Irving H. Shames-Engineering Mechanics (Statics and Dynamics).(1996)
Short Description
For Combined Statics and Dynamics courses.This edition of the highly respected and well-known book for Engineering Mecha...
Description
Engineering Mechanics Statics and Dynamics
Irving H. Shames Professor Dept. of Civil, Mechanical and Eirvirorrmenrul En,qirierring The George Washington Uiiiver.yiQ
Prentice Hall, Upper Saddle River, New Jersey 07458
Acqui\iiionr Editor: William Stenquiit Editor in Chic1 Marcia Hoiton I’mduclion Editor: Ro\e Krrnao Text I l e s p c r : Christme Wull Covcr Dc\igsei: Amy Roam Editorial Ashiavant: Meg Wci.1 Manufacturing Buyer: Lhnna Sullivan
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Contents
Preface
$2.3 Addition and Subtraction of Vectors 24 2.4 Resolution of Vectors; Scalar Components 30 t2.5 Unit Vectors 33
ix
1 Fundamentals of Mechanics Review I 3
tl.1 t1.2
Introduction 3 Basic Dimensions and Units of Mechanics, 4 71.3 Secondary Dimensional Quantities 71.4 Law of Dimensional Homogeneity 71.5 Dimensional Relation between Force and Mass 9 1.6 Units of Mass 10 1.7 Idealizations of Mechanics 12 11.8 Vector and Scalar Quantities 14 1.9 Equality and Equivalence of Vectors 17 t1.10 Laws of Mechanics 19 1.11 Closure 22
2 Elements of vector Algebra Review II 23
t2.1 72.2
Introduction 23 Magnitude and Multiplication of a Vector by a Scalar 23
7 8
2.6
Useful Ways of Representing Vectors 35
2.7
Scalar or Dot Product of Two Vectors 41
2.8
Cross Product of Two Vectors 47
2.9 Scalar Triple Product 5 1 2.10 A Note on Vector Notation 54 2.11 Closure 56
3 Important vector Quantities 3.1 3.2 3.3 3.4 3.5 3.6
61
Position Vector 61 Moment of a Force about a Point 62 Moment of a Force about an Axis 69 The Couple and Couple Moment 77 The Couple Moment as a Free Vector 79 Addition and Subtraction of Couples 80
IV
CONTENTS
3.7 3.8
Moment of a Couple About a Line Closure 89
4 Equivalent Force systems 4.1 4.2 4.3 4.4 4.5 4.6
82
93
Introduction 93 Translation of a Force to a Parallel Position 94 Resultant of a Force System 102 Simplest Resultants of Special Force Systems 106 Distributed Force Systems 107 Closure 143
5 Equations of Equilibrium
151
lntroduction 15 I The Free-body Diagram 152 Free Bodies Involving Interior Sections 154 *5.4 Looking Ahead-Control Volumes 158 5.5 General Equations of Equilibrium 162 5.6 Problems of Equilibrium I 164 5.7 Problems of Equilibrium 11 183 5.8 Two Point Equivalent Loading 199 5.9 Problems Arising from Structures 200 5.10 Static Indeterminacy 204 5.11 Closure 210
5.1 5.2 5.3
6 introduction to structural Mechanics 221 Part A: Trusses 221 6.1
6.2 6.3 6.4 6.5
The Structural Model 221 The Simple Truss 224 Solution of Simple Trusses 225 Method of Joints 225 Method of Sections 238
6.6
Looking Ahead-Deflection of a Simple, Linearly Elastic Truss 242
Part B: Section Forces in Beams 247 6.7 6.8 6.9
Introduction 247 Shear Force, Axial Force, and Bending Moment 247 Differential Relations for Equilibrium 25Y
Part C: Chains and Cables 266 6.10 Introduction 266 6.11 Coplanar Cables; Loading is a Function ofx 266 6.12 Coplanar Cables: Loading is the Weight of the Cable Itself 270 6.13 Closure 277
7 Friction FOrCeS 7.1 7.2 *7.3 7.4 7.5 7.6 7.7 *7.8 7.9
281
Introduction 281 Laws of Coulomb Friction 282 A Comment Concerning the Use of Coulomb’.: Law 284 Simple Contact Friction Problems 284 Complex Surface Contact Friction Problems 299 Belt Friction 301 The Square Screw Thread 3 17 Rolling Resistance 319 Closure 323
8 Properties of surfaces 8.1 8.2 8.3 8.4
331
Introduction 331 First Moment of an Area and the Centroid 331 Other Centers 342 Theorems of Pappus-Guldinus
347
CONTENTS
8.5
Second Moments and the Product of Area of a Plane Area 355 8.6 Tranfer Theorems 356 8.7 Computations Involving Second Moments and Products of Area 357 8.8 Relation Between Second Moments and Products of Area 366 8.9 Polar Moment of Area 369 8.10 Principal Axes 370 8.11 Closure 375
9
Moments and Products oflnertia 379 9.1 9.2 9.3
9.4 “9.5 *9.6 *9.7 9.8
Introduction 379 Formal Definition of Inertia Quantities 379 Relation Between Mass-Inertia Terms and Area-Inertia Terms 386 Translation of Coordinate Axes 392 Transformation Properties of the Inertia Terms 395 Looking Ahead-Tensors 400 The Inertia Ellipsoid and Principal Moments of Inertia 407 Closure 410
V
10.5 Looking Ahead-Deformable Solids 424
Part B: Method of Total Potential Energy 432 10.6 10.7
Conservative Systems 432 Condition of Equilibrium for a Conservative System 434 10.8 Stability 441 10.9 Looking Ahead-More on Total Potential Energy 443 10.10 Closure 446
11 Kinematics of a Particle-simple Relative Motion 451 11.1 Introduction
45 1
Part A: General Notions 452 11.2
Differentiation of a Vector with Respect toTime 452
Part B: Velocity and Acceleration Calculations 454 11.3
10 *Methods of virtual work and stationary Potential Energy 413 10.1 Introduction 413
Part A: Method of Virtual Work 414 10.2 Principle of Virtual Work for a Particle 414 10.3 Principle of Virtual Work for Rigid Bodies 415 10.4 Degrees of Freedom and the Solution of Problems 418
Introductory Remark 454 11.4 Rectangular Components 455 11.5 Velocity and Acceleration in Terms of Path Variables 465 11.6 Cylindrical Coordinates 480
Part C: Simple Kinematical Relations and Applications 492 11.7 11.8
Simple Relative Motion 492 Motion of a Particle Relative to a Pair of Translating Axes 494
11.9
Closure 504
Vi
12
CONTENTS
13.3 13.4
Particle Dynamics 511 12.1
Introduction
51I
Conservative Force Field 594 Conservation of Mechanical Energy 598 Alternative Form of Work-Energy Equation 603
Part A: Rectangular Coordinates; Rectilinear Translation 512
13.5
12.2
Part B: Systems of Particles 609
12.3 12.4
Newton's Law for Rectangular Coordinates 5 12 Rectilinear Translation 5 12 A Comment 528
13.6 13.7 13.8
Part B: Cylindrical Coordinates; Central Force Motion 536
13.9
Work-Energy Equations 609 Kinetic Energy Expression Based on Center of Mass 614 Work-Kinetic Energy Expressions Based on Center of Mass 619 Closure 631
12.5
Newton's Law for Cylindrical Coordinates 536 12.6 Central Force MotionAn Introduction 538 *12.7 Gravitational Central Force Motion 539 "12.8 Applications to Space Mechanics 544
Part C: Path Variables 561 12.9
Newton's Law for Path Variables 561
Part D: A System of Particles 564 12.10 The General Motion of a System of Particles 564 12.11 Closure 571
13
Energy Methods for Particles 579 Part A: Analysis for a Single Particle 579 13.1 13.2
Introduction 579 Power Considerations
14
Methods of Momentum for Particles 637 Part A: Linear Momentum 637 14.1
Impulse and Momentum Relations for a Particle 637 14.2 Linear-Momentum Considerations for a System of Particles 643 14.3 Impulsive Forces 648 14.4 Impact 659 "14.5 Collision of a Particle with a Massive Rigid Body 665
Part B: Moment of Momentum 675 14.6 14.7 14.8 *14.9
585
14.10
Moment-of-Momentum Equation for a Single Particle 675 More on Space Mechanics 678 Moment-of-Momentum Equation for a System of Particles 686 Looking Ahead-Basic Laws of Continua 694 Closure 700
CONTENTS
15 Kinematics of Rigid Bodies:
*16.9 Balancing 838 16.10 Closure 846
Relative Motion 707 15.1 Introduction 707 15.2 Translation and Rotation of Rigid Bodies 707 15.3 Chasles’ Theorem 709 15.4 Derivative of a Vector Fixed in a Moving Reference 71 1 15.5 Applications of the Fixed-Vector Concept 723 15.6 General Relationship Between Time Derivatives of a Vector for Different References 743 15.7 The Relationship Between Velocities of a Particle for Different References 744 15.8 Acceleration of a Particle for Different References 755 15.9 A New Look at Newton’s Law 773 15.10 The Coriolis Force 776 15.11 Closure 781
16 Kinetics of Plane Motion of Rigid Bodies 787 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8
Introduction 787 Moment-of-Momentum Equations 788 Pure Rotation of a Body of Revolution About its Axis of Revolution 791 Pure Rotation of a Body with Two Orthogonal Planes of Symmetry 797 Pure Rotation of Slablike Bodies 800 Rolling Slablike Bodies 810 General Plane Motion of a Slablike Body 816 Pure Rotation of an Arbitrary Rigid Body 834
vii
17
Energy and Impulse-Momentum Methods for Rigid Bodies 853 17.1
Introduction
853
Part A: Energy Methods 853 17.2 17.3
Kinetic Energy of a Rigid Body Work-Energy Relations 860
853
Part B: Impulse-Momentum Methods 878 17.4
17.5 17.6 17.7
Angular Momentum of a Rigid Body About Any Point in the Body 878 Impulse-Momentum Equations 882 Impulsive Forces and Torques: Eccentric Impact 895 Closure 907
18 *Dynamics of General Rigid-Body Motion 911 18.1 18.2 18.3 18.4
18.5 18.6 18.7 18.8
Introduction 91 1 Euler’s Equations of Motion 914 Application of Euler’s Equations 916 Necessary and Sufficient Conditions for Equilibrium of a Rigid Body 930 Three-Dimensional Motion About a Fixed Point; Euler Angles 930 Equations of Motion Using Euler Angles 934 Torque-Free Motion 945 Closure 958
...
VI11
CON'lt.NlS
I9 Vibrations 19.1
19.2 19.3 *19.4 *IY.S 19.6
19.7
19.8
* 19.9 19.10
961
Introduction 961 Frce Vihratioii 961 Torsional Vibration 973 Examples of Other Free-Oscillating Motions 9x2 Energy Methods 984 Linear Restoring Force and a Force Varying Sinusoidally with Time 900 Linear Restoring Force with Viscous Damping 999 I.inelrr Rehtoring Force, Viscous Damping, and a Harmonic Disturhance IO07 Oscillatory Systems with Multi-Degrees of Freedom IO I4 Closurc 1022
APPENDIX I
Integration Formulas xvii APPENDIX II
Computation of Principal Moments of Inertia xix APPENDIX 111
Additional Data For the Ellipse xxi APPENDIX IV
Proof that Infinitesimal Rotations Are Vectors xxiii Projects xxv
Index Ixxvii
Preface
With the publication of the fourth edition, this text moves into the fourth decade of its existence. In the spirit of the times, the first edition introduced a number of “firsts” in an introductory engineering mechanics textbook. These “firsts” included
a) the first treatment of space mechanics b) the first use of the control volume for linear momentum considerations of fluids c ) the first introduction to the concept of the tensor Users of the earlier editions will be glad to know that the 4th edition continues with the same approach to engineering mechanics. The goal has always been to aim toward working problems as soon as possible from first principles. Thus, examples are carefully chosen during the development of a series of related areas to instill continuity in the evolving theory and then, after these areas have been carefully discussed with rigor, come the problems. Furthermore at the ends of each chapter, there are many problems that have not been arranged by text section. The instructor is encouraged as soon as hekhe is well along in the chapter to use these problems. The instructors manual will indicate the nature of each of these problems as well as the degree of difficulty. The text is not chopped up into many methodologies each with an abbreviated discussion followed by many examples for using the specific methodology and finally a set of problems carefully tailored for the methodology. The nature of the format in this and preceding editions is more than ever first to discourage excessive mapping of homework problems from the examples. And second, it is to lessen the memorization of specific, specialized methodologies in lieu of absorbing basic principles. ix
X
PREFACE
A new feature in the fourth edition is a series of starred sections called “Looking Ahead . . . .” These are simplified discussions of topics that appear in later engineering courses and tie in directly or indirectly to the topic under study. For instance, after discussing free body diagrams, there is a short “Looking Ahead” section in which the concept and use of the control volume is presented as well as the system concepts that appear in tluid mechanics and thermodynamics. In the chapter on virtual work for particles and rigid bodies, there is a simplified discussion of the displacement methods and force methods for deformable bodies that will show up later in solids courses. After finding the forces for simple trusses, there is a “Looking Ahead” section discussing brietly what has to be done to get displacements. There are quite a few others in the text. It has been found that many students find these interesting and later when they come across these topics in other courses or work, they report that the connections so formed coming out of their sophomore mechanics courses have been most valuable. Over 400 new problems have been added to the fourth edition equally divided between the statics and dynamics books. A complete word-processed solutions manual accompanies the text. The illustrations needed for problem statement and solution are taken as enlargements from the text. Generally, each problem is on a separate page. The instructor will be able conveniently to select problems in order to post solutions or to form transparencies as desired. Also, there are 30 computer projects in which, for a number of cases, the student prepares hidher own software or engages in design. As an added bonus, the student will be able to maintain hidher proficiency in programming. Carefully prepared computer programs as well as computer outputs will be included in the manual. I normally assign one or two such projects during a semester over and above the usual course material. Also included in the manual is a disk that has the aforementioned programs for each of the computer projects. The computer programs for these projects generally run about 30 lines of FORTRAN and run on a personal computer. The programming required involves skills developed in the freshman course on FORTRAN. Another important new feature of the fourth edition is an organization that allows one to go directly to the three dimensional chapter on dynamics of rigid bodies (Chapter 18) and then to easily return to plane motion (Chapter 16). Or one can go the opposite way. Footnotes indicate how this can be done, and complimentary problems are noted in the Solutions Manual.
PREFACE
Another change is Chapter 16 on plane motion. It has been reworked with the aim of attaining greater rigor and clarity particularly in the solving of problems. There has also been an increase in the coverage and problems for hydrostatics as well as examples and problems that will preview problems coming in the solids course that utilize principles from statics. It should also be noted that the notation used has been chosen to correspond to that which will be used in more advanced courses in order to improve continuity with upper division courses. Thus, for moments and products of inertia I use I, I,, lxzetc. rather than ly l,, ,P etc. The same notation is used for second moments and products of area to emphasize the direct relation between these and the preceding quantities. Experience indicates that there need be. no difficulty on the student’s part in distinguishing between these quantities; the context of the discussion suffices for this purpose. The concept of the tensor is presented in a way that for years we have found to be readily understood by sophomores even when presented in large classes. This saves time and makes for continuity in all mechanics courses, particularly in the solid mechanics course. For bending moment, shear force, and stress use is made of a common convention for the sign-namely the convention involving the normal to the area element and the direction of the quantity involved be it bending moment, shear force or stress component. All this and indeed other steps taken in the book will make for smooth transition to upper division course work. In overall summary, two main goals have been pursued in this edition. They are
1. To encourage working problems from first principles and thus to minimize excessive mapping from examples and to discourage rote learning of specific methodologies for solving various and sundry kinds of specific problems. 2. To “open-end” the material to later course work in other engineering sciences with the view toward making smoother transitions and to provide for greater continuity. Also, the purpose is to engage the interest and curiosity of students for further study of mechanics. During the 13 years after the third edition, I have been teaching sophomore mechanics to very large classes at SUNY, Buffalo, and, after that, to regular sections of students at The George Washington University, the latter involving an international student body with very diverse backgrounds. During this time, I have been working on improving the clarity and strength of this book under classroom conditions
XI
xii
PREFACE
giving it the most severe test as a text. I believe the fourth edition as a result will be a distinct improvement over the previous editions and will offer a real choice for schools desiring a more mature treatment of engineering mechanics. I believe sophomore mechanics is probably the most important course taken by engineers in that much of the later curriculum depends heavily on this course. And for all engineering programs, this is usually the first real engineering course where students can and must be creative and inventive in solving problems. Their old habits of mapping and rote learning of specific problem methodologies will not suffice and they must learn to see mechanics as an integral science. The student must “bite the bullet” and work in the way he/she will have to work later in the curriculum and even later when getting out of school altogether. No other subject so richly involves mathematics, physics, computers, and down to earth common sense simultaneously in such an interesting and challenging way. We should take maximum advantage of the students exposure to this beautiful subject to get h i d h e r on the right track now so as to be ready for upper division work. At this stage of my career, I will risk impropriety by presenting now an extended section of acknowledgments. I want to give thanks to SUNY at Buffalo where I spent 31 happy years and where I wrote many of my hooks. And I want to salute the thousands (about 5000) of fine students who took my courses during this long stretch. I wish to thank my eminent friend and colleague Professor Shahid Ahmad who among other things taught the sophomore mechanics sequence with me and who continues to teach it. He gave me a very thorough review of the fourth edition with many valuable suggestions. I thank him profusely. I want particularly to thank Professor Michael Symans, from Washington State University, Pullman for his superb contributions to the entire manuscript. I came to The George Washington University at the invitation of my longtime friend and former Buffalo colleague Dean Gideon Frieder and the faculty in the Civil, Mechanical and Environmental Engineering Department. Here, I came back into contact with two well-known scholars that I knew from the early days of my career, namely Professor Hal Liebowitz (president-elect of the National Academy of Engineering) and Professor Ali Cambel (author of recent well-received book on chaos). 1 must give profound thanks to the chairman of my new department at G.W., Professor Sharam Sarkani. He has allowed me to play a vital role in the academic program of the department. I will be able to continue my writing at full speed as a result. 1 shall always be grateful to him. Let me not forget
PREFACE
the two dear ladies in the front office of the department. Mrs. Zephra Coles in her decisive efficient way took care of all my needs even before I was aware of them. And Ms. Joyce Jeffress was no less helpful and always had a humorous comment to make. I was extremely fortunate in having the following professors as reviewers. Professor Shahid Ahmad, SUNY at Buffalo Professor Ravinder Chona, Texas A&M University Professor Bruce H. Kamopp. University of Michigan Professor Richard E Keltie, North Carolina State University Professor Stephen Malkin, University of Massachusetts Professor Sudhakar Nair, Illinois Institute of Technology Professor Jonathan Wickert, Camegie Mellon University I wish to thank these gentlemen for their valuable assistance and encouragement. I have two people left. One is my good friend Professor Bob Jones from V.P.I. who assisted me in the third edition with several hundred excellent statics problems and who went over the entire manuscript with me with able assistance and advice. I continue to benefit in the new edition from his input of the third edition. And now, finally, the most important person of all, my dear wife Sheila. She has put up all these years with the author of this book, an absent-minded, hopeless workaholic. Whatever I have accomplished of any value in a long and ongoing career, I owe to her.
To my Dear, Wondeijiil Wife Sheila
...
XI11
About the Author
Irving Shames presently serves as a Professor in the Department of Civil, Mechanical, and Environmental Engineering at The George Washington University. Prior to this appointment Professor Shames was a Distinguished Teaching Professor and Faculty Professor at The State University of New York-Buffalo, where he spent 31 years. Professor Shames has written up to this point in time 10 textbooks. His first book Engineering Mechanics, Statics and Dynamics was originally published in 1958, and it is going into its fourth edition in 1996. All of the books written by Professor Shames have been characterized by innovations that have become mainstays of how engineering principles are taught to students. Engineering Mechanics, Statics and Dynamics was the first widely used Mechanics book based on vector principles. It ushered in the almost universal use of vector principles in teaching engineering mechanics courses today. Other textbooks written by Professor Shames include:
Mechanics of Deformable Solids, Prentice-Hall, Inc. Mechanics of Fluids, McGraw-Hill * Introduction to Solid Mechanics, Prentice-Hall, Inc. * Introduction to Statics, Rentice-Hall, Inc. * Solid Mechanics-A Variational Approach (with C.L. Dym), McCraw-Hill Energy and Finite Element Methods in Structural Mechanics, (with C.L. Dym), Hemisphere Corp., of Taylor and Francis Elastic and Inelastic Stress Analysis (with F. Cozzarelli), PrenticeHall, Inc.
-
XVI
ABOIJTTHF AI'THOK
In recent ycars, I'rofesor Shalne\ has expanded his teaching xtivitics and t i a h held two suiiiiner fiicully workshops in mechanics \ponsored by the State (if Ncw York, and one national workshop sponsorcd by the National Science Foundation. The programs involved the iiitegr:ition both conceptually and pedagogically 0 1 mechanics from the sophomore year on through gt-aduate school.
Statics
REVIEW I*
Fundamentals of Mechanics +l.l Introduction Mechanics is the physical science concerned with the dynamical behavior (as opposed to chemical and thermal behavior) of bodies that are acted on by mechanical disturbances. Since such behavior is involved in virtually all the situations that confront an engineer, mechanics lies at the core of much engineering analysis. In fact, no physical science plays a greater role in engineering than does mechanics, and it is the oldest of all the physical sciences. The writings of Archimedes covering buoyancy and the lever were recorded before 200 B.C. Our modem knowledge of gravity and motion was established by Isaac Newton (1642-1727), whose laws founded Newtonian mechanics, the subject matter of this text. In 1905, Einstein placed limitations on Newton's formulations with his theory of relativity and thus set the stage for the development of relativistic mechanics. The newer theories, however, give results that depart from those of Newton's formulations only when the speed of a body approaches the speed of light ( I 86,000 mileslsec). These speeds are encountered in the largescale phenomena of dynamical astronomy. Furthermore for small-scale phenomena involving subatomic particles, quantum mechanics must be used rather than Newtonian mechanics. Despite these limitations, it remains nevertheless true that, in the great bulk of engineering problems, Newtonian mechanics still applies.
*The reader is urged 10 be sure that Section 1.9 is thoroughly understood since this Section is vital for a goad understanding of statics in panicular and mechanics in general. Also, the nutation t before the titles of certain sections indicates thal specific queslions concerning the contents of these sections requiring verbal answers are presented at the end of the chapler. The instructor may wish to assign these sections as a reading asignment along with the requirement to answer the aforestated asssiated questions as the author routinely daes himself.
3
4
CHAPTER I
FUNDAMENTALS OFMECHANICS
t1.2
Basic Dimensions and Units of Mechanics
To study mechanics, we must establish abstractions to describe those characteristics of a body that interest us. These abstractions are called dimensions. The dimensions that we pick, which are independent of all other dimensions, are termed primary or basic dimensions, and the ones that are then developed in terms of the basic dimensions we call secondary dimensions. Of the many possible sets of basic dimensions that we could use, we will confine ourselves at present to the set that includes the dimensions of length, time, and mass. Another convenient set will he examined later. Length-A Concept for Describing Size Quantitatively. In order to determine the size of an object, we must place a second object of known size next to it. Thus, in pictures of machinery, a man often appears standing disinterestedly beside the apparatus. Without him, it would be difficult to gage the size of the unfamiliar machine. Although the man has served as some sort of standard measure, we can, of course, only get an approximate idea of the machine's size. Men's heights vary, and, what is even worse, the shape of a man is too complicated to be of much help in acquiring a precise measurement of the machine's size. What we need, obviously, is an object that is constant in shape and, moreover, simple in concept. Thus, instead of a threedimensional object, we choose a one-dimensional object.' Then, we can use the known mathematical concepts of geometry to extend the measure of size in one dimension to the three dimensions necessary to characterize a general body. A straight line scratched on a metal bar that is kept at uniform thermal and physical conditions (as, e.g., the meter bar kept at Skvres, France) serves as this simple invariant standard in one dimension. We can now readily calculate and communicate the distance along a cettain direction of an object by counting the number of standards and fractions thereof that can be marked off along this direction. We commonly refer to this distance as length, although the term " length could also apply to the more general concept of size. Other aspects of size, such as volume and area, can then be formulated in terms of the standard by the methods of plane, spherical, and solid geometry. A unit is the name we give an accepted measure of a dimension. Many systems of units are actually employed around the world, but we shall only use the two major systems, the American system and the SI system. The basic unit of length in the American system is the foot, whereas the basic unit of length in the SI system is the meter. Time-A Concept for Ordering the Flow of Events. In observing the picture of the machine with the man standing close by, we can sometimes tell approximately when the picture was taken by the style of clothes the man is 'We are using the word "dimensional" here in its everyday sense and not as defined above.
SECTION 1.2 BASIC DIM!3”SIONS AND UNITS OF MECHANICS
wearing. But how do we determine this? We may say to ourselves: “During the thirties, people wore the type of straw hat that the fellow in the picture is wearing.” In other words, the “when” is tied to certain events that are experienced by, or otherwise known to, the observer. For a more accurate description of “when,” we must find an action that appears to he completely repeatable. Then, we can order the events under study by counting the numher of these repeatable actions and fractions thereof that occur while the events transpire. The rotation of the earth gives rise to an event that serves as a good measure of time-the day. But we need smaller units in most of our work in engineering, and thus, generally, we tie events to the second, which is an interval repeatable 86,400 times a day. Mass-A Property of Matter. The student ordinarily has no trouble understanding the concepts of length and time because helshe is constantly aware of the size of things through hisher senses of sight and touch, and is always conscious of time by observing the flow of events in hisher daily life. The concept of mass, however, is not as easily grasped since it does not impinge as directly on our daily experience. Mass is a property of matter that can be determined from two different actions on bodies. To study the first action, suppose that we consider two hard bodies of entirely different composition, size, shape, color, and so on. If we attach the bodies to identical springs, as shown in Fig. 1.1, each spring will extend some distance as a result of the attraction of gravity for the hodies. By grinding off some of the material on the body that causes the greater extension, we can make the deflections that are induced on both springs equal. Even if we raise the springs to a new height above the earth’s surface, thus lessening the deformation of the springs, the extensions induced by the pull of gravity will he the same for both bodies. And since they are, we can conclude that the bodies have an equivalent innate property. This property of each body that manifests itself in the amount of gravitational attraction we call man. The equivalence of these bodies, after the aforementioned grinding operation, can be indicated in yet a second action. If we move both bodies an equal distance downward, by stretching each spring, and then release them at the same time, they will begin to move in an identical manner (except for small variations due to differences in wind friction and local deformations of the bodies). We have imposed, in effect, the same mechanical disturbance on each body and we have elicited the same dynamical response. Hence, despite many obvious differences, the two bodies again show an equivalence. mpcs, thn, Chomcrcrke8 a body both in the action of na1 a n r a c k and in tlu response IO a mekhnnicd
The pcoperry of
To communicate this property quantitatively, we may choose some convenient body and compare other bodies to it in either of the two above-
Body A
Body B
Figure 1.1. Bodies restrained by identical springs.
