Analytical Elements of Mechanics Volume 2: Dynamics By Thomas R. Kane, Ph.D. Professor Engineering Mechanics Stanford University ?2005 by the author Previously Published by Academic Press New York and London 1961 Now published by The Internet-First University Press This manuscript is among the initial offerings being published as part of a new approach to scholarly publishing. The manuscript is freely available from the Internet-First University Press repository within DSpace at Cornell University at http://dspace.library.cornell.edu/handle/1813/62 The online version of this work is available on an open access basis, without fees or restrictions on personal use. A professionally printed and bound version may be purchased through Cornell Business Services by contacting: digital@cornell.edu All mass reproduction, even for educational or not-for-profit use, requires permission and license. We will provide a downloadable version of this document from the Internet-First University Press. For more information, please contact dcaps@cornell.edu. Ithaca, N.Y. March 2005 ANALYTICAL ELEMENTS OF MECHANICS Volume I ANALYTICAL ELEMENTS OF MECHANICS Volume 2 DYNAMICS THOMAS R. KANE, Ph.D. Professor of Engineering Mechanics, Stanford University Stanford, California ACADEMIC PRESS New York and London 1961 COPYRIGHT ? 1961, BY ACADEMIC PRESS INC. ALL RIGHTS RESERVED NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS. ACADEMIC PRESS INC. Ill FIFTH AVENUE NEW YORK 3, N. Y. United Kingdom Edition Published by ACADEMIC PRESS INC. (LONDON) LTD. 40 PALL MALL, LONDON S.W. 1 Library of Congress Catalog Card Number 59-13826 PRINTED IN THE UNITED STATES OF AMERICA PREFACE This book is the second of two volumes intended for use in courses in classical mechanics. The first deals with analytical elements of statics; the present one is concerned with dynamics. As before, the symbolic language used is vector analysis. Proofs and derivations are given in considerable detail, while dis- cursive material is omitted, leaving the teacher free, on the one hand, and obligating him, on the other, to explain the physical significance of mathematically defined quantities and to discuss related topics of practical, philosophical, or historical interest. Initially, physics and mathematics are kept separate, in the be- lief that this ultimately facilitates the task of relating the two. More than one hundred illustrative examples and problems are worked out in complete detail in the text, and practice problems, closely correlated with matters treated in the text, are provided in the form of Problem Sets, in a number regarded as both neces- sary and sufficient to insure adequate coverage without expendi- ture of excessive amounts of time. As for the "level" of the book, I have used it with under- graduates, both at the University of Pennsylvania and at the Manchester College of Science and Technology, spending approxi- mately ninety lecture hours during a period of thirty weeks, and with graduate students not previously exposed to this material, covering the same ground in fifty to sixty lecture hours. In the preface to Volume I, I expressed the opinion that one of the principal sources of difficulty in teaching mechanics has *>een the attempt to deal with forces intuitively. To this I now add the contention that the greatest obstacles standing in the way of effective solution of particle and rigid body dynamics Problems are inadequate understanding of kinematics and of the VI PREFACE theory of moments and products of inertia of rigid bodies. For this reason, both topics have received very full treatment in the present book, and this treatment differs considerably from those found elsewhere. Specifically, consideration of kinematics is preceded, in Chap- ter 1, by an extensive discussion of differentiation of vectors, particular emphasis being placed on the importance of reference frames. Fundamentals of the differential geometry of space curves are introduced in a form rendering the results practically useful, and, incidentally, the opportunity is thus provided to use differentiation theorems developed earlier. Chapter 2 begins with the definition of "rate of change of orientation of a rigid body," which is a generalization of the concept of angular velocity and makes it unnecessary to discuss finite rotations. The definition of and theorems dealing with angular velocity, which follow, serve to shed light on a number of questions which have not always been answered satisfactorily. For example, the matter of "simultaneous angular velocities" and such related questions as "Does a rigid body possess various angular velocities with respect to various points and lines, or is its angular velocity unique?" can now be disposed of readily (see Sections 2.2.7, 2.2.8, 2.2.1). Next, reversing the usual order, rela- tive velocities and accelerations are defined before absolute ve- locities and accelerations are mentioned, this making it possible to treat the latter as special cases of the former, which both simplifies and unifies the presentation. The fact that angular velocity and acceleration of a rigid body have been dealt with previously permits immediate consideration of velocity and ac- celeration relationships applicable to points fixed on a rigid body (see Sections 2.4.5, 2.5.9, and 2.5.10) or free to move in a moving reference frame (see Section 2.5.13). Two major departures from more conventional presentations of the topic of inertia occur in Chapter 3: Inertia properties of sets of particles and rigid bodies are considered only after similar properties of purely geometric entities (i.e., points, point sets, curves, surfaces, and solids) have been studied, and certain vec- tor quantities, here called "second moments" and having much in common with the vector quantity "traction" of elasticity PREFACE Vll theory, are taken as the fundamental building blocks of the entire theory. This leads to many relationships applicable both to geometric entities and to material bodies, and permits one without the use of second rank tensors to define quantities and to derive expressions independent of particular coordinate sys- tems. Furthermore, it is precisely these second moment vectors which later quite naturally turn out to characterize the inertia properties of rigid bodies in expressions such as those for inertia torques, angular momenta, and kinetic energy (see Sections 4.1.3, 44 11, and 4.5.4). The study of kinetics is undertaken in Chapter 4, where force concepts previously discussed in Volume I are brought together with ideas developed in the preceding chapters, in a rather broad statement of D'Alembert's principle, which then becomes the Point of departure for the examination of various other prin- clPles. In this portion of the text and of the related Problem kets, work done earlier in the course is presumed to justify immediate consideration of topics generally considered abstruse, and the attempt is made not only to state and illustrate the use ?f each principle, but also to examine it from the point of view ?* its suitability for the solution of specific classes of problems, the intention being both to increase the student's effectiveness as an analyst and to set the stage for later study of advanced Mechanics. T. R. KANE Philadelphia, Pennsylvania March, 1961 CONTENTS PREFACE V 1 DIFFERENTIATION OF VECTORS 1.1 Vector junctions of a scalar variable 1 1-1.1 Definition of a vector function of a scalar variable in a refer- ence frame; independence of a variable in a reference frame; vectors fixed in a reference frame. 1.15 dependence on a variable in one reference frame, independence of the same variable in an- other reference frame. 1.1.3 Measure numbers characterize the be- havior of a vector function. 1.1.4 Constant measure numbers. 1.1.5 Values of a vector function. 1.1.6 Equality. 1.1.7 Dependence of results on the reference frame in which an operation is performed. 1-1.8 Independence of results of the reference frame in which an operation is performed. 1.1.9 Notation. 1.1.10 Expression of results m terms of unit vectors fixed in any reference frame. 1.1.11 Func- tional character of results of operations involving vector functions. 1 .2 The first derivative of a vector function 9 1-2.1 Definition of the first derivative. 155 First derivatives equal to zero. 1^.3 Dimensions of the first derivative. 15.4 Expression of the first derivative in terms of unit vectors fixed in any refer- ence frame. 15.5 Equality of first derivatives of equal vector functions. 15.6 Notation. 15.7 The first derivative as a limit. 15.8 Properties of the derivative. 18 The second and higher derivatives of vector functions 12 1.O.1 Definition of the second derivative of a vector function; definition of higher derivatives of a vector function. 1-4 Derivatives of sums 13 14.1 Equality of the first derivative of a sum and the sum of the first derivatives. 1.45 Applicability of 1.4.1 to second and higher derivatives. 1.5 Derivatives of products 14 15.1 First derivative of the product of a scalar and a vector function. 1.55 First derivative of the scalar product of two vectors. 1.5.3 First derivative of the vector product of two vectors. ix X CONTENTS 1.5.4 First derivative of the continued product of any number of vector and scalar functions. 1.6 Derivatives of implicit functions 19 1.6.1 First derivative of an implicit function of a scalar variable. 1.7 The first derivative of a unit vector which remains per- pendicular to a line fixed in a reference frame 20 1.7.1 Expression for the first derivative. 1.7.2 Interpretation of one of the terms appearing in 1.7.1, as a rate of rotation. 1.7.3 Convenience of 1.7.1 when limited information available. 1.8 Taylor's theorem for vector functions 26 1.8.1 Statement of the theorem. 1.8.2 The use of Taylor's theorem for purposes of computation and in connection with functions not specified explicitly. 1.9 Vector tangents of a space curve 29 1.9.1 Vector tangents expressed in terms of the first derivative of a position vector. 1.9.2 Sense of the vector tangents obtained by using various scalar variables. 1.9.3 Expression for the vector tangent in terms of the derivative of the position vector with respect to arc-length displacement. 1.9.4 Definition of the normal plane at a point of a space curve. 1.10 Vector binormals of a space curve 33 1.10.1 Vector binormals expressed in terms of derivatives of a position vector. 1.10.2 Perpendicularity of vector tangents and vector binormals. 1.105 Sense of vector binormals obtained by using various scalar variables. 1.10.4 Simplification introduced by the use of arc-length displacement as independent variable. 1.105 Definition of the plane of curvature or osculating plane at a point of a space curve. 1.11 The vector principal normal of a space curve 38 1.11.1 Definition of the vector principal normal. 1.115 Expression for the vector principal normal in terms of derivatives of a posi- tion vector. 1.115 Expression for the vector principal normal in terms of the second derivative with respect to arc-length dis- placement. 1.11.4 Definition of the rectifying plane at a point of a space curve. 1.12 The vector radius of curvature of a space curve 39 1.12.1 The vector radius of curvature expressed in terms of de- rivatives of a position vector. 1.12.2 The vector radius of curva- ture as the product of a scalar and the vector principal normal. 1.12.3 Expressions in terms of derivatives with respect to arc- length displacement. 1.13 The Serret-Frenet formulas 42 1.13.1 Derivatives of vector tangents, binormals, and principal CONTENTS XI normal with respect to arc-length displacement. 1.13.2 The torsion of a space curve, expressed in terms of derivatives of a position vector. 2 KINEMATICS &-1 Rates of change of orientation of a rigid body 49 2.1.1 Definition of the rate of change of orientation of a rigid body in a reference frame with respect to a scalar variable. 2.1.2 Im- portance of rates of change of orientation as analytical tools. 2.1.3 Symmetry of the expression for rates of change of orienta- tion. 2.1.4 The relationship between the first derivatives of a vector function in two reference frames. 2.1.5 The derivative in two reference frames of the rate of change of orientation. 2.1.6 Interchange of reference frames. #?# Angular velocity 54 2.2.1 Definition of the angular velocity of a rigid body in a refer- ence frame. 222 Expression for the angular velocity as a product of an angular speed and a unit vector. 2.23 Pictorial representa- tion of angular velocity. 25.4 Angular velocity of fixed orienta- tion. 2.2.5 Omission of qualifying phrases in the description of frequently encountered systems. 2.2.6 Application of 2.2.4 to the motion of bodies possessing no fixed point. 22.7 Addition of angular velocities. 22.8 Resolution of angular velocities into components. 2.2.9 Kinematic chains. 2.2.10 Reference frames hav- ing no physical counterparts. W Angular acceleration 68 23.1 Definition of the angular acceleration of a rigid body in a reference frame, 2.3.2 The relationship between measure numbers of components of angular velocity and angular acceleration vec- tors. 2.33 Interchange of reference frames. 2.3.4 Expression for the angular acceleration as the product of a scalar angular ac- celeration and a unit vector. 2.3.5 The relationship between angu- lar speed and scalar angular acceleration. 2.3.6 Pictorial repre- sentation of angular acceleration. 2.3.7 Angular acceleration of a body having an angular velocity of fixed orientation. 2.3.8 Plane linkages. 23.9 Graphical method for the determination of the scalar product of unit vectors. 2.3.10 Applicability of 2.3.8 to linkages containing sliding pairs. 23.11 Addition of angular ac- celerations. 2-4 Relative velocity and acceleration 80 ? 1 Definitions of velocity and acceleration of one point relative to another. 242 The relationship between the velocity of P relative to Q and the velocity of Q relative to P. 2.4.3 Relative velocity and acceleration of two points fixed in a reference frame. Xll CONTENTS 2.4.4 Addition of relative velocities; addition of relative accelera- tions. 2.4.5 Relative velocity and relative acceleration of two points fixed on a rigid body. 2.5 Absolute velocity and acceleration 84 2.5.1 Definitions of absolute velocity and absolute acceleration of a point. 2.5.2 Velocity and acceleration of a point fixed in a reference frame. 2.5.3 Expression for the velocity as the product of a speed and a vector tangent. 2.5.4 Tangential and normal accelerations; scalar tangential and scalar normal accelerations. 2.5.5 Convenience of normal and tangential accelerations. 2.5.6 Velocity and acceleration of a point fixed on a rigid body which is rotating about a fixed axis. 2.5.7 Rectilinear motion. 2.5.8 Rectilinear motion as a limiting case of curvilinear motion. 2.5.9 Velocity and acceleration of points fixed on a rigid body possessing no fixed point. 2.5.10 Rolling; pure rolling contact; pivoting. 2.5.11 The instantaneous axis of a rigid body; minimum velocity. 2.5.12 Plane motion of a rigid body; instantaneous centers. 2.5.13 The relationship between the velocities of a point in two refer- ence frames; the relationship between the accelerations of a point in two reference frames; Coriolis acceleration. 2.5.14 Equality of the lengths of contact arcs during rolling. 2.5.15 Expressions for relative velocities and relative accelerations in terms of absolute velocities and absolute accelerations. 3 SECOND MOMENTS S.I Second moments of a point 113 3.1.1 Definition of the second moment of one point with respect to another. 3.1.2 Expression of the second moment of a point for one direction in terms of the second moments of the point for three mutually perpendicular directions. 3.1.3 Parallelism of the second moment for a direction with that direction. 3.1.4 Defini- tion of the second moment of one point with respect to another for a pair of directions. 3.1.5 Alternative expression for the second moment of one point with respect to another for a pair of di- rections. 3.1.6 Symmetry of second moments for a pair of direc- tions. 3.1.7 Definition of the second moment of a point with respect to a line. 31.8 Expression for the second moment of one point with respect to another in terms of second moments for three pairs of directions. 3.1.9 Expression for the second moment of one point with respect to another for a pair of directions in terms of second moments for three pairs of mutually perpendicular directions. 3.1.10 Expression of the second moment of a point CONTENTS Xlll with respect to a line in terms of second moments for three pairs of directions. 3.2 Second moments of a set of points 120 32.1 Definition of the second moment of a set of points with respect to a point. 3.2.2 Definition of the second moment of a set of points with respect to a point for a pair of directions. 3.2.3 Definition of the second moment of a set of points with respect to a line. 3.2.4 Applicability to second moments of sets of points of relationships discussed in Sees. 3.1.2-3.1.10. 3.2.5 Definition of the radius of gyration of a set of points with respect to a line. 35.6 Parallel axes theorems. 3.2.7 Parallel axes theorem for second moments with respect to a line. 3.2.8 Parallel axes theorem for radii of gyration. 3.2.9 Determination of all second moments of a set of points by successive use of Sees. 3.2.4, 3.2.6, and 3.2.7. 3.3 Principal directions, axes, planes, second moments, and radii of gyration of a set of points 126 3.3.1 Definition of principal directions, principal axes, principal planes, principal second moments, principal radii of gyration. 3.3.2 Necessary and sufficient condition that a unit vector define a principal direction of a set of points. 3.3.3 Zero second moment for a pair of perpendicular directions when one is a principal di- rection. 3.3.4 Location of a principal direction by consideration of zero second moments for two pairs of directions. 3.3.5 Non- zero second moments. 3.3.6 Planes of symmetry. 3.3.7 Coplanar points. 3.3.8 Location of two principal axes in a principal plane. 3.3.9 Location of three principal axes for any set of points. 3.3.10 Use of principal axes and principal second moments in the deter- mination of second moments for arbitrary directions. 3.3.11 Determination of second moments by successive use of Sees. 3.2.6, 3.2.7, 3.3.10. 3.3.12 Centroidal principal directions. 3.3.13 The momental ellipsoid of a set of points. 3.3.14 Lines of maximum or minimum second moment. 3.3.15 Minimum second moment of a set of points. 3.4 Second moments of curves, surfaces, and solids 142 3.4.1 Definition of the second moment of a figure with respect to a point. 3.4.2 Integral expression for the second moment of a figure with respect to a point. 3.4.3 Definition of the second mo- ment of a figure with respect to a point for a pair of directions. 3.4.4 Definition of the second moment of a figure with respect to a line. 3.4.5 Definition of the radius of gyration of a figure with respect to a line. 3.4.6 Principal directions, axes, planes, second moments, radii of gyration of a figure. 3.4.7 Use of relationships discussed in Parts 3.2 and 3.3 for the solution of problems involving curves, surfaces, and solids. 3.4.8 Polar second moments. 3.4.9 XIV CONTENTS Explanation of the Appendix. 3.4.10 Decomposition of complex figures. 3.4.11 Figures obtained by subtraction. 3.4.12 Radii of gyration obtained by regarding a surface as a limiting case of a solid. 3.5 Second moments of sets of particles and continuous bodies 159 3.5.1 Definitions of the second moment of a set of particles with respect to a point, products of inertia, and moments of inertia. 3.52 Definition of the second moment of a continuous body with respect to a point. 3.53 Definition of the second moment of a continuous body with respect to a point for a pair of directions. 3.5.4 Moment of inertia of a continuous body about a line. 3.5.5 Definition of the radius of gyration of a continuous body with re- spect to a line. 3.5.6 Principal directions, axes, planes, second moments, moments of inertia, and radii of gyration. 3.5.7 Ap- plicability of relationships discussed in Parts 3.2, 3.3, 3.4. 3.5.8 The relationship between second moments of a uniform body and second moments of the figure occupied by the body. 3.5.9 Co- incidence of principal direction, axes, and planes of a uniform body and of the figure occupied by the body. 4 LAWS OF MOTION 4.1 Inertia forces and force systems 171 4.1.1 Definition of the inertia force acting on a particle in a refer- ence frame. 4.12 Definition of the inertia force and the inertia couple acting on a continuous body in a reference frame. 4.13 Expressions for the inertia force and inertia torque acting on a rigid body. 4.1.4 Resolution of the inertia torque acting on a rigid body into any mutually perpendicular components. 4.1.5 Resolu- tion of the inertia torque acting on a rigid body into components parallel to a right-handed set of mutually perpendicular principal directions. 4.1.6 Resolution of the inertia torque acting on a rigid body into components parallel to a right-handed set of mutually perpendicular principal directions for the mass center. 4.1.7 Ex- pression for the inertia torque acting on a rigid body which has an angular velocity of fixed orientation. .J^.2 D'Alembert's principle 182 42.1 Statement of D'Alembert's principle. 4.22 Necessary and sufficient condition that a reference frame be a Newtonian refer- ence frame. 423 Equations of motion and free-body diagrams. 42.4 Evaluation of the earth as a Newtonian reference frame. 425 Foucault's pendulum. 42.6 Motion of a particle in the neigh- CONTENTS XV borhood of the earth's surface. 4.2.7 Motion of ballistic missiles and earth satellites. 4.2.8 Approximately Newtonian reference frames. 4.3 Motions of rigid bodies 207 4.3.1 Equations of motions for a rigid body. 4.35 Evaluation of contact forces acting on a rigid body whose motion is specified. 4.3.3 Plane free-body diagrams. 4.3.4 Several body problems. 43.5 The law of action and reaction. 44 Linear and angular momentum 230 4.4.1 Definition of the linear momentum of a set of particles rela- tive to a point in a reference frame. 4.42 Linear momentum of a continuous body relative to a point in a reference frame. 4.4.3 Linear momentum of a body relative to a point in a reference frame. 4.4.4 The linear momentum of a body relative to the mass center of the body. 4.4.5 The absolute linear momentum of a body in a reference frame. 4.4.6 The linear momentum principle. 4.4.7 The principle of conservation of linear momentum. 4.4.8 Defini- tion of the angular momentum of a set of particles relative to a point in a reference frame. 4.4.9 Angular momentum of a con- tinuous body relative to a point in a reference frame. 4.4.10 Angular momentum of a body relative to a point in a reference frame. 4.4.11 The angular momentum of a rigid body relative to a point fixed on the body. 4.4.12 Absolute angular momentum of a body in a reference frame. 4.4.13 The angular momentum principle. 4.4.14 Modified form of the angular momentum princi- ple. 4.4.15 The principle of conservation of angular momentum. 4.4.16 Relative advantages and disadvantages of the angular mo- mentum principle. 4.5 Activity and kinetic energy 242 4.5.1 Definition of the kinetic energy of a set of particles relative to a point in a reference frame. 4.55 Kinetic energy of a continu- ous body relative to a point in a reference frame. 4.5.3 Kinetic energy of a body relative to a point in a reference frame. 4.5.4 Kinetic energy of a rigid body relative to a point in a reference frame. 4.5.5 Absolute kinetic energy of a body in a reference frame. 4.5.6 The activity-energy principle for a particle. 4.5.7 Relative advantages and disadvantages of the activity-energy principle for a particle. 4.5.8 The activity-energy principle for a rigid body. 4,5.9 Elimination of contact forces by means of the activity-energy principle for a rigid body. 4.5.10 The activity- energy principle for a set of rigid bodies. 4.5.11 Rigid bodies con- nected to each other by light, helical springs. 4.5.12 The use of the activity-energy principle for a set of rigid bodies when the number of bodies is large. XVI CONTENTS PROBLEM SETS Problem Set 1 (Sections 1.1.1-1.7.3) 281 Problem Set 2 (Sections 1.8.1-1.13.3) 283 Problem Set 3 (Sections 2.1.1-2.3.11) 285 Problem Set 4 (Sections 2.4.1-2.5.8) 289 Problem Set 5 (Sections 2.5.9-2.5.15) 293 Problem Set 6 (Sections 3.1.1-3.2.9) 299 Problem Set 7 (Sections 3.3.1-3.5.9) 303 Problem Set 8 (Sections 4.1.1-4.1.7) 307 Problem Set 9 (Sections 4.2.1-4.2.8) 312 Problem Set 10 (Sections 4.3.1-4.3.5) 314 Problem Set 11 (Sections 4.4.1-4.4.16) 319 Problem Set 12 (Sections 4.5.1-4.5.12) 321 APPENDIX Curves 327 Surfaces 328 Solids 330 Index 333 DIFFERENTIATION OF VECTORS 1.1 Vector functions of a scalar variable 1.1.1 If one or more of the characteristics (magnitude, ori- entation, sense) of a vector v in a reference frame R depends on a scalar variable z> v is called a vector function of z in R; otherwise, v is said to be independent of z in R. In particular, if v is inde- pendent of every scalar variable in R} one speaks of v as fixed in R. Example: In Fig. 1.1.1,* R represents a reference frame con- D, ,C sisting of a rigid rectangle A BCD. 0 is the midpoint of line AB, P a point free to move between A and B on line AB, p the position ?Each figure has the number of the section in which it is first cited. When more than one figure appears in a section, the letters a, b, c, etc. are appended to the section number as in, for example, Fig. 1.1.3a, Fig. 1.1.3b. 1 SEC. 1.1.2 [Chapter 1 vector of P relative to 0, and q the position vector of D relative to A. Let z be the displacement of P relative to 0, regarding z as positive when P is between 0 and A and as negative when P is between 0 and B. Then the magnitude of p depends on the scalar variable z; the orientation of p in R is independent of z; and the sense of p in R depends on z. Thus, as two of the characteristics of p in R depend on 2, p is a vector function of z in R. On the other hand, q is independent of z in R. In fact, q is fixed in R. 1.1.2 A vector may be a vector function of a given scalar variable in one reference frame while being independent of this variable in another reference frame. Example: Fig. 1.1.2 shows two reference frames, R and R'9 FIG. 1.1.2 consisting of rigid rectangles, ABCD and A'B'C'D'', respectively, the two rectangles being free to move in their common plane. Let z be the angular displacement of line A'B' relative to line AB, regarding z as positive when the displacement is generated by a clockwise rotation of A'B1 relative to AB. Then q, the position vector of D relative to A, is independent of z in R (see Example 1.1.1), but is a function of z in R', as the orientation and sense of Diff. of Vectors] SEC. 1.1.3 q in Rf depend on z. For example, when z = TT/2, q has the sense B'A'\ while when z = -T/2, q has the sense A'B1. 