockwise at 2000 r p m
w h e n v i e w e d from t h e rear. Find t h e gyroscopic m o m e n t
on t h e propeller shaft w h e n t h e p l a n e is at t h e b o t t o m of a
vertical loop of 2000-ft r a d i u s w i t h a speed of 500 m p h . In
w h i c h direction will t h e tail of t h e p l a n e t e n d to m o v e
because of this m o m e n t ?
O
7.54 S h o w t h a t if t h e solution
of Example 7.6 is
projected into t h e xy p l a n e , t h e tip of t h e projection vector
travels o n a circle of r a d i u s
at t h e frequency
7.55 A result of t h e e a r t h ' s bulge is t h a t I/J = 0.997.
Use this result to c o m p u t e t h e period of o n e revolution of
the earth's a n g u l a r velocity vector ( N o r t h Pole!) a b o u t its
axis of s y m m e t r y . (The a n s w e r , o b t a i n e d b y Euler in
1752, is a b o u t 4 m o n t h s less t h a n t h e actual p e r i o d first
7.59 T h e blades of a fan t u r n at 1750 r p m , a n d t h e fan
oscillates a b o u t t h e vertical axis z (Figure P7.59(a)) at the
rate of o n e cycle every 10 sec. A s s u m i n g t h a t t h e fan
travels at t h e constant a n g u l a r velocity of 0.2 r a d / s e c
except w h e n it is reversing direction (Figure P7.59(b)),
find t h e m o m e n t exerted by t h e b a s e o n t h e a r m section A
at t h e -cycle p o i n t d u e to gyroscopic action. For t h e cal­
culation (only!) consider the blades (Figure P7.59(c)) to be
4-in.-diameter circular a l u m i n u m plates all in t h e s a m e
plane and
in. thick. Use a density of 0.1 l b / i n . .
3
Page 489
7.64 A disk D rolls a r o u n d in a circle w i t h its p l a n e verti­
cal a n d its center traveling at constant s p e e d V . Find t h e
tension in the string, a n d t h e friction force exerted o n D
by t h e floor. (See Figure P7.64.)
C
7.65 There is a relationship a m o n g V , g, r, R a n d such
that t h e disk can roll a r o u n d in a circle as s h o w n in Fig­
ures P7.65(a) a n d (b), w i t h V a n d r e m a i n i n g constant.
Find this relationship.
C
C
(a)
(c)
(b)
Figure P7.59
Figure P7.61
Figure P7.60
Figure P7.64
7.60 Disk D in Figure P7.60 turns in bearings at C at
angular rate
a b o u t t h e light r o d R, a n d b o t h precess
about axis z at angular rate
as s h o w n . S h o w w i t h a
free-body diagram h o w it is possible for t h e m a s s center C
to remain in a horizontal plane. T h e n find t h e reactions
exerted o n t o R b y t h e socket at O. Is there a n y difference
in t h e solution if D a n d R are rigidly connected?
a
7.61 In t h e grinding mill of Problem 7.5, s u p p o s e that
the wall is absent. (See Figure P7.61.) Find, for a given
(constant angular speed of S ) , t h e angle that t h e axis of
the grinder D will m a k e w i t h t h e vertical. Observe that
with t h e wall present a n d fixed, larger s p e e d s t h a n this
will allow t h e grinder to work. In particular, s h o w that
the following set of p a r a m e t e r s is satisfactory: r = 2.5 ft,
rad/sec, and
. Neglect t h e m a s s
of b o d y in comparison w i t h t h e h e a v y grinding disk D.
(a)
7.62 Find t h e grinding force N p r o d u c e d at t h e wall of
the grinding mill of P r o b l e m s 7.5 a n d 7.61 for t h e given
parameters.
7.63 Find t h e m a g n i t u d e a n d direction of t h e force
a n d / o r couple exerted o n disk D b y t h e shaft S in P r o b ­
lem 7.4.
(b)
Figure P7.65
Page 490
7.66 Obtain the results of Example 7.9 by using the
Euler equations (7.42). Hint: This time the axes are bodyfixed, and the
part of
changes direction in D; there­
fore the components
and
have derivatives that
were formerly zero in
The
component of
is
where
is the angle of roll as s h o w n in Figure P7.66.
Differentiate this expression, substitute
and
b. If the ship o n a straight course in rough seas
pitches sinusoidally at
amplitude with a
6-s period, what are the maximum bearing re­
actions then?
A heavy disk D of mass m and radius r spins at the
angular rate
with respect to the rigid, but
light, bent bar
(See Figure P7.68.) Body turns at rate
about a vertical axis through O, a point of
then
take
Finally,
to Equations (7.42) and
both and
the inertial frame G.
Find the force
and go
couple
stitute your results.
7.67 A ship's turbine has a mass of 2500 kg and a radius
of gyration about its axis (y in Figure P7.67) of 0.45 m .
It is mounted on bearings as indicated and turns at
5000 rpm clockwise w h e n viewed from the stern (rear)
of the boat.
c
a. If the ship is in a steady turn to the right of
radius 500 m and is traveling at 15 knots, what
are the reactions exerted on the shaft by the
bearings? (1 knot = 1.15 m p h = 1.85 k m / h r )
Figure P7.68
Figure P7.66
(a)
(b)
Figure P7.67
Figure P7.69
Page 491
HKIU, length
that m u s t b e acting o n at O to p r o d u c e a m o t i o n of t h e
system for w h i c h
and
are constants. Both sets of
axes in t h e figure are fixed in
a n d n o t e t h a t (X , Y , Z )
are always principal axes for D at C e v e n t h o u g h t h e y are
not fixed in D.
C
C
mass m,
radius R
C
7.69 C o m p u t e t h e m o m e n t M applied to the shaft S in
Figures P7.69(a) a n d (b), o n t h e preceding page, as a
function of t h e angle if
and
are constants.
7.70 A bike rider enters a turn of radius R at a constant
speed of v . (See Figure P7.70.) O t h e r quantities are d e ­
fined below:
c
Figure
P7.71
r=radiusof w h e e l
d=distanceb e t w e e n axle a n d C
I , I = principal m o m e n t s of inertia of entire bike
plus rider w i t h respect to
and
directions t h r o u g h C
1
2
i = m o m e n t of inertia, w i t h respect to
direction, of o n e w h e e l a b o u t its axis
of s y m m e t r y
m = total m a s s
7.72 In P r o b l e m 6.24, if
and
are constants,
find t h e resultant m o m e n t exerted o n D at its m a s s center,
w h e n is pointing straight u p . T h e b o d y D is symmetri­
cal, w i t h centroidal principal m o m e n t s of inertia J along
a n d I n o r m a l to
• 7.73 T h e b o d y in Figure P 7 . 7 3 is a n ellipsoid of revo­
lution w i t h m a s s = 1 slug, a n d semiminor a n d semimajor
axis lengths a a n d 2a, w i t h a = 1 ft. T h u s
= angle s h o w n
Solve for t h e resultant force
and moment
in
terms of these quantities. C o m p a r e t h e effects of t h e
D'Alembert force (— m a ) a n d t h e inertia torque
in
righting t h e bike w h e n is small. Neglect t h e p r o d u c t s of
inertia.
and
c
T h e s h a d e d light frame A is driven a r o u n d the fixed post
p , w i t h angular velocity
b y a m o t o r torque T applied
at P. A n o t h e r m o t o r (neither is s h o w n ) b e t w e e n A a n d
applies a torque
which causes t h e b o d y to spin in the
frame. The axes a n d unit vectors s h o w n are fixed in A.
During a n interval of motion,
rad/sec and
2
= 2t r a d / s e c . Find all forces a n d couples applied onto A
at P w h e n t = 1 sec. T h e distance from P to t h e x axis is
2 ft.
Figure P7.70
• 7.71 Disk D in Figure P7.71 turns in bearings a r o u n d
r o d R as it rolls o n t h e g r o u n d . Find all t h e forces acting o n
D, a n d contrast t h e solution w i t h t h a t of Example 7.9.
The rod is free to slide along the vertical post about w h i c h
it turns at constant rate.
Figure P7.73
Page 492
made to turn relative to arm about the axis of
and
is made to turn with respect to the ground (inertial frame)
about the vertical through point O. At the given instant,
the angular velocity of D in
expressed in terms of unit
vectors
fixed in
is
In addition, at this instant
w e are given
and
For the disk, m = 10 slugs, I = 1.4 slug-ft , and I = 0.7
slug-ft . Find all forces and couples that are acting on
C
2
xx
C
yy
2
• 7 . 7 5 In the preceding problem, use
and their
derivatives to compute the right-hand sides of the Euler
equations (7.40). Explain w h y these results are not the
components of the moments of external forces acting on
D at C.
Figure P7.74
7.74 By means of appropriately mounted control motors
(not shown), the disk D in Figure P7.74 is made to turn
about its axis of symmetry relative to the arm
is
7.6
• 7 . 7 6 In the preceding two problems, at the given in­
stant find
and
for i = 1, 2, and 3, where
is the
angle of rotation of D with respect to
is the angle
of rotation of
with respect to , and
is the angle of
rotation of
with respect to
Gyroscopes
W e n o w return our attention to the gyroscope w h o s e orientation w a s
e x a m i n e d i n S e c t i o n 6 . 8 . First w e s h a l l d e r i v e t h e e q u a t i o n s of r o t a t i o n a l
m o t i o n of s u c h a g y r o s c o p e G.* W e b e g i n b y e x p r e s s i n g its a n g u l a r v e l o c ­
ity
i n t h e f r a m e F ; s e e F i g u r e s 7.7, 7.8 a n d 7.9, r e p e a t e d f r o m
2
Section 6.8.
(7.44)
Figure 7.7
First rotation.
Figure 7.8
Second rotation.
Figure 7.9
Third rotation.
* We are taking G to be the rotor and are considering it as heavy with respect to the
inner and outer gimbals, w h o s e mass w e then neglect. We also assume Gtobe symmet­
ric about its axis. Of course, a gyroscope does not have to possess any gimbals; the earth
is a massive gyro, as will be seen in an example to follow.
Page 493
The axes (x ,y ,z ) of the inner gimbal are not fixed inGbecause of its
2
spin
2
2
— but they are nonetheless
permanently
principal.
This important
fact allows u s to write the angular momentum of the gyroscope in the
frame f as
2
(7.45)
Euler's second law, together with Property (6.20) of the angular v e ­
locity vector, then gives the equations of motion of the gyroscope as
follows:
(7.46)
In the preceding calculations w e have used the addition theorem to
observe the following:
(7.47)
so that
is the same vector as in Equation (7.44) if
equations of motion of G are therefore
is omitted. The
(7.48)
where
is the component of
about the spin axis of symmetry of the
gyroscope. Note that it is made u p of part of the precession speed as well
as all of the spin.
The gyroscope equations are seen to be nonlinear, including not only
products of the angles' derivatives but also trigonometric functions of
them. Their general solution is an unsolved problem; however, there are
two special solutions that are quite worthy of study. The first of these is
steady precession; the second is torque-free motion. We shall have a look
at each in turn.
Steady Precession
Steady precession is defined by the nutation angle
the precession
speed
and the spin speed each being constant throughout the mo­
tion. Let us call these constants
and
and substitute them into
Equation (7.48) to obtain:
(7.49)
Page 494
W e s e e t h a t o n l y a m o m e n t a b o u t t h e y axis is n e e d e d t o s u s t a i n s t e a d y
precession. Also
means that
is a c o n s t a n t . *
In t h e case in w h i c h
, w e h a v e t h e precession a n d spin axes
o r t h o g o n a l ; t h e s i t u a t i o n is s h o w n i n F i g u r e 7 . 1 0 . If t h e g y r o is s p i n n i n g
a n d t h e t o r q u e is a p p l i e d , t h e r e w i l l s i m u l t a n e o u s l y o c c u r a p r e c e s s i o n
t h a t t e n d s t o t u r n t h e s p i n v e c t o r t o w a r d t h e t o r q u e v e c t o r . T h i s is s o m e ­
t i m e s c a l l e d t h e l a w of g y r o s c o p i c p r e c e s s i o n . T h e t o r q u e i n t h i s c a s e ,
2
f r o m E q u a t i o n ( 7 . 4 9 ) , is
(7.50)
a n d it is s e e n t o b e t h e p r o d u c t of t h e s p i n m o m e n t u m
precessional angular speed
and the
W e t u r n n o w t o a n i l l u s t r a t i o n of t h e l a w of g y r o s c o p i c p r e c e s s i o n .
W e h a v e j u s t s e e n t h a t w h e n a f r e e l y s p i n n i n g b o d y is t o r q u e d a b o u t a n
axis n o r m a l t o t h e s p i n a x i s , it p r e c e s s e s a b o u t a t h i r d axis t h a t f o r m s a n
o r t h o g o n a l t r i a d w i t h t h e s p i n a n d t o r q u e v e c t o r s . T h e d i r e c t i o n of t h e
p r e c e s s i o n is s u c h t h a t it t u r n s t h e s p i n v e c t o r t o w a r d t h e t o r q u e v e c t o r .
T h i s l a w of g y r o s c o p i c p r e c e s s i o n is r e s p o n s i b l e f o r t h e l u n i s o l a r p r e c e s ­
s i o n of t h e e q u i n o x e s .
W h a t is t h e l u n i s o l a r p r e c e s s i o n ? B e c a u s e of t h e b i l l i o n s of y e a r s of
g r a v i t a t i o n a l p u l l f r o m t h e s u n a n d m o o n , t h e e a r t h is s l i g h t l y b u l g e d
i n s t e a d of r o u n d . It is i n f a c t a b o u t 2 7 m i l e s s h o r t e r a c r o s s t h e p o l e s t h a n it
is a c r o s s t h e e q u a t o r . T h i s b u l g e , p l u s t h e fact t h a t its a x i s is t i l t e d
to
t h e ecliptic, c a u s e s t h e s u n ( a n d m o o n ) to t o r q u e t h e e a r t h i n a d d i t i o n to
the gravity pull, as s h o w n in Figure 7.11.
Figure 7.10
Daily spin
Earth
Sun
Sun pulls
harder on
shaded half
(because it is closer), giving a
torque (T) as well as a force (F)
Figure 7.11
W e s e e , t h e n , t h a t w e l i v e o n t h e s u r f a c e of a s p i n n i n g g y r o s c o p e t h a t
is c o n s t a n t l y b e i n g a c t e d o n b y a n e x t e r n a l t o r q u e , a n d a p r e c e s s i o n is
t h u s o n g o i n g . T h i s p r e c e s s i o n , s h o w n i n F i g u r e 7 . 1 2 , t u r n s t h e s p i n axis
of t h e e a r t h o u t of t h e p l a n e of t h e p a p e r t o w a r d t h e t o r q u e v e c t o r . T h i s
m o t i o n r e s u l t s i n c o u n t e r c l o c k w i s e m o v e m e n t , o n t h e c e l e s t i a l s p h e r e , of
t h e c e l e s t i a l p o l e t o w h i c h t h e e a r t h ' s r o t a t i o n a x i s is d i r e c t e d . ( T h i s p o i n t
* N o t e that z and z are the same axis.
2
Now
Figure 7.12
13,000 years
from now
The spin axis of the earth is aligned with its angular velocity vector
The point
where the
vector, placed at C, cuts !he surface of the earth is the real meaning of the
North Pole. The North Pole wanders about the geometric pole (on the symmetry axis)
as time passes; it has remained within a few feet of it in this centurv.
Page 495
is c u r r e n t l y c l o s e t o P o l a r i s , t h e N o r t h S t a r . ) T h e p e r i o d of t h i s r o t a t i o n is
a b o u t 2 6 , 0 0 0 y e a r s , a n d it i s i n t e r e s t i n g t h a t t h e m o o n ' s effect is 2 . 2 t i m e s
t h a t of t h e s u n ' s b e c a u s e it is m u c h c l o s e r .
Torque-Free Motion
W e n o w t a k e u p t h e o t h e r e x a m p l e of a s o l u t i o n t o t h e g y r o s c o p e e q u a ­
t i o n s : t h e c a s e of torque-free m o t i o n . " T o r q u e - f r e e " m e a n s t h a t
v a n i s h e s , s o t h a t H is a c o n s t a n t . T h i s f o l l o w s f r o m E u l e r ' s s e c o n d l a w :
c
W e s h a l l c o n v e n i e n t l y l e t t h e d i r e c t i o n of t h e Z a x i s ( s e e F i g u r e 7 . 1 3 ) ,
w h i c h is a r b i t r a r y , c o i n c i d e w i t h t h e c o n s t a n t d i r e c t i o n of H . T h e n Z
b e c o m e s t h e p r e c e s s i o n a x i s of t h e m o t i o n , a n d t h e ( x , y , z ) a x e s a p p e a r
a s s h o w n i n F i g u r e 7 . 1 3 . it is s e e n t h a t s i n c e
c
2
2
2
(7.51)
a n d since H is s e e n a l w a y s t o lie i n t h e x z
2
C
2
plane, then
must vanish:
(7.52)
Note that
lies i n t h e x z p l a n e a l o n g w i t h H , s i n c e i t s y c o m p o n e n t ,
vanishes. Furthermore, w e see that
2
2
c
2
Figure 7.13
(7.53)
But also, a s w e c a n s e e f r o m Figure 7.13,
(7.54)
T h e r e f o r e , e q u a t i n g t h e first of E q u a t i o n s ( 7 . 5 3 ) a n d ( 7 . 5 4 ) f o r
, we
obtain
(Since H ,
a n d I are constants)
c
(7.55)
(7.56)
a n d w e see that
(a c o n s t a n t )
S i m i l a r l y , e q u a t i n g t h e t w o p r e c e d i n g v a l u e s of
(7.57)
gives
(7.58)
(7.59)
• Note again that the (x , y , z ) axes are permanently principal, even though x and y
are not body-fixed, and this lets us write H in terms of the
along these axes.
2
2
2
2
C
2
Page 496
so that
(7.60)
T h e r e f o r e all c o n d i t i o n s a r e s a t i s f i e d f o r t h e t o r q u e - f r e e b o d y t o b e i n a
s t a t e of s t e a d y p r e c e s s i o n a b o u t t h e z a x i s fixed i n !
Dividing E q u a t i o n (7.55) b y (7.58) leads to
(7.61)
a n d Figure 7.14 s h o w s t h a t
(7.62)
where
is t h e a n g l e b e t w e e n z a n d
2
. Therefore
Figure 7.14
a n d w e see that t h e a n s w e r to w h e t h e r
is l a r g e r o r s m a l l e r t h a n
d e p e n d s o n t h e r a t i o of J t o I. If J < I, a s i n t h e e l o n g a t e d s h a p e
in Figure 7.14, t h e n
a n d t h e a n g u l a r v e l o c i t y v e c t o r lies i n s i d e of
H a n d z , making a constant angle with each. T w o cones m a y be imag­
i n e d — o n e fixed t o t h e b o d y , t h e o t h e r i n s p a c e
. T h e b o d y c o n e is
s e e n t o roll o n t h e fixed s p a c e c o n e a s its s p i n a n d p r e c e s s i o n v e c t o r i ally a d d t o t h e v e c t o r
, w h i c h c h a n g e s only in direction.
C
2
T h i s p r e c e s s i o n ( F i g u r e 7 . 1 5 ) is c a l l e d direct b e c a u s e
and
have
the same counterclockwise sense w h e n observed from the
vector out­
s i d e t h e c o n e s . If, h o w e v e r , J > I, t h e n
and
lies outside t h e a n g l e
Space cone
(negative!)
Space
cone
Body cone
Body cone
Figure 7.15
Figure 7.16
Page 4 9 7
ZCz. T h i s s i t u a t i o n is h a r d e r t o d e p i c t , b u t j u s t a s i m p o r t a n t . T h i s t i m e t h e
b o d y c o n e r o l l s a r o u n d t h e o u t s i d e of t h e e n c l o s e d , fixed s p a c e c o n e
(Figure 7.16), a n d t h e t w o r o t a t i o n s
and
h a v e opposite senses. This
p r e c e s s i o n is c a l l e d
retrograde.
A final n o t e o n t h e t h e o r y of t h e t o r q u e - f r e e b o d y : If t h e c o n s t a n t
v a l u e of
is e i t h e r 0 o r 9 0 ° , t h e r e is no p r e c e s s i o n a n d t h e g y r o s c o p e is
s i m p l y i n a s t a t e of p u r e r o t a t i o n , p l a n a r m o t i o n :
Here z = 2, so that
and
If
the
body simply spins about its axis. In this case, the rates
and
cannot be distinguished
Here w e have
and
. Thus:
If
, the body spins about a transverse axis without
precessing
Question 7.5
(7 60) to get
W h v can w e not use Equations (7.57) and
and in this case?
It r e q u i r e s a g r e a t m a n y t e r m s t o d e s c r i b e t h e c o m p l e x m o t i o n of t h e
e a r t h . W e h a v e s e e n o n e e x a m p l e of t h i s i n t h e l u n i s o l a r p r e c e s s i o n
caused b y the gravity torque exerted o n the earth b y the sun a n d m o o n .
T h i s m o t i o n is a n a l o g o u s t o a d i f f e r e n t i a l e q u a t i o n ' s p a r t i c u l a r s o l u t i o n ,
w h i c h it h a s w h e n e v e r t h e e q u a t i o n h a s a n o n z e r o r i g h t - h a n d s i d e . T h e
c o m p l e m e n t a r y , o r h o m o g e n e o u s , s o l u t i o n is a n a l o g o u s t o t h e t o r q u e free p a r t of t h e s o l u t i o n t o t h e e a r t h ' s r o t a t i o n a l m o t i o n . T h i s p a r t , c a l l e d
t h e free p r e c e s s i o n of t h e e a r t h , is i n fact r e t r o g r a d e . B o t h t h e s p a c e a n d
b o d y cones are very thin as the
, H , a n d z a x e s a r e all q u i t e c l o s e
t o g e t h e r ; e a c h lies a b o u t
off t h e n o r m a l t o t h e e c l i p t i c p l a n e .
C
Answer 7.5 In deriving (7.57), if
w e h a v e divided both sides of an equation by
zero. This result is then u s e d in getting in (7.60).