5
6
CHAPTER I
FUNDAMENTALS OF MECHANICS
mentioned actions. The two basic units commonly used in much American engineering practice to measure mass are the pound mass, which is defined in terms of the attraction of gravity for a standard body at a standard location, and the slug, which is defined in terms of the dynamical response of a standard body to a standard mechanical disturbance. A similar duality of mass units does not exist in the SI system. There only the kilugmm is used as the basic measure of mass. The kilogram is measured in terms of response of a body to a mechanical disturbance. Both systems of units will he discussed further in a subsequent section. We have now established three basic independent dimensions to describe certain physical phenomena. It is convenient to identify these dimensions in the following manner: length
[L]
time mass
[tl
[MI
These formal expressions of identification for basic dimensions and the more complicated groupings to he presented in Section 1.3 for secondary dimensions are called “dimensional representations.” Often, there are occasions when we want to change units during computations. For instance, we may wish to change feet into inches or millimeters. In such a case, we must replace the unit in question by a physically equivalent number of new units. Thus, a foot is replaced by 12 inches or 30.5 millimeters. A listing of common systems of units is given in Table 1.1, and a table of equivalences hetween these and other units is given on the inside covers. Such relations between units will he expressed in this way: 1 ft
12 in.
= 305 mm
The three horizontal bars are not used to denote algebraic equivalence; instead, they are used to indicate physical equivalence. Here is another way of expressing the relations above: Table 1.1 common systems of
units SI
c!P
Mass
Length Time FOKC
Gram Centimeter Second Dyne
English
Mass Length Time Force
Pound mass Foot Second Poundal
Mass Kilogram Length Meter Time Second Force Newton American Practice Mass Length Time Force
Slug or pound mass Foot Second Puund force
SECTION 1.3 SECONDARY DIMENSIONAL QUANTITIES
The unity on the right side of these relations indicates that the numerator and denominator on the left side are physically equivalent, and thus have a 1:l relation. This notation will prove convenient when we consider the change of units for secondary dimensions in the next section.
t1.3
Secondary Dimensional Quantities
When physical characteristics are described in terms of basic dimensions by the use of suitable definitions (e.g., velocity is defined2 as a distance divided by a time interval), such quantities are called secondary dimensional quantities. In Section 1.4, we will see that these quantities may also be established as a consequence of natural laws. The dimensional representation of secondary quantities is given in terms of the basic dimensions that enter into the formulation of the concept. For example, the dimensional representation of velocity is [velocity] = [Ll
[/I
That is, the dimensional representation of velocity is the dimension length divided by the dimension time. The units for a secondary quantity are then given in terms of the units of the constituent basic dimensions. Thus, [velocity units] = [ftl [secl A chunge of units from one system into another usually involves a change in the scale of measure of the secondary quantities involved in the problem. Thus, one scale unit of velocity in the American system is 1 foot per second, while in the SI system it is I meter per second. How may these scale units he correctly related for complicated secondary quantities? That is, for our simple case, how many meters per second are equivalent to 1 foot per second? The formal expressions of dimensional representation may he put to good use for such an evaluation. The procedure is as follows. Express the dependent quantity dimensionally; substitute existing units for the basic dimensions; and finally, change these units to the equivalent numbers of units in the new system. The result gives the number of scale units of the quantity in the new system of units that is equivalent to 1 scale unit of the quantity in the old system. Performing these operations for velocity, we would thus have l(&)
I(*)
>A more precise definilion will be given
= .305(2) in the chapters on dynamics.
7
8
CHAPTER I
FUNDAMENTALS OF MECHANICS
which means that ,305scale unit of velocity in the SI system is equivalent to I scale unit in the American system. Another way of changing units when secondary dimensions are present is to make use of the formalism illustrated in relations 1.1. To change a unit in an expression, multiply this unit by a ratio physically equivalent to unity, as we discussed earlier, so that the old unit is canceled out, leaving the desired unit with the proper numerical coefficient. In the example of velocity used above, we may replace ft/sec by mlsec in the following manner:
It should he clear that, when we multiply by such ratios to accomplish a change of units as shown above, we do not alter the magnitude of the actual physical quantity represented by the expression. Students are strongly urged to employ the above technique in their work, for the use of less formal methods is generally an invitation to error.
t1.4
l a w of Dimensional Homogeneity
Now that we can describe certain aspects of nature in a quantitative manner through basic and secondary dimensions, we can by careful observation and experimentation leam to relate certain of the quantities in the form of equations. In this regard, there is an important law, the law of dimen.siona1 homogeneity, which imposes a restriction on the formulation of such equations. This law states that. because natural phenomena proceed with no regard for manmade units, basic equations representing physical phenomena must be valid f o r all systems of units. Thus, the equation for the period of a pendulum, 7 t = 2 x , / ~ / g , must be valid for all systems of units, and is accordingly said to be dimensionally homogeneous. It then follows that the fundamental equations of physics are dimensionally homogeneous; and all equations derived analytically from these fundamental laws must also be dimensionally homogeneous. What restriction does this condition place on an equation? To answer this, let us examine the following arbitrary equation: x=ygd+k
For this equation to be dimensionally homogeneous, the numerical equality between both sides of the equation must he maintained for all systems of units. To accomplish this, the change in the scale of measure of each group of terms must be the same when there is a change of units. That is, if the numerical measure of one group such as ygd is doubled for a new system 0 1 units, so must that of the quantities x and k. For 1hi.r to occur under all systems of units, it is necessary that everj grouping in the eyuution have the .same dimensirmal representation. In this regard, consider the dimensional representation of the above equation expressed in the following manner:
SECTION 1.5 DIMENSIONAL. RELATION BETWEEN FORCE AND MASS [XI
= b g 4 + [kl
From the previous conclusion for dimensional homogeneity, we require that [XI
= [yg4 = [kl
As a further illustration, consider the dimensional representation of an equation that is not dimensionally homogeneous:
[LI = [fl’ + [rl
When we change units from the American to the SI system, the units of feet give way to units of meters, but there is no change in the unit of time, and it becomes clear that the numerical value of the left side of the equation changes while that of the right side does not. The equation, then, becomes invalid in the new system of units and hence is not derived from the basic laws of physics. Throughout this book, we shall invariably be concerned with dimensionally homogeneous equations. Therefore, we should dimensionally analyze our equations to help spot errors.
Dimensional Relation Between Force and Mass
t1.5
We shall now employ the law of dimensional homogeneity to establish a new secondary dimension-namely force. A superficial use of Newton’s law will be employed for this purpose. In a later section, this law will be presented in greater detail, but it will suffice at this time to state that the acceleration of a particle3 is inversely proportional to its mass for a given disturbance. Mathematically, this becomes a = -1 (1.2) m where is the proportionality symbol. Inserting the constant of proportionality, F, we have, on rearranging the equation,
-
F=ma (1.3) The mechanical disturbance, represented by F and calledforce, must have the following dimensional representation, according to the law of dimensional homogeneity: [ F ] = [ M I -[Ll
[fIZ
(1.4)
The type of disturbance for which relation 1.2 is valid is usually the action of one body on another by direct contact. However, other actions, such as magnetic, electrostatic, and gravitational actions of one body on another involving no contact, also create mechanical effects that are valid in Newton’s equation. ‘We shall define panicles in Section 1.7.
9
10
CHAPTER I
FLNDAMENTALS OF MECHANICS
We could have initiated the study of mechanics by consideringfiirce as a basic dimension, the manifestation of which can he measured by the elongation of a standard spring at a prescribed temperature. Experiment would then indicate that for a given body the acceleration is directly proportional to the applied force. Mathematically, F
m
a; therefore, F = mu
from which we see that the proportionality constant now represents the property of mass. Here, mass is now a secondary quantity whose dimensional representation is determined from Newton's law:
As was mentioned earlier, we now have a choice between two systems
of basic dimensions-the MLt or the FLr system of basic dimensions. Physicists prefer the former, whereas engineers usually prefer the latter.
1.6
Units of Mass
As we have already seen, the concept of mass arose from two types of actions -those of motion and gravitational attraction. In American engineering practice, units of mass are based on hoth actions, and this sometimes leads to confusion. Let us consider the FLt system of basic dimensions tor the following discussion. The unit of force may he taken to be the pound-force (Ihf), which is defined as a force that extends a standard spring a certain distance at a given temperature. Using Newton's law, we then define the slug as the amount of mass that a I-pound force will cause to accelerate at the rate of I foot per second per second. On the other hand, another unit of mass can he stipulated if we use the gravitational effect as a criterion. Herc. the pound muxs (Ihm) is defined as the amount of matter that is drawn by gravity toward the earth by a force of I pound-force (Ihf) at a specified position on the earth's surface. We have formulated two units of mass by two different actions, and to relate these units we must subject them to the sumt. action. Thus, we can take 1 pound mass and see what fraction or multiplc of it will be accelerated 1 ft/sec2 under the action of I pound afforce. This fraction or multiple will then represent the number of units of pound mass that are equivalent to I slug. It turns out that this coefficient is go, where g, has the value corresponding to the acceleration of gravity at a position on the earth's surface where the pound mass was standardized. To three significant figures, the value of R~ is 32.2. We may then make the statement of equivalence that I slug
= 32.2 pounds mass
SECTION 1.6 UNITS OF MASS
To use the pound-mass unit in Newton’s law, it is necessary to divide by go to form units of mass, that have been derived from Newton’s law. Thus,
where m has the units of pound mass and &go has units of slugs. Having properly introduced into Newton’s law the pound-mass unit from the viewpoint of physical equivalence, let us now consider the dimensional homogeneity of the resulting equation. The right side of &. 1.6 must have the dimensional representation of F and, since the unit here for F is the pound force, the right side must then have this unit. Examination of the units on the right side of the equation then indicates that the units of go must be (1.7) How does weight tit into this picture? Weight is defined as the force of gravity on a body. Its value will depend on the position of the body relative to the earth‘s surface. At a location on the earth’s surface where the pound mass is standardized, a mass of 1 pound (Ibm) has the weight of 1 pound (Ibf), but with increasing altitude the weight will become smaller than 1 pound (Ibf). The mass, however, remains at all times a I-pound mass (Ibm). If the altitude is not exceedingly large, the measure of weight, in Ibf, will practically equal the measure of mass, in Ibm. Therefore, it is unfortunately the practice in engineering to think erroneously of weight at positions other than on the earth‘s surface as the measure of mass, and consequently to use the symbol W to represent either Ibm or Ibf. In this age of rockets and missiles, it behooves us to be careful about the proper usage of units of mass and weight throughout the entire text. If we know the weight of a body at some point, we can determine its mass in slugs very easily, provided that we know the acceleration of gravity, g, at that point. Thus, according to Newton’s law, W(lbf) = m(s1ugs) x g(ft/sec*) Therefore, (1 3 )
Up to this point, we have only considered the American system of units. In the SI system of units, a kilogram is the amount of mass that will accelerate 1 m/sec2 under the action of a force of 1 newton. Here we do not have the problem of 2 units of mass; the kilogram is the basic unit of mass. However, we do have another kind of problem-that the kilogram is unfortunately also used as a measure of force, as is the newton. One kilogram of force is the weight of 1 kilogram of mass at the earth‘s surface, where the acceleration of gravity (Le., the acceleration due to the force of gravity) is
11
12
CHAPTER I
FUNDAMENTALS OF MECHANICS
9.81 m/sec2. A newton, on the other hand, is the force that causes I kilogram of mass to have an acceleration of 1 m/sec2. Hence, Y.8 1 newtons are equivalent to I kilogram of force. That is, 9.81 newtons
1 kilogram(force)
= 2.205 Ibf
Note from the above that the newton is a comparatively small force, equaling approximately one-fifth of a pound. A kilonewton (1000 newtons), which will be used often, is about 200 Ib. In this text, we shall nor use the kilogram as a unit of force. However, you should he aware that many people do." Note that at the earth's surface the weight W o1a mass M is:
(1.9)
W(newtons) = [M(kilograms)](Y.81)(m/s2) Hence: M(kilograms) =
W(newtons) _ _ _ 9.81 (rnls')
_
~
~
~
Away from the earth's surfxe, use the acceleration of gravity 9.81 in the above equations.
1.7
~
x
(1.10)
rather than
Idealizations of Mechanics
As we have pointed out, basic and secondary dimensions may sometimes be related in equations to represent a physical action that we are interested in. We want to represent an action using the known laws of physics, and also to be able to form equations simple enough to he susceptible to mathematical computational techniques. Invariably in our deliberations, we must replace the actual physical action and the participating bodies with hypothetical, highly simplified substitutes. We must he sure, of course, that the results of our substitutions have some reasonable correlation with reality. All analytical physical sciences must resort to this technique, and. consequently, their coniputations are not cut-and-dried but involve a considerable amount of imagination, ingenuity, and insight into physical behavior. We shall, at this time, set forth the most fundamental idealizations of mechanics and a hit of the philosophy involved in scientific analysis.
Continuum. Even the simpliI"ica1ion of matter into molecules, atoms, electrons, and so on, is too complex a picture for many problems of engineering mechanics. In most problems, we are interested only in the average measurable manifestations of these elementary bodies. Pressure, density, and temperature are actually the gross effects of the actions of the many molecules and atoms, and they can be conveniently assumed to arise from a hypothetically continuous distribution of matter, which we shall call the continuum, instead of from a conglomeration of discrete, tiny hodies. Without such an 'This is particularly true in the marketplace where the word "kilos" is often heard
SECTION 1.7 IDEALIZATIONS OF MECHANICS
artifice, we would have to consider the action of each of these elementary bodies-a virtual impossibility for most problems. Rigid Body. In many cases involving the action on a body by a force, we simplify the continuum concept even further. The most elemental case is that of a rigid body, which is a continuum that undergoes theoretically no deformation whatever. Actually, every body must deform to a certain degree under the actions of forces, hut in many cases the deformation is ton small to affect the desired analysis. It is then preferable to consider the body as rigid, and proceed with simplified computations. For example, assume that we are to determine the forces transmitted by a beam to the earth as the result of a load P (Fig. 1.2). If P is small enough, the beam will undergo little deflection, and we can carry out a straightforward simple analysis using the undefomed geometry as if the body were indeed rigid. If we were to attempt a more accurate analysis-even though a slight increase in accuracy is not required-we would then need to know the exact position that the load assumes relative to the beam afrer the beam has ceased to deform, as shown in an exaggerated manner in Fig. 1.3. To do this accurately is a hopelessly difficult task, especially when we consider that the support must also “give” in a certain way. Although the alternative to a rigid-body analysis here leads us to a virtually impossible calculation, situations do arise in which more realistic models must be employed to yield the required accuracy. For example, when determining the internal force distribution in a body, we must often take the deformation into account, however small it might be. Other cases will be presented later. The guiding principle is to make such simplifications as are consistent with the required accuracy ojthe results.
Point Force. A finite force exerted on one body by another must cause a finite amount of local deformation, and always creates a finite area of contact between the bodies through which the force is transmitted. However, since we have formulated the concept of the figid body, we should also be able to imagine a finite force to be transmitted through an infinitesimal area or point. This simplification of a force distribution is called a point force. In many cases where the actual area of contact io a problem is very small but is not known exactly, the use of the concept of the point force results in little sacrifice in accuracy. In Figs. 1.2 and 1.3, we actually employed the graphical representation of the point force. Particle. The particle is defined as an object that has no size but that has a mass. Perhaps this does not sound like a very helpful definition for engineers to employ, but it is actually one of the most useful in mechanics. For the trajectory of a planet, for example, it is the mass of the planet and not its size that is significant. Hence, we can consider planets as particles for such computations. On the other hand, take a figure skater spinning on the ice. Her revolutions are controlled beautifully by the orientation of the body. In this motion, the size and distribution of the body are significant, and since a
Figure 1.2. Rigid-body assumption-use original geometry.
Figure 1.3. Deformable body.
13
14
CHAPTER 1 FUNDAMENTALS OF MECHANICS
particle, by definition. can have no distribution. i t i s patently clear that a particle cannot represent the skater in this case. If; however, the skater should he hilled as the “human cannonball on skates” and he shot out of a large air gun. i t would be possible to consider her as a single particle i n ascertaining her Lrajectory, since arm and leg movements that werc significant while she was spinning on the ice would have l i t t l e effect o n the arc traversed by the main portion of her body. You will learn later that the wiitri- ofnirrss 01-muss w i i f r r i s a hypiithelical point at which one can concentrate thc mass ot the body for ccrliiin dynamics calculations. Actually i n the previous cxamplcs of thc planet and the “human cannonball on skates,” the particle wc reler to i s actually the mass center whose motion i s sufficient for the desired infiirmation. Thus, when the motion of the mass center o f a body suffices for thc information desired, we can replace the body by a particle. n m e l y the mass center. Many other simplifications pervade mechanics. The perfectly elastic body, the frictionless fluid, and so on. will become familiar as you study various phases nf mechanics.
11.8
Vector and Scalar Quantities
We have now proposed sets of basic dimensions and secondary dimensions to describe certain aspects uf nature. However. more than just the dimensional identification and the number (if units arc often necdcd to convey adequately the desired information. For instance, to specify fully the motion o f a car, which we may represent as ii particle at this tiine. we must answer the lollowing questions:
1. How fast? 2. Which way? The concept o f velocity entails the information desired in questions I and 2. The first question, “How fast?”, i s answered hy the speednmeter reading, which gives the value o f the velncity in miles per hour or kilometers per hour. The second question, “Which w a y ~ y i,s more complicated. hecause two separete factors arc involved. First, we must specify the angular orientation of the velocity relative to a reference Srame. Second, we [nust speciSy the sense (if the velocity, which tells us whether we are moving r o w i r d o r uw’ay.from ii given point. The concepts o f angular orientation of the velocity and sense o f the velocity are often collectively denoted as the dire&irr of the velocity. Graphically, we may use a directed linr, .rrgmr’nt (an arrow) to describe the velocity of the car. The length o f the directed line segment gives information as to “how fast” and i s the magnitude o f the velocity. The angular orientation of the directed line segment and the position of’the arrowhead give inturmation as to “which way”-that is, as tii the direction o f the velocity. The
SECTION 1.8 VECTOR AND SCALAR QUANTlTlES
15
directed line segment itself is called the velocity, whereas the length of the directed line segment-that is, the magnitude-is called the speed. There are many physical quantities that are represented by a directed line segment and thus are describable by specifying a magnitude and a direction. The most common example is force, where the magnitude is a measure of the intensity of the force and the direction is evident from how the force is applied. Another example is the displacement vecior between two points on the path of a particle. The magnitude of the displacement vector corresponds to the distance moved along a straighr line between two points, and the direction is defined by the orientation of this line relative to a reference, with the sense corresponding to which point is being approached. Thus, pae (see Fig. 1.4) is the displacement vector from A to B (while p,, goes from B to A). 7
Path of a particle
*-(. \-
I
1.4
\
Figure 1.4. Displacement vector pAB.
Certain quantities having magnitude and direction combine their effects in a special way. Thus, the combined effect of two forces acting on a particle, as shown in Fig. 1.5, corresponds to a single force that may be shown by experiment to be equal to the diagonal of a parallelogram formed by the graphical representation of the forces. That is, the quantities add according to the parallelogram law. All quantities that have magnitude and direction and that add according to the parallelogram law are called vector quaniities. Other quantities that have only magnitude, such as temperature and work, are called scalar quantities. A vector quantity will be denoted with a boldface italic letter, which in the case of force becomes F.5 The reader may ask Don’t all quantities having magnitude and direction combine according to the parallelogram law and, therefore, become
.iYour inslmclm on the blackboard and you in your homework will not be able lo use bld; face notation lor vcctors. Accordingly, you may choose IO use a superscript arrow or bar, e.&. F or F (E or E are other possibilities).
e F, + F2
Figure 1.5. Parallelogram law.
16
CHAPTER I
FUNDAMENTALS OF MECHANICS
vector quantities? No, not all of them do. One very important example will he pointed out after we reconsider Fig. 1.5. In the construction of the parallelogram it matters not which force is laid out first. In other words, “ F , combined with F,” gives the same result as “F, combined with F,.” In short, the combination is commutative. If a combination is not commutative, it cannot in general he represented by a parallelogram operation and is thus not a vector. With this in mind, consider the finite angle of rotation of a body about an axis. We can associate a magnitude (degrees or radians) and a direction (the axis and a stipulation of clockwise or counterclockwise) with this quantity. However, the finite angle of rotation cannot he considered a vector because in general two finite rotations about different axes cannot he replaced by a single
~
I
90”
Figure 1.6. Successive rotations are not
commutative.
SECTION I .Y
finite rotation consistent with the parallelogram law. The easiest way to show this is to demonstrate that the combination of such rotations is not commutative. In Fig. I .6(a) a book is to he given two rotations-a 90" counterclockwise rotation about the x axis and a 90" clockwise rotation about the i axis, both looking in toward the origin. This is carried out in Figs. 1.6(b) and (c). In Fig. 1.6(c), the sequence of combination is reversed from that in Fig. 1.6(b), and you can see how it alters the final orientation of the hook. Finite angular rotation, therefore, is not a vector quantity, since the parallelogram law is not valid for such a ~ o m b i n a t i o n . ~ You may now wonder why we tacked on the parallelogram law for the definition o f a vector and thereby excluded finite rotations from this category. The answer to this query is as follows. In the next chapter, we will present cemin sets of very useful operations termed w c t u r algebra. These operations are valid in general only if the parallelogram law is satisfied as you will see when we get to Chapter 2. Therefore, we had to restrict the definition of a vector in order to he able to use this kind of algebra for these quantities. Also, it is to he pointed out that later in the text we will present yet a third definition consistent with our latest definition. This next definition will have certain advantages as we will see later. Before closing the section, we will set forth one more definition. The /ine (,faction of a vector is a hypothetical infinite straight line collinear with the vector (see Fig. 1.7). Thus, the velocities of two cars moving on different lanes of a straight highway have different lines of action. Keep in mind that the line of action involves no connotation as to sense. Thus, a vector V' cnllinear with V in Fig. 1.7 and with opposite sense would nevertheless have the same line of action.
1.9
17
EQUALITY AND EQUIVALENCE OF VECTORS
+==
/
-m
Figure 1.7. Line of action of B vector.
Equality and Equivalence of Vectors
We shall avoid many pitfalls in the study of mechanics if we clearly make a distinction between the equality and the equivalence of vectors. Twjo L'ecfors are equal if they have the .same dimcmsions, rnugnirudc,, and direction. In Fig. 1.8, the velocity vectors of three particles have equal length, are identically inclined toward the reference xyr, and havc the samc sense. Although they have different lines of actinn, they are nevertheless equal according to the definition. Two vectorr are equivalent in a certain c a p a c i y if each prodnces the vev ,same e f t k t in this capacity. If the criterion in Fig. I .8is change of elevation of the particles or total distance traveled by the particles, all three vectors give the same result. They are, in addition to being equal, alsu "However. wriishingly rniull rotations can be considered a i YCE~UIS since thc commutative law applies for the combiniltiun of such rotations. A proof of this assertion is presenlcd in Appcndin IV. The tbct that infinitesimal rotations are vectors i n accordance with our definition w i l l be an irnpoltant consideration when we discuss angular velocity in Chapter 15.
Figure 1.8. Equal-velocity vectors.
18
CHAPTER I
~ U N U A M L N T A L Sor MECHANICS equivalent fur thcsc capacities. I f the absolute height u l the parlicles above the .cy plane i s the quesliiin i n piiint, these vectors w i l l no1 he equivalenl despite their equality. Thus, i t must be cinphasizcd that cquol 1wtor.s need t i u f uIwri?,s bP ryuivulent: i f deprid.s cririrelv oii / h e situuriori ut hund. Furthermore, vectors that itre n o t equal may s t i l l hc cquivalcnt i n soine capacity. l'hus, i n the beam i n Fig. 1.9, forces F , and are unequal, since their magnitudes are IO Ih and 20 Ih, respectively. However, it i s clear from elementary physics cliat their mnments ahiiul the hase 01 the heani are equal, and su the forces liave the same "turning" action at the fixed end of the heain. I n that capacity. the forces are equivalenl. If. however, we are interested i n the dellcction of the free end of the heam resulting from each force, there i s no longer ;in equivalcncc hclwcen thc force.;, since each w i l l give a different dcllcctiun.
C;
,
I_
,]'
~~~~~
Figure 1.9. F , and I.?equivalent Tor iiioriient ahw1 A.
To sum up. the ryrcnli~ynf two vecturs i s determined by the vectors themselves. and thc equivuleurp hctwecn two vccturs i s dctcrniined hy the task involving the vectors. In probleins o f mech;mics. we can prufitehly delineate three classes of situatiuns cunccrning equivalence of veckirs: 1. 5'irirution.s it[ M h i d ~vw/o,-.v miry he p . s i r i o w d unywherr in .spuce wirlwur 1o.u or (.huuKr r,/meuriinp providrd thuc mu,ynilurlr und dir(,..
SECTION 8.7
COMPUTATION INVOLVING SECOND MOMENTS AND PRODUCT OF AREA
Example 8.9 Find the centroid of the area of the unequal-leg Z section shown in Fig. 8.29. Next, determine the second moment of area about the centroidal axes parallel to the sides of the Z section. Finally, determine the product of area for the aforementioned centroidal axes.
Figure 8.29. Unequal-leg Z section.
We shall subdivide the Z section into three rectangular areas, as shown in Fig. 8.30. Also, we shall insert a convenient reference xy,as shown in the diagram. To find the centroid, we proceed in the following manner: A, (in.?) 1. (2)(1) = 2 2. (8)(1) = 8 3. (4)(1) =A
EA, = I4
?, (in.)
I 2.50 S
V,. (in.)
A,rj(in.')
7.50 4
2 20
.so
20 A,,?, = 42
Figure 8.30. Composite area.
A;?; (in.') 15
32 2 A,Y, = 49
361
362
CHAPTER x
PRO PERTIE S OF SURFACES
Example 8.9 (Continued) Therefore,
We h w e shown the centroidal axes .xc);. in Fig. 8.31. We now find lxcxc and ly,vt, using the parallel-axis theorem and the lormulas &hh’ and &hb3 for the second moments of area about centroidal axes of symmetry of a rectangle.
+[(&1(4)(13)+ (4)(3*)] =
4
0 I?? . = [(&)(1)(23 + ( 2 ) ( 2 2 ) ]+ [(/2)(8)(13) + (8)(fP]
0
0 +[(&)(I)(49 + (4)(22)] = i
Figure 8.31. Centroidal axe,
32.