1.1.3 If v is a vector function of a scalar variable z in a reference frame R and n?, i = 1, 2, 3, are unit vectors (not parallel to the same plane) fixed in R, the n?, i = 1, 2, 3, measure numbers of v (see Vol. I, Sec. 1.10.2) are scalar functions of z which char- acterize the behavior of v in R. The unit vectors may be vector functions of z in a second reference frame R'. If this is the case, it does not affect the behavior of v in R, but it does affect v's be- havior in Rr. Problem: Referring to Example 1.1.2,* let ru and n2 be unit D FIG. 1.1.3a vectors fixed in R and having the directions AB and AD, respec- tively, as shown in Fig. 1.1.3a. A force F is defined as = 2 (l - ?) n, + 3 sin (2z)n2 lb Draw sketches showing F in R and Rr, for z = 0, z = x/2, and z = -T/4. * Illustrative examples and problems are designated by the number of the section in which they appear. For example, Problem 1.1.3 means "the problem appearing in Section 1.1.3." 4 SEC. 1.1.3 Solution: From the given expression for F, [Chapter 1 F|W2 = 0 Fig. 1.1.3b shows ni and n2 in R for z = 0 and z = ? 7r/4. (AS IU and n2 are fixed in Ry ni|,?o and nil*.-^ are identical; similarly, n2|??o and n^*.-^.) F|z=0 and F^.-,.^ thus appear in R as shown in Fig. 1.1.3c. F|z?x/2, being a zero vector, is omitted from the sketch. FIG. 1.1.3b FIG. 1.1.3C Diff. of Vectors] SEC. 1.1.4 Figure 1.1.3d shows nt and n2 in Rf for z = 0 and z ? ? T/4. (Note that ni|,?o now differs from nil,..-^; similarly, n2|z-o dif- fers from n2|?--ir/4.) In R', F|z?0 and F|za,_r/4 thus appear as shown in Fig. 1.1.3e. FIG. 1.1.3e 1.1.4 If v is independent of a variable z in a reference frame R (see 1.1.1), the measure numbers of v, when v is expressed in terms of unit vectors fixed in R, are independent of z. Conversely, if the measure numbers of v, when v is expressed in terms of unit vectors fixed in a reference frame Ry are constants, then v is 6 SBCS. 1.1.5-1.1.8 [Chapter 1 independent of z in R. This follows from the fact that the charac- teristics of v in R determine these measure numbers, and vice versa. 1.1.6 WTien v is a vector function of a scalar variable z in a reference frame R, the vector obtained by assigning to z a particu- lar value, say z*, is called the value of v at z*. For example, Figs. 1.1.3c and 1.1.3e each show the values of the vector function F at z = 0 and z = ? TT/4. 1.1.6 Two vector functions vi and v2 of the same variable z are said to be equal in the interval Z\ ^ z ^ z2 if and only if the values of vi and v2 have identical characteristics for every z in this interval. When two equal vector functions are each expressed in terms of unit vectors fixed in any reference frame whatsoever, the measure numbers of corresponding components are equal. Conversely, when the measure numbers of corresponding components of two vector functions are equal for all z in the interval Z\? ^ z ^ 22, the two vector functions are equal in this interval. 1.1.7 The result of an operation involving two or more values of a vector function?or the values of several vector functions of the same variable, corresponding to distinct values of this variable ?depends on the reference frame in which the operation is per- formed. Example: For the force F denned in Problem 1.1.3, the scalar product of F|*,?o and F|,?_,/4 in R is given by (see Fig. 1.1.3c) 2(3V2) cos - = 6 lb2 while in R' the scalar product of these two values of F is equal to zero, F|?_o and F|*? ?*-/4 being perpendicular to each other in Rf (see Fig. 1.1.3e). 1.1.8 The result of an operation involving only one value of a vector function?or the values of several vector functions of the same variable, corresponding to one value of this variable?is in- dependent of the reference frame in which the operation is per- formed. Diff. of Vectors] SECS. 1.1.9-1.1.10 7 Example: Referring to Problem 1.1.3, and letting E be the vector function E = nilb the value of E at z = ? T/4 is seen to be Figures 1.1.3b and 1.1.3c thus show E|,?-,/4 and F\gam-w/4 in Ry while Figs. 1.1.3c and 1.1.3d show these two vectors in R'. It appears that in both reference frames the scalar product of the values of E and F at z = ? a-/4 is equal to 3 lb2. 1.1.9 In order to be unambiguous, any combination of mathe- matical symbols which indicates an operation involving two or more values of a vector function?or values of several vector functions of the same variable, corresponding to distinct values of this variable?must include specific mention of the reference frame in which the operation is performed. For example, the two scalar products evaluated in Example 1.1.7 can be denoted by F|.-0*F|..-ir/4 = 61b2 and by Rf pi pi n 1.1.10 Although the result of an operation involving vector functions may depend on the reference frame in which the opera- tion is performed (see 1.1.7 and 1.1.8), this result, if it is a vector, can be expressed in terms of unit vectors fixed in any reference frame whatsoever. Problem: In Fig 1.1.10, R and Rf represent two reference frames (coplanar rectangles), m and n2 represent unit vectors fixed in R'. Two vector functions, a and b, are defined as Q = "~ Zlti. b ? ~" Zfl2 T T where z is the angular displacement of R relative to R\ measured 8 SEC. 1.1.10 [Chapter 1 in radians, z being positive for the configuration shown in Fig. 1.1.10. R Express < in terms of Solution From Fig. FIG. 1.1.10 each of the vectors R a|??ir/2 + b|,?j (a) in and n2, i i (a):* II bU-3 R 0|**ir/2 + b|#-j ft' 1.1.10, n2' f,r/2 and ( {b) ni and i T/2 ~ 311!^ = 1 Ck /I,/2 = 12n 21 jx/2 = 3ni - W2 = 3n2' ? = sin z ni -+ OU-?/2 + n2'. 'w/2 ? (F 1.1.10 *?3ir/2 = (Fl. - 12ni ? +- 12n2' = ? cos z n2 b|^3l 3n2' ) 1.10) 15n2' R'Hence a|*?,r/2 + b(*-31r/2 = 15(sin z in + cos z n2) Results (a): ? 9iu, 15(sin z nx + cos z n2). * Numbers beneath equal signs refer either to corresponding sections of the text or to equation numbers appearing in the section under consideration When a section number is preceded by P, E, or F, this indicates a reference to a specific problem, example, or figure, respectively. For example, (P 1.2.1) is to be read "see the problem discussed in Section 1.2.1." Difi. of Vectors] SECS. 1.1.11-1.2.1 Solution (b): From Fig. 1.1.10, ni == cos z ni + sin z n2' From Solution (a), R Hence R' a|*-?/2 + b|?-3?/2 = ? 9(coszni' + si Results (b): ? 9(cos z n/ + sin z n2'), 15 n2'. 1.1.11 The vectors and/or scalars resulting from operations performed with vector and/or scalar functions of a variable z are, in general, vector or scalar functions of z (see, for example, Prob- lem 1.1.10.) 1.2 The first derivative of a vector function 1.2.1 Given a vector function v of a scalar variable z in a reference frame R (see 1.1.1), theirs/ derivative of v with respect to z in R is denoted by and is defined as Rdv~-r or Rdv/dz Rd* dz where (see Fig. 1.2.1) n?, i = 1, 2, 3, are unit vectors (not parallel FIG. 1.2.1 10 SECS. 1.2.2-1.2.3 [Chapter 1 to the same plane) fixed in R and Vi is the n? measure number of v. Problem: Referring to Problem 1.1.10, evaluate the first de- rivative of a with respect to z in (a) R and (b) R'. Solution (a): Solution Hence a (b): 6 = - z a = Rdo dz '' ni = cos z n^ 6= - Z 111 T = i(l cos z ni ?) / + sin sin z n2' R'd* d /6 \ , , d /6 . \ , 7 = 7"^ cos z ) ni + ? I - z sin 2 ) n2 ? = - (cos z ? z sin z)in' + - (sin z + z cos z)n2'oz ^ ^r 1.2.2 If a vector v is independent of a variable z in a reference frame R (see 1.1.1), the first derivative of v with respect to z in R (see 1.2.1) is equal to zero. In particular, the first derivative of a zero vector (see Vol. 1, Secs. 1.5 and 1.10.3) with respect to any- scalar variable in any reference frame is equal to zero. Conversely, if Rdv/dz = 0, v is independent of z in R. This follows from Secs. 1.1.4 and 1.2.1. 1.2.3 As the derivative of v with respect to z in R (see 1.2.1) is a vector, the dimensions of the derivative are those of the mag- nitude of this vector (see Vol. 1, Sec. 1.1.5) and are, therefore, found by determining the quotient of the dimensions of v and of z. Problem: The position vector p (see Vol. 1, Sec. 2.1) of a point P relative to a point 0 which is fixed in the reference frame R shown in Fig. 1.1.10 is given by p = 5t2 nx ft where t is the time in seconds. Determine the first derivative of p with respect to t in R at t = 2 sec. Diff. of Vectors] SECS. 1.2.4-1.2.6 11 Solution: - = 10, j = 1, 2, 3, be unit vectors fixed in R (and not parallel to the same plane), va the n> measure number of vt. Then and the resultant of the vectors vt, i = 1, 2, . . . , n, is given by n n/3 \ 3/n\E v< = Z (E ??>">) = Z (E ??) n> i?l t-l\;-I / j_l \t-l / The first derivative of the resultant thus becomes (see 1.2.1) But, from the calculus of scalar functions, 14 SECS. 1.4.2-1.5.1 [Chapter 1 Hence y d.2.1) fr[ dz and this is the resultant of the first derivatives with respect to z in R of the vectors v,, i = 1,2,..., n. Problem: For a certain value z* of a variable z the first deriva- tives with respect to z in a reference frame ft of two vector func- tions a and b have the values ni + 2n3 and 3rii ? n2 + n3. Deter- mine the first derivative of a +? b with respect to z in R, at z = z*. Solution: *d ,, *da , Bd\o5 2(a + b)= Jz+ Iz *d . , ,,"1 *da ( + b)U= ~dz *db dz = ni + 2n3 + 3ni ? n2 + n3 = 4ni + 3n3 ? n2 1.4.2 The statement made in Sec. 1.4.1 applies also to second and higher derivatives of vector functions. 1.5 Derivatives of products 1.6.1 Given a variable z, a vector function v of z in a reference frame R, and a scalar function s of 2, Rd , N ds Rdv Proof: Let nt, i = 1, 2, 3, be unit vectors fixed in R (and not parallel to the same plane), t\ the nt measure number of v. Then Diff. of Vectors] SEC. 1.5.2 15 and -r (?v) = 2^az ci.2.1) fzV* d2 as Problem: A vector p, referred to mutually perpendicular unit vectors nt, i =? 1, 2, 3, fixed in a reference frame R, is given by p = -6"3tn2 + n3 ft where t is the time in seconds. Evaluate the magnitude of the first time-derivative of p in a reference frame Rf for t = 0 if, at this time, the first time-derivatives of the unit vectors in Rr have the values 4n2 sec"1,""^! sec"1, and zero. Solution: Rfdp R'd , Xl , , R'd , ,. , . R'dnz dt (i.2.5) dt (1.4.D at at = 3n2 + 4ni ft sec"1 Result: 5 ft sec"1 1.5.2 Given two vector functions vx and v2 of a variable z in a reference frame R, d , v fldvi-(V1.V2)= ^?v. + vx where Rr is any reference frame whatsoever. 16 SEC. 1.5.3 [Chapter 1 Proof: Resolve vi and v2 into components parallel to mutually perpendicular unit vectors fixed in some reference frame R', then proceed as in Sec. 1.5.1. Problem: The derivative of a vector v with respect to a vari- able z in a reference frame R is perpendicular to v. Show that the derivative with respect to z of the magnitude of v is equal to zero. Solution: (v ? v)1'2 But dz - dzK* T/ ~2V' " dz _ 11 |-i (Rdl~ 2 |v| \ dz' _1 ,,_,/?.. *dv\ -r = 0 dz if v is perpendicular to Rdv/dz. 1.6.3 Given two vector functions vi and v2 of a variable 2 in a reference frame R, Proof: The proof can be carried out by resolving vi and V2 into components parallel to mutually perpendicular unit vectors fixed in R, etc. However, a shorter proof, based on the alternative defi- nition of the derivative discussed in Sec. 1.2.7, is given below, to illustrate the use of this definition. (1.2.7) A*->0 AZ X (V2 + *AV2) - Vl X V2 r Avx X V2 + vi X *Av2 + gAVl X *Av2= lim A Diff. of Vectors] SEC. 1.5.4 17 + (lim *$&) x Aim 1.5.4 The product Pn of n scalar and/or vector functions Fiy i = 1, 2, . . . , n, of a scalar variable 2 can always be regarded as the product of only two functions, say Pn-i and Fn, where Pn-i is the product of the n ? 1 functions Fif i = 1, 2, . . . , n ? 1. That is, Pn = Pn^Fn From the laws governing the differentiation of scalar functions, together with the theorems in Sees. 1.5.1-1.5.3, it follows that RdPn_?dL(p F,_ *dP^ *dFn After replacing Pn-i with Pn-2^n-i, etc., one obtains where the letter R may be omitted whenever the quantity being differentiated is a scalar. The order in which the terms in any one product occur is the same as that in which these terms appear in Pn) and all symbols of operation, such as dots, crosses, parentheses, brackets, etc., must be kept in place. Problem: The line of action of a 10 lb force F passes through two corners of a rectangular parallelepiped, as shown in Fig. 1.5.4. A line L, lying in the top face of the parallelepiped, passes through one corner of this face and makes an angle 6 with one edge. Letting R be a reference frame fixed in the parallelepiped, de- termine the magnitude of the derivative in R with respect to 6 of the moment MF/L of F about L, for 0 = 0. Solution: Let nt, i = 1, 2, 3, be unit vectors parallel to the edges of the parallelepiped; p the position vector of a point on the 18 [Chapter 1 FIG. 1.5.4 line of action of F relative to a point on L; n a unit vector parallel to L, as shown in Fig. 1.5.4. Then n = cos 6 ih+ sin 6 n2 p = 4ni + 3n2 ft F = 6n2 + 8n3 lb In accordance with the definition of the moment of a vector about a line (see Vol. 1, Sec. 3.2), M^ = [n, p, F]n Differentiate: T=[> ']?+[?? '%? ?] Now Rdx\JT = ?sin 6 ni + cos 6 n 2 (IV (1.2.1) Rdp = Rdl ^ Q dd dd (i.2.2) Difi. of Vectors] SEC. 1.6.1 Thus ? = [n2, 4fi! + 3n2, 6n2+ 8n3- and 19 3n2, 6n2 + 8n3]n2 = -32n1 + 24n2ftlbrad"1 dd 40 ft lb rad-1 1.6 Derivatives of implicit functions 1.6.1 If v is a vector function of a scalar variable y in a refer- ence frame Rf and y is a function of the scalar variable 2, then Rdv _ Rd^(h[ dz dy dz This follows from the definition of the derivative of a vector function (see 1.2.1), together with the laws governing the differ- entiation of scalar implicit functions. Problem: Referring to Problem 1.5.4, and supposing that line L revolves uniformly, performing one revolution per second, with 0 increasing, determine the first time-derivative in R of the moment of F about line L, for 6 = 0. Solution: Let t be the time in seconds. Then 6 = 2jrt rad and so that Now and dt dt dd dt , d6 dd -0 de = 0 (P 1.5.4) -32n! + 24n2 ft lb rad"1 2T rad see"1 'w-o 20 Hence dt 0-0 SEC. 1.7.1 [Chapter 1 = 27r(-32n! + 24n2) ft lb sec"1 1.7 The first derivative of a unit vector which remains perpendicular to a line fixed in a reference frame 1.7.1 If a unit vector n is a function of a scalar variable z in a reference frame R1', and n remains perpendicular to a line K fixed in R' as z varies, then Rrdn , __ d$~7~ = k x n , dz dz where k and 6 are denned as follows (see Fig. 1.7.1a): k is a unit vector parallel to line K. 6 is the angular displacement of a line L -R1 /K (fixed in R1) / ^^V^; ' PerPend LP0f 'CU/< and *r *? K) FIG. 1.7.1a which is parallel to n, relative to a line U which is fixed in R1 and perpendicular to line K, 6 being regarded as positive when this dis- placement is generated by a k rotation of L relative to L', a k rota- tion being one during which a right-handed screw whose axis is parallel to k advances in the k direction. Proof: Let ru be a unit vector parallel to line L', n2 a unit vec- tor perpendicular to both k and ih, choosing ni and n2 such that Diff. of Vectors] SEC. 1.7.1 i X n2 = k 21 Then n; referred to the mutually perpendicular unit vectors ni, n2, k (all of which are fixed in R'), is given by (see Fig. 1.7.1b) and But That is Hence k FIG. 1. n = cos 0 nx Rfdn R'dnd6 dz (1.6.1) d$ dz (1.2.1) k X n = k X (cos I (A) = cos 0 k X = cos 0 n2 (Fl.7.1b) k X n = ?sin 0ih Rfdn dz (B,C) 7.1b + sin 0 n2 -sin ni + cos n2) 9 ni + sin 0 n2) nx + sin 0 k X n2 + sin 0( ? ni) + cos 0 n2 ^0 L1 de (A) (B) (C) (Note that k X n is a unit vector perpendicular to both Ic and n. It follows that R'dn/dz is perpendicular to both k and ?h but is not, in general, a unit vector.) 22 SEC. 1.7.1 [Chapter 1 Problem: In Fig. 1.7.1c, A and B represent two rectangular FIG. 1.7.1C plates connected by hinges. , the angular displacement of A relative to B, regarded as positive for the configuration shown, is given by where t is the time in seconds, a is a unit vector fixed in A, b a unit vector fixed in J5, c a unit vector fixed in both A and B. For t = 2 sec, determine the time-derivative (a) of a in B, and (b) of bin A. Solution (a): As 4> is positive when the angular displacement of A relative to B is generated by a ? c rotation of A relative to B, Bda d and Bda dt -7 =7b sec-i Solution (b): If is regarded as the angular displacement of B relative to A, then is positive when this displacement is generated by a c rotation of B relative to A. Hence Diff. of Vectors] SEC. 1.7.2 23 Ado ? d | x^ ir and 1.7.2 When a function /(x) has a positive first derivative at x = x*, then /(:r* + Ax) is algebraically larger than /(z*) for every positive and sufficiently small Ax; or, in other words, f(x) is in- creasing algebraically at x*; and df/dx measures the rate at which f(x) is increasing. Applying this to the angular displacement d(z) defined in Sec. 1.7.1, one concludes that dd/dz measures the rate at which L is performing a k rotation relative to L'. Problem: In Fig. 1.7.2a, OA represents a line which is perpen- 0 FIG. 1.7.2a /77//////E7/ dicular to the surface of the earth E. OP is a rigid rod, attached to a support at 0 in such a way that it can rotate in a vertical plane fixed in E. n is a unit vector parallel to OP and having the sense OP. Supposing that the rod rotates clockwise (as seen by the reader), completing 10 revolutions per second, and that the rota- tion is uniform, that is, that the angle through which OP turns during any time interval is proportional to this time interval, draw sketches showing the first time-derivative of n in E for the instants at which the rod passes through its two vertical and two horizontal positions. Solution: Let k be a unit vector perpendicular to the plane determined by the points A, 0, P, and 6 the angular displacement of OP relative to OA, regarded as positive for the configuration shown in Fig. 1.7.2a. Then 24 SEC. 1.7.3 [Chapter 1 Edn at ^ ddX n - at and d0/d? is the time-rate at which OP is rotating counterclockwise. Accordingly, ?dd/dt is the rate at which OP is rotating clockwise and, as 6 changes by an amount of 10 X 2TT radians during each second of the motion, Hence ?? = 207r rad sec"1at ^ = -20* k X ndt As V is perpendicular to n, k X n is a unit vector. Its direction can be determined by inspection. See Fig. 1.7.2b. dt EdnI* dt Id? Edn dt f///77T FIG. 1.7.2b 1.7.3 The use of the expression (see 1.7.1) for the evaluation of R'dn/dz is particularly convenient when infor- mation regarding the manner in which 6 depends on z is confined to knowledge of dd/dz for a specific value z* of z. To determine Rfdn/dz at z* it is then only necessary to know n at z*. Problem: In Fig. 1.7.3a, D represents a circular disc which rolls on a circular track C in such a way that at a certain instant t* dd dt = ? 5 rad sec"1 Difj. of Vectors] SEC. 1.7.3 25 where 6 is the angular displacement of a line d fixed in Z>, relative to a line c fixed in C, 0 being regarded as positive when generated -C FIG. 1.7.3a by a clockwise (as seen by the reader) rotation of d (or D) relative to c (or C). A and B are points fixed in D. Draw a sketch showing the time-derivative in C at t* of the position vector p of B relative to A. Solution: Let n be a unit vector parallel to line AB. Then p is given by p = 3n ft and cdp cd dtt* (i.2.5) dt vv>i/1* dt Let k be a unit vector perpendicular to the plane of the disc D, choosing the sense of k such that D rotates clockwise during a k rotation of D relative to C (see Fig. 1.7.3a). Then cdn dt so that t* (1.7.1) cdp dt = -5(k X lO^ = -15(k X n)*ft sec"1 26 SEC. 1.8.1 [Chapter 1 (Note that line d is not parallel to n, but that 1.7.1 is neverthe- less used to evaluate cdn/dt. The justification for this is that any angle between line c and a line parallel to n can be expressed as the sum of 0 and s^n angle which is independent of t, so that the time-derivative of such an angle is equal to dB/dt.) Result: See Fig. 1.7.3b. FIG. 1.7.3b 1.8 Taylor's theorem for vector functions 1.8.1 If v is a vector function of a scalar variable z in a reference frame R (see 1.1.1), the difference in R between the values (see 1.1.5) of v at 2* + h and at 2*, where z* is a particular value of z and h a scalar having the same dimensions as 2, can be expressed in terms of values of derivatives of v with respect to z in R at 2*, as follows: /} " v'** = T~! dz** 2! d22 Proof: Let nt, i = 1, 2, 3, be unit vectors (not parallel to the same plane) fixed in R. Then i\, the n? measure number of v, is a scalar function of z (see 1.1.3) and, by Taylor's theorem for scalar functions, i\t*+h tV 1!& ^-r 2!^2 The values of v at z* and at 2* + A, expressed in terms of their nifi = 1, 2, 3, components, are 3 3 t=i Diff. of Vectors] SEC. 1.8.1 27 Subtract the first of these from the second, in R: R ^ \ 2 \ USubstitute: 22 ,* + 2\dzT n, + . . . (i.2.i,i.3.i) I-' dz 2! Problem: In Fig. 1.8.1, R and 72' represent two reference frames (coplanar rectangles), ih and n2 represent unit vectors fixed R FIG. 1.8.1 in R, and n/ and n2' are unit vectors fixed in R'. A vector function v is defined as v = z2 n/ where z is the angular displacement of R relative to Rf. Determine approximately the difference between the values of v at z = 1.00 rad and z = 1.01 rad in (a) R and (b) R'. Solution (a): Using only two terms of the Taylor series, Rdy = (o.oi) Jz (0.01)2 1.00 dt2 1.00 28 Next SEC. 1.8.2 [Chapter 1 and Rd2v5~ = 2n/ - 2zn 2' - 2zn2' -az 1' (2 - z2)n/ - 4zn2' Hence Rdv dz dz' 1.00 ? n2 1.00 = m' - 4n2' Substitute (and note that the contribution of the second term of the series is much smaller than that of the first): - v i.oo 0.02005ni' - 0.01020n2' Solution (b): In the present case, use of only the first two terms of the series gives an exact result, because the third and all higher derivatives of v in Rf are equal to zero: Rfdv R'd2v R' R'dv - vU = (0.01) ? 1.00 = O^n/ + O.OOOln/ That is, R' v|,01 - v|,00 = 0.0201*' dz2 1.00 1.8.2 Problem 1.8.1 illustrates how Taylor's theorem may be used for purposes of computation. It should be noted, however, that the same results can be obtained, sometimes more conven- iently, without the use of this theorem. For example, as v is given explicitly in terms of z, Solution (b) of Problem 1.8.1 can be re- placed with |101 - |x 00 = (l.oi)V - 0.020W Diff. of Vectors] SEC. 1.9.1 29 The full power of Taylor's theorem becomes apparent in situa- tions involving vector functions which are not specified explicitly. 1.9 Vector tangents of a space curve 1.9.1 If a curve C is fixed in a reference frame R (see Fig. 1.9.1) and T is the tangent to C at a point P of C, a unit vector r parallel to T is called a vector tangent of C at P and is given by where p is the position vector of P relative to a point 0 fixed in R and primes denote differentiation with respect to z in R, z being FIG. 1.9.1 any scalar variable such that the position of P on C depends on z. Proof: Let Abea scalar having the same dimensions as z, Pi the point of C with which P coincides when z is increased by an amount h, and Pi the position vector of Pi relative to 0. By definition, the tangent T is the line which a line passing through Pi and P approaches as Pi approaches P. Hence, if Ci is the position vector of Pi relative to P, cx tends to become parallel to T when h approaches zero, r is thus the limit approached by a unit vector parallel to Ci, that is, 30 SEC. 1.9.1 [Chapter 1 Now Hence r = limT^T /l?O |Cl| R Ci = Pi ? p (1.8.1) li ^ and (see Vol. 1, Sec. 1.14.11) |ci| = (c!2)1/2 = (A2p'2 + Vp' ? p" + . . .) so that p' + hhp" + . . . 1/2 ^Substitute: T = (P'2 + hp' ? P" + . . O1'2 pM^ip- (P'2 + *p' ? p" + ? . -)1/2 (P'2)1/2 Problem: Referring to Problem 1.5.1, suppose that there exists a point 0 which is fixed in both R and R' and that p is the position vector of a point P relative to 0. Then P traces out a curve C in R and curve C in R'. Determine the cosine of the angle a between the tangents to C and C at point P, for t = 0. Solution: Let r and r' be vector tangents of C and C at P. Then COS a = r|t = 0 ? T'I^Q and T = dt > T' = dt Now so that /\*'dp "t - Rd dtL = 3n2 and, from Problem 1.5.1, R'dp dt = 3n2 Difi. of Vectors] SECS. 1.9.2-1.9.3 31 Hence and = n2, r'\t=0 = |(3n2 COS a = 1.9.2 The sense of a vector tangent r, when r is obtained by using the expression (see 1.9.1) depends on the choice of the scalar variable which governs the position of P on C: r points in the direction in which P moves when the variable increases algebraically. This is the reason for speaking of "a," rather than "the," vector tangent of a curve. Proof: Let z and z' be two scalar variables, each of which can be used to describe the position of P on C, and let r and r1 be the corresponding vector tangents, that is, dz/ \ dz T,_Rdp/\ dz'/ Idz' z may be regarded as a function of z', so that Thus Rdp __ Rdp dz_ dz1 (l.e.i) dz dzf Rdp dz_ Rdp dz_ dzdz1 dz dzf dz_ dz' Rdpdz_ dzdz' Rdp\\dz_ dz\ \dz' (1.9.1) dz_ dz' But (dz/dz')/\dz/dz'\ is equal to plus or minus one, according as dz/dz' is positive or negative. Hence 1.9.3 If the scalar variable governing the position of P on C is s, the arc-length displacement of P relative to a point Po fixed on C (that is, s is equal to plus or minus the length of the arc of C joining Po to P, the sign depending on the initial direction of motion of a point proceeding from Po to P), a vector tangent r of 32 SEC. 1.9.3 [Chapter 1 C at P (see 1.9.1), which points in the direction in which P moves when s increases algebraically, is given by Rdp T= Ts Proof: If Pi is the point to which P moves (see Fig. 1.9.3) when s is increased algebraically by an amount h, the arc PPi has isi FIG. 1.9.3 the length h and, when h approaches zero, the magnitude of the vector Ci joining P to Pi approaches h. This follows from the definition of arc-length. Thus lim ^ h-+0 (A) Now Hence and Thus Substitute: 1/2 (B) = 1 I (A.B) Difi. of Vectors] SECS. 1.9.4-1.10.1 Next 33 (1.9.1) Hence ds/ I ds ds While this expression for r appears to be simpler than the one given in Sec. 1.9.1, it is frequently less convenient, because the functional dependence of p on s is often more complicated than p's dependence on some other variable. 