Page 498
PROBLEMS
•
Section 7.6
7.77 Find the angular acceleration in
for the case of steady precession.
of the gyroscope
7.78 The spinning top (Figure P7.78) is another example
of a gyroscope. Show that if the top's peg is not moving
across the floor, the condition for steady precession is
given by
axis of the 7.79
body isAalways
outside the
space cone.
top steadily
precesses
about the fixed direction Z
at
60
rpm.
(See
Figure
P
7
.
7
9
.
)
Treating
7.83 The graph in Figure P7.83a depicts the stability ofthe top as a cone of
radius 1.2 in. and height 2.0 in., find the rate of spin of
the top about its axis of symmetry.
7.80 Cone C in Figure P7.80 has radius 0.2 m and height
0.5 m. It is processing about the vertical axis through the
ball joint, in the direction shown, at the rate of
If the angle
is observed to be 20° and
unchanging, what must be the rate of spin of the cone?
7.81 In the preceding problem, suppose that is given to
be 400 r a d / s in the same direction as given in the figure
and that the cone's height H is not given. Find the value of
H for which this steady precession will occur.
7.82 Using the fact that the sum of any two moments of
inertia at a point is always larger than the third (Problem
7.14), s h o w that for a torque-free axisymmetric body un­
dergoing retrograde precession,
and that the z
symmetrical satellites spinning about the axis z normal to
the orbital plane. The abscissa is the ratio of
to the
moment of inertiaI ,about any lateral axis (they are all the
same for what is called a "symmetrical" satellite—it need
not be physically symmetric about z ) . The ordinate is the
ratio of the spin speed
(about z ) in the orbit to the
orbital angular speed
.
c
1
c
c
a. For a satellite equivalent to four solid cylinders
each of mass m, radius R, and height 3R, find
and I . The distance from C to any cylinder's
center is 2R, and the connecting cross is light.
The cylinders' axes are normal to the orbital
plane. (See Figure 7.83b.)
1
Moment of
inertia =
Figure P7.78
Stable
Unstable
C (5 kg; radius 0 2 m,
height 0.5 m)
Figure P7.79
Unstable
Stable
Figure P7.80
Figure P7.83a
Page 499
b. Determine w h e t h e r t h e station is stable for t h e
following cases:
i. T h e station's orientation is fixed in inertial
space.
ii. T h e station travels a r o u n d t h e earth as t h e
m o o n does.
iii. T h e station h a s twice t h e angular velocity of
t h e orbiting frame.
iw. T h e s a m e as (iii), b u t t h e spin is opposite in
direction to t h e orbital angular speed.
is also zero. N o n e of t h e particles of t h e t o p n o t o n
t h e Z axis are in equilibrium, h o w e v e r , because t h e y all
h a v e (inward) accelerations
. A b o d y is in equilibrium
if a n d only if all its particles are in equilibrium, so the
sleeping t o p c a n n o t b e in equilibrium. Explain this state­
m e n t in light of
and
, which were the
equilibrium equations for a b o d y in statics. Hint: If O is a
fixed point of rigid b o d y in a n inertial frame
then
C
Therefore s h o w that just because
and
n e e d n o t b e zero. Use t h e t o p as a counterexample
a n d explain w h y t h e first t w o terms o n t h e right side of
t h e preceding equation vanish. T h u s
are
necessary b u t n o t sufficient conditions for equilibrium of a
rigid body.
Figure P7.83b
T h e following five p r o b l e m s are a d v a n c e d looks at statics
of rigid bodies that d e p e n d o n o u r s t u d y of dynamics.
7.84 It is possible for a spinning t o p to " s l e e p , " m e a n i n g
.
that its axis remains vertical a n d its p e g stationary as it
spins o n a floor. (See Figure P7.84.) In t h e absence of a
friction couple a b o u t t h e axis of t h e top, n o t e t h a t t h e spin
speed
is constant a n d that t h e equations of m o t i o n
t h e n reduce to
and
. It t h u s follows t h a t
7.85 In t h e sleeping top counterexample of P r o b l e m
7.84, t h e terms
and
b o t h vanish i n d e p e n ­
dently. S h o w t h a t there are m o r e complicated counterex­
amples in w h i c h
is n o t constant in direction in a n d
a n d in w h i c h t h e t w o terms add to zero. Hint:
W h a t is
for t h e torque-free
7.86 S h o w that if
t h e converse true?
at all times, t h e n so is
7.87 If a frame
is m o v i n g relative to a n inertial
frame , it can be s h o w n that is also a n inertial frame
if a n d only if
at all times and t h e acceleration
in of at least o n e point of is zero at all times. Use this
t h e o r e m to s h o w that if a rigid b o d y is in equilibrium
in a n inertial frame
then
is itself a n inertial frame.
Is t h e converse true?
7.88 S h o w that a rigid b o d y
inertial f r a m e . if a n d only if (a)
fixed in a n d (b)
at all
m u m n u m b e r of constraints o n
(b)? Describe o n e set of physical
sure equilibrium.
Figure P7.84
7.7
. Is
is in equilibrium in an
at least o n e point of is
times. W h a t is t h e mini­
that will satisfy (a) a n d
constraints t h a t will as­
Impulse and Momentum
A s w e d i d i n C h a p t e r 5 f o r t h e c a s e of p l a n e m o t i o n , w e c o u l d a p p l y t h e
p r i n c i p l e s of i m p u l s e a n d m o m e n t u m a n d t h o s e of a n g u l a r i m p u l s e a n d
a n g u l a r m o m e n t u m t o t h e t h r e e - d i m e n s i o n a l m o t i o n of a r i g i d b o d y
As
Page 500
w e s a w in Section 5.3, h o w e v e r , these applications are really n o t h i n g
m o r e t h a n t i m e i n t e g r a t i o n s of t h e e q u a t i o n s of m o t i o n .
T h e r e is o n e t y p e of p r o b l e m , h o w e v e r , i n w h i c h t h e s e t w o p r i n c i p l e s
f u r n i s h u s w i t h a m e a n s of s o l u t i o n — p r o b l e m s i n v o l v i n g i m p a c t . S o m e
t h r e e - d i m e n s i o n a l a s p e c t s a r e sufficiently d i f f e r e n t f r o m t h e p l a n a r c a s e
t o w a r r a n t a n e x a m p l e . B u t first w e u s e t h e i n t e g r a l s of t h e E u l e r l a w s t o
derive the n e e d e d relations:
(7.63)
(7.64)
A n a l t e r n a t i v e t o t h e r o t a t i o n a l e q u a t i o n ( 7 . 6 4 ) is t o i n t e g r a t e t h e e q u a l l y
general equation
(7.65)
w h e r e O is n o w a fixed p o i n t of t h e i n e r t i a l f r a m e
(7.66)
To u s e either E q u a t i o n (7.64) or (7.66) in a n i m p a c t situation, w e u s e
E q u a t i o n ( 7 . 1 1 ) for t h e b o d y ' s a n g u l a r m o m e n t u m before the
deformation
starts ( a t t ) a n d t h e n a g a i n after it ends (at t ). T h e f o l l o w i n g e x a m p l e
illustrates t h e procedure.
t
t
EXAMPLE 7 . 1 1
The b e n t b a r , of Example 4.16 is d r o p p e d from a height H a n d strikes a rigid,
s m o o t h surface o n o n e e n d of as s h o w n in Figure E 7 . l l . If the coefficient of
restitution is e, find the angular velocity of as well as t h e velocity of C, just after
the collision.
Solution
Using t h e y c o m p o n e n t equation of (7.63) yields
(1)
in w h i c h t h e impulse of t h e gravity force is neglected as small in comparison w i t h
the impulsive u p w a r d force exerted by the surface over the short time interval
Next w e write the c o m p o n e n t equations of (7.64); w e first n e e d the inertia
properties of t h e body, w h i c h can be c o m p u t e d to be
Figure E7.11
(2)
Page 5 0 1
We t h e n obtain, from Equation (7.64),
(3)
in w h i c h t h e initial angular velocity c o m p o n e n t s vanish a n d t h e desired final
c o m p o n e n t s are
T h e c o m p o n e n t equations of (3) are
(4)
(5)
(6)
At this point w e h a v e four equations in t h e five u n k n o w n s
a n d t h e impulse .
W e get a fifth equation from t h e definition of t h e coeffi­
cient of restitution together w i t h t h e y c o m p o n e n t of t h e rigid-body velocity
relationship b e t w e e n P a n d C:
(7)
and
(8)
w h i c h h a s t h e y - c o m p o n e n t equation
(9)
Using Equation (7), w e obtain
(10)
T h e solution to t h e five equations ( 1 , 4, 5, 6, 10) is
(11)
Returning to Equation (8), w e find t h a t t h e x a n d z c o m p o n e n t s of v , vanish:
c
(12)
T h e results in Equations (12) are obvious, since if t h e r e is n o friction at t h e point of
Page 502
contact t h e r e can b e n o impulsive forces in t h e horizontal p l a n e to c h a n g e the
m o m e n t u m (from zero) in t h e x or z directions.
It is seen that t h e single n o n z e r o p r o d u c t of inertia causes a coupling
between
and
(see Equations (4) a n d (6)), w h i c h prevents
from v a n ­
ishing—even t h o u g h t h e only m o m e n t c o m p o n e n t w i t h respect to C is about
the x axis!
W e s h a l l n o w s e e w i t h a n o t h e r e x a m p l e t h e a d v a n t a g e s of E q u a t i o n
(7.66), w h i c h m a y b e u s e d to eliminate u n d e s i r e d forces f r o m m o m e n t
e q u a t i o n s , j u s t a s w a s d o n e i n o u r s t u d y of s t a t i c s .
EXAMPLE 7 . 1 2
Rework the preceding example by using Equation (7.66) instead of the combina­
tion of Equations (7.63) a n d (7.64). Find t h e value of
after impact.
Solution
Equation (7.66) allows u s to eliminate t h e impulse
about the point (P') of impact:
by summing moments
We still h a v e to u s e t h e coefficient of restitution a n d relate v a n d v exactly as
before; m a k i n g this substitution for , leads to the following three scalar c o m p o ­
n e n t equations:
P
These equations, of course, h a v e the s a m e solution as
ing example.
PROBLEMS
•
c
in the preced­
Section 7.7
7.89 Bend a coat h a n g e r or pipe cleaner into the s h a p e of
the b e n t bar of Example 7.11. E>rop it o n t o t h e edge of a
table as in t h e example a n d observe t h a t t h e angular v e ­
locity direction following impact agrees w i t h t h e results of
the example.
Page 503
• 7 . 9 0 T h e equilateral triangular d i n n e r bell in Fig­
u r e P7.90 is struck w i t h a horizontal force in t h e y direc­
tion that imparts a n impulse
to t h e bell. Find the
angular velocity of t h e bell immediately after t h e b l o w is
struck. Is t h e a n s w e r t h e s a m e if t h e bell is a n equilateral
triangular plate of t h e s a m e mass? W h y or w h y not?
• 7 . 9 1 In t h e preceding problem, s u p p o s e t h e h a m m e r is
replaced b y a bullet of m a s s m a n d s p e e d v that r e b o u n d s
straight back w i t h a coefficient of restitution e = 0 . 1 . De­
termine t h e resulting angular velocity of the bell.
• 7.94 A diver D leaves a diving b o a r d in a straight, s y m ­
metric position w i t h angular velocity a n d a n g u l a r m o ­
m e n t u m vectors each in t h e x direction as indicated in
Figure P7.94a. Since
is zero, there will b e n o change
in t h e angular m o m e n t u m H e in t h e inertial frame (the
s w i m m i n g pool) as long as t h e diver is in t h e air. T h e r e ­
fore, as long as h e r e m a i n s in t h e straight position, his
constant angular m o m e n t u m is expressed b y
b
(1)
7.92 Repeat P r o b l e m 7.90, b u t this time s u p p o s e t h e bell
h a n g s from a string instead of from a ball a n d socket joint.
7.93 T h e b e n t b a r of Figure P7.93 h a s t h e inertia p r o p ­
erties listed b e l o w . It is i n m o t i o n i n a n inertial frame
a n d at a certain instant h a s angular velocity
. Use t h e angular impulse a n d angular
m o m e n t u m principle to a n s w e r t h e following question:
Is it possible to strike
at point Q w i t h a n impulse
that reduces
to zero
after t h e impulse? If so, find t h e c o m p o n e n t s of t h e
impulse in terms of m,
and
. If not, s h o w w h y
not.
(2)
(3)
w h e r e w e a s s u m e t h e b o d y to b e sufficiently internally
symmetric so t h a t t h e p r o d u c t s of inertia all vanish. N o w
s u p p o s e t h e diver instantaneously m o v e s his a r m s as
s h o w n in Figure P 7 . 9 4 b to initiate a twist. Following t h e
m a n e u v e r , h e m a y again b e treated as a rigid b o d y a n d w e
m a y u s e t h e s a m e body-fixed axes as before. (Note that
t h e m a s s center c h a n g e s very little.)
a. From Figure P7.94c a r g u e t h a t t h e indicated
changes in the products of inertia occur. (Only
the s h a d e d a r m s contribute to t h e p r o d u c t s of
inertia.) A r g u e also t h a t I is smaller t h a n I
a n d also less t h a n
, O b s e r v e t h a t all three
C
C
yz
Figure
P7.90
Figure
Figure P7.93
xz
P7.94a
Figure P7.94b
Page 5 0 4
C
I <0
C
I >0
C
I >0
xy
xz
yz
Figure P7.94c
p r o d u c t s of inertia are small c o m p a r e d w i t h t h e
three m o m e n t s of inertia a n d that I < I
< I , with I , b e i n g very m u c h smaller t h a n t h e
other t w o m o m e n t s of inertia. N o t e that
(x, y, z) are n o longer principal, b u t this d o e s
n o t m a t t e r since w e are n o t m a k i n g u s e of prin­
cipal axes h e r e .
C
C
yy
C
xx
C
zz
yy
As t h e diver's b o d y begins to twist a n d
turn, t h e right sides of Equations (1) to (3) will
change a n d none of t h e quantities o n t h e left
will r e m a i n zero. But t h e right sides will consti­
tute t h e c o m p o n e n t s in the body frame D of t h e
vector H , w h i c h will still vectorially a d d to
b. After t h e r a p i d twist m a n e u v e r , b u t before t h e
diver begins to twist, his axes are still instan­
taneously aligned w i t h those of t h e frame
Use Equations (1) to (3), w i t h t h e right sides
a n d t h e n o w n o n z e r o p r o d u c t s of
inertia, t o s h o w that:
i. There will b e a small (compared t o t h e
original
) angular velocity d e v e l o p e d
a b o u t t h e — Z direction (negative
).
ii. There will b e a n angular velocity of twist
d e v e l o p e d a b o u t Y (positive
).
iii. There will b e a n increase in t h e somer­
saulting angular velocity c o m p o n e n t
C
C
c
w h e r e is t h e original direction in of H
after t h e diver leaves t h e diving b o a r d (to t h e
right in t h e first sketch).
7.8
c
In arguing statements (i) to (iii), a s s u m e n o t h i n g a b o u t t h e
following t h e m a n e u v e r except that
is still in t h e
s a m e direction as before.
Work and Kinetic Energy
A s p e c i a l i n t e g r a l of t h e e q u a t i o n s of m o t i o n of a r i g i d b o d y
yields a
r e l a t i o n s h i p b e t w e e n t h e w o r k of t h e e x t e r n a l f o r c e s ( a n d / o r c o u p l e s )
a n d t h e c h a n g e i n t h e k i n e t i c e n e r g y of
To d e v e l o p this relation, w e
m u s t first e x p l o r e e x p r e s s i o n s f o r t h e k i n e t i c e n e r g y of t h e r i g i d b o d y .
Kinetic e n e r g y i s u s u a l l y d e n o t e d b y t h e l e t t e r T a n d i s d e f i n e d b y ( s e e
Section 5.2)
(7.67)
i n w h i c h v i s t h e d e r i v a t i v e of t h e p o s i t i o n v e c t o r f r o m O (fixed p o i n t i n
the inertial frame
i n F i g u r e 7 . 1 7 ) t o t h e d i f f e r e n t i a l m a s s e l e m e n t dm. I n
this section all t i m e derivatives, velocities, a n d a n g u l a r velocities a r e
taken in
u n l e s s o t h e r w i s e specified.
Since
c e n t e r C of
Figure 7.17
i s a r i g i d b o d y , w e m a y r e l a t e v t o t h e v e l o c i t y v of t h e m a s s
c
:
(7.68)
Page 505
in w h i c h
is
a n d r is t h e p o s i t i o n v e c t o r f r o m C t o dm a s s h o w n i n
Figure 7.17. Substituting E q u a t i o n (7.68) into (7.67), w e get
(7.69)
where v and
do not vary over the body's volume a n d can thus be
t a k e n o u t s i d e t h e i n t e g r a l s . T h e i n t e g r a l i n t h e l a s t t e r m i s z e r o b y v i r t u e of
t h e d e f i n i t i o n of t h e m a s s c e n t e r :
c
(7.70)
T h e i n t e g r a l i n t h e first t e r m o n t h e r i g h t s i d e of E q u a t i o n ( 7 . 6 9 ) is of
c o u r s e t h e m a s s m of
T h e i n t e g r a n d of t h e r e m a i n i n g t e r m m a y b e
simplified b y t h e vector identity:*
(7.71)
Therefore E q u a t i o n (7.69) b e c o m e s
(7.72)
A s w e h a v e a l r e a d y s e e n i n S e c t i o n 7.2, t h e i n t e g r a l i n E q u a t i o n
( 7 . 7 2 ) is t h e a n g u l a r m o m e n t u m ( m o m e n t of m o m e n t u m ) of t h e b o d y
w i t h r e s p e c t t o C, a n d t h u s w e c a n w r i t e
(7.73)
It is s e e n t h a t t h e k i n e t i c e n e r g y c a n b e r e p r e s e n t e d a s t h e s u m of t w o
terms:
1.
A part
t h a t t h e b o d y p o s s e s s e s if its m a s s c e n t e r is
in motion.
2.
A part
t i e s of t h e p o i n t s of
t h a t is d u e t o t h e d i f f e r e n c e b e t w e e n t h e v e l o c i ­
a n d t h e v e l o c i t y of i t s m a s s c e n t e r .
The term
c a n b e i n t e r p r e t e d q u i t e s i m p l y if a t a n i n s t a n t w e let
t h a t is, if w e a l i g n t h e r e f e r e n c e axis x w i t h t h e a n g u l a r v e l o c i t y
vector at t h a t i n s t a n t . In this case, u s i n g E q u a t i o n s (7.11), w e o b t a i n
(7.74)
* Which is nothing more than interchanging the dot and cross of the scalar triple prod­
uct
where E is the vector C X D.
Page 506
so that
(7.75)
T h i s m e a n s t h a t t h e " r o t a t i o n a l p a r t " of T is instantaneously
of t h e s a m e
f o r m a s it w a s f o r t h e p l a n e c a s e i n C h a p t e r 4 . T h e d i f f e r e n c e , of c o u r s e , is
t h a t t h e d i r e c t i o n of t h e a n g u l a r v e l o c i t y v e c t o r
c h a n g e s in t h e general
(tluee-dimensional) case.
Suppose the body
h a s a p o i n t P w i t h z e r o v e l o c i t y . ( T h i s is n o t
a l w a y s t h e c a s e i n g e n e r a l m o t i o n a s w e h a v e a l r e a d y s e e n i n C h a p t e r 6.)
T h e n if v i n E q u a t i o n ( 7 . 6 7 ) is r e p l a c e d b y
, where
r ' e x t e n d s f r o m P t o t h e m a s s e l e m e n t dm, w e o b t a i n
(7.76)
T h e i d e n t i c a l s t e p s t h a t p r o d u c e d t h e s e c o n d t e r m of E q u a t i o n ( 7 . 7 3 )
f r o m t h e m i d d l e t e r m of ( 7 . 6 9 ) t h e n g i v e
(7.77)
a n d t h e t w o t e r m s of E q u a t i o n ( 7 . 7 3 ) h a v e c o l l a p s e d i n t o o n e if H is
e x p r e s s e d r e l a t i v e t o a p o i n t of z e r o v e l o c i t y i n s t e a d of C.
I n S e c t i o n s 2 . 4 a n d 5.2 w e d e m o n s t r a t e d o n e w o r k a n d k i n e t i c e n ­
e r g y p r i n c i p l e t h a t r e m a i n s t r u e for t h e g e n e r a l c a s e . T h i s r e s u l t c a m e
from integrating
(7.78)
A s e c o n d principle will n o w b e d e d u c e d from t h e m o m e n t equation*
(7.79)
b u t first w e n e e d t o p r o v e t h e n o n - o b v i o u s r e s u l t t h a t :
T o d o t h i s , w e first r e c a l l t h a t
(7.80)
If
is t h e d e r i v a t i v e of H t a k e n i n t h e b o d y
relative to t h e inertial frame can b e written
c
, then the derivative
(7.81)
* Derivatives such as are taken in the inertial frame in this section unless the letter
appears to the left of the dot, in which case the derivative is taken in the body.
Page 507
Dotting
w i t h b o t h s i d e s of E q u a t i o n ( 7 . 8 1 ) s h o w s t h a t
(7.82)
a n d s i n c e r is c o n s t a n t i n t i m e r e l a t i v e t o b o d y
E q u a t i o n (7.80) there a n d o b t a i n
w e c a n differentiate
(7.83)
I n E q u a t i o n ( 7 . 8 3 ) w e h a v e u s e d t h e p r o p e r t y of
and
a r e t h e s a m e ; t h a t is,
t h a t its d e r i v a t i v e s i n
S u b s t i t u t i n g E q u a t i o n (7.83) i n t o (7.82) t h e n gives
Hence
(7.84)
W e are n o w in a position to observe that
(7.85)
Integrating E q u a t i o n (7.85), w e h a v e
(7.86)
N o t e t h a t t h e right s i d e s of E q u a t i o n s ( 7 . 7 8 ) a n d ( 7 . 8 6 ) e a c h r e p r e s e n t s
the change, occurring in the time interval
, of p a r t of t h e kinetic
e n e r g y of t h e b o d y . T h e left s i d e s of t h e s e e q u a t i o n s a r e u s u a l l y c a l l e d a
f o r m of work.