<
a
Finally, we consider the product of area Cc? : ,
1%
, Figure 9.4. ?: i\ plane 0 1 cymmrlry.
each half w i l l give ii crintrihuti(in ofthe same magnitude hut ol'oppositc sign. We can most readily hcc that this i s so by loohing along the y axis tiiwai-d thr origin. The plane of symmetry then appears as a line coinciding with thc z i i x i c (ECC Fig. 9.5). We can cnnsidcr the body to he composed of pairs of inass elements diii which are mirror imagcs of each other with respect to pnsition and shape ahout the planc of symmetry. The product (if inertia Sor such ii pair i s then I:
dm
Ifor h e thin homogcneous hrxrp of mass M.
B
Figure P.9.6. Figure P.9.0.
compute I r < , I.,, I:?, and I , , for the ~lomoyrneous lar parallelepiped. 9.4.
9.7. Find 1 . ~and l l vlor the homueencous right circular cylinder of maw M. r
i
Figure P.Y.7. Figure P.9.4. 3 . 5 A wire having the shape of a parabola is shown. The curve s in the yz plane. If the mass of the wire is .3 N/m,~whatare I,,, ind lrz'? [Hinr: Replace dr along ihe wire by l ( d y 1 & ) 2 + I dz. 1
9.8. For the cylindzr in Prohlem Y.7, the density increases lincarly in the 2 direction from a value o f . 100 gramsimm' at the left end to a value of ,180 pramsimm' at the right end. Take I = 30 and I.. . mm and I = IS0 mm. Find
9.9.
+
Show that I, for the homogeneous right circular cone is
Y
I
MR~.
Figure P.9.9.
Figure P.9.13.
9.10. In Problem 9.9, the density increases as the square of z in the z direction from a value of ,200 gramslmm’ at the left end to a value of ,400 grams/mm3 at the right end. If r = 20 mm and the cone is 100 mm in length, find lz,. 9.11. A body of revolution is shown. The radial distance r of the boundary from the x axis is given as r = Z.?. m. What is I, for a uniform density of 1,600kg/m3?
9.14. Find the second moment of area about the x axis for the front surface of a very thin plate. If the weight of the plate is .02 N/mm2, find the mass moments of inertia about the x and y axes. What is the mass product of inertia I=?? x
y
I
x
Figure P.9.14.
Figure P.9.11. 9.12. A thick hemispherical shell is shown with an inside radius of 40 mm and an outside radius of 60 mm. If the density p i s 7,000 kglm,’ what is
*9.15. A uniform tetrahedron is shown having sides of length a, b, and c, respectively, and a mass M. Show that lyz = &Mac. (Suggestion: Let z mn from zero to surface ABC. Let x run from zero to line AB. Finally, Let y run from zero to B. Note that the equation of a plane surface is z = a* + py + 1: where a,p, and y are constants. The mass of the tetrahedron is pabc16. It will he simplest in expanding ( I - xlb - y/c12 to proceed in the form [(I - ylc) - (x/b)I2, keeping (1 - ylc) intact. In the last integration replace y by [ - c(1 ylc) + cl, etc.) ~
Figure P.9.12.
9.13. Find the mass moment of inertia I, for a very thin plate forming a quarter-sector of a circle. The plate weighs .4N. What is the second moment of area about the x axis? What is the product of inertia? Axes are in the midplane of the plate.
Figure P.9.15.
391
9.18. Can you identify hy inspection any o i i h r principii1 iixe\ 01 inertia iit A'! At R' Explain The dcnsitj oithu matcriiil i\ oniil-ni.
392
Figure P.9.19.
SECTION 9.4 TRANSLATION OF COORDINATE AXES
393
Note that the quantities bearing the subscript c are constant for the integration and can be extracted from under the integral sign. Thus,
where p dz: has been replaced in some terms by dm, and the integration
JIS p V
in the first integral has been evaluated as M, the total mass of the body. The origin of the primed reference being at the center of mass requires of the first moments of mass that I j j x ’ d m = I I J y ’ d m = / / j z ’ d m = 0. The middle two terms accordingly drop out of the expression above, and we recognize the last expression to be Jz,;,. Thus, the desired relation i s I.. ‘. = I,. .,.,
+
M(x:
+ y):
(9.7)
By observing the body in Fig. 9.12 along the z and z’ axes (Le., from directly above), we get a view as is shown in Fig 9.13. From this diagram, we can see that y: + x,Z = d.’ where d is the perpendicular distance between the z‘ axis through the center of mass and the z axis about which we are taking moments of inertia. We may then give the result above as
Izz =
i Md2
Let us generalize from the previous statement.
The momea of inenia of a body any a i s the inenia of the body about a parallel a i s that goes tkough the center of mass, plus the total mass times the perpendicular db a e s squared. We leave it to you to show that for products o f inertia a similar relation can he reached. For I t +for example, we have
Here, we must take care to put in the proper signs of xc and y, as measured .from the xyz reference. Equations 9.8 and 9.9 comprise the well-known parallel-uxir lhc~oremsanalogous to those formed in Chapter 8 for areas. You can use them lo advantage for bodies composed of simple familiar shapes, as we
now illustrate.
Y’
Figure 9.13. View along L direction (from above).
394
CHAPTER 0 MOMENTS AND PROOLJCTS OF INERTIA
Example 9.4 Find I,, and I,, for the body shown in Fig. 9.14. Take pas constant for the body. Use the formulations for moments and products of inertia at the center of mass as given on the inside front cover page. We shall consider first a solid rectangular prism having the outer dinlensions given in Fig. 9.14, and we shall then subtract the contribution of the cylinder and the rectangular block that have been cut away. Thus, we have, for the ovcriill rectangular block which we consider as body I,
From this, we shall take away the contribution of the cylinder, which we denote as body 2. Using the formulas from the inside front cover page, (Ixr)2 =
M(3rZ + h z ) + Md’
&[pn(1)2(1S)][3(l’) = 5,243~ =
+ IS2] + [ p ~ ( l ) ~ ( l S j ] [ 6+’ 7.5’1
(b)
Also, we shall take away the contribution of the rectangular cutout (bod) 3): ( l x x )= 3 & M ( u 2 + h 2 ) + Md2 = &[P(8)(6)(4)](4*
+ 6 2 ) + [P(8)(6)(4)](2’ + 3’)
(c)
= 3,328~
The quantity lrAfor the body with the rectangular and cylindrical cavities is then l a x= (231,200 - 5,243 3,328)~ ~
I,,
=
2
(d)
We follow the same procedure to obtain lxy.Thus, for the block as a whole, we have (IkV), = (I,,,, + Mx,yc At the center of mass of the block, both the (x’), and ( y ’ ) ] axes are normal to planes (if symmetry. Accordingly. (I,,),. = 0. Hence,
C~),
= 0 + [P(20)(8)(1S)1(-4~(-IO) = 96,nnop
(e)
For the cylinder, we note that both the ( x ‘ ) and ~ (.y’),- axes at the center of mass are normal to planes of symmetry. Hence, we can say that (I,>), = 0 + 1p(~)(I2)(15)1(-8)(~6) =
2,262~
(f)
IS’
Figure 9.14. Find lu and ilv.
SECTION 9.5 TRANSFORMATION PROPERTIES OF THE INERTIA TERMS
Example 9.4 (Continued) Finally, for the small cutout rectangular parallelepiped, we note that the ( x ' ) ~and cy'), axes at the center of mass are perpendicular to planes of symmetry Hence, we have
(I& The quantity is then
I
/Iy
= 0 + [~(8)(6)(4)1(-2)(-16) = 6,144~
(g)
for the body with the rectangular and cylindrical cavities
If p is given in units of lbdft,) the inertia terms have units of Ibm-ft.z
"9.5
Transformation Properties of the Inertia Terms
Let us assume that the six independent inertia terms are known at the origin of a given reference. What is the mass moment of inertia for an axis going through the origin of the reference and having the direction cosines 1, m, and n relative to the axes of this reference? The axis about which we are interested in obtaining the mass moment of inertia is designated as kk in Fig. 9.15.
Y
x
Figure 9.15. Find lhh
From previous conclusions, we can say that (9.10)
395
396
CHAPTER 9 MOMENTS AND PRODUCTS OF INERTIA
where @ is the angle between kk and r . W e shall now put sin2 @ i n t o a more uscful form by considering the right triangle Sormed by the position vector r and the axis kk. This triangle is shown enlarged in Fig. 9.16. The side a of the triangle has a magnitude that can be given hy the dol pniduct of r and the unit vector E, along kk. Thus. 0 =
r
-
Ek
= (.xi
+ ~f +
zkj
(/i
+
mj
+
nk)
(9. I 1)
Figure 9.16. Right triangle Sormcd by rand kk
Hence 11
= /.r
+
niy
+
IIZ
Using the Pythagorean theorem, we can now givc side b as h? =
lr12
-
02
= (.r2 + y2 + .?1 -
(/Y + m2j2 +
Thc lerm sin2 $may next be givcn
ii2?
+
?hry
+
2lrzxz
+
2mnyz)
iis
Substituting hack into Eq. 9.10. wc get, on canceling kernis.
-(Pr’
+ m’j’
+iiZ;’
+ 2 / n ~ r ?+ 2/nrz + ~ m n y z j l p d i ,
Since I’ + in2 + ti2 = I , we can multiply the tirst bracketed exprcssion in thc integral by this sum:
SECTION 9.5 TRANSFORMATION PROPERTIES OF THE INERTIA TERMS
397
Carrying out the multiplication and collecting terms, we get the relation
Refemng back to the definitions presented by Eqs. 9.1, we reach the desired transformation equation:
We next put this in a more useful form of the kind you will see in later courses in mechanics. Note first that 1 is the direction cosine between the k axis and the x axis. It is common practice to identify this cosine as ab instead of 1. Note that the subscripts identify the axes involved. Similarly, m = ub and n = akz.We can now express Eq.9.13 in a form similar to a matrix etc. array as follows on noting that Izy =
6,
This format is easily written by first writing the matrix m a y of I‘s on the right side and then inserting the a’s remembering to insert minus signs for off-diagonal terms. Let us next compute the product of inertia for a pair of mutually perpendicular axes, Ok and Oq, as shown in Fig. 9.17. The direction cosines of Ok we shall take as I, m, and n, whereas the direction cosines of Oq we shall take as ,‘l m’, and n’. Since the axes are at right angles to each other, we know that Ek.€
Y
k
=o Y
Therefore,
II’
+ mm’ +
nn’ =
0
(9.15)
Noting that the coordinates of the mass element p dv along the axes Ok and Oq are r * ek and r * eq,respectively, we have, for Ikq:
Using xyz components of r and the unit vectors, we have =
jjj[(i+ yj + ~ V
k * )(1i + n ~ +‘ A)]
x
4
398
CHAPTER 9
MOMENTS . \ i W PKODIK'TS OF I N L K I I A
Hcnce.
+ y n l ' + yrmn' + wi/' + -\.nni')p
+xdin' t .x:lii'
(11'
Collecting k i l n s and bringiiig ~ I i ciiirrctii)n cosines outside the inlegratiolis. wc gel
Noting lhc definitions i n Eil. 9 I . we
li,, =
~
iiitlir/~r
~
,illr/ ~
+
+ (Id
ciiii
statc Ihc desired Lransforinalion:
+
(/n1' In/')/,, d ! I ,+ ~
+
(111li'
+
Ii?d!/),
(9.19)
We can now rewrilc the p r e \ i w s cquarion in ii more u\cIuI mil simple torin using U ' Y iis direction cosine\. Thus. noting !hat I' = (I,,).ctc.. we proceed as in Eq. [I. 14 to uhtain
SECTION 9.5 TRANSFORMATION PROPER'IlES OF THE INERTIA TERMS
399
EXaI'rIple 9.5 Find Iz,7, and I,,,, for the solid cylinder shown in Fig. 9.18. The reference x'y'z' is found by rotating ahout they axis an amount 30", as shown in the diagram. The mass of the cylinder is 100 kg. It is simplest to first get the inertia tensor components for reference xyz. Thus, using formulas from the inside front cover page we have
z
2
Ja = -IM r 2 = L(lOO)(!$) 2 2
=
21.13 kg-mz
-Ew =
85.56 kg-m2
30"
Noting that the xyz coordinate planes are planes of symmetry, we can conclude that l x z = IY l = I YZ
=o
Next, evaluate the direction cosines of the z' and the x' axes relative to nyz. Thus, For z' axis: 60" = ,500 cos 90" = 0 Z'Y ai.? = cos 30" = ,866
a:,= = a
COS
=
For .c' axis: aXlr = COS
30" = ,866
a*Y = cos90° = 0 a~c,z = cos 120" = -SO0
First, we employ Eq. 9.14 to get J,,,,. = (85.56)(.500)z
+ (21.13)(.866)'
= '37.h
Finally, we employ Eq. 9.20 to get I,;,. -Ix,,, = (85.56)(.500)(.866) + (21.13)(.866)(-,500)
Therefore,
x'
b l . 3 m ~ +
Figure 9.18. Find lz.,,and ldz,
400
CHAPTER 9
MOMWTS AND PROIlI!('TS OF INEXI'IA
SECTION 9.6 LOOKING AHEAD: TENSORS
401
You will leam that because of the common transformation law identifying certain quantities as tensors, there will be extremely important common chardcteristics for these quantities which set them a p a t from other quantities. Thus, in order to leam these common characteristics in an efficient way and to understand them better, we become involved with tensors as an entity in the engineering sciences, physics. and applied mathematics. You will soon he confronted with the stress and strain tensors in your courses in strength of materials. To explore this point further, we have shown an infinitesimal rectangular parallelepiped extracted from a solid under load. On three orthogonal faces we have shown nine force intensities (Le., forces per unit area). Those with repeated indices are called normal stwsses while those with different pairs of indices are called shear stresses. You will leam, that knowing nine such stresses, you can readily find three stresses, one normal and two onhogonal shear stresses, on any interface at any orientation inside the rectangular parallelepiped. To find such stresses on an interface knowing the stresses shown in Fig. 9.19, we have the .same fran.~jbnnationeyuutions given by Eqs. 9.21 and 9.22. Thus stress is a second-ordijr tensor.
/
/
-.x
r1.r
Figure 9.19. Nine stresses o n three orthogonal interfxes at a point. A two-ilimensionalsimplification of r!, involving the quantities T ~ T,,~., , and T~~ (= T,J as the only nonzero stresses is called plune .stress. This occurs in a thin plate loaded in the plane of symmetry as shown in Fig. 9.20. Plane stress is the direct analog of second moments and products ofurea, which is a two-dimensional simplification of the inertia tensor. Clearly, plane stress and second inoments and products of area have the same transformation 5,. 5,. T~;.replacing equations, which arc Eqs. 8.26 through 8.28 with I h, (dZVld8Z), = is positive, and so this is the desired requirement for stable equilibrium. Thus for stable equilibrium, R > h.
R-(R-h)
Figure 10.21. Plate with circular
bottom edge.
10.9
Figure 10.22. One degree of
freedom.
Looking Ahead: More on Total Potential Energy
When we have conservative forces acting on pruticles and rigid bodies, we found earlier that for establishing necessary and sufficient conditions of equilibrium we could extremize the potential energy Vassociated with the forces. That is, we could set 6V = 0 to satisfy equilibrium. Recall that this result was derived from the method of virtual work. We have a similar formulation for the case of an elastic (not necessarily linearly elastic) body wbeEby we can guarantee equilibrium. This more-
COS
0
Figure 10.23. Position of C.G,
443
444
CHAPTER I O
METHODS OF VIRTUAL WORK AND STATIONARY POTEiYTI,\L liNFR(;Y
general principle is derivable from the more-gencral virtual work principle mentioned earlier i n section lO.5. Hcre. we 11iu\t extrenii7.e an exprcssinn inore complicated (as you might expect) than V . This expi-ession is denotcd as nwith no relation to the number 3.1416.. . . The expression n is what we CJII a.fim.tiond, whercin for the substitution of each functim such as v(~r),into the functional a nunihcr is established. A simple example of ii functional I is as follows:
where F is a function o f r (the so-called independent \iiriahIe), y, and clyldr. Subtilutinn of. a function y(.C into F fiillnwed hy iiii integration hetween the fixed limils. yields for this funclion y(.i) :I number h r 1. Fonctionals per\,adr the field of mechanicc and most other analytic Ciclds o l knowledge. A vital step is to find the function ~ ( n that ) w i l l cxtremix I . This function then becomes knnwn aq the urrrrriral/un,.fion.'The ciiIcuIus Ir,r doing this is called the w l d u s 0 1 variations. Thc particular functional for the method of tntill potential cncrgy is h' riven as
n
=
-111B
* u 0'1. -
Ii
8
7
u d.4
+ L'
.\
The function to he adjusted to cxtremize ii is u(.x,J, z) taking the placc of ~ ( ~ 1 ) in the preceding functional. and iinw the independcnt \,;iriablcs are .x. y. and 2 in place of just .r in thc preceding functional. The expression 11 is the energy of defnrmation that you will study l a t a in your solids ciii~rseand presented cai-lier in Section 10.5 aTi,&,, 2K1, we have unstable equilibrium. The value W = 2KI is Called il cririrul load for reasons that are explained i n Problem 10.5X.
Figure P.10.58.
10.58. I n Problem 10.57, apply a small transverse force F to body A as shown. Compute the horizontal deflection 6of paint A for a position of equilibrium by using ordinary statics as developed i n earlier chapters. Now show that when W = 2K1 @e.,the critical weight), the deflection 6mathemdtically blows up to infinity. This shows that, even if W < 2KI and we have stable equilibrium with I.’ = 0, we get increasingly very large dcflections as the weight W approaches its critical value and a side load F , however small, is introduced. The study of stability of equilibrium configuration therefore is an important area of study in mechanics. Most of you will encounter this topic in your strength of materials course.
10.10
10.59. Cylinders A and B have semicircular cross-sections. Cylinder A supports a rectangular solid shown as C. If p, = 1,600 kg/mz and p< = 800 kg/m’, ascertain whether the arrangement shown is in stable equilibrium. [ H i m Make use of point 0 in computing V.1
I 6m -
t
Closure
In this chapter, wc have taken an approach that differs radically from the approach used earlier in the text. In earlier chapters, we isolated a hody for the purpose of writing equilibrium equations using all the forces acting on the body. This is the approach we often call vectorial mechanics. In this chapter, we have mathematically compared the equilibrium configuration with admissible neighboring configurations. We concluded that the equilibrium configuration was one from which there is zero virtual work under a virtual displacement. Or, equivalently for conservative active forces, the equilibrium configuration was the configuration having stationary (actually minimum) potential energy when compared to admissible configurations in the neighborhood. We call such an approach variational mechanics. The variational mechanics point of view is no doubt strange to you at this stage of study and far more subtle and mathematical than the vectorial mechanics approach. Shifts like the one from the more physically acceptahle vectoriul mechanics to the more abstract variational mechanics take place in other engineering sciences. Variational methods and techniques are used in the study of plates and shells, elasticity, quantum mechanics, orbital mechanics, statistical thermodynamics, and electromagnetic theory. The variational methods and viewpoints thus are important and evcn v i t d in more advanced studies in thc engineering sciences, physics, and applied mathematics. 446
10.60. At what position must the operator of the counterweight crane locate the 50-kN counterweight when he lifts the IO-kN load of steel?
10.63. The spring is unstretched when 0 = 3LT. At any position of the pendulum, the spring remains horizontal. If the spring constant is 50 Iblin., at what position will the system be in equilibrium?
20 m
Figure P.10.63.
10.64. If the springs are unstretched when 0 = 0,. find the angle 0 when the weight W is placed on the system. Use the method of stationary potential energy.
Figure P.10.60.
10.61. What is the relation between P and Q for equilibrium?
Figure P.10.61.
10.62. A 50 Ih-ft torque is applied to a press. The pitch of the screw is .5 in. If there is no friction on the screw, and if the base of the screw can rotate frictionlessly in a base plate A, what is the force P imposed by the base plate on body B?
-ft
Figure P.10.62.
Figure P.10.64.
10.65. A mass M of 20 kg slides with no friction along a vertica rod. Two internal springs K, of spring constant 2 Nlmm and ai external spring K2 of spring constant 3 Nlmm restrain the weigh W. If all springs are unstrained at 6' = 3LT, show that the equilib rium configuration corresponds to 0 = 27.8'.
Figure P.10.65.
10.66. When rod AH i h in thc vaticill position, the spring attached to the wheel by a flexible cord is unstretched. Determinc all the possible angles L? for equilibrium. Show which are stable and which are not stable. The spring has a spring constant 01 8 Ib/in.
Figure P.10.67.
Figure P.10.66.
10.67. Two identical rods are pinned together :it H and arc pinned at A and C . At H there i s a torsinnal spring requiring SO0 N-mlrad of rotation. What is the maximuin weight W thal ciich Ii,d can have for a c11w uf stable equilibrium when thc ~riids arc collinear?
448
IO.6X. A rectangular xdid body (11 height h rests on ii cylinder with a semicircular scctiun Set u p criteria for \lahlc and unstable riluilihrium iii trrnis n i ii and X 1 1 1hc ~ pasilion shoun
Figure P.10.68.
Dynamics
Kinematics of a ParticleSimple Relative Motion 11.1 Introduction Kinemutics is that phase of mechanics concerned with the study of the motion of particles and rigid bodies without consideration of what has caused the motion. We can consider kinematics as the geometry of motion. Once kinematics is mastered, we can smoothly proceed to the relations between the factors causing the motion and the motion itself. The latter area of study is called dynamics. Dynamics can be conveniently separated into the following divisions, mnst of which we shall study in this text:
1. Dynamics of a single particle. (You will remember from our chapters on statics that a particle is an idealization having no volume but having mass.) 2. Dynamics of a system of particles. This follows division 1 logically and forms the hasis for the motion of continuous media such as fluid flow and rigid-body motion. 3. Dynamics of a rigid body. A large portion of this text is concerned with this important part of mechanics. 4. Dynamics of a system of rigid bodies. 5. Dynamics of a continuous deformable medium. Clearly, from our opening statements, the particle plays a vital role in the study of dynamics. What is the connection between the particle, which is a completely hypothetical concept, and the finite bodies encountered in physical problems? Briefly the relation is this: In many problems, the size and shape of a body are not relevant in the discussion of certain aspects of its motion; only the mass of the object is significant for such computations. For example, in towing a truck up a hill, as shown in Fig. 11.1, we would only be concerned
45 I
452
('HAPTRK I I
K1NEM:YI'ICS OF A PARTICLE-SIMPLE REl.AT1VE MOTION
W
i+
Figure 11.1. 'liwck uimsidcrrd :is u pxiiclc.
with the mass of the truck and 1101 with i t s shape or size (if we neglect force!, from the wind, etc.. and the riitatiiinal cffects (if the wheels). The truck can just a, well he considered ii p;uticle i n computing the necessary towing force. We car present this relationship more precisely i n the following ntaiiner. As will he learned in tlie next chapter (Section 12.10). the equation iii inotioii o f the center of nias\ of any hody can he foriiied by:
1. C(mccntriiting thc cntirc mass ill thc miss cciitcr of thc h d y .
2. Applying the
total rc\LiIIant lorce acting [in thc hiidy t o this hypothetical
particle. When the iniition of the mass ccntcr charactcrizcs a l l w e nced to know ahout the motiiin (if tlie hody. we employ the particle concept (i.e,, we find the motion of the m i s s center). Thus. if all points of a hody have the same vclucily at airy tinie I ([hi\ i s callcd rrun.rl~iror~ i n o l i o i i ) . we necd iinly know the miition o i t l i e mass center to liilly ch;ir;~cterize the motion (This was the case fix thc truck. where the rotational inertia of the wheels was neglected.) If. additiiinally, the sizc i i f a hiidy i \ \mall compared to itr trajectory (as i n plarie l x y niiitiiin. for example). the motion of the center o f n i x \ i s all that might he needed. and so again we can use the particle conccpt ior such hodics.
Part A: General Notions 11.2
Differentiation of a Vector with Respect to lime
In the study of statics. we dealt with vector quantities. We found i t con\cnient to incorporale the directional naturc of thcsc quantities i n ii certain n~itatiiin and set of opevatiiiii\. We called the tolalily iiithese \ery useful foi-intilalions "\ector algehra." We s h a l l again enparid our thinking iron1 scalars to vectors-hi\ tinic for the operations of differenti;ition and intcgration with rcspcct t o any scalar variable I (such as timr).
SECTION I I .2 DIFFERENrIATION OF A VECTOR WITH RESPECT TO TIME
For scalars, we are concerned only with the variation in magnitude of some quantity that is changing with time. The scalar definition of the time derivative, then, is given as (11.1) This operation leads to another function of time, which can once more be differentiated in this manner. The process can he repeated again and again, for suitable functions, to give higher derivatives. In the case of a vector, the variation in time may he a change in magnitude, a change in direction, or both. The formal definition of the derivative of a vector F with respect to time has the same form as Eq. I I . I : (11.2)
If F has no change in direction during the time interval, this operation differs little from the scalar case. However, when F changes in direction, we find for the derivative of F a new vector, having a magnitude as well as a direction, that is different from F itself. This directional consideration can be somewhat troublesome. Let us consider the rate of change of the position vector for a reference xyz of a particle with respect to time; this rate is defined as the vrlucily vector, V , of the particle relative to q z . Following the definition given by Eq. 11.2, we have
The position vectors given in brackets are shown in Fig. 11.2. The subtrxtion
I
..
Path of
Figure 11.2. Particle at times f and
f
+ Al
between the two vectors gives rise to the displacement vector Ar, which is shown as a chord connecting two points As apart along the trajectory of the particle. Hence, we can say (using the chain rule) that
453
wherc M'C have multiplied and di\'idrd hy As i n thc liist cxpression As A/ goes to Lerci. the direction ol Ar approaches tangency t o the trajectory at position r ( / )and apprixiches A,\ i n magnitude. Consequently. i n the limit. ArlAs hecoines a unit w c t w e,. tangent to the traieclor>. Tliat i \
( I 1.3)
We c i ~ then i sii~
Therefore. drlih leads to a vector having n magnitudc equal to the speed 01 the pailicle and a dii-ection ~aiigentto the trajectory. Kccp i i i mind that there can he any ansle hetueen the p(isition Vector and the velocity vector. Students seem to \'ant t o limit this anglc to 90c',which actually restricts you to ii cit-culiir p a t h The acceleration ~ e c t o (il r a Ixirtick ciin then be given as
The dillcrcntiation and integration of vector( r, V . and a will concern thniuphout tlic text.