1.9.4 The plane passing through P and normal to r is called the normal plane of C at P. 1.10 Vector binormals of a space curve 1.10.1 If a curve C is fixed in a reference frame R (see Fig. 1.10.1a) and B is the binormal to C at a point P of C, a unit vector p parallel to B is called a vector binormal of C at P and is given by p P' x where p is the position vector of P relative to a point 0 fixed in R and primes denote differentiation with respect to z in R, z being any scalar variable such that the position of P on C depends on z. FIG. 1.10.1a 34 SEC. 1.10.1 [Chapter 1 Proof: Let h be a scalar having the same dimensions as z, Px the point of C with which P coincides when z is increased by an amount h, P2 the point of C with which P coincides when z is de- creased by an amount h, pi the position vector of Pi relative of O, and p2 the position vector of P2 relative to 0. By definition, the binormal B is the line which a line passing through P and perpendicular to both PPi and PP2 approaches as Pi and P2 approach P. Hence, if Cx and c2 are the position vectors of Pi and P2 relative to P, Ci X c2 tends to become parallel to B when h approaches zero. P is thus the limit approached by a unit vector parallel to Ci X c2, that is, h^0 |d X c2| Now R h , . h2 ? (1.8.1) It *! and Evaluate Ci X c2 and |ci X c2|: ci X c2 = hY Xp" + |ci X c,| = \h*\ \p' X p"| Substitute and let h approach zero: Problem: A straight line AC is drawn on a rectangular sheet of paper ABCD having the dimensions shown in Fig. 1.10.1b. The paper is then folded to form a right-circular cylinder of radius a/*-, the line AC thereby being transformed into the (right-wound) cir- cular helix H shown in Fig. 1.10.1c. Determine the cosine of the angle \p between a unit vector k parallel to the axis of the cylinder and a vector binormal ft of H at a point P of line AC. Solution: Let P' be a point on AB, such that line PPf is paral- lel to AD. Let z be the distance between P and P', 0 the intersec- Diff. of Vectors] SEC. 1.10.1 35 tion of the axis and base of the cylinder, n a unit vector parallel to line OP', 0 the radian measure of angle AOP', p the position FIG. 1.10.1C vector of P relative to O, and p' and p" the first and second deriv- atives of p with respect to z in a reference frame in which H is fixed. Then cos \// = k ? fi 36 SEC. 1.10.1 [Chapter 1 where Now ^ = IP'XP"| p = - n + zV Differentiate with respect to z: From Fig. 1.10.1c and from Fig. 1.10.1b Hence and Thus 1 Differentiate again: P" arc a arc AP' -- 6 = T dd dz a -5kx A ?1* z/l b y m\ t (k ia/b **2 But (see Fig. 1.10.1c) k X (k X n) = -n Hence Next p' X p" = -^ fx (k X n) X n + k X nl0 1_6 J and (as k is perpendicular to n X k and both are unit vectors) Diff. of Vectors] SECS. 1.10.2-1.10.4 37 Thus and -1/2 cos \f/ = k ? P = 1.10.2 The vectors r and ft (see 1.9.1 and 1.10.1) are perpen- dicular to each other, as can be seen by evaluating r-fi. 1.10.3 The sense of a vector binormal 0, when fi is obtained by using the expression given in Sec. 1.10.1, depends on the choice of the scalar variable which governs the position of P on C. To prove this, proceed as in Sec. 1.9.2. 1.10.4 Once again (see 1.9.3) a simplification is introduced by choosing for the scalar variable governing the position of P on C the arc-length displacement s of point P relative to a point Pa fixed on C (see Fig. 1.9.3). fi is then given by Rdp Rd?p ds ds2 ds2 Proof: In Sec. 1.9.3 it was shown that RdP _1 ds ~ Squaring, RdPi = (*t ds \ ds) Differentiating with respect to s, 2. = 0 ds ds2 Hence, when RcPp/ds2 is not equal to zero, Rdp/ds is perpendicular to Rd?p/ds2. (Rdp/ds is never equal to zero, for it is equal to r, a unit vector, as was shown in Sec. 1.9.3.) Thus, whether or not RcPp/ds2 is equal to zero, 38 SECS. 1.10.5-1.11.3 [Chapter 1 Rdp RcPp ds A ds2 = "dp ds dS = ds2 and ds A ds2 ?dp ds ds* (1.10.1) *dpas Rd?p ds2 1.10.5 The plane passing through P and normal to p is called (for reasons which will appear in subsequent sections) the plane of curvature or the osculating plane of C at P. 1.11 The vector principal normal of a space curve 1.11.1 If r and P are a vector tangent (see 1.9.1) and a vector binormal (see L10.1) of a curve C at a point P, the unit vector p given by V = P X T is called the vector principal normal of C at P, provided r and ? have the senses obtained by using the same variable z in the ex- pressions given in Secs. 1.9.1 and 1.10.1. The sense of p, as con- trasted with those of r and P (see 1.9.2 and 1.10.3), is unique. 1.11.2 As the vector p' X p" is perpendicular to p', the product of the magnitudes of p' X p" and p' is equal to the magnitude of (p' X p") X p'. From Secs. 1.9.1, 1.10.1, and 1.11.1 the following is thus an alternative expression for the vector principal normal: _ (p' X p") X p' v |(P' x p") x p'l 1.11.3 In terms of derivatives of p with respect to s, P is given by p = ds2 Rd?p ds2 This follows from Secs. 1.11.1, 1.9.3, and 1.10.4, together with the facts that RcPp/ds2 is perpendicular to Rdp/ds and Rdp/ds is a unit vector. Diff. of Vectors] SECS. 1.11.4-1.12.1 39 1.11.4 The plane passing through P and normal to v is called the rectifying plane of C at P. 1.12 The vector radius of curvature of a space curve 1.12.1 The position vector p of the center of curvature of a space curve C at a point P of C, relative to P, is called the vector radius of curvature of C at P, and is given by where p is the position vector of P relative to a point 0 fixed in a reference frame /2 in which C is fixed and primes denote differen- tiation with respect to z in Rf z being any scalar variable such that the position of P on C depends on z. Proof: Let h be a scalar having the same dimensions as 2, Pi the point of C with which P coincides when z is increased by an FIG. 1.12.1 amount h (see Fig. 1.12.1), pi the position vector of Pi relative to 0, r a vector tangent of C at P, and T the tangent to C at P. By definition, the center of curvature of C at P is the point approached, as Pi approaches P, by the center Q of a circle which is tangent to T at P and passes through Px. Letting r be the position vector of Q relative to P, p is thus given by 40 SEC. 1.12.2 [Chapter 1 p = lim r (A) Now r is (by construction) perpendicular to both r and r X Ci, where Ci is the position vector of P\ relative to P. Hence there exists some scalar, say X, such that r = XT X (T X ci) (B) On the other hand, as Q lies on the perpendicular bisector of line PPh r may be regarded as the sum of Ci/2 and a vector perpen- dicular to both Ci and r X Ci. Hence there exists some scalar, say (JL, such that r = | + MCi X (r X d) (C) Dot-multiply each of Eqs. (B) and (C) with Ci (in order to eliminate /x), equate the resulting expressions for r ? Ci, solve for X, and substitute into Eq. (B): X (T X cQ r = Use the relationships (1.9.1) and let h approach l Note that p'X while and use Eq. (A). |p'f zero im r (P' 2ei-r Ci = (1.8.1) : (P')2P X (r X cx) ? D/# -1- ? D" -4-1! 2! ' X (p' X p") " p" ? p' X (p' X p") Xp") = -(p'Xp") Xp' :(P'XP") = -(P'XP")2 1.12.2 The vector radius of curvature p of a curve C at a point P (see 1.12.1) has the same direction as the vector principal normal v of C at P (see 1.11.1) and can therefore be expressed in the form P = pv (1) where p is an intrinsically positive quantity, p is called the radius of curvature of C at P, and is given by Diff. of Vectors] SEC. 1.12.3 41 n - 9 ~ >' X p"| K ' Proof: Multiply the numerator and denominator of the ex- pression forp given in Sec. 1.12.1 with |p'|: |p? (P' X p") X p' [p?P = ' x P"| |P' x P"| P' x P" Problem: Determine the radius of curvature at point P of the helix H described in Problem 1.10.1. Solution: From the solution of Problem 1.10.1, and Hence P = + (aW)lw _ a (. V\ 1.12.3 In terms of derivatives of p with respect to s, p and p (see 1.12.1 and 1.12.2) are given, respectively, by (1)dsi and p = Proof: From Section 1.12.1, Now /*dp *^p ( ~f x \ as X ~f = 1 (1.9.3) (2) (A) 42 and ds X dsy SEC. 1.13.1 Rdp ds ds2 (1.10.4) [Chapter 1 Wi (B) Furthermore V ds ds2/ X & \ ds) ds2 ds2 ' ds ds Rd2p Hence Next (1.9.3) - 0 (1.10.4) P = ds2 _ ds2 P = (1.12.2) c ds !E ?s2 2 ds >s ( 3 Is* 2 1 (A,B) "dtp ds1 1.13 The Serret-Frenet formulas 1.13.1 The derivatives of the vectors r, 0, and ? (see 1.9, 1.10, 1.11) with respect to s are given by Rdr _ p ds p ds Rdv ds = -\v (2) (3) where p is the radius of curvature (see 1.12.2) and X, called the torsion of the curve C at point P, is given by Difi. of Vectors] SEC. 1.13.1 43 The expressions (1), (2), and (3) are called the Serret-Frenet formulas. Proof (1): Differentiate with respect Rdr ds(i r Rdp (1.9.3) ?S to s in 72: *d?p .2.5) ds2 11 Q\ Rd2p ~ds2 V y = Ci 19 Q\ D Proof (2): From Sec. 1.10.4, fi can be expressed in the form Rdp Rd P " dsX d Differentiate with respect to 5 in 72: dfi = Rdp Rd*p f /WpVl-1'2 ds (i.5.4) ds X ds*l\ ds2) J _ Rdp R is equal to zero (see 1.11.1), 44 Hence where or, expressing 7 Proof (3): SEC. ~ds X = ?pr ? and P in terms < x - * H^>[_ d?S j; = (1.11 1.13.1 = -X* ? CSx 0 of derivatives of p, *d p *d3p~| 0 X T ? 1) Differentiate with respect to s in R: Use (1) and (2) But and Hence ds (1.5.3) d? , above: R? - -xds PXT = (0 (1.11.1) (1.11.1) Rdp Xr + /JX Js v P X r) X r = -P X (0 X r) = -r r [Chapter 1 Problem: When a vector tangent r of a curve C at a point P is multiplied by a constant r having the dimensions of length, and the resulting vector p is regarded as the position vector relative to a point 0 of a point P on a curve C, then U lies on a sphere of radius r, center at 0, and is called a spherical indicatrix of C (see Fig. 1.13.1). Letting j5 be the radius of curvature of ?? at P, express j5 in terms of the radius of curvature p and the torsion X of C at P. Solution: The position of P on f? may be regarded as depend- ing on the arc-length displacement s of P relative to a point Po fixed on C. (s is not the arc-length displacement of P relative to a Diff. of Vectors] SEC. 1.13.1 45 FIG. 1.13.1 point Po fixed on 27.) If primes denote differentiation with respect to s, p is then given by - _ IPT By definition Hence (1.12.2) |p' X P"| p = rr , d = yP =ds dr r== r ? = - v ds (i) p where v is the vector principal normal of C at P. Next, -? d (r \ I 1 dp , 1 Hence / 1 dp , X r (3) \ p2 as p p P' X P" = h (\v X /3 - ^p or, replacing y with & X r (see 1.11.1) 46 SEC. 1.13.2 [Chapter 1 Substitute: (VP2)]1/2 (1 + XV)1/2 1.13.2 The torsion X (see 1.13.1), expressed in terms of deriv- atives of the position vector p with respect to a variable z other than s, is given by x = P2[P', P", P"'] IPI"6 Proof: Express the first, second, and third derivatives of p with respect to s in terms of derivatives of p with respect to z: (1.6.1) as ( (dzV dz' and Substitute into the expression for X given in Sec. 1.13.1. Problem: Determine the torsion X of the curve H described in Problem 1.10.1. Solution: From the solution of Problem 1.10.1, Diff. of Vectors] SEC. 1.13.2 47 and Differentiate p" with respect to z: ,? _ wad* __ w2a . p " ~&2 &- ?6rkXn Evaluate |p'| and [p;, p", p'"]: IP'I = (I + |) , [P', P", P"'] = ^ The radius of curvature p was found in Problem 1.12.2: Hence x = p'[P', p", p'"] ip'i- = ^^ 2 KINEMATICS 2.1 Rates of change of orientation of a rigid body 2.1.1 If the orientation of a rigid body R in a reference frame R' depends on only a single scalar variable z, there exists for each value of z a vector R'zR is called the rate of change of orientation of R in Rf with respect to z, and is given by K do K do D/ ? dz dz R'da (2) 49 2 KINEMATICS 2.1 Rates of change of orientation of a rigid body 2.1.1 If the orientation of a rigid body R in a reference frame R' depends on only a single scalar variable z, there exists for each value of z a vector R'a)zR such that the derivative with respect to z in Rf of every vector c fixed in R (see Fig. 2.1.1) is given by FIG. 2.1.1 R'dc R, X c (1) The vector R'wzR is called the rate of change of orientation of R in R' with respect to z, and is given by Rfdo R'db dz dz R'da , (2) 49 50 SEC. 2.1.1 [Chapter 2 where a and b are any two nonparallel vectors fixed in R. R/wzR is a, free vector; that is, it is in no way associated with any particular point of either R or R''. Proof: If a and b are fixed in R, their magnitudes and the angle between them are independent of z. Hence a <*' xb'z = / , a' and b' are seen to be given by a' = *V* X a, b' = ?'?.* X b (A) Next, any vector c fixed in R can be expressed in the form c = aa + 0b + 7a X b (B) where a, 0, and y are independent of z. Differentiating with re- spect to z in R', and using Eqs. (A) to eliminate a' and b', c' = a R'uzR X a + 0 R'o>zR X b + 7[(*V X a) X b + a X (R'ozR X b)] = a RfaR X a + 0 R'ta* X b + y(R'cozR ? ba ? a ? b R' dvi , ^ R'dn{ Rdv . ^ p, p v. (1.2.1) = -7- + ^Wz22 X VdZ (A) Problem: Referring to Problem 2.1.3, and letting v be a vector given by v = ni ft sec"1 determine the derivative of v with respect to z in P, for z = 2 ft. Solution: Rdv pdv , p P v. Tz=Tz+ "?' Xv Hence Now pd? dz 2-2 Rdv dz 2-2 360v| f-2 = "^7~ ni = 180ni ft sec""1 54 SECS. 2.1.5-2.2.1 [Chapter 2 Rdv dz ?-?4 = ? 90ni sec"1 and (see Problem 2.1.3) WL-2 = ^ (~6n! + 3n2 - n3) ft"1 Thus pdv 180 dzz=2 X 45 = ? 90ni + 4n2 + 12n3 sec""1 2.1.5 The derivatives of Rf|fSte = -5k rad sec"1 which can, alternatively, be expressed as = 5( ?k) rad sec"1 Result (a): ?5 rad sec"1. Result (b): 5 rad sec"1. 2.2.3 Like other vectors, angular velocities are sometimes de- picted most conveniently by means of straight or curved arrows accompanied by measure numbers (see Vol. I, Secs. 1.2.1 and 1.7.5). The angular velocity in question is then regarded as having the direction indicated by the arrow if the measure number is posi- tive, and the opposite direction if it is negative. For example, Fig. 2.2.3a shows four representations of the same angular velocity. FIG. 2.2.3a Kinematics] SEC. 2.2.4 57 Problem: Referring to Problem 2.2.1, draw a sketch of D showing two representations of c and 6 are the angular displacements of AP and BP relative to AB, both being regarded as positive when the system is in the con- figuration indicated in Fig. 2.2.5b. Furthermore, as 0 decreases algebraically during counterclockwise rotation of Z>, dd dt? = -3 X 2TT = -6TT rad min"1 (C) 60 SEC. 2.2.6 [Chapter 2 and d/dt (or ?d/dt) is the angular speed (see 2.2.2) to be determined. FIG. 2.2.5b Express in terms of 0: From Fig. 2.2.5b, TQ = r sin 0 and Hence so that and cotan + r cos 0 = r V2 sin 0 cotan + cos 0 = V2 . - - ? cos 6\ = arc cotan 2 - cos 0\ \ sin 0 / When P crosses line AB, Thus, using Eq. (C), = 6TTdt V2 cos 0 - 1 3 - 2V2 cos 0 0 = 0 - V2 ? 45.5 rad min" 1 (D) 3 - 2V2 (The minus sign means that the follower is rotating clockwise at the instant in question, as may be seen either by substitution into Eq. (B) or by noting that an algebraic decrease in corresponds to clockwise rotation of line AP relative to line AB.) 2.2.6 The theorem stated in Sec. 2.2.4 applies not only to mo- tions of rotation about axes fixed both in a body and in a reference Kinematics] SEC. 2.2.7 61 frame (as in Problem 2.2.5), but also to motions of a body in a reference frame in which no point of the body remains fixed. Problem: Referring to Problem 1.7.3, determine the angular velocity cu>D of the disc D in the track C for the instant t*. Solution: Let k be a unit vector fixed in both C and Z>, as shown in Fig. 1.7.3a. Then (2.2.4) and 7/J ccoD\t* = -rat k = ? 5k rad sec"1 t* (Compare this solution with that of Problem 2.2.1.) 2.2.7 Given n reference frames Rif i = 1, . . . , n, the angular velocity R** of D and the point of contact C of D and P, n2 parallel to the tangent T to D at C (T lies in plane P), and n3 perpendicular to D and hence to ni and n2. X is a line fixed in P and L' a line fixed in D. The angular velocity p cos 6 ni ? sin ^) n3 2.3 Angular acceleration 2.3.1 The first time-derivative in reference frame R' of the angular velocity R'wR of a rigid body R in #' (see 2.2.1) is called the angular acceleration of R in JB' and is denoted by R'aR: Kinematics] SEC. 2.3.2 69 R'd R'ctR = ? R'o)Rat Problem: Referring to Problem 2.2.7, determine the angular acceleration of the blade. Solution: It is implied that the angular acceleration in ques- tion is the angular acceleration of the blade in a reference frame whose orientation relative to the earth is fixed. Using the notation introduced in the solution of Problem 2.2.7, this angular accelera- tion is denoted by RictB and is given by R*WB = ? (0.2 cos \p ki + k2) ~rr I L. ?I ? (0.2 cos \p ki + k2) T" + (0.2 cos \p ki + k2) ~nr :i.5.i) L at Jat or, as by RtctB ? I ?0.2 sin ^ ? ki + 0.2 cos \f/ -77 ) ? (1.5.1) \ at at I at = (-0.2 sin ^^k +0.2 cos ^k2 X kx %?) ^ \ at (1.7.1) dt / at = -0.2 (^)2(ki sin ^ + ki X k2 cos ^) 2.3.2 Both the angular velocity R'vR of a rigid body R in a reference frame Rf (see 2.2.1) and the angular acceleration R'otR of R in Rr (see 2.3.1) can always be resolved into components re- spectively parallel to unit vectors nt*, i = 1, 2, 3, fixed in any reference frame R*. When this is done, the three measure numbers of the components of R'ctR are equal to the time-derivatives of the measure numbers of the corresponding components of RfwR if and only if R' cos 6) = ?<)> cos 6 + d>6 sin 6at Kinematics] SECS. 2.3.3-2.3.4 71 ? (^ + sin 0) = $ + 0 sin 0 + 00 cos 0 The angular acceleration pctD is given by Pfj Riff (2.3.1) ?* (2.1.4) ?? = 37 P?D + (*to* + pa>?') X at (2.2.7) = m di ^""^ C?S ^ ~*~ n2di (~? ~^~ n3dt^ ~^~ ^ Sin ^ + [-6*2 + 0(-cos 0ni + sin 0n3)] X puP That is, PaP = (-* cos 6 + *0 sin 6 - 0^)ih + ( ?ff + 0^ cos 0)n2 + ($ + 0 sin 0 + 00 cos 0)n3 and only the n3 measure number of potD is equal to the time- derivative of the corresponding measure number of pwD. 2.3.3 RfotR and RotRt are related as follows: R'agR = ? RQ^R' Proof: R> R R'd R' R *d R> R *<**"*' R R> RctR = ? w = ?R (jjR = ? = ?RQ^R (2.3.1) dt (2.1.5) ^ (2.1.6) rf< (2.3.1) 2.3.4 The angular acceleration RfctR can always be expressed in the form R'ctR = R'aRna where na is a unit vector parallel to R'ctR, and R'aR is a scalar, called the scalar angular acceleration of R in 72' for the direction na, which is positive when R'otR and no have the same sense, vanishes when R'OLR is equal to zero, and is negative when the sense of R'OLR is opposite to that of na. Problem: Referring to Problem 2.2.5, determine the scalar an- gular acceleration of F for the l< direction (see Fig. 2.2.5b). Solution: From Eqs. (B), (C), and (D) of the solution of Problem 2.2.5, the angular velocity of F in R is given by a 1 ~ ^2 cos 0 i , . . = OTT 7= k rad mm"1 3 - 2V2 cos 0 72 SECS. 2.3.5-2.3.7 [Chapter 2 Let RctF be the angular acceleration of F in R. Then RctF is given by Rd *?" = Tt (2.3.1) Ut fi V2 sin 0 de, = O7T 7= ~ K or, as by dd = -36TT2 (3 - 2V2 cos ey dt ?A = ? 6x rad min~ V2 sin ^ U rad min~ 2(3 - 2V2 cos ey Thus the scalar angular acceleration of F for the Ic direction is equal to 00 V2 sin e? oow 1= (3 - 2V2 cos ey and is negative when P is above line A B, positive when P is below line AB. 2.3.5 Referring to Secs. 2.2.2 and 2.3.4, note that, in general, dt 2.3.6 Once again (see 2.2.3) a pictorial representation employ- ing straight or curved arrows accompanied by measure numbers is sometimes convenient. For example, Fig. 2.3.6 shows four repre- FIG. 2.3.6 sentations of the same angular acceleration. 2.3.7 When a rigid body R moves in a reference frame R' in such a way that there exists a unit vector V whose orientation in Kinematics] SEC. 2.3.7 73 both R and Rf is independent of time t (see 2.2.4), the angular acceleration RfotR remains parallel to k and is given by where 0 is defined as in Sec. 2.2.4. Furthermore (compare with 2.3.5), where R'aR and R'uR are the scalar angular acceleration (see 2.3.4) and the angular speed (see 2.2.2) of R in Rr for the k direction. And -- - % <*> Proof: (2.3.1) dt (2.2.4) dt \dt (1.5.1) dt1 dt dt (i.2.2) d*2 Next d& = dt \di) (2I4) ~dT Hence ID/ D kdt so that (see 2.3.4) d*6 d^ (2.2.4) dt2 Problem: Referring to Problem 2.2.9 and Fig. 2.2.9b, and assuming that the angular speed of S in R for the n direction is constant, express the scalar angular acceleration (Ras') of S' in R for the n' direction as a function of 0 and a. Solution: R, as> . * (1) (2) (3) where (see Fig. 2.3.8a) ft is a reference frame in which the plane R FIG. 2.3.8a of the linkage is fixed, B is a bar of the linkage, k is a unit vector perpendicular to the plane of the linkage, n is a unit vector parallel to By n' is the unit vector Ic X n, and ROJB and Ra* are the angular Kinematics] SEC. 2.3.8 75 speed and scalar angular acceleration (see 2.2.2 and 2.3.4) of B in R for the V direction. Example: Figure 2.3.8b shows the configuration at time t* of a plane four bar linkage, one bar of which, BA, is fixed. At time t* the angular velocity of bar Bz and the angular acceleration of bar Bi have the values indicated in the sketch. The angular velocities of bars Bi and B^ and the angular accelerations of bars B* and B% at time t* are to be determined. Let k be a unit vector perpendicular to the plane of the linkage, and introduce unit vectors n, and n/ respectively parallel and per- pendicular to Bi, i = 1, 2, 3, 4, choosing n/ such that n/ = kx n,. Next, note that the position vector of D relative to A is equal both to ADnA and to lOih ? 9n2 ? 4n3 ft, so that lOih - 9n2 - 4n3 = ZDn4 Differentiate with respect to t, and let 3n3' = 0 (A) (i) Now, at time t*y o?3 = ?6 rad sec""1 Hence, at time t*, - 4(-6)n,' = 0 (B) This vector equation can be solved for the two scalar unknowns wi and w2, either analytically or graphically. One analytical method 76 SEC. 2.3.8 [Chapter 2 consists of eliminating one unknown at a time by dot multiplica- tion of the equation with a unit vector perpendicular to the unit vector associated with this unknown. For example, to eliminate o>2, dot multiply with n2: lOwini' ? n2 + 24n3' ? n2 = 0 (C) From Fig. 2.3.8b, n/ ? n2 = cos (ni;, n2) = ?- n3' ? n2 = cos (n3', n2) = ? 1 h24(-l) = 0 and Hence and o>i = ?3 rad sec"1 Similarly, dot multiplying Eq. (B) with ni to eliminate o>i, w2 = ? 2 rad sec"1 The angular velocities of bars B\ and B2 at time t* are shown in Fig. 2.3.8c. Note that the sense of each of these vectors is op- \ 2 rad sec"[_ / \ 3rad sec"' B, FIG. 2.3.8C posite to that of Ic. Next, differentiate Eq. (A) (not Eq. (B), which is valid only at time t*), letting a, be the scalar angular acceleration of bar Bi for the k direction: + - 9a2n2' ) (2,3) Kinematics] SEC. 2.3.9 77 At time t* the known values of coi, w2, w3, and <*i then give 1/ - 90ni - 9a2n2' + 36n2 - 4a3n3' + 144n3 = 0 Find a2 and a3 by dot multiplication of this equation with n3 and n2, respectively: a2 = 11.33 rad sec"2 ?3 = 14.5 rad sec~2 The angular accelerations of B2 and Bz are shown in Fig. 2.3.8d. B2 ^ n-33 rod sec"2 B, 2.3.9 When the geometry of a linkage is complicated, the nec- essary dot products of unit vectors (see Eqs. (B) and (C), Example 2.3.8) cannot be evaluated so readily as they were in Example 2.3.8. The following graphical procedure yields these dot products with considerable accuracy: Draw a circle, choosing for the radius a length which can readily be divided into small, equal parts. (The larger the number of such parts, the better.) Through the center of the circle draw a line parallel to each unit vector. Taking the sense of each unit vector into account, label one point of inter- section of each such line with the circle with a number designating the corresponding unit vector. In Fig. 2.3.9 this is done for the unit vectors used in Example 2.3.8. If the radius of the circle is now regarded as defining a unit of length, the absolute value of the dot product of any two unit vectors is equal to the number of such units contained in the line segment joining the center of the circle to the foot of the perpendicular dropped from the labeled point corresponding to one unit vector on the line corresponding to the other unit vector; and the dot product is positive when the foot of this perpendicular falls on the labeled side of the line, nega- 78 [Chapter 2 FIG. 2.3.9 tive when it falls on the unlabeled side. For example (see Fig. 2.3.9), ni' . n2 = -0.8 and n3' ? n2 = ?1 2.3.10 The method discussed in Sec. 2.3.8 applies not only to plane linkages containing bars of fixed lengths, but also to those containing sliding pairs. Problem: Solve Problem 2.2.5 by using the method discussed in Sec. 2.3.8. Solution: Let z be the (variable) distance between A and P. Then (see Fig. 2.3.10) Kinematics] SEC. 2.3.11 79 and, differentiating, dz -7 nF + z Ro)F nF' ? r V nD' = 0 where Ro)D = 6ir rad min"1 Figure 2.3.10 shows that, when 6 = 0, nF = ? nD, n/ = ? nD\ z = (A/2 ? l)r Hence ^1 (-nz>) + (V2 - l)r /2^|^0(-ni>/) - r(6w)nD' = 0 Thus n w-, I O1T ^ f. 1 - A/2 2.3.11 Given n reference frames Riy i = 1, 2, . . . , n, the angu- lar acceleration R~ctR of a rigid body # in a reference frame /2n is not, in general, equal to the sum (Compare with 2.2.7.) Problem: Referring to Problem 2.2.7 and Fig. 2.2.7b, deter- mine the angular accelerations RlctB and R*aRl, then compare their sum with the angular acceleration RtctB found in Problem 2.3.1. Solution: (2.3.7) (2.3.7) Hence while = -0.2 f^V fk, sin * + k X k4 cos (P2.3.D \aOPIQ = _* = (2) at (A) dt (2.1.4) SECS. 2.4.2-2.4.4 to zero. Hence R\P/Q\t* -. R'WR X ^v^ - Rto)R X r) 70 _|_ R'aR X r (2.3.1) 81 + R'wR X (*vp/<> + R'?* X r) (A) = RoPIQ + 2 R'u>R X Bvp^ + ?'?* X r + R'u>R X (*?* X r) Hence, at time ?*, ?V/0|e* = ^a^^lt* + 2R'u>R X V*^!^ 2.4.2 The velocity of a point P relative to a point Q in a refer- ence frame R differs from the velocity of Q relative to P in R only in sense; that is, RyP/Q = ?RyQIP (1) Similarly, the accelerations Rap/Q and RoQIP are related by RQPIQ = ?RoQiP (2) This follows from Sec. 2.4.1 and the fact that the position vector of P relative to Q differs from that of Q relative to P only in sense. 2.4.3 The velocity and acceleration of a point P relative to a point Q in a reference frame R (see 2.4.1) are both equal to zero if P and Q are fixed in R, because the position vector of P relative to Q is then fixed in R (see 1.1.1) and all of its time-derivatives are equal to zero (see 1.2.2). The converse is not necessarily true. For example, if P and Q move on opposite sides of a rectangle fixed in R, in such a way that line PQ remains parallel to one side of the rectangle, the velocity and acceleration of P relative to Q in R are equal to zero although P and Q are not fixed in R. 2.4.4 Given n points Pif i = 1,2,..., n, the velocity and acceleration of the first point relative to the nth in a reference frame R can be expressed as 82 SEC. 2.4.5 [Chapter 2 RyPl/Pn -- RyPl/Pt -[- RyPt/P, + . . . + R^Pn-l/Pn (1) and ?aPi/p? = ?aPi/P? + ?aiVP. + . . . + RQPn-i/p* (2) Proof: The position vector rPl/p? of Pi relative to Pn (see Fig. 2.4.4) can be expressed as FIG. 2.4.4 Hence RyPl/P* (2.4.1) tU ? RyPl/Pt _|- # -|_ and (2.4.1) (2.4.1) (2.4.1) (1.4.1) d.4.1) dt ' ' ' dt 2.4.5 If two points P and Q are fixed on a rigid body Ry the velocity R'vPIQ and acceleration R'oPIQ of P relative to Q in a refer- ence frame Rf are given by (see Fig. 2.4.5a) R'vPIQ = R'uR x r (1) R'QPIQ = ?'a? x r + R'wR X ft'vp/Q (2) Kinematics] SEC. 2.4.5 83 Proof: R\P/Q FIG. 2.4.5a (2.4.1) dt (2.1.1,2.2.1) in/ lRy (2.4.1) dt (1.5.3)*'a* X r + (2.3.1) X (2.4.1) Problem: Regarding the earth as a rigid sphere R whose angu- lar velocity RR X R'vp/Q (2) 2.5 Absolute velocity and acceleration 2.5.1 Given a point P moving in a reference frame R, and a point 0 fixed in R, the velocity and acceleration of P relative to 0 in R (see 2.4.1) are independent of the position of 0. This rela- tive velocity and acceleration are called, respectively, the absolute velocity of P in R and the absolute acceleration of P in R, or, for short, the velocity of P in R and the acceleration of P in R, and are denoted by V and Rap. It follows from Sec. 2.4.1 that (see Fig. 2.5.1) v _ *? (i) Kinematics] SEC. 2.5.1 85 RJRyP dt (2) Proof: Let 0 and O' be two points fixed in R\ r the position vector of 0' relative to 0; p the position vector of P relative to 0; p' the position vector of P relative to 0', as shown in Fig. 2.5.1. FIG. 2.5.1 It is to be shown that RyP/0 _ RyP/O' and From Fig. 2.5.1, p = r + p' Differentiate with respect to time t in R: Rdp = Rdr W dt dt dt Now = dt (2.4.1) Rdr 0, Hence Differentiate: tW (2.4.4) RyPIO = RyP/O' ~dt ~dt (2.4.1) 86 But Hence SECS. 2.5.2-2.5.3 [Chapter 2 dt (2.4.1) dt (2.4.1) RQP/O = RQP/O' 2.5.2 If a point P is fixed in a reference frame R, the velocity and the acceleration of P in R (see 2.5.1) are equal to zero. The converse applies to velocities, but not necessarily to accelerations. That is, if the velocity of P in R is equal to zero during a certain time interval, P is fixed in R during this time interval; but the acceleration of P in R may be equal to zero while P is moving in R. This occurs, in fact, whenever P moves in such a way that Rvp is independent of tin R (see 1.2.8). 