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Table of contents :
Intro
ENGINEERING MECHANICS AN INTRODUCTION
CONTENTS
PREFACE
Chapter 1
ENGINEERING MECHANICS AN INTRODUCTION
1 KINEMATICS OF MATERIAL POINTS OR PARTI
1.1 Introduction
1.2 Reference Frames and Vector Derivati
1.3 Position, Velocity, and Acceleration
1.4 Kinematics of a Point in Rectilinear
1.5 Rectangular Cartesian Coordinates
1.6 Cylindrical Coordinates
1.7 Tangential and Normal Components
SUMMARY • Chapter 1
REVIEW QUESTIONS • Chapter 1
Chapter 2
Chapter 3
3 KINEMATICS OF PLANE MOTION OF A RIGID
3.1 Introduction
3.2 Velocity and Angular Velocity Relati
3.3 Translation
3.4 Instantaneous Center of Zero Velocit
3.5 Acceleration and Angular Acceleratio
3.6 Rolling
3.7 Relationship Between the Velocities
3.8 Relationship Between the Acceleratio
SUMMARY Chapter 3
REVIEW QUESTIONS Chapter 3
Chapter 4
4 KINETICS OF A RIGID BODY IN PLANE MOTI
4.1 Introduction
4.2 Rigid Bodies in Translation
4.5 The MassCenter Form of the Moment E
4.6 Other Useful Forms of the Moment Equ
4.7 Rotation of Unbalanced Bodies
Chapter 5
Untitled
5 SPECIAL INTEGRALS OF THE EQUATIONS OF
5.1 Introduction
5.2 The Principle(s) of Work and Kinetic
5.3 The Principles of Impulse and Momentum
Chapter 6
6 KINEMATICS OF A RIGID BODY IN THREEDI
6.1 Introduction
6.2 Relation Between Derivatives/The Ang
6.3 Properties of Angular Velocity
6.4 The Angular Acceleration Vector
6.5 Velocity and Acceleration in Moving
6.6 The Earth as a Moving Frame
6.7 Velocity and Acceleration Equations
6.8 Describing the Orientation of a Rigi
6.9 Rotation Matrices
SUMMARY • Chapter 6
REVIEW QUESTIONS • Chapter 6
Chapter 7
7 KINETICS OF A RIGID BODY IN GENERAL MO
7.1 Introduction
7.2 Moment of Momentum (Angular Momentum
7.3 Transformations of Inertia Propertie
7.4 Principal Axes and Principal Moments
7.5 The Moment Equation Governing Rotati
7.6 Gyroscopes
7.7 Impulse and Momentum
7.8 Work and Kinetic Energy
SUMMARY • Chapter 7
REVIEW QUESTIONS • Chapter 7
Chapter 8
8 SPECIAL TOPICS
8.1 Introduction
8.2 Introduction to Vibrations
8.3 Euler's Laws for a Control Volume
8.4 Central Force Motion
REVIEW QUESTIONS
Appendices
APPENDICES CONTENTS
A UNITS #573,0,32768 AMPLES OF NUMERICAL ANALYSIS / THE NEWTO
C MOMENTS OF INERTIA OF
D ANSWERS TO ODDNUMBERED PROBLEMS
INDEX
•
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 474041626
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 ParallelAxis Theorems, 229 The MassCenter 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 IMPULSEMOMENTUM 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 THREEDIMENSIONAL 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 ODDNUMBERED 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 twothirds 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 truefalse 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 oddnumbered 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 "pointmass" 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 pointmass 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 (threedimensional, 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 freebody 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 freebody 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, Wanlee 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 thirdedition 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 vt 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 forcemotion 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 rateofchange of position, a n d acceleration is rateofchange of velocity. It is the acceleration in N e w t o n ' s Second Law a n d the positionvelocityacceleration 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 1.41.6) 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 constantmagnitude b u t varyingdirection 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 1.11.8, 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.