LI'I
Part 6: Velocity and Acceleration Calculations 11.3
Introductory Remark
As you know lroni statics. w c ciin expres? ii vector i n many ways. For illstance, we can use rectangular compiinents. or. as we udl hhortly explain, we can use cylindrical components. In evaluating dcriviitivcs of vectors with respect 10 time. we mu51 pi-oceed i n accordance with the nianner in which the vector has heen expre\rcd. I n Part B of this chapter. wc will therefore exaiiiine certain diffcrcntiation processes tliiit iirc used extensively i n mcchanics. Other diftkrenti;ition proccsses will he exaniined l a t ~ at r appropriate tiiiies. We have already carried out a derivative operation in Section 11.2 directly on the vector r . You will scc in Section 11.5 that llic approach u x d gives tlic dcrivative iii terms 01' potti vorilrblr~c.This approach will he one o S several that we shall iiow examine with some care.
SECTION 11.4 RECTANGULAR COMPONENTS
11.4
Rectangular Components
Consider first the case where the position vector r of a moving particle is expressed for a given reference in terms of rectangular components in the following manner: r ( t ) = x(r)i + y ( r ) j + z(t)k
( I 1.6)
where x ( f ) ,y ( t ) , and z(t) are scalar functions of time. The unit vectors i, j , and k are fixed in magnitude and direction at all times, and so we can obtain drldt in the following straightforward manner:
dr = V ( t ) = -dx(t) ' + dy(t)j t dt
dt
k = i ( t ) i + y(t)j + i(r)k
+ wdt
( I 1.7)
A second differentiation with respect to time leads to the acceleration vector:
df2
=a =
i ( t ) i + j i ( t ) j + Z(t)k
(11.8)
By such a procedure, we have formulated velocity and acceleration vectors in terms of components parallel to the coordinate axes. Up to this point, we have formulated the rectangular velocity components and the rectangular acceleration components, respectively, by differentiating the position vector once and twice with respect to time. Quite often, we know the acceleration vector of a particle as a function of time in the form a(t) = X ( t ) i
+ s ( t ) j + Z(r)k
( I 1.9)
and wish to have tor this particle the velocity vector or the position vector or any of their components at any time. We then integrate the time functionx(t), y ( t ) , and i ( r ) , remembering to include a constant of integration for each integration. For example, considerx(t). Integrating once, we obtain the velocity component V,(r) as follows: Vx(t) =
1
f ( t ) dr
+ C,
(11.10)
where C, is the constant of integration. Knowing \ at some time to, we can determine C, by substituting to and (V,),, into the equation above and determining C , . Similarly, for x(r) we obtain from the above: x(t)
= j[jf(r)dt]dt+C,t+C?
(11.11)
where C, is the second constant of integration. Knowing x at some time f , we can determine C , from Eq. 1 1.1 I . The same procedure involving additional constants applies-to the other acceleration components. We now illustrate the procedures described above in the following series of examples.
455
456
('tIA1Tl:tt
II
K
Example 11.1 Pins A and H must aluayh remain in the vrrtical sIo1 01yrAc C'. which tno\'e\ ttr tlic right at a c ~ t i s t i i n tspccd o f 6 I t l s e c i n Fig. I I ..?. titrthcrmorc. the pins cannot leave the elliptic slot. (a) What is the speed at which the pins apprmach each whet- wlieti the yoke s l o t is at ~r = 5 ft? ( h ) What i s the rate olch;nrge of \peed k i w m l ciicli nlhcr when the yoke \lot i\ at .i~ = 5 ft'! Thc cqu:ltioti of Ihc clliptic path iii which the pins iniiict ii~ovci \ seen by inspection to he
.\z ~~~~
+ >r- _ I
IO'
j
t
! t
i j
(:I)
6'
clciirly. iIcoordinates (.1.y ) are to represent tlic cii(irdinatcs of pin R. tiicy m u ~he t time functions iiich Ih:it for any tinie i llie v;iliies .r!i) and ? ( I ) sillisfy Eq. (a). Also..i(/) andy(i) tiiust hc such that piti II t i i m e s at all times i n the elliptic path. We cati s i t i d y thesc rcquircnictits hy first dilrcrcnlialiiif Eq. (a) with respect to tiiiic. Cancelins (lie factor 2. we ohtain
N o w ~ ~ (Y(II..~(I), 0. andy(i) ~ i i i i ssati\fy ~ Ikl. ( h ) fur iill viilucs o f i 1 0 cnsurc that H rciniiiiis iii the elliptic path We can now proceed to calve par1 ( a ) of this pi-ohleln. We !wow that pili H must have a velocity .i = h fllscc hcciiiisc (IItlic yokc. Furthcrniiirc. when ~c = 5 It. we know from E[]. (a) that
:.
v = 5.20 ft
N o w going tu Eq. (b). we can solve f o r i at the
iii\tiiiil
01 intcrcit.
! ii spccd of ?.OX ftlccc. Clcarly. pili A n i u s t move u p w i r d with the s m i c speed nf 2.08 f t l x c . Thc pins apprmacli each olhcr at the itistatit o f intcrcsl at ii speed 014.10 I'tlscc.
Thus. pin H tiioves downward with
;
Figure 11.3. Pin \ l i d o in \IDI ; ~ n d)rihc.
SECTION 11.4
Example 11.1 (Continued) To get the acceleration? of pin H , we first differentiate Eq. (b) with respect to time.
The accelerations.? and? must satisfy the equation above. Since the yoke moves at constant speed, we can say immediately that x = 0. And using for x, y , k, and j known quantities for the configuration of interest, we can solvc for? from Eq. (d). Thus,
o+65 + 5 . 2 0 +~ 2.0X2 = 0 ~~
102
62
Therefore ji = -3.32 ft/sec*
Pin B must be accelerating downward at a rate of3.32 ft/sec2 while pin A accelerates upward at the same rate. The pins accelerate toward each other, then, at a rate of 6.64 ftt/secz at the configuration of interest.
In the motion of particles near the earth’s surface, such as the motion of shells or ballistic missiles, we can often simplify the problem by neglecting air resistance and taking the accelcration of gravity g as constant (32.2 ftlsec? or 9.XI mlsec’). For such a case (see Fig. I1.4), we know immediately that y(r) = -g andi(r) = i ( t ) = 0 . On integrating these accelerations, we can often determine for the particle useful information as to velocities or positions at certain times of interest in the pmhlem. We illustrate this procedure in the following examples.
L
X
Figure 11.4. Simple ballistic motion of a shell
RECTANGULAR COMPONENTS
457
458
('IIAPTER I I
K I N I M A l I C S OF A PARTICLE- Slh.1I't.ti KEIATIVI'. !KITION
Example 11.2 Ballistics Problem 1.
A shcll i s fired from a hill 500 I t ahove n plain. Ttic angle (1 of firing (see Fig. I I . S I i s 15' ahovc thc horimntal. and the muizle velocity i s 3.000 ftisec. At what horizontal distance. d, w i l l lhc shcll hit the plain if wc ncglcct frictioti of the air'?What i s the nhaximun~ hcighl of the shell ahovc LIE plain? l:iti;illy, determine the trajectory of thc shell 1i.e.. find =.f(.xll.
v)
v
We know imnicdiately ttiat
We need not hothcr with ?(f). since the motion i s coplanar with i(11 = := 0 at all times. We iicxt separate the velocity variahles froni the timc viiriahles hy bringing rlf to Ihe right sidc.: 0 1 the previous cqoations. Thus
dVv
=
"V, =
-32.2dr Odt
Inlegrating the ahove equations. we get
V > [ J )= -32.21 V)(I) =
+
i,ii
Figure P. I 1.104.
Figure
P.II.IUh.
ll.lU7. Pilots of tighter planes war special suits designed til prevent blackouts during a severe maneuver. These suits tend to keep the blood from draining out of the head when the head i s accelerated in a dil-ectiuo from shoulders to head. With this suit, B flier can take 5 ~ ' so f acceleration in the aforementioned direction. If a tlicr is diving at a speed of 1,001) kmlhr, what i s the minimum radius of curvature that he can manage at pullout without suffcrillg had physiological effects'?
11.108. A particle mwes with constant speed uf 1.5 mlyec along
11.110. A mechanical ''aim" for handling radioactive matcriiils i s shown. The distance Fcan he varied by telescoping action of the
arm. The ann can he rotated about the vertical axis A- A. Finally, the arm can he raised or lowered by a worm gear drive (not shown). What i s the velocity and acceleration of the object C if the end of the arm mvves U U radially ~ at a rate of 1 ftJsec while the arm turns at a speed w of 2 r a d s e c ? Finally, the arm i s raised at a rate of 2 ftJsec. The distance T at the instant of interest i s 5 ft. What i s the acceleration in the direction E = .Xi ff!
+
a path given as x = j2 In y m. Give the acceleration vector of the particle in terms o f rectangular components when the partick i s at position y = 3 m. Do the problem by using path coordinate techniques and then by Cartesian-component techniques. How many x ' s of acceleration i s the particle subject to'!
L
~
I A
ll.109. A submarine i s moving in a translatory manner with the following velocity and acceleration relative to an inertial refercnce: V
=
hi
+ 7.5j + 2k
knots
a
=
.2i
~
.24j
+ .X?k knotils
A device inside the submarine consists of an arm and a mdSb at die end of the arm, At the instant of interest, the arm i s rotating in a vertical plane with the following angular speed and angular acceleration:
w = It) r a d s
dJ = 3 rad/s*
The ann i s vertical at this instant. Thc mass at the end of the m d may he ctrnsidered t o he a particle having a mass of 5 kg. What are the velocity and acceleration vectors for the motion of the particle at this instant relative to the inertial reference? Use units of meters and seconds. What must he the force vector from the arm onto the particle at this instant?
x
/ " Figure P.ll.llO.
11.111. A top-section view of a water sprinkler i s yhown. Water four pas-
cnteis at thc center Srom helaw and then goes thrvugh
~agewayiin an impeller. The impeller is rotating at constant speed w of 8 rpm. As xzii from the impeller. the water ICBVCS at a speed o f I O ft/sec at an angle uf 3U" relative to r. What is the velucity and acceleration as seen from the ground of the water as i t leaves the impeller and becomes free of the impeller? Give results in the radial, axial, and transverse directions. Use one reference only.
Y I
Figure P.11.109.
Figure P . l l . l l l .
507
11.112. A luggdgc dispcnscr at ;maiiyort rcxmblch a pyramid with iix flat scgrncnts as sidcs as \liowri in the diagram The S Y ~ ~ ~ IOLLLI~S ITI with ail angular spced w u l 2 rpm. Luggagc is dropped from above ind slidzb duwii lhc lilccs t u bz pickcd up by I w c I e r s at the ~ J W A piecc (if luggage IS shmw on a f x r . It ha\ j u c t heeri Jropped at the position indicaled. If has at thir instant ieru vclocity as \een from the rotating face but has at this instmt and thereiftci an acceleration crf .2g along the f i c c . What i s the tovill iccrlcriltiun, as seen from the ground, of the l u ~ ~ n ea> t .i t r c a c l i c ~ he hasr
ill H'I
Use one reference only.
Figure P.11.113. 11.1 14. A jet uf wiltet Ihiu 3 s p e d ai the n o u l e of 20 mi\. At what position dues i t lhit thc parabolic hill? What i s i t & speed at lliilt puiril'! Dc IOI include I l i c t i w i .
Sidc view
Figure P.11.112. A landing craft i s in the pn,ce\r 01 landing o n M: i.herr the accclcratmn of gravity i s . 3 X S times that 01 the c a I'he craft has thc followmg :iccelerntiiin relative 10 the landing 11.11
i u r f k x at the instant 01interest:
a
=
.2gi
i .4gj
~
2xk rni\cc'
1s the :icceleretirm 01gravity o n thc with. At this ~iisliliit. astnrnaut is raising a harid c i m c w weighing 3 N on the earth. I1 he i \ giving the ciliiirril an upward :I Icrution 0 1 3 mdscc' relalive tu the landinp craft, whiit forcc i i i u ~the t igtimnilut cxcrt (in [he :aniera at the instant of intercct'.'
where fi XI
iO8
11.119. A tube, must of whose centerline is that of an ellipse given as ?2
?I
-+--I 1.8’ .122
Figure P.11.116. *11.117. A particle has a variable velocity V(t) along a helix wrapped around a cylinder of radius e. The helix makes a constant angle a with plane A perpendicular to the z axis. Express the acceleration a of the particle using cylindrical coordinates. Next, express 6, using cylindrical unit vectors and note that the sum of the transverse and axial components of a (lust computed) can be given simply as VE,. Next, express the acceleration of the particle using path coordinates. Finally, noting that E , , = --E-, show that the radius of curvature is given as R = ?-cos2 a
has a cross-sectional diameter D = 100 mm. The tube has the following rotational motion at the instant of interest:
w = . I 5 radis
W = ,036 radls’
Water is flowing through thc tube at the following rate at the instant of interest:
Q = . I 8 Lis
Q = ,025 L/s’
The tube is in the vertical plane at the instant of interest. What is the acceleriltion of the water particles at the centerline of the tube at point C using cylindrical coordinates and cylindrical components’! Assume over the cross-section of the tube that the water d o c i t y and acceleration are uniform.
’I’ Figure P.11.119. Figure P.11.117. 11.118. An eagle is diving at a constant $peed of 40 ftlsec to catch a IO-ft snake that is moving at a constant speed of 15 fl/sec. What should ff be so that the eagle hits the small head of the snake‘? The eagle and the snake are moving in a vertical plane.
11.120. A World War I fighter plane is in level tlight moving at a speed of60 k d h r . At time lo it has an acceleration given as: a = .2gi
~
3gj
+2
k m/s2
Also at this time, the co-pilot is raising a camera upward with an acceleration of 0. IRrelative to the plane. If the camera has il m a s of .01 kg, what forcc must thc co-piloi exert on the camera to give it the desired motion at time t,,? Note that the plane never rotates during this action. Take g = Y.81 d s ’ .
X’ Figure P.ll.118.
Figure P.II.120.
509
Particle Dynamics 12.1
Introduction
In Chapter 1 I , we examined the geometry of motion-the kinematics of motion. In particular, we considered various kinds of coordinate systems: rectangular coordinates, cylindrical coordinates, and path coordinates. In this chapter, we shall consider Newton’s law for the three coordinate systems mentioned above, as applied to the motion of a particle. Before embarking on this study, we shall review notions concerning units of mass presented earlier in Chapter 1. Recall that a pound mass (Ibm) is the amount of matter attracted by gravity at a specified location on the earth’s surface by a force of I pound (Ibf). A slug, on the other hand, is the amount of matter that will accelerate relative to an inertial reference at the rate of I ft/secz when acted on by a force of 1 Ibf. Note that the slug is defined via Newton’s law, and therefore the slug is the proper unit to be used in Newton’s law. The relation between the pound mass (Ibm) and the slug is M (Ibm) (12.1) M (slugs) = 32.2 Note also that the weight of a body in pounds force near the earth’s surface will numerically equal the mass of the body in pounds mass. It is vital in using Newton’s law that the mass of the body in pounds mass be properly converted into slugs via Eq. 12. I In SI units, recall that a kilogram is the mass that accelerates relative to an inertial reference at the rate of I meter/sec2 when acted on by a force of I newton (which is about one-fifth of a pound). If the weight W of a body is given in terms of newtons, we must divide by 9.81 to get the mass in kilograms needed for Newton’s law. That is,
(12.2) M ( k g ) = W (N) 9.81 We are now ready to consider Newton’s law in rectangular coordinates. ~
511
Part A: 12.2
Rectangular Coordinates: Rectilinear Translation
Newton's l a w for Rectangular Coordinates
In iect,ingul,ii c i i ~ i i c I i n ~ i l cwc \ ciin expic\\ Ncuton
\
I,\\\ d\ tollow\
If the notion i s hnown rclalivc IC) mi incrtiill rclercncc. we can easily ~ I v for c the rcckingulai- ciiiiiprincnts oc (lie r c s t t l ~ i i nforce ~ on the palticlc. The eqoatii)ns IO he s o l v e d arc j u s 1 algebraic cqiiations. The iiii.er.re o f this problem. wherein the forces iirc known ovcr ii l i m e inlcrviil and thc motion i s desired during this iiitcr\~~iI, i s no1 s o himplc. For the in\,crsc case. wc tnu\t get inwilved generally with intcgrxtiiin procedures. I n the next section. w e s l i i i l l considcr siiuatioiis in which the resultiiiit lorce on a piisiiclc hiis thc \amc dirccliiin and line 0 1 d o n tit all times. Thc resulting niotiiiii i s then confined to ii sll-ai:hc line and i s usually called r w l i ~ llmYrr fr',~l.~l~~fio~~.
12.3
Rectilinear Translation
For reclilinciir tran&lalion. wc ma) considcr llic line of action 01the tnotion to be collinew with one a x i s 01a rectilinear coordiiiale \yhlein. Newton'\ law i \ then one of the cquatiiins (11 the set 12.3. Wc shall use the .r axi.; to coincide with the line 01 iiclion (11the nici1ioii. The reculinnt lorce I- (we shall not bother with the .t siihxript hcrc) can hc a conslaiit. a functioii of time, ii function 01 speed, ii luticlion o f piisitioii. or any comhinalion of these. At thic time, u'c shall enamine some 0 1 these c a e s . leiiving others to Chapter 19. where. with the aid of the \tudcnts' knowlcdgc (if differential equations,' wc s h a l l he iniorc prcparcd l o crinsidcr Ihcni. Case 1. Force Is a I'unction ofl'imc or a Constant. A piirticlc of mass t,z acted oii by ii limc-\arying force F ( r ) i s shown i n Fig. 12.1. The plane on which the hod? iiio\cc\ i\ lrictiiinlcss. Tlic Ihrcc 01 gravity i s equal and opposite
SECTION 12.1 RECTILINEAR TRANSLATION
Figure 12.1. Rectilinear translation.
to the normal force from the plane so that F(t) is the resultant force acting on the mass. Newton’s law can then be given as follows:
Therefore, d2.x -
dr2
F(t) m
~~
(12.4)
Knowing the acceleration in the x direction, we can readily solve for F(f). The inverse problem, where we know F ( t ) and wish to determine the motion, requires integration. For this operation, the function F(t) must be piecewise continuous? To integrate, we rewrite Eq. 12.4 as follows:
Now integrating both sides we get
where C, is a constant of integration. Integrating once again after bringing dt from the left side of the equation to the right side, we get (12.6) We have thus found the velocity of the particle and its position as functions of time to within two arbitrary constants. These constants can be readily determined by having the solutions yield a certain velocity and position at given times. Usually, these conditions are specified at time t = 0 and are then termed initial conditions. That is, when t = 0,
V
=
V, and
.x = xo
(12.7)
These equations can be satisfied by substituting the initial conditions into Eqs. 12.5 and 12.6 and solving for the constants C, and C,. Although the preceding discussion centered about a force that is a function of time, the procedures apply directly to a force that i s a constant. The following examples illustrate the procedures set forth. >That is, the function haa only a finite number d finice discontinuities.
5 13
5 14
CHAPTER 12 PARTICLE DYNAMICS
Example 12.1 A 100-lb body is initially stationary on a 45" incline as shown in Fig. 12.2(a). The coefficient of dynamic friction pd between the block and incline is .5. What distance along the incline must the weight slide before it reaches a speed of 40 fthec? A free-body diagram is shown in Fig. 12.2(b). Since the acceleration is zero in the direction normal to the incline, we have from equilibrium that 100~0~4 = 5N~= 70.7 Ib
(a)
Now applying Newton's law in a direction along the incline, we have
Therefore,
2
= 11.38
Rewriting Eq. (b) we have d ( 2 ) = I1.38dt
Integrating, we get
2
+ C,
- = I1.38f
t2
s = 11.38-
2
+ C,t + C,
(C)
(4
When f = 0, s = d d d t = 0, and thus C, = C, = 0. When d d d i = 40 ft/sec, we have for t from Eq. (c) the result 40 = I1.38t
Therefore, f
= 3.51 sec
Substituting this value off in Eq. (d), we can get the distance traveled to reach the speed of 40 ftlsec as follows:
Figure 12.2. Body slides on an incline.
SECTION 12.3 RECTILINEAR TRANSLATION
rn
Example 12.2
"
A charged particle is shown in Fig. 12.3 at time f = 0 between large parallel condenser plates separated by a distance d in a vacuum. A time-varying voltage V (notation not to he confused with velocity) given as V = 6 sin ut
(a)
is applied to the plates. What is the motion of the particle if it has a charge q coulombs and if we do not consider gravity'? As we learned in physics, the electric field E becomes for this case
The force on the particle is qE and the resulting motion is that of rectilinear translation. Using Newton's law we accordingly have dZx z =
q
6sinot m
d
dx 6 sin wt df x) = 4 7-
Integrating, we get d x - -- 6q c o s w t + c ,
dt
wmd
Applying the initial conditions x = b and dx/dt = 0 when f = 0, we see that C, = hq/mwd and C, = b. Thus, we get
The motion of the charged particle will he that of sinusoidal oscillation in which the center of the oscillation drifts from left to right.
Case 2. Force Is a Function of Speed. We next consider the case where the resultant force on the particle depends only on the value of the speed of the particle. An example of such a force is the aerodynamic drag force on an airplane or missile. We can express Newton's law in the following form: ~dV ~
dr
F(V) m
-
Figure 12.3. Charged particle
(C)
Rewriting Eq. (c). we have d(
y
(12.8)
between condenser plates.
5 15
516
CHAPTEK I? PAKTICLL DYNAMICS
where F ( V ) is a piecewise continuous function reprcsenting the force in the positive x direction. If we rearrange the equation in the following manner (this is called separnrion of vuriuhlrs):
we can integrate to obtain (12.9)
The result will give I as a function of V. However, we will generally prefer to solve for V in terms o f f . The result will then have the form
v
= H(t,
c,)
where H is a function of t and the constant of integration C,. A second integration may now he performed by first replacing V by dxldt and bringing dt over to the right side of the equation. We then get on integration x =
j H ( r , C,
rir
+ c2
( 12. I O )
The constants of integration are determined from the initial conditions of the problcm.
rn
Example 12.3 A high-speed land racer (Fig. 12.4) is moving at a speed of 100 mlsec. The resistance to motion of the vehicle is primarily due to aerodynamic drag. which for this specd can he approximated as .2V2 N with V in mlsec. If the vehicle has a mass of 4,000 kg, what distance will it coast before its speed is reduced to 70 mlsec? We have. using Newton's law for this case.
1
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.3 (Continued) Integrating, we have
- _I - -5 x lo-% + c, V
(c)
Taking f = 0 when V = 100, we get C, = -1/100. Replacing V b y dx/dt, we have next
df
5 x IO-%
V - & =
I +100
(4
Separating variables once again, we get dt - dx 5 x 10-5t + (1/100)
To integrate, we perform a change of variable. Thus
lJ = 5 x IO-% + (1/100)
:.
dlJ = 5 x Io-sdf
We then have as a replacement for our equation
3 = 5 x 10-3 dx II
Now integrating and replacing lJ, we get
(
)'
= 5 x lo-%+
In 5 x 1 0-5 f + l m
c,
When t = 0, we take x = 0 and so C, = In (l/lOO), We then have on combining the logarithmic terms: ln(5 x
+ I ) = 5 x 10-5.r
(e)
Substitute V = 70 in Eq. (d); solve for f. We get t = 85.7 sec. Finally, find x for this time from Eq. (e). Thus, In[(5 x 10-3)(85.7) + I] = 5 x l 0 - h Therefore,
The distance traveled is then 7.13 km.
5 17
518
CHAPTER I 2
I'AKI'ICLb; IUYNAMICS
Example 12.4 A conveyor i s inclined 20" from the horizontal ah shown in Fig. 12.5. As a i-esult o l spillagr ciluil (in l l i z belt. Ihcrc i s a viscous friction Scirce hctwzen body 11 and the helt. This foi-ce equals I). I Ihf per unit relative velocity between body L, and thc hell. Thc helt moves at il conhtant speed VNup the conveyor while inilially hrdy I ) hac a speed (V,,),, = 2 Il/scc relative 10 the ground i n :I direction doum the conveyor. What \peed V;, should the belt have in order for hody 1) l o be ahlc to cvciitually approach a Lero velocity rclatiw t o tlic ground? For hell speed VI,. and Sor the giuen initial speed o l body 0, namely IV,,),, = Z ftlsec, detcmiinc the time when body 1) attains a !,peed of I ftlsec relative to the ground. l h c mass of ll i s 5 I b m
I.'iFure 12.5. A I h d y \lides ilvwn a convcyoi~hell
WCI
with oil.
We hegin hy assigning axes for the prohlem a s I i l l i i u , s (see Fig. 12.6):
Y
Figure 12.6. Friction force f hrtwzcn hody D and the hrlr.
From kinematics we can say
15
0.1 limes the relati\c vclority
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.4 (Continued) For the friction forcefwe have
f
=
-(.MVD)qz
=
-(.lW,
+ VB)xni
We may now use Newton's law for body D in the x direction. Since all velocities from here on will be relative to the ground, we can dispense with the reference subscripts. Thus
When body D attains a theoretical permanent zero velocity relative to the ground, V, and dV, are equal to zero. This gives us (noting that V, now dr becomes V i )
Now determine the time for body D to attain a velocity of 1 ft/sec relative to the ground for a belt speed of 17.10 ft/sec. For this we go back to Eq. (a). dVD dt
:.
- (-.l)(V, + 17.10) + 5sin20'
In V, = - -r3.22 5
=
-.lVD
+ C,
When f = 0,
V, = 2 ftlsec,
.:
Ct = l n 2
Hence, on combining log terms3 In
[$1
=
-.644r
Set V, = 1 ftlsec. Solve for f . f
'
= --] n(S00) = ,644
'Note from this equation that V , = 2e4.M4' and that VD = -1.288e4.w' and so we see that as t approaches infinity both of these quantities approach zero. Thus, theoretically body D could appmach a permanent zero velacity relative to the gmnnd.
5 19
Case 3. Force Is a Function of Position.
As the final case of this series. we now consider the rectilinear motion of a body under the action of a force that i s exprcssihlc cis ii function of pocilion. Pcrhaps the simplest example of such a case i s Ihc frictionless miss-spi-ing system shown in Fig. 12.7. The body i s shown at a position where the spring i s unstrained. The horizontal force froni the spring at d l positions of the body clcarly w i l l be a function of position .c,
Figure 12.7. Mn\s-spring systcm.