2.5.3 The velocity (*vp) of a point P in a reference frame R (see 2.5.1) is parallel to the tangent at P to the curve C on which P moves in R. It can, therefore, be expressed in the form V = vTr (1) where r is a vector tangent of C at P (see 1.9) and vT is a scalar, called the speed of P in R for the T direction. vT is positive when Rvp and r have the same sense, vanishes when Rvp is equal to zero, and is negative when the sense of Rvp is opposite to that of r. Furthermore, if s is the arc-length displacement of P relative to a FIG. 2.5.3 Kinematics] SEC. 2.5.3 87 point Po fixed on C, and r points in the direction in which P moves when s increases algebraically, then Proof: Let p be the position vector of P relative to a point 0 fixed in R (see Fig. 2.5.3). Then V = = = r (2.5.1) dt (i.e.!) ds dt (1.9.3) dt This shows that Rvp is parallel to the tangent to C at P. Rvp can, therefore, be expressed in the form Ryp = VTT where r may be either of the two vector tangents of C at P (see 1.9.2). Furthermore, ds V' = Jt if (see 1.9.3) r points in the direction in which P moves when s increases algebraically. Problem: Referring to Problem 1.10.1 and Fig. 1.10.1b, and supposing that point P oscillates on line AC in such a way that the displacement x of P relative to point C, regarded as positive when P is between A and C, is given by x = 1 + sin 2rt ft where t is the time in seconds, determine the speed v of P for the direction AC at t = ? sec and t = 1 sec, noting that P occupies the same position at these two instants. Then, supposing that the sheet ABCD has been folded to form the helix H (see Fig. 1.10.1c), determine the velocity v of P at t = ? sec, expressing it in terms of the unit vectors n and Ic. Solution: As the curve on which P moves is a straight line, x is equal to the arc-length displacement of P relative to C; and, as x decreases algebraically when P moves in the direction AC, v is given by dxv = ?37 = ?2T COS 2rt(2) dt 88 SEC. 2.5.4 [Chapter 2 Hence v\immk = 2TT ft sec"1, v\t=l = ?2TT ft sec"1 When the sheet has been folded, x is still equal to the arc- length displacement of P relative to C. The velocity v of P is, therefore, given by dx V = T7T (i,2) at where r must be the vector tangent which points in the direction in which P moves when x increases algebraically. Using the nota- tion introduced in the solution of Problem 1.10.1, this vector tan- gent is given by (q/b)lc X n + k_ _ |p'| [(a7&2) Hence o (a/b)lc X n + and (o/b)lc Xn+k ''-* [(a2/^2) + 1]1/2 2.5.4 The acceleration (*ap) of a point P in a reference frame R (see 2.5.1) is parallel to the osculating plane at P of the curve C on which P moves in R (see 1.10.5). It can, therefore, be expressed in the form Rap = aTr + avv (1) where r and v are, respectively, a vector tangent and the vector principal normal (see 1.9 and 1.11) of C at P, and aT and a, are scalars. The vectors aTr and avv are called, respectively, the tan- gential acceleration and the normal acceleration of P in R. aT is called the scalar tangential acceleration of P in R for the r direction, and may be positive, negative, or zero, a, is called the scalar nor- mal acceleration of P in R, and is never negative. Furthermore, if s is the arc-length displacement of P relative to a point Po fixed on C, and r points in the direction on which P moves when s increases algebraically, then ar~ dt * Iff (2) Kinematics] SEC. 2.5.5 89 and p p\dt) where vT is the speed of P in R for the r direction (see 2.5.3) and p is the radius of curvature of C at P (see 1.12.2). Proof: RQP = *^fvp = Rd_ ^^ (2.5.1) a^ (2.5.3) dt dvT . Rdr= -J7 r + v T ?(1.5.1) at at __ a\ , Rdr ds dt T ds dt (1.6.1) "~ dt T VT p VT (1.13.1),(2.5.3) or R P __ dVr i^L a " dt r p V This shows that Rap is parallel to the osculating plane of C at P (see 1.10.5). Rap can, therefore, be expressed in the form Rap = aTr + avv and it follows that dvT d2s ar ~ 17 = 17i at (2.5.3) ttf a, = ? = - P (2.5.3) P 2.5.5 Whether or not it is convenient to work with tangential and normal accelerations depends, in part, on the ease with which the radius of curvature of the curve on which the motion takes place can be determined. One curve for which this determination presents no difficulties is the circle: The radius of curvature of a circle is equal to the circle's radius. Problem: Starting from rest, an automobile reaches a speed of 90 miles per hour in one minute while travelling with uniformly increasing speed on a circular track having a radius of 500 ft. 90 SEC. 2.5.5 [Chapter 2 Determine the magnitude of the acceleration of a point P on the automobile (a) at the initial instant of the motion, (b) 30 sec- onds later, and (c) one minute after starting. Solution: Let r be the radius of the circle C on which P moves, r a vector tangent of C at P, v the speed of P for the r direction, a the acceleration of P. Then (2.5.4) T " (2.5.4) L\dt/ r* J The phrase "with uniformly increasing speed" is ambiguous, because it does not contain any mention of the variable as a func- tion of which the speed is increasing. However, this phrase is gen- erally regarded as referring to the time-rate of change of the speed. That is, it means that dv . . ?; = c, a constant Hence Let t = 0 From Eq. Thus and, again at the (A), i from v = ct + c'y initial instant v\u v\t. c Eq. (A), of -0 -0 t = c' a constant the motion. Then = 0 = c' ?? 0 (A) v = ct (B) At t == 1 min, P has a speed of 90 mile hr"1: t>|t?imin = 90 milehr"1 From Eq. (B), t>|R X Rvp/Q in the expression for the acceleration are then, respectively, the tangential and the normal acceleration of P in R' (see 2.5.4). Problem: Referring to Example 2.3.8, determine the tangen- tial acceleration aT and the normal acceleration a, of point B at time t*. Solution: Fig. 2.5.6 shows bar B\) the angular velocity a>i (as found in Example 2.3.8) and angular acceleration en of this bar; and the unit vectors iii, n/ and k, previously shown in Fig. 2.3.8b. FIG. 2.5.6 92 SEC. 2.5.7 [Chapter 2 Let r be the position vector of B relative to A. Then, as A is fixed on the axis of rotation of Bh Qr = ttl x r = 5k X (10nO = 50ni' ft sec"2 and a, = Wl x v* = -3k X vB where v* = w X r = ?3k X (lOnO = ? 3(W ft sec"1 Hence a, = -3k X (-30ni') = -90m ft sec~2 2.5.7 The motion of a point P is said to be rectilinear in a reference frame R when the curve on which P moves in R is a straight line L. The velocity and acceleration of P in R are then parallel to this line, and are given by and ? P dx v =^ tfx a? (1) (2) where (see Fig. 2.5.7) n is a unit vector parallel to L and x is the R FIG. 2.5.7 displacement of P relative to a point 0 fixed on L, x being regarded as positive when the position vector p of P relative to 0 has the same sense as n. Kinematics] SEC. 2.5.8 93 Proof: As x is positive when p has the same sense as n, p = xn Now /? P *^P ^ / N dx v T/T = ^(xn) = 17n (2.5.1) ?* (1.2.5) ?t (1.5.1.1.2.2) ?^ Next = n (2.5.1) dt (1.2.5) d< \d< / (1.5.1,1.2.2) dt% 2.6.8 The expressions given in the preceding section can be obtained directly from Sees. 2.5.3 and 2.5.4 by noting that n is a vector tangent of the curve (L) on which P moves and regarding the radius of curvature of this curve as infinite. Furthermore, Sees. 2.5.3 and 2.5.4 lead to the following expressions, applicable to rectilinear motion: V = vn, Rap = an (1) where dx dv /ox v - * ? ? - Jt (2) and v and a are then called, respectively, the speed of P in R for the n direction and the scalar acceleration of P in R for the n direction. Each of the four quantities t} x, v, a may be plotted on rectan- gular axes as a function of any one of the remaining three; and these plots, called motion curves, can be used to solve certain prob- lems by graphical methods. The following properties of motion curves form the bases of such solutions: 1. The slopes of the x-t and v-t curves are proportional to v and a. This follows from the definitions of v and a, Eqs. (2), above. 2. The areas under any portions of the v-t and a-t curves are proportional to the corresponding algebraic increases in x and v} for these areas are proportional to I v dt and f adt or, replacing v with dx/dt and a with dv/dt, to L Ttdt and L Jtdt 94 SEC. 2.5.8 [Chapter 2 so that, integrating, one obtains x\tl - x\tl and v\tt - v\h 3. The product of the slope at, and ordinate of, a point on the v-x curve is proportional to a, for this product is given by dvdtdv dxV dt dx V dt dx/dt v= a - v 4. The area under any portion of the a-x curve is proportional to one-half of the algebraic increase in v2, for this area is propor- tional to fXt i fXtdv, [xtdvdx, fxtdv .I adx = I -ydx = I T~~lA x = \ Tv dxJXI JXl dt JXI dx dt JXl dx -L? (&*) dxdx - 0>U)2] As the relationship between the quantities ty x, v, a is precisely the same as that between the quantities t, s, vT, aT in Sees. 2.5.3 and 2.5.4, the above theorems apply to the latter quantities when x is replaced with s, v with vT, and a with aT. Problem: Figure 2.5.8 shows the aT-s curve for a point P which moves on a helix. It is known that P has a velocity of 12 ft sec"1 when s = 7 ft. Determine the speed of P for the r direction when s = 1 ft. 12 10 8 1 c rr sec V z \ \ \ 1 1 \ \ \ 3 4 S (ft) FIG. 2.5.8 Kinematics] SEC. 2.5.9 95 Solution: Let (t>T)i and (vT)2 be the speeds of P for the r direc- tion when s = 1 ft and s = 7 ft, respectively, and let A be the area under the aT-s curve between s = 1 ft and s = 7 ft. Then (t>r)2 = it 12 ft sec"1 and so that A = KW22 - W12] = J[144 - W12] (tv)i = db(144 - 2Ayi* Find A by counting squares in Fig. 2.5.8: A = 20 squares = 40 ft2 sec"2 Then (t>T)i = ?(144 - 80)1/2 = ?8 ft sec"1 2.5.9 The velocity R'vp and acceleration R'ap of every point P of a rigid body R in a reference frame /?' can be found as soon as the angular velocity R'o)R, angular acceleration R'aR, velocity R'vQ, and acceleration R'aQ are known, Q being any point fixed on R: R\P = v^ + R'^R X r (1) R'OP = R'QQ + ?'<** x r + * o>* X (* ?^ X r) (2) where r is the position vector of P relative to Q. (Q is called a base-point.) Proof: Let O be a point fixed in Rf (see Fig. 2.5.9a), p the position vector of P relative to 0, and q the position vector of Q relative to O. Then FIG. 2.5.9a 96 or Next R>QP SEC. 2.5.9 R'dp *'d, (2.5.1) d< (1.2.5) d? (2.5.1) (2.4 R'dR'yp R'd (PV, (2.5.1) d< (A) dt ft //ft v^ s/ /,|/? d.4.1) d? d^ = ^a^ + *'?* X r +? (2.5.1) (2.3.1) d.4.1) dt )R X r .1,2.4.5) (1.5.3) '?* X (^'to1 (2.4.1,2.4.5) ) X 2 X [Chapter 2 R'dr (A) R'dr dt r) Problem: Referring to Example 2.3.8, and using the results obtained in Problem 2.5.6, determine the acceleration ap at time t*, P being a point fixed on bar B2, four feet to the left of point B. Solution: Figure 2.5.9b shows bar B2, the angular velocity ?2 and angular acceleration ot2 of this bar (previously shown in Figs. 2.3.8c and 2.3.8d), and the acceleration aB of point B, resolved into the two components aT and a, found in Problem 2.5.6. (The unit vectors k, ni, n/, etc., are those previously shown in Fig. 2.3.8b.) # Kinematics] SEC. 2.5.10 97 The position vector r of P relative to B is given by r = ?4n2 ft Hence, using B as base point, aF = afl + a2Xr+tf2X(?2X r) = -90n! + 5(W + 11.33k X (-4n2) + (-2k) X [(-2k) X (-4n,)] = -90ni + 5(W - 45.32n2' + 16n2 ft sec"2 2.5.10 The velocity relationship of Sec. 2.5.9 furnishes a con- venient means for the determination of the velocity of a point P fixed on a rigid body R which is rolling on a surface S. For, by definition of rolling, there exists at every instant during such a motion at least one point PR of R (see Fig. 2.5.10) which is in FIG. 2.5.10 contact with a point Ps of S and whose velocity (in any reference frame) is equal to that of Ps> (If suR is parallel to the tangent plane of S at Ps, R and S are said to be in pure rolling contact. If s, \p (and their time-derivatives), the radius of D being r. 98 SEC. 2.5.11 [Chapter 2 Solution: Using C as base point, V7 = 0 and X r (2.5.9) . where r = ?rni and piaD = ? cos 0ni ? 0n2 + (^ + sin 0)n3 (E2.2.10) Hence Next PQD* = ?J1 = ^- PV (2.5.1) ?C (2.1.4) dt = 3: v + ('o>* + a)') X v dt (2.2.7) = ? r[(# + sin (? + <^ cos ^)n2 + 0n3] + [-^n2 + <^(-cos 0iii + sin 0n3)] X pvD* That is ^a^* = r[62 + * sin 0(^ + 6 cos 0]n2 + r[-0 + 0 cos 0(^ + sin 0)]n3 2.5.11 At any instant at which the angular velocity R'm)R of a rigid body R in a reference frame Rr is equal to zero, the velocities of all points of R in R' are equal to each other (see 2.5.9). If R'* X r* (2.5.9) 100 SEC. 2.5.11 [Chapter 2 that is, the equation w J R'^R = R'yQ + R'^R X r* or, subtracting R'vQ from both sides, R'WR >< Q and cannot be equal to R'v*f because R'coR X a is not equal to zero. Problem: Given a rigid body R and a reference frame R\ show that (a) when R moves in such a way that one point of R remains fixed in R', the instantaneous axis of R in R1 passes through this point, and (b) when R moves in such a way that a line fixed in R remains fixed in R'y the instantaneous axis of R in Rf coincides with this line. Solution (a): Let the fixed point be the base point Q. Then R\Q = o Kinematics] SEC. 2.5.12 101 and the position vector r* of one point on the instantaneous axis, relative to the fixed point, is equal to zero, which shows that the instantaneous axis passes through the fixed point. Solution (b): Let any point of the fixed line be the base point Q. Then, as in (a) above, it follows that the instantaneous axis passes through this point. Hence the instantaneous axis passes through every point of the fixed line, that is, it coincides with it. 2.5.12 If at any instant t* during the motion of a rigid body R in a reference frame R' the velocity of one point of R (or R ex- tended) is perpendicular to R'wR, the velocity of any other point of R is either also perpendicular to R'o>R at this instant, or it is equal to zero. (This follows from the fact that the velocities of all points of R have equal resolutes parallel to R'wR.) R is then said to be in a state of plane motion in Rf at this instant. In accordance with Sec. 2.5.11, the velocity in Rf of every point of R lying on the instantaneous axis of R in Rf is equal to zero whenever R is in a state of plane motion in Rr. For this reason every point of the instantaneous axis is called an instantaneous center of R in Rf. Furthermore, the velocity R'vp (but noty in gen- eral, the acceleration ^'a**) of any point P of R has at this instant the value which it would have if the instantaneous axis were fixed both in R and in R' and R were revolving about this axis with an angular velocity equal to the actual angular velocity at this in- stant; for, at time t*, R'vp is given by *y = R'wR X r (2.5.9) where r is the position vector of P relative to any instantaneous center, and, from Sec. 2.5.6, the velocity of P during the fictitious motion described above is given by the same expression. To locate the instantaneous axis, and thus all instantaneous centers, of a body in plane motion, it is sufficient to know the ori- entations of the velocity vectors R'vPl and R'vPt of any two points Pi and Pi of R, provided these velocities be nonparallel: The in- stantaneous axis is the intersection of two planes passing through Pi and P2, each plane being perpendicular to the velocity vector of the point through which it passes. 102 SEC. 2.5.12 [Chapter 2 Proof: Let P* be a point on the instantaneous axis. Then the position vector ri* of P* relative to Pi is given by 1 W(2.5.11) ( RQ)R)2 where Xi is a certain scalar, and, similarly, the position vector r2* of P* relative to P2 is given by '2 Now, ri* is perpendicular to R'vPl, and r2* to fl/vPt, as r2* . *'vp? = Xj^w72 ? ?vPs and R/?R is perpendicular to both RfvPl and ftVP2 when 72 is in a state of plane motion in 7?'. Thus P* lies in a plane which passes through Pi and is perpendicular to R'vPl, and P* also lies in a plane which passes through P2 and is perpendicular to R'vPt. Accordingly, P* lies on the intersection of these two planes. Problem: Referring to Example 2.3.8, locate an instantaneous center of bar 7?2 at time t*, and use it to determine the angular velocity of this bar at this instant. FIG. 2.5.12 Kinematics] SEC. 2.5.13 103 Solution: Point C (see Fig. 2.5.12) moves on a circle, center at D. Hence the velocity of C is perpendicular to CD, that is, line CD lies in a plane which passes through C and is perpendicular to the velocity vector of C. Therefore, line CD (or line CD ex- tended) passes through the instantaneous axis of bar B2. Similarly, point B moves on a circle, center at A, whence line AB (or line AB extended) passes through the instantaneous axis of bar B2. It follows that point Q, the intersection of lines CD and AB, is an instantaneous center of bar B2. The velocity of C, when C is regarded as a point of bar Bz, is given by vc = 24n2 ft sec"1 (2.5.6) Alternatively, when C is regarded as a point of bar B2?and B2 as a rigid body rotating about a line which passes through Q, is parallel to k, and is fixed in bar B2?then vc is given by vc = -12o>2n2 (2.5.6) where co2 is the angular speed of B2 for the k direction. Hence 24 = -12a>2 and the angular velocity of B2 at time t* has the value ? 2k rad sec"1. 2.5.13 Given two reference frames R and Rr, the velocities and R'yp of a point P at an instant t* are related as follows: R\P = RyP + R\P* (1) where P* is that point fixed in R which coincides with P at time t*. The accelerations Rop and R'ap at time t* are related by R V = *aF + *'ap* + 2 *'?* X *vp (2) BVF* and K'a1** are called, variously, the drag-, vehicle-, transport-, or coincident point velocity and acceleration for point P in the refer- ence frames R and Rf. The vector 2R'?* X Rvp is called the Coriolis acceleration of P for the reference frames R and R'. Proof: Let 0 be a point fixed in R, Or a point fixed in Rf, p the position vector of P relative to 0, p' the position vector of P rela- 104 SEC. 2.5.13 [Chapter 2 tive to 0', and r the position on vector of 0 relative to 0', as shown in Fig. 2.5.13a. Then R'yP* FIG. 2.5.13a fl'yO _[_ R'^R >< (2.5.9) 'ap* = * 'a0 + R'CLR X p + R'?R X (2.5.9) X p) (2.5.1) dt " (2.5.1) ?? (2.1.4) (C.D.E) (F.A) Next p _ *'d^vp _ ft/d_ (2.5.1) d< (F) d< *** + R'R X Rvp + R'OLR X p (2.5.1) (2.5.1) (2.3.1) X + (E) *'?* X p) (B) Problem: An airplane P flies due south at a constant (low) altitude above a meridian of longitude, with a constant speed of 600 mile hr"1. At a certain instant the line joining P to the center O of the earth makes an angle of 45 degrees with the earth's north- south axis. Letting R be a reference frame fixed in the earth, and Rr a reference frame in which the earth's center is fixed and whose motion is such that R't?)R = ? k rad hr"1 where k (see Fig. 2.5.13b) is a unit vector fixed in both R and R', determine the east-west, north-south, and up-down components of the acceleration of P in R' at this instant. FIG. 2.5.13b Solution: Let P* be the point of the earth which coincides with P at the instant under consideration, p the position vector of P (or P*) relative to 0, ni a unit vector perpendicular to line 106 SEC. 2.5.13 [Chapter 2 OP and parallel to the meridian plane passing through P (ni points southward at P), n2 a unit vector perpendicular to the meridian plane passing through P (n2 points eastward at P), n3 a unit vector parallel to line OP (n3 points upward at P). In ft, P moves on a circle, center O, radius 3960 miles. See Sees. 2.5.5 and 2.5.4: The tangential acceleration of P in R is equal to zero, because the speed of P in R for the ni direction (see 2.5.3) is constant (600 mile hr~l). The normal acceleration of P in R has the value (600)2 , 3960 ^n^ = ~"91n3 mile hr~2 ? n3 being the vector principal normal of the curve on which P moves in R. Hence Rop = -91n3 mile hr~2 Using 0 as base point (see 2.5.9), find the transport accelera- tion of P for the reference frames R and R': R'QP* = R'Qo + R'^R x p + R'u*R X (* '?* X p) (2.5.9) where *'a? = 0 (2.5.2) /?' * R'd UK' * R'dk n, (2.3.1) dt 12 dt (1.2.2) *?* X (R'uR Xp)=^x[~kX (3960n3)] = -136(ni + n3) mile hr~2 Thus R a^ = -136(11! + ns) milehr"2 The velocity of P in R is given by RyP = 600lh (2.5.3) Hence the Coriolis acceleration of P for the reference frames R and QP (2) RQP R>QP* x = -91n3 - 136(11! + n3) + 222n2 = - 136nx + 222n2 - 227n3 mile hr"2 Result: The acceleration of P in R' has the following compo- nents: 136 mile hr~2, northward; 222 mile hr~2, eastward; 227 mile hr~2, downward. 2.5.14 When a rigid body R rolls on a surface S (see 2.5.10) in such a way that only one point of R is in contact with S at any instant, the points PR of R which successively come into contact with S form a curve CR on R, and the points Ps of S which come into contact with R form a curve Cs on S (see Fig. 2.5.14a). These curves have a common tangent at their point of contact; and if P% and P% are the points of the two curves which coincide with each other at an instant ?*, and sR and ss are the arc-length dis- FIG. 2.5.14a placements of PR relative to P% and of Pf relative to Pf, then sR = ss Proof: There exists a point P which remains at all times on both CR and Cs. (P coincides with PR and Ps.) In accordance with Sec. 2.5.3, 108 SEC. 2.5.14 [Chapter 2 where rR and rs are vector tangents of CR and Cs at PR and Ps. Now SyP _- rtyP _^_ SyP* and, by definition of rolling (see 2.5.10), SyPR ?- SyPs But, Ps being a point of S} Hence and JvPs = 0 SyP = t dss which shows that CR and Cs have a common tangent at their point of contact and that dsR _ d?s dt " dt Consequently SR = ss + const. and, as both SR and ss vanish at t*9 it follows that SR = ?s FIG. 2.5.14b Kinematics] SEC. 2.5.15 109 Problem: A block R, placed on a right-cylindrical surface S of radius r (see Fig. 2.5.14b), performs oscillations during which R rolls on S. For an instant at which R occupies the position shown in Fig. 2.5.14c, determine the distance sR between points P% and PR. Solution: While R is not a body only one point of which is in contact with S at any instant, the face of R which contains points P% and PR may be regarded as such a body. Hence where ss is the length of the circular arc P*Ps (see Fig. 1.5.14c). As the angle P%-0-Ps is equal to 0, Hence ss = rd sR = rd 2.5.15 The definitions of absolute velocity and acceleration (see 2.5.1) are such that the absolute velocity of a point P in a reference frame R is a special case of a relative velocity of P in R (see 2.4.1); similarly for accelerations. Conversely, the velocity RvPIQ and acceleration *aF/0 of a point P relative to a point Q in a reference frame R can be expressed in terms of the absolute velocities Rvp and RvQ and absolute accelerations Rap and RaQ of P and Q in R: 110 SEC. RyPIQ ? RQPIQ = 2.5.15 RyP __ RyQ RaP _ RQQ [Chapter 2 (1) (2) Proof: Let 0 be a point fixed in /?. Take n = 3 in Sec. 2.4.4, and let Pi = P, P2 = 0, P3 = Q. Then and RyP/Q = RyPIO (2.4.4) (2.4.4) _ /2VP _ RyQ (2.5.1) (2.4.2,2.5.1) ? RQP __ ?aiz?, the sec- ond moment of P with respect to 0 for the pair of directions ih, n3. Solution: %? = #f/? . n3 = (160ni + 120n3) ? n3 = 120 slug ft2 3.1.5 The following is an alternative expression for ^? (see 3.1.4): 4>T = N(na X p) ? (n6 X p) Proof: % nb = N[p, na X p, n?] (3.1.4) (3.1.1) = N[na X p, n6, p] = AT(na X p) ? (n6 X p) 3.1.6 The expression for U? (see 3.1.4) given in Sec. 3.1.5 shows that because the expression remains unaltered when na and n& are interchanged. 3.1.7 When the unit vector n6 is equal to the unit vector na the expressions given in Secs. 3.1.4 and 3.1.5 become, respectively, C/o = *aP/0 ? na (1) and t%? = #(na X p)2 (2) From the second of these it follows that if ll/o is the distance from point P (strength N) to a line La parallel to na, and 0 is any point on La (see Fig. 3.1.7), then 4%? = N(C?)2 (3) aa? is called the second moment of P with respect to line La. 116 SEC. 3.1.8 [Chapter 3 Problem: Referring to Example 3.1.1, determine the second moment of P with respect to (a) line OR and (b) line OQ. P(N) FIG. 3.1.7 Solution (a): Let ^? be the second moment of P with respect to line OR, and 1%/O the distance from P to this line. Then = 4?i? ? n2 = (250n2) ? n2 = 250 slug ft2 (1) (E3.1.1) Alternatively, %? = 10(Z^/o)2 = 10(25) = 250 slug ft2 (3) Solution (b): Let na be a unit vector parallel to line OQ, 4>ll? the second moment of P with respect to line OQ, and r the position vector of P relative to Q. Then 3ni ? 5n2 ? 4n3 na = =t 7= V50 r = 5n2 ft V50 and *? = 10(no X r)? = 10(25)g>+16) = 125 slug ft*(2) 50 3.1.8 Given three mutually perpendicular unit vectors in, n2, n3 and the six associated second moments 4>v/Q, i, j = 1, 2, 3, of a Second Moments] SEC. 3.1.9 117 point P with respect to a point 0 (see 3.1.4 and 3.1.6), the second moment #?/o of P with respect to 0 can be expressed as where a? is,the nt measure number of no. Proof: Y0 (A) (3.1.2) ?rj If riy, j = 1, 2, 3, are mutually perpendicular unit vectors, the following is an identity (see Vol. I, Sec. 1.14.9): 3 *f/o = V* *f/? . n n (B) y-iBut ^f/? ? ny = %/o (C) (3.1.4) Hence 3 a/? of P with respect to 0 for a pair of directions no, n& (see 3.1.4) can be expressed as where a, and 6? are the n, measure numbers of no and i Proof: (3.1.4) (3.1.8) \i-iy-l 118 SEC. 3.1.10 [Chapter 3 i\b, resolved into components parallel to n?, i = 1, 2, 3, is given by nb = V 6?nt Hence n, ? n6 = nt fly ? flt = 0 3 3 But unless i ? jy in which case Hence and 3.1.10 Given three mutually perpendicular unit vectors Hi, n2, n3 and the six associated second moments v/o, i, j = 1, 2, 3, of a point P with respect to a point 0 (see 3.1.4 and 3.1.6), the second moment of P with respect to a line La passing through 0 (see 3.1.7) can be expressed as 51 0 FIG. 3.1.10a .P/O i 1 2 3 1 80 0 60 j 2 0 125 0 3 60 0 45 FIG. 3.1.10b Second Moments] SEC. 3.1.10 119 where a? is the nt- measure number of a unit vector na parallel to Lo. This is an immediate consequence of Sec. 3.1.9 and the definition of the second moment of a point with respect to a line, Sec. 3.1.7. Problem: In Fig. 3.1.10a, nh n2, n3 are mutually perpendicular unit vectors. The six second moments f/?y i, j = 1, 2, 3, of a point P (not shown) with respect to point 0 are recorded in Fig. 3.1.10b. Determine the second moment of P with respect to line OQ. Solution: Let no be a unit vector parallel to line OQ (see Fig. 3.1.10a), a< the n? measure number of no. Then 3fh ? 