While the relationships b e t w e e n w o r k a n d kinetic energy that h a v e
been developed are important, another relationship that combines t h e m
is o f t e n m o r e u s e f u l . W e c a n d i f f e r e n t i a t e E q u a t i o n ( 7 . 7 3 ) a n d g e t
Using Euler's l a w s a n d E q u a t i o n (7.84), this m a y b e p u t into t h e f o r m
(7.87)
Page 508
If w e n o w l e t F ! , F
2
, . . represent t h e external forces acting o n t h e b o d y ,
a n d C , C , . . . r e p r e s e n t t h e m o m e n t s of t h e e x t e r n a l c o u p l e s , t h e n
1
2
(7.88a)
(7.88b)
w h e r e P , P , . . . a r e t h e p o i n t s of w h e r e F , F , . . . a r e r e s p e c t i v e l y
applied a n d where
r = r , r = r , a n d s o forth, a s s h o w n i n Fig­
u r e 7 . 1 8 . Recall from statics t h a t a c o u p l e h a s t h e s a m e m o m e n t a b o u t
a n y point in space, so that t h e C 's are simply a d d e d into t h e m o m e n t
equation (7.88b).
t
2
1
1
CP1
2
2
CP2
i
S u b s t i m t i n g E q u a t i o n (7.88) into (7.87), w e o b t a i n
Figure 7.18
(7.89)
However,
so that
(7.90)
W e note that
is t h e v e l o c i t y v of p o i n t P ,
a p p l i c a t i o n of F . T h e r e f o r e
1
l
t h e p o i n t of
1
(7.91)
E q u a t i o n ( 7 . 9 1 ) l e a d s u s t o d e f i n e t h e p o w e r , o r rate of work, a s f o l l o w s :
(7.92)
Therefore
Integrating E q u a t i o n (7.91), w e get
(7.93)
T h e i n t e g r a l o n t h e left s i d e of E q u a t i o n ( 7 . 9 3 ) is c a l l e d t h e w o r k d o n e o n
b e t w e e n t a n d t b y t h e external forces a n d couples. H e n c e
1
2
(7.94)
T h a t is, t h e w o r k d o n e o n
e q u a l s i t s c h a n g e i n k i n e t i c e n e r g y . It is
left a s a n e x e r c i s e f o r t h e r e a d e r t o s h o w t h a t E q u a t i o n ( 7 . 9 4 ) is i n fact
t h e s u m of t h e t w o " s u b e q u a t i o n s " ( 7 . 7 8 ) a n d ( 7 . 8 6 ) .
Page 509
EXAMPLE 7 . 1 3
Find t h e w o r k d o n e on the b e n t bar of Example 7.11 b y a m o t o r that brings it u p to
speed
from rest. (See Figure E7.13.)
Solution
The mass center C does n o t m o v e , so that Equation (7.73) gives, in this case,
(1)
Figure E7.13
Since h a s only a
a n d get
component,
w e m a y substitute Equation (7.11) into (1)
(2)
C
W e n o t e t h a t e v e n t h o u g h I is n o t zero, it h a s n o effect o n t h e kinetic energy of
since it is multiplied by
w h i c h is forced to vanish by the bearings aligned
with z.
T h u s the w o r k d o n e b y the motor o n is given simply by Equation (7.94):
xz
(3)
when
from Example 7.11. The motor w o u l d , of course, h a v e to
do additional w o r k besides that given by (3) to overcome its o w n a r m a t u r e
inertia, bearing a n d belt friction, a n d air resistance.
W e n o w consider a n e x a m p l e in three dimensions in w h i c h the
p r o d u c t s of i n e r t i a d o p l a y a r o l e i n t h e k i n e t i c e n e r g y c a l c u l a t i o n .
EXAMPLE 7 . 1 4
Find t h e kinetic energy lost b y the b e n t bar of Example 7.10 w h e n it strikes the
table t o p as s h o w n in Figure E7.14a.
Solution
During the impact w i t h the table top, t h e bodies d o n o t b e h a v e rigidly. The
kinetic energy lost b y b a r is transformed into noise, heat, vibration, a n d b o t h
elastic a n d p e r m a n e n t deformation. In Example 7.11 w e found v a n d
just
before a n d after impact; w e n o w u s e these vectors to find t h e kinetic energy lost
by
Just after impact w e h a v e
c
Figure E7.14a
Page 510
The term
can b e written just after impact, using Equation (7.11), as
follows. (Note t h a t
a n d t w o of the products of inertia are zero here.)
Using this result a n d Equations (11) a n d (12) from Example 7.14, w e obtain
which, after simplification, equals
The initial kinetic energy (just prior to t h e collision) w a s
Thus the change in kinetic energy of t h e b e n t bar is given by
W e see that if e = 1 (purely elastic collision), n o loss in kinetic energy occurs a n d
h e n c e n o w o r k is d o n e in changing T. T h e energy lost is seen in Figure E7.14b to
vary quadratically, w i t h a m a x i m u m percentage loss ( w h e n e = 0) of
in this case. Because t h e point of striking is the e n d of the bar, 83.9 percent of the
kinetic energy is retained. If t h e m a s s center of t h e b a r w e r e t h e p o i n t t h a t struck
the table, h o w e v e r , all t h e kinetic energy w o u l d h a v e b e e n lost if e = 0.
Figure E7.14b
PROBLEMS
7.95
•
Section7.8
Find the kinetic energy of disk
in P r o b l e m 6.27.
7.96 The center of m a s s C of a gyroscope G is fixed.
S h o w that the kinetic energy of G is
where
are the Eulerian angles a n d A, A, C are the
principal m o m e n t s of inertia of G at C.
Page 511
7.97 Find the kinetic energy of the w a g o n wheel in
Problem 6.49 and use it to deduce the work done by the
boy in getting it up to its final speed from rest.
7.98 A disk D of mass 10 kg and radius 25 cm is welded
at a 45° angle to a vertical shaft S. (See Figure P7.98.) The
shaft is then spun up from rest to a constant angular speed
• 7.100 The rigid body in Figure P7.100 consists of a disk D
and rod R, welded together perpendicularly as s h o w n in
the figure. If the body is spun up to angular speed
about the z axis, h o w much work was done o n it (exclud­
ing the overcoming of frictional resistance)?
a. H o w much work is done in bringing the assem­
bly up to speed?
b. Find the force and couple system acting on the
plate at C after it is turning at the constant
speed
Figure P7.100
* 7.101 Figure P7.101 shows a thin homogeneous triangu­
lar plate of mass m, base a, and height 2a. It is welded to a
light axle that can turn freely in bearings at A and B.
Given:
a. If the plate is turning at constant angular speed
find the torque that must be applied to the
axle, and find the dynamic bearing reactions.
Figure P7.98
b. Find the principal axes at A and the principal
moments of inertia there. Draw the axes o n a
sketch.
7.99 A thin rectangular plate (Figure P7.99) is brought
up from rest to speed
about a horizontal axis Y.
a. Find the work that is done.
b. If two concentrated masses of m/2 each are
added on the x axis, one on each side of the
mass center, find their distances d from the
mass center that will eliminate the bearing reac­
tions.
c. If possible, give the radius of a hole that, w h e n
drilled at C, will eliminate the bearing reactions,
Give the answer in terms of m and pt (density
times thickness) of the plate.
d. Find the work done in bringing the plate up to
speed
from rest.
c
Bearing
Figure P7.99
Figure P7.101
Page 512
• 7.102 A thin equilateral triangular plate P of side s is
w e l d e d to t h e vertical shaft at A in Figure P7.102. The
shaft is b r o u g h t u p to speed
from rest by a motor.
a. H o w m u c h w o r k is d o n e in bringing the system
u p to speed?
b. Find the force a n d couple system acting o n the
plate at A after it is t u r n i n g at t h e speed
and
the motor is t u r n e d off.
• 7.105 A ring is w e l d e d to a r o d at a point A as s h o w n in
Figure P7.105. T h e cross sections a n d densities of t h e rod
a n d ring are the s a m e . The c o m b i n e d b o d y is released
w i t h a gentle n u d g e w i t h e n d B of the r o d connected to the
s m o o t h p l a n e b y a ball joint a n d w i t h point A at its highest
point as s h o w n . At the instant w h e n A reaches its lowest
point, find the relationship b e t w e e n the horizontal a n d
vertical angular velocity c o m p o n e n t s of the body.
7.106 If in the preceding p r o b l e m t h e p l a n e is r o u g h
e n o u g h to p r e v e n t slipping, find the m a g n i t u d e of the
angular velocity w h e n A reaches the floor.
Figure P7.102
7.103 T w o concentrated masses m = 10 kg a n d m = 20
kg are connected by a 15-kg slender rod m of length
1.5 m . As s h o w n in Figure P 7 . 1 0 3 ,
are unit vectors
fixed in direction in t h e inertial frame a n d
are
parallel to principal axes fixed at C in t h e c o m b i n e d body.
At t w o times t a n d t , the velocities of C a n d t h e angular
velocities of the combined b o d y are
1
2
3
1
Figure P7.103
2
Find t h e total w o r k d o n e o n t h e system b e t w e e n t a n d t .
1
2
7.104 Find t h e kinetic energy of t h e grinder in Problem
7.62. Is this equal to the w o r k d o n e b y a m o t o r on S w h i c h
brings the system u p to speed? (Neglect t h e masses of S
and )
COMPUTER PROBLEM
•
Chapter 7
• 7 . 1 0 7 Use a c o m p u t e r to generate data for a plot of maxi­
m u m values of
versus
in Example 7.10, for
Figure P7.105
Page 513
SUMMARY
•
Chapter 7
I n t h i s c h a p t e r w e h a v e d e v e l o p e d e x p r e s s i o n s f o r a n g u l a r m o m e n t u m of
a rigid b o d y in general t h r e e - d i m e n s i o n a l m o t i o n . W i t h respect to t h e
m a s s c e n t e r , it is
A n d if P is t h e l o c a t i o n of a p o i n t of t h e b o d y w i t h z e r o v e l o c i t y ,
T r a n s f o r m a t i o n p r o p e r t i e s of m o m e n t s a n d p r o d u c t s of i n e r t i a i n ­
clude the parallel-axis t h e o r e m s
t o g e t h e r w i t h f o r m u l a s for o b t a i n i n g t h e m o m e n t s a n d p r o d u c t s of i n e r ­
tia a s s o c i a t e d w i t h a x e s t h r o u g h a p o i n t w h e n t h o s e p r o p e r t i e s a r e k n o w n
for o t h e r a x e s t h r o u g h t h e p o i n t :
In these t w o equations
a n d n , n , n are r e s p e c t i v e l y t h e d i r e c t i o n
c o s i n e s of x' a n d y' r e l a t i v e t o a x e s x, y, a n d z.
P r i n c i p a l a x e s of i n e r t i a a r e v e r y i m p o r t a n t a n d h a v e t h e k e y p r o p ­
erty t h a t w e r e a b o d y to r o t a t e a b o u t a principal axis at a p o i n t P , t h e n t h e
a n g u l a r m o m e n t u m w i t h respect t o P w o u l d b e in t h e s a m e direction as
the a n g u l a r velocity, or
x
y
2 ,
w h e r e J is t h e m o m e n t of i n e r t i a a b o u t t h e p r i n c i p a l a x i s , a n d is c a l l e d a
p r i n c i p a l m o m e n t of i n e r t i a .
A l l p r o d u c t s of i n e r t i a a s s o c i a t e d w i t h a p r i n c i p a l axis v a n i s h , a n d a t
a n y point there are three mutually perpendicular principal axes. The
l a r g e s t a n d s m a l l e s t of t h e p r i n c i p a l m o m e n t s of i n e r t i a a r e t h e l a r g e s t
a n d s m a l l e s t of all t h e m o m e n t s of i n e r t i a a b o u t a x e s t h r o u g h t h e p o i n t .
Page 514
S o m e i m p o r t a n t special cases are:
1.
If P lies i n a p l a n e of s y m m e t r y of t h e b o d y , t h e n t h e axis t h r o u g h P
a n d p e r p e n d i c u l a r t o t h e p l a n e is a p r i n c i p a l a x i s .
2.
If P lies o n a n a x i s of s y m m e t r y of t h e b o d y , t h e n t h a t a x i s a n d e v e r y
l i n e t h r o u g h P a n d p e r p e n d i c u l a r t o it is a p r i n c i p a l a x i s . F u r t h e r ­
m o r e , t h e m o m e n t s of i n e r t i a a b o u t t h e s e t r a n s v e r s e a x e s t h r o u g h a
g i v e n p o i n t a r e all t h e s a m e .
3.
If P is a p o i n t of s p h e r i c a l s y m m e t r y , e.g., t h e c e n t e r of a u n i f o r m
s p h e r e , t h e n e v e r y l i n e t h r o u g h P is a p r i n c i p a l axis a n d a l l of t h e
c o r r e s p o n d i n g p r i n c i p a l m o m e n t s of i n e r t i a a r e e q u a l .
T h e m o s t c o n v e n i e n t f o r m of E u l e r ' s s e c o n d l a w ,
to use in a
p a r t i c u l a r p r o b l e m is o f t e n d e p e n d e n t o n t h e p r o b l e m . W h e n b o d y - f i x e d
principal axes a r e u s e d for reference, t h e n w e h a v e w h a t a r e u s u a l l y
referred to as t h e Euler equations:
H o w e v e r , it is v e r y o f t e n m o r e c o n v e n i e n t t o e x p r e s s t h e a n g u l a r m o ­
m e n t u m i n t e r m s of i t s c o m p o n e n t s p a r a l l e l t o r e f e r e n c e a x e s a s s o c i a t e d
w i t h s o m e i n t e r m e d i a t e f r a m e of r e f e r e n c e , s a y f, w h i c h is n e i t h e r t h e
b o d y itself n o r t h e i n e r t i a l f r a m e
so that
J u s t a s i n t h e c a s e of p l a n e m o t i o n ( C h a p t e r 5 ) , t h e w o r k of e x t e r n a l
f o r c e s e q u a l s t h e c h a n g e i n k i n e t i c e n e r g y f o r rigid b o d i e s i n g e n e r a l
m o t i o n . I n t h r e e - d i m e n s i o n a l m o t i o n t h e k i n e t i c e n e r g y , T, c a n b e w r i t t e n
in g e n e r a l as
T h e s e c o n d t e r m m a y b e c o m p a c t l y w r i t t e n as
w h e r e I is t h e m o m e n t of i n e r t i a a b o u t t h e a x i s , t h r o u g h C, t h a t is i n s t a n ­
taneously aligned with
REVIEW QUESTIONS
•
Chapter 7
True or False?
1. P r o d u c t s of i n e r t i a a s s o c i a t e d w i t h p r i n c i p a l a x e s a l w a y s v a n i s h , b u t
only at t h e mass center.
2. If t h e p r i n c i p a l m o m e n t s of i n e r t i a a t a p o i n t a r e d i s t i n c t , t h e n t h e
p r i n c i p a l a x e s of i n e r t i a a s s o c i a t e d w i t h t h e m a r e o r t h o g o n a l .
Page 515
3. T h e m a x i m u m m o m e n t of i n e r t i a a b o u t a n y l i n e t h r o u g h p o i n t P of
rigid b o d y
is t h e l a r g e s t p r i n c i p a l m o m e n t of i n e r t i a a t P .
4. G e n e r a l m o t i o n is a m u c h m o r e difficult s u b j e c t t h a n p l a n e m o t i o n . A
m a j o r r e a s o n f o r t h i s i s t h a t n e i t h e r t h e k i n e m a t i c s n o r k i n e t i c s dif­
f e r e n t i a l e q u a t i o n s g o v e r n i n g t h e o r i e n t a t i o n m o t i o n of t h e b o d y a r e
linear.
5. If w e s o l v e t h e E u l e r e q u a t i o n s ( 7 . 4 0 ) , w e i m m e d i a t e l y k n o w t h e
o r i e n t a t i o n of t h e rigid b o d y i n s p a c e .
6. T h e s u n a n d t h e m o o n e x e r t g r a v i t y t o r q u e s o n t h e e a r t h , a n d t h e y
c a u s e t h e a x i s of o u r p l a n e t t o p r e c e s s .
7. If a t a c e r t a i n i n s t a n t t h e m o m e n t of i n e r t i a of t h e m a s s of b o d y
a b o u t a n axis t h r o u g h C p a r a l l e l t o t h e a n g u l a r v e l o c i t y v e c t o r is I,
t h e n t h e k i n e t i c e n e r g y of
at t h a t i n s t a n t is
8. T h e e a r t h ' s l u n i s o l a r p r e c e s s i o n is t h e r e s u l t of both t h e b u l g e a t t h e
e q u a t o r and t h e tilt of t h e a x i s .
9. T h e k i n e t i c e n e r g y l o s t d u r i n g a c o l l i s i o n of t w o b o d i e s d o e s n o t
d e p e n d o n t h e a n g u l a r v e l o c i t i e s of t h e b o d i e s p r i o r t o i m p a c t .
10. T h e w o r k - e n e r g y a n d i m p u l s e - m o m e n t u m p r i n c i p l e s a r e g e n e r a l
i n t e g r a l s of t h e e q u a t i o n s of m o t i o n for a rigid b o d y .
11. S o m e t i m e s it is b e t t e r t o u s e t h e p r o d u c t s of i n e r t i a i n
t h a n t o t a k e t h e t i m e t o c o m p u t e p r i n c i p a l m o m e n t s a n d a x e s of
inertia so as to b e able to utilize Euler's e q u a t i o n s (7.40).
12. I n s t e a d y p r e c e s s i o n w i t h t h e n u t a t i o n a n g l e e q u a l i n g 9 0 ° , t h e s p i n
vector always precesses a w a y from the torque vector.
Aimran: 1. F 2. T 3. T 4. T 5. F 6. T 7. T 8. T 9. F 10. T 11. T 12. F
8
SPECIAL TOPICS
8.1
8.2
8.3
8.4
Introduction
Introduction to Vibrations
Free Vibration
Damped Vibration
Forced Vibration
Euler's Laws for a Control Volume
Central Force Motion
R E V I E W QUESTIONS
Page 5 1 6
Page 5 1 7
8.1
Introduction
In this chapter, we examine three subjects which are of considerable
practical importance in Dynamics. In the first of these special topics, we
introduce the reader to the subject of vibrations, limiting the presentation
to a single degree of freedom (in which the oscillatory motion can be
described by just one coordinate).
The second special topic deals with problems in which mass is con­
tinuously leaving and / or entering a region of space known as a control
volume. A rocket is a good example: as the fuel is burned and the com­
bustion products are ejected from a control volume enveloping the
rocket, its momentum changes and it is propelled through the atmo­
sphere. Euler's laws still apply, though the resulting equations are a bit
more complicated than they were for the "constant mass" particles and
bodies of earlier chapters.
The final topic in the chapter is central force motion, the most com­
mon example of which is that of orbits—such as the motions of planets
around the sun, and of the moon and of man-made satellites around the
earth.
The topics of Sections 8.2, 3, and 4 all stand alone, and can be read
and understood after the reader has mastered Chapters 1 and 2, except
for some of the problems in Section 8.2 in which the moment equation for
rigid bodies in plane motion from Chapter 4 is also needed.
8.2
Introduction to Vibrations
Vibration is a term used to describe oscillatory motions of a body or
system of bodies. These motions may be caused by isolated disturbances
as when the wheel of an automobile strikes a bump or by fluctuating
forces as in the case of the fuselage panels in an airplane vibrating in
response to engine noise. Similarly, the oscillatory ground motions re­
sulting from an earthquake cause vibrations of buildings. In each of these
cases the undesirable motion may cause discomfort to occupants; more­
over, the oscillating stresses induced within the body may lead to a
fatigue failure of the structure, vehicle, or machine.
Free Vibration
Figure 8.1
Figure 8.2
For perhaps the simplest example of a mechanical oscillator consider the
rigid block and linear spring shown in Figure 8.1. The block is constrained
to translate vertically; thus a single parameter (scalar) is sufficient to
establish position and hence the system is called a single-degree-offreedom system. We choose z to be the parameter and let z = 0 corre­
spond to the configuration in which the spring is neither stretched nor
compressed.
Using a free-body diagram of the block in an arbitrary position (Fig­
ure 8.2), Euler's first law yields
mz = mg — kz
Page 518
or
(8.1)
mz + kz = mg
which is a second-order linear differential equation with constant coeffi­
cients describing the motion of the block. The fact that the differential
equation is nonhomogeneous (the right-hand side is not zero) is a conse­
quence of our choice of datum for the displacement parameter z. For if we
make the substitution y = z — m g / k , the governing equation (8.1) be­
comes
(8.2)
my + ky = 0
which is a homogeneous differential equation. It is not a coincidence that
this occurs when the displacement variable is chosen so that it vanishes
when the block is in the equilibrium configuration—that is, when the
spring is compressed mg / k.
Motion described by an equation such as (8.2) is called a free vibra­
tion since there is no external force (external, that is, to the spring-mass
system) stimulating it.
Rewriting Equation (8.2), we obtain
or, defining w „
n
(8.3)
which has as its general solution
(8.4)
or
(8.5)
where
Whether expressed in the form of (8.4) or (8.5), y is called a simple
harmonic function of time, w is called the natural circular frequency, C
is called the amplitude of the displacement y, and (p is said to be the
phase angle by which y leads the reference function, sin co„t. The simple
harmonic function is periodic and its period is r„ = 2n / w . Another
quantity called frequency is f = 1 / T W / 2n, which gives the num­
ber of cycles in a unit of time. When the unit of time is the second, the unit
for f is the hertz (Hz); 1 Hz is 1 cycle per second.
The constants A and B in (8.4), or equivalently C and (p in (8.5), are
determined from initial conditions of position and velocity. Thus if
n
n
=
n
n
n
n
Page 519
and
then
and
Now let us investigate what might seem an entirely different situa­
tion—that of a rigid body constrained to rotate about a fixed horizontal
axis (through O as in Figure 8.3). Since the only kinematic freedom the
body has is that of rotation, a single angle is sufficient to describe a
configuration of the body. Let the angle be 9 as shown, where we note
that when
the mass center C is located directly below the pivot O.