Nrwfi,,!',T In),. for position-dependent forces can hc giber] as
We caiiiiol scpur;ilc the variables for this form o l thc cquiltion as in previous c:iscs since there are three variables (V. f. and.r). However, by using the chain rulc of differentiation. w e can change thc left side of the equation 10 a inore desirable tnrrn in the Ibllowinp manncr:
We ciin now scparate the variable? in
Eq. 12.1 I as follows:
r17VdV = F ( s ) r l . r Intcgrating. we pet
(12. I ? ) Solving 1,r Vand using rl.rldt i n i t s pliicc, we get
SECTION 12.3 RECTILINEAK TRANSLATION
Separating variables and integrating again, we get
For a given F(x), V and x can accordingly be evaluated as functions of time from Eqs. 12.12 and 12.13. The constants of integration C, and C, are determined from the initial conditions. A very common force that occurs in many problems is the lineur restoring,forct!. Such a force occurs when a body W is constrained by a linear spring (see Fig. 12.7). The force from such a spring will be proportional to x measured from a position of W corresponding to the undefiirmed configuration of the system. Consequently, the force will have a magnitude of IKxI, where K, called the .spring consfunf, is the force needed on the spring per unit elongation or compression of the spring. Furthermore, when x bas a positive value, the spring force points in the negative direction, and when x is negative, the spring force points in the positive direction. That is, it always points toward the position x = 0 for which the spring is undeformed. The spring force is for this reason called a restoring force and must be expressed as -Kx to give the proper direction for all values of x . For a nonlinear spring, K will not be constant but will be a function of the elongation or shortening of the spring. The spring force is then given as
SpmE
(12.14)
In the following example and in the homework problems, we examine certain limited aspects of spring-mass systems to illustrate the formulations of case 3 and to familiarize us with springs in dynamic systems. A more complete study of spring-mass systems will be made in Chapter 19. The motion of such systems, we shall later learn, centers about some stationary point. That is, the motion is vibratury in nature. We shall study vibrations in Chapter 19, wherein time-dependent and velocity-dependent forces are present simultaneously with the linear restoring force. We are deferring this topic so as to make maximal use of your course in differential equations that you are most likely studying concurrently with dynamics. It i s important to understand, however, that even though we defer vibration studies until later, such studies are not something apart from the general particle dynamics undertaken in this chapter.
521
522
CHAPTER 1 2 PARTICLE DYNAMICS
Example 12.5 A cart A (see Fig. 12.8) having a mass of 200 kg is held on an incline so as to just touch an undeformed spring whose spring constant K is SO N/mm. If body A is released very slowly, what distance down the incline must A move to reach an equilibrium configuration'? If body A is released suddenly, what is its speed when it reaches the aforementioned equilibrium configuration for a slow release'!
1
u
-1
Figure 12.8. Cart-\pring \y\tem
As a first step, we have shown a free body of the vehicle in Fig. 12.9. To do the first part of the problem, all we need do is utilize the
(200)'(Y
xI
Figure 12.9. Free-hody diagram of cart
definition of the spring constant. Thus, if. 6represents the compression of the spring, we can say:
SECTlON 12.3 RECnLlNEAK TRANSLATION
Example 12.5 (Continued) Therefore,
Thus, the spring will be compressed ,01962 m by the cart if it is allowed to move down the incline very slowly. For the case of the quick release, we use Newton’s law. Thus, using x in meters so that K is (SO)( 1,000) N/m: 200x = (200)(9.81)sin 30” - (5O)(l,OOOJ(x)
Therefore, f
= 4.905 - 250x
Rewritingi, we have dV V; i ;= 4.905 - 250x
Separating variables and integrating,
-v_2 - 4.9051 2
~
1 2 5 ~ ’+ C,
To determine the constant of integration C , , we set x = 0 when V = 0. Clearly, C; = 0. As a final step, we set x = ,01962 m and solve for V. V = {2[(4.905)( ,01962) - (125)( .01962)’]\”
’
The following example illustrates an interesting device used by the U S . Navy to test small devices for high, prolonged acceleration. Hopefully, the length of the problem will not intimidate you. Use is made of the gas laws presented in your elementary chemistry courses.
523
524
CHAPTER 1 2 PARTICLE DYNAMICS
Example 12.6 An air gun is used to test the ability of small devices to withstand high prrr longed accelerations. A “floating piston” A (Fig. 12.10).on which the device to he tested is mounted, is held at position C while region I> is tilled with highly compressed air. Region E is initially at atmospheric pressure hut is entirely sealed from the outside. When “fired,” a quick-release mechanism releases the piston and it accelerates rapidly toward the other end of the gun, where the trapped air in E “cushions” the motion so that the piston will begin eventually to return. However, as it starts back. the high pressure developed in E is released through valve F and the piston only returns a short distance. Suppose that the piston and its test specimen have a combined mass of 2 Ihm and the pressure initially in the chamber D is 1,000 psig (above atmosphere). Compute the speed of the piston at the halfway point of the air gun if we make the simple assumption that the air in D expands according t o p = constant and the air in E is compressed also according to p t > = ~ o n s t a n t .Note ~ that I ’ is the specific volume (Le., the volume per unit mass). Take / ‘ o f this fluid at D to he initially ,207 ftg/lbm and oin E to he initially 13.10 ft’llbm. Neglect the inertia of the air. The force on the piston results from the pressures on each face, and we can show that this force is a function o f x (see Fig. 12.10 for refercnce axes). Thus, examining the pressure plj first for region D , we have, from initial conditions, (pf,~’fj= ) c [ci,cno , + 14.7)(144)](.~07) =m
oo
(a)
Furrherrnore, the mass of air D given as MI, is determined from initial data as
where (V,,), is the volume of the air in D initially. Noting that p7) = const. and then using the right side of Eq. (a) for pf,tiL, as well as the first part of Eq. (h) for cD, we can determine pf, al any position x of the piston:
‘YOU should r ~ dliom l your earlier work 10 physics and chemistry thilr “U we using 11em the isothermal form of the quiltion of state for a perfect gas. Two factors of caution should be pointed out ~ ~ I a t i vtoethe use of chis expression. First. at the high pressures involved in p‘uf ofthe expansion, the pedcct gas m o d i is only an approximation for the gas. and so the yuarion ol state of a perfect gas that gives uspri = Consrant is only apprurimale. Fnnhemorr. the assumption 01 isotherm4 expansion gives only an approximatim of the actual process. Perhaps il better approximation is to ilssumr an adiaharic expansion (i.e. no heat wmler). This is dune in Prohlem 12.130,
Val
Figure 12.10. Air gun.
SECTION 12.3 RECTILINEAR TRANSLATTON
Example 12.6 (Continued) We can similarly get p, as a function of x for region E . Thus, ( p E u E ) o= (14.7)(144)(13.10) = 27,700 and
Hence, at position x of the piston 27,700 - 27,700 27,700 vE V E / M , - (rr/4)(Iz)(50 - x)/2.88
PE=---
Therefore, 101,600
PE
=
5O-x
Now we can write Newton’s law for this case. Noting that V without subscripts is velocity and not volume, M V -dV
zl2 --
50 - x
(4
where M is the mass of piston and load. Separating variables and integrating, we get M V Z = $[293,00OInx+ 101,6001n(50-x)]+C,
2 To get the constant C,, set V = 0 when x = 2 ft. Hence, C, = -
$ (293,000 In 2 + 101,600In 48)
Therefore,
C, = 468,000 Substituting C, in Eq. (e), we get
(6) { ~ [ 2 9 3 , 0 0 0 l n x + 1 0 1 , 6 0 0 l n ( 5 0 ~ x ) ] - 4 6 8 , 0 0 0 In
V =
We may rewrite this as follows noting that M = 2 Ibdg: V = 566[23 In x
+ 7.98 ln(50 - x) - 46.81”’
At x = 25 ft, we then have for V the desired result: V = 566(231n 25 + 7.98 In 25 - 46.8)”’
(e)
525
526
CHAPTER I ?
PARTICLE DYNAMICS
*Example 12.7 A light stiff riid i s pinned at it and i s constrailled by two linear springs. K, = 1.000 Nitn and K? - = 1.2(X) Nim. The spring are urirtretched when the rod i s horimntal. At the right end o f the rod. ii mass M = 5 kg i s attached. If Ihc rod i s rotated 12" doc.kwi.sc front a horizontal ciintiguration and thcn released, what i s the spccd of the mass when the rod returns to a position corresponding to the .s/& ~ ~ q ~ r i l i h r iposilion i r n ~ with miss M attached'!
Fieure weightless
A free-hody diagram 01 the systcin for pi.si/iw H i s shown i n Fig. 12.12(;1) end a free-hody diagram o f the pailicle M i s shown i n
F . mI
F.B.U. I I
(a,
(hl
Figure 12.12. Frcc-body iliagrams of thc system and the pnniclr il,r positive c)
Fig. I2.12(b). The spring forces for small positive ro1iitions H :
4
=
F;
aiid F,
011
tlic rod arc givrn as follows
( . 3 ) ( H ) ( K ,= l 3000N
F2 = - ( . I ) ( B ) ( K ? )= - 1 2 0 O N
(ai
where 8 is in radians. I n the first free hody. we w i l l think 01 the rod as a illassless perfectly irigid Icvcr iih sludied i n high school or perhaps even
SECTION 12.3 RECTILINEAR TRANSLATION
Example 12.7 (Continued) earlier. Then we can say for the forces on the secund of our free bodies stemming from the springs5 (fromF;) =
-4
= -300ON 1 (fromFz) = ? ( F , ) = 4 O N
F; F;
We can now give Newton's Law for M as follows using y for the vertical coordinate of the particle: 5 y = -5g-300e-40e
... j i =
- g - 3 4 50
e
(b)
Next, from kinematics, we can say for small rotation
Now going back to Eq.(b) we replace y by
in order to be able to separate variables. Also replace O by yl.3. We then may say ydy=(-227y-g)dy Integrating 2 When O = - 12" = -(&)(2n) rad = - ,2094 rad,8 = y = 0. We can 360 then solve for the constant of integration using y = .3 8.
L
J
Hence,
SNote that a positive B gives negative values for F ; and F; on M and vice versa. It is for this reason that we require the minus signs.
521
528
C H A PTER 1 2 PAKTICL~.U Y N A M I ~ S
Example 12.7 (Continued) For the stiitic equilihrium configuratior of the rod. we require firoin
Fig. 12.12(al
Suhstituting values irim Eqs. (a) and noting that we are only using the magnitudes of the forces ahiive ii)r the required ncSativc ino~iiciitswc get
-(SI(9,Xl)(.3)
-
(300B,/ j(.3) - ~ 1 2 0 B , ~ , l l ~= . l l0
Solving for (I,,(,
BEllowingdata apply: I
Figure P.12.55. :mica1 pendulum of Icngth 1 i s > w nThe nadc tn rotate ill a constant angular spccd of wahoo axis. Compute the tension in the cord if thc pcndulum huh ha!
56.
5
M , = 2 kg
M,j = 3 kg
M,. = 2 kg
If thr spi~inpi\ \trctchcd h) an a m r u n t .(I25 m. dcterminc the total lurce components :sting o n thc miisr at ('and the tenqile force in mcmher /),4 [ I l i n i ('onwicr a single panicle. then a \y\tem 0 1 pmticIc\; I;\ and OH arc pin connec1ed.l
12.61. A platform rotates at 2 radhec. A body C weighing 450 N rests on the platform and is connected by a flexible weightless cord In a mass weighing 225 N, which is prevented from swinging out by pan of the platform. For what range of values o f x will bodies C and B remain stationary relative to the platform? The static coefficient of friction fix a11 surfaces is .4. x-
_-- .', Figure P.12.58.
12.59. In the preceding problem consider that member DA is welded to the sphere at A . Now, at the instant of interest, there is also an angular acceleration of the system having the value of .28 rad/sec* counterclockwise. What are the force components acting on panicle C. and what are the force components from rod AD acting on panicle A'? See hint given in the preceding problem. 12.60. A device called a flyhall governor is used to regulate the speed o f such devices as steam engines and turbines. As the governor is made to rotate through a system of gears by the device to he controlled, the halls will attain a configuration given by the angle 6'. which is dependent on both the angular speed w of the governor and the force P acting on the collar hearing at A. The upand-down mntion ofthe bearing at A in response to a change in w is then used to open or close a valve to regulate the speed of the device. Find the angular velocity required to maintain the configuration of the flyhall governor for 8 = 30". Neglect friction.
wL,=.4 for 811 surfac Figure P.12.61.
12.62. A particle moves under gravitational influence about a body M, the center of which can he taken as the origin of an inertial reference. The mass of the particle is 50 slugs. At time t, the particle is at il position 4,500 mi from the center of M with direction cosines I = .5, m = -3. n = ,707. The panicle is moving at a speed of 17,000 milhr along the direction E, = .Xi + .2j + .Shhk. What is the direction of the normal t o the plane of the trajectory?
Figure P.12.62.
Figure P.12.60.
12.63. If the position of the particle in Problem 12.62 were to reach a distance of 4,300 mi from the center of hody M, what would the transverse velocity V, of the particle be? 551
12.64. Use Eqs. 12.38h and 12.40 t o show' that i f t h c eccentricity is x r u , the trujectoq nmsl he that of a circle.
12.65. A satellite ha\ at m c t i m e during its flight around thc caith a radial crrmpment 01velocity 3.200 krnlhl- and a tlans\.cr\e cm~poneril0 1 2h.XIO km/hr. I1 the s;ilrllitc i \ at a distanvc 01 7.040 krn frtmm tlic cciitzr 01 the eimrth. u'hal is i l \ arral velocity'! 12.66. Compute the c s u p e velocit) ;it :I pnsition X.000 In, f i r m the center of the sarlh. What q x c d i s needcd to inninlain il circw tar orhit at that di.;tance from the earth's center? Derive Ihc c q u a ~ tiaii for the speed riccdcd fur il circular orhit direclly from Neuston's law without ucing infill-malioo iibrrut ecccnlricilieh. ctc.
z
Figure P.12.69.
sonnection
I
Figure P.12.67.
i5X
12.72. Consider a satellite of mass m in a circular orbit around the earth at a radius R, from the center of the earth. Using the universal law of gravitation (Eq. 1.1 I ) with M as the mass of the eanh and using Newron's law in a direction normal to the path, show that -
IGM
"ctrc.
12.78. The satellite Hyperion about the planet Saturn has a motion with an eccentricity known to he ,1043. At its closest distance from Saturn, Hyperion is 1.485 x IO6 km away (measured from center to center). What is the period of Hyperion about Saturn? The acceleration of zravitv of Saturn is 13.93 m/recz at its surface. The radius of Saturn is 57,600 km "
I
= '-
4 Ro
for a circular orbit. Now at the earth's surface use the gravitational law again and the weight ,f the body, to show that CM = gRE.ar,,, where y is the acceleration of gravity.
12.73. The acceleration of gravity on the planet Mars is about ,385 times the acceleration of gravity on earth, and the radius of Mars is about ,532 times that of the earth. What is the escape velocity from Mars at a position 100 mi from the surface of the planet?
12.79. Two satellite stations, each in a circular orbit around the earth, are shown. A small vehicle is shot out of the station at A tangential to the trajectory in order to "hit" station B when it is at a position E 120" from the x axis as shown in the diagram. What is the velocity of the vehicle relative to station A when it leaves? The circular orbits are 200 miles and 400 miles, respectively, from the earth's
12.74. In 1971 Mariner 9 was placed in orbit around Mars with an eccentricity of .5. At the lowest point in the orbit. Mariner 9 is
320 km from the surface of Mars. (a) Compute the maximum velocity of the space vehicle relative to the center of Mars. (b) Compute the time of one cycle. x
Use the data in Problem 12.73 for Mars.
12.75. A man is in orbit around the earth in a space-shuttle vehicle. At his lowest possible position, he is moving with a speed of 18,500 mihr at an altitude of 200 mi. When he wants to come back to earth, he fires a retro-rocket straight ahead when he is at the aforementioned lowest position and slows himself down. If he wishes subsequently to get within SO mi from the earth's surface during the first cycle after firing his retro-rocket, what must his decrease in velocity be? (Neglect air resistance.)
12.76. The Pioneer 10 space vehicle approaches the planet Jupiter with a trajectory having an eccentricity of 3. The vehicle comes to within 1,000 mi of the surface of Jupiter. What is the speed of the vehicle at this instant? The acceleration of gravity of Jupiter is 90.79 ftlsec' at the surface and the radius is 43,400 mi.
12.77. If the moon has a motion about the earth that has an eccentricity of ,0549 and a period of 27.3 days, what is the closest distance of the moon to the earth in its trajectory?
Figure P.12.79.
12.80. In Problem 12.79, determine the total velocity of the vehicle as it arrives at E as seen by an observer in the satellite B . The values of C and D for the vehicle from Problem 12.79 are mi-', respectively. 7.292 x IO' m i 2 h and 7.373 x 12.81. The Viking I space probe is approaching Mars. When it is 80,650 km from the center of Mars, it has a speed of 16,130 kmihr with a component (V,) toward the center of Mars of 15,800 kmJhr. Does Viking I crash into Mars, go into orbit, or have one pass in the vicinity of Mars? If there is no crash, how close to Mars does it come? The acceleration of gravity on the surface of Mars is 4.13 m/sec2, and its radius is 3,400 km. Do not use formula for D as given by Eq. 12.54, but work from the trajectory equations.
559
12.83.
110 I'mhlcm 12.U2 u'ilh the
ilitl
of Eq. 12.14. Figure l'.l2.87.
12.85.
Do Prohleni 12.84 with the aid of Eq. 12.54.
Figure P.12.88.
SECTION 12.9 NEWTON’S LAW FOR PATH VARIABLES
Part C: 12.9
Path Variables
Newton’s Law for Path Variables
We cm express Newton’s law lor path variables as follows:
< = m , d2s dt
(12.57a)
Notice that the second of these equations is always nonlinear, as discussed in Section 12.5.’’ This condition results from both the squared term and the radius of curvature R. It is therefore difficult to integrate this differential equation. Accordingly, we shall be restricted to reasonably simple cases. We now illustrate the use of the preceding equations. ‘%quation 12.57a could also he
nonlinear, depending on Lhe nillure of the function $.
Example 12.14 A portion of a roller coaster that one finds in an amusement park is shown in Fig. 12.27(a). The portion of the track shown is coplanar. The curve from A to the right on which the vehicle moves is that of a parabola, given as (y -
=
loox
(a)
,’ , ,
v
I:1
loop
, , , , / / I ’
n n
x
(a) Figure 12.27. Roller coaster trajectory.
P, (b)
561
562
CHAPTER 12 PARTI(’1.E DYNAMICS
Example 12.14 (Continued) with x and y in feet. If the train of cars is moving at a speed of 40 ft/sec when thc front car is 60 fl above the ground, what is the total normal force exerted hy a 200-1h occupant nf thc front car on the seat and floor of lhe car’! Since we requirc only the force F normal to the path, we need only he concerned with u,~.Thus, we have
Wc can compute R from analytic geometry as follows:
wherein from Eq. (a) we have
d\
SO
&=m
Substituting into Eq. ( c ) ,we have
AI the p o w o n of intere\t. we get
SECTION 12.9 NEWTONS LAW FOR PATH VARlABLES
Example 12.14 (Continued) Accordingly, we now have for y,, as required by Newton's law:
Nnte that I;; is the total force component normal to the trajectory needed on the occupant for maintaining his motion on the given trajectory. This force component comes from the action of gravity and the forces from the seat arid floor of the car. These forces have been shown in Fig. 12.27(h), where P, and p a r e the normal and tangential force components from the car acting on the occupant. The resultant of this force system must, accordingly, have a component along n equal to 94.6 Ib. Thus,
-200 cos
To get
+ P,
= 94.6
(h)
p, nnle with the help of Eq. (d) that
Therefore. /3 = 51.3" Suhstituting into Eq. (h) and solving for P,. we get
9, =
200 COS S I.30
+
94.6
This is the force component from the vehicle onto the passenger. Tne reaction to this force is the force component from the passenger onto the vehicle.
563
564
CHAPTER I ?
PARTICLE DYNAMICS
Part D: A System of Particles 12.10
The General Motion of a System of Particles
I.et us examine a system of n particles (Fig. 12.28) that has inlcractions between thc particles fiir which Newton '.s thit-d lobe of motion (action equals
Figure 12.28. Force\ on ith pairiclr of the system
reaction) applies. Newror~'.ssrt.ond /ai?.for any particle (lcl us say thc ith particle) is then
(12.58) ,*I
wherc4.;. is the force on particle i from particle j and is thus considered an internal force for the system of particles. Clearly. the,, = i term of thc sunimation musf be deleted since tlic ith particle cilnnoi exert force iin itself. The force F, reprcscnts the resultant force on the ilh particle from the forccs < w c , i nrrl to thc syslem of paniclcs. If thehe equations are added for all n particlcs, we have
(12.59) Carrying out the double summation and excluding terms with rcpeakd indexes, such as&,,&,. etc., we find that lor each term with any one set o f indexes there will he a term with Ihe reverse of thcse indexes present. For example. lor the force.(,, it lorcef2, will ex is^. Considering the ineming iifthe indexes. we see thatf..,.I andJ, represent action and reaction lorccs between a pair of particles. Thus, as a result of N e w o n ' s third Irrw. thc double summation i n Eq. 12.59 should add LIPt u zero. N < w t r n i ' .srmrtd ~ law lor a system iif paiticlcs then becomes: I2.60 1
where F now rcprehenls the vector sum of all the r,rtr,ruol forccs acting on all the panicles of the system.
SECTION 12.i o THE GENERAL MOTION OF A SYSTEMOF PARTICLES
To make further useful simplifications, we use the first moment of mass of a system of n particles about a fixed point A in inertial space given as first moment vector =
C miri i=l
where 7 represents the position vector from the point A to the ith particle (Fig. 12.29). As explained in Chapter 8, we can find a position, called the
Figure 12.29. Center of mass of system
center of mass of the system, with position vector r,, where the entire mass of the system of particles can be concentrated to give the correct first moment. Thus,
Therefore, (1 2.61)
Let us reconsider Newton's law using the center-of-mass concept. To m,r( by Mr, in Eq. 12.60. Thus, do this, replace (12.62) We see that the center ofmass of any aggregate ofparticles has a motion that can he computed by methodr already ser,forth, since fhis i.7 a problem involving a single hypothetical particle of mass M . You will recall that we have alluded to this important relationship several times earlier to justify the use of the particle concept in the analysis of many dynamics problems. We must realize for such an undertaking that F is the total external force acting on all the particles.
565
566
CHAPTER 12 PARTlCLt DYNAMICS
Example 12.15 Three charged particles i n a vacuum are shiiwn in Fig. 12.30. Particle I has a mass of 1W5 kg and a charge o f 4 x IO-' C (coulomhs) and i s at the origin at the instanl (ifinterest. Particlcs 2 and 3 each have a mass of 2 x 1 0 kg and a charge o f 5 x 10 C and are located. respectively, at the
rP? Figure 12.30. Char&
particles in field E .
inslant of interest I ni d i m g t h e y axis and 3 m along the :axis. An electric field E given a s
E = 2.ti
+ 3 r j + 3(r + .?)k
NIC
(a)
i s imposed from thc nutsidc. Compute: (a) the position [ifthe center 01 mass for the system. (h) the acceleration o f the center o f mass, and (c) the acceleratinn o f particle I
To get the position [if the ccnter of mass, we merely equate moments of the masses ahnut the origin with that of R particle having a mass equal to the sum o f masses of the system. Thus. (I+ 2
+ 2)x
IO ' r , = ( 2 x 10 i ) j+ ( 2 x 1Wi)3k
Therefore,
rc = .4j+ 1.2km
(h)
To get the acceleration of the mass center. we must find the sum of the extrmal forccs acting on thc particles. T w o cxternal forccs act on each particle: the force o f gravity and the electrostatic force from the external field. Recall from physics the1 this electrostatic force i s given as 4E, where 4 i s the charge on the particle. Hcnce, the total exlemal force for each particle i s given a s Sollows:
Fl
=
-(9.X1)(11i~7)k+ 0 N
Fz =-(9,Xlj(2x 1 0 ~ ~ j k + ( 5 ~ 1 0 ~ ~ j ( 3 k ) N F3 = -(9.81)(2 x
jk
+ (5 x I O s ) ( 9 j + 27k) N
(C)
(d)
(e)
SECTION 12.10 THE GENERAL MOTION OF A SYSTEM OF PARTICLES
Example 12.15 (Continued) The sum of these forces FT is
F~ = 45 x
io-5j
+ 100.9 x 10-5k N
(f)
Accordingly, we have forw5:
=
i9ilOV:
cg)
And according to Eq. IC). \/
.I
L \/ ~I
R
Suhslituting thc results Ironr Eqs. ( h ) and (i) into (gl. MC gel
(i)
SECTION 13.2 i
Example 13.3 (Continued) Adding Eqs.(f) and (j), we can eliminate T to form the following equation with V, as the only unknown: (981)(3) = f ( V i ) ( l O O
+ F)
Therefore,
V,, = 6.14 d s e c downward Hence,
V, = 1.534 d s e c to the left
13.2
Power Considerations
The rate at which work is performed is calledpowrr and is a vely useful concept to represent work, we have for engineering purposes. Employing the notation
pace \rehick is in a circular "parking" orhit ( I aruund the earth 200 k m above the earth's hurfacr. I t i \ lo lransfrr t u another circular orbit (2) 500 k m ahow the earth's surface. The transfer t o the second orhit i s dune in two stage,.
x
Figure P.14.111.
I
14.112. Do Problem 14.1 I I for an inelastic impact with
t
= .h.
14.113. A small elastic sphere i s dropped fiom pusitiun (2, 3, 30) f t onto a hard soherical body havine - a radiub of 5 ft positiuned so that the I axis of the reference shown i s along a diameter. Far a perfectly elastic collision, give the speed of the s m d 1 sphere
1. Fire rucheth so the vehicle has an upoper cquill tu the r d i u 0 1 the sccond circular urhit. Whal uhangc of speed i\ required l o i ~ this
2. At u ~ i o p wrockets are (ired again t i l get
into the second circular orhit. Whal i s thi\ srcund change of hpced?
directly after impact.
500 k m
0 I
,, ~-
.
\,',_---..~. .,. ~~
-
\
\
1 I I I
I I I
200km-
~. ~~
~
,
,
~~
Figure P.14.117.
*14.118. A tugbwdt weighing 100 tuns is moving toward a statiuiiay buge weighing 200 tom and c u y i n g a load C wcighing 50 ton&. / The tug ib moving at 5 knots and its propellers arc devcloping a thrust of 5,000 Ib when it cwlacts the huge. Ai a result of thc soft padding Figure P.14.113. at the nose of the tug. consider that there i s plastic impact. If the load 14.114, D~ problem 14,113 for an inelastic impact with = ,6, C I S not tied in any way to the huge and has B dynamic coefticient of friction o f . I with the slippery deck of the harge, what i\ the sped V 14.115. A bullet hits a smooth, hard, massive two-dimensional of the barge 2 sec after the tug first contacts the barge'! 'The load C body whose boundary has been shown as a parabola. If ihe bullet slips during a I-see internal starting at the beginning ofthe contact.
x
3i + 1j mlsec. After impingement three droplets are formed moving parallel to the ry plane. We have the following information:
i
D , = .6 mm,
V, = 2 d s e c ,
8, = 45’
D, = 1.2 mm,
V, = I mlsec,
0, = 3W
Find D,, V,, and 8,.
I
Figure P.14.118.