5n2 ? 4n3 3 -5 JZ? ai " vf>' ?2 = Vw az ~ Vso and, letting U? denote the second moment of P with respect to line OQ, t?l;-l t-1 ? = 62.5 120 SECS. 3.2.1-3.2.4 [Chapter 3 3.2 Second moments of a set of points 3.2.1 Given a point O, a set S of points P?, i = 1, 2, . . . , n, of (dimensionally homogeneous) strengths Nif and a unit vector no, the vector *f/o, defined as 1=1 where &%i/o is the second moment of F? with respect to 0 for the direction na (see 3.1.1), is called the second moment of S with respect to 0 for the direction na. (In general, #?/0 is not parallel to na. See 3.1.3.) 3.2.2 Given a point 0, a set S of points Pi} i = 1, 2, . . . , n, of (dimensionally homogeneous) strengths AT?, the scalar <^?, de- fined as where ^?/o is the second moment of Pi with respect to 0 for the pair of direction na, iu (see 3.1.4), is called the second moment of S with respect to 0 for the pair of directions no, n&. 3.2.3 When the unit vector nb is equal to the unit vector na, the expression given in Sec. 3.2.3 becomes and (see 3.1.7) aa? is the second moment of P? with respect to the line La which passes through 0 and is parallel to na. Accord- ingly, aa? is called the second moment of S with respect to line La. 3.2.4 As a consequence of the definitions given in Secs. 3.2.1, 3.2.2, and 3.2.3, the relationships between second moments of a point, discussed in Secs. 3.1.2-3.1.10, are equally valid for second moments of a set of points. That is, ?> (l) (3.1.2) frf = *?/0-nfc (2) (3.1.4) Second Moments] SEC. 3.2.5 121 i-S/O A.S/O /Q\H>ab ? aa? of a set ? with respect to a line La (see 3.2.3) can be expressed as the product of a quantity N, called the strength of S, and defined as # = |>< a) and the square of a positive, real quantity ki/o (which has the dimensions of length), that is, k?/o exists such that then ka/o is called the radius of gyration of S with respect to line La, and is equal to the distance between line La and any point of strength N whose second moment with respect to La (see 3.1.7) has the value f/?, i, j = 1, 2, 3, with respect to point 0 are tabulated in Fig. 3.2.5b. Determine the radius of gyration of S with respect to line OQ. Solution: Let no be a unit vector parallel to line OQ (see Fig. 3.2.5a), k%/o the desired radius of gyration, N = 10 slug the strength of S, and aa? the second moment of S with respect to line OQ. Then kS/O = and, letting a, be the n, measure number of no, 122 Hence SEC. 3.2.6 3 3 (3.2.4) t-li-1 kS/o = (6.25)1/2 = 2.5 ft [Chapter 8 id, = 62.5 slug ft2 (P3.1.10) 31 slug i o ft2 1 2 3 1 80 0 60 3 2 0 125 0 3 60 0 45 FIG. 3.2.5a FIG. 3.2.5b 3.2.6 Given a set S of n points, a point 0, and two directions no and n&, the second moments of S with respect to 0 and with FIG. 3.2.6 Second Moments] SEC. 3.2.6 123 respect to the centroid P* of S (see 3.2.1 and 3.2.2) are related as follows: (1) where ^T/o and ^/0 are second moments of the point P* (see 3.1.1 and 3.1.4) regarded as having a strength equal to the strength N of S (see 3.2.5). Proof: Let Pt, i = 1, 2, . . . , n, be the points of S, N* the strength of P?, p? the position vector of P? relative to 0 (see Fig. 3.2.6), p* the position vector of P* relative to 0, and r, the position vector of P? relative to P*. Then Pi = P* + r, (A) *f/? = iVp*X(naXp*) (B) (3.1.1) N = ViV, (C) (3.2.5) ^Ti and (see Vol. I, Sec. 2.4) ? iV.f .? = 0 (D) Hence 4^1/o = V Of'70 = y\ NiPi X (na X Pi) (3.2.1) ft-J (3.1.1) fTi (A) /r j2 N<) p*x (na x p*) + p*x (n-x E if>) X ^ X p*) + X) ^* X (n. X r*) + o + o + f (B.C) (D) (D) (3.1.1,3.2.1) Next 4U? = ^f/o ? n6 = (3.2.4) (1) (3.1.4) (3.2.4) 124 SBCS. 3.2.7-3.2.8 [Chapter 8 3.2.7 Given a set S of n points, the strength N of S (see 3.2.5), a line La parallel to a unit vector no and passing through a point 0, a line LJ parallel to na and passing through the centroid P* of S, and the distance l?/o between the lines La and LJ (see Fig. 3.2.7), the second moments of ? with respect to the lines La and L; (see 3.2.2) are related to each other as follows: where 0 is any point on line Lo. u FIG. 3.2.7 Proof: Let nb be a unit vector equal to no. Then (3.2.6) The distance from P* to line La is equal to Va/O. Hence, when P* is regarded as having the strength N, (3.1.7) Thus 3.2.8 If ka/o is the radius of gyration of a set S with respect to a line La passing through a point 0 (see 3.2.5), kl/p* the radius of gyration of S with respect to a line LJ which is parallel to La and passes through the centroid P* of Sf and l%*/0 the perpendicular distance between the lines Lo and LJ (see Fig. 3.2.7), then Second Moments] SEC. 3.2.9 125 Proof: Using the notation of Sees. 3.2.5 and 3.2.7, (3.2.5) ? N (3.2.7) N +(3.2.5) 3.2.9 By successive use of expressions given in Sees. 3.2.4, 3.2.6, and 3:2.7 all second moments of a set S of points can be found whenever the following are known: (a) The strength of S. (b) The location of the centroid of S. (c) The six second moments of S with respect to one point for three mutually perpendicular directions. Problem: In Fig. 3.2.9a, nh n2, n3 are mutually perpendicular unit vectors. P* is the centroid of a set S of points. S has a strength N of 25 ft2, and the second moments of S with respect to point 0 have the values shown in Fig. 3.2.9b. Determine the second moment of S with respect to line AB. 41 (ft4) i 1 2 3 1 25 -8 -75 3 2 -8 10 -9 3 -75 -9 50 FIG. 3.2.9a FIG. 3.2.9b Solution: Let no be a unit vector parallel to line AB, and the desired second moment. Then (3.2.6) = (3.2.7) (3.2.4) 126 SEC. 3.3.1 [Chapter 3 Z/A = N(na X AP*)2 (3.1.7) Hence where ai = 1, a2 = 0, a3 = ? i UP* = 2 ft, AP* = 3m + 2n2 ft so that 3 3 ()(*t) = 113ft4 N(OP*Y = 25(4) = 100 ft4 N(na X ZP*)2 = 25[(lnx - ?n3) X (3nx + 2n2)]2 = 244 ft4 and *faM = 113 - 100 + 244 = 257 ft4 3.3 Principal directions, axes, planes, second moments, and radii of gyration of a set of points 3.3.1 Given a set S of points, a point 0, and a unit vector nz, n, is called a principal direction of S for 0 if and only if *f/o (see 3.2.1) is parallel to nz or equal to zero. When nz is a principal direction of S for 0, the line Lz parallel to n* and passing through 0 is called a principal axis of S for 0, the plane Pz passing through 0 and per- pendicular to nz a principal plane of S for 0, the second moment 4>i\/o a principal second moment of S for 0, and the radius of gyra- tion Af/O a principal radius of gyration of S for 0. In the above definitions the point 0 plays an important part: Only with reference to a specific point is it meaningful to speak of principal direction, axes, etc. Problem: Fig. 3.3.1 shows a set S of two points Pi and P2 of equal strength 2V. Determine whether or not the second moment of S with respect to line AB is a principal second moment of S for Second Moments] SEC. 3.3.1 127 (a) point A and (b) point B, and show that line PXP2 is a principal axis of S for every point on this line. A FIG. 3.3.1 Solution: Let na and n& be unit vectors parallel to AB and P1P2, respectively, and let ai, 02, bi, b2 be the position vectors of Pi and Pi relative to A and B, as shown in Fig. 3.3.1. Then #f/A = Na! X (nB X ai) + Na2 X (no X a2) (3.2.1,3.1.1) = N(5na - 3n6) This shows that na is not a principal direction of S for point A. Consequently, line AB is not a principal axis of S for point A, and the second moment of S with respect to line AB is not a prin- cipal second moment of S for point A. Next, *sa/B = Nhx X (n. X bx) + Nb2 X (n. X b,) = 5Mia Accordingly, na is a principal direction of S for point B} line AB is a principal axis of ? for point 5, and the second moment of S with respect to line A B is a principal second moment of *S for point B. Finally, the second moment of S with respect to every point of line P\P2 for the direction n6 is equal to zero. n& is thus a prin- cipal direction of S for every point of this line, and the line is a principal axis of S for every point on the line. 128 SECS. 3.3.2-3.3.3 [Chapter 8 3.3.2 A unit vector nz is a principal direction of a set S for a point 0 if and only if Proof: If *?/o = *?'?n. then *f/o is either parallel to n2 or equal to zero. Hence (see 3.3.1) n, is a principal direction of S for 0. Next, suppose that *?/o ^ <*?/on2 (A) Then n2 cannot be a principal direction of S for 0. For, #?/o cannot be parallel to n2, because any #f/o parallel to n2 can be expressed as which, after But so that and *?/o = Xn2 dot-multiplication with n2, gives *f/o * n2 = Xn2 ? n2 = X *f/o-n2 = ?/?(3.2.4) X = fz/onz and this again contradicts the hypothesis, Eq. (A). 3.3.3 If nz is a principal direction of a set S for a point 0 (see 3.3.1), the second moment yZ? is equal to zero for all directions ny perpendicular to nz. Proof: ?/ / ny = sJ?(nz ? ny) = 0 (3.2.4) (3.3.2) Second Moments] SECS. 3.3.4-3.3.6 129 3.3.4 If the second moments 4>xl? and 4>U? of a set S with respect to a point 0 for the two pairs of directions nx, nz and ny, n, are equal to zero, and nx and ny are each perpendicular to n* (and not parallel to each other), then nz is a principal direction of S for point 0. Proof: *&'? = *f/o ? nx, 4%? = *?/o ? ny (3.2.4) (3.2.4) Hence, by hypothesis, #f/o ? nx = 0, #f/o - ny = 0 and #f/o is either equal to zero or perpendicular to both nx and ny, that is, parallel to their common normal, nz. In either case, nz is a principal direction of ? for point O (see 3.3.1). 3.3.5 If n* is a principal direction of S for a point 0, and nx and ny are unit vectors perpendicular to nZ} the second moment xi? is not, in general, equal to zero. (Compare with 3.3.3 and 3.3.4.) 3.3.6 In certain situations, principal directions, axes, and planes can be located by inspection. For example, if the points of a set S are placed such that corresponding to every point on one side of a plane Pz there exists a point of the same strength on the other side, the two points lying on the same normal to, and being equidistant from, PZy then Pz is a principal plane of S for every point of Pz. Pz is then called a plane of symmetry of Sy and one may say, in short, that a plane of symmetry is a principal plane for each of its points. Proof: S may be divided into three sets of points: Say consist- ing of points lying in Pz; Sb, consisting of points lying on one side of Pz\ and Sc, containing the points lying on the opposite side of Pz. Consider the second moments of Sa> Sb, and Sc with respect to a point 0 lying in plane Pz, for the direction nz perpendicular to Pz. Sa'- Let a be the position vector relative to 0 of a typical point P of strength N. Then NQ X (nz X a) is parallel to n*. Hence #f/o is parallel to nz. Sb and Sc: Let b be the position vector relative to 0 of a typ- ical point of Sb, and c the position vector of the corresponding point of ?c (see Fig. 3.3.6). Then b and c can be expressed as 130 SEC. 3.3.7 [Chapter 3 r FIG. 3.3.6 b = d + /inz c = d ? /in2 and, as the two points have the same strength, say N, the contri- bution of this pair of points to i'2/o is equal to zero. Solution: The plane determined by the points is a principal plane of S' for 0. Hence n2 is a principal direction of S' for 0. As ni is perpendicular to n2, f2/o = 0 (3.3.3) Second Moments] SEC. 3.3.8 131 3.3.8 When a plane A is a principal plane of a set S for a point O (see Fig. 3.3.8a), there exist at least two principal directions of S for 0 which are parallel to A and perpendicular to each other. If nz is one of these principal directions, and in and n2 are unit vectors perpendicular to each other and parallel to A, the corre- sponding principal second moment of S for 0 is given by and the ratio of the ih and n2 measure numbers of nz has the value S = *&?-^? (2) The principal second moment corresponding to the principal direc- tion which is perpendicular to nz and parallel to A is given by FIG. 3.3.8a Proof: Let n3 be a unit vector perpendicular to A (and thus a principal direction of S for 0). Then, for any direction na, (3.2.4) t where a? is the nt measure number of nfl. Let na be perpendicular to n3, so that a3 = 0, and note that (see 3.3.3) U? = ?{? = 0 132 SEC. 3.2 It follows that that i and I /J.S *v = \yn "i -r via u2jni -p ^vi . is a principal direction of S for 0 if (3.3.2) is, using Eq. (A) to eliminate i al(^? - ? 4 ps/o - a2U and only if *i + a2n2) ? = 0 [Chapter 3 (A) (B) - 4>SaL?) = 0 (C) As na is a unit vector, ai and a2 must satisfy the further con- dition ai2 + a22 = 1 (D) which shows that ai and a2 cannot vanish simultaneously. From this it follows that Eqs. (B) and (C) can, in general, be satisfied simultaneously only if the determinant of the coefficients of ai and a,2 is equal to zero: or, expanding, - (f(? + 'MO)Sa?? + *fl/O^? ~ (U?)2 = 0 This quadratic equation has two real roots, which may be called yy? and 4>&?, and which give the values of two principal second moments of S for 0: Note that (E,F) so that Let 2/1 and y2 be the values of ai and a2 corresponding to 4>yy?, and ^i and 22 the values of ax and a2 corresponding Second Moments] SEC. 3.3.8 133 to aa? = ?z?- Replace ax with yh a2 with y2, and U? with 4>%? in Eq. (B), and solve for 1/1/2/2: 111 +&? (Q) 2/2 " *&'? - f{? ? Similarly, replace ai with Z\, a-i with z2, and aL? with 0?/o in Eq. (C), and solve for zi/z2: ?/? - 444? This shows that, provided 4>U? 1* 0, there exist two and only two principal axes of S for 0 which lie in plane A. One is parallel to the principal direction ny defined as and the other parallel to the principal direction n, defined as Hj = Z\f\i -p z2n2 That these principal axes are perpendicular to each other follows from . /8s*i_i_i\ nv # n* = 2/1*1 "I" 2/2*2 = I T" * I 2/2*2 \2/2 *2 / and ^2 *2 (GfH) ^yy0 "" ^22? (E,F) which give nv ? ng = 0 If ^f^ = 0, Eqs. (B) and (C) reduce to (? - *sai?) = 0 ? 0f/?) = 0 and, provided f{? -^ il?, these two equations, as well as Eq. (D), are satisfied only by either ai = ?1, a2 = 0, sai? = 4>f(? or ax = 0, a2 = zbl, Accordingly, zbni and ztn2 are principal directions, and the only principal directions parallel to A, when ^f^0 = 0 and f{? ?* 4>U?. 134 SEC. 3.3.8 Finally, if U? = 0 and < satisfied by taking [Chapter 3 = 44L?, Eqs. (D) and (I) can be and letting a,i and a2 have any values consistent with Eq. (D). Consequently, all lines passing through 0 and lying in plane A are then principal axes of S for 0, and the second moment of S with respect to any such line is a principal second moment of S for 0. Problem: The strengths of the four points 0, P, Q, R of the rectangle shown in Fig. 3.3.8b are 1, 2, 10, and 20 lb. P(2) Q(i0) R(20) FIG. 3.3.8b Locate three principal axes of this set S of points, and evaluate the corresponding principal second moments of S for point R. Solution: Let nh n2, n3 be the unit vectors shown in Fig. 3.3.8b. Then (see 3.3.7) n3 is a principal direction of S for R; a line passing through R and parallel to n3 is a principal axis of S for R; and the corresponding principal second moment of S for R is given by (3.2.3) = 1(16) + 2(20) + 10(4) + 20(0) = 96 lb ft2 (3.1.7) Next, to locate two principal axes in the plane of the rectangle, evaluate f(Ry UR, and 4>f{*: 4>f{* = 48 lb ft2, UR = 48 lb ft2 3317) = -16 1b ft2 (3.2.3,3.1.5) Second Moments] SEC. 3.3.9 135 Then, letting n* be one of the principal directions parallel to the plane of the rectangle, the corresponding principal second moment is given by = 48 + 16 = 64 lb ft2 and the ratio Zi/z2 of the ih and n2 measure numbers of nz by fi 64 - 48 -16 = -1 The corresponding principal axis Lz thus has the orientation shown in Fig. 3.3.8c. A second principal axis of S for 0, Lv, is perpen- FIG. 3.3.8c dicular to L2, and the corresponding principal second moment is given by = 48 + 48 - 32 = 64 lb ft2 3.3.9 There exist at least three mutually perpendicular princi- pal directions of every set of points for every point in space. Proof: Let S be a set of points and 0 a point in space. It suf- fices to show that there exists one principal direction of S for 0, since, from Sec. 3.3.8, it then follows that there exist at least two more, perpendicular to each other and to the first. 136 SEC. 3.3.9 [Chapter 3 Let n?, i = 1, 2, 3, be mutually perpendicular unit vectors and a, the n, measure number of a unit vector no. Then no is a principal direction of S for 0 whenever 2 (3.3.2) (3.2.4) fT or, expanding, = 0 ? = 0 I (A) - 4>SJ?) = 0 J where ai, a2, a3 must satisfy the condition a!2 + a22 + a32 = 1 (B) which shows that ah a2, a3 cannot vanish simultaneously. From this it follows that Eqs. (A) can in general be satisfied simul- taneously only if the determinant of the coefficients of ah a2, and a3 is equal to zero; that is, + A2(?fa/O) - A3 = 0 (C) where Al - f{? + 4>U? + *U? Equation (C), being a cubic equation, has at least one real root. (Ai, A2, and A$ are real quantities.) This root furnishes the value of one principal second moment of S for 0, tod, once it has been found, any two of Eqs. (A) can be used together with Eq. (B) to determine the measure numbers defining the corresponding prin- cipal direction. Problem: Figure 3.3.9a shows a set S of points, a point 0, and three mutually perpendicular unit vectors n?, i = 1, 2, 3. The six second moments g/o, i, j = 1, 2, 3, are tabulated in Fig. 3.3.9b. Second Moments] SEC. 3.3.9 137 (ft1) i 1 2 3 1 15 -2V3 i 2 3V3 21 2 3 -2V3 2 12 FIG. 3.3.9a FIG. 3.3.9b Determine three principal directions of S for 0. Solution: Ax = 15 + 21 + 12 = 48 ft4 A2 = 315 + 252 + 180 - (27 + 4 + 12) = 704 ft8 Az = 3780 - 72 - (60 + 252 + 324) = 3072 ft12 Hence (%?)z - 48(<#L/O)2 + 704%? (by trial and error), calling the roots ^?y %?, f/?: 4>U? = 8 ft4, H? = 16 ft4, f/? = 24 ft4 Replace a? with xif i = 1, 2, 3, H? with xx?f and use the first two of Eqs. (A): 7xi + 3V3 x2 = 2V3 xz 3V3zi + 13x2 = -2x3 Solve for xx and x2 (in terms of xz): 1 Use Eq. (B): Thus (1 + i + 1W = 1 138 SECS. 3.3.10-3.3.12 [Chapter 3 and V6 V2 Xi = ??i X2 = ? ? Let nz be the principal direction corresponding to aL? Then _/V($ V2 , V2 nx = ?^"4" ni "" T" "2 + ~2~ Similarly, principal directions nv and n*, corresponding to and *fa/o = *?/o, are given by /V6 V2 V2 \ I , V3 3.3.10 Knowledge of three mutually perpendicular principal axes and the corresponding principal second moments (or principal radii of gyration) of a set S for a point 0 simplifies the determina- tion of second moments of S with respect to 0 for arbitrary direc- tions na and n6. For, if n^, i = 1, 2, 3, are mutually perpendicular principal directions of S for 0, then U?j ${? are equal to zero (see 3.3.3), and Eqs. (5), (6), and (7) of Sec. 3.2.4 reduce, respectively, to (1) (2) and UW + +UW + +UW (3) 3.3.11 By successive use of expressions given in Secs. 3.2.6, 3.2.7, and 3.3.10, all second moments of a set S of points can be found whenever the following are known (compare with 3.2.9): (a) The strength of S. (b) The location of the centroid of S. (c) Three mutually perpendicular principal axes and the cor- responding principal second moments (or principal radii of gyra- tion) of S for one point. 3.3.12 Principal directions, axes, etc., for the centroid of a set Second Moments] SEC. 3.3.12 139 S are called centroidal principal directions, centroidal principal axes, etc. Problem: In Fig. 3.3.12a, lines CB, CD, CE represent mutually perpendicular centroidal principal axes of a set S of points. The H D \\ \ G 41 A 61 B FIG. 3.3.12a FIG. 3.3.12b corresponding centroidal principal radii of gyration are equal to 2, 3, and 4 ft. Determine the radius of gyration of S with respect to line AH. Solution: Let no and nt, i = 1, 2, 3, be the unit vectors shown in Fig. 3.3.12b, and k%/A the desired radius of gyration. Then = (ksa/cy + (ity (3.2.8) - . (3.2.5) N (3.3.10) N N (3.2.5) ) V Hence (ksa/Ay = where = (c x nay (F3.3.12b) (ki/c)W + (c X na)2 na =and Q>i = 0, a2 = -f, az = i k?/c = 2 ft, fcf/c = 3 ft, ki/c = 4 ft c = -6ih - 3n2ft 140 so that and 9(9) 25 SEC. 3.3 16(16] + 25 .13 I 36(29) 25 1381 25 [Chapter 3 k*'A 7.46 ft 3.3.13 The locus E of points P whose distance R from a point 0 is inversely proportional to the square-root of the second mo- ment of a set S of points with respect to line OP is an ellipsoid. It is called a momental ellipsoid of S for 0. Proof: Let ni, n2, n3 be mutually perpendicular principal direc- tions of S for 0 (see Fig. 3.3.13a), Lh L2, L3 the corresponding principal axes, Xi, x2, x3 the coordinates of point P when Lh L2, Lz are regarded as cartesian coordinate axes whose positive senses are defined by ih, n2, n3, and na a unit vector parallel to OP. FIG. 3.3.13a It is to be shown that xh x2, x3 satisfy the equation of an ellip- soid whenever R = k(sa??)-v* (A) where k is a constant. Now (3.3.10) (B) where at is the nt measure number of na and f/? is a principal second moment of S for 0. Furthermore, as the position vector Second Moments] SEC. 3.3.14 141 of P relative to 0 can be expressed either as Rna or as Xitii + ?3n3, at can be replaced with Xi/R, so that Hence, using Eq. (A) to eliminate or where Ai = k(4%?)-u*, i = 1, 2, 3 (D) Equation (C) is the equation of an ellipsoid whose center is at 0, whose axes coincide with the principal axes Lh L2, Lz of S for 0, and whose semidiameters are equal to Ait i = 1, 2, 3, as shown in Fig. 3.3.13b. \ FIG. 3.3.13b 3.3.14 Of all lines passing through a point 0, those with respect to which a set S of points has a larger, or smaller, second moment than it has with respect to all other lines passing through 0 are principal axes of S for 0. Proof: The numbering of three mutually perpendicular prin- cipal directions n,, i = 1, 2, 3, of S for 0 can always be accom- plished in such a way that 142 SECS. 3.3.15-3.4.1 [Chapter 3 or, referring to Eq. (D), Sec. 3.3.13, in such a way that the lengths Ah A2, A3 of the, semidiameters of a momental ellipsoid of S for 0 satisfy Ai ^ A2 ^ A3 The distance R from the center 0 to a point P of the ellipsoid (see Fig. 3.3.13b) is never smaller than the smallest semidiameter, and never larger than the largest. That is, Ax ^ R ^ Az or, using Eqs. (A) and (D), Sec. 3.3.13, so that ?{? > 4>saL? > 446? which asserts that the second moment of S with respect to a line which passes through 0 and is not a principal axis of S for 0 can- not be smaller than the smallest principal second moment of S for 0 or larger than the largest principal second moment of S for 0. 3.3.15 Given a set S of points with positive strengths, no sec- ond moment of S with respect to a line (see 3.2.3) is smaller than the smallest centroidal principal second moment of S (see 3.3.12). The smallest centroidal principal second moment is therefore called the minimum second moment of S. Proof: The second moment ai? of S with respect to a line La passing through a point 0 is greater than the second moment 4>aLP* of S with respect to a parallel line passing through the centroid P* (see 3.2.7); and ?aP* is larger than or equal to the smallest cen- troidal principal second moment of S (see 3.3.14). 3.4 Second moments of curves, surfaces, and solids 3.4.1 Given a point 0, a figure (curve, surface, or solid) F, and a unit vector nn, a vector #a/o, called the second moment of the figure F with respect to point 0 for the direction no, is obtained by perform- ing the following operations: (1) Divide F into n elements of arbitrary size and shape. (2) Select a point in each element. Second Moments] SEC. 3.4.1 143 (3) Assign to each such point a strength equal to the length, area, or volume of the corresponding element. (4) Evaluate the second moment of this set of points with re- spect to O for the direction na (see 3.2.1). (5) Determine the vector approached by this second moment as n tends to infinity and each of the elements chosen in (1) shrinks to a point. Problem: In Fig. 3.4.1a, 0 is one endpoint of a straight line S of length L, and no is a unit vector perpendicular to S. Determine the second moment #?/o FIG. 3.4.1a Solution: (1) Division of S into n elements: Choose elements of equal length, L/n. Number these as shown in Fig. 3.4.1b. I 2 H 1 H?I h FIG. 3.4.1b (2) Selection of a point in each element: Use the right-most point in each element, calling these points Pi, P2, ? ? ? , Pn, as shown in Fig. 3.4.1c. P? is a typical point of this set of points. FIG. 3.4.1C (3) Assigning of strengths to the points P?, i = 1, 2, . . . , n: 144 SEC. 3.4.1 [Chapter 3 Let Ni be the strength of P*. As the length of each element is equal to L/n, Ni = L/n,i = 1,2, . . . ,n (A) (4) Evaluation of the second moment of the set of points Pi, . . . , Pn: From Sees. 3.2.1 and 3.1.1 it follows that the second moment is given by ? NiPi X (nfl X Pi) (B) t = i where p* is the position vector of P? relative to 0. Let n be a unit vector parallel to line S, as shown in Fig. 3.4.1c, and note that the distance from 0 to P? is equal to iL/n. Then and Pi X (no X P?) = ?;? n X (no X n) = ? nan so that iPi X (na X pt) = ? na and (B) can be replaced with The sum appearing in this expression has the value V * = n(n+l)(2n + l) (C) t-i ^ Hence the second moment of the set of points Pi, . . . , Pn with re- spect to 0 for the direction no is given by (n + l)(2n + l) (5) Determination of the vector approached by the second mo- ment of the set of point Pi, . . . , Pn: As the elements into which S was divided were chosen in such a way that each element auto- matically shrinks to a point as n tends to infinity, *f/o is given by Second Moments] SEC. 3.4.2 145 ^ lim (2 + - + i;) 6 n_ao \ n n2/6 L3 3.4.2 In a manner similar to that employed in connection with the location of centroids (see Vol. I, Sec. 2.5.1), the limiting process described in Sec. 3.4.1 can be replaced with the evaluation of an integral: *r/o = /fpX(n,X P) dr where (see Fig. 3.4.2a) p is the position vector of a typical point FIG. 3.4.2a P of the figure F relative to point 0, and dr is the length, area or volume of a differential element of F. FIG. 3.4.2b 146 SEC. 3.4.3 [Chapter 3 Problem: Solve Problem 3.4.1 by integration. Solution: Let P be a typical point of S (see Fig. 3.4.2b), p the position vector of P relative to 0, n a unit vector parallel to S, s the distance from 0 to P, and dr the length of a differential element oi S. Then p = sn P X (no X p) = s2 no dr = ds L s2na or, as the characteristics of the unit vector na are independent of s, na / JoJO 6 3.4.3 Given a point 0, a figure F, and two unit vectors no and n&, the scalar Zb?, denned as Fd? = *l/o - nb (1) where &Fa/O is the second moment of F with respect to 0 for the direction no (see 3.4.1), is called the second moment of F with respect to 0 for the pair of directions na, nb. It follows from Sec. 3.4.2 that FJ>? can be expressed asf F (na X P) ? (lib X p) dr (2) Problem: no and n6 in Fig. 3.4.3 are unit vectors parallel and perpendicular to the diameter OQ of a semsicrcular curve C of 0 FIG. 3.4.3 Second Moments] SEC. 3.4.4 147 radius R. (n& is parallel to the plane determined by C.) Eval- uate ?>/o. Solution: (see Fig. 3.4.3 for notation): p = J?