1.91.16 If the vectors enumerated in Problems 1.11.8 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 forcemotion 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 forcemotion 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
1.231.27 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 onedimensional 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 vt 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 vt 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 vt 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 vt 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 vt 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 vt 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 vt 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 10s 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 lefthand 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 onedimensional (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 1.351.37, 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 vt diagram. 1.39 Ben Johnson set a world record of 9.83 seconds in the 100meter dash on August 31, 1987. He had also set the record for the 60meter 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 3mi 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 wedgeshaped 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 threedimensional 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 (threedimensional) 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 xcomponent 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 1.831.86 (see Figures P1.83P1.86), 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 Firstquadrant branch of rectangular hyperbola 1.86 Firstquadrant branch of semicubical parabola 1.871.90 Find the respective acceleration vectors at f = 10 s of the points whose motions are described in Problems 1.831.86.
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 xcomponent 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 timedependent 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 xcomponent 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 fourleaf 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, ki, 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 twodimensional (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 xcomponent 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 SerretFrenet 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 FreeBody 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 freebody 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 selfequilibrating 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: SpringerVerlag, 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 particlesystem 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 crosssectional centroids, find the coordinates of the mass center in the deformed configuration. Solution In the first configuration the centerofmass 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 particlelike 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 FreeBody 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 freebody 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 freebody 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 freebody 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 freebody 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 freebody 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 freebody 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 freebody 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 actionreaction 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 onesixth that at the surface of the earth. 0
EXAMPLE 2 . 3
Figure E2.3a
If the 20kg 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 freebody 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
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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 freebody 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 freebody 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 NewtonRaphson 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 freebody 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 10fthigh 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 highflying 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.
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Figure P2.17
2.17 A baseball pitcher releases a 90mph 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 popup 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'seye?
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'seye 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 onesixth 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 liquidrocket 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 200lb block is at rest on the floor ( = 0.2) before the 50lb 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 200lb 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 (4000mile 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 100Newton 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 250Ib 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 5lb 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
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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 camfollower 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, 90minute 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
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•
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 spinspeed i , the bodies be have as one rigid body (meaning remains constant). Find the relationship. Hint: Use separate freebody 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 velocitysquared 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 76lb cannonball of Figure P2.88, find the maximum height it reaches. Compare your result with the answer neglecting air resistance. D
2
*
A 50lb 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 onetenth 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 160lb parachutist in the "freefall spreadstable position" (Figure P2.90) reaches a velocity of 174 f t / s e c
Figure P2.94
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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 ftlb is the unit of work and energy. The N • m and lbft 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 forcestretch relationship?
Answer 2.4 (k) . (stretch) gives the equalinmagnitude but oppositeindirection 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 forcestretch 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.
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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 controlmoment 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 hlbft. 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 lunisolar precession is presented in Chapter 7.
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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 forceacceleration 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.
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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 masscenter 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 freebody 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
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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
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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 6lb 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 singlevalued, 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
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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 80lb child rides a 10lb 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
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2.126 The system in Figure P2.126 consists of the 12lb body the light pulley the 8lb "rider" and the 10lb 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 5lb 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
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• 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.
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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 varyingmass 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.
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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.
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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 highspeed 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.
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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 freebody 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 onehalf 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 .
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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
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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
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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 postcollision 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.
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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 onesixth 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 50lb 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
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2.138 An unattached 2.2lb 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 threecar 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 20ft 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
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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 headon 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 4in.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 beforecollision 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 10kg 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 10lb 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 16kg body and the 32kg 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 workenergy 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
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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 freebody 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 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 onedimensional 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
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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
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OR
w h i c h is the particlesystem 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
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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 nonrigid 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
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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 freebody diagram (Figure E2.18b) of the pulleyropegymnasts 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 freebody 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.
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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, crosssectional area A, and length L undergoes smallamplitude, 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
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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 30lb 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 problemsolving process w a s the freebody 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,
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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 rigidbody 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 righthand 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 rigidbody 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 crosssectional 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 (threedimensional) 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(ac)
(See next page for df)
(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(df)
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 righthanded 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 bodyfixed 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 finitesized 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 (threedimensional) 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 30ft 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
3.73.11 in the following five problems involving a "fourbar 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 15cm 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 slidercrank 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 bodyfixed 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 5inch (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 5inch 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 10ft bar in Figure P3.68 is sliding down the 13ftradius 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 xaxis. 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 contactpoint 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 = 0. Therefore when
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 torquefree 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 torquefree 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 rigidbody 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
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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.