3 mlsec 7
1 m/sec
(a) Figure P.14.121.
(b)
14.122. If the coefficient of restitution is .8 for the two spheres, what are the maximum angles from the vertical that the spheres will reach after the first impact? Neglect the mass of the cables.
I
Figure P.14.119.
14.120. A body A weighing 2 tons is allowed to slide down an incline o a barge as shown. Body A moves a distance of 25 ft along the incline before it is stopped at E . If we neglect water resistanc how far does the baree shift in the horizontal direction?
1
i
14.121. phere at t droplet o The vel
\/
25’
-4
. ..... .
..- -.
2.5 kg Figure P.14.122. 123. Thin discs A and B slide along a :tior s surface. Each disc has a radius of 25 mm. Disc ~A has a mass of 85 g, whereas disc B has a mass of 227 g. What are the speeds of the discs after collision for c = .7? Assume that the discs slide on a frictionless surface.
I
. ,
.
,
_____~
Figure P.14.120.
A water droplet of diameter 2 mm is falling in the atmose rate of 2 d s e c . As a result of an updraft, a second water diameter 1 mm impinges on the aforementioned droplet. ity of the second droplet just prior to impingement is
I
_
; Figure P.14.123.
703
14.124. A BB i\ shot at the hard, rigid surface. The speed o l l l i c pellet is 3110 fisec a\ i t strikrs the suriace. If the directimi of the velocity for the prllel i s @\en hy thc i d l o w i n g unit vector: t = -.6i - .Xk what is thc final velocity \'ectoi of thc pcllet lor ii collision havinz E
14.127. Cmipnle thc angular cmonicntiim uhout 0 01a uniform Irrd, of Icngth L = 3 rii iind n i t w pcr unit Icnglh nr (11 7.5 hs/rn. ill l l i e i n a i i i f Nhcn i t i i \erticitl iind h a \ a n a n g ~ i l aspeed ~ ~ (0 (6 3 i-adlscc.
= .7?
0
7 -
I
k ,/
w
\
A
5'
Figure P. 14.124. !5. A chain 0 1 wrriught i1on, with length o f 7 111 and il n i a v of 100 kg, is held so thal i t jus1 touches the \upport AB. I l t l i e ~ h i i i r i i s rclcased, determine the total impulse driririg 2 sec in thc wrtic:il directinn cxpcrienced by thc support if the impiicl i\ plastic (i.u., the chain does not bounce up) arid i f we m w t : thz support \(I that thc links land UI the platfimn and 1not 011 each othcr'! IHirit: Notc that any chain wrsrin*. on AH dclivcrc a vcrtiCal impulsc. Also check 10 see if thc entire chaiii lands 01, A H hcfiire 2 sec.]
Figure P.14.125. 14.126.
Two trucks x e shown moving up ii I O incliiir. Tmck ,A weighs 26.7 kN and i s developing il 13.30-kN driving lorce on thc road. Truck R wzighs 17.8 hN and i \ curinccled with an incxtensihle cable to truck A. By operating a winch h, truck B approaches truck A with a conmnt acceleratim oi 3 m/xc.'llill lime f = 0 both r l u c k ~ have a speed of 10 mlsrc. what arc their ipeeil\ a1 timer = 15 scc?
Figure P.14.126.
704
Figure P.14.127.
Fieurc P.14.12X.
i
14.130. container stationary you will I ity of the as 101r f momentu
closed container is full of water. By rotating the or some time and then suddenly holding the container we develop a rotational motion of the water, which, am in fluid mechanics, resembles a vunex. If the velocuid elements is zero in the radial direction and is given sec in the transverse direction, what is the angular of the water?
1
14.132. A spacecraft has a burnout velocity Vo of 8,300 mlsec at an elevation of 80 km above the earth's sulface. The launch angle a is 1 5 ~What . is the maximum elevation h from the earth's surface for the spacecraft?
Figure P.14.132,
Ill'
Figure P.14.130. 14.131. dentical thin masses A and B slide on a light horimntal rod that i' attached to a freely turning light vertical shaft. When the masse; are in the position shown in the diagram, the system mtates at ;Ispeed w o f 5 radlsec. The masses are released suddenly from this position and move out toward the identical springs. which have a spring constant K = 800 Iblin. Set up the equation for the CO npression 6 of the spring once all motions of the bodie? relative tu the rod have damped out. The mass of each body ii 1 0 Ibm. Neglect the mass of the rods and coulombic friction Show that S = .OX361 in. satisfies your equation.
14.133. A set of particles, each having a mass o f 112 slug, rotates about axis A--A. The masses are moving out radially at a constant speed of 5 ft/sec at the same time that they are rotating about the A-A, axis. When they are I ft from A-A, the angular velocity is 5 radlsec and at that instant a torque is applied i n the direction of motion which varies with time f in seconds as torque = ( 6 1 + ~ l o t ) lb-ft What is the angular velocity when the masses have moved out radially at constant speed to 2 f t ?
Torque
Figure P.14.133. 14.134. A torpedo hoat weighing 100,000 Ib moves at 40 knots ( I knot = 6,080 ftlhr) away from an engagement. To gv even faster, all four 50-caliber machine guns are ordered to fire simultaneously toward thc rear. Each weapon fires at a muzzle veloc-
I
I
Figure P.14.131.
ity of 3,000 ftlsec and fires 500 rounds per minute. Each slug weighs 2 0%.How much is the average force o n the boat increased by this action'? Neglect the rate of change of the total mass of boat.
705
14.135 A devicc to be detunatcd with a small charge ir suspended in space [ser ( a ) ] .Dircctly after detonation. lour fragments are formed moving away from the paint of suspension. The following information is known ahout thcse fragments:
I Ihm v, = 200;
m , = 2 lhm
123
=
+
l0O.j f l k c IXOj
~
m , = 1.6 lhm V , = 200; + l S O j ~
=
100k Wsec
+
action i T it occurs in I rec'!
*14.138. I n the ,fission procebs i n a iiuclziir reactor, a risU nucleus first ahsorhs (11 captures a neutron lsee (a)]. A short tinic later. the "'U nucleu5 hreakr up into lission products plus ncu-
m, =
~
t o the war of 10 Wsec relative tu the initial spccd uf thc octopus. What horicpowcr is hcing dcvckrped hy the octopus in thc ahme
1X0k fUsec
troiis, which may suhsequently he captured by ulher 23'U nuclei and inaintain a :+ =
=
R + 2w x
v,,:+ 6 x p + w x
( w x pl
(1.7i- 2.023j + . 2 k ) t (2i + ? j - . 5 k ) + 2 ( . i k ) X (-.075i- i j + . S k ) + 0 x 11 + (.RkJ x ( . 3 k x . 2 5 j ) 5 . 5+ .909?j - .3k m/s2
Now we apply Newton's law to the inass ccntcr G at the imtmt of intercsl. Denoting the force fr0m the rod AD onto
u j = 5 radlsec'
e w , = IOradIsec o2= 15 radsec
R = 100jm/sec2 w = 101' radsec 6 = -8k rad/sec2 If a force F given as
F = 500i + 200j -3OOk N Figure P.15.121.
acts on a particle of mass I kg at position
+
p = .Si
Ij m
15.122. In Problem 15.91, what is the total external force acting on the cylinder for the CdSe when
what is the ac~elerationvector relative to the probe'? The panicle has a velocity V relative to xyz of V = IOi
+
v
= 5 ftlsec V = -2 ft/sec2 w, = 2 radlsec
20j d s e c
h,
= 1
rad/sec2'?
The mass of the cylinder is 100 Ibm 15.120. In the space prohe of Prohlem 15.1 19, what must the velocity vector Vr,,:of the particle be to have the acceleration a,&\:= 495.21'
+
lO0j dsec'
if all other conditions are the same? Is there a component of that can have any value for this problem?
15.123. A truck is at speed v of 10 ,,,i1hr. A crane AB is at time t at 0 = 45" with 0 = I radlsec and = .2 rad/sec2. Also at time f, the base of AB rotates with speed 0 , = I radlsec relative to the truck. If AB is 30 ft in length, what is the axial force along AB as a result of mass M of 100 Ihm at B?
e
yvz
15.121. A mash A weighing 4 oz is made to rotate at a constant angular speed of wz = 15 radkec relative to a platform. This motion is in the plane of the platform, which, at the instant of interest, is rotating at an angular speed 0 , = I O radlsec and decelerating iit a rate of 5 radisec2 relative to the ground. If we ncglect the mass of the rod supporting the mass A, what are the axial force and shear force at the base of the rod (i.e., at O)? The rad at the instant of interest is shown in the diagram. The shear force is the total force acting on a cross-section of the member in a direction trtngent io the section.
Figure P.15.123.
715
" .,," .I
..
.
......___^__I.
15.124. An cxplaratory pmhc shot from the earth ii rettoning 10 the earth. On cntzring the earth'\ atmnrpherr, i t ha* a conqtanl angw lar velocity coniponent (ti, of IO radlscc ahorit an axis nonnnl I n thr page and a conslatit comprment n i , of SO radlscc ahout the Lertical axis. The velocity of the pinbe at thc timc of interest i s I.300 mlsec vertically downward with a decclcration of I60 mlsec'. A wm11 sphcre i s rotating ill w, = 5 radlsec inside the pnrhc, as s h o w n At the time of intcrect, the prnhe i s orientrd sn that the trajectory [ifthe sphere in the probe is in the plane of the page and the arm i s \icrticaI. What are the axial force 111 the arni and the hending mirinenr nl i t \ h a w (neglect the mass of the arm) at thi\ instant of time, if the sphel-c has a mass of 300 p'! (The hending moment i s the cniiple rniinicnt acting on the c m s ~section of the hear".)
Figure P.15.124. A river flows at 2 ftlsec avei~agev r l x i t y i n Ihe Norlhi.iii Hemisphere at a latitude 0140" in the north -roulh direction. What i s the Coriolis acceleration of the walrr relative to the ccntcr of the earth'?What i s the Coriolir: force on I lhin 01 watcr?
15.125.
t ~ ~ ~ ~rzmh m ~= 7,'JLO ~ ~ ,111 c r ~ ~ l
Figure 1'.15.125. 15.126. A clutch aswmhly i s shown. R o d < A H are pinned Iu a disc at H, which Tolate\ at an a n p l n r speed w , = I radlsrc and i i , = 2 radlsec' at time I . .These rods extend lhnwgh a roil EF, which rotates with the rods and at the same time i s moving to the left with a speed 1'01 I mlsec. At the instant ?hewn. corresprmrl~ ing to time f, what i s the axial Iurce o n the niemher AH a c a result of the motion of particle A having a niass o f .h kg'?
7x0
I
SECTION 15.11
15.11 closure In this chapter, we first presented Chasles' theorem for describing the motion of a rigid body. Making use of Chasles' theorem for describing the motion of a reference xyz moving relative to a second reference XYZ, we presented next a simple but much used differentiation formula for vectors A .fixed in the reference xyz or a rigid body. Thus,
(--) dA
= w X A XYZ
where w is the angular velocity of xyz or the rigid body relative to XYZ. We next considered two points,fi.red in a rigid hody in the presence of a single reference. We can relate velocities and accelerations of the points relative to the aforementioned reference as follows:
yz
'1,
PCih
+
a,, = a BOUIES
P
Example 16.4 (Continued)
!
Part B. We next express the momenl-of-momentum equation Cor the rod at ;iny arhitrary position 0. Observing Fig. 16.14, we gel =
2 i
--,Lo 1 w 3
7.-
s
Therefore.
j
lcigure 16.14. (a) Rod
1
1'
ih)
(ill
a1 pmition
Consequently. at H = 45" we have = ( I .5)(.707l
Wc shall also nced
d, and
:
follows:
i
Separating variables, we get
f
i
dde
'L
H: (h) rod a1 0 -
= I .O6O
f
=
3 s .
2 -LoSHd8 1'
~
wlilch we integrate to get
:
When 8 = 0,8 = 0 ; accordingly. C = 0. We then liave
0'
(el
accordingly we now rewrite Eq. (d) as
!
i !
35".
=
3K Lc ' in8
(g)
i
: At the instant o f interest. H = 45" and we get for6': = 3 " ( 707) = 2.12 8
I.
i
ccnlcr
L
(h)
For H = 45", wc can now givc the acceleriltion component u I of the (if mass directed ~iormal10 the rod and coiriponent [I? directed i~long
SECTION 16.5 PURE ROTATION OF SLABLIKE tlOD1E.S
Example 16.4 (Continued) the rod [see Fig. 16.14(b)]. From kinematics we can say. using Eqs. (e) and (h): a, = 7 0 =
L"
7
L(
1.0601' = , 5 3 0 , ~
Now, employing Newton's law for the mass center, we have on noting that xy is no longer collinear with X Y .
A
( - a , sin 45" - a, cos 45') x - g
Therefore, using Eqs. (i)
A, = -1.124W Also, from Newton's law
-A,
+W
=
~
W (-a2 sin 45" + a, cos 45") g
Therefore, again using Eq. (i)
A,. = 1.375W
(j)
Consider next the case of a body undergoing pure rotation about an axis which, for some point A in the body (or massless hypothetical extension of the body), is a principal axis (see Fig. 16.15). For a reference xyz fixed at A with z collinear with the axis of rotation, it is clear that = I?, = 0, and hence the moment of momentum equations simplify to the exact same forms as presented here for the rotation of slablike bodies. p,,-Pfincipd
axis
r,, Cross ,section
X Figure 16.15. Axis z is a principal axis for point A.
803
h
I'ieure F.lh.5. Figure P.16.3.
Figure P.16.9.
16.10. Two cylinders and a rod are oriented in the vertical plane. The rod is guided by hearings (not shown) to move vertically. The following arc the weighls of the three bodies:
W,
= 1,000 N
W, =
300 N
W, =
200 N
What is the acceleration if the rod? There is rolling without slipping.
w Figure P.16.7.
16.8. A pulley A and its rofating accessories have a mass of 1,000 kg and a radius of gyration of .25 m. A simple hand brake is applied a$ shown using a force P. If the dynamic coefficient of friction between belt and pulley is .2, what must force P he to change w from 1,750 rpm to 300 rpm in 60 sec?
Figure P.16.10.
16.11. A har A weighing 80 Ih is supported at one end by rollers that move with the bar and a stationary cylinder B weighing 68 Ib that rotates freely. A 100 Ib force is applied to the end of the bar. What i s the friction force between the bar and the cylinder as a function of x'? Indicate the ranges for nrm-slipping and slipping conditions. Take p < = . S and ut, = .3.
P
1
"
l a 0 0 mm
~3 1
Figure P.16.8.
16.9. A tlywheel is shown. There is a viscoub damping torque due to wind and bearing friction which is known to he -.04w N-m, where w is in radisec. If a torque T = 100 N-m is applied, what i s the speed in 5 min after starting from rest'' The mass of the wheel is 500 kg, and the radius of gyration is .SO m.
Figure P.16.11.
805
I
Figure P.lh.16.
16.17. Do Problem 16.16 for the case where at the instant of interest wD = 2 rad/sec counterclockwise. 16.111. A torque 7of 50 N-m is applied to the device shown. The bent rods are of mass per unit length 5 kg/m. Neglecting the inertia of the shaft, how many rotations does the system make in I O sec? Are there forces coming onto the bearings other than from the dead weights of the system?
Figure P.16.20.
/-6OO
*16.21. In Problem 16.20, consider that the mass per unit length varies linearly from 5 Ibm/ft at I = 1 ft (at the bottom of the rod) to 6 Ibm/ft at I = 3 ft (at the top of the rod). Find T,, at any poiition r and then compute I,, for I = 1.5 ft.
mm
Figure P.16.18. 16.19. An idealized torque-versus-angular-speed curve for a shunt, direct-current motor is shown as cuwe A. The motor drives a pump which has a resisting torque-versus-speed curve shvwn in the diagram as curve H. Find the angular speed of the system as a function of time, after starting, over the range of speeds given in the diagram. Take the moment of inertia of motor. connectine shaft, and pump to be I .
16.22. A plate weighing 3 Ib/ft* is supported at A and H. What are the force components at H at the instant support A is removed'?
Torque
ie
Figure P.16.22. 0 Angular speed
8
Figure P.16.19. 16.20. Rods of length L have been welded onto a rigid drum A . The system is rotating at a speed w of 5,000 rpm. By this lime, you may have studied stress in a rod in your strength of materials class. In any case, the stress is the normal force per unit area of cross section of the rod. If the cross-sectional area of the rod is 2 in? and the mass per unit length is 5 Ibm/ft, what is the normal stress f, on a section at any position r'?The length L of the rods is 2 ft. What is 5,at I = 1 . 5 ft? Consider the upper rud when it is vertical.
16.23. When the uniform ngid bar is horizontal, the spring at C is compressed 3 in. If the bar weighs 50 Ib, what is the force at B when support A is removed suddenly'? The spring constant is 50 Ib/in.
A A
C
> fFigure P.16.23.
807
Figure P.Ih.27
16.28. A hollow cylinder A of mass 100 kg can rotate over a sedtionary solid cylinder B having a mass of 70 kg. The surface of cmtilct is lubricated s? that there is,a resisting torque between the bodies given as 0.28,4 N-m with 0, in radians per second. The outer cylinder is connected to a device at C which supplies a force equal to -SO?N. Starting from a counter-clockwise angular speed of I radlsec, what is the angular speed of cylinder A after the force at C starts to move downward and moves 0.7 m'!
a = 45"? The weight of the member A B is W. What is the angular acceleration of the bar for these conditions at the instant of interest'?
Y
\
Figure P.16.30. *16.31. A rod AB is welded t o a rod CD, which i n turn is welded to a shaft as shown. The shaft has the following angular motion ill time I : w = 10 radlsec h = 40 rad/sec:! Wire
Figure P.16.28.
I
What are the shear force, axial force, and bending moment along CD at time f as a function of r'! The rods have a mas? per unit length of S kgim. Neglect gravity.
-50 Y
16.29. A hox C weighing 150 Ib rests on a conveyor belt. The driving drum H has a mass of 100 Ibm and a radius of gyration of 4 in. The driven drum A has a mass of 70 Ibm and a radius of gyration of 3 in. The helt weighs 3 Ib/ft. Supporting the belt on the top side is a set of 20 rollers each with a mass of 3 Ihm, a diameter of2 in., and a radius of gyration of .8 in. If a torque T of SO ft-lb is developed on the driving drum, what distance does C travel in I sec starting from rest? Assume that no slipping occurs.
Figure P.16.31. 16.32. A four-bar linkage is shown (the ground is the Courth linkage). Each memher is 300 mm long and has a mass per unit length of I O kglm. A torque T o f 5 N-m is applied to ililch of bars AB and DC. What is the angular acceleration of hm AH and Cl)? R
I-
20'
Figure P.16.29. 16.30. A uniform slender member is supported by a hinge at A . A force P is suddenly applied at an angle a w i t h the horizontal. What value should t' have and at what distance d should it be applied to result in zero reactive forces at A at the configuration shown if
Figure P.lh.32
809
16.6
Rolling Slablike Bodies
We now consider the rolling witlinul slipping i i S slahlike hodies such as cylinders. spheres, nr plane gears. As wc have indicated i n Chapter 15, the point OS contact of the hody has insrantaneiiusly 3uro vrlociry, and we havc pure i i i . S I N I I I ~ I ~ P O rol(rriort I~S nhoirr t h i s uinfrrcr point. We pointed out that Sor getling velocities of points o n such a rolling body, we could imagine that there i s ii h i r r p at the piiinl 111 contacI. Also, the iiccclciatioti i i l the center (if a riilling withiiut slipping sphere or cylinder can he computed using the simple formula-KB. Finally. yini can rcadily show that if the angular speed i s zerii, wc c u i coinpiire the ;icceler;itinn CIS a n y pnint i n the cylinder o r sphere hy again imagining a h i n p at the point (IS cnntiict. Fix other cases. we must use more dctailcd kinematics, as discussed i n Chapter I S . A vcry important cnncIu~ionwe reached i n Chapter 15 Sor cylinders and spheres was (hat for rolling without slipping the acceleration of the colitact point o n the cylinder o r Yphere i s /(iwwrd thu ~ r o m ~ , . f i criitri-ii. ($ the i:diti(li,r 0 ) ,sph(,ri,. If the centel- of mas.; OS the hody lies anywhere along the linc A 0 froin the contact point A to the geometric center 0. then clearly we can usc Eq. 16.6 h r the point A . 'This action i h justified since point A i s thcn iiti example 01 case 1 i n Section 16.2 (A accelerates toward the mass center and i s part of the cylinder). Thus, for the body i n Fig. 16.16 t'nr n o slipping wc can use T = In iihiiot the point (if contact A nf the cylinder at the instant shown. However. in Fig. 16.17 we c;inniit do this hecause the point of contact A of the cylinder i s not accelerating toward the center ol.niass as i n the prcvitius case. We can use T = la about the i'eiili'r nf'nz(i.s.s i n the latter casc.
Figure 16.16. Point A ~ ~ C C C I C ~ C toward C
center of
rnasc.
We shall now examinc a problem involving rolling without slipping. The equations of motim, you can readily dcducc. arc ths same as i n the previous cection.
Figure 16.17. F'oinl A doer n o t a ~ ~ ~ l c r i towmrd iti. center of miass
SECTION 16.6 ROLLING SLABLIKE BODIES
Example 16.5 A steam roller is shown going up a 5' incline in Fig. 16.18. Wheels A have a radius of gyration of 1.5 ft and a weight each of 500 Ih, whereas roller B has a radius of gyration of I f t and a weight of 5,000 Ib. The vehicle, minus the wheels and roller but including the operator, has a weight of 7,000 Ib with a center of mass positioned as shown in the diagram. The steam roller is to accelerate at the rate of I ftlsec'. In part A of the problem, we are to determine the torque T,, from the engine onto the drive wheels.
Figure 16.18. Steam roller moving up incline.
Part A. In Fig, 16.19, we have shown free-body diagrams of the drive wheels and the roller. Note we have combined the two drive wheels into a single 1.000-lb wheel. In each case, the point of contact on the wheel accelerates toward the mass center of the wheel and we can put to good
Figure 16.19. Free-body diagrams of driving wheels and roller.
use thc moment-of-momentum equation (16.9) for the points of contact on the cylinders. Accordingly, we fix xyz to cylinder A and we fix another
___
81 1
8 12
CH.\PTEI< 16 K I N I T K ' S O F PI.ANF: hlO'IlON 0 1 ' RIGID RODIES
Example 16.5 (Continued) reference~iy: to cylinder N :it their respective points of contiict a s has heen shown." Hence. liir cylindcr A wc have
) ( I . s= ~ . ' --(I' ~
R
+ I.+~G,#
( h)
W e havc hcrc two equations and no fcwei- than five unknown?. By ciinsidcring the free hody of the vehicle minus wheels shoun diagrammatically in Fig. 16.20, we can say friini Newton's law noting once again that A, A , , etc., hecausc of the parallel orientation of axes XYZ with axes .ry: of Fig. I f i l Y ( a ) and Fig. I6.lY(h) at titile i
A ! - E x - 7,011(lsin 5" =
1-
7,000
s
(1)
(e)
~-1,
I'
Figure 16.20. Free-hdy diagram o i vehicle without w~heelsand mllcr. Finally. from kinematics wc can say:
x .1
-
= I1
= -
F,,
x
-~
r,#
= ..
'
2
=
-.s
riidlulc'
(d)
I = -.hh7 r a d k c '
1.5
where ,y = I ftlsec' i s the acceleration o f the vchicle up the incline. W e can now readily mIve thc eqoalirrnc. Wc get l3\ directly from Eq. (h) on
SECTION Ih.6 ROLLING SLAFLIKE BODIES
Example 16.5 (continued) replacing 8, by -.667. Next, we get AA from Eq. (c). Finally, going back to Eq. (a), we can solve for .and Trig, we have
(1,487)(2)- (660)(2.5)+ 7 , 0 0 0 c o ~ 5-~Ex,+ Bv(l I)+ 3,250 = 0 Therefore, E, = 1.155 Ib
Now from Eq. (g) we get A?: A v = 7,000cos.5"
+ 1,155 = 8,12X Ib
Finally, from Eqs. (e) and (0 we get N , and N2 N, = 8,128
+
N2 = - I , I55
1,000cos5" =
+ 5,000 COS 5" =
9,1201b
3,830 Ib
Hence, on each wheel wc have a normal force of 4,560 Ib, and for the roller we have a normal force of 3.830 Ib.
8 13
8 14 '4
CHAPTI;K 16
KINI:I'IC5
01; PLANK MiYIION 01: Rl(;lD r30011:.<
*Example 16.6 A gear A weighing 100 N i s connected to a stcpped cylinder H (scc Fig. 16.21) hy a light rod / I C . Thc stepped cylinder weighs 1 hN and has a radius of gyration or 250 iiiin aliing i t s centerline. Tile g e ~ hiis A a radius of gyration of 120 niin ;ilong l i s ccnterlinr. A fol-ce F = 1,500 N i s applied t o the gear ill I ) . What i s thc coniprcc\ive force i n mcmhcr D c i l , at the i i i s f a i i t that F i s applicd, Ihc system i s cr;ition;iry?
I
i! l
Noting that I ) ( ' is ;Itwo-Sorce compressiw mcmhcr. we drau the lrcc -body diagrams for the gear and the stepped cylindcr i l l Fip. 16.22.
x
Figure 16.22. R c e - h d y d i q x m o f f c a r and cylinder.
' The moment-of-momentum equations about tlie c o i i t x r points Sor both : hodie\ (points ( I and h. respectively) iire
:
.-.
.
.
,
...._I_^x_" ....
.... ... . .
~-
.
-
.
~
SECTION 16.6 ROLLING SLABLIKE BODIES
Example 16.6 (Continued) -(IO0
+ I,SOO)(.lS) + DC(sin20")(.15)
(DCsin20"
=
(g)[(.12)2
+ (.15)2]8A
+ I,000)(sin30")(.10) - (DCcos20°)(.30 + .10cos30°)
These equations simplify to the following pair: ,0513DC - 240 = ,3768,
(a)
+ SO = 1.398,
(b)
-.3462DC
Clearly, we need an equation from kinematics at this time. Considering rod DC, we can say:9 ac = a,
+ h j n cx PDC +
wIlc x (wDc x P i x )
e,k x [(.3O + .1Ocos3O")j - (.lOsin30")i] =
FDj+ &,,k
X
(cos 20"i - sin 20"j)
+0
The scalar equations are
-38668, -.058,
= =
.342hDc y, + .940h,,
(C) (d)
Also, from kinematics we can say. considering gear A :
..