(cos 0 ? l)na + R sin 6 no (n? X p) ? (n6 X p) = R2(l - cos 6) sin $ dr = Rdd 4>CJ? = = #3 f (n. X p) ? (lit X p) dr (1 - cos 6) sin 6 dd 2/23 3.4.4 When the unit vector n6 is equal to the unit vector na, the expressions given in Sec. 3.4.3 become 4%? = *f/? ? na (1) 4%o = fF (na X p)2 dr (2) and <#w/o is called the second moment of F with respect to line La, La being the line which passes through 0 and is parallel to na. As (na X p)2 is equal to the square of the distance la/o from La to a typical point P of F, li? can also be expressed as (3) FIG. 3.4.4a 148 SEC. 3.4.5 [Chapter 8 Problem: Determine the second moment of (a) a spherical sur- face A of radius R and (b) a spherical solid B of radius R, with re- spect to a line La passing through the center 0 of the figure. Solution (a) (see Fig. 3.4.4a for notation): 1PJ? = R sin + dr = (ft aa? (see 3.4.4) can always be ex- pressed as the product of r, the length, area, or volume of the figure F, and the square of a positive, real quantity fca/O, called the radius of gyration of F with respect to line La: Problem: Determine the radius of gyration of (a) a spherical surface A of radius R and (b) a spherical solid B of radius R with respect to a line La passing through the center 0 of the figure. Second Moments] SECS. 3.4.6-3.4.7 149 Solution (a): The area T of a spherical surface A of radius R is given by (see Solution (a), 3.4.4 for notation) r = R2 I** d$ fj sin ^ dtfr From Problem 3.4.4, Hence Solution (b): The volume r of a spherical solid B of radius R is given by (see Solution (b), 3.4.4 for notation) T = P M So d+ So r' sin * dr From Problem 3.4.4, Hence 3.4.6 Given a figure F, a point O, and a unit vector n,, n, is called a principal direction of F for 0 if and only if *f/o (see 3.4.1) is parallel to n, or equal to zero. When nt is a principal direction of F for 0, the line Lt parallel to n, and passing through 0 is called a principal axis of F for 0; the plane Pt passing through 0 and perpendicular to n, is called a principal plane of F for 0; the second moment R/? = y v Ag/? (3.2.4) i-1 ; = 1 = *f/?a!2 + . . = 11.52 ft4 OP, = i .+ 4>aa the desired second mo- a3 = 0 2(*f/?ala, + . . .) Problem (b): Determine the second moment of a spherical sur- face of radius R with respect to a tangent to the surface. Solution: Let A be the surface, Lo a line passing through the center 0 of A and parallel to the tangent in question, and aaP the desired second moment (see Fig. 3.4.7b). Then (3.2.7) Second Moments] 151 FIG. 3.4.7b where r = 4wR2 is the area of A and 1S/P tween the tangent and line La. Hence R is the distance be- (P3.4.4) Problem (c): Determine the radius of gyration of a spherical solid with respect to a tangent to the solid. Solution: Let B be the solid, La a line passing through the center 0 of B and parallel to the tangent in question, and fcf/p the desired radius of gyration (see Fig. 3.4.7c). Then (3.2.8) FIG. 3.4.7C 152 SEC. 3.4.7 [Chapter 3 where 1%/p = R is the distance between the tangent and line La. Hence (fc?/p)2 = %R2 + R2 = iR2 (P3.4.5) and fcB/p _ nyi*R Problem (d): In Fig. 3.4.7d, no is a unit vector parallel to the diameter OQ of a semicircular curve C of radius R. FIG. 3.4.7d Show that no is not a principal direction of C for point 0. Solution: Let nb be any unit vector perpendicular to na. If no were a principal direction of C for 0, U? is equal to 2#3, as was shown in Problem 3.4.3. Hence no cannot be a principal direction of C for 0. Problem (e): Locate three centroidal principal axes (see 3.3.12) of the rectangular parallelepiped shown in Fig. 3.4.7e. Solution: The three planes passing through the centroid P* and parallel to the faces of the parallelepiped are planes of sym- metry (see 3.3.6) and, therefore, principal planes for all points in these planes. In particular, they are principal planes for their FIG. 3.4.7e point of intersection, P*. Consequently, their normals passing through P* are centroidal principal axes of the parallelepiped. Second Moments] SEC. 3.4.7 153 Problem (f): Figure 3.4.7f shows a plane curve C. 0 is a point of C. c- FIG. 3.4.7f Locate one principal axis of C for point 0. Solution: The plane in which C lies is a principal plane of C for all points in this plane (see 3.3.7). A line passing through 0 and normal to this plane is thus a principal axis of C for 0. Problem (g): Referring to Problem 3.4.7a, locate three prin- cipal axes and determine the corresponding principal second mo- ments of R for point 0. Solution: The plane determined by R is a principal plane of R for all points in this plane (see 3.3.7), hence for point 0. A line passing through 0 and parallel to n3 is thus a principal axis of R for 0, and the corresponding principal second moment of R for 0 is equal to 100 ft4. Let be a unit vector parallel to a principal axis Lt of R for 0, this axis lying in R. Then the corresponding principal second moment is given by WO A.R/OX 2 11/2 . .. { Z ^ ) +(*f/?H =88.6 ft* (3.3.8) and so that Jt has the orientation shown in Fig. 3.4.7g. The remaining principal axis is perpendicular to Lt, and the corresponding prin- cipal second moment has the value 154 SEC. 3.4.7 [Chapter 3 4>il? + %? - l/o = 11.4 ft4 Note that this second moment is smaller than the second mo- FIG. 3.4.7g ment of R with respect to line OP (see Problem 3.4.7a), in agree- ment with Sec. 3.3.14. Problem (h): The centroidal radius of gyration of a solid cube with respect to a line parallel to an edge of the cube (as found by integration) is equal to L/(6)1/2, where L is the length of any edge. Determine the second moment of a 2 X 2 X 2 ft solid cube C with respect to a diagonal of a face of C. Solution (see Fig. 3.4.7h for notation): nh n2, n3 are centroidal C. FIG. 3.4.7h Second Moments] SEC. 3.4.8 155 principal directions of C (see 3.3.6). The corresponding centroidal principal radii of gyration are given by Jfi"" = kc2/p* = kg"" = 2/V6 and, as the volume r of C is equal to 8 ft3, the centroidal principal second moments of C have the values (3.2.5) 8(2/V6)2 = ^ ft5 Hence (3.3.10) (This result could have been obtained, alternatively, by not- ing that the momental ellipsoid (see 3.3.13) of C for P* must be a sphere, because the three centroidal principal second moments are equal to each other. From this it follows that C has the same sec- ond moment with respect to all lines passing through P*.) Finally, the desired second moment is given by (3.2.7) 3.4.8 The second moment of a plane figure F with respect to a line Lc which is normal to F and intersects the plane of F at a point 0 is called the polar second moment of F with respect to 0, and is denoted by JF'?. That is, Lc 156 SECS. 3.4.9-3.4.10 [Chapter 3 JFI? = VJ0 (1) If U? are the second moments of F with respect to two orthogonal lines La and Lb which lie in the plane of F and inter- sect at point 0, then the polar second moment of F with respect to 0 is given by JF,0 = tf/O + ^F/O (2) Proof (see Fig. 3.4.8 for notation): JFIO = +r/o = f {lcy dT (3.4.4) JF L2 + h2) dr = fF I* <2r + /F ^2 dr (3.4.4.) 3.4.9 The Appendix contains sketches of ten curves, surfaces, and solids. For each figure the centroid and one or more centroidal principal axes (see 3.4.6) are shown, and the squares of the corre- sponding centroidal principal radii of gyration, as well as the length, area, or volume of the figure, are listed. Where less than three principal axes are shown, the orientation of those not shown can be found either by symmetry considerations (see 3.4.7 and 3.3.6) or by recalling that there exist at least three mutually per- pendicular principal axes of every figure for every point in space (see 3.4.7 and 3.3.9). In the case of plane figures, Sec. 3.4.8 can be used to evaluate the third centroidal principal radius of gy- ration. The information given in the Appendix, together with the the- orems stated in the sections which follow, makes it possible, with- out integration, to find second moments (or radii of gyration) of a great variety of figures. 3.4.10 Given a point 0, two directions no and n6, and a figure F composed of (dimensionally homogeneous) figures Fif i = 1, 2, . . . , n, the second moments &F/O and 4>Fl? can be expressed as = i>r/o a) Second Moments] SEC. 3.4.10 157 This is a consequence of the fact that &FJ? (see 3.3.1) is defined in terms of the second moment of a set of points, that this second moment is given by a sum (see 3.2.1), and that this sum obeys the associativity law for vector addition (see Vol. I, Sec. 1.9.2). Problem: Figure 3.4.10a represents the cross section of a struc- tural steel member. 2" 3" 3" 2"' 4" \ \ \ \ \ \ \ i" \ I"' 4" FIG. 3.4,10a Determine the polar second moment JSI? (see 3.4.8) of this surface S with respect to point O. Solution: Regard S as being composed of rectangles Si, S2, S3 and let in and n2 be unit vectors as shown in Fig. 3.4.10b. Next, construct the table shown in Fig. 3.4.10c. 0 p; ?p; 11 FIG. 3.4.10b 158 SEC. 3.4.11 [Chapter 3 Quantity Ai (*?fi/o ii/o Units in.2 in.2 in.2 in.2 in.2 in.2 in.2 in.4 in.4 i - 1 12 4 3 16 4 7 196 84 i - 2 12 3 i 0 0 3 i 36 4 i - 3 12 4 3 16 4 ?V 7 196 84 Reference App. Fig. 3 App. Fig. 3 App. Fig. 3 Fig. 3.4.10 Fig. 3.4.10 3.4.7, 3.2.8 3.4.7, 3.2.8 3.4.5 3.4.5 FIG. 3.4.10C Then 3 f{? = V* 4>f{/? = 196 + 36 + 196 = 428 in.4 (2) t=i 3 U? = Y) %? = 84 + 4 + 84 = 172 in.4(2) f=i and jsio = C2. Then, from Fig. 9 of the Appendix, and the volumes Vi and F2 of Ci and C2 are given by Fx = irRSh, F2 = 7rR22h so that (see 3.4.5) and (2) Next, the volume F of T is given by and ka/p*f the desired radius of gyration, by V2 U22 - R12) " \ 2 1/2 3.4.12 Radii of gyration of a surface can sometimes be ob- tained by regarding the surface as the limiting form of a solid. Problem: Determine the radius of gyration of a right-circular cylindrical surface S of radius R with respect to the axis of the surface. Solution: Regard S as the limiting form of the solid T consid- ered in Problem 3.4.11, when Rx = R and R2 approaches R: = R 3.5 Second moments of sets of particles and continuous bodies 3.5.1 Given a point 0, a direction na, and a set S of particles situated at points Pif i = 1, 2, . . . , n, and having masses rat, the 160 SEC. 3.5.2 [Chapter 3 second moment of the set of points Pi of strengths iVt = mt with respect to 0 for the direction no (see 3.2.1) is called the second mo- ment of the set of particles with respect to 0 for the direction no. Similarly, second moments of S with respect to 0 for a pair of directions na, n6, second moments of S with respect to lines, radii of gyration of S, principal directions of S, etc., are denned in terms of the corresponding properties of the set of points P? of strengths Ni = mi. Consequently, all of the material in Parts 3.2 and 3.3 is directly applicable to sets of particles. The second moment of a set S of particles with respect to point 0 for the pair of directions na, nb is called, alternatively, the product of inertia of S with respect to O for the pair of directions na, n&; and the second moment of S with respect to a line La is called the moment of inertia of S about line La. 3.5.2 Given a point 0, a unit vector na, and a continuous body C regarded as occupying a figure F, a vector #?/o, called the second moment of C with respect to 0 for the direction na, is obtained by per- forming the operations described in Sec. 3.4.1, after replacing (3) with " Assign to each such point a strength equal to the mass of the material occupying the corresponding element." As in the case of curves, surfaces, and solids (see 3.4.2), the required limiting process can be replaced with the evaluation of an integral: = /F pX(naXp)pdr where (see Fig. 3.5.2a) p is the position vector of a typical point Second Moments] SEC. 3.5.2 161 P of the figure F relative to point 0, p is the mass density of the body C at point P (expressed in units of mass per unit of length, area, or volume, according as F is a curve, surface, or solid), and dr is the length, area, or volume of a differential element of F. Problem: In Fig. 3.5.2b, 0 represents one end of a thin straight wire W which may be regarded as occupying a straight line S of length L. The mass density p of the wire is given by P = 10-3 (l + ||) slug ft"1 where s is the distance from 0 to a typical point P of AS. r fr ?*? i L_ P - S ?- P (IS L- FIG. 3.5.2b Determine the second moment of W with respect to 0 for a direction no perpendicular to S. Solution: Let n be a unit vector parallel to S (see Fig. 3.5.2b), p the position vector of P relative to O, and dr the length of a dif- ferential element of S. Then p = sn P X (naXp) = ?*n? dr = ds and *"'? = fcPX (naXp)pdr = 10~~3L3 na slug ft2 (L expressed in ft) 162 SECS. 3.5.3-3.5.4 [Chapter 3 3.5.3 Given a point 0, two unit vectors no and nft, and a con- tinuous body C regarded as occupying a figure F, the scalar 0?>/o, defined as AJC/O jftP/O _ /1 \ ab? can be expressed as cal? = fF (n? X p) ? (n? X p)p dr (2) 3.5.4 When the unit vector nb is equal to the unit vector na, the expressions given in Sec. 3.5.3 become "a (1) X p)2p dr (2) and aa? is called either the second moment of body C with respect to line L^ or the moment of inertia of body C about line La, La being the line which passes through 0 and is parallel to na. As (na X p)2 is equal to the square of the distance la /o from La to a typical point P of F, aa? can be expressed as \la ) P UT \*J) Problem (a): Referring to Problem 3.5.2, determine the mo- ment of inertia of W about a line which passes through 0 and is parallel to na. Solution: = *f/0 ? na = 10~3L3na ? na (1) (P3.5.2) = 1O"3L3 slug ft2 (L expressed in ft) Problem (b): Figure 3.5.4 shows the cross section of a steel shell S whose inner surface H is hemispherical and whose wall thickness t varies linearly with the radian measure yp of the angle between line OP and the axis La of the shell. The shell has a mass of m slug. Determine the moment of inertia of S about line La- Second Moments] SEC. 3.5.4 163 FIG. 3.5.4 Solution: Regard the shell S as matter distributed with vari- able density on the hemispherical surface H of radius R = 12 in. (This is an approximation which may be expected to give good results if the thickness of the shell is sufficiently small in com- parison with the radius.) As / varies linearly with ip, where a and 0 are constants which can be evaluated by noting that t = 0.1 for ^ = 0 and t = 0.2 for ^ = w/2: ^=o = a = 0.1 in. t|*-?/2 = a + frr/2 = 0.2 in. Hence - (0.2 - a) = - (0.2 - 0.1) = ? in. 7T IT T t = 0.1 + ? 7T Assume that the steel of which the shell is made has the same properties at all points of the shell, and take p, the mass of the shell per unit of area of surface H at point P, proportional to the thickness t at P: where k is a constant of proportionality, which can be expressed in terms of the mass m of the shell S (see Fig. 3.4.4a for 0): 164 SEC. 3.5.5 [Chapter 8 m= / pdr = / d$ / fc(0.1 + ?J// Jo Jo V ^ = 2TT/C#2[0.1 + (0.2/TT)] 2irR?[0.l + (0.2/TT)] Let la/o be the distance from line La to point P. Then or (-2* rir/2 / o saL? = / (la/o)2P dr = / <& / /b/24 (0.1 +(3) JH (A) Jo Jo \ x in3 and, with ft = 12 in., <*>?a/o = 102m slug in.2 3.5.5 The second moment <#^? (see 3.5.4) can always be ex- pressed as the product of the mass m of the body C and the square of a positive, real quantity k%/o, called the radius of gyration of C with respect to line La. In general, if F is the figure which C is regarded as occupying, kca/o is not equal to K/o (see 3.4.5). Problem: Determine the radius of gyration of (a) the wire W and (b) the straight line S described in Problem 3.5.2, with respect to a line passing through 0 and parallel to na. Solution (a): kw,o _ 4>U? = 10-3L3 slug ft2 (P3.5.4a) m = f p dr = 10-' fL (l + ||) ds = | lO-'L slug kw/o = (3),?L Solution (b): ks/o = /f\"! = /f/o (3.4.5) \ L / (3.4.4) \ (P3.4.2) \ 3 Second Moments] SECS. 3.5.6--3.5.8 165 3.5.6 Given a continuous body C, a point 0, and a unit vector n*, nz is called a principal direction of C for 0 if and only if *f/o (see 3.5.2) is parallel to n, or equal to zero. When nz is a principal direction of C for 0, the line Lz parallel to nz and passing through 0 is called a principal axis of C for 0; the plane Px passing through 0 and perpendicular to nz is called a principal plane of C for 0; the second moment Z/o is called a principal second moment or principal moment of inertia of C for 0; and the radius of gyration kz/O is called a principal radius of gyration of C for 0. In general, if F is the figure which C is regarded as occupying, principal directions of C for 0 are not principal directions of F for 0, and vice versa. 3.5.7 As a consequence of the definitions given in Secs. 3.5.2- 3.5.6, many of the relationships discussed in Parts 3.2, 3.3, and 3.4 have counterparts in the theory of second moments of continuous bodies. After minor changes in wording and notation these re- lationships can, therefore, be used for the solution of problems involving continuous bodies. 3.6.8 If the mass density p of a continuous body C is the same at all points of the figure F which C is regarded as occupying, both FIG. 3.5.8 the density p and the body C are said to be uniform. Second mo- ments and radii of gyration of a uniform body C can be expressed in terms of p (or the mass m of C and the length, area, or volume 166 SEC. 3.5.9 [Chapter 3 T of F) and corresponding second moments and radii of gyration of F: = ? *?/o (l) = f 2b/o is equal to zero (see 3.5.7 and 3.3.3). If ab/o is also equal to zero, nb is a principal direction of A for 0 (see 3.5.7 and 3.3.4) and axis B-B is a principal axis of A for 0 (see 3.5.6). Thus K/o = 0 (A) guarantees dynamic balance. Let Hi and i/2 be the cylindrical bodies removed from S when the holes are drilled. Then ti? = sJ? + cJ? - *S1/O - 4>&'? (B) (3.5.7,3.4.10,3.4.11) Both na and nb are principal directions of S for 0 (see 3.5.7 and 3.3.6). Hence sJ? = 0 (C) (3.5.7,3.3.4) Regard C as occupying a surface F obtained by removing from a circular surface Fx an elliptical surface F2. The principal direc- tions of F for 0 are in, n2, n3 (see 3.4.7, 3.3.7 and 3.3.6). Hence T = (3.4.7,3.3.10) where a2 = 0, a3 = - | b2 = 0, 63 = i are the nt, i = 1, 2, 3, measure numbers of na and nb. Thus = 1, But 4U? = (3.4.8) Hence Second Moments] SEC. 3.5.9 169 or, expressing ll? in terms of second moments of the circular surface Fi and the elliptical surface F2y (3.4.11) io r O2 O ^2~1= -?i (* X 3 2) f - 2(TT X 2.5 X 2/2) ^(3.4.5) ^O L 4 4 J (App.F5) (App.F6) That is 4>%? = -5.977rin.4 and, letting pc be the mass of C per unit of area of F, Cal? = (3.5.8) Now, as C is 1/16 in. thick, and the mass density of the material of jvhich C is made is 1.2 times as great as the mass density ps of the shaft material, Thus Let Pi* be the mass center of the cylindrical body Hi. Then *Sl'? =(3.5.7,3.2.6) 0 + mx(na X Pi) ? (n* X Pi) (3.57,3.3.6,3.3.4) (3.1.5) where mx is the mass of Hh and px is the position vector of Pi* relative to 0; that is, mi = TT( 1/4)2 dps (App.F9) and Pi = na H Thus Similarly - -Z:d((>-i)ps (F) 170 SEC. 3.5.9 [Chapter 3 Substitute from Eqs. (C), (D), (E), and (F) into Eq. (B), then use Eq. (A): d2 - I2d + 7.164 = 0 Solve for d: d = 0.63 in. or 11.37 in. 4 LAWS OF MOTION 4.1 Inertia forces and force systems 4.1.1 If RQP is the absolute acceleration of a particle P in a reference frame R (see 2.5.1), and m the mass of P, the bound vector RFP applied at P and given by is called the inertia force acting on P in R. Problem: Referring to Problem 2.5.13, and regarding the air- plane P as a particle of mass m = 3000 slug, determine the mo- ment about the earth's north-south axis of the inertia force acting on P in (a) R and (b) R'. Solution (a): As the acceleration of P in R is parallel to n3 (see Fig. 2.5.13b), the line of action of the inertia force RFP intersects the earth's axis. Hence the moment of this force about this line is equal to zero. Solution (b): R'fP = -m R'op = 3000(136ih - 222n2 + 227n3) slug mile hr"2 (P2.5.13) The distance s from line NS to the point of application of R'?p is given by s = 3960 sin 45? = 2800 miles The moment of RTP about line NS thus has the value sm(-222)k = -2800(3000)(222)k slug mile2 hr~2 = _2800(3000)(222)(5280)2 fc ft2 2 (ooOu) = -40,000 k ft lb 171 172 SEC. 4.1.2 [Chapter 4 4.1.2 Given a continuous body C regarded as occupying a fig- ure F, and a reference frame R, divide F into n elements of arbi- trary size and shape; select a point of C in each element, and, regard- ing this point as a particle whose mass is equal to that of the material in the element, determine the inertia force (see 4.1.1) acting on this particle in R. The system of n forces thus obtained can be replaced with (see Vol. I, Sec. 3.5) a force applied at the mass center P* of C, together with a couple. When n tends to infinity and each of the elements shrinks to a point, this force and couple approach limits called the inertia force acting on C in R and the inertia couple acting on C in R. The inertia force *FC (applied at the mass center P* of C) and the torque Rlc of the inertia couple acting on C in R are given by and (1) (2) where RQP is the acceleration in R of a generic point P of C (see Fig. 4.1.2a), p is the mass density of C at P, r is the position vector of P relative to P*, and dr is the length, area, or volume of a dif- ferential element of F. FIG. 4.1.2a FIG. 4.1.2b Proof: Let Pt, i = 1, 2, . . . , n, be the points selected in the n elements into which F is divided (see Fig. 4.1.2b). The mass Laws of Motion] SEC. 4.1.3 173 rrii of a typical element can be expressed as the product of a quan- tity pi and the length, area, or volume r, of the element: Hence, letting *ap* be the acceleration of Pt in R, the inertia force R?Pi acting on a particle situated at P? and having the mass m, is given by RfP< = -miR*p> = -Rap< (*'?* X r)p dr(E) / r X [* ?* X (* ?* X r)] p dr} (F) Laws of Motion] SEC. 4.1.3 175 If na is a unit vector parallel to R'ctRy and n0 is a unit vector parallel to R'wR, then the following are identities: R'OLR ? na ? R'ctR na R'wR = no ? R'R X (*'?* X f)] = no ? ?'?*r X [*?* X (n. X r)] which, after using the identity r X [*?* X (no X r)] = R'<*R X [r X (n0 X r)] becomes r X [*?* X (^?* X r)] = no ? R'**R R'wR X [r X (no X r)] (H) Thus R'a* jp r X (na X r)p rfr (F.G.H) + n0 ? *'?**'?* X ffX(noX r)p. = - [na ? R'ocR *R/P* + no ? ?'?* *'?? X f] (3.5.2) Problem: A uniform solid sphere S of mass m and radius r has an angular acceleration Rcts in a reference frame R. Determine the torque RJS of the inertia couple acting on S in R. Solution: It follows from symmetry considerations (see 3.5.7 and 3.3.6) that every direction is a principal direction of S for the , mass center P* of ?. Hence, in particular, unit vectors na and no, parallel to Ras and Rws respectively, are principal directions of S for P*, and &S/P* _ S/P*n*a ? S>aa ??a (3.5.7,3.3.2) 4*s/p* _?*o ? (3.5.7,3.3.2) Thus and = 0 176 SEC. 4.1.4 [Chapter 4 Furthermore "" (3.5.5) ? (3.5.8,App.F8) Hence RjS = ? |mr 2 /2aS (2) 4.1.4 If nt, i = 1, 2, 3, are ani/ mutually perpendicular unit vec- tors, and RfctiR and ^'co** are the nt measure numbers of the angular acceleration R'otR and angular velocity R'a>R of a rigid body R in a reference frame R'', the torque fiT* of the inertia couple acting on ft in ft' (see 4.1.3) is given by Proof: If na is a unit vector parallel to R'ctR, no is a unit vector parallel to R'{RR'?>R X n.) (A) i = i Now, if nt, i = 1, 2, 3, is a right-handed set of unit vectors, R'wR X ni = R'wzR n2 ? R'o>2R n3 RfwR X n2 = ^'wi* n3 ? R'wzR nx R'u)R X n3 = R'o)2R ni ? ^'coi12 n2 Expand the right-hand member of Eq. (A) and substitute. (Note that it was not assumed that nt, i = 1, 2, 3, are fixed in either R or R'. In general, there exist only three mutually per- pendicular principal directions of R for P*, and these are fixed in R. However, certain rigid bodies possess infinitely many prin- cipal directions for their mass center. It is in the study of the motions of such bodies that the expression for R1R given above is most useful.) Problem: Referring to Example 2.2.10, and supposing that the disc D is uniform (see 3.5.8), has a mass m and radius r, and is rolling on plane P (see 2.5.10), determine the inertia force ''F^ and the torque ^T^ of the inertia couple acting on D in P. Solution (see Fig. 2.2.10a for notation): pfD = -mPQD* = _mr[# + ^ sin ey + ^ sin 0)]ni (4.1.3) (P2.5.10) + mr[$ + $ sin 6 + 2^6 cos 0]n2 + mr[8 - cos d(\J/ + sin 0)]n3 178 SEC. 4.1.6 [Chapter 4 Next (see Example 2.2.10) coi == ? (f) cos 0, o?2 == ? 0. C03 s= ^ ~f~ sin v and (see Example 2.3.2) pa\D = ? # cos 0 + <$ sin 6 ? d\{/ P(X2^ = ?0 ~4~ ^ cos 0 Fa3D = 1// + $ sin 0 + 00 cos 0 Furthermore T^II ^~ V22 "^ TMvrCi ) ?" ^~~^ (3.5.5) (3.5.8,App.F5) 4 and (3.4.8) ^ Hence, as in, n2, n3 are principal directions of D for Z>* (although they are not fixed in D), + (0 ? 2 ^ cos 6 ? Rio)2B = sin (3 37* ^ and their time-derivatives (keeping in mind that - cos p dt and that * 180 SEC. 4.1.7 [Chapter 4 as a consequence of the definition of 0) are given by dt = 0.2 cos 0 cos \f/( 37 The principal moments of inertia of 2? for P* are *fi/J" = mfc<2, t = 1, 2, 3 (3.5.5) Thus *iT* = -0.2m ^V {[-/ci2 sin ^ + (/c32 - /c22) sin 0 cos 0]ih + [(/ci2 + /c22 - fc32) cos 0 cos + [(-fcr2 + k22 - /c32) sin 0 cos ^]n3} 4.1.7 When a rigid body R has an angular velocity of fixed orientation in a reference frame R' (see 2.2.4), the torque R'1R of the inertia couple acting on R in R' (see 4.1.3) is given by where P* is the mass center of fi, nt, i = 1, 2, 3, is a right-handed set of mutually perpendicular unit vectors with n3 parallel to the angular velocity of R in R', and R'uR and ft'afi are the angular speed and the scalar angular acceleration (see 2.2.2 and 2.3.4) of R in R' for the direction n3. Proof: The angular velocity R'$r R'aR - ^S'-f*To*)']* where n3 is parallel to R'&p* and &*" must be equal to zero, and it follows that n3 is a principal direction of R for P (see 3.5.7 and 3.3.4) and that R' the angle between n* and k. FIG. 4.2.3a The string tension W (see Vol. I, Sec. 4.8.1), which is called the weight of B (see Vol. I, Sec. 4.4.4), and the angle 0 between n* and n are to be expressed in terms of , m, g, and c, where Laws of Motion] SEC. 4.2.3 185 and G (3.42 X 10~8 lb ft2 slug"2) is the gravitational constant (see Vol. I, Sec. 4.2), on the assumption that (a) E and (b) A is a Newtonian reference frame. Solution (a): Figure 4.2.3b shows a free-body diagram of a body B' consisting of B and a portion of the string, drawn on the basis of the following considerations : FIG. 4.2.3b (1) The system of all gravitational forces acting on B' may be regarded as equivalent to the single force of magnitude Gmm!'