Neglecting any friction at the axis of rotation, the free-body diagram
appropriate to an arbitrary instant during the motion is shown in Figure
8.4. Summing moments about the axis of rotation, we get
Figure 8.3
(8.6)
where I is the mass moment of inertia about the axis of rotation. Equa­
tion (8.6) is a nonlinear differential equation because sin 9 is a nonlinear
function of 9, but if we restrict our attention to sufficiently small angles so
that sin
Equation (8.6) becomes
0
(8.7)
That is,
is a simple harmonic function:
Figure 8.4
where now
The two preceding examples have an important feature in common:
Motion near the equilibrium configuration is governed by a homoge­
neous, second-order, linear differential equation with constant coeffi­
cients, and in each case the motion is simple harmonic. A point of differ­
ence is that in the block-spring case the gravitational field plays no role
other than establishing the equilibrium configuration; in particular the
natural frequency does not depend on the strength (g) of the field. In the
second case where the body is basically behaving as a pendulum, the
gravitational field provides the "restoring action" and the natural fre­
quency is proportional to
Page 520
EXAMPLE 8 . 1
Find the natural frequency of small oscillations about the equilibrium position of
a uniform ball (sphere) rolling on a cylindrical surface.
Solution
Let m be the mass of the ball, let R be the radius of the path of its center, and let
be the polar coordinate angle locating the center as shown in Figure E8.1a. Thus
and the angular acceleration of the ball is a = —
because of the no-slip
condition. We shall now use a and a in the equations of motion:
c
(1)
Figure
Figure E 8 . 1 a
Hence, from the free-body diagram shown in Figure E8.lb, the
nent equations of (1) are:
E8.1b
and
compo­
(2)
and
(3)
Also, from summing moments about C, we have
or
(4)
Eliminating the friction force F between Equations (2) and (4), we obtain the
differential equation
For small
so that sin
Page 521
from which we see that
or
(w = 0.845
n
Damped Vibration
The simple harmonic motion in our examples of free vibration has a
feature that conflicts with our experience in the real world; that is, the
motion calculated persists forever unabated. Intuition would suggest
decaying oscillations and finally the body coming to rest. Of course the
problem here is that we have not incorporated any mechanism for energy
dissipation in the analytical model. To do that, we shall return to the
simple block-spring system and introduce a new element: a viscous
damper (Figure 8.5). The rate of extension of this element is proportional
to the force applied, through a damping constant c, so that the force is c
times the rate of extension.
Referring to the free-body diagram in Figure 8.5 and letting y = 0
designate the e q u i l i i u m position as before, we have
Damper
or
(8.8)
Figure 8.5
The appearance of the cy term in (8.8) has a profound effect on the
solution to the differential equation and hence on the motion being
described. Solutions to (8.8) may be found from
y = Ae
(8.9)
n
where A is an arbitrary constant and r is a characteristic parameter.
Substituting (8.9) into (8.8), we obtain
2
(mr + cr + k)Ae
rt
= 0
(8.10)
which is satisfied nontrivially (i.e., for A = 0) with
2
mr + cr + k = 0
(8.11)
This characteristic equation has two roots given by
(8.12)
2
Except for the case in which (c / 2m) = k/m, the roots are distinct; if we
call them r and r , then the general solution to (8.8) is
x
2
Page 522
2
In the exceptional case (c / 2m) = k / m, there is only the one re­
peated root r = — c / 2m, but direct substitution will verify that there is a
solution to (8.8) of the form te
so that the general solution in that
case is
-(c/2m)t
(8.13)
With initial conditions
and
we find that
A = yo
1
and
Since
the solution is
(8.14)
Displacements given by (8.14) are plotted in Figure 8.6 for several
representative sets of initial conditions (positive y„ but positive and nega­
tive v ). Two features of the motion are apparent:
0
1.
y
2.
The motion is not oscillatory; the equilibrium position is "overshot"
at most once and only then when the initial speed is sufficiently large
and in the direction opposite to that of the initial displacement.
0 (the equilibrium position) as t
In the case we have just studied, the damping is called critical
damping, because it separates two quite different mathematical solu-
Figure 8.6
Motion of a critically damped system.
Pag523
tions: For greater damping the roots of the characteristic equation (8.11)
are both real and negative, and for small damping the roots are complex
conjugates. If we let the critical damping be denoted by
then we have
seen that
(8.15)
Now let us consider the case for which c > c^; the mechanical
system is then said to be overdamped or the damping is said to be
supercritical. In this case the roots given by (8.12) are both real and
negative since (c / 2m) > k/m; if we call these roots — a and — a , with
a > fli > 0, then the general solution to the differential equation of
motion is
2
x
2
2
(8.16)
The motion described here is in no way qualitatively different from that
for the case of critical damping, which we have just discussed. For a given
set of initial conditions, Equation (8.16) yields a slower approach to y = 0
than does (8.13). That is, the overdamped motion is more "sluggish" than
the critically damped motion as we would anticipate because of the
greater damping.
Finally we consider the case in which the system is said to be underdamped or subcritically damped; that is, c < c^. The roots given by
(8.12) are the complex conjugates
where i =
It is possible to express the general solution to the gov­
erning differential equation as
y =
(8.17)
where co =
A typical displacement history corre­
sponding to (8.17) is shown in Figure 8.7. We note that, just as in the
preceding cases, y
0 as f
; however, here the motion is oscillatory.
We see that the simple harmonic motion obtained for the model without
d
Displacement curve
"Envelope"
Figure 8.7
Motion of an underdamped system.
Page 524
damping is given by (8.17) with c = 0. Moreover, we see that with light
damping (small c) the analytical model that does not include damping
adequately describes the motion during the first several oscillations. It is
this case—subcritical damping—that is of greatest practical importance
in studies of vibration.
EXAMPLE 8 . 2
Find the damping constant c that gives critical damping of therigidbar executing
motions near the equilibrium position shown in Figure E8.2a.
Solution
We are going to restrict our attention to small angles 9, and thus we may ignore
any tilting of the damper or the spring. However, it may help us develop the
equation of motion in an orderly way if we assume that the upper ends of the
spring and damper slide along so that each remains vertical as the bar rotates
through the angle 6. Without further restriction, if we sum moments about A with
Figure E 8 . 2 a
(1)
b
where
is the spring stretch at equilibrium. Thus for small O (that is, sinO
1) we linearize Equation (1) and obtain
Figure E 8 . 2 b
Of course,
is the equilibrium configuration so that
The linear governing differential equation is then
and for critical damping we get, associating the coefficients of
with those of
or
Any c less than this critical value will result in oscillations of decreasing ampli­
tude.
Forced Vibration
Fluctuating external forces may have destructive effects on mechanical
systems; this is perhaps the primary motivation for studying mechanical
vibration. It is common for the external loading to be a periodic function
of time, in which case the loading may be expressed as a series of simple
Page 525
k
m
Psin
harmonic functions (Fourier series). Consequently it is instructive to con­
sider the case in which the loading is simple-harmonic. For the massspring-damper system shown in Figure 8.8, the differential equation of
motion is
mx + cx + kx = P sin cot
Figure 8.9
(8.18)
The general solution is composed of two parts: a particular solution
(anything that satisfies the differential equation) and what is called the
complementary solution (the general solution to the homogeneous differ­
ential equation). A particular solution of the form x = X sin(cof — <p) may
be found. If we substitute this expression in (8.18) we obtain
or
or
For this to be satisfied at every instant of time,
(8.19)
and
(8.20)
From (8.20) we get
(8.21)
so that
and
Substituting these expressions for sin
and cos
into (8.19), we obtain
so that
(8.22)
Page 526
We may now write the complete solution to the differential equation
(8.18):
(8.23)
where x is the complementary solution and is one of the three cases
enumerated in the preceding section. That is, the form of x depends on
whether the system is overdamped, critically damped, or underdamped.
However, in each of these cases the negative exponent causes the func­
tion to approach zero as time becomes large. Thus for large time x tends
to zero and x(t) tends to the particular solution. For this reason the
simple-harmonic particular solution is called the steady-state displace­
ment, since it represents the long-term behavior of the system.
We note that the steady-state motion is a simple harmonic function
having amplitude X and lagging the excitation (force) function by the
phase angle
We may put these in a convenient form by dividing
numerator and denominator of (8.21) and (8.22) by k, so that
c
c
c
(8.24)
tan
and
(8.25)
Investigating the dimensionless quantity ceo / k, we find
But we know that 2mco„ = c ^ , , the critical damping, so if we let be the
damping ratio ( c / c ^ ) ,
(8.26)
and
(8.27)
tan
and
(8.28)
The phase angle and the dimensionless displacement amplitude
kX/P are plotted against the frequency ratio w/w in Figures 8.9 and
8.10, respectively, for various values of the damping ratio We see that,
with small damping, large amplitudes of displacement occur when the
excitation frequency to is near the natural frequency w . This phenome­
non is called resonance, and the desire to avoid it has led to the develop­
ment of methods for estimating natural frequencies of mechanical sysn
n
Page 527
Figure 8.9
Figure 8.10
terns. Note that the steady-state response curves are insensitive to the
damping for sufficiently small damping (say < 0.1) provided that we
are not in the near vicinity of w / w„ = 1. This is an important observa­
tion because often in engineering practice we have reason to believe that
the damping is small but we do not have accurate quantitative informa­
tion about it.
We close this section by discussing the usual source of a simple
harmonic external loading—an imbalance in a piece of rotating ma­
chinery. Let the machine be made up of two parts. The first, of mass m , is
a rigid body constrained to rotate about an axis fixed in the second body
(mass m ), which translates relative to the inertial frame of reference. Let
the mass center of the rotating body lie off the axis of rotation a distance e
and let the body rotate at constant angular speed to. (See Figure 8.11(a).)
Referring to the free-body diagram in Figure 8.11(b), we have
x
2
If we denote the total mass of the machine by m, then m = m + m and
1
Figure 8.11
2
Page 528
Thus the amplitude of the apparent "external" sinusoidal loading is
m ew
and its frequency is the angular speed of the rotating element.
1
2
EXAMPLE 8 . 3
A piece of machinery weighing 200 lb has a rotating element with imbalance
and an operating speed of 1200 rpm. There are four springs, each of stiffness
1500 lb/in., supporting the machine whose frame is constrained to translate
vertically. The damping ratio is = 0.3. Find the steady-state displacement of the
frame.
Solution
The effective spring stiffness is
k = 4(1500) = 6000 lb/in.
so that
The effective external force amplitude is
From Equation (8.28) we get
The phase angle
is given by
tan
so that
= 205 rad (117°).
(m e
1
Page 529
EXAMPLE 8 . 4
The machine of Example 8.3 (weight = 200 lb, imbalance = 5 Ib-in., operating
speed = 1200 rpm) is to be supported by springs with negligible damping. If the
machine were bolted directly to the floor, the amplitude of force transmitted to
the floor would be
2
(m e)W = 206 lb
1
What should the stiffness of the support system be so that the amplitude of the
force transmitted to the floor is less than 20 lb?
Solution
The force exerted on theflooris transmitted through the supporting springs and
is of amplitude kX, where X is the amplitude of displacement of the machine.
From Equation (8.28) we have
or with negligible damping
Thus for
2
2
2
2
it is clear that 1 — w /w is negative. Note that only w h e n w / w > 2 is
n
n
Therefore we inquire into the condition for which
or
or
since w= 126 rad/sec. But
Thus to satisfy the given conditions the support stiffness must be less than
725 lb/in.
If the only springs available give a greater stiffness, the problem may be
solved by increasing the mass; particularly we might mount the machine on a
Page 530
block of material, say concrete, and then support the machine and block by
springs. For example, if the only springs available were those of Example 8.3
for which the weight is
(4.29)(386) = 1660 lb
Therefore we need a slab or block weighing
1660 - 200 = 1460 lb
EXAMPLE 8 . 5
Find the steady-state displacement x(t) of the mass in Figure E8.5 if y(t) = 0.1
cos 120t inch, where f is in seconds, m = 0.01 lb-sec /in., k = 100 lb/in., and
c = 2 lb-sec/in. In particular: (a) What is the amplitude of x(t)? (b) What is the
angle by which x(t) leads or lags y(t)l
2
Figure E 8 . 5
Solution
The differential equation of motion of the mass is seen to be
or
where Y = 0.1 in. and co = 120 rad/sec. Using Equation (8.18), we see thatkYis
or
The phase angle is
= 76.9° or 1.34 rad
(lagging)
Thus the steady-state motion is
= 0.0406 cos(120t- 1.34) in.
Page 531
PROBLEMS
•
Section 8.2
8.1 Find the frequency of small vibrations of the round
wheel C as it rolls back and forth on the cylindrical surface
in Figure P8.1. The radius of gyration of C with respect to
the axis through C normal to the plane of thefigureis k .
Verify the result of Example 8.1 with your answer.
c
B . 2 - 8 . 4 Find the equations of motion and periods of vi­
bration of the systems shown in Figure P8.2 to P8.4. In
each case, neglect the mass of the rigid bar to which the
ball (particle) is attached.
8.6 A uniform cylinder of mass m and radius R is float­
ing in water. (See Figure P8.6.) The cylinder has a spring
of modulus k attached to its top center point. If the specific
weight of the water is y,findthe frequency of the vertical
bobbing motion of the cylinder. Hint: The upward
(buoyant) force on the bottom of the cylinder equals
the weight of water displaced at any time (Archimedes'
principle).
8.5 The cylinder in Figure P8.5 is in equilibrium in the
position shown. For no slipping, find the natural fre­
quency of free vibration about this equilibrium position.
Figure P8.6
Figure P8.1
8.7 It is possible to determine experimentally the mo­
ments of inertia of large objects, such as the rocket shown
in Figure P8.7. If the rocket is turned through a slight
angle about z and released, for example, it oscillates with
a period of 2.8 sec. Find the radius of gyration k .
c
Figure P 8 . 2
2c
Figure P8.3
Figure P8.7
Figure P 8 . 4
8.8 In the preceding problem, when the rocket is.
caused to swing with small angles about axisttas shown,
the period is observed to be 8 sec. Find from this informa­
tion the value of k .
Xc
Figure P8.5
8.9
Prove statements (1) and (2) on page 522.
Page 532
8.10 Find the frequency of small amplitude oscillations
of the uniform half-cylinder near the equilibrium position
shown in Figure P8.10. Assume that the cylinder rolls on
the horizontal plane.
8.11 A particle of mass m is attached to a light, taut
string. The string is under tension, T, sufficiently large
that the string is, for all practical purposes, straight when
the system is in equilibrium as shown in Figure P8.ll.
Find the natural frequency of small transverse oscillations
of the particle.
• 8.12 The masses in Figure P8.12 are connected by an
inextensible string. Find the frequency of small oscilla­
tions if mass m is lowered slightly and released.
*
The solid homogeneous cylinder in Figure P 8 . 1 3
weighs 200 lb and rolls on the horizontal plane. When the
cylinder is at rest, the springs are each stretched 2 ft. The
modulus of each spring is 15 lb/ft. The mass center C is
given an initial velocity of ft / sec to the right.
a. How far to the right will C go?
b. How long will it take to get there?
c. How long will it take to go halfway to the ex­
treme position?
Mass m
Figure
8.14 A sack of cement of mass m is to be dropped on the
center of a simply supported beam as shown in Figure
P8.14. Assume that the mass of the beam may be ne­
glected, so that it may be treated as a simple linear spring
of stiffness k. Estimate the maximum deflection at the
center of the beam.
• 8 . 1 5 A particle Pof mass m moves on a rough, horizontal
rail with friction coefficient m. (See Figure P8.15.) It is
attached to a fixed point on the rail by a linear spring of
modulus k. The initial stretch of the spring is 7 figm / k.
Describe the subsequent motion if it is known that the
particle starts from rest. Show that the mass stops for
good when
Hint: The differential equa­
tion doesn't have quite the form found in the text; also,
every time the particle reverses direction, so does the fric­
tion force—thus the equation needs rewriting with each
stop.
A spring with modulus 120 lb/in. supports a
200-lb block. (See Figure P8.16.) The block is fastened to
the spring. A 400-lb downward force is applied to the top
of the block at t = 0 when the block is at rest. Find the
maximum deflection of the spring in the ensuing time.
• 8 . 1 7 A block weighing 1 lb is dropped from height
H = 0.1 in. (See Figure P8.17.) If k = 2.5 lb/in., find the
time interval for which the ends of the springs are in
contact with the ground.
P8.10
Figure
P8.14
Figure P 8 . 1 1
Figure
Figure
g = 386 in./sec
Figure P 8 . 1 6
P8.12
Figure
P8.13
P8.15
Figure
P8.17
1
533
8.18 Assume that the slender rigid bar Ein Figure P8.18
undergoes only small angles of rotation. Find the angle of
rotation 6{t) if the bar is in equilibrium prior to f = 0, at
which time the constant force P begins to traverse the bar
at constant speed v.
8.19. Refer to the preceding problem: (a) Find the work
done by P in traversing the bar B; (b) show that this work
equals the change in mechanical energy (which is the
kinetic energy of B plus the potential energy stored in the
spring).
' 8.20 The turntable in Figure P8.20 rotates in a horizontal
plane at a constant angular speed to. The particle P
(mass = m) moves in the frictionless slot and is attached
to the spring (modulus k, free length as shown.
8.23 If k= 100 lb/in. and the mass of the uniform,
slender, rigid bar in Figure P8.23 is 0.03 lb-sec /in., what
damping modulus c results in critical damping? Compare
with the c from Problem 8.22. For this damping, find (t)
if the bar is turned through a small angle
and then
released from rest. If the dashpot were removed, what
would be the period of free vibration?
2
8.24 A cannon weighing 1200 lb shoots a 100-lb cannonball at a velocity of 600 ft/sec. (See Figure P8.24.) It
then immediately comes into contact with a spring of
stiffness 149 lb / ft and a dashpot that is set up to critically
damp the system. Assuming that there is no friction be­
tween the wheels and the plane, find the displacement
toward the wall after sec has elapsed.
a. Derive the differential equation describing the
motion y(t) of the particle relative to the slot.
b. What is the extension of the spring such that P
does not accelerate relative to the slot?
c. Suppose the motion is initiated with the spring
unstretched and the particle at rest relative to
the slot. Find the ensuing motion y(f).
Find the value of c to give critical damping of the
pendulum in Figure P8.21. Neglect the mass of the rigid
bar to which the particle of mass m is attached.
Figure P8.21
8.22 If k = 100 lb/in. and the mass of the uniform,
slender, rigid bar in Figure P8.22 is 0.03 lb-sec / in., what
damping constant c results in critical damping?
2
Figure P 8 . 2 2
Figure P 8 . 1 8
Figure P 8 . 2 3
Figure P 8 . 2 0
Figure P 8 . 2 4
534
8.25 Consider free oscillations of a subcritically damped
oscillator. Do local maxima in the response occur periodi­
cally?
In Figure P8.26, find the steady-state displacement
x(t) if y(t) = 0.1 sin lOOf inch, where Ms in seconds,
m = 0.01 lb-sec /in., k = 100 lb/in., and c = 2 lb-sec/
in. In particular:
2
a. What is the amplitude of x(t)7
b. What is the angle by which x(t) leads or lags y(t)l
Figure P 8 . 2 9
Figure P 8 . 3 0
• 8.30 In Figure P8.30 find the response x (t) for the initial
conditions x (0) = x (0) = 0 if
t
Figure P 8 . 2 6
1
1
k = 100 lb/in.
2
m = 0.01 lb-sec /in.
8.27 In Figure P8.27 find the steady-state displacement
x(t) if y(t) = 0.2 sin 90f inch, where f is in seconds,
m = 0.01 lb-sec /in., = 501b/in.,andc = 1 lb-sec/in.
In particular:
c = 1.0 lb-sec/in.
2
a. What is the amplitude of x{t)l
b. What is the phase angle by which x(t) leads or
lags y(0?
X = 0.05 in.
2
w = 100 rad/sec
• 8.31 Repeat the preceding problem if (a) c = 2.0 lbsec/in.; (b) c = 0.5 lb-sec/in.
Optical equipment is mounted on a table whose
four legs are pneumatic springs. If the table and equip­
ment together weigh 700 lb, what should be the stiffness
of each spring so that the amplitude of steady, simpleharmonic, vertical displacement of the table will not be
greater than 5 percent of a corresponding motion of the
floor? The forcing frequency is 30 rad/sec. Neglect
damping in your calculations.
Figure P 8 . 2 7
The cart in Figure P 8 . 2 8 is at rest prior to t = 0, at
which time the right end of the spring is given the motion
y = vt, where v is a constant. Find x(t).
8.28
• 8.33 The block of mass m in Figure P8.33 is mounted
through springs k and damper c on a vibrating floor. De­
rive an expression for the steady-state acceleration of the
block (whose motion is vertical translation). Show that
the amplitude of the acceleration is less than that of the
floor, regardless of the value of c, provided that co
where co„ is the frequency of free undamped
vibrations of the block. Show further that if
then the smaller the damping the better the isolation.
Figure P 8 . 2 8
B.29 The block in Figure P8.29 is at rest in equilibrium
prior to the application of the constant force P = 50 lb at
f = 0. If k = 100 lb/in., m = 0.01 lb-sec /in., and the
system is critically damped,findx(t).
y = Y
2
Figure P 8 . 3 3
sin wt
535
8.3
Euler's Laws for a Control Volume
Euler's laws describe the relationship between external forces and the
motion of any body whether it be a solid, liquid, or gas. Sometimes,
however, it is desirable to focus attention on some region of space (con­
trol volume) through which material may flow rather than on the fixed
collection of particles that constitute a body. Examples of this sort are
abundant in the field of fluid mechanics and include the important prob­
lem of describing and analyzing rocket-powered flight. Our purpose in
this section is to discuss the forms taken by Euler's laws when the focus of
attention is the control volume rather than the body.
We take as self-evident what might be called the "law of accumula­
tion, production, and transport" — that is, the rate of accumulation of
something within a region of space is equal to the rate of its production
within the region plus the rate at which it is transported into the region.*
Thus, for example, the rate of accumulation of peaches in Georgia equals
the rate of production of peaches in the state plus the net rate at which
they are shipped in. This idea can be applied in mechanics whenever we
are dealing with a quantity whose measure for a body is the sum of the
measures for the particles making up the body. Thus we can apply this
principle to things such as mass, momentum, moment of momentum,
and kinetic energy.