Y~= ,
.neA
(e)
Multiply Eq. (c) by .940/.342 and rewrite Eq. (d) below it with %replaced by using 1%. (e): - I ,0638, = .940hlJ, -.OSOB - . 15eA = .940d1,~:
Subtracting, we get -1.0138,
+
=0
Therefore,
(fl
= 6.1Se,
Solving Eq. (a), (b), and
(0simultaneously gives us for DC the result
Also, note that Y cnmes out negative indicating that D accelerates downward. Wote that because cylinder B has zero angular velocity. we can imagine i t to be hinged at b for computing a? Also, we do not know the sign of and so we leave i t as positive and thus let the mechanics yield the correct sign at the end of the CdCUlationS.
FD
8 15
x
/
SECTION 16.7 GENERAL PLANE MOTION OF A SLABLIKE BODY
Example 16.7 Find the acceleration of block B shown in Fig. 16.24. The system is in a vertical plane and is released from rest. The cylinders roll without slipping along the vertical walls and along body B. Neglect friction along the guide rod. The I 50 N-m torque MA is applied to cylinder A . We first draw free-body diagrams of the three bodies comprising the system as shown in Fig. 16.25 where it will he noticed that we have deleted the horizontal forces since they play no role in this problem. As usual, X Y Z is our inertial reference. Points a,b, d, and e in F.B.D. I and F.B.D. 111, respectively, are contact points in the respective bodies where, we repeat, there is rolling without slipping. Furthermore, it should be clear that points u and b are accelerating toward respective mass centers. Accordingly we fix xyz to the cylinders at these points.
0 Guide rod
I
Data W, = 100 N W , = 300 N W,.=S0N MA= IS0 N-m F.U.D. I
F.B.D. 111
P.R.D. I1
Figure 16.25. Free-body diagrams rrf the system elements with horizontal forces
deleted. We may immediately write the moment-of-momentum equations for the two cylinders about their respective points of contact a and h. Thus
F.B.D. I
.. :. -31; - 165 = ,3448, F.B.D. 111 (f2)(.2)+(50)(.1) =
50 +--(.l)2pc R
(a)
Figure 16.24. A block and two c: in a vertical plane.
8 17
81 8
CHAPTER I6
KINETICS OF PLANE MOTION OF 131G11) RODlES
Example 16.7 (Continued) Next going to F.R.D. I1 we employ Newton's law since we have simplc translation. Refelmirig now Io tlic incrtid reference XYZ w e have
F.H.D. I1
Since the three bodies are interconnected hy nonslip rolling conditions we must next consider the kinematics of the system. Thus
..
38, =
.. Y,
.2 ti,. = - Y;, Using the ahove rcwlts to replace
G,, and gc.in Eqs. (a) and (b) wc get
N o w solve for.( and/z in thc above equations and suhstitutc into Eq. ( c ) We get
:.
= -24.10m/sz
Now from Eqs. (d) and (c) we can determinef, and f,.
,f, = -457.9 N
,f, = 21.07
N
Thus. cylinder A fcirccs hiirly H downward whilc cylinder C resists this motion. Notice, unless we want tu determine the friction fiirces at the walls therc i s no need to use Newton's law for the cylinders. Also note that we could not
use the moment-of-momentum equations for points ( ' and d of the cylinders even lhough there i s rolling without slipping there. The reason for thiq. :is you must know. i s that these points d o not rrccckmte towurd o r u w m f r o m t h mn,m cenrer,~of thc r$iriden.
.... . .. .
. .,..
__
,
.. .., . .
.
,
..
SECTION I6.7 GENERAL PLANE M O TIO N OF A SLABLIKE B O DY
Example 16.8 A stepped cylinder having a weight of 450 N and a radius of gyration k of 300 mm is shown in Fig. 16.26(a). The radii R , and R, are, respectively, 300 mm and 600 mm. A total pull Tequal to 180 N is exerted on the ropes attached to the inncr cylinder. What is the ensuing motion'? The coefficients of static and dynamic friction hetween the cylinder and the ground are, respectively, .I and .OX.
Y, Y
R . = 300 rnrn
I
Figure 16.26. (a) Stepped cylinder; (b) free-body diagram of cylinder. XYis
stationary.
A free-body diagram of the cylinder is shown in Fig. 16.26(h). Let us assume first that there is no slipping at the contact surface. Of course, we will have to later check this supposition. We have then pure instantaneous rotation ahout contact point A . Fix x y z to the body at A. X Y Z as usual is stationary. We can then say for the moment-of-momentum equation ahout the axis of contact: T ( / $ - R,)=
("s k 2 +;.:)e
wherein we have used the parallel-axis theorem for moment of inertia. Inserting numerical values. we can solve directly for 0 at the instant that the force T i s applied. Thus,
Therefore,
0 = 2.62 radlsec'
(b)
8 19
+
1x0
j
i ‘
!
7
wx
IC1
s
Thus. Sor no \lipping. we
n i t t \ t he ;ihlc t n ( k w l o p ;I friction Iorcc of 1117.9 N. Thc miixiiniini lriclion Irircz lh;it u’e can have. however. is. according
10
Coulomb’s law. = Wjl,
[
/
~
(450)(.11 = 45 N
Accordingly, u’c niii\r cnncliidc: that the cylindrr h m slip. and examine the prohlem ;IS :I , q m 6 , u d ~ ~ i ( ~ i ~ ~ ~prohlem. - i i i ~ ~ i ;
le! UK
nimt re-
~ ~ i ~
Using AI,, = .OX. wc now takc.jto hc 36 N and employ the momentof-momentum eqtl;ition for t h r rentrr of mas4 with w: iinw fixed ar the center of mass. We thcn h a w IF]?. l6.26(h)i
..
X = -3.14 rnlsec*
11)
SECTTON 16.7 GENERAL PLANE MOTION OF A SLABLIKE BODY
821
Example 16.9 A 4.905-kN flywheel rotating at a speed o of 200 rpm (see Fig. 16.27) breaks away from the steam engine that drives it and falls on the floor. If the coefficient of dynamic friction between the floor and the flywheel surface i s .4, at what speed will the flywheel axis move after 2 sec? At what speed will it hit the wall A? The radius of gyration of the flywheel is 1 m and its diameter is 2.30 m. Do not consider effects of bouncing in your analysis. Neglect rolling resistance (Section 7.7) and wind friction losses.
A
Figure 16.27. Runaway flywheel at initial position We assume slipping occurs when the flywheel first touches the floor (see Fig. 16.28).Newton’s law for the center of mass of the flywheel is (.4)N = (4;’ki:5j --
x ”
Therefore,
I N = 500g X = 3.92 mlsec,
Figure 16.28. xv fixed at initial position.
Integrate twice: X = 3.921
+
C,
X = 1.962t2 + C,t
+
C,
(a) (b)
At r = 0, .i‘ = 0 and X = 0. Hence, C, = 0 and C, = 0. The moment-ofmomentum equation for axes fixed to the body at the center of mass is next given.
Therefore,
8 = 4.51 rad/sec2 Integrate twice:
e = 4.51t + c, 0 = 2.261’
+ C,t + C,
(C) (dl
822
('HAPTER I6
KIVI:'I'ICS OF PL.hNk. blUrlON OF RI(;III tl0l)lF.S
Example 16.9 (Continued)
I-.*-
When I = 0. 6 = 0. ;ind 6 = -(200)(2n/f/hO) = -2O.W r;idlsec. llcncc. C., = -20.94 and C,; = 0. We now ask when does the slipping stop? Clearly, i t stops whcn thcrc i s :PTO w l w i l ~ 01 the point of contiicI 01' the c)linder."' Friinn kinematics we havc for this condilirm:
y Substituting from Eq. Eq. (e):
(:I)
3.92/
+
(2$!)8
and ( c j 111-k and
+ (2.$)(4.5 -
(el
=
d,
respectively. w~have 111-
I, -- 20.041
0
:
Thrrclore. I =
2.64 scc
Since we get a Lime tiers greater than Lcro. we ciin be iissurcd that (tic iiiiill ltic lime o i iiiitiiil no-slipping i s deduced f h n i Eq. (h). Thus.
tial slipping aswmption i s valid. Thc position X,
i I
9
X, s, = ( I .Oh2ji2.64)' = 13.67 111
Accordingly, thc flywheel hits the wall ((ficri t stiirts rolling withoof dipping. At f = 2 sec, thcrc i s s l i l l slipping. and U c ciiii L I ~ C Eq. (;I) t o find X at this instant. Thus.
(a,,,,,= (3.92)(2) = 7.85 mlsec The spred. once ihcre i s no further slipping. i s cwulaiit, ;ind a t the wall i s f o u n d hy using i = 2.64 scc in k1. ( i l l . Thus, (&,
= (3.92)(2.64) = 10.35 mlsec
5 0
the s p n d
16.33. A stepped cylinder is released from a rest configuration where the spring is stretched 200 mm. A constant force F of 360 N acts on the cylinder, as shown. The cylinder has a mass of 146 kg and has a radius of gyration of I m. What is the friction force at the instant the stepped cylinder is released? Take p > = .3 for the coefficient of friction. The spring constant K is 270 N/m.
16.35. The cylinder shown is acted on by a 100-lb force. At the contact point A , there is viscous friction such that the friction force is given as
f = .0sv, where % is the velocity of the cylinder at the contact point in ft/sec. The weight of the cylinder is 30 Ih, and the radius of gyration k is l ft. Set up a third-order differential equation for finding the position of 0 as a function of time.
Figure P.16.33. 16.34. A stepped cylinder is held on an incline with an inextensible cord wrapped around the inner cylinder and an outside agent (not shown). If the tension T o n the cord at the instant that the cylinder is released hy the outside agent from the position shown is 100 Ib, what is the initial angular acceleration'? What is the acceleration rd'the mass center? Use the following data: W = 300lb k = 3ft R, = 2ft R, = 4 f t p = .I
Figure P.16.35.
16.36. The cylinder shown weighs 445 N and has a radius of gyration of .27 m. What is the minimum coefficient of friction at A that will prevent the body from moving'? Using half of this coefficient of friction, how far d does Doint 0 move in 1.2 sec if the cylinder is released from rest?
x
45" Figure P.16.34.
Figure P.16.36.
82:
16.37. Thc >cIocitic\ B , are
~ W O point\
Vi = 6
111 /\
0 1 a cylindcr, namcly A and
y j = 2 111I\
What i s the velocity 01 pninl /I? Ii tlre cylinder hila il mass vi 4.2 kc, what i \ tlir ancolar iiccelcriition for il dynamic cucfficient
16.39. A light rod AH connect\ a platc C with il cylinder 0 which may roll without slipping. A torque 7 of \ d u e XI ft-I17 i s applied t o plate C'. What i s the angular accclcratiiin of cylinder I I when the torque i s applied'! The plate ueiphs 100 Ih arid the cyliiidcr weighs 200 Ih. I'
I'
Figure P.lh.39. 16.40. A semicircular cylinder A i s shou'tr. 'ihe iliiiniclcr ut A I C I fl. ;and the wright is 100 Ih. What i s thc mgukir :iccelcmtim of A at the QC,\ilihaft after I O 5ec i f the \y\trni i \ initially at rest'! The bent rnd has a milss pcr unit lcnplh of 5 hplm. Will there hc forces :an the hcil-ings 0 1 the shali other than tliuw from gravity'? Why?
16.56. A cable is wrapped around two pulleys A and U . A force T i s applied tu the end of the cables at G. Each pulley weighs 5 Ib and has a radius of gyration of 4 in. The diameter of the pulleys is 12 in. A body C weighing 100 Ib i s supported by pulley B. Suspended from body Cis il body /I weighing 25 Ib. Body 1) i s Iowered from body C s o as tu accelerate at the rate of 5 ftlsec' relative to body C. What force T i s then needed to pull the cable downward at C at the increasing rdte of 5 ftlsec'?
1 -
IhOmm
- 1
Figure P.16.53.
16.54. A tractor and driver has a mass of 1,350 kg. I f a total torque T of 300 N-m i s developed on the two drive wheels by the motor, what is the acceleration of the tractor"? The large drive wheels each have a mass of YO kg. a diameter of I m, and a radius of gyrativn of 400 nim. The small wheels each have a mass 20 kg and have a diameter of 300 mm with a radius of gyration of 100 mm.
A Figure
Figure P.16.54. 16.55. A hlock weighing 100 Ih rides [in two identical cylind e n C and D weighing 50 Ib each as shown. On top of block U is a block A weighing 100 Ib. Block A is prevented from moving to the left by a wall. If we neglect friction between A and U and between A and the wall and we consider no slipping at the contact surfaces of the cylinders, what is the angular sptcd ufthe cylinders after 2 sec for P = X0 Ib'!
Figure P.16.55.
16.57. A cylinder A is acted on by a torque T of 1,000 N-m. The cylinder has a mass of 75 kg and a radius of gyration of 400 mm. A light rod CD connects cylinder A with a second cylinder tJhavins a mass of 50 kz and a radius of I, ZWdtlOn of 200 mm. What is c the force in member CD when torque T i\ applied'! The system is stationari at the Assume no slipping the torque is of cylinder along the illcline.
Figure P36.57.
A
400 N Figuru P.lh.58.
Figure P.lh.61.
500 N 450
A A
Im Figure P.16.59.
S Bigure l'.lh.62.
828
16.67. In the preceding problem the disc is in the vertical plane and is held up by a horimrital surface where the coefficients ot frictian are ps = .005 and p,, = ,003. What are the initial linear acceleration of the center of niav and the initial angular acceleration of the disc? Start by assuming 110 slipping. 16.68. A cylinder A slide5 off the flatbed of a truck (Fig. P. 16.68.) onto the r o d with zero angular velocity. The niasi of the cylinder is 100 kg; its redius is I m; and its radius ofgyretion about the axis through the center of mass is 0.75 m. The coefficient of friction 1,) betwccn the cylinder and the pavement is 0.6. If marks on the pavemetit from the cylinder while it is sliding extend over a distance along the road of 3 rn. and if the axis of the cylinder remains perpendicular to the sides of the road during the action, what was the approximate s p e d of the truck when the cylinder slid off. Ncglcct the apeed of the cylinder relative to the truck whcn it slides off. [Hinr; What do thc pavcmcnt marks signify'?]
Figure P.16.64.
Neglect the mass of the slider at C, hut do not consider it to he frictionless. A strain gauge informs us that there is a torque of 20 ft-lb actine on rod AB at A .
Figure P.16.68.
16.66. A circular disc is shown with a circuhr hole. It rests on a frictionless surface and the view shown is from above. A force F = .MIb acts on the disc. The thickness of the disc is 2 in and the specific weight is 350 Ih/ft3. What is the initial linear acceleration of the center of mass and the angular acceleration of the disc'!
16.6Y. Three forces act on a plate resting on a frictionless surface, They are F, = )()() Ib, p2 = 200 Ih, and a fr,rce T~ = 35" (h ,,,hose direction @ i s t,> be determined so that the ,,late has a c O u n ~ terclockwice angular acceleration of ,204 red/st.ct. ~~~~~~~i~~ aiso the aCcelerillion for the centel of mBbS of the ,,late, The mass of the ,,late is 178 Ihnl.
Y
Y
I
t=2in 350 Iblfl'
Y=
tI4
6' = 2W Ib
Figure P.16.69.
Figure P.16.66.
16.70. In Pruhleni 16.63, what force P is needed to uniformly accelerate the cart so that the cylinder A tmoves 1 m i n 2 sec rclative to the cart. Cart R has a mass of 10 kg. Neglecl the inertia o f the small rollers wppolling the a r t , and assume there is nn slipping.
829
16.71. A wcdgc H i s shown u d h ii cylinder A of ~ r i i i i s20 kg and diameter SO0 mm o n the incliLw The wedge i s givcn a cimsLm1 acceleration of 20 mlsrc2 t o the right. How far d doe\ lhc cylinilci muve in scc relative tu Ihc incline i t there i\ 110 lipp pin^'! 'TI,? systcin starts froin rest.
Figure P.16.71. 16.72. A block weighing 100 N i s hcld h> I h i ~ e rincxlemihli. guy wires. What arc the toroes i n wires ,AC and H I ) at lhc i i i s l m l 1 1 ~ 1 wire EC i s cut?
'1 Figure P.16.72.
16.73. A buwlrr relzilscs his ball at il speed nf 3 m l c c c . If thc ball has a diameter ot 250 m n , what spin (I) h > u l d he ~ U on I ihc bull 50 there is 110 blipping? It he puts un unly half of this spin w. keeping the same speed of 3 mlscc, w h t l i\ thc final (teinminitl) speed of the ball? What i s thc speed after 3 x c ' ? .lhe ball hiis a "lass of 1.8 kg and a dynamic coefficient of triclioii with thr 1 1 , ~ ~ of .I.Neglect wind resistance :ind rolling resi\tance (a\ d i s c w \ e d in Section 7.7).
Figure P.16.73.
Figure P.16.76.
16.77. The great English liner, the Queen Elizabeth (QE 11). is the last transatlantic luxury ship left. All the others have been either scrapped or made into pleasure excursion ships. The QE I1 is 700 ft long and weighs 60,000 tons. (a) At a top speed of 4o knots, with the engines producing I 103000 what is the thrust coming from the propellers? (b) In a harbor, two tugboats are turning an initially stationary QE I1 as shown in the diagram. Determine an approximate value for the angular acceleration of the ship. Consider the ship to be a long uniform rod. Include an additional one fourth of the mass of the ship to account for the water which must be moved to accommodate the movement of the ship. Each tug develops a force of 5,000 Ib. (c) If the tugs remain perpendicular to the QE 11, and if we assume constant angular acceleration, how many minutes are required to turn the ship IO degrees?
16.79. A cylinder of mass 20 kg can rotate inside a block B whose mass is 35 kg. There is a constant resisting torque for this rotation given as 200 N-m. A horizontal 1,000-N force is applied to a cable firmly wrapped around the cylinder. What is the velocity of the block after moving 0.5 m'? What is the angular velocity of the cylinder when the block reaches this The coefficient ofdyn&c friction between block B and the flour is 0.3. The system is shown in a vertical orientation.
Y
Figure P.16.79.
Lx
16.80. A rod AB of length 3 m and weight 445 N is shown immediately after it has been released from rest. Compute the tension in wires EA and DB at this instant.
16.78. A cable supports cylinder A of mass 40 kg and then wraps around a light cylinder B and finally supports cylinder C of mass 20 kg. If the system is released from rest, what are the accelerations of the centers of A and C?There is no slipping.
Y
Figure P.16.80. 16.81. Rod AB is released from the configuration shown. What are the supporting forces at this instant if we neglect friction'! The rod weighs 200 Ib and is 20 ft in length.
Lx
Figure P.16.78.
Figure P.16.81.
83 I
Figure P.lh.Xh.
832
Figure P.16.88. Figure P.16.90.
16.89. We are looking down from above at a baseball player swinging his bat in a horizontal plane in the process of hitting a baseball at point B. The player is holding the bat such that the resultant of the force from his hands is at point A. At what distance d should the ball hit the bat to render as zero the normal force component from the batter's hands onto the centerline of the bat'? This point B is called the retrfrr of percussion. Because this kind of hit feels effortless to the batter it is often called the "sweet cpot" by athletes in both baseball and tennis. The radius of gyration ahout the center of mass is k..
16.91. In Prohlcm 15.48, wjc determined the following results from kinematics: wAx = -509 rads
V, = 4.3X mis
hA,,= 14.71 radlszc'
&!
= -31.28 d s 2
The mass rif rod AB is 1 0 kg. If we neflect the muses of the sliders, what are the forces coming onto the end pins of the rod'! The horizontal slot in which slider A is moving is frictionless.
x
Y
Figure P.16.89.
16.90. A torque 7 = 10 N-m is applied to body C. If there is no slipping, how many rotations does cylinder B make in I sec if the system starts from rest'? A is stationary at all times. The system starts from rest. We are observing the system from above. The following data apply. Mc = 50 kg
k, = .2m
MB = 30 kg k,* = . I m
A
Figure P.16.91.
16.92. In the preceding problem. the ilider at A no longer moves in the slot without friction and we do not know the friction f k c e there. However, we have a strain gage mounted nn rod AB giving data indicating a 200 N compressive axial force at A . Using the data of the previous problem, compute the force components at the ends of the rod.
x33
16.8
Pure Rotation of an Arbitrary Rigid Body
We now consider ii body having an arhitriiry d i h h u t i o n of mass rotating a b ~ u t an axis of rotation fixed i n inertial space. We consider this axis to he the : axis fixed i n the body a\ well a h being an incrtial ciiordiiiate axis %. We can take the origin ofr:~;anywhere along the :axis since all such points are fixed i n i n c l t i d space. The , I I O n ? P , 1 1 - 0 f l f l O I n r l l f l ( n l equations tu be used will now hc the general equations 16.6 hilice I, nnd I ~w> i l l generally 1101 equal zerti. I f the center of inass i s along the :axis. then i t oh\,iously has no acceleration. and 50 we can then apply tlic rulcs 01 statics to the ccntcr of mass. For othei- cases we shall often need t u use N(WIOII'SI r i w iur tlic cciiter of niass. I n this !regard i t w i l l he helpful to nole lroni the definition 01 the centcr of mass that ibr a \ysteiii of rigid bodies such ii\ i\ i h o w n i n Fig. 16.29 ~
where
,)I,
i s the mass iif the ith rigid body,
r, i s the positioii vectnr to the ceni \ the position vecon differrentiatiiig:
ter of mass of the ith ]rigid hndy, M i s the tnlal mass, and tnr LO Ihe ~ e i i l e rof n i a h h of the system. We can then say
<
111 Ncw~on'.,Iuw fiir the mass czntrr (if ii bystcni of rigid hodics. wc coiicludc that we can use the centers nf inass 01 the component parts of the system as given on h e right side of Eq. 16.18 rather than the ccntcr n l inash of the Iutiil
Inas\.
SECTION 16.8 PURE ROTATION OF AN ARBITRARY RlGlD BODY
Example 16.10 A shaft has protruding arms each of which weighs 40 Nlm (see Fig. 16.30). A torque T gives the shaft an angular acceleration h of 2 rad/sec2. At the instant shown in the diagram, o is 5 radlsec. If the shaft without x.
x
Figure 16.30. Rotating shaft with arms
arms weighs 180 N, compute the vertical and horizontal forces at bearings A and 5 (see Fig. 16.31j. Note that we have numbered the various arms for convenient identification. x,
x
I
Figure 16.31. Supporting forces
We first fix a reference xyz to the shaft at A. Also at A we fix an inertial reference XYZ to the ground. We can directly use Eqs. 16.6a and 16.6b about poitit A . For this reason, we shall compute the required products of inertla of the shaft system for reference xyz. Accordingly, using the parallelaxis theorem, we have: (/xz
=0
+ 0(.3[(.6)(.3)] 8
Hence, for the system, I,, = ,440 kg-m2. We next consider I.,. Accordingly, we have
= .440 kg-m2
835
836
CHAVIEK 1 6 KINETICS OF PLANE MOTION OF R I G I D HOI)I~.S
Example 16.10 (Continued)
KJ,,,~,,= n + o
=
o
Hence, for the system, 5,. = 4.62 kg-in'. We can now cmploy the moment of momentum equations (Eqr. 16.6) to get the required moment:, M, and M,,about point A needed fol-the motion we are considering. Thus, we have MA = -(2)(.440)
+ (S'J(4.62)
M, = -(21(4.62)
-
= 114.7 N-m
(5')(.4401 = - 20.2 N-III
Summing moments of all the forces acting on the system ;ihout the at A , we can say (see Fig. 16.3I!:
(a)
(hi axis
M3 = -20.2 = -~40)(.60J(.601 - (40)(.60)(1.9)
-(40)(.60)(2.7)- (180)(1.6) + ( B , l(3.2)
Therefore, we require
i
Summing moments about the s axis at A. we can say M r = 114.7 = -B>(1.?)
Therefore, we require
Wc next usc Newton's law considering the three arms to be three particles at their mass centers a s ha\ heen shown in Fig. 16.32. In the ~i
SECTION 16.8 PURE ROTATION OF AN ARBIlRARY RIGID BODY
Example 16.10 (Continued)
Figure 16.32. Arms replaced by mass centers.
direction at time f we have, using Eq. 16.18 and noting that each of the aforementioned particles has circular motion: 118.9 + Ax - 180 - (3)[(40)(.60)] = - (40)(.60) (,30)(02) 8
Setting w = 5 radsec and Ci, = 2 radsec*, we get
In they direction, we can say similarly at time
f
Av - 35.8 = 0"0)(.30)(h) R
- --(.30)(w2) (40K.60) R
- (40)(.60) (,60)(02
R
Therefore,
The forces acting on the shaft are shown in Fig. 16.33. The reactions to these forces are then the desired forces on the bearings. In the z direction it should be clear that there is no force on the bearings. 118.9 N
llON
35.8N
17.78 N A
B
Figure 16.33. Forces on shaft.
837
Ii.i n the last cxamplc. wc had ignored the constant lorcch o i p i v i l y . wc would lriivc dclcrmincd lorcc\ at hearings A and /I that are due entirely 1 0 the nitrlion 01 the hody. Forces c~~nipiited in thi< \\':I)' ai-c ciillcd h.iwnri~.,fiin.~,.\. It tlrc hody a c r e rotating with constant speed w. lhcse lorce\ would clearly h i i constililt ~ viilucs in the .c and y directions. Since the i~ axes are riitaling with the hody rclativc to the ground 1-efercncc XYZ. such dynamic forces inust d s o rotate relative to the ground ahout the a x i s 01rotation with the spccd (oot thc hody. This iiieans that. i n any l i r r i l direcfioii iroriiiiil 10 the h i l i ill ii hcariiig, llicre w i l l be a . ~ ; I I ~ . ~ ~ i~w ;r i~< /i i~i with ~~~~~ a~ , frequency / ~ J ~ ~ correaponding ~ ~ ~ to the angular rotiitioii 0 1 the shaft. Such force\ can induce vibrations 11i Iwge ;iiiiplitudc i n lhc s~rucxiircor support i l ii natural irequency or multiplc 01 a natural frequency i s reached in these hodie?.' I Wlrcn a shall creates rotating iorces on the hcxings hy virtue 01' i t s own rotation, the shah i s m i d 111 h e uiibalanced. We shall set up critcriii lor halancing ii rotating hody i n Ihe next scclion
Balancing
"16.9
We shall iiow \el iorth lhc critcriii ior the condilion of dynamic bal;incc i n ii I-otaling body. 'l'lic~i. w e h l l XI iorth llic rcq~iirc~iicnts needed to achie\c halance i n a rotating body. Consider thcn ~ m i arhitl-ary c rigid body rotating with angular speed (0and ii iriitc of change 0 1 angular speed ahout axis AH (Pig. 16.34).We shall set up general cqualioiis lor dctcnnining the supporling iot-ces at the hearings. Considci- point ( i on llic i i x i s o f rotation at the hc;iring A :urd establish a \et o l iixcs I.!: l i x r d 10 the rotating hody wilh the :iixis corresponding ttr !he u i h o l rotation Thc .( nnd !axes are chosen lor convcnience. Axes XYZ are, iis usi1~1.inertial iixcs. Ilsing the i i i ( , l l i n , i ~ ( 1 - i ~ i , i I 1 P l i f i i l ) i equations ( a ) and ( h ) i n Eq. 16.6 and including only (Iyiiarriic lorces. we get ior p i n t
1,:
/?>I= -I$ H /
rz
+ /$
-1 v: (i, . ~I.I; (01
( I 6. I%I I
(Ih.lc)b)
SECTION 16.9 BALANCING
If the axis of rotation is a principal M ~ S at hearing A, then {,z = 5, = 0 in accordance with the results of Chapter 9. The dynamic forces at hearing B are then zero. Next, we shall show that if in addition the center of mass lies alonx the axis of rotation, this axis is a principal axis for all points on it. In Fig. 16.35 a set of axes x’y’r’ fixed to the body and parallel to the xyz axes has been set up at an arbitrary point E along the axis of rotation. We can see from the arrangement of the axes that for any element of the body dm:
z‘= D+z
x’ = x,
y‘ = y.