/R2 (see Vol. I, Sees. 4.4.2 and 4.4.3) shown in Fig. 4.2.3b. (2) The system of all contact forces acting on Bf is represented by the single force of magnitude W (see Vol. I, Sec. 4.8.1). (Con- tact forces exerted by the air surrounding B1 are neglected.) (3) As every point of Bf is at rest in E, no inertia forces act on B' inE (see 4.1.1). Assuming that E is a Newtonian reference frame, the following equation of motion, called a force equation, is justified (see 4.2.1): Gmm! It follows that and Wn - W n R2 n Gmm' R2 = n* 186 SEC. 4.2.3 [Chapter 4 Hence and W = mg 6 = (n, n*) = 0 Solution (b): In addition to the forces shown in Fig. 4.2.3b, the free-body diagram shown in Fig. 4.2.3c includes two vectors which, together, represent the system of inertia forces acting on Bf in reference frame A. These are the inertia force AFB, given by (4.1.3) = -m(?k) X [(?lc) X (fin*)] (2.5.9) = mRa)2 sin n' where n' is a unit vector perpendicular to k and parallel to the FIG. 4.2.3C plane determined by lines NS and CP*, and the torque A1B of the inertia couple acting on B in i. (As the string is regarded as "light," no inertia forces acting on the string are shown.) The following force equation is justified by the assumption that A is a Newtonian reference frame: Wn - sin n' (A) Laws of Motion] SEC. 4.2.3 187 This equation shows that n is parallel to the plane determined by lines NS and CP*. Noting that the angle between n and n' (see Fig. 4.2.3d) is given by */2 FIG. 4.2.3d (B) cross-multiply Eq. (A) with n (in order to eliminate W), then solve for 6: 6 = arc tan where /e sin $ cos / ? Gm'/R* "" g To find W, dot-multiply Eq. (A) with n: JW (D) W = D2 cos 0 ? ra/2o>2 sin (cos ^ sin ? sin ^ cos ) (A,B) ^ or W = W(7[(l ? 6 sin2 0)cos 6 + 6 sin cos 0 and 1 ? 6 sin2 C0S (l-2e sin2 + 62 sin2 ?)1/2 Hence W = m^(l - 26 sin2 + 62 sin2 <*>)1/2 (F) (E) 188 SEC. 4.2.4 [Chapter 4 (Note that the results of Solution (a) can be obtained from Eqs. (C) and (F) by setting e equal to zero, that is, by setting co equal to zero (see Eq. (D)). 4.2.4 The results obtained in Example 4.2.3 do not furnish a satisfactory basis for an experiment intended to reveal whether E or A (if either) is a Newtonian reference frame. For with R = 3960 miles, & = 2TT rad day1, and g = 32.2 lb slug"1, e has the value 0.0035(D) and the maximum angle between the plumb line PQ and line CP*y obtained by taking = x/4 rad (and regarding A as a Newtonian reference frame) has the very small value 0max = 0.0017 rad (C) while the minimum weight of B, corresponding to 4> = TT/2 rad, is Wmin = 0.997 mg (F) which differs by less than one part in one hundred from the value obtained by regarding E as a Newtonian reference frame. What the results do show is that for a body at rest relative to the earth the difference between regarding A, on the one hand, and E, on the other, as a Newtonian reference frame is small. FIG. 4.2. 5a Laws of Motion] SEC. 4.2.5 189 4.2.5 The hypothesis that A (see Example 4.2.3) is a good approximation to a Newtonian reference frame is supported by an analysis of the motion of FoucauWs pendulum, described as follows: In Fig. 4.2.5a, P represents a particle of mass m suspended from a point Q by means of a light string of length L, Q being fixed relative to E. (E and A are defined as in Example 4.2.3.) O is the point of intersection of line CQ and the surface of E, the angle between lines OC and NS, nh n2, n3 unit vectors pointing southward, eastward, and upward at 0, r the position vector of P relative to 0, and n a unit vector parallel to the string PQ. The gravitational force G exerted on P by E is given by (see Vol. 1, Sec. 4.4.2) G = -Groro'[(fln, + r)2]-3/2(#n3 + r) For motions during which |r| is sufficiently small in comparison with /?, this reduces to G ? -Gmm'[(Rnzy]-*i*(Rnz) = ^g^ n3 or, letting Gm! to G tt ? mgnz (A) The inertia force AFP acting on P in A is given by AfP = -mAop = -m[EQp + *<**" + 2(wlc) X Evp] (4.1.1) (2.5.13) where P* is the point of E which coincides with P, so that ^* = (?lc) X [(cJc) X (ftn3 + r)] (2.5.9) or, again confining attention to motions during which |r| is much smaller than R, Aap* ? (ulc) X [(a*) X (Bn,)] = RoA X (Ic X n,) Hence AFP is given by Afp ? -m[Eop + RtfU X (Ic X ns) + 2cok X Evp] (B) The system of all contact forces acting on the body consisting of P and a portion of the string is regarded as equivalent to the Laws of Motion] SEC. 4.2.5 191 FIG. 4.2.5O and Now and h (2.5.1) Furthermore Thus, dropping all n n X Hind n X (Ic X E r ? LOW + (h n ^ ? eW - EVP = E^-^L{i (2.5.1) dt (E) J>(lEyP at (o l< = ?sin <^> ni (F4.2.5a) terms containing X n3 ? 0n" (F) ^ap & L[($ - 0i/ (F.H) vp) ^ ? L cos <^ (F.G.I) -L)n3 t- n3 (1.5.1,1.7.1) (1.5.4,1.7.1) + cos n3 02 and 00, >(0n "4" 0^n ) (E) (F) (G) >"] (H) (I) In'] Equation (D) therefore leads to the following scalar equations for motions of P during which 6 remains small: 6$ + 26(+ + o) cos 0) ? 0 (J) cos One solution of these equations is obtained by taking \p + u) cos = 0 192 SEC. 4.2.5 [Chapter 4 that is, yp = ? 0) COS (L) which satisfies Eq. (J) for all 6 and reduces Eq. (K) to '6 + %? B & 0 (M) where a>2 cos2 has been dropped in comparison with g/L in order to preserve consistency with the approximation made in the deri- vation of Eq. (D). The general solution of Eq. (M) is 6 = d sin (^L t^j + C2 cos (^jL ^ (N) where d and C2 are constants. Equations (L) and (N) may be interpreted as follows: The assumption that A is a Newtonian reference frame leads to the prediction that there exists a motion of P during which the string QP oscillates in a plane passing through line OQ and rotating through \2TT COS \ rad per day, the rotation being clockwise as seen by an observer looking from Q toward 0 when Q is in the northern hemisphere, counterclockwise when Q is in the southern hemisphere. The oscillations have a period T = 2w(L/g)112. The above predictions are in close agreement with observations of actual motions. Consequently they strengthen the hypothesis that A is a good approximation to a Newtonian reference frame. The motion of P predicted by assuming that E is a Newtonian reference frame (and retaining all of the approximations made so far) can be found by setting a> equal to zero in Eqs. (J) and (K): = 0 The oscillatory motion described by Eq. (N) corresponds to a solu- tion of the second of these if ^ = 0, which also satisfies the first. Hence the assumption that E is a Newtonian reference frame leads to a prediction which is incorrect as regards \f/ but correct as re- gards 6. Note, however, that the error in yp is proportional to time (see Eq. (L)), and that the constant of proportionality has the value |o) cos \, which suggests that for motions taking place dur- Laws of Motion] SEC. 4.2.6 193 ing sufficiently short time intervals, particularly near the equator, E may be a satisfactory approximation to a Newtonian reference frame. 4.2.6 When a particle P moves freely near the surface of E (see 4.2.5), the time interval during which the motion takes place, the locality in which P moves, and the initial velocity of P deter- mine whether or not one may regard E as a Newtonian reference frame when analyzing the motion of P. This is shown as follows: In Fig. 4.2.6a, P represents a particle of mass m, free to move subject to two restrictions: (1) P does not come into contact with E and (2) the magnitude of the position vector r of P relative to a point 0 fixed on E's surface remains small in comparison with the radius R of the earth. (All other symbols appearing in Fig. 4.2.6a have the same meaning as in Example 4.2.3 and Sec. 4.2.5.) A ?E(m')R 7^ 1 1 ' P(m) FIG. 4.2.6a FIG. 4.2.6b The free-body diagram shown in Fig. 4.2.6b is constructed by proceeding as in Sec. 4.2.5, the gravitational force G and the inertia force AFP acting on P in A again being given by G tt ?mgnzand AFP ? -m[EQp + Rrfic X (k X n,) + 2a>k X Evp] Set the sum of AFP and G equal to zero (see 4.2.1), dropping Rw2k X (k X n3) in comparison with gn3 (see 4.2.5) and expressing 194 SEC. 4.2.6 [Chapter 4 Evp and EQP in terms of time-derivatives of r (see 2.5.1). The assumption that A is a Newtonian reference frame then leads to the following equation of motion: This equation may be solved by noting that r = r?-o + * T, (1.8.1) ,-0 2! dt* t-o (B) and using Eq. (A) to obtain the third and all higher time-deriva- tives of r. That is, (A) X n3 + 4 = 0 in Eq. (C)) is r = r0 + v0* - ^- n3 (D) The difference between the descriptions of the motion of P obtained by using Eqs. (C) and (D) depends on the time interval during which the motion takes place, because the (dimensionless) Laws of Motion] SEC. 4.2.7 195 factor o)t is proportional to this time interval. Furthermore, for a given value of t, this difference is seen to depend on the locality in which the motion occurs, which determines n3, and on v0, the initial velocity of P in E. 4.2.7 As Eq. (C), Sec. 4.2.6, applies only to motions of P dur- ing which |r| remains small in comparison with 72, it furnishes no information about certain motions of interest, for example, those of ballistic missiles and earth satellites. An approximate descrip- tion of such motions is obtained as follows. (Approximations are introduced by neglecting the oblateness of the earth, gravitational forces exerted by bodies other than the earth, and all contact forces.) In Fig. 4.2.7a, E represents the earth, regarded as a sphere FIG. 4.2.7a having a radius R, a mass ra', and a mass density which at any point of E depends only on the distance from the point to the center 0 of E. A is a reference frame in which 0 and the earth's north-south axis, line NS, are fixed, and such that Au>E = o>k where o) = 2v rad day"1 and Ic is a unit vector parallel to line NS. B is a rigid body of mass m, whose largest dimension is small in comparison with R, P* is the 196 SEC. 4.2.7 [Chapter 4 mass center of B, and p* is the position vector of P* relative to 0. Assume that gravitational forces exerted on B by bodies other than E are negligible in comparison with those exerted by E. Then the system of all gravitational forces acting on B is equivalent to the force G shown in the free-body diagram Fig. 4.2.7b, and (see Vol. I, Sec. 4.4.2) G = -Gmm'(p*2)"3/2P* The vectors A1B and AFB appearing in Fig. 4.2.7b represent the torque of the inertia couple and the inertia force acting on B in A (see 4.1.3). AFB is given by Assume that all contact forces acting on B are negligible and that A is a satisfactory approximation to a Newtonian reference frame. Then AfB + Q = 0 (4.2.3) from which it follows that AOP* = -GW(p*2)-3>2p* Now (2.5.1) at Hence where fc2 = gR> (B) and Laws of Motion] SEC. 4.2.7 197 Cross multiplication of Eq. (A) with p* shows that p* X ?~ = 0 Now = Avp* X Avp* + P* X ^ (2.5.1) ?^ Hence Is (P* X V) = 0 and (see 1.2.2) the vector a, defined as a - p* X Avp* (D) is independent of time t in reference frame A (see 1.1.1). As a is at all times perpendicular to p*, it follows that point P* moves in a plane (or on a straight line) which passes through the earth's center and whose orientation in A is independent of time. Further- more, Eq. (D) is valid for all values of t, so that the vector a can be found as soon as any pair of values (pj, AVo**) of p* and Avp* cor- responding to the same value of t is known: a = pg x Avopm (E) Next, cross-multiply Eq. (A) with a: a X ^~ = -/c2(p*2)-3/2a X P* or, as a is independent of tin A, AJt(a x AvP*} = -^(P*2)"3720 x P* = -/c2(p*2)-3/2(p* X ^v^*) X P* (D) (P) (1.5.1,2.5.1) 198 SEC. 4.2.7 [Chapter 4 That is, AJt [? X AyP* + fc2(P*2)"1/2P*] = ? from which it follows that the vector b, defined as b = aX Avp* + /c2(p*2)-1/2P* (F) is independent of t in A. b can thus be found as soon as any pair of values (pj, AVo") of p* and Avp* corresponding to the same value of t is known: b = a X Avop' + fc2(po*2)-1/2Po* (G) Suppose, temporarily, that a^O. Then, as a ? Ayr* = 0 (D) the following is an identity: ,vP, _ (q X V) X a a Solve Eq. (F) for a X Avp* and substitute: AVP* = ? ?i?J P. x a (a ^ o) (H) A P. = = P* X (b X q) - fc'tp*2)-1'^* X (p* X q) V Q Cross-multiply with p*: P s\ V Q _ (D) (H> a Expand the right-hand member, keeping in mind that p* is per- pendicular to a: -p* - ba + fc2(p*2)1/2a a = or q[fc2(P*2)1/2 - p* ? b - q2] = 0 Asa^O, this implies that fcj(p?)i/2 - p* . b - q2 = 0 (I) If q = 0, Eq. (F) gives b = fc2(p?)-l'2p* which, after dot multiplication with p*, becomes p* ? b = fc2(p*2)1/2 Laws of Motion] SEC. 4.2.7 199 As this is precisely the relationship to which Eq. (I) reduces when a = 0, Eq. (I) is, in fact, valid for all values of a. Equation (H), on the other hand, leads to an indeterminate expression for Avp* when a = 0. Hence, an alternative expression, applicable when a = 0, is required. Equation (F) shows that p* is parallel to b for all values of t when a = 0. As the orientation of b in A is independent of t, this means that P* moves on a straight line fixed in A and passing through 0. p* can therefore be expressed as p* = xn* where n* (see Fig. 4.2.7c) is a unit vector fixed in A and pointing from O toward P*, and x is an intrinsically positive scalar, because FIG. 4.2.7C P* is prevented by E from ever reaching 0. ^v^* is now given by A r>* dX + and Hence, with Eq. (A) gives dt (2.5.1) at 1. These curves and their orientations relative to 0 and c are shown in Figs. 4.2.7f, g, h, and a summary of results is given in Table 4.2.7. i L L r k0 * ? L L i FIG. 4.2.7f A complete description of the motion of P* cannot be given unless one knows not only the curve C on which P* moves, but Laws of Motion] SEC. 4.2.7 203 TABLE 4.2.7 ? ? - 0 0 < ? < 1 ? = 1 ? - 1 ? > 1 L L > L > L > L = L > 0 0 0 0 0 Curve Circle Ellipse Parabola Straight line Hyperbola Figure 4.2.7f 4.2.7g 4.2.7d 4.2.7h also the relationship between the time t and the position of P* on C. This is obtained as follows: Given a pair of values (pj, Avf*) of p and ^v^, corresponding tQ the same value of t, find a, b, L, and e (see Eqs. (E), (G), (L), (M)). Then, provided L ^ 0, p* remains in a plane perpendicular to a, and it is possible to describe the motion of P* in terms of polar coordinates p* and 6, where p* is the distance between 0 and P* (see Fig. 4.2.7i) and 6 is the angular displacement of line OP* relative to a line which passes through 0 and is parallel to c, 6 being regarded as positive when the displacement is generated by an a rotation of p* relative to this line. If n* is a unit vector directed from 0 toward p*, then FIG. 4.2.7i and so that a = p*X (2.5.1) p* = p*n* (1.5.1) (0) |a| d.7.1) at >* X (o X n' dt 204 SEC. 4.2.7 [Chapter 4 or (as a^O) %\Jt=l (P) Now, it is always possible to choose 0 such that p* ? p*e cos 0 ? L = 0 (K.O) or ^ 1 ? e cos 0 And a = (L) Hence dt L3'2 dd (P) A;(l ? e cos 0)2 and where 0i and d2 are the values of $ when t is equal to 0, 7 i ? ? cos e v } _ 1 + cos 0 + sin2 0 " 3 sin 0(cos 0?1) For c > 1, \ / To obtain the corresponding expressions for the case L = 0, note that d< rj) L ? \dt )o xo J Laws of Motion] SEC. 4.2.7 205 so that t _ f = , [" dx 2 L U?/*) + (dx/dt)f - (2fc2Ao)]1/2 = ? [(?(*,) - (?(?,)] where Xi and x2 are the values of x when < is equal to h and <2, respectively, and the function G(x) is defined as -g log C = I T- I ? G(x) = with Problem: The perigee and apogee of an earth satellite's orbit are the points nearest to and furthest from the earth's center. The corresponding distances d and D between the center of the earth and the satellite's mass center are called the perigee distance and the apogee distance. Express the time T required for one passage of a satellite from apogee to perigee in terms of rf, D, and k (see Eq. (B)). Solution: In Fig. 4.2.7J, constructed by reference to Fig. 4.2.7f, the sense of Avp* has been chosen arbitrarily. For this choice of FIG. 4.2.7J 206 SEC. 4.2.7 [Chapter 4 *, a must, in accordance with Eq. (D), have the sense shown in the figure, and 0 must be regarded as positive. Let 0i and 02 be the values of 0 corresponding to |p*| = D and |p*| = d. Then 0! = 0, 02 = T and F(0i) = F(0) = 0 (R) while F(02) = F(w) From Eq. (R) j?( \ _ c>S(7r) + arc sin S(w) W "" (1 - ?2)3'2 and from Eq. (S) 1 ? ? COS 7T Hence v J (1 - e2)*2 and, letting t\ and t2 be the values of t corresponding to 0i and 02, (Q) From Figs. 4.2.7f and 4.2.7J, and 1 - ?2 ' 1 + ? l-? or, solving for L and ?, D+d' f~D Laws of Motion] SECS. 4.2.8-4.3.1 207 Consequently z (d + D\i* k\ 2 ) 4.2.8 Astronomical observations show that there exist bodies (for example, the planets) whose motions in reference frame A (see 4.2.3-4.2.7) conflict with the hypothesis that A is a Newtonian reference frame. To obtain descriptions of these motions analyti- cally, one regards as Newtonian a reference frame A' in which the sun is fixed, the earth's center (that is, a point fixed in A) moves on a certain elliptical path (described once per year), and the angu- lar velocity of A (A'wA) is equal to zero. Re-examination of the motions discussed in Secs. 4.2.3-4.2.7 then reveals that the results there obtained remain essentially unaltered because the inertia forces which must now be added are small in comparison with forces already considered. A' is therefore a better approximation to a Newtonian reference frame than is either A or E. These considerations lead to two conclusions: (1) The search for a truly Newtonian reference frame requires exploration of mo- tions taking place at ever greater distances from the earth. (2) Whether or not a given reference frame may be regarded as New- tonian for the purpose of solving a particular problem depends on the accuracy required of the solution. Except where indicated otherwise, the earth E is regarded as a Newtonian reference frame throughout the sections which follow. Other reference frames are regarded as Newtonian if and only if their motions relative to E satisfy the requirements set forth in Sec. 4.2.2, and the word "fixed" means "at rest in a Newtonian reference frame." 4.3 Motions of rigid bodies 4.3.1 Infinitely many equations of motion (see 4.2.3) can be written in order to describe any motion of a rigid body. However, at most six such (scalar) equations are both independent of each other and nontrivial (see Vol. I, Sec. 3.6.10). To obtain these six (or, in many cases, a smaller number) with a minimum amount of labor, it is necessary to use those properties of a zero system which 208 SEC. 4.3.1 [Chapter 4 permit one to take advantage of specific features of the problem under consideration. Problem: Referring to Example 2.2.10, and supposing that P is fixed and horizontal, that the disc D is uniform, has a mass m and radius r, and that D is rolling on P, find three equations governing 0, , and if/, and use these to discuss the stability of a spinning disc. FIG. 4.3.1a Solution: The vectors appearing in the free-body diagram of D, Fig. 4.3.1a, are described as follows: ni, n2, n3, and n are unit vectors which were described in Ex- ample 2.2.10. W represents the system of all gravitational forces acting on D, and has a magnitude mg. R represents the reaction (see Vol. I, Sec. 4.6.1) of P on D, and has a line of action passing through point C. (The basis for representation of this reaction by means of a single force is discussed in Vol. I, Sec. 4.8.3.) PFD and PJD are the inertia force and the torque of the inertia couple acting on D in reference frame P. (Expressions for these in terms of m, r, 0, , $ were obtained in Problem 4.1.5.) To find the desired equations, set the moment about C of all forces acting on D equal to zero, noting that the unknown force R is thus eliminated from further consideration: (W + P?D) X (rm) + I* = 0 or, writing the corresponding scalar equations of motion, # cos 6 + 26x1/ = 0 | ^ L (A) ? + 0 sin 6) + 56 cos 0 = 0 Laws of Motion] SEC. 4.3.1 209 The disc is said to be "spinning" when 0 = 0 j = o + 4>ot I (B)J where all symbols with the subscript zero denote constants. In particular, 0 gives the angular speed (for the direction n) with which D is spinning. That Eqs. (B) describe a motion which satis- fies Eqs. (A) may be verified by substitution. A motion described by 0 = 0* = <*>o + ht + * [ (C)J where 0*, *, ^* are functions of t, differs only slightly from that of spinning, provided 0*, *, \p* which remain arbitrarily small and which, when 6, , \p as given in Eqs. (C) are substituted into Eqs. (A), permit Eqs. (A) to be satisfied, at least approximately. Substitute from Eqs. (C) into Eqs. (A), expressing sin 6* and cos 6* in terms of powers of 6*: + 56*- 6(A> + i>*)i* (l - Y + ? ? ? + 5(4>o + *)6* (l - Y + ???)= 0 210 SEC. 4.3.1 [Chapter 4 Drop all terms of second or higher degree in the starred quan- tities: $* = 0 (D) 50* - 6o2 + j)o* = 0 (E) = 0 (F) Equation (D) is satisfied by * = o* + o*t where o* and o* are constants. Hence * can be kept arbitrarily small by taking o* = 0 and assigning a sufficiently small value to o*. Equations (E) and (F) may be solved simultaneously. The latter is satisfied whenever 3^* + 5)nS (U) (Q) and the motion persists until the instant t\ at which Vs* attains the value vf* defined as Xn (V)CD r2 + k2 where b = rn X vf + fc'wo3 (W) (i) The time interval t\ ? to is given by h _ <0 . K - vf) ? rf (u> tig Subsequent to h, S* moves with velocity vf* on the straight line parallel to vf* and passing through the point whose position vector Pi relative to 0 is Pi - Po + (h ~ fc)vf -?(*i- toY nS7 (Y) (T) 4 (During both phases of the motion the angular velocity ws has the value <*s = ?jf + m X (vf - v5*) (Z) (I.W) where Vs* is given by Eq. (U) prior to h and is equal to vf* at and subsequent to h.) 216 SEC. 4.3.2 [Chapter 4 If Vo (see Eq. (S)) is equal to zero, S* moves with velocity vjf on the straight line parallel to Vo * and passing through the position of S* at time t0. (%R*a + <*>3W)n2 (A) Finally, the system of all contact forces acting on R is represented by the known force C and torque y, together with an unknown force F, applied (arbitrarily) at R*, and the torque T of an unknown couple. FIG. 4.3.2b F and T may be found by setting the resultant and the moment about R* of all forces represented in Fig. 4.3.2b equal to zero (see 4.2.3): F + W* + C = 0 Thus T + T* + y = 0 F = _(W* + C) T= -(T* + ?) 218 SEC. 4.3.2 [Chapter The system of all unknown contact forces acting on R has now been reduced to a known force and a couple of known torque. But these do not furnish detailed information about forces exerted on S by the bearings. The system of unknown contact forces acting on R may be regarded as consisting of four force systems: SA and SB, comprised of forces exerted on S across the bearing surfaces at A and B; and SA and SB', containing the forces exerted on S by the portions of the shaft contiguous to S at A and B. In the free-body diagram of R shown in Fig. 4.3.2c, these four force systems are represented by four forces (A, B, A', B') and four torques (a, 0, a', 0') of couples. (WR, T*, C, and *y have the same meaning as in Fig. 4.3.2b.) Note that the choice of points of application for the forces A, A', B, B' is an entirely arbitrary one. FIG. 4.3.2C Two equations relating the eight unknown vectors appearing in Fig. 4.3.2c may be obtained by setting moments about points A and B equal to zero: n3 X [a(W* + C) + (a + b)(B + B')] + a + a! + fi + 0' + y + 1* = 0 (B) -n3 X [6(W* + C) + (a + 6)(B + B')] + a + a! + p + fi' + y + T* = 0 (C) Laws of Motion] SEC. 4.3.2 219 In order to extract useful information about the force systems SA and SB from these equations, one must make a number of assumptions. If Ai9 B{, etc., are the nt, i = 1, 2, 3, measure num- bers of A, B, etc., these assumptions-may take the following form: The bearing surfaces are so smooth that A, B, a, and fi are perpendicular to the shaft axis (see Vol. I, Sec. 4.8.2). Thus Az = Bz = <*3 = ft = 0 (D) The bearings are so short that contact between the shaft and the bearing surfaces takes place only along two circles, one having its center at A, the other at B (see Vol. I, Sec. 4.8.4). Thus ax = a2 = ft = ft = 0 (E) SA' and SB' can each be reduced to a force whose line of action coincides with the shaft axis, together with a couple whose torque is parallel to the shaft axis. Thus Ax' = At' = Bx' = B2' = ax' = at' = ft' = ft' = 0 (F) (In order to justify the last assumption, one must consider forces acting on portions of the shaft not shown in Fig. 4.3.2a. As nothing has been said about these forces, the assumption should be re- garded as tenuous.) The scalar equations corresponding to Eqs. (B) and (C), solved for A\y A*, Bh and B2y now give (after using Eq. (A)) Ci) - 72 a b(W* + a a(W* + a a(W$ + + b -Ct) + + b ? Ci) + + b 7i 72 7i a + b *V b(W$ + Ct) + 7i+ (G) , 4>UR*? + 4>?rf/0, ?/O, 2 (and is equal 222 [Chapter 4 FIG. 4.3.4a to i/?), while b3 X (W? + F?) + T^1 + T* + y] - bi - 0 which, after using Eqs. (A)-(D) and noting that (see Fig. 4.3.4a) W? = ?Mgc2 = ? Mg(cos y b2 ? sin y b3) can be expressed as zM(g cos 7 + a%) - [^/oaf + 224 SEC. 4.3.4 [Chapter 4 where a? is the b2 measure number of aQ. Furthermore, as b2 is principal direction of R for 0, %? = %? = 0 (E) (3.5.7,3.3.3) Hence the equation of motion becomes zM(g cos 7 + a?) - [&? - using Eqs. (A)-(E), (G), and (H), and noting that c2 = cos 7b2 ? sin 7b3 (I) the following equation is obtained: sin - cos y[zMa2 cos2 7 + i sin2 7 + Mz2 cos2 7 + 4%i?)i where h 2, 03 are defined as /o/3) sin 7] = 0 (Q) (R) (S) and can be recognized as moments of inertia of the body consisting of A and R about lines passing through 0 and parallel to bi, b2, b3, respectively. (The motivation for expressing the second equation of motion in the form given in Eq. (Q) is explained by the paren- thetical comment at the end of Example 4.4.14.) The gyroscope is said to be in a state of steady precession when ft = a>,, 7 = 70, B = o)p where ?? 70, and o>p are constants, co, is called the spin velocity, o)p the precession velocity. Substitution into Eqs. (P) and (Q) shows that this motion is possible only either when To = ~ Laws of Motion] SEC. 4.3.4 227 or when a)8, y0, and wp are related as follows: &?u.wp + (02 - 03 + Mz2)a>p2 sin 70 - zMg = 0 (U) Hence for given values of o>p and 70 (70 ^ v/2) there exists a unique value of a>,; and when a>, and 70 are given, wp must have either of the two values sin 7o]I/2 - 03 + Mz2) sin 70 The corresponding motions are called a sZow steady precession and a/as? steady precession. Note that there exists only one nonzero precession velocity when 70 ?* 0 and the block Q is removed, that is, M = 0; that when Af 5^ 0 one of the values of o)p approaches infinity as 70 approaches zero; and that both values of wp vanish when both M and 70 are equal to zero. A motion described by 4 = o>,, 7 = 7o + 7*, 5 = a>p + 8* (V) where co, and wp are constants, and 7* and 5* are periodic functions of time, is called a periodic nutation. One such motion takes place subsequent to the application of small disturbing forces when the gyroscope is in one of the states of steady precession described above. An approximate description of this nutation is obtained by substituting from Eqs. (V) into Eqs. (P) and (Q) and dropping all terms of second or higher degree in starred quantities. Re- stricting the discussion to the case 70 = 0 this gives + <(?2 + Mz2 - and - [(02 + MZ2 + 02T)(O>P + ?*) - *3V%7*] (Q=v) 0 the second of which requires that 228 SEC. 4.3.4 [Chapter 4 where C is a constant, so that the first becomes (01 + Mz>)r + [^(2 + M* . ^ + ^ ^f/i2^o] 7* (X) As Eqs. (W) and (X) were obtained by dropping second and higher degree terms in 7* and $*, they are valid only when their solutions are functions which can be kept arbitrarily small. In the case of Eq. (X) this is assured by taking zMg in which case this equation has the general solution 7* = A sin (?n? + 0) (Y) where A and 6 are arbitrary constants, and a>n, the circular fre- quency of nutation, is given by wrTM? L" 11 (*2 + Mz ~ *3) + + + M* + tfru Furthermore, when 70 = 0, Eq. (U) reduces to so that C = and i* *&?