Suppose that at an instant a closed region V (control volume) con­
tains material (particles) making up body B. Let m denote the mass of
body and m denote the mass associated with V (that is, the mass of
whatever particles happen to be in V at some time). Instantaneously
m = m , but because some of the material of is flowing out of V and
some other material is flowing in, m # m . In fact by the accumulation
principle stated above
e
v
v
e
v
B
(net rate of mass flow into V)
(8.29)
since clearly m represents the rate of buildup (accumulation) of mass in
V and since m , the rate of change of mass of the material instantaneously
within V, represents the production term. Of course a body, being a
specific collection of particles, has constant mass; thus m = 0 and (8.29)
becomes m = (rate of mass flow into V), which is often called the conti­
v
B
B
v
nuity equation.
For momentum L, the statement corresponding to Equation (8.29) is
(net rate of flow of momentum into V)
(8.30)
But Euler's first l a w applies to a b o d y (such as ) so that
where
EF is t h e resultant o f t h e external forces on ? — or, in other words, the
resultant o f t h e external forces acting o n t h e material i n s t a n t a n e o u s
The mathematical statement of this is known as the Reynolds Transport Theorem.
in
V. T
Page 536
which is the control-volume form of Euler's first law. The momentum
flow rate on the right of (8.31) is calculated by summing up (or integrat­
ing) the momentum flow rates across infinitesimal elements of the
boundary of V, where the momentum flow rate per unit of boundary area
is the product of the mass flow rate per unit of area and the instantaneous
velocity of the material as it crosses the boundary.
A similar derivation produces a control-volume form of Euler's
second law, for which the result is
(net rate of flow into V of moment
of momentum with respect to O)
where O is a point fixed in the inertial frame of reference.
It is important to realize that nothing in our derivations here has
restricted the control volume except that it be a closed region in space. It
may be moving relative to the frame of reference in almost any imagin­
able way, and it may be changing in shape or volume with time. We
conclude this section with examples of two of the most common applica­
tions of Equation (8.31).
EXAMPLE 8 . 6
A fluid undergoes steady flow in a pipeline and encounters a bend at which the
cross-sectional area of pipe changes from A to A . At inlet 1 the density is p and
the velocity (approximately uniform over the cross section) is v .At outlet 2 the
density is p . Find the resultant force exerted on the pipe bend by thefluid.(See
Figure E8.6a.)
1
2
1
t
2
Figure E 8 . 6 a
Solution
Let the velocity of flow at the outlet be given by u (cos 6i + sin 6)). Then for
steady flow the rate of mass flow at the inlet section is the same as that at the
outlet section:
2
so that
Control
volume V
Figure E 8 . 6 b
Let the control volume be the region bounded by the inner surface of the
pipe bend and the inlet and outlet cross sections. (See Figure 8.6b.) A conse-
Page 537
quence of the condition of steady flow is that within the control volume the
distributions of velocity and density are independent of time. Thus the total
momentum associated with V is a constant and
But
(net rate of momentum flow into V)
Therefore
(net rate of momentum flow out of V)
And if P is the force exerted on the fluid by the bend, then
where p and p are the inlet and outlet fluid pressures respectively. Therefore
1
2
and the force exerted on the bend by the fluid is — P with
EXAMPLE 8 . 7
To illustrate how the control volume form of Euler's first law is used to describe
the motion of a rocket vehicle, consider such a vehicle climbing in a vertical
rectilinear flight. Let v) be the velocity of the vehicle from which combustion
products are being expelled at velocity — v j relative to the rocket. Further let
M(t) be the mass at time t of the vehicle and its contents, let p be the rate of
mass flow of the ejected gases, and let p be the gas pressure at the nozzle exit of
cross section A.
e
Solution
Force D in the free-body diagram (Figure E8.7), representing the drag or resist­
ance to motion, is the resultant of (1) all the shear stresses acting on the surface of
Figure E 8 . 7
Free-body diagram of rocket
Page 538
terms have been separated so that we may point out that the force pA remains
At this point we must approximate L by Mvj, this is an approximation because
v
or
But of course
so that
which is of the form of force = mass X acceleration, where one of the "forces" is
the "thrust" pv .
e
This term may be neglected if exhaust gases have expanded to atmospheric pressure or
nearly so.
PROBLEMS
•
Section 8.3
8.34 Let dm /dt and dm /dt be the respective rates at
which mass enters and leaves a system. Show that Equa­
t
0
8.36 A child aims a garden hose at the back of a friend.
(8.31)weight
may be62.4
expressed
in terms of these r
(See Figure P8.36.) If the watertion
(specific
lb/
ft ) stream has a diameter of {in. and a speed of 50 ft / sec,
estimate the force exerted on the "target" if: (a) he is
stationary; (b) he is running away from the stream at a
speed of 10 ft/ sec. Assume the flow in contact with the
3
where mv = L and it is assumed that all the incoming
particles have a common velocity v,- (in an inertia] frame)
and that all the exiting particles have a common velocity
v
8.35 Liquid of specific weight w flows out of a hole in the
the jet is v, determine the force exerted on the tank by the
supporting structure that holds the tank at rest. Note that
the pressure in the jet will be atmospheric pressure.
side of a tank in a jet of cross section A. If t
Figure P8.36
Page 539
boy's back to be vertical relative to him; that is, neglect
any splashback.
8.37 A steady jet of liquid is directed against a smooth
rigid surface and the jet splits as shown in Figure P8.37.
Assume that each fluid particle moves in a plane parallel
to that of the figure and ignore gravity. Ignoring gravity
and friction, it can be shown that the particle speed after
the split is still v as depicted. Estimate the fraction of the
flow rate occurring in each of the upper and lower
branches. Hint: Use the fact that no external force tangent
to the surface acts on the liquid.
(with torque M applied to the drum on the right) at con­
stant speed v . Find the power that the motor must de­
liver, neglecting friction in the shaft bearings and assum­
ing the belt does not slip. Hint: Use the control volume
indicated by the dashed lines to compute the difference in
belt tensions, neglecting any sag of the belt due to the
weight of the rocks.
B
8.41 Sand is being dumped on aflatcarof mass M at the
constant mass flow rate of q. (See Figure P8.41.) The car is
being pulled by a constant force P, and friction is negligi­
ble. The car was at rest at t = 0. Determine the car's accel­
eration as a function of P, M, q, and t.
Figure P 8 . 3 7
Figure P8.41
8.38 Air flows into the intake of a jet engine at mass flow
rate a (slug/sec or kg/s). If v is the speed of the airplane
flying through still air and u is the speed of engine exhaust
relative to the plane, derive an expression for the force
(thrust) of the flowing fluid on the engine. Neglect the
fact that the rate of exhaust is slightly greater than q be­
cause of the addition of fuel in the engine.
8.39 Revise the analysis of the preceding problem to ac­
count for the mass of fuel injected into the engine. Let/be
the mass flow rate of the fuel, and assume that the fuel is
injected with no velocity relative to the engine housing.
8.40 In a quarry, rocks slide onto a conveyor belt at the
constant mass flow rate k, and at speed v , relative to the
ground. (See Figure P8.40.) The belt is driven by a motor
re
8.42 The pressure in a 90° bend of a water pipe is 2 psi
(gauge). The inside diameter of the pipe is 6 in. and water
8.43 The reducing section in Figure P8.43 connects a 36in. inside-diameter pipe to a 24-in. inside-diameter pipe.
Water enters the reducer at 10 ft /sec and 5 psi (gauge)
and leaves at 2 psi. Find the force exerted on the reducer
by the steadily flowing water.
10 ft/sec
Figure P 8 . 4 3
Figure
P8.40
Page 5 4 0
8.44 If the plane of Figure P8.44 is horizontal, find the
force and moment at O that will allow the body B to
remain in equilibrium when the open stream of water
impinges steadily on it as shown. The stream's velocity is
weight is 62.4 lb/ft .
3
speed relative to the car of 2500 ft / sec. The bullets origi­
nally comprised 2 percent of an initial total mass of m
= 20 slugs. If the system starts from rest at t = 0, find:
60 ft / (b)
sec, how
and its constant area i
(a) the maximum speed of Bonnie and Clyde;
long it takes to attain this speed.
0
1 ft
0.5 ft
30
1 ft
Figure P 8 . 4 6
Figure P 8 . 4 4
8.45 A coal truck weighs 5 tons when empty. Itispushed
under a loading chute by a constant force of 500 lb. The
chute, inclined at 60° as shown in Figure P8.45, delivers
100 lb of coal per second to the truck at a velocity of 30
ft / sec. When the truck contains 10 tons of coal, its
velocity
is 10 ft/sec to the right, (a) What is its acceleration
Figure P 8 . 4 7
8.47 A black box with an initial mass of m (of which 10
percent is box and 90 percent is fuel) is released from rest
on the inclined plane in Figure P8.47. The coefficient of
friction is fi between the box and the plane.
0
60°
500 lb
a. Show that with tan
the box will begin to
slide down the plane.
b. Assume now that tan
and that a mecha­
nism in the box is able to sense its velocity and
eject particles rearward (up the plane) at a con­
stant mass flow rate of k , and at a relative ve­
locity always equal to the negative of the veloc­
ity of the box. Find the velocity of the box at
the time t when the last of the fuel leaves.
0
Figure P 8 . 4 5
8.46 Bonnie and Clyde are making a getaway in a cart
with negligible friction beneath its wheels. (See Fig­
ure P8.46.) Clyde is killing two birds with one stone by
using his machine gun to propel the car as well as to ward
off pursuers. Hefires500 rounds (shots) per minute with
each bullet weighing 1 oz and exiting the muzzle with a
f
c. Show that the box is going 5.5 times faster at t
= t than it would have gone if no fuel had
been ejected.
f
8.48 Santa Claus weighs 450 lb and drops down a 20-ft
chimney (Figure P8.48). He gains mass in the form of
Page 541
ashes and soot at the rate of 3 slugs / sec from a very dirty
chimney.
a. Find Santa's velocity as a function of time. (Ne­
glect friction.)
b. Calculate the velocity v and the time t at
which he would hit bottom without adding
mass and then compare v with his "ashes and
soot velocity" at the same t .
b
b
b
b
which the pressure force pA and the drag D are negligible,
the initial mass of the rocket is m , and the gravitational
acceleration g, the rate p, and the relative velocity v are all
constants. Initially, the rocket is at rest.
0
t
8.52 Repeat the preceding problem, but this time in­
clude a drag force of —to,where k is a positive constant.
8.53 (a) Extending Problem 8.51, find the height of the
rocket as a function of rime, (b) If the fraction of m which
is fuel is /, find the rocket's "burnout" velocity and posi­
tion when all the fuel is spent.
0
8.54 Spherical raindrops produced by condensation are
precipitated form a cloud when their radius is a. They fall
freely from rest, and their radii increase by accretion of
moisture at a uniform rate k. Find the velocity of a rain­
drop at time f, and show that the distance fallen in that
time is
' 8.55 A chain of length L weighing y per unit length
begins to fall through a hole in a ceiling. (See Fig­
ure P8.55.) Referring to the hint in Problem 8.50:
Figure P8.48
8.49 A small rocket is fired vertically upward. Air resist
ance is neglected. Show that for the rocket to have con
stant acceleration upward, its mass m must vary with time
t according to the equation
a. Find v(x) if v = 0 when x and t are zero.
b. Show that the acceleration of the end of the
falling chain is the constant g/ 3.
c. Show that when the last link has left the ceil­
ing, the chain has lost more potential energy
than it has gained in kinetic energy, the differ­
ence being yL /6. Give the reason for this loss.
2
where a is the acceleration of the rocket and u is the
velocity of the escaping gas relative to the rocket.
B.50 The end of a chain of length L and weight per unit
length w, which is piled on a platform, is lifted vertically
by a variable force P so that it has a constant velocity v.
(See Figure P8.50.) FindP as a function of x. Hint: Choose
a control-volume boundary so that material crosses the
boundary (with negligible velocity) just before it is acted
on by the moving material already in the control volume.
That is, there is no force transmitted across the boundary
of the control volume. The solution will be an approxima­
tion to reality because of assuming arbitrarily small indi­
vidual links; but the more links having the common ve­
locity of the fully engaged links within the volume, the
better the approximation will be.
8.51 Solve the final equation of Example 8.7 for the ve­
locity if j of the rocket as a function of time in the case in
constant
Figure PB.50
Figure P8.55
Page 542
8.56 The machine gun in Figure P8.56 has mass M ex­
clusive of its bullets, which have mass M ' in total. The
bullets arefiredat the mass rate of Kg "slugs" per second,
with velocity u relative to the ground. If the coefficient of
friction between the gun's frame and the ground is fi, find
the velocity of the gun at the instant the last bullet is fired.
in the gravitational attraction, (a) show that the greatest
upward speed is attained when the mass of the rocket is
reduced to M, and determine this speed, (b) Show also
that the rocket rises to a height
0
8.60 With the same notation and conditions as in Prob
lem 8.34, show that Equation (8.32^may be written as
where r, and i are position vectors for the mass centers of
the incoming and exiting particles.
0
Coefficient of friction
Figure P 8 . 5 6
•
A pinwheel of radius a, which can turn freely about
a horizontal axis, is initially of mass M and moment of
inertia I about its center. A charge is spread along the rim
and ignited at time t = 0. While the charge is burning, the
rim of the wheel loses mass at a constant rate m mass
units per second, and at the rim a mass m of gas is taken
up per second from the atmosphere, which is at rest. The
total mass m, + m is discharged per second tangentially
from therim,with velocity v relative to the rim. Prove that
if 8 is the angle through which the wheel has turned after
t sec, then
x
2
" 8 . 5 7 A particle of mass m, initially at rest, is projected
with velocity v at an angle a to the horizontal and moves
under gravity. (See Figure P8.57.) During its flight, it
gains mass at the uniform rate k. If air resistance is ne­
glected, show that its equation of motion is
0
2
and that the equation of its path is
where
• 8.62 A wheel of radius a starts from rest and fires out
matter at a uniform rate from all points on the rim (Fig­
ure P8.62). The matter leaves tangentially with relative
speed v and at such a rate that the mass decreases at the
rim by m mass units per second. Show that the angle 8
turned through by the wheel is given by
Figure P 8 . 5 7
in which I is the initial moment of inertia of the wheel
about its axis.
0
•8.58 If in the preceding problem the air offers a resist­
ance —ci, determine the equation of the path.
•
From a rocket that is free to move vertically up­
ward, matter is ejected downward with a constant rela­
tive velocity gT at a constant rate 2M/T. Initially the
rocket is at rest and has mass 2M, half of which is avail­
able for ejection. Neglecting air resistance and variations
Figure P 8 . 6 2
Page 543
8.4
Central Force Motion
In Chapter 2 we defined a central force acting on a particle pas one which
(1) always passes through a certain point O fixed in the inertial reference
frame J and (2) depends only on the distance r between O and p. (See
Figure 8.12.) In this section we are going to treat the central force in more
detail. We shall go as far as we can without specializing F(r) — that is,
without saying how F depends on r. In the second part of the section we
shall study the most important of central forces: gravitational attraction.
If the central force F is the only force acting on the particle, then
F = ma; and since the central force always passes through point O, r X F
is identically zero. These two facts allow us to write:
Figure 8.12
or, since r= v,
Therefore for a particle acted on only by a central force,
(8.33)
= constant vector in
Dotting this equation with r, we find, since r X v is perpendicular to r,
and we see that r is always perpendicular to a vector that is constant in J;
therefore Amoves in a plane in
Using polar coordinates to then de­
scribe the motion of pm this plane, the governing equations are:
(8.34)
and
(8.35)
From Equation (8.35) we see immediately that
(8.36)
constant
where h is the magnitude of the constant vector h of Equation (8.33)
because, expressing r and v in polar coordinates, we find
Q
o
constant
so that
constant
(8.37)
Equation (8.37) is a statement of the conservation of the moment of
momentum, or angular momentum of P; the constant h is the magnitude
of the angular momentum H of p divided by its mass m. Thus we shall
call h the angular momentum (magnitude) per unit mass.
0
o
0
Page 544
We can use the previous pair of results to show that the second of
Kepler's three laws of planetary motion is in fact valid for any central
force. This law states that the radius vector from the sun to a planet
sweeps out equal areas in equal time intervals. From Figure 8.13 the
incremental planar area A swept out by p between 8 (at t) and 6 + A6
(at i + t) is approximately given by the area of the triangle OBB'*
(see Figure 8.14):
base
sin
ura 8.13
height
Dividing by the time increment t and taking the limit as
have
we
or
Figure 8.14
2
(a constant that is r 8/ 2)
(8.38)
Thus the rate of sweeping out area is a constant. This is why a satellite
or a planet in elliptical orbit (Figure 8.15) has to travel faster when it is
near the perigee than the apogee—the same area must be swept out in the
same period of time.' We emphasize again that this result is valid for all
central force trajectories, not just elliptical orbits and not just if the central
force is gravity.
Perigee P*
A pogce
Figure B,15
Next we focus our attention on the most important central force:
gravitational attraction. If G is the universal gravitation constant and M
and m are the masses of what we are considering to be the attracting and
attracted bodies,* then the central force acting on m for this case is
* " A p p r o x i m a t e l y ' ' b e c a u s e t h e a r e a b e t w e e n a r c a n d c h o r d i s o u t s i d e t h e triangle.
W e u s e perigee a n d apogee in a g e n e r a l s e n s e ; t e c h n i c a l l y t h e s e w o r d s refer t o t h e n e a r ­
est a n d f a r t h e s t p o i n t s , respectively, f o r t h e m o o n a n d artificial satellites. F o r t h e orbits
of p l a n e t s , t h e p r o p e r t e r m s a r e perihelion a n d aphelion.
Actually, of course, b o t h a r e attracting a n d b o t h a r e attracted — e a c h t o t h e other! T h e
c o n s t a n t G M is, f o r t h e s u n , 4 . 6 8 X 1 0 f f / s e c .
2 1
3
J
Page 545
(8.39)
and Equation (8.34) becomes
(8.40)
2
Canceling m and inserting h for r 8 gives
a
(8.41)
Multiplying Equation (8.41) by r will allow us to integrate it:
(8.42)
Integrating, we get
(8.43)
2
If we multiply Equation (8.43) by m and replace h by r O, we see that
Q
(8.44)
and the left side of Equation (8.44) is seen to be the total energy of p,
kinetic plus potential. Thus we shall replace C by E, the energy of p-pex
unit mass, and obtain
t
(8.45)
This equation will be helpful to us later. But now we are interested in
studying the trajectory of particle p— that is, in finding r as a function of
By the chain rule,
and since
from Equation
(8.46)
We need the second derivative of r in Equation (8.41), so we apply the
chain rule once more:
(8.47)
Page 5 4 6
Substituting into Equation (8.41), we get
or
(8.48)
The following simple change of variables will make the solution to this
differential equation immediately recognizable:
(8.49)
Substituting Equation (8.49) into (8.48) along with
(8.50)
gives
(8.51)
or
(8.52)
The solution to Equation (8.52), from elementary differential equations,
consists of a homogeneous (or complementary) part plus a particular
part:
(8.53)
Switching variables back from u to r by Equation (8.49), we obtain
(8.54)
This solution for r(6) is the equation of a conic; it can be put into a
more recognizable form after a brief review of conic sections. For every
point P on a conic, the ratio of the distances from P to a fixed point (O: the
focus) and to a fixed line (l: the directrix) is a constant called the eccen­
tricity of the conic:
(8.55)
Therefore, in terms of the parameters in Figure 8.16,
Figure 8 . 1 6
(8.56)
Page 547
or, solving for r,
(8.57)
The conic specified by Equation (8.57) is a:
Hyperbola if | e \ > 1
Parabola if e = 1
Ellipse if - 1 < e < 1
(8.58)
The ellipse becomes a circle if e = 0. It is an ellipse with perigee (closest
point t o O ) a t
i if 0 < e < 1 and an ellipse with apogee (farthest
point from O) at
if — 1 < e < 0; this latter type is called a subcircular ellipse.
Returning to our solution (8.54) for
, it is customary to select one
of the constants
and B j so that, as suggested by Figure 8.16, dr/d
= 0 when
. This condition easily gives B = 0, as the reader may
wish to demonstrate using calculus. The result simply means that we are
measuring 8 from the perigee of the conic. At this point we should
compare Equations (8.57) and (8.54) with B = 0:
1
t
(8.59)
and
(8.60)
By direct comparison of these two expressions for r, we see that
It is more customary, however, to express the constant A (as well as the
eccentricity) in terms of the energy E of the orbit. To do this, Equa­
tions (8.45) and (8.46) give
1
(8.61)
At the point r where 8 = 0 and dr /dd = 0, we see that
P
(8.62)
Thus not all of h , r , and E are independent. We shall eliminate r .
Multiplying Equation (8.62) by r , we get
0
P
P
(8.63)
Solving via the quadratic formula, we have
(8.64)
Page 5 4 8
in which we use the plus sign since we need the smaller root for closed
conies (£ < 0). The positive sign also ensures a positive r for open conies
(E > 0).
Returning to our solution (8.59), when 6 = 0 then
P
(8.65)
Equating the two expressions for r>, Equations (8.64) and (8.65), we can
solve for A .
We see by comparing Equations (8.59) and (8.60) that the eccentric­
ity e of our conic will be (h} /GM)A .
Equating the right sides of Equa­
tions (8.64) and (8.65) and solving for this quantity, we get
t
)
1
(8.66)
Therefore
(8.67)
which expresses r as a function of 9, the constant GM, the energy E, and
the angular momentum per unit mass h . Note that by again comparing
Equations (8.59) and (8.60) we can obtain the distance a between the
focus O and the directrix /:
Q
(8.68)
Thefirstof Kepler's three laws of planetary motion states that the
planets travel in elliptical orbits with the sun at one focus.* These ellipses
are very nearly circular for most of the planets; the eccentricity of earth is
e = 0.017. To obtain the third of Kepler's laws, we return once more to
our equations and obtain for elliptic orbits, from (8.67), the distance r
when 8 = 90°:
(8.69)
This distance, the semilatus rectum, may be used to express the distance r
between the focus O and apogee A*, and the distance r between O and
the perigee P*. (See Figure 8.17.) At apogee, 9 — n and Equations (8.59),
(8.60), and (8.69) give
A
P
(8.70)
Figure 8.17
Kepler's laws, based on his astronomical observations and set forth in 1609 and 1619,
were studied by Newton before the Englishman published the Principia, which con­
tained his own laws of motion.