(16.20)
Figure 16.35. Reference x’v’z’ fixed at E.
Also, we know for the xyz reference that lvz= I M y z d m = 0
xzdm = 0,
(16.21)
And if the center of mass is along the centerline, we can say:
My, = 0
I M y d m = I,y’dm
=
I,xdm
= Mxc = 0
= I,x’dm
(16.22)
We shall now show that all products of inertia involving the z’axis at E are zero under these conditions and, consequently, that the z’axis is a principal axis at E . Substituting from Eqs. 16.20 into 16.21, we get Djdm = 0
I,x‘(z’-
jMy,(z’-Djdm=0
(a) (b)
If we cany out the multiplication in the integrand of the above Eqs. (a) and (b), we get
1,
x’z’ dm
~
DIMx’ dm = 0
y’z’dm - D J y’dm = 0 M
(C)
(d)
839
As a result of Eq. 16.22. thc sccond integrals of Eqs. (c) and (d) are zero, and w e conclude that the product5 of inertia I1..and I>,,.,arc Lero. Now tlie.ty axes and hence the ~ n i ' iixcs can have irny orientiition as long a s they are norinal to the axis of rotiitim. This means that at I , If the weights are olaced in these olanes at a distance of I ft from the axis of rotation, determine the value of these weights and their position relative to the q z reference. We have two unknown weights and two unknown angles, that is, four unknowns [see Fig. 16.37(b)], to evaluate in this problem. The condition that the mass center he on the centerline yields the following relations:" ~~
L
When the numerical values of r,, r2,etc., are inserted, these equations become
w,coso, +
W , ~ O S ~ ,=
13.18
(a)
W, sin 6, - W, sin 6, = -6.77
(b)
Now we require that the products of inertia ivzand lxz be zero for the in the balancing plane B.
xyz refere,nce positioned so that xy is
I_ = 0 : W !R ( h ) ( - r , sin 20") + -2WK ( 2 ) ( r zsin 450) + 3 (9)(r3s i n e 3 ) = o R
(c)
ly7= 0 : 1W- ( 6 ) ( r ,cos20")+--?(2)(rzcos45")+-(Y)(-r,cos8,) W w1 K R 6
=0
(d)
Equations (c) and (d) can be put in the form 9r% sine, = -6.71
9w1 COS^, = 45.2 '>We are considering the weights to he pilnicles in this dircursion. In solile homework problems you will he asked to halance rotating systems fhc which the partick model will not he proper. You will then have to carry out integrations andlor employ thc formulas and trani~ fer theorems for first moments of m a s and products of inrrlia.
n
Balancing
A
,/
planes
,-
y'wi
~
'I
1- -r;,
I
...
3'
>
I _ / . +w4 2'
(a)
(h)
Figure 16.37. Rotating system tu he balanced.
842
CHAPTER 16
K1NI:TICS OF PI.ANF. MOTION 01: Kl(ill1 ROD1F.S
Example 16.11 (Continued) Dividing Eq.
(flinto Eq. (e).
wc gct
Ian H3 = -. 1486
e,
= 171.~0 or 351.6"
a n d so. from Eq. i t ) .
In order to have I positive weight W,. we chose 0, to he 35 I .6" rather than 171.5". Now we return 10 Eqs. (a) and (h). We can then say. on suh\tituling known values of W, and 8,: W ~ C ~ , S C ) ,=
W, sin
13.18-5.01 - 8 . 1 6
= 6.77 - .75 = 6.02
ipl
ihJ
Dividing Eq. (g) into Eq. (hJ. we get tan 0, = ,738
e,
= 36.4" or 216.4"
Hence. from b q (g), we have
where we iise b,' = 36.4" rather than 216.4" to prevcnt a negative W,. Thc linal orientation of the halanced system is shown in Fig. l6.3X.
Figure 16.38. Balanced sysit.m.
16.93. A shaft shown supported by bearings A and B i s rotating at a speed w of 3 radisec. Identical blocks C and D weighing 30 Ih each are attached to the shaft by light structural members. What are the hearing reactions in the x and y directions if we neglect the weight of the shaft'?
X
2'
16.95. Do Problem 16.94 for the case i n which w = 20 rad/sec and h = 38 rad/icc2 at the iiiitant of interest as shown. 16.96. A uniform wooden panel is shown supported by hearings 4 and 8.A 100-1h weight i s connected with an inextensihle cable to the panel at point G over a light pulley I). If the system is released from rest at the configuration shown, what is the angular acceleration of the panel, and what are the forces at the bearings? The panel weiFhs 60 Ib. x
F 3 ' 4 /
12,,
Figure P.16.93.
16.94. Shaft A B is rotating at a constant speed w of 20 radlsec. Two rods having a weight of I O N each are welded to the shaft and suppolr a disc D weighing 30 N. What w e the suppolting forces at the instant shuwn'! 100 Ih
Figure P.16.96.
I
I
16.97. Do Problem 16.96 when there is a frictional torque at the hearings of 10 ft-lh and the pulley has a radius of I ft and a moment of inertia of IOm-fl'. 16.98. A thin rectangular plate weighing SO N is rotating about its diameter at il speed w of 25 radlsec. What are the supporting forces in the .r and s directions at the instant shown when the plate is parallel to t h e y plane?
.r
I
300 mm
4
YO0 mm
y
2
Figure P.16.94.
Figure P.16.98.
84:
16.99. A shaft is shown rotating at a speed of 20 radlsec. What are the supporting forces at the bearings? The rods welded to the shaft weight 40 Nim. The shaft weighs XO N.
16.101. A bent shaft has applied to it a torque 7 including grav~ ity piven iis
7 = IO
+
5 r N-m
where 1 is in scconds. What are thc supporting forces at the bearings i n the r and? directions when i = 3 S C C ? The shaft is made from a rod 20 inin in diamctcr and weighing 70 Nlm. At i = 3 sec. the pnsilion of the .;haft i h a b shown.
i
Fipure P.16.99.
Figure P.16.101.
16.100. A cylinder is shown mounted at an angle 0130" to a shaft. The cylinder weighs 400 N. If a torque T of 20 N-m is applied, what is the angular acceleration of the system? What arc the supporting forces in the x and j directions at the configuration shown wherein the system is stationary! Neglect the mass of the shaft. The centerline of the cylinder is in the x7 plane at the instant yhown.
[he right thz bearing at A ' ? ~ l what~ arc ~ , ,nr,mcnt a,,d shear r,,rces ~h~ \haft and rr,ds have it diameter 0120 nim and a weight per unit lrnpih of50 N / m
Figure P.16.100.
Pigure P.16.102.
*l6.102. A ?halt has an angolar velocity o of 10 rad/scc and an accclcratlOn 0 "1 5 rad/scc: at the instant ],,
(17.12)
For angu1;ir nioiiietiluiii. we CUI say 1.111- the dcl(irmalion period using I- a s the distancc Iron? the center of m a s s to PI-11:
ri
= i(d)lj - i(O,
(17.13)
Similarly. for the period (11 resliltition:
Ij K i l t
=: I ( O ,
-
/(o,j
(17.34)
Now, suhstitutc thc rifht \ides of Eqs. 17.3 I and 17.32 into Eq. 17.29. We gel (in cimcclliitioii 0 1 M :
Next. suhslitiite lor lhc inipulscs cet 011 canceling oiily I :
ill
Eq. 17.29 iihing Eqh. 17.33 iind 17.34. We
c
(17.36)
SECTION 17.6 IMPULSIVE FORCES AND TORQUES: ECCENTRIC IMPACT
Adding the numerators and denominators of Eqs. 17.35 and 17.36, we can then say on rcaranging the terms:
We pause now to consider the kinematics of the motion. We can relate the velocities of points A and C o n the body (see Fig. 17.36) as follows:
VA =
v,. + w x pcA
(17.38)
Note that the magnitude of pcA is R as shown in the diagram. Since w is normal to the plane of symmetry of the body and thus to pCa, the value of the last term in Eq. 17.38 is Rw with a direction normal to R as has been shown in Fig. 17.36. The components of the vectors in Eq. 17.38 in direction n can then he given as follows:
(v,),, = (y.)n+ o R c o s 0 Since R cos
e= r (see Fig.
(17.39)
17.36), we conclude that
(V,),,
=
(V,), + rw
(17.40)
With the preceding result applied to the initial condition (i), the final condition 0, and the intermediate condition ( D ) , we can now go hack to Eq. 17.37 and replace the expressions inside the braces (( )) by the left side of Eq. 17.40 as Follows: ( I 7.41)
A similar procers for the body having contact point B will yield the preceding equation with subscript B replacing subscript A:
Now add the numerators and denominators of the right side of Eq. 17.41 and the extreme right side of Eq. 17.42. Noting that
we get Eq. 17.30, thus demonstrating the validity of that equation. Let us next consider the case where one or both bodies undergoing impact is constrained to mrate about a ,fued axis. We have shown such a body in Fig. 17.37 where 0 is the axis of rotation and point A is the contact point. If an impulse is developed at the point of contact A (we have shown the impulse during the period of deformation), then clearly there will be an
Figure 17.36. Slab with lpcAl as R.
897
898
CHAPTER 17
ENERGY AND IMPULSE-MOMENTUM METHODS FOR RIGID BOL>II?S
impulsive force at 0, as shown in the diagram. We shall employ the uiigulor inipulsr~momriirumequation ahout the fixed point 0. Thus. we have for the period of delormatioir and the period of restitution: r j D ~ =I /,,wf, -
/p,
r j K d f = /,,of- I,,o,,
Now solve for the impulses in the equation ahove, and substitute into Eq. 17.29. Canceling I(,, the moment of inertia about the axis of rotation at 0, we get
Figure 17.31. impact for a body under constraint at 0 .
In Fig. 17.38 we see that V,, = Rw
Thcrcfore. (V,, I,, = Rw cos 0 =
IUO
Using thc above result in Eq. 17.44, we get
Figure 17.38. Velocity of point A is R w
But this expression is identical to Eq. 17.41. And by considering the second hody. which is either frce or constrained to rotate, we get an equation corresponding to Eq. 17.42. We can then conclude that Eq. 17.30 is valid for impact where one or both bodies are constrained to rotate about ii fixcd axis. In a typical impact problem, the motion of the bodies preimpact is given and the motion of the bodies postimpact is desired. Thus, thcrc could bc four unknowns-two velocities nf the mass centers of the bodies plus t w o angular velocities. The required equations for solving the problem are formed from linear and angular momcntum considerations of the bodies taken separately or taken its a system. Only impulsive forces are taken into considefiitioii during the time interval spanning the impact. I S the bodics are considered separately, we simply use the formulations of Section 17.5, rcmenibering to observe Newtiin’s third law at the point of impact between the two bodies. Furthermore, we must use the coefficient of restitution equation 17.30. Generally. kinematic considerations are also needed to solvc the problcm. Whcn there are n o other impulsive forces other than those occurring at the point of impact, i t might he profitable to consider the hodies as one system. Then. clearly, as a result of Newton’s third law, we must have cons~rvufionof lifteur momenrum relative to an inertial reference, and also we must have cuiiservafion of angular mumrnfum about any one axis fixed in incrtial space. We now illustrate these rcmarks in thc following scries of examples.
SECTION 17.6 IMPULSIVE FORCES AND TORQUES: ECCENTRIC IMPACT
Example 17.11 A rectangular plate A weighing 20 N has two identical rods weighing I0 N each attached to it (see Fig. 17.39). The plate moves on a plane smooth surface at a speed of 5 mlsec. Moving oppositely at 10 mlsec is disc B , weighing 10 N. A perfectly elastic collision (E = I ) takes place at G. What is the speed of the center of mass of the plate just after collision (postimpact):’ Solve the problem two ways: consider the bodies separately and consider the bodies as a system
5 dsec
Figure 17.39. Colliding bodies.
We can consider the disc to be a particle ce this tiun only translate. Let point G be the point of contact on the rod. U have, for Eq. 17.30:
will then
Therefore, (VGIi - ( V B ) , = 15
(a)
Now, consider linear and angular momentum for each of the bodies. For this purpose we have shown the bodies in Fig. 17.40 with only impulsive forces acting. We might call such a diagram an “impulsive freebody diagram.” We then see that C, must move in the plus or minus y direction after impact. We can then say for body A, using linear impulse-momentum and angular impulse-momentum equations (the latter about the center of mass):
899
Y
I For body /I we
Iiii\,c
Sur the linear impulse-momentum equation. tdi
We have iiinv solvc. We get:
ii
coiiiplcte set of equations which wc can
iv,. Solution 2.
j
1 1
Equation ( a ) ahovc
= ~-.305lll/sCc
Stoiii
the coellicieiit of rcstitiilioii and Eq.
( e )ahove Srmi kirieinaticc a l s o apply to solutioii 2. Suhctitiitiiig 111!Q, iii Eq. (a) of scilutiun I using Eq. ( e ) of soliiticvi I .
iv,
)#
,I
+ .130Jl -(L')#),
M.C
=
get lor snlutioii 2:
1s
!ai
Conservation US linear niornentuin for the system leads to thc I-equire~iientiii t i l e y dircction that
Wiercforc.
-uL;.l,i,
i iL,#I,
-
~111
(hl
Alsu angular momentum is conserved about any fixcd 2. axis. We choose the axis at the positioii corresponding lo (1, at LIic lime (if itlipact. Noting that I for the plate and i i i m h i s . ( M Y 6 kg-311~ f'ioin s o l u t i o n I (sec liq. ( e ) ) and noting that R caii he considcicd a s a paiticle. w e have
SECTION 17.6 IMPULSIVE FORCES AND TORQUES: ECCENTRIC IMPACT
Example 17.11 (Continued) ~(IO)(.l3= ) (.0496)0, R _
c
_
-
+ -(Vflj,(.13j 10 R
""gUIU
mgular
mumenium
mOmenfYm p"stimpacr
prrirnpacr
Therefore.
w A = -2.67(Vfl),
+ 26.7
Solving Eqs. (a), (b), and (c) simultaneously we get:
We leave it to you to demonstrate there has been ronservutiun .f mechanicd e r i e r p during this perfectly elastic impact. We could have used this iact in place of Eq. (a) in solution I .
Note that there was some saving of time and labor in using the system approach throughout for the preceding problem wherein a rigid body, namely the plate and its arms, collided with a body, the disc, which could he considered as a particle. In problems involving two colliding rigid bodies neither of which can be considered a particle, we must consider the bodies separately since the system approach does not yield a sufficient number of independent equations as you can yourself demonstrate, In the preceding example, the bodies were not constrained except to move in a plane. If one or both colliding bodies is pinned, the procedure for solving the problem may be a little different than what was shown in Example 17.1 1 . Note that there will he unknown supporting impulsive forces at the pin of any pinned body. If we are not interested in the supposting impulsive force for a pinned body, we only consider angular mowirrilurn about the pin for that pinned body; in this way the undesired unknown supporting irnpulsive forces at the pin do not enter the calculations. Other than this one factor, the calculations are the same as in the previous example. You will recall from momentum considerations of particles that we considered the collision of a comparatively small body with a very massive one. We could not use the conservation of momentum equation for the collision between such bodies since the velocity change of the massive body went to zero as the mass (mathematically speaking) went tu infinity, thus producing an indeterminacy in our idealized formulations. We shall next illustrate the procedure for the collision of a very massive body with a much smaller one. You will note that we cannot consider a system approach for linear or angular momentum conservation for the same reasons set foi th in Cliaptcr 14.
901
In cithcr casc, lhc vclocity 0 1 cnd B preimp;ut is Vlj = ,\?,y/I =
t
( 2 ) ( 3 2 . 2 ) ( 2 )= 11.35 ft/scc
,,
i j we now consider each cake sepal-;rrcly.
I
Case A.
I i
I
Equation 17.30 can be uscd here. ' I h u \ .
: Thcrcforc. iil I
[ ( v , ~ ) , ] = I 1.3s it/\ti.
Ncxl, considering rod AB i n Fig. 17.42 we 1 1 3 ~l1)r ~ . linear impulse- ;Ind angular impulse-momentum considerations (the latter ahiiut an axi\ at the center of mass).
pi/,10i I!,.)( h'
-(2)(cos30")Jmr From kinematics.
(h)
(-I 1.3511
= ' '
-
=
(
i I2 20
L Fieure 17.12. Inlpuisivr lirce-hud) diagrmi
=
(V(.Il
+ w,
i
I
)(qu,
we have" lVl$J,
I
?
x I)(.)+
[ ( V , , i ~ ] l i + 1 1 . 3 S=j i V , j , j + m f k x (ZJ(-.X66i-.5j)
[1 V ) ] i + I I3 S . j H
I
/
= ( V,, I
j
-
,
I , 7 3 2 ~ 0j
+ (u
,i
i t
903
SECTION 17.6 IMPULSIVE FORCES AND TORQUES: ECCENTRIC IMPACT
Example 17.12 (Continued) We now have a complete set of equations considering F dt as an unknown. Solving, we have for the desired unknowns:
wI
=
A
-9.07 radlsec
[ ( V B ) x ] l= -9.07 ftlsec
You can demonstrate that energy has been conserved in this action. We could have used this fact in lieu of Eq. (a) for this problem.
Here we have no slipping on the rough surface and zero vertical movement of point E . Accordingly, we have shown rod EA with vertical and horizontal impulses in Fig. 17.43. The linear momentum equations for the center of mass then are Case B.
The angular impulse-momentum equation about the center of mass is -2(cos30°)j
F, dt + 2(sin30°)1 F2 dt
=
From kinematics, noting that at postimpact there is pure rotation about point E , we have [(VC)J,=
C COS ~O")(W,) =
1.732~~
[(V,),], = -2(sin 30")(0,) = -a, Substitute for F, dr and
1F2
(i)
dr from Eqs. (0 and (8) into Eq. (h). We get
on then employing Eqs. (i) and (i): (1.732~,
Therefore,
0)
+ 11.35) + (2)(.5)(?)(-0,)
=
S Figure 17.43. Impulsive free-body diagram.
0
-1 Y
I
H
600 mm
4
G
I)
I
x
&
Figure P.17.71).
17.72. A stiff bent rod is dropped so that end A strikes a heavy table 11. If the impact is plastic and there is no sliding at A, what is the postimpact speed of end R'? The rod weighs per unit length
mass of the drumstick is 200 g. Idealize the drumstick as a uniform slender rod.
I 300 mm I M
'I
~
Figure P.17.72.
70 mm
(h)
Figure P.17.74.
17.73. Solve Problem 17.72 for an elastic impact at A . Take the surface of contact to be smooth. Demonstrate that energy has been conscrved.
17.74. A drumstick and a wooden learner's drum are shown i n (a) of the diagram. At the beginning of a drum roll, the drumstick is horizontal. The action of the drummer's hand is simulated by form components I;; and F,. which make A a stationary axis of rotation. Also, there is a constant couple M of .5 N-m. Ifthe action starts from a stationary position shown, what i \ the frequency of the drum roll far perfectly elastic collision hetween the drumstick and the learner's drum? The
17.75. A hlock of ice I It x I ft x $ ft slides along a surface at a speed of I O ftlsec. The hlock strikes slop D in a plastic impact. Does the block turn over after the impact? What is the highest angular speed reached? Take y = 62.4 Ib/ft2.
I'
r) Figure P.17.75.
905
17.76. A hnrimntal rigid rod i s dropped from a height IO fi above a hcavy tahle. The end of the rod collides with the tahle. II the coefficient nf impact t hetwcen the end 1 the rod and the cor ncr of the tahle i s .h, what i s the postimpact angular velocity of the rod? Also. what i i the velocity n1 the center n l mass postimpact The rod i s I It in length and weighs 1.5 Ih.
-I
1,-1
Figure P.17.78 Figure P.17.76.
17.77. A rod A 1s dnrpped from a height of 120 mm ahove H. II an elastic cnllisinn takes place at 8, at what timc A i kiter doe\ a second cnllision with support 11 take placc'! Thc rod weighs 40 N .
m m I-
l---400 ~
I.
I20 mm
A
?n mm
17.79. An annw mwing esientially horirmtally a1 a speed ot 2 0 mAec impinges on ii statinnary w’z’ has an angular velocity 0 relative to XYZ. Clearly, one can say at time t:
However, the rime drrivutives of the corresponding components will not be equal 10 each other at time I . Accordingly, in the next section, whereHA is treated, we must properly account for f L
X
/ Figure 18.3. .r?z and x’y’z’ coincide at time 1.
INTRODUCTION
913
914
CHAPTER 18
OYNAMICS O r GENTRAI RIGID BODY MOTlOh
18.2
Euler's Equations of Motion
We shall now restrict point A of Section 18. I further, as we did i n Chapter 17. by ciinsidcring only Ihohe points for which thc equation M, = If,i s valid:
1. .The inass center.
2. Points fixed or miiving with constant V
at time I in inertial space (i.e., points having zero acceleration at time f relative to inertial reference X Y a . 3. A point accelerating toward or away Sroni thc mas$ centel-.
We learned i n Chapter 15 that derivatives of a vector as seen from different references could he related as follows:
where w i h the angular velocity 0 f . x ~ : rcliitivc 10 X Y Z . We shall employ this equation in the hasic mi,munf-ifl~n,,niPnliimuq11urion to shili the obscrvalion reference fnim X Y Z to .x?: i n the following way where now R represents thc mgular vclocity o f .ryz. Thus
W
Figure 18.4. Axis of rotation goes throueh fixed point A .
The idea tiow i s to choiisc the angular velocity R o f reference .zS: at A i n such a way that (dH,,/dr),, i s most easily evaluated. With this accomplished. the next qtep is to attempt the integration of the resulting differential equation. With regard t~ attempts at integration, we point out at this early stage that Eq. 18.3 i s valid only a s king as point A i s one of the three qualified points just discussed. Clearly. if A i s the mass ccntcr, then Eq. 18.3 i s valid at all times and can he integrated with respect l o time provided that the mathematics are not es (2) and ( 3 ) point A qualifies only at time f. too difticult. However. if for then Eq. 18.3 i s valid only at tinic 1 and accordingly c ~ n ~ ihe o t integratcd. If. on the lither hand. liir case (2). there i s an axih o f rotation fixed i n inertial space. then Eq. 18.3 i s valid at a11 times for any point A along the axis ofriitatioii and accordingly can hc intrgrated. We have already done this in Chapter 16. If, furthermore. the axis of r ~ t a t i o nalways p e s thriiugh the fixed point A hut does not have a fixed orientation i n inertial space (see Fig. 18.4). we can again use Eq. 18.3 at all times and attcmpt to integrate it with respect to time. The carrying out 01 huch inlegrations may he quite dillicult, however.' Returning to Eq. 18.3, we can work directly with this equation selecting a rcfcrcncc xyz for each problcm tu yield thc siinplcst working equation. On the other hand, we can develop Eq. 18.3 further for certain classes o f refkrences xy;. For example, we could have .vy: translate relative to X Y Z . This would
SECTION 18.2 EULER'S EQUATIONS OF MOTION
mean that = 0 so that Eq. 18.3 would seem to be more simple for such cases.' However, the body will be rotating relative to x y and the moments and products of inertia measured about xyz will then he time functions. Since the computation of these terms as lime functions is generally difficult, such an approach has limited value. On the other hand, the procedure offixing xyz in the body (as we did for the case of plane motion i n Chapter 16) does lead to very useful forms of Eq. lX.3, and we shall accordingly examine these equations with great care. Note first that the moments and products of inertia will be constants for this case and that 0 = w. Hence. we have
MA=(%)
+OXHA M
'We shall liilcr find il advanPagcous not to fin xyc to the body for ccrlilin problems
(18.4)
915
Thi, i\ iiideed a frirmidable set [if equations. For the impiirtant special case of pliinr nioliori the :axis i h always iimnal to the X Y planc which is the plane o l niotioti. Hence w inusi he iiormal to plane XY and thus collinear with the r iixic. Thus replacing < by (0and setting the other components equal l o zero. we get the w m w i f ofmiiii i i ~ t i ~qi ii d o i u for jioimii pliriir iii~i~im. Thus we haye
(MA).T = -I,$
(MAJ y =
+ I&
-lYza - I,:W2
(MAJ= y I$ The reader n a y no\+ or iil any tinie liiter go h;ick to Chapter 16 ior a raiher c~irefuI\tiid), 01 tlic use < I ( !lie ahovc plane motion w m e , i I a / nioimvifuiii equations. Wc now continue n'ith !lie I1iree~diiiirnsi~)naI approach. Note ncw that if we choose reierence S:K 111 he pr;iic;pol axes of the hody iit point 11. ltieri i t i s clear thei the products of inertia iirc all zero i n the system 01' cqu;itions 18.5. and this iw enables u s 111 siinpliiy the equations considerably. The rcsultinF equations given helou' are the fiinious Eider- q u o t i o m of tilotion. Niitc tliiit these equatiiins relate the 1iigul;ir velocity and thc angular wceleraiion l o lhc !nomen1 01 ihc cxtcrml fiirces ;ahout the point A .
',,",
MR= + m p r ( f z : - I>?) M,, = I>$,, + ap,(IU . . M: = lzpjz+ m p s ( l v x-, I x x J
118.h;~) (
I 8.6h)
(18.hc)
111 hoth set\ d Eq\. 18.5 iiiid 18.h. we l i i i ~ cthree siniultaiieous tint-order difkrenti;il equiiiicms. It the inorion ( i t Ihe hody ahout point A i s known. we can easily compute the required tiioiiien1s ahoul point A . On the other hand. if the iiioineiils are knowii function.; 01 iinic and the angular velocity i s desired, w e have the difficult prohleiii of siilvinf siiiiultancous lioiilincar differential equation\ for the uiiknowni (05. coy. and (or. However. in practical problem?. we oitcii kiiow soiiic of Ihc angular velocity and accclcratioii componcnh ircm constrilitits or Fiveti data. so. with the restrictions mentioned earlier. we ciin s
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