<*- * A44?a. . which shows that 5* can be kept arbitrarily small. Note that Eq. (Z) may be used to express wn in the alternative form which, when M = 0, reduces to COn = Laws of Motion] SEC. 4.3.5 and, when a)8 is sufficiently large, to 229 Mz* both of which are of the form where (see Eqs. (R) and (S)) h is the sum of the moments of in- ertia of A, R, and Q about the line passing through 0 and parallel to bi, while 72 is the moment of inertia of all moving parts about the axis of rotation of gimbal ring C when 7 = 0. 4.3.5 The analysis of contact forces exerted by & body B on a body B' across a surface ?/?sin7oC3 4.4 Linear and angular momentum 4.4.1 Given a set S of n particles situated at points P?, i = 1, 2, . . . , n, having masses rat, and moving with velocities Laws of Motion] SECS. 4.4.2-4.4.4 231 relative to a point Q in a reference frame R (see 2.4.1), the linear momentum of S relative to Q in R, denoted by RLS/Q, is defined as 4.4.2 Given a continuous body C regarded as occupying a fig- ure F, and a point Q moving in a reference frame R, the linear momentum of C relative to Q in R, defined as the limit of the corre- sponding linear momentum of a set of particles constructed in a manner similar to that employed in Sec. 4.1.2, can be expressed as = f where Rvp'Q is the velocity of a generic point P of C relative to Q in R and p is the mass density of C at P. 4.4.3 The linear momentum of a body B (any collection of matter) relative to a point Q in a reference frame R is defined as the sum of the corresponding linear momenta of the particles (see 4.4.1) and continuous bodies (see 4.4.2) comprising B. Given two points Q and Q' moving in a reference frame R, the linear momenta RLBIQ and RLB/Q/ are related to each other as follows: R^B/Q _ R\_B/Q' _|_ R\_Q'/Q (1) where RLQf/Q is the linear momentum relative to Q in R of a par- ticle situated at Q' and having a mass m equal to that of the body B; that is, RIQ'/Q = mRyQ'/Q (2) Proof: Use Sec. 2.4.4 and the definitions given in Secs. 4.4.1 and 4.4.2. 4.4.4 The linear momentum RLB/P* of a body B relative to the mass center P* of B is equal to zero. Consequently the linear momentum of B relative to any point Q is given by where m is the mass of B. Proof: If B is a set S of n particles, and rt, i = 1, 2, . . . , n, are the position vectors of the particles relative to P*, then 232 SECS. 4.4.5^1.4.6 [Chapter 4 (4.4.1) ?Ti (2.4.1) ~ ?l (1.4.1) ai J^-J and n by definition of P*. Use Sec. 4.4.3. Similarly, if B contains con- tinuous bodies. 4.4.5 If 0 is a point fixed in a reference frame R, the linear momentum RlmBi? of a body B (see 4.4.3) is independent of the position of 0 in Ry because the velocity RvPI? of any point P of B is independent of the position of 0 (see 2.5.1). Accordingly, the linear momentum of B in R relative to any point fixed in R is called the absolute linear momentum of B in /?, or, for short, the linear momentum of B in R; it is denoted by RLB and involves only absolute velocities, that is, RLS = V miRvp< (1) (4.4.1) ?"f ?LC = f p?vpdr (2) (4.4.2) JP and can be expressed as RLB = m*vp* (3) (4.4.4) where m is the mass of B and P* is the mass center of B. 4.4.6 If F is the resultant of all gravitational and contact forces acting on a body B, and R is a Newtonian reference frame (see 4.2.1), F is related to the linear momentum RLB of B in R (see 4.4.5) as follows: This equality is known as the linear momentum principle. Proof: If B is a set S of particles situated at points Pi, i' = 1, 2,. . . , n, Laws of Motion] SECS. 4.4.7-4.4.9 233 R^*L* = *-'Ym*vp* = Ymi dt (4.4.5) dt f^\ % (1.4.1) fr{ * dt (2.5.1) {?{ (4.1.1) fT\ But r + YRFPt = o ft^ (4.2.1) Hence and dt L " Similarly, if 5 contains continuous bodies. 4.4.7 If the resultant of all gravitational and contact forces acting on a body B is equal to zero, the linear momentum RLB of B in a Newtonian reference frame R (see 4.4.5) is time-independent in R (see 1.1.1). This follows from Secs. 4.4.6 and 1.2.2, and is known as the principle of conservation of linear momentum. 4.4.8 Given a set S of n particles situated at points F?, i = 1, 2, . . . , n, having masses m?, and moving with velocities Rvp;P; = 0 (A) t = i from which it follows (use 1.4.1 and 2.4.1) that = 0 (B) Laws of Motion] SEC. 4.4.11 Then, if B is a set of n particles, 235 (4.4.8) f m.x. r* + Pi) X (F4.4.10) r* X (2.4.4) + ? miPi X t=i ? mt\ r* X R 2 = 0 + ?Afl/^ + mr* X Rvp*/Q + 0 (B) (4.4.8) (A) Similarly,if B contains continuous bodies. 4.4.11 The angular momentum RtARf<> of a rigid body R relative to a point Q fixed on R can be expressed as = no ? *'? (1) or as (2) where n0 is a unit vector parallel to the angular velocity (R'?*) of R in the reference frame R' (see 2.2.1) and n;, j = 1, 2, 3, are mu- tually perpendicular unit vectors. Proof (see Fig. 4.4.11 for notation): FIG. 4.4.11 236 SEC. 4.4.11 [Chapter 4 R'AR'Q = f pr X R'yp'Q dr = f pr X (* ?* X r) dr (4.4.9) JF (2.4.5) -/F If no is a unit vector parallel to R'uR, the following is an identity: Hence (3.5.2) 'A*'<* = n0 ? ?'?* f pr X (n0 X r) dr = no ? ?' JF (3.5.2) Next E5 (3.5.7,3.2.4) ^Ti y^ where ot is the nt measure number of no, that is, Oi = n o ? R'taR Thus Problem: Referring to Example 4.3.4, determine the angular momentum ASI? of the system S of bodies consisting of A, B, P, P', and Q. Solution: A5/o = AA/0 + A^/o + AiQI? (4.4.10) (2) = ^/?7bi + ?{?8 sin Mz2)8 cos T b2 + ($3/o0 + 4>%?y ~ ? n = jt (RAB?> ? n) - where n is any unit vector, and MQ and RABIQ are defined as in Sec. 4.4.13. Proof: M n = I ~JT I' n (4.4.13) \ at I Now d fRKRio \ (RdRABI(>\ , RABIO Rdnj { RAB^ ? n) = ( ?? ) ? n + RAB^ ? -fi-at (i.5.2) \ at / at Hence Laws of Motion] SEC. 4.4.15 239 and MQ . n = jt (RAB?* ? n) - *A*'? ? ^ Example: Referring to Example 4.3.4 and Problem 4.4.11, and letting M? be the moment about point 0 of all gravitational and contact forces acting on S (the body consisting of A, B, P, P', and Q), M? ? bi is seen to be given by M? ? bi = zMg cos y Hence zMg cos 7 = j (A*'* ? bO - As?/?8 sin y (P4.4.11) so that jt (As>? ? bi) = (0i + Mz2)y - R{?(b sin 7+57 cos 7) Also ?jj- = ? 5c3 = ?5(sin 7 b2 + cos 7 b3) so that As,o . ^1 = _^(02 + Mz2)i sin CQS ?^ (P4.4.11) + (*^oj8 + *f/?7 - R{ab sin 7 + 5[(02 + Mz2 - 03)5 sin 7 + &?$] cos 7 in agreement with Eq. (P) of Example 4.3.4. (Equation (Q) of Example 4.3.4 may be obtained in a similar manner, c2 playing the part of n.) 4.4.15 If Q is any point fixed in a Newtonian reference frame R (see 4.2.1), or if Q is the mass center of a body B, and the sum of the moments about Q of all gravitational and contact forces acting on B is equal to zero, then the angular momentum RABtQ (see 4.4.10) is time-independent in R (see 1.1.1). This follows from 240 SEC. 4.4.15 [Chapter 4 Sees. 4.4.13 and 1.2.2, and is known as the principle of conservation of angular momentum. Example: When the system of all gravitational and contact forces acting on a rigid body R is equivalent to a single force whose line of action passes through the mass center P* of R it is possible for R to move in such a way that the angular velocity R'uR of R in a Newtonian reference frame Rr remains fixed in both R and Rf: R'u)R must be parallel to a principal axis of R for P*, and the motion is stable if and only if the moment of inertia of R about this axis is either larger or smaller than R's moment of inertia about any other line passing through P*. This is shown as follows: Let nt, i = 1, 2, 3, be mutually perpendicular principal direc- tions of R for P*, and let these be fixed in R (see 3.5.7 and 3.3.9). Then, if cot is the nt measure number of Rfo)R, R'fiji/p* = dWih + *2*2/P*co2n2 + *gr?,n, (A) (4.4.11) and 'AR/P*Jp- + R'^R X R>AR/P* = ut (2.1.4) = (A) so that the principle of conservation of angular momentum gives C*>2 = CO3CO1C2 (B) C03 = CO1W2C3 where the constants Ch C2, and C3 are defined as c2 = Equations (B) show that two of the measure numbers of R'o>R must vanish if all three are to remain constant, which means that Laws of Motion] SEC. 4.4.15 241 R'i? is a constant and the starred quantities are functions of time. Substitute into Eqs. (B) and drop all terms of the second degree in the starred quantities: Differentiate the second of these equations and use a>3* as given by the third; then differentiate the third and use o>2* as given by the second. The following differential equations are then seen to govern w2* and a>3*: "2* - (CO!?)2C2C3CO2* = 0 "3* ~ (a>i?)2C2C3co3* = 0 Hence, in order that the motion COi = COi?, CO2 = CO3 = 0 be stable, it is necessary and sufficient that C2C3 < 0 or, from Eqs. (C), that wr - tfnwr - %n > o This inequality can be satisfied in only two ways: Either *?r>*%p* and tfr>4>Rr or *Rr < ^r and Rr < *Rr In the first case, the moment of inertia of R about the principal axis passing through P and parallel to R'?oR is larger, and in the second case itissmaller,than,R'smomentof inertia about any other line passing through P*, since R's largest and smallest principal moments of inertia are respectively larger and smaller than R's moment of inertia about any line which passes through P* and is not a principal axis of R for P* (see 3.5.7 and 3.3.14). 242 SECS. 4.4.16-4.5.3 [Chapter 4 4.4.16 The angular momentum principle (see 4.4.13 and 4.4.14) furnishes a convenient means for the derivation of differential equations governing the motion of a system when M? does not involve any unknown contact forces (see, for instance, Example 4.4.14). In particular, this is the case whenever the principle of conservation of angular momentum (see 4.4.15) is applicable. When use of the angular momentum principle necessitates the introduction (and subsequent elimination by application of, for example, the linear momentum principle) of contact forces which do not appear in a solution based on D'Alembert's principle (see 4.2.1, 4.2.3, and 4.3.1), the latter principle yields results more readily. For the purpose of determining contact forces when the motion of a system is known (see 4.3.2), D'Alembert's principle is fre- quently more convenient than the angular momentum principle, because of the restrictive character of the point Q (see 4.4.13). 4.5 Activity and kinetic energy 4.5.1 Given a set S of n particles situated at points P?, i = 1, 2, . . . , n, having masses mif and moving with velocities RvpR)2 (1) or, as O Q = i where R'wR is the angular velocity of R in R', o/Q is the moment of inertia of R about a line passing through Q and parallel to R'KR,Q = J(*'w*)Next (3.5.7,3.2.4) tZlfz = 7 (C) (no tf/Q *'<***'&,* Substitute into Eq. (D). Problem: A uniform solid sphere S of mass m performs a pure rolling motion (see 2.5.10) on a surface fixed in a reference frame R. A particle P of the same mass m moves in such a way that its velocity in R is at every instant equal to the velocity in R of the center C of ?. Determine the ratio of the kinetic energy of S relative to 0 and the kinetic energy of P relative to 0, 0 being a point fixed in R. Solution: Let r be the radius of S, r the position vector of C FIG. 4.5.4b 246 SEC. 4.5.5 [Chapter 4 relative to the point of contact between S and the surface on which S rolls (see Fig. 4.5.4b). Then RKSIO = RKSIC + im(/2vC/O)2 (4.5.3) = h%c(Ras)2 + im(Ru>s X r)2 (1) (2.5.10) The moment of inertia of S about all lines passing through C has the value 2mr75 (see Appendix, Fig. 8); and for pure rolling (see 2.5.10), Rws is perpendicular to r, so that (*?s X r)2 = r2(??s)2 (A) Hence RKSIO = ?EL (?WS)2 + ? (^) = Next RKpf? -= im(Rvpi?y (4.5.1) By hypothesis RyPIO = RyCIO = R^S X T (2.5.10) Thus and RKPIO . Im(?ws X r)2 = ^ 2 (A) 2 RKPI? 4.5.5 If 0 is a point fixed in a reference frame #, the kinetic energy RKBI? of a body B (see 4.5.3) is independent of the position of 0 in R, because the velocity Rvpi? of any point P of B is inde- pendent of the position of 0 (see 2.5.1). Accordingly, the kinetic energy of B in R relative to any point fixed in R is called the abso- lute kinetic energy of B in R, or, for short, the kinetic energy of B in R. It is denoted by RKB and involves only absolute velocities, that is, RKS = *X>t(*v^)2 (1) (4.5.1) t = l = \ fp(Rypydr (2) (4.5.2) JF Laws of Motion] SEC. 4.5.6 247 and can be expressed as RKB = RKB/p* + RKp* (3) (4.5.3) where RKp* = Jm(V*)f (4) (4.5.3) and m is the mass of B, while P* is B's mass center. Problem: Referring to Example 4.3.4, determine the kinetic energy DKS of the system S of bodies consisting of Ay B, C, P, P\ and Q, in a reference frame in which D is fixed. Solution: DKS = DKA + DKR + DK<* + DKC (4.5.3) [ (4.5.4) 2 + ^/?52 cos2 7 + 3W - 5 sin 7)2] (F4.3.4d) DK* = (4.5.4) sin 7]7 + ?&??2 cos2 r + [-*?/?7 + *&?? sin 7]? sin -y} X zb,)2J() (4.5.1) (2.4.5) = \Mz*(h cos -y b, - 7b2)2 = ?Afz2(?2 cos2 7 + 72) DKC = (4.5.4) Substitute: + [(*i + Mz1) cos2 7 + &?y8 sin 7 + &?0 ~ 2? sin 7)18} 4.5.6 The kinetic energy ^K7" of a particle P in a Newtonian reference frame R (see 4.5.5), the velocity 'V, and the resultant F of all gravitational and contact forces acting on P are related as follows: 248 SEC. 4.5.6 [Chapter 4 = *AP where RAP, called the activity in R of the gravitational and contact forces acting on P, is given by RAP = RVP . p (2) Proof: The inertia force *FP acting on P in R (see 4.1.1) can be expressed as = m (2.5.1) at Dot-multiply with Rvp: Ryp . Rfp = -mRyp . 9LJLat In accordance with Sec. 4.2.1, F + flp = 0 or Rfp = -F Hence Rvp ? F = mRvp ? -3?(A) at and, defining /2Ai> as RAP = Rvp ? F it follows that Next ?1 = RAP dt = 3: o m( v ) = rnRwp ? ?jr = ^ |2 V J (1.5.2) ^ (B) j, j I ^ f?lfl T J I ~~" lit/ V j, ^A at (4.5.5) at |_* J (1.5.2) at (B) Problem: In Fig. 4.5.6, R represents a Newtonian reference frame, Pi and P2 are particles which are fixed in R and have masses mi and m2, and P is a particle of mass m, which moves under the action of gravitational forces exerted on it by Pi and P2. Express the magnitude of the velocity Rvp in terms of the grav- itational constant G, the masses, mi and m2, and the distances X\ and x2. Solution: The resultant F of all gravitational and contact forces acting on P is given by Laws of Motion] 249 FIG. 4.5.6 Y ~ ~^Fni ~ ~^~ "2 where ih and n2 are the unit vectors shown in Fig. 4.5.6. Hence *A* = ?VP . F = -Gm (& *** ? n, + ^f *vF ? n2) (A) \Xi X2 / Rvp can be expressed either as R p d v = T/ (2.5.1) tit (1.5.1) or as Thus vF = ? (x2n2) = x2n2 + x2 -JT (2.5.1) <" (1.5.1) at (A) (B) (C) But and, similarly, 250 SEC. 4.5.7 [Chapter 4 Consequently RAP Next Thus ttt |_^ J (D,E) \ X\ Xi / Note that ~ (jYi\X\ 1712^2 1 /-r>i\ = ~Gm W + ~xT) (D) (E) _Gm (*& + ?&\* \Gm (na + na \ Xi2 x22 / dt I \xi x2Hence and 2 v ' ' " Vxi where C is a constant. As it follows that 4.5.7 The law of motion stated in Sec. 4.5.6 has the following advantages over those discussed previously: It obviates the neces- sity to consider accelerations and frequently leads to a readily integrable differential equation (see, for example, Eq. (F), Problem 4.5.6). However, as this law furnishes only one equation, it must at times be used in conjunction with those considered earlier. Problem: Figure 4.5.7a represents a parcel chute: A slowly A Laws of Motion] SEC. 4.5.7 251 moving belt carries the parcels from A toward B, from which point they slide toward C on a right-circular cylindrical surface. Regarding the parcels as particles, determine the range of per- missible values of the coefficient of friction p. Solution: The kinetic energy Kp of a parcel P of mass m is Kp = \m(ypY (4.5.5) where the velocity (vp) of P, expressed in terms of the quantities shown in Fig. 4.5.7b, is given by B FIG. 4.5.7b Hence (2.5.3) Kp = i (A) (B) The resultant F of all gravitational and contact forces acting on P is equivalent to the system of three forces shown in Fig. 4.5.7c. Thus B 0 FIG. 4.5.7C F = {mg sin 6 ? N)ni + (mg cos 6 ? and the activity Ap is given by Ap = vp ? F = R6(mg cos 6 - (4.5.6) (A) (C) 252 SEC. 4.5.7 [Chapter 4 Consequently, in accordance with Sec. 4.5.6, ? ( - mR262) = RB(mg cos 0 ? /xiV) (D) at \z / (B,C) Before this equation can be used, the quantity N, which is an as yet unknown function of 0 and 0, must be eliminated. Figure 4.5.7d is a plane free-body diagram (see 4.3.3) of P. (It differs from Fig. 4.5.7c only in the addition of the inertia forces ifiRftfii and ?mR$n2-) Set the sum of the rii resolutes of all forces equal to zero (see 4.2.3): mRB2 + mg sin 0 - N = 0 (E) Solve for N and substitute into Eq. (D): B 0 FIG. 4.5.7d mF mgf \mR* f I W = [<7(cos 6 - M sin 9) - Regard 6 as a function of 0. Then (F) and Eq. (F) can be written, alternatively, as J = ^ (cos 0 - M sin 0) This first-order, linear, nonhomogeneous differential equation has the general solution 62 ^rTT\ t3^ cos 0 + (1 - 2M2) sin 0] + Ce~2^ (G) Laws of Motion] SEC. 4.5.7 253 where C is a constant whose value is found by taking vp = 0 when 0 = 0 so that which gives 6\e=o = 0 (A) ^ = ?rfc)(3M) from which it follows that [3M(cos 6 - e-2**) + (1 - 2M2) sin 6]4 M2) Now, unless 0 will stop before it reaches point C. Hence M must be such that -Sfie-^ + 1 - 2/x2 > 0 In Fig. 4.5.7e, the function /(/x) defined by I.O 0.8 0.6 ? 0.4 0.2 0 -0.2 0.1 0.2 0.3 0.4 0.5 0.6\0.7 FIG. 4.5.7e 254 SEC. 4.5.8 [Chapter 4 is plotted against JU, for 0 < /x < 0.7. This graph shows that satis- factory performance may be expected for 0 < n < 0.6. (Note that a second force equation (resolutes parallel to n2) could have been used in place of Eq. (D): ? mR$ ? nN + mg cos 0 = 0 which after elimination of N (see Eq. (E)) becomes + = & (cos 6 - /x sin 0) This is precisely the equation obtained when the differentiation indicated in the left-hand member of Eq. (F) is carried out.) 4.5.8 When the system of all gravitational and contact forces acting on a rigid body R is equivalent (see Vol. I, Sec. 3.5) to n forces Ft, i = 1, 2, . . . , n, applied at points Pt, i ? 1,2,..., n, of Ry the kinetic energy (R'KR) of R in a Newtonian reference frame R' (see 4.2.1) is related to these forces and to the velocities R'vPi, i = 1, 2, . . . , n, as follows: dR'KR dt = R'AR (1) where R'AR, called the activity in R' of the gravitational and contact forces acting on R, is given by R'AR = (2) Proof: In the free-body diagram of R shown in Fig. 4.5.8a, P* R1 FIG. 4.5.8a Laws of Motion] SEC. 4.5.8 255 is the mass center of R, and *'F* and R1R represent the system of inertia forces acting on R in Rf (see 4.1). These can be expressed as (4.1.3,2.5.1)* and (see 3.5.7 and 3.3.4) ? -, + W --? X n,) where ny, j = 1, 2, 3, are mutually perpendicular principal direc- tions of R for P*9 fixed in R. Dot-multiply the first of these equations with * V*, the second with R' [RLJ*Z?/ L t = l (D) [Chapter 4 *, u, F J Thus, defining ^'A72 as it follows that Next R'AR = ? R'AR (4.5.5) Hence dR'KR dt \ Jp f/p\R'oijR)2 (4.5.4,3.3.3) 3 (A) (B) RtAR (F) (F) FIG. 4.5.8b Laws of Motion] SEC. 4.5.8 257 Problem: In Fig. 4.5.8b, B represents a uniform rectangular plate of mass M} which is suspended by two light strings. During one possible motion of this system, both strings remain taut while the plate rotates about a vertical axis passing through the mass center B* of B, and B* moves upward and downward on this axis. Determine the angular velocity of B and the tension T in each of the strings at the instant when h = 0 (see Fig. 4.5.8b), assuming that the plate is released from rest when h = a/6. Solution: Let no be a unit vector pointing vertically upward (see Fig. 4.5.8b). Then v** = Ano, (2.5.7) (4.5.5) = 6no (2.2.4) (A) (4.5.4) (4.5.5) or (A) 2 12 (3.5.8,3.4.9) M (B) The system of all gravitational and contact forces acting on B is equivalent to the three forces shown in Fig. 4.5.8c, where F and F f FIG. 4.5.8C F', representing forces exerted on B by the strings, are parallel to the strings. In order for the strings to remain taut (but un- 258 SEC. 4.5.8 [Chapter 4 stretched), the velocities vp and vp/ of the corners P and P' must be perpendicular to the strings. Hence sp ? F + vp' ? F' = 0 (C) and the activity AB is given by A* = vp ? F + vp/ ? F' + vB* ? (-Mgno) That is AB = -Mgh (D) (C.A) Consequently ? ? (b2d2 + I2h2) = ?Mghat L24 (B> J (i) (D) or from which it follows that ^ 12A2) +gh = C (E) where C is a constant whose value is found by noting that 6 and h are each equal to zero when h = a/6, so that ga/b = C (F) (E) and (E,F) When h = 0, h attains its minimum value (see Fig. 4.5.8b). Hence AU-o = 0 (H) and 24 (G) 6 or *U-o = ?| (gaY" (I) Thus Laws of Motion] SEC. 4.5.8 259 As the forces F and F' exerted on the plate by the strings have so far entered the solution of the problem only in a form making it impossible to evaluate them (see Eq. (C)), some principle other than the one presently under consideration must be employed for their determination. The linear momentum LB of B is given by LB = MvB* = Mhn0 (4.4.5) (A) so that, differentiating, dLB n?l ? = Mhn0 The resultant of all gravitational and contact forces acting on B is F + F' - Mgn0 Consequently F + F' - Mgn0 = Mhn0 and, letting it follows that (4.4.6) F|A-C \h=0 Tno 2T - Mg = Mh\h-o Now, from Fig. 4.5.8d, 6 (J) a* = (a - h)* + b2 sin2 g a-h FIG. 4.5.8d 260 SEC. 4.5.9 [Chapter 4 so that, differentiating twice with respect to time, 2h? - 2(a - h)h + ~ (62 cos 6 + 0 sin 0) = 0 Thus, as 6 = 0 when A = 0, and from Eqs. (H) and (I), h\h=o = g and 2T - Mg = Mg (J) or Note that Eq. (1), Sec. 4.5.8, while not sufficient for the deter- mination of T, provided useful information in the form of Eq. (I). 4.6.9 One of the advantages of the law of motion stated in Sec. 4.5.8 over those discussed previously is that it facilitates elimina- tion of certain contact forces (see, for example, Eq. (C), Problem 4.5.8). In particular, when a rigid body R rolls on a surface S fixed in a Newtonian reference frame R' (see 4.2.1), the system of forces exerted on R by S contributes nothing to the activity R'AR (see 4.5.8), because the velocity of every point of R which is in contact with S is equal to zero (see 2.5.10). Problem: Eeferring to Problem 2.5.14, and assuming that R is a uniform rectangular parallelepiped having the dimensions shown in Fig. 4.5.9a, obtain a differential equation governing 0, and use it to discuss motions of R during which 6 remains small. R FIG. 4.5.9a Solution (see Fig. 4.5.9b for notation): K* = (4.5.5,4.5.4) (A) Laws of Motion] 261 w" = ? <0n3 (2.2.4) = o)R X (s*n2 (2.5.10) \ (B) X (P2.5.14) V^ = (B) 3 (3.5.5) 1 KR = - (3.5.8) 12 (App.F7) (B) (C) (D) (E) (A,B,C,D) The system of all gravitational forces acting on R is equivalent to the force F shown in Fig. 4.5.9b, and F = ? + sin 6 n2) As the only contact forces acting on R are those exerted on R by S, the activity AR is given by AR = vp* ? F = -mg6 (rS cos 6 - 5 sin 0) (G) (C.F) V ^ / 262 SEC. 4.5.9 [Chapter 4 Hence, in accordance with Sec. 4.5.8, For motions during which 0 remains small, ql V 2*2 ~?L . hl12 + 3 "+"r "l2 + 3 and rd cos 0 ? - sin 0Z Hence, when 6 remains small, Eq. (H) can be replaced with a?--*- m or, equivalently, with 6 + p26 = 0 (J) where P a2/12 + 62/3 The solution of Eq. (J) assumes one of three forms according as p2 is positive, negative, or equal to zero: Forp2 >0 (r > 6/2), 6 = 0O cos p? + ? sin p? (L) where 0O and ^0 are the values of 6 and 6 at ? = 0. This describes a harmonic oscillation about the equilibrium position (0 = 0), with circular frequency p and amplitude A, A = [0o2 + (6o/p)2]112 and A can be kept arbitrarily small by appropriate choice of 0O and 60. For p2 < 0 (r < 6/2), let p2 = -p2. Then (M) The corresponding motion of R is one during which 0 increases without limit (so that Eq. (M) must be regarded as describing only the initial stage of the motion), unless Laws of Motion] SEC. 4.5.10 263 A t\ ? "o = ?UoP In the latter case, Eq. (M) becomes S = 0oe-p< which shows that R "drifts" toward the equilibrium position, taking infinitely long to reach it. (When 6 = 6 = 0, R is then said to be in unstable equilibrium, because any disturbance, how- ever small, causes a large departure of R from the equilibrium position.) For p2 = 0 (r = 6/2), 0 = 0O + fat (N) which again describes a motion during which 6 increases without limit, unless fa = 0, in which case Eq. (N) asserts that a state of rest in a position other than that corresponding to 6 = 0 is possible. However, as Eq. (N) furnishes only an approximate description of motions of 72, such states of rest do not, in fact, exist. 4.5.10 When the system of all gravitational and contact forces acting on the body R} of a set S of N rigid bodies Ri} j = 1, 2, . . . , N, is equivalent to n> forces R, i = 1, 2, . . . , n>, applied at points P{} i = 1, 2, . . . , n>, of these bodies, the kinetic energy (R'KS) of S in a Newtonian reference frame R' is related to these forces and to the velocities Rfypiy t = 1, 2, . . . , n;, as follows: R'AS (1) at where R'AS, called the activity in R' of the gravitational and contact forces acting on the bodies of S, is given by R'AS = V J RV* ? R (2) The contribution to ^'^l5 of all forces exerted by Rj, j = 1, . . . , N, on each other is equal to zero when all gravitational forces exerted by the bodies of S on each other can be neglected (see Vol. I, Sec. 4.4.3) and every contact between the bodies JR>, j = 1, 2, . . . , AT, either is a rolling contact (see 2.5.10) or takes place across a smooth surface (see Vol. I, Sec. 4.8.2). 264 SEC. 4.5.10 [Chapter 4 Proof (see Fig. 4.5.10a): FIG. 4.5.10a Fj From Sec. 4.5.8, dR'KRi R>AR ,A. = * AR* (A)dt where R\H ? R (B) Let j take on the values 1, 2, . . . , N in Eq. (A), and add the N equations thus obtained: y-i Now .if- R>K*> d*'KSfz[ dt dt fr{ (4.5.3) dt and, letting AT n, y-it=i it follows that Suppose that all gravitational forces exerted by J?/, j = 1, 2, . . . , AT, on each other can be neglected and that N = 2, that is, that S consists of only two bodies; further, that these are in contact at only one point, as shown in Fig. 4.5.10b, where Ci rep- resents the contact force exerted on i?i by R2, Ci's point of applica- tion being Ch and C2, applied at C2, represents the contact force Laws of Motion] SEC. 4.5.10 265 exerted on R2 by Ri. n is a unit vector perpendicular to the surface of Ri at &. The contribution of Ci and C2 to R'AS is given by i ?Cx + are in R'y* FIG. 4.5.10b R'yCi . { c2 fl'yCl . rolling R'yCl d + ?VC' ? = -Cx (4.3.5) C2 = (*'?* contact (2.5.10) + * vc? ? d d ? (C) *VCi) ? d 0 But Hence d + 2 ( ) (C) If /2i and Consequently If Ri and R2 are in contact across a smooth surface, d is parallel to n, and the right-hand member of Eq. (C) vanishes for one of two reasons: Either R'vCl ? R\Cr is perpendicular to n, and hence to d, or R'vCl ? R'vCt is not perpendicular to n, in which case Ri and R2 either penetrate each other, which is impossible in view of their rigidity, or they lose contact with each other, which leads to d = 0. Thus, when two bodies are in rolling contact at one