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• 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 bodyfixed 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
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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 (tlueedimensional) 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 15kg 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 parallelaxis 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 manmade 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 singledegreeoffreedom 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 freebody 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 secondorder linear differential equation with constant coeffi cients describing the motion of the block. The fact that the differential equation is nonhomogeneous (the righthand 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 springmass 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 freebody 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, secondorder, 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 blockspring 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 noslip
condition. We shall now use a and a in the equations of motion: c
(1)
Figure
Figure E 8 . 1 a
Hence, from the freebody 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 blockspring 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 freebody 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 simpleharmonic. For the massspringdamper 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 — 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 steadystate 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 lbsec /in., k = 100 lb/in., and c = 2 lbsec/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 steadystate 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 halfcylinder 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 200lb block. (See Figure P8.16.) The block is fastened to the spring. A 400lb 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 lbsec /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 100lb 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 lbsec / 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 steadystate displacement x(t) if y(t) = 0.1 sin lOOf inch, where Ms in seconds, m = 0.01 lbsec /in., k = 100 lb/in., and c = 2 lbsec/ 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 lbsec /in.
8.27 In Figure P8.27 find the steadystate displacement x(t) if y(t) = 0.2 sin 90f inch, where f is in seconds, m = 0.01 lbsec /in., = 501b/in.,andc = 1 lbsec/in. In particular:
c = 1.0 lbsec/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 lbsec/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 steadystate 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 lbsec /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 rocketpowered 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 selfevident 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 controlvolume 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 controlvolume 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 crosssectional 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 freebody 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
Freebody 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. insidediameter pipe to a 24in. insidediameter 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 20ft 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 controlvolume 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 ppex 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
865 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 (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 steadystate 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 onedimensional controlvolume 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 ODDNUMBERED PROBLEMS Appendix E ADDITIONAL MODELBASED 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 meterkilogramsecond (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 lbsec /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 poundfoot (lbft). Work and energy have this same dimension; the U.S. unit is the ftlb 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 ftlb/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 slugft slugft/sec slugft /sec lbsec Ibftsec 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 lbft. Since we know there to be 3.281 ft per meter and 4.448 N per pound, 1 lbft = 1
= 1.356 N • m
Note that if the undesired unit (such as lb in this example) does not cancel, the conversion fraction is upsidedown! 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 poundmass (Ibm) slug (lbsec /ft)
kilogram (kg) kg
0.45359 14.594
2.20462 0.068522
Force pound (lb) or poundforce (Ibf)
newton (N)
4.4482
0.22481
2
2
2
2
2
2
2
2
Density poundmass/inch (lbm/in. ) poundmass/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 footpound or poundfoot (ftlb) (lbft)
joule (J) or newton • meter (N • m)
1.3558
0.73757
Power footpound/minute (ftlb/min) horsepower (hp) (550 ftlb/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 slugfoot (slugft or lbftsec ) 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) slugfoot/second (slugft/sec)
kg • m / s
4 4482
0.22481
Impulse (or linear impulse) poundsecond (lbsec)
N • s (or kg • m/s)
4 4482
0 22481
1 3558
0 73756
1 3558
0 73756
Moment of momentum (or angular slugfoot /second (slugft /sec) 2
2
Angular impulse poundfootsecond (Ibftsec)
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 poundmass (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 8oz 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 lbf^/slug . 1
2
2
2
B
B EXAMPLES OF NUMERICAL ANALYSIS / THE NEWTONRAPHSON 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 NewtonRaphson 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 NewtonRaphson 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
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= 2.607668744
Appendix B / Examples of Numerical Analysis/The NewtonRaphson Method
This,
distance
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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 NewtonRaphson as previously described. Figure B. 7 suggests that might serve as a good first guess at the root. A NewtonRaphson 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
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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 = 2rrRtH
(0, 0, 0) V = r{R
2

2
r )H
M o m e n t s of Inertia About Indicated Axes
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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
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Object
2. If a = 2 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
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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 parallelaxis theorem, C is on zaxis. 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 kgm ; 0.0957 kem ; 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 crossproduct 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)