Page 549
and at perigee
= 0),
(8.71)
The semimajor axis length of the ellipse is
(8.72)
and the seinirninor axis length is, from analytic geometry,
(8.73)
An ellipse has area
or
(8.74)
With these results in hand, we shall now prove Kepler's third law.
Since dA / dt is constant,
(8.75)
where we take A = 0 when t = 0, say at the perigee. Over one orbit we
have, with T being the orbit period,
area of ellipse
(8.76)
Since (from Equations (8.69) and (8.72))
(8.77)
we obtain the following from Equation (8.76):
(8.78)
so that
or
(8.79)
Page 550
Equation (8.79) states the third of Kepler's laws: The squares of the
planets' orbital periods are proportional to the cubes of the semimajor
axes of their orbits.
EXAMPLE 8 . 8
Calculate the semimajor axis length of an earth satellite with a period of 90 min.
Solution
We can solve this problem using Kepler's third law. The weight of a particle
(mass m) on the earth's (mass M) surface is both mg and GMm
thus
and we see that the product of the unwieldy constants G and M is
Therefore
which is about 170 mi above the earth's surface.
We shall present one more example on elliptical orbits under gravity,
but first we need two equations relating the velocities v and v at any two
points P and P with radii r and r on the orbit. The first of these comes
from Equation (8.37), which states that
= constant. From Fig­
ure 8.18, since the velocity v is always tangent to the path, we see that if
is the angle between r and v, then in cylindrical coordinates
1
1
2
1
v sin
2
2
= (transverse component of v) =
so that
constant
or, for two points P and P on the orbital path,
1
2
(8.80)
Figure 8 . 1 8
Note that at apogee and perigee, = 9 0 ° . Thus letting P and P be these
two points, we get from Equation (8.80)
1
2
(8.81)
and the two velocities are inversely proportional to the radii, with v being
faster at perigee as we have already seen from Kepler's second law.
The other equation relating v and v comes from the potential for
gravity, which from Equation (2.27) and Example 8.8 is
1
2
Page 551
Using conservation of energy between P and P ,
1
2
(8.82)
If we let point P represent the perigee P*, as suggested in Figure 8.18,
then Equation (8.80) becomes
2
(8.83)
=
n
where sin
s i 90° = 1. Now if r v and (p^ are initial (launch)
values of r, v, and
then we may consider these as given quantities.
Substituting Equation (8.83) into (8.82), we can obtain an equation for
the perigee radius
2
1 #
Multiplying through by
l t
and rearranging, we get
(8.84)
We see that Equation (8.84) is simply a quadratic equation in the ratio
and that [2GM /
is a nondimensional parameter of the
orbit. We now illustrate the use of this important equation in an example.
EXAMPLE 8 . 9
A satellite is put into an orbit with the following launch parameters: H
= 1000 mi, i>! = 17,000 mph, and 0 = 100° (see Figure E8.9). Find the apogee
and perigee radii of the resulting orbit.
Solution
3
2
We need GM in mi /hr ; therefore
Figure E 8 . 9
The parameter 2GM
in Equation (8.84) is therefore
Page 552
Equation (8.84) becomes
The quadratic formula gives
= 0.911 and 1.46
Therefore
0.911(4960) = 4520 mi
The other root corresponds to the apogee. (Since the starting condition of sin
= 1 is the same for apogee and perigee, both answers are produced by the
quadratic formula!)
1.46(4960) = 7240 mi
The altitudes are
Perigee height = 4520 - 3960 = 560 mi
Apogee height = 7240 - 3960 = 3280 mi
To pin down the orbit in space, we need to know the angle to the perigee point
from the launch point and also the orbit's eccentricity. Problems 8.81 and 8.82
will be concerned with finding these two quantities given initial values of r,v, and
PROBLEMS
•
Section 8.4
8.63 Show that a satellite in orbit has a period T given by
T=
8.64 Show that, for a body in elliptical orbit (Fig­
ure P8.64), b
sages have been recorded since 240 B.C.!) What is the
approximate semimajor axis length of Halley's comet?
(Use 76 years as the period.)
8.66 Find the minimum period of a satellite in circular
orbit about the earth. Upon what assumption is your an­
swer based?
8.67 Repeat the preceding problem if the satellite orbits
the moon. Assume
&moon
r
Figure P 8 . 6 4
8-65 Halley's comet's latest return to Earth was in 1986.
The comet orbits the sun in an elongated ellipse every 74
to 79 years; the period varies due to perturbations in its
orbit caused by the four largest (Jovian) planets. (Its pas­
moon
earth
—
0.27r
earth
8.68 Show that for circular orbits around an attracting:
body of mass M, rv = GM. Then use the 93 X 10 mi
average orbital radius of earth, and the fact that its orbit is
nearly circular, to find the constant GM, for the sun as
attracting center (heliocentric system).
2
6
8.69 The first artificial satellite to orbit the earth was the
Russians' Sputnik I. Following insertion into orbit it had a
Page 553
period of 96.2 min. Find the semimajor axis length. If the
initial eccentricity was 0.0517, find the maximum and
minimum distances from earth following its injection into
orbit.
8.70 Show that if a satellite is in a circular orbit at radius r
around a planet of mass M, the velocity to which it must
increase to escape the planet's gravitational attraction is
given by
200 mi
2800 mi
Find Vnayc if "to which" is replaced by "by which."
Figure P8.77
8.71 Show that if the launch velocity in Example 8.9 is
15,000 mi/hr, the satellite will fail to orbit the earth.
8.72 Using Equation (8.54), show that B, = 0 follows
from the condition dr/d0=O when 0=0.
8.73 Prove that Equation (8.66) follows from (8.64) and
(8.65).
8.74 Find the form of the central force F(r) for which all
circular orbits of a particle about an attracting center O
have the same angular momentum (and the same rate of
sweeping out area).
8.75 Show that, in terms of the radiusrp.and speedvp. at
perigee, the energy and eccentricity of the o
Figure P8.78
will send the satellite to position A, at radius R (<R )<
180° away. Then find the second negative velocity incre­
ment, this time applied at A, that will put the satellite in a
circular orbit of radius R . Hint: Use Problem 8.76.
2
and
1
2
8.76 Use Equations (8.81) and (8.82) to show that the
velocities at apogee and perigee, in terms of the known
radii r . and rp., are:
A
B.79 A satellite has
= 8000 mi and tp. = 5000 mi. If
it was launched with a velocity of 15,000 mi/hr, what
was its launch radius? What was the angle between r
and v at launch? Hint: Use Problem 8.76.
8.80 What is the period of the satellite in the preceding
problem?
Show that if the launch parameters, T ,V , and ,
are known, then the angle from perigee to the launch
point of a satellite in orbit is given by
•8.81
and
• 8.77 A rocket is in a 200-mi-high circular parking orbit
above a planet. What velocity boost at point P will result
in the new, elliptical orbit shown in Figure P8.77? HintUse the results of Problem 8.75.
• 8.78 A satellite is in a circular orbit of radius R . (See
Figure P8.78.) Find the (negative) velocity increment that
1
1
tan
GM
C3
1
GM
• B.82 Show that if the launch parameters r,, v , and
are known, then the eccentricity of the resulting conic is:
t
Page 554
* 8.83 A large meteorite approaches the earth. (See Fig­
ure P8.83.) Measurements indicate that at a given time it
has a speed of 8000 mph at a radius of 100,000 mi. Will it
orbit the earth? If so, what is the period? If not, what is the
maximum velocity v that would have resulted in an orbit?
Hint: Use the result of the preceding problem.
8.84 Classify the various orbits according to values of
the dimensionless parameter GM/(r i>o) for a satellite
launched with the conditions of Figure P8.84.
0
Find the kinetic energy increase needed to move a
satellite from radius R to nR(n > 1) in circular orbits. Hint:
Use Problem 8.76.
•8.85
A particle of mass m moves in the xy plane under
the influence of an attractive central force that is propor­
tional to its distance from the origin (F(r) = kr). It has the
same initial conditions as Problem 8.84. Find the largest
and smallest values of r in the ensuing motion.
*8.86
8.87 A satellite has apogee and perigee points 1000 and
180 mi, respectively, above the earth's surface. Compute
the satellite's period.
Figure P 8 . 8 3
REVIEW QUESTIONS
Figure P 8 . 8 4
•
8.88 In the preceding problem find the speeds of the
satellite at perigee and at apogee. Hint: Use Problem 8.76.
Chapter 8
True or False?
1. Frequencies of vibration of periodic motion are associated only with
small amplitudes.
2. Natural frequencies of vibration are associated only with translational motion.
3. Natural frequencies of vibration of bodies in a gravitational field do
not have to depend on "g."
4. Free vibrations will always decrease in time due to "real world
damping."
5. There are three types of damped vibrations, and the subcritical case
has the greatest practical importance.
6. In forced vibration, the steady-state part of the response dies out due
to damping inherent in the physical system.
7. In general, the rate of accumulation of momentum within a region of
space equals the rate at which it is transported into that region.
In using the control volume form of Euler's Laws, the control vol­
ume:
B. has to be fixed in the inertial frame of reference;
9. may change in shape or volume with time;
10. has to be a closed region of space.
11. In a one-dimensional control-volume problem, Euler's first law be­
comes, in general,
Page 555
12. The law of accumulation, production, and transport applies only to
scalar quantities.
13. A central force depends only on the distance r between the attracted
and ataacting particles.
14. There are central forces besides gravity.
15. In every central force problem the angular momentum about the
attracting point O is a constant.
16. All three of Kepler's laws apply to motions of a particle under the
influence of any type of central force.
17. All central force problems result in paths which are conies.
18. In a gravitational central force problem, the type of conic is deter­
mined by the eccentricity.
Answers
1.F 2. F 3.T 4.T
16. F 17. F 18. T
5. T
6. F
7. F
8. F
9. T
10. T
11. F
12. F
13. T
14. T
15. T
•
•
•
APPENDICES CONTENTS
Appendix A UNITS
Appendix B EXAMPLES OF NUMERICAL ANALYSIS/THE NEWTONRAPHSON METHOD
Appendix C MOMENTS OF INERTIA OF MASSES
Appendix D ANSWERS TO ODD-NUMBERED PROBLEMS
Appendix E ADDITIONAL MODEL-BASED PROBLEMS FOR
ENGINEERING MECHANICS
Page 557
A
UNITS
The numerical value assigned to a physical entity expresses the relationship of
that entity to certain standards of measurement called units. There is currently an
international set of standards called the International System (SI) of Units, a
descendant of the meter-kilogram-second (mks) metric system. In the SI system
the unit of time is the second (s), the unit of length is the meter (m), and the unit
of mass is the kilogram (kg). These independent (or basic) units are defined by
physical entities or phenomena. The second is defined by the period of a radia­
tion occurring in atomic physics; the meter is defined by the wavelength of a
different radiation; the kilogram is defined to be the mass of a certain body of
material stored in France. Any other SI units we shall need are derived from these
three basic units. For instance, the unit of force, the newton (N), is a derived
quantity in the SI system, as we shall see.
Until recendy almost all engineers in the United States used the system
(sometimes called British gravitational or U.S.) in which the basic units are the
second (sec) for time, the foot (ft) for length,* and the pound (lb) for force. The
pound is the weight, at a standard gravitational condition (location), of a certain
body of material stored in the United States. In this system the unit of mass, the
slug, is a derived quantity. It is a source of some confusion that sometimes there is
used a unit of mass called the pound (the mass whose weight is one pound of
force at standard gravitational conditions); also, particularly in Europe, the term
kilogram is also sometimes used for a unit of force.f Grocery shoppers in the
United States are exposed to this confusion by the fact that packages are marked
by weight (or is it mass?) both in pounds and in kilograms. Throughout this text,
without exception, the pound is a unit of force and the kilogram is a unit of mass.
The United States is currently in the painful process of gradual changeover
to the metric system of units after more than 200 years of attachment to the U.S.
system. The new engineers who begin practicing their profession in the 1990s
will doubtless encounter both systems, and thus it is crucial to master both
(including thinking in terms of the units of either) and to be able to convert from
one to the other. The units mentioned here are summarized in Table A. 1 for the SI
and the U.S. systems.
* Sometimes, particularly in the field of mechanical vibrations, the inch is used as the
unit of length; in that case the unit of mass is 1 lb-sec /in., which equals 12 slugs.
| A kilogram was a force unit in one of two mks systems, compounding the misunder­
standing.
2
Page 558
Page 559
Table A.1
Quantity
SI (Standard International
or "Metric") Unit
U.S. Unit
force
mass
length
time
newton (N)
kilogram (kg)
meter (m)
second (s)
pound (lb)
slug
foot (ft)
second (sec)
We now examine how the newton of force is derived in SI units and the slug
of mass is derived in U.S. units. Let the dimensions of the four basic dimensional
quantities be labeled as F (force), M (mass), L (length), and T (time). From the first
law of motion (discussed in detail in Chapter 2), F = ma, we observe that the four
basic units are always related as follows:
F=
This means, of course, that we may select three of the units as basic and derive the
fourth. Two ways in which this has been done are the gravitational and the
absolute systems. The former describes the U.S. system; the latter describes SI.
(See Table A.2.) Therefore, in U.S. units the mass of an object weighing W lb is
W / 3 2 . 2 slugs. Similarly, in SI units the weight of an object having a mass of M kg
is 9.81M newtons.
Table A.2
Gravitational System
Absolute System
The basic units are force, length, and
time, and mass is derived:
The basic units are mass, length,
and time, and force is derived:
This system has traditionally been
more popular with engineers
As an example, in the U.S.
system of units the pound, foot, and
second are basic Thus the mass
unit, the slug, is derived:
This system has traditionally been
more popular with physicists.
As an example, in the SI
(metric) system of units the
kilogram, meter, and second are
basic. Thus the force unit, the
newton, is derived:
This is s u m m e d up by: A slug is the
quantity of mass that will be
accelerated at 1 f t / s e c w h e n acted
upon by a force of 1 lb
2
This is s u m m e d up by: A newton is
the amount of force that will
accelerate a mass of 1 kg at 1 m / s
2
In the SI system the unit of moment of force is the newton • meter (N • m);
in the U.S. system it is the pound-foot (lb-ft). Work and energy have this same
dimension; the U.S. unit is the ft-lb whereas the SI unit is the joule (J), which
equals 1 N • m. In the SI system the unit of power is called the watt (W) and
equals one joule per second ( J / s ) ; in the U.S. system it is the ft-lb/sec. The unit of
Page 560
2
pressure or stress in the SI system is called the pascal (Pa) and equals 1 N / m ; in
the U.S. system it is the l b / f t , although often the inch is used as the unit of length
so that the unit of pressure is the l b / i n . (or psi). In both systems the unit of
frequency is called the hertz (Hz), which is one cycle per second. Other units of
interest in dynamics include those in Table A.3.
2
2
Table A3
Quantity
SI Unit
U.S. Unit
velocity
angular velocity
acceleration
angular acceleration
mass m o m e n t of inertia
momentum
moment of m o m e n t u m
impulse
angular impulse
mass density
specific weight
m/s
rad/s
m/s
rad/s
kg . m
kg • m / s
kg • m / s
N • s ( = kg • m/s)
N • m • s ( = kg • m / s )
kg/m
N/m
ft/sec
rad/sec
ft/sec
rad/sec
slug-ft
slug-ft/sec
slug-ft /sec
lb-sec
Ib-ft-sec
slug/ft
lb/ft
2
2
2
2
2
2
2
2
2
3
3
3
3
Moreover, in the SI system there are standard prefixes to indicate multiplica­
tion by powers of 10. For example, kilo (k) is used to indicate multiplication by
1000, or 1 0 ; thus 5 kilonewtons, written 5 kN, stands for 5 X 1 0 N. Other
prefixes that commonly appear in engineering are shown in Table A.4. We reemphasize that for the foreseeable future American engineers will find it desirable to
know both the U.S. and SI systems well; for that reason we have used both sets of
units in examples and problems throughout this book.
3
3
Table A.4
tera
giga
mega
kilo
hecto
deka
deci
T
G
M
k
h
da
d
12
10
10
10
10
10
10
109
6
3
2
1
1
centi
milli
micro
nano
pico
femto
atto
c
m
µ
n
P
f
a
-2
10
1C
1C
10
10
10
10
-3
-6
-9
-12
-15
-8
We turn now to the question of unit conversion. The conversion of units is
quickly and efficiently accomplished by multiplying by equivalent fractions until
the desired units are achieved. Suppose we wish to know how many newtonmeters (N m) of torque are equivalent to 1 lb-ft. Since we know there to be
3.281 ft per meter and 4.448 N per pound,
1 lb-ft = 1
= 1.356 N • m
Note that if the undesired unit (such as lb in this example) does not cancel, the
conversion fraction is upside-down!
For a second example, let us find how many slugs of mass there are in a
kilogram:
Page 561
lkg =
= 0.06852 slug
Inversely, 1 slug = 14.59 kg. A set of conversion factors to use in going back and
forth between SI and U.S. units is given in Table A.5.*
Table A.5
To Convert From
To
Reciprocal (to Get
from SI to U.S. Units)
Multiply By
Length, area, volume
foot (ft)
inch (in.)
statute mile (mi)
f o o t (ft )
inch (in. )
f o o t (ft )
inch (in. )
meter (m)
m
m
meter (m )
m
meter (m )
m
0.30480
0.025400
1609.3
0 092903
6.4516 X
0.028317
1 6387 X
Velocity
feet/second (ft/sec)
f e e t / m i n u t e (ft/min)
knot (nautical mi/hr)
mile/hour (mi/hr)
mile/hour (mi/hr)
meter/second (m/s)
m/s
m/s
m/s
kilometer/hour (km/h)
0.30480
0.0050800
0.51444
0.44704
1.6093
3.2808
196.85
1 9438
2.2369
0 62137
2
2
2
3
3
2
2
2
3
3
2
3
3
3
10
-4
-5
10
3.2808
39.370
6.2137 X 1 0 10.764
1550.0
35.315
61024
Acceleration
f e e t / s e c o n d (ft/sec )
i n c h / s e c o n d (in./sec )
m e t e r / s e c o n d (m/s )
m/s
0.30480
0.025400
3.2808
39.370
Mass
pound-mass (Ibm)
slug (lb-sec /ft)
kilogram (kg)
kg
0.45359
14.594
2.20462
0.068522
Force
pound (lb) or
pound-force (Ibf)
newton (N)
4.4482
0.22481
2
2
2
2
2
2
2
2
Density
pound-mass/inch (lbm/in. )
pound-mass/foot (lbm/ft )
s l u g / f o o t (slug/ft )
kg/m
kg/m
kg/m
Energy, work, or moment of force
foot-pound or pound-foot
(ft-lb)
(lb-ft)
joule (J)
or newton • meter (N • m)
1.3558
0.73757
Power
foot-pound/minute (ft-lb/min)
horsepower (hp) (550 ft-lb/sec)
watt (W)
W
0.022597
745.70
44.254
0 0013410
Stress, pressure
p o u n d / i n c h ( l b / i n . or psi)
p o u n d / f o o t (lb/ft )
N / m (or Pa)
N / m (or Pa)
6894.8
47.880
1.4504 X
0.020886
1.3558
0.73756
3
3
3
3
3
3
2
2
2
2
2 7680 X 1 0
16.018
515.38
3
3
2
2
2
Mass moment of inertia
slug-foot (slug-ft or lb-ft-sec )
2
3
2
kg • m
2
4
3.6127 X 10
0.062428
0.0019403
4
-5
-4
10
• Rounded to the five digits cited. Note, for example, that 1 ft = 0.30480 m, so that
(Number of feet) X
= number of meters
Page 562
Table A.5
Continued
To Convert From
To
Reciprocal (to Get
from SI to U.S. Units)
Multiply By
Momentum (or linear
momentum)
slug-foot/second (slug-ft/sec)
kg • m / s
4 4482
0.22481
Impulse (or linear impulse)
pound-second (lb-sec)
N • s (or kg • m/s)
4 4482
0 22481
1 3558
0 73756
1 3558
0 73756
Moment of momentum (or angular
slug-foot /second (slug-ft /sec)
2
2
Angular impulse
pound-foot-second (Ib-ft-sec)
momentum)
kg • m / s
2
2
N • m • s (or kg • m / s )
Note that the units for time (s or sec), angular velocity (rad/s or 1/s), and
angular acceleration ( r a d / s or 1/s ) are the same for the two systems. To five
digits, the acceleration of gravity at sea level is 32.174 f t / s in the U.S. system
and 9.8067 m / s in SI units.
We wish to remind the reader of the care that must be exercised in numerical
calculations involving different units. For example, if two lengths are to be
summed in which one length is 2 ft and the other is 6 in., the simple sum of these
measures, 2 + 6 = 8, does not of course provide a measure of the desired length.
It is also true that we may not add or equate the numerical measures of different
types of entities; thus it makes no sense to attempt to add a mass to a length.
These are said to have different dimensions. A dimension is the name assigned to
the kind of measurement standard involved as contrasted with the choice of a
particular measurement standard (unit). In science and engineering we attempt
to develop equations expressing the relationships among various physical enti­
ties in a physical phenomenon. We express these equations in symbolic form so
that they are valid regardless of the choice of a system of units, but nonetheless
they must be dimensionality consistent. In the following equation, for example, we
may check that the units on the left and right sides agree; r is a radial distance, P is
a force, and dots denote time derivatives:
2
2
2
2
2
P — mg cos 6 = m(r — r0 )
Dimensions of
P
are
F
mg cos d
mr
2
-mr8
Therefore the units of (every term in) the equation are those of force. If such a
check is made prior to the substitution of numerical values, much time can be
saved if an error has been made.
Page 563
PROBLEMS
•
Appendix A
A.1 Find the units of the universal gravitational con­
stant G, defined by
A.6
Is the following equation dimensionally correct?
(v = velocity;
a = acceleration)
in (a) the SI system and (b) the U.S. system.
A.2
Find the weight in pounds of 1 kg of mass.
A. 3
Find the weight in newtons of 1 slug of mass.
A.4 One pound-mass (lbm) is the mass of a substance
that is acted on by 1 lb of gravitational force at sea level.
Find the relationship between (a) 1 lbm and 1 slug; (b)
1 lbm and 1 kg.
A. 5 The momentum of a body is the product of its mass
m and the velocity v of its mass center. A child throws an
8-oz ball into the air with an initial speed of 20 mph. Find
the magnitude of the momentum of the ball in (a) slugft/sec; (b) kg • m / s .
c
A. 7 The equation for the distance r from the center of
the earth to the geosynchronous satellite orbit is
5
(i
= angular speed of earth;
= earth radius)
a. Show that the equation is dimensionally correct.
b. Use the equation to find the ratio of the orbit
radius to earth radius.
A.8 The universal gravitational constant is G = 6.67
X 10 N • m / k g . Express G in units of lb-f^/slug .
-1
2
2
2
B
B EXAMPLES OF NUMERICAL ANALYSIS / THE NEWTON-RAPHSON METH
There are a few places in this book where equations arise whose solutions are not
easily found by elementary algebra; they are either polynomials of degree higher
than 2 or else transcendental equations. In this appendix we explain in brief the
fundamental idea behind the Newton-Raphson numerical method for solving
such equations. We shall first do this while applying the method to the solution
for one of the roots of a cubic polynomial equation that occurs in Chapter 7.
To solve the cubic equation of Example 7.5,
we could, alternatively, use the Newton-Raphson algorithm. This procedure
finds a root of the equation
= 0 (it need not be a polynomial equation,
however) by using the slope of the curve. The algorithm, found in more detail in
any book on numerical analysis, works as follows. If
is an initial estimate of a
root , then a better approximation is
Figures B. 1 and B.2 indicate what is happening. The quantity
causes
a backup in the
approximation — in our case from the initial value of 3 to the
improved estimate ,:
1342
Actual root
= 3 - 0.408602150
w h i c h w e seek
= 2.591397850
where
-152
so that/'(3) = —372. Repeating the algorithm, we get
Figure B.1
= 2.591397850 + 0.016270894
Page 564
= 2.607668744
Appendix B / Examples of Numerical Analysis/The Newton-Raphson Method
This,
distance
Page 565
is
Actual root
which
we seek
Figure B.2
Old
estimate
Root
And one more time:
Figure B.3
= 2.607668744 + 0.000026410
= 2.607695154
This algorithm is easily programmed on a computer. After doing this, the results
(with the same initial guess
= 3) are:
Old
Root
estimate
=3
= 2.591397850
= 2.607668744
= 2.607695154
= 2.607695156
= 2.6076951531
= 2.607695153
Figure B.4
convergence!
= 2.607695153J
Rool
which is in agreement with the results in Example 7.5.
Incidentally, note from Figures B.3 to B.5 that adding
to form the
new estimate works equally well for the three other sign combinations of f andf'.
Note also that if the estimate is too far from the root, such as P in Figure B.4, the
procedure might not converge; the tangent at Q in this case would send us far
from the desired root.
We next consider the equation from Problem 5.140 when M = 4m:
(B.l)
Old estimate
Figure B.5
with the derivative of / being
Page 566
functions or
Root
There is but one root of Equation (B. 1) for > 0, as can be seen from Figure B.6,
which shows the two functions making up
. To find this root, we can use
Newton-Raphson as previously described. Figure B. 7 suggests that might serve
as a good first guess at the root. A Newton-Raphson program shows that it is, and
yields the answer below very quickly:
3.141592654
2.094395103
Figure B.6
1.913222955
1.895671752
1.895494285
1.895494267'
1.895494267
convergence!
1.895494267-
Figure B.7
The last example in this appendix will be to solve the equation
from Example 2.6. We write this equation as
with
f(q)
0.707
The rough plot in Figure B.8 shows a few points which indicate that
is fairly
close to the root. Here are the results of a program, which uses the NewtonRaphson method as in the first two examples, to narrow down on the root quickly
and accurately:
1.570796327
0.121-
1.683007224
1.679300543
1.679296821'
-0.879
1.679296821
1.679296821-
-1
Figure B.8
convergence!
c
C MOMENTS OF INERTIA OF
MASSES (SEE ALSO SECTION 4 . 3 )
Object
slender rod
Mass Center Coordinates
and Volume V
M o m e n t s of Inertia
About Indicated Axes
10, 0, 0)
V =
At
(A = area of cross section)
slender
circular rod
bent
slender rod
Page 567
Page 568
APPENDIX C / MOMENTS OF INERTIA OF MASSES
Object
Mass Center Coordinates
and Volume V
rectangular
solid
(0, 0, 0)
hollow
cylinder
(0, 0, 0)
Four Special
Cases
(0, 0 0)
V =
abc
2
V =
2
TT(IR -
i )H
;
V =
2
irR H
1. If r = 0:
solid
cylinder
[0, 0, 0)
V = 2-rrRtH
(0, 0, 0)
V = r{R
2
-
2
r )H
M o m e n t s of Inertia
About Indicated Axes
Page 569
Object
thin right
triangular
plate
thin elliptical
plate
thin
paraboloidal
plate
thin circular
sector
plate
Mass Center Coordinates
and V o l u m e V
M o m e n t s of Inertia
About Indicated Axes
Page 570
Object
2. If a = 2<rr:
circular
plate
thin circular
segment
plate
rectangular
tetrahedron
hollow sphere
solid ellipsoid
Mass Center Coordinates
and Volume V
M o m e n t s of Inertia
About Indicated Axes
Page 571
Object
solid
spherical cap
paraboloid of
revolution
ei iptic
paraboloid
solid cone
solid right
rectangular
prism
Mass Center Coordinates
and Volume V
M o m e n t s of Inertia
About Indicated Axes
Page 572
Object
solid toroid
frustum
of cone
Mass Center Coordinates
and V o l u m e V
M o m e n t s of Inertia
About Indicated Axes
D
D ANSWERS TO
ODD-NUMBERED PROBLEMS
In the solutions to problems in Chapters 1 - 5 , unless identified otherwise below,
and are unit vectors in the respective directions
, and out of the page. In
Chapters 6 - 8 , the unit vectors are respectively parallel to axes defined in the
problems.
C H A P T E R
1
z
a,
30.8
5
2
1.17 Answer given in problem.
11.3
2
5
I
2
5
D
1.33 15 sec to return (21 seconds total elapsed time)
a,
5 10 15 20
- 125
-1125
20
,
-10
Page 573
Page 574
APPENDIX D / ANSWERS TO ODD-NUMBERED PROBLEMS
2
1.129 6.5f + C , where C is a constant of integration
1.97 For x within the intervals (290, 1200) ft and
(1800, 2700) ft
1.155 Answers given in problem.
C H A P T E R
2
2.1 Answer eiven in problem.
measuring r from pulley to
bumper, is the velocity of the shingles; it is also the
component of
along the rope.
2.47 Answer given in problem. 2.49 0.788 sec
2.51 (a) 13.1 m (b) smaller because now
resists
the motion of _
2.53 0.032 lb 2.55 136 N
2.59 Answer given in problem.
2.61 (a) a (vertical) component of the string tension;
1.125 Answer given in problem.
1.127 Answer given in problem.
Page 575
(c) another component of the string tension, this one m
the
direction.
2.63 170 miles 2.65 Answer given in problem.
2.67 3.13 rad/sec 2.69 p = a / b ; 52.0 mph, so yes.
2.71 0.5, at
=0
2.173 11.5 ft upward 2.175 rf/4 upward
2.177 Answer given in problem.
2.179 Answer given in problem.
2
C H A P T E R
3
2.79 Answer given in problem.
3.1 a,c,d,e
2.97 Answer given in problem.
2 . 9 9 Answer given in problem. (Set N = 0 to find the
leaving point.) 2.101 76,300 l b / f t
2.103 20.7 ft/sec down the plane.
2.105 (a) 12 lb (b) 2.84 ft/sec 2.107 72 l b / f t
3.27 The plots can be constructed from the answer to
3.31 In each case,
is at the intersection of the radial
line OA and the normal to the slot at B;
2.129 Distance between them is ( — 0.02) m, where
= unstretched length. They are 0.22 m closer
together. Final spring force = 1 lb (compressive).
2.131 1.41 m i l e s / s e c 2.133 7.45 sec
2.135 36.2 — f t / s e c 2.137 7.86 sec; 491 f t / s e c
2.147 Answer given in problem. (If e > 1, there would
be an energy gain!) 2.149 0.446
2.165 Answers given in problem.
2.169 The derivation of Equation (2.36) nowhere
requires that the point be the mass center.
2.171 Answer given in problem.
Page 576
3.141 Answer given in problem.
3.153 Answer given in problem.
3.157 Answer given in problem.
C H A P T E R
4
3.131 Let x and y respectively be directed down and
toward the plane, with origin at the center of the disk.
Then the point has (x, y) = (4.80, 3.60) ft.
4.33 Answer given in problem.
2
3.137 21.6 m / s (It is the highest point of p.)
3.139 (a) Answer given in problem,
(b) Curve is concave downward.
Page 577
4.95 (a) Answer given in problem.
4.41 From the comet,
4.55 Answer given in problem
4.61 Answer given in problem; only (b) starts without
4.63 2 g / 3 ; 5 g / 7 ; g / 2
4.65 (a) Wally, Sally, Carolyn, Harry;
approximation.
4.133 Answers given in problem.
4.143 On the section to the left of the cut,
10.4
8,27
4.149 Answer given in problem.
2.07
4.93 (a)
3
(a)
4.157 For rolling on fixed surface, is normal to the
surface, hence toward geometric center of round body;
thus
since geometric center is mass center.
10.4
2.07
(b)
3
lb]
13.8
2.85
9.51
1.48
3
to
Page 578
5 . 5 9 It starts out to the right, the spring goes slack, and
then it leaves on the right. (It would need one more
foot of plane to stay on.)
up on left bearing and down on right bearing.
down on left and up on
right, onto shaft and turning with it.
4 . 1 9 1 By parallel-axis theorem,
C is on z-axis. Thus using the theorem again,
Same arguments for
since
= 0.
5 . 7 3 Answer given in problem.
C H A P T E R
5
5 . 4 3 8 in.; the two points are the intersections of the
perimeter of
(in the starting position) with a circle of
radius 12 in. and center at
(in the final position!
5 . 1 2 1 Answer given in problem.
5.123
final motion is given by (a).
Page 579
6.43
5.133 Answer given in problem.
2
5.137 0.545 m from left end; 0.562 kg-m ;
0.0957 ke-m ; 0.657 m from left end
2
is a "force" that will change the
particles' velocity directions relative to the earth so as
to produce ccw rotation in the northern hemisphere.
The effect is opposite in the southern hemisphere.
6.47 Answer given in problem.
CHAPTER
6
6.1 Answer given in problem.
6.15 To the components in 6.13, respectively,
and the
cross-product is not generally zero this time.
6.63 1.36 rad/s, directed from O through the line of
contact between C and 2>.
6.67 Answer given in problem.
Also,
so right side is (6H - 260)<a«
6.71 (a) Answer given in problem;
6.75 (a) Answer given in problem.
(b) Answer same if is replaced by at.
6.41 Answer given in problem.
Page 580
6.83 Let point A be displaced from its original to its
final position. Then, using Euler's Theorem, all other
points of the body may be placed in their final
positions via a single rotation about an axis through A.
with
6 . 9 3 Be sure the cross is ngid and planar!
C H A P T E R
in any direction normal to
working with six digits
and rounding at the end
7
7.1 With along the axle from O through the wheel
center, and out of the page,
7.5 With
and
parallel to x and y of the figure,
7.19 If
, the ellipsoid
cannot be an ellipsoid of inertia, for then it would
represent a body having one moment of inertia
sum
of other two, a physical impossibility.
Note: There is a precision problem here because
(1)
and
are so much larger than
and (2)
and
are nearly equal.
at B: same magnitude but
the direction is different.
so that z is a principal axis for every point on that axis
7.55 332 days
7.63 With x along S from Q and y upward,
PAGE 5 8 1
7.95 815m ft-lb, where m is the mass in slugs
7.69
(c) no hole is physically possible;
7.103 358 1
7.71 With
of v , F on
c
from O toward C, and j in the direction
7.107 Some check values:
by the bearing at C,
from gravity, and mgk from the ground;
C H A P T E R
8
and C on 3 by bearing =
The difference is that in this problem the normal
force from the ground is just the weight.
7.73
7.75 RHS of x-eqn = 5.6 (the same);
RHS of y-eqn = 5.6
7.7; RHS of z-eqn = 2.8
1.4;
Axes used for Euler Equations must be body-fixed,
not just permanently principal.
8.9 Answer given in problem.
8.13 (a) 0.278 ft
(b) 0.876 sec
(c) 0.291 sec
B.I5 (1) moves to left with x
(2) moves to right with x
(i) is unstable and the other three are stable.
7.85 For a torque-free body in general motion in an
inertial frame
with
not parallel to
,
we have
and the two terms add to zero.
7.87 If is in equilibrium in
all its points are
stationary there; thus
for all these points,
and also
Hence is an inertial frame.
But if is an inertial frame, it can at most translate
at constant velocity with respect to another
inertial frame . Thus it need not be stationary in
i.e., need not be in equilibrium in
even though none of its points accelerates in
7.89 Answer agrees with Example 7.11.
7.93 No. Two different results are obtained for
(3) moves to left with x
to left of unstretched position.
stops for good
and total distance
Time in each interval is
traveled
8.25 Yes, they do. (Start with
and investigate when
Page 582
(c) Mechanical energy is lost (to heat, deformation,
vibration, etc.) as the links suddenly join the falling
part of the chain. 8.57 Answers given in problem.
8.59 (a)0.193gT. (b) Answer given in problem.
8.61 Answer given in problem.
8.63 Answer given in problem. (Use Kepler's laws!)
8.67 107 min ("No air resistance" needn't be assumed
this time!) 8.69 4320 mi; 583 mi and 137 mi
the radicand is
, and for these co's,
the radicand is smaller for smaller values of c.
8.73 Answer given in problem. (Isolate the radical and
square both sides.) 8.75 Answer given in problem.
where W is the weight of tank plus fluid.
8.47 (a, c) Answer given m problem;
A P P E N D I X
8.49 Answer given in problem.
valid until fuel
gone
8.53
A
INDEX
Absolute system, 559
Acceleration, 6, 8
angular, 163-164, 398-399
Coriolis, 207, 405-406
in different frames, 207-208
equations of, 414-415
in moving frames of reference, 401,
405-406
radial and transverse components of,
32
tangential and normal components of,
43
Action-reaction principle, 64
Addition theorem for angular velocities,
385-386,428
Amplitude, 518
Angle
Eulerian, 428-433
friction, 66
nutation, 493
phase, 518
Angular acceleration, 163-164
Angular acceleration vector, 398-399
Angular impulse, 344
Angular momentum, 118, 227-228
conservation of, 350
relative, 124
in three dimensions, 446
Angular speed, 136, 184
Angular velocity, 134-135,136,
380-381, 383, 432
addition theorem and, 385-386
differentiation of, 387
properties of, 384-388
simple, 386-388
Apogee, 544
Ardength, 43
Axes of inertia
non-principal, 473
principal, 455, 456, 462, 466
Axis of rotation, 277-278, 422
Balance, static, 292
Balancing a rotating body, 295
Bearing reactions, 291-294
Binormal vector, unit, 51
Body, 56
cone, 496
mass center of, 62-64
rigid. See Rigid body
unbalanced, rotation of, 291 —
293
Body extended, 130
Cardan suspension, 431
Cartesian coordinates, 24, 381
Center
of curvature, 49
of mass, 58-60, 62,219
of percussion, 362
of zero velocity, 149-151
Central force, 89-90
Central force motion, 543
Central impact, 107
Chasle's theorem, 434
Circular frequency, 518
Circular orbit, 547
Coefficient of friction, 65
Coefficient of restitution, 106-108, 355,
361
Cones, body and space, 496
Conic, 546
Conservation
of angular momentum, 350
of energy, 92
of mechanical energy, 92, 330-331
of moment of momentum, 120-121
of momentum, 102-103, 350
Conservative forces, 91-92, 329331
Constant force, 89
Continuity equation, 535
Page 583
Page 584
Control volume
Euler's laws for, 5 3 5 - 5 3 6
Coordinates
Cartesian, 24, 381
cylindrical, 31
polar, 31
spherical, 43
Coriolis acceleration, 207, 4 0 5 - 4 0 6
Coulomb law of friction, 249
Couple, 309, 311
Cramer's rule, 457
Critical damping, 522
Curvature, 4 5 , 149
center of, 49
radius of, 45
Curvilinear translation, 148
Cylindrical coordinates, 31
Damped vibration, 521
Damping
critical, 522
subcritical, 523
supercritical, 523
Deformation, 6 0 , 1 1 6
Derivative, 399
of a vector, 3
in different frames, 1 9 8 - 2 0 0
Derivative formula, 384
Differential equations
complementary solution, 5 2 6
energy integral, 6 8
particular solution, 526
of plane motion, 246
Direct impact, 107
Direct precession, 4 9 6
Displacement, 522
steady-state, 526
Drag, 74, 85
Earth
as a moving frame, 4 1 0 - 4 1 1
gravity and, 91
lunisolar precession of, 494, 497
motion near the surface, 410
orbit around sun, 440
satellites, geosynchronous, 81, 439
Eccentricity, 5 4 6
Eigenvalue, 457
Eigenvector, 4 5 7
Ellipsoid of inertia, 454
Energy
conservation of, 92
kinetic. See Kinetic energy
mechanical, 3 3 0 - 3 3 1
potential, 3 2 9 - 3 3 1
Energy integral, 6 8
Equations of motion, 2 4 9 - 2 5 0 , 2 7 2 - 2 7 3 ,
274
for fixed-axis rotation, 2 7 7 - 2 7 8
Escape velocity, 85, 553
Euler equations, 432, 4 7 2 - 4 7 3
Eulerian angles, 4 2 8 - 4 3 3
Euler's first law, 5 6 - 5 8 , 62
control-volume form of, 536
momentum forms of, 1 0 1 - 1 0 2
Euler's second law, 1 1 7 - 1 1 8 , 495
control-volume form of, 5 3 6
momentum forms of, 1 1 9 - 1 2 0
External forces and couples, 5 8
Fixed-axis rotation, 2 7 7 - 2 7 8
Fluid flow, 535
Focus of conic, 546
Foot, 558
Force, 355
conservative, 9 1 - 9 2 , 3 2 9 - 3 3 1
nonconservative, 331
work and, 8 9 - 9 0 , 309, 311, 313
Forced vibration, 5 2 4 - 5 2 5
Frame
earth as moving, 4 1 0 - 4 1 1
intermediate, 4 7 4
of reference, 3, 7, 401, 4 0 5 - 4 0 6
acceleration and, 2 0 7 - 2 0 8 , 4 0 1 ,
405-406
inertial, 57
velocity and, 1 9 8 - 2 0 0 , 401
Free vibration, 5 1 7 - 5 1 8
Free-body diagram, 6 2 - 6 4
Frequency, 518
Friction, 249
Gears, 170, 1 8 3 - 1 8 4
Geosynchronous orbit, 81, 439
Gravitation, 57
Gravitational system, 559
Gravity, 9 1 , 351
Gyration, 2 3 9 - 2 4 0
Gyroscopes, 4 9 2 - 4 9 5
Gyroscopic moment, 481
Gyroscopic precession, 494
Harmonic motion, simple, 518
Helix, 35
Hertz, 560
Hooke's joint, 391
Impact, 103, 3 5 5 - 3 5 6
central, 107
direct, 107
energy loss, 113, 356
restitution coefficient for, 1 0 6 - 1 0 8
Impulse, 1 0 2 - 1 0 3 , 344, 4 9 9 - 5 0 0
Impulsive force, 3 5 5
Inertia
ellipsoid of, 455
Page 5 8 5
moments of, 229, 235-236, 455, 457,
458, 466
maximum and minimum, 468
principal axes of, 455, 456, 462, 466
products of, 240-241
torque (moment), 488
Inertia properties, 228-229
transformations of, 448-449
at a point, 449
Inertial frame of reference, 57
Instantaneous axis of rotation, 149-151,
422
Instantaneous center of zero velocity,
149-151
Intermediate frame, 474
Intrinsic components of velocity and
acceleration, 43
Jerk, 52
Joule, 88, 359
Kane, T. R„ 383, 432
Kepler's laws of planetary motion, 544,
548, 550
Kilogram, 558
Kinematics, 2, 7
of a point, 8
Kinetic energy, 504
alternative form of, 308
mass center, 87, 306
of a particle, 88
rate of change of, 309
ofrigidbody, 305-306
rotational, 308, 506
translauonal, 308,506
work and, 87-89, 97
principle of, and, 305-306, 316,
327-328
Kinetics, 2
Levinson, D. A., 432
Linear impulse, 102-103, 344, 499-500
Linear momentum. See Momentum
Linear spring, 90
Local vertical, 411
Lunisolar precession, 494, 497
Mass center, 58-60, 62, 219, 246-247
Matrices, 434-435
Maximum moments of inertia, 468
Mechanical action, 56
Mechanical energy, 92, 330-331
Meter, 558
Minimum moments of inertia, 468
Moment
of inertia, 229, 235-236, 455, 457,
458, 466
of masses, 567-572
maximum and minimum, 468
of momentum, 118-119,120-121,
227-228
conservation of, 120-121
in three dimensions, 446
work of, 313
Moment equations of motion, 249-250,
272-273, 274
forfixed-axisrotation, 277-278
governing rotational motion, 472-473
Momentum, 343,499-500
angular, 118, 227-228, 350, 446
conservation of, 102-103, 350
and Euler's second law, 119-120
linear, 101
moment of, 118-119,120-121,
227-228, 446
net rate offlowof, 535
of a particle, 101-102
Motion, 176, 305-306
central force, 543
equations of, 249-250, 272-273,
274, 277-278
of mass centers, 58-60, 62
of particles, 62
planetary, Kepler's laws of, 544,548,
550
rectilinear, 6
rotational 492
simple harmonic, 518
torque-free, 473,495, 497
translation^, 147
Moving a derivative, 399
Moving frames of reference, 401,405-406
Natural circular frequency, 518
Net rate offlowof momentum, 535
Newton, 558
Newton, I., 56
Newton frame of reference, 57
Newton-Raphson method, 564
Newton's laws, 56-58
Nonconservative force, 331
Nonimpulsive force, 355
Non-principal axes, 473
Normal component of acceleration, 46
Normal, principal unit, 46
Normal and tangential components of
velocity and acceleration, 43
Nutation angle, 493
Orbit
apogee of, 544
eccentricity of, 546
perigee of, 544
period of, 549
Orientation of arigidbody, 132, 428
Orthogonal components, 4
Page 586
Orthogonality of principal axes, 466
Overdamped system, 523
parallel-axis theorem for, 241
Projectile motion, 65
Parallel-axis theorem
for moments of inertia, 235-236
for products of inertia, 241
Particle, 2,136
momentum of, 101-102
motion of, 62
work and kinetic energy for, 87-89,
97
Particular solution, 526
Path, 6
Percussion, 362
Perigee, 544
Period, natural, of vibration, 518
orbit, 549
Phase angle, 518
Pivot, 152-153, 277-278, 473
Plane, reference, 131-133
Plane motion, 130
equations of
impulse and momentum for rigid
body in, 343-345
motion of, 246-247
kinetic energy of rigid body in,
305-306
Point, 8
Poisson equations, 433
Polar coordinates, 31, 249
Position vector, 6
Potential energy, 92, 329-331
of a central force, 90, 544
of a constant force, 90, 330
of a linear spring, 90, 330
Pound, 558
Power, 309, 311, 508
Precession, 91, 493-495, 496, 497
direct, 496
gyroscopic, 494
lunisolar, 494, 497
retrograde, 497
Primitive, 56
Principal axes of inertia, 294, 455, 456,
462, 466
Principal directions, 458-459
Principal moments of inertia, 455, 457,
458, 466
Principal unit normal vector, 46
Principle of
action and reaction, 64
impulse and momentum, 343-345,
500
Principle of
work and kinetic energy, 305-306,
316, 327-328
Products of inertia, 240-241
Quantity of matter, 56
Radial components of velocity and
acceleration, 32
Radius
of curvature, 45
of gyration, 239-240
Rate
of change of kinetic energy, 309
of work of a couple, 311, 508
of work of a force, 81, 311,508
Rectangular Cartesian coordinates, 24
Rectilinear motion, 8
Rectilinear translation, 148
Reference, frame of. See Frame of reference
Reference plane, 131-133
Relative
acceleration, 207, 405
angular momentum, 124
velocity, 198, 402
Resonance, 526
Restitution, coefficient of, 106-108, 355,
361
Resultant force, 57, 248
Resultant moment, 248
Retrograde precession, 497
Reversed effective forces, 488
Reynolds transport theorem, 535
Rigid body(ies), 3,130,198
angular acceleration of, 163-164
kinetic energy of, 305-306
orientation of, 132, 428
points on
acceleration of, 163-164
velocities of, 134-135
principle of impulse and momentum
for, 343-345
in translation, 219-220
translational motion of, 147, 348
velocity and acceleration equations
for, 414-415
work and kinetic energy principles
for, 311-312
Rigid extension of the body, 130
Robot, 397
Rocket, 537
Rolling, 170-172, 176
Rotation
of axes (transformation of inertia
properties), 448-449
axis of, 277-278
instantaneous, 149-151, 422
of unbalanced bodies, 291-293
Rotation matrices, 434-435
Page 587
Rotational motion, 492
Satellite, 81, 439, 474, 498
Second, 558
Serret-Frenet formulas, 52
SI units, 559
Simple angular velocity, 386-388
Simple harmonic motion, 518
Single-degree-of-freedom system, 517
Sliding friction, 66
Slug, 558
Space cone, 496
Speed, 6
angular, 136
Spring, linear
modulus, 90
work done by, 90, 314-315
Steady-state displacement, 526
Subcritical damping, 523
Supercritical damping, 523
Tangent vector, unit, 44
Tangential and normal components of
velocity and acceleration, 43
Thrust, rocket, 537
Torque-free motion, 473, 495, 497
Total energy, 92
Trajectory, ballistic, 64
Transfer theorem. See Parallel-axis
theorem
Transformation of inertia properties, 449
Translation, 147-149
rigid bodies in, 219-220
Translational motion, 147, 348
Transverse components of velocity and
acceleration, 32
Truesdell, C, 57
Unbalanced bodies, 291-293
Underdamped system, 523
Unit, 558-562
Unit conversion, 560-562
Unit tangent, 44
Universal gravitation, 57
Universal joint, 391
Vector, 472
angular acceleration, 398-399
angular velocity, 384-385
derivatives of, 3,384
position, 6
principal unit normal, 46
Velocity, 6,107
angular. See Angular velocity
in different frames, 198-200, 401
of points in a rigid body, 134-135
radial and transverse components of,
32
tangential and normal components of,
43
zero, 149-151,176, 422-423
Velocity equations for rigid bodies,
414-415
Vibration, 517
damped, 521
forced, 524-525
free, 517-518
v-t diagram, 11
Wave propagation, 116
Work, 309, 311, 313
of a couple, 311
of a force, 89-90, 309, 311, 313
of gravity, 91
and kinetic energy, 87-89,97,
305-306, 316, 327-328,504
rate of, 508
of a spring, 90
Zero velocity, 149-151,176, 422-423
Page 589
Rationale for the New Tear-Out Section
This new set of tear-out problems is designed to assist the student in summarizing and synthesizing
the more important equations and procedures in a first course in Dynamics. The goal of each problem is
stated at its outset.
These problems lie somewhere between the book's examples and homework problems. They are
like homework problems (and unlike examples) in that they ask the student to do the work; but they
are like the examples (and unlike homework problems) in that they lead the way.
Thus we view these problems as a learning step between (a) classroom lectures/textbook examples
(in which the student is passive) and (b) homework problems (in which he/she is active). Should the
professor wish to use them, the tear-out feature and "fill-in-the-blanks" format make submission and
grading relatively easy.
Page 591
1:
To demonstrate understanding of both
rectilinear motion and motion on a circle.
GOAL
Point P is projected at
from point A along the circle shown,
with initial velocity
and with
. One sec­
ond later, point Q is projected from point B along BC with initial
speed
and with
. Determine
so that P and Q
will reach point C simultaneously.
1.
Integrate
twice and determine the constants using initial conditions on the motion of P:
2.
Substitute
3.
Now, integrate
twice and determine the constants using initial conditions on the motion
of Q; use a new time t for Q:
for 0 and determine the time at which P arrives at C:
2
Page 592
4.
Substitute t-1 for t in ( 3 ) :
5,
Finally, substitute the value of XQ at point C, and of f from (2), to determine the required value
of
:
2
Note that at any speed above this value, Q will arrive too soon; at any speed below it, Q will get there
too late.
PAGE 5 9 3
2: To demonstrate an understanding of the use of cylindrical
coordinates to determine the velocity and acceleration of a point.
GOAL
Point P travels on the spiral path around the surface of a cone as shown.
Points on the path of P obey the relationship
Find
and
when
, if the angular rate is a constant.
1. Use the sketch below to write r as a function of z, and then incorporate
a function of :
2.
Use Eq. (1.31) to obtain
as a function of
to write r as
and the radius R and height H of the cone:
Page 5 9 4
3.
Using your answer to (2), give the reason why the speed of P cannot be constant in
this problem:
4.
Use Eq. (1.37) to obtain ap as a function of the same quantities:
Page 595
To demonstrate understanding of tangential and normal
components of velocity and acceleration of a particle.
G O A L 3:
A particle P starts at rest at point A, and accelerates uniformly
for 10 seconds, when it reaches B. Find the tangential and
normal components of the velocity and acceleration of P just
before and just after P reaches B. If now at B the particle begins
to decelerate uniformly and reaches C 5 seconds later, find the
acceleration components of P when it reaches C.
1. From A to B, write an expression for
2. Integrate to obtain expressions for
3.
in terms of the unknown acceleration magnitude of P:
and s:
Substitute s = 1000 m at t = 10 sec to obtain the acceleration magnitude of P between A and B, and
then the speed and velocity of P at B (noting that the velocity is always tangent to the path):
Page 596
4.
The velocity in (3) is the same just before, when, and just after P reaches B. However, the
acceleration is not. Use (3) to express just before P reaches B:
5.
Now find
6.
Now, between B and C, express
7.
Integrate twice, starting time over at B and noting that
8.
Substitute
of P at C:
9.
Using (8) find a at point C:
just after P passes B:
as a new constant, with S2 measured from B:
at
P
sec and obtain
of P between B and C, and then the speed
Page 597
GOAL 4: To demonstrate ability to use rectangular Cartesian coordinates to compute
Vp and a , and to correlate the results with tangential and normal components.
P
Y
A particle P moves on the indicated circular path in a counterclock­
wise direction after starting from the origin of coordinates shown.
The speed of P is
= constant. As a function of x, find the velocity
and acceleration of P by respectively differentiating
and
in
rectangular Cartesian coordinates. Check the results using tangential
and normal components for the three dotted points.
2
(0,0)
l 2
X
1.
Write
in terms of x, by solving for y from the given equation of the circle:
2.
Differentiate
to obtain
in terms of x and :
3.
Use
to express
in terms of
Note from your answer that
and
and x:
cannot both be constant.
Page 5 9 8
4.
Eliminate from (2) using (3), and thereby obtain
5.
Check by verifying that
6. Now differentiate
as a function of x (and
from (4) and obtain a in terms of x (and
7. Check the answers for
P
and
):
):
at (0, 0) using tangential and normal components:
Page 599
8.
Repeat (7) at the point (2, 2):
9.
Repeat (7) at the point (1,
):
Page 601
GOAL 5:
To demonstrate understanding of kinematics
and kinetics of rectilinear motion.
The small boxes
and
(each of mass m) are
connected by an inextensible cord that passes over
a small, smooth peg. Man
is running at constant
speed
to reach
before
reaches man
,
who is pulling down on a second cord with the
constant force Kmg. The friction coefficient
between
and its floor is and both boxes start
from rest in the positions shown. Find the smallest
value of man
speed
such that he will reach
in time.
1. Complete the sketches below, with arrowheads and names of forces, to form fbd's of
2.
Noting the X and y directions in the figure, write theX-equationof motion for
tion for :
3.
Eliminate the cord tension by combining your two equations:
1
and
and the y-equa-
Page 602
4.
You should now have one equation in two unknowns,
and . Write a kinematic equation
expressing constancy of the rope length, and differentiate it twice to obtain in terms of :
5. Substitute your result into ( 3 ) and integrate twice to obtain y(t) as a function of t (and the
constant K):
6. Set y = H to get the time t when
2
7. When
8. Integrate
reaches
, how far will
=V
2
0
reaches man M :
2
have moved?
to obtain x as a function of time:
603
9.. Compute the time t for M to reach
1
10. Set the time
1
and answer the original question:
11. Show that your answer makes sense, namely that (a) the smaller is K, (b) the smaller is g, (c) the
smaller is H, and (d) the larger is then the smaller will be the required value of V .
0
12. Note that in (11) the answers for (a) and (d) are obvious. But (b) and (c) would seem to have
cancelling effects on and
. Explain why they do not.
Page 605
To demonstrate understanding of E q . ( 3 . 8 ) .
G O A L 6:
The block moves to the left
with a prescribed motion
mm. It is pinned
at A to rod , which is itself pinned to the center of gear at B. A cord
wrapped around a slot near the periphery of is fixed to block at Q.
Find
at the given instant, at which t = 3 s. (The teeth on the gear and
wall are not shown.)
1. First, show that
= 6ti mm/s, so that at t = 3,
2.
Next, use Eq. (3.8) to relate
(from (1)) to
and
incorporate the constraint on the motion of B:
3.
Solve and obtain V = 24 mm/s and
B
=18 mm/s:
on the rod. In your equation, be sure to
Page 606
4.
Now, again using Eq. (3.8), relate
(from (3)) to
and
on the gear, where T is the
"tooth point" in contact with the rack (a rack is a "straight line gear," in this case forming
the vertical fixed wall):
5.
Solve and obtain
6.
Next, use Eq. (3.8) for a third time to relate
directly above Q and to the left of B:
and
to
, where D is the point on the gear
7. Solve and obtain
8.
Finally, state why
Note the rigid adherence to the three rules accompanying Eq. (3.10) each and every time Eq. (3.8)
is used.
Page 607
To demonstrate understanding of
the instantaneous center of zero velocity.
G O A L 7:
Repeat the preceding tear-out problem, using the concept of instantaneous center instead of Eq. (3.8).
1. Show with a sketch that
— the instantaneous center of zero velocity of
below A and to the right of B:
2.
Using
, show first that
3.
State why T is
4.
Using
5.
Using
and
and
from above, show that
— is the point directly
= 0.2O rad/s and then that
= 0.6O rad/s:
from above, find
And as in the previous example,
. Note that in using the instantaneous center, we
avoid the use of vectors and are usually finding one unknown at a time. Arrows (such as
) are
more helpful in this approach than unit vectors, and you must assign the proper directions to velocities
and angular velocities as you work through the problems.
Page 609
GOAL
8: To demonstrate understanding of Eq. (3.19)
and the rolling result Eq. (3.26).
In tear-out problem (6), find the acceleration of point Q.
1. Show that
2.
Next, use Eq. (3.19) on
3.
Solve and obtain
to relate
and
;
and
:
Note that if you tried to use
to obtain
and , you would obtain incorrect answers.
The instantaneous center of zero velocity is just what it says; it does not work for accelerations
4.
Next, use Eq. (3.19) again, this time on
to relate
and
(above) to
Page 6 1 0
At this point, note that the above vector equation, which comprises two scalar component equations,
contains three unknowns:
and the x and y components of . But the gear is considered to be
rolling on the wall, so:
5.
Use Eq. (3.26) to determine that
6.
Solve the equation from (4) on the previous page and obtain
7.
Use Eq. (3.19) a third time to relate
8.
Solve and obtain
9.
Obtain
from
Explain why
and
from above with
Page 6 1 1
GOAL
9: To demonstrate proficiency with Eq. (3.51).
The cylinder C rolls on the plane
to the right, with its center C at
the instant shown, determine the
angular velocity of rod in and
the velocity, relative to
of the
point P of O in contact with point
P'of
1.
Write Eq. (3.51), relating
2.
Compute
3.
Compute
(in terms of
of P' and O on
4.
Express
to
using Equation (3.8) to relate the velocities of P and C on C:
) by using Equation (3.8) again, this time to relate the velocities
in terms of an unknown magnitude
and a known direction (unit vector):
co
Page 6 1 2
5. Substitute the results of Steps (2), (3), and (4) into (1):
6.
Collect the separate and coefficients and solve the two resulting equations for
7. Using the results of (6), write the required vectors
8.
and
and
:
:
In this problem, the directions of all three vectors in Step (1) —
and
— are known
at the outset. Complete the sketch below by drawing arrows representing the three vectors; note
that the second and third must add up vectorially to the first!
Page 613
GOAL
10: To demonstrate understanding of Eq. (3.55).
Bars
and
are symmetrically placed, pinned
at their lower ends to a floor at 0 and 0 . Two
collars C and C are connected together by means
of a ball and socket at P, and they slide on
and
1
1
1. Relate
to
the given instant:
2
2
using Equation (3.55); let the point V be the point of
coincident with P at
1
2.
In your equation you will need
. Write Equation (3.51), relating
to
3.
Note that by the symmetry, P must move vertically (in ). Note further that
is an unknown
magnitude multiplied by a unit vector along
. Incorporate these results into your Step (2)
equation and solve for
resp
Page 614
4.
We remark that it is not necessary to use symmetry as above; but without it, one must also relate
to
and end up with 4 equations in 4 unknowns. What would be these four unknowns?
5.
In the Equation in Step ( 1 ) , use the same two notes given in Step (3), this time for accelerations
instead of velocities, to reduce the vector equation to two scalar equations in the required
unknowns. Then solve for them.
Page 615
GOAL
11: To demonstrate use of the transfer (parallel axis)
theorem for mass moments of inertia.
Y
Use the results of Appendix C to compute
for the bodyconsisting of a pair of identical homogeneous solid cones
and C as shown. Then compute
where X is an axis parallel
to x through the point P at (x, y, z) = (3R, R, H).
X
2
H
1. The equation
in the appendix is the moment of inertia for one cone about an
axis such as x
in the figure. We cannot transfer this result from
to x. because point O is not
the mass center of the lower cone C . First we must obtain
for C . Use the transfer theorem to
do this:
b a s e
2
2.
2
Now transfer your result from C to C, and double it to account for the other cone, symmetrically
placed:
2
616
3.
Since C is the mass center of the combined body comprising C and C , transfer
obtain
:
1
2
to P and
Page 617
GOAL
12: To demonstrate understanding of Eq. (4.15)
and of the steps listed on pp. 249-250.
Cylinder O rolls on the board
A
cord wrapped around a slot near the
periphery of O passes around a light
pulley and is attached to the weight
at A. Each of the three bodies
1.
Complete the free-body-diagrams below by adding arrows and names of forces:
Note that because the wheels beneath are light,
we can neglect their masses. Thus in a freebody of one of them:
we see that
But also I ' = 0 if mass is negligible, so that
C
Thus there is no friction force exerted on the bottom of
by the wheels.
has ma
Page 618
2.
Using the sign convention
fill in the two equations below for O:
3.
Now fill in this equation for
4.
And this equation for
5.
Now list the unknowns in your four equations thus far. There should be six — four kinematic
quantities (accelerations, angular accelerations) and two forces:
6. State why none of: the y-equation for O and
and
need be written.
7.
the x-equation for
and the moment equation for
As a result of (5), we need two more equations. Turn to kinematics and on O, relate
and
In doing so, it is crucial to note that
and
are acceleration components relative to tne ground
(inertial frame); show that
, and number this equation
:
Page 619
8.
Next, relate
9.
And now relate
to the acceleration component
of T, the point at the top of C:
to
10. Combine the results of (8, 9) to show that
; call this equation
11. Now we have six equations in six unknowns. Solve them and obtain
12. State why this result is dimensionally correct and also why it is plausible:
Page 621
To demonstrate proficiency with the work
and kinetic energy equation ( 5 . 1 5 ) .
GOAL 1 3 :
Cylinder
(mass
2m)
The cylinder O is pinned smoothly to the ground at C
Ring , positioned between C and the fixed circular track
rolls on both these surfaces. The spring is attached to
spring has unstretched length 3r/2, and the system is
released from rest in the position shown. In terms of m,
r, and g, find the spring modulus k that will result in
coming to rest at its lowest (dashed) position P . If k is
half that value, find the angular velocity of O when
passes through the dashed position.
2
1. Complete the work done by gravity between P and P :
t
2
2.
Calculate the work done by the spring between P and P :
3.
From
here, note that W > 0 for there to be K.E. remaining at P2, and from this
determine the value of k for which will come to rest at P :
1
2
2
Page 622
4.
Write the kinetic energy of C:
5.
And the kinetic energy of
6. Relate, now, the angular speeds of C and
using Eq. (3.8):
7. Use (4-6) to express the total K.E. in terms of
8. Now, with k = half the value found in (3), use
to find
in position P :
2
Page 623
14: To demonstrate an understanding of the impulse/
momentum principles for a rigid body in plane motion.
GOAL
The two rods and disk are all rigidly attached, and they rotate along with
the solid ball in a vertical plane as indicated. Just after the position
shown, point T3 hits the obstruction and suddenly stops without rebound.
The 1.5 lb solid ball is flung outward by the impact. Find the impulse
imparted to the rod at B by the obstruction. Neglect the effect of the
1. Complete the fbd of the system comprising all the above bodies:
2. Write the angular impulse, about the fixed point O, imparted to the above system during
the impact:
Page 624
3.
Write the angular momentum of the ball
before and after impact:
4.
Express
5.
Use the principle of angular impulse and momentum to find
6.
To find the other impulse, use linear impulse and momentum; first write the total linear impulse
on the system:
for the body comprising rods plus disk:
Page 625
7. Next write the initial linear momentum:
8.
And the final linear momentum:
9.
Using the principle of linear impulse and momentum, find
10. Now, check by showing that
separate bodies:
use the equation
for the
I
Page 627
To demonstrate understanding of important properties of
the angular velocity vector in general motion of a rigid body.
GOAL 15:
are
fixed in
frame 6
Shown is a gyroscope consisting of:
(a) Frame C which turns in at rate
about
which is fixed in and
(b) Frame which turns in C at rate
about
which is fixed in C and
(c) Body a which turns in at rate ro about u
which is fixed in and a.
3
It is given that
and
are constants.
Note that is shown aligned with the direction
of at t = 0, but will not generally be parallel to
Write
in frame O, and find
1. Use the addition theorem to express
in O, i.e., in terms of the unit vectors
direction in O; note that the angle in the small sketch above is
2.
Find
by "moving the derivative" to frame C where the unit vectors are fixed:
fixed in
Page 628
3.
As a check, write
differentiate the first term in
derivative" in the second and third terms to
directly, and "move the
Page 629
GOAL 1 6 : To
combine kinematics and kinetics in solving
a problem involving general motion of a rigid body.
Use Eq. (2.45) to investigate the motion of a uniform sphere of mass m and radius R rolling under
the influence of gravity on a fixed horizontal plane
In particular, show that the mass center C of
must travel on a straight line at constant speed (i.e., that it has zero acceleration). Thus C cannot move
on a curved path without slipping regardless of the amount of friction available.
1. Complete the fbd of by adding arrowheads and naming the four forces that act on
vectors
are fixed in with being vertical:
The unit
2.
Use Eq (7.11) to express
. Note that all axes through C are principal and that the moment of
inertia for each such axis is
: Note also that will serve as the inertial frame for this problem:
3.
Now use Eq. (2.45), taking moments about the base point B of. Note that we choose this form,
of the moment equation because
for the external forces acting on
Page 630
4.
since
Note that, using (2),
are constant in
Also, note that an
is always
using Eq. (6.56) and the rolling condition that
is needed. Write
expression tor
zero in this problem smce is rolling on the fixed plane:
5.
Now differentiate your equation from (4) to obtain
6.
Substitute
is constant:
and
in terms of
from (4,5) into (3), and conclude that
7. Using = constant in (4), conclude finally that
in a straight line at constant speed:
and thus that
is also constant, which means that C must move
Page 631
The previous result is fascinating because in fact the center of a rolling sphere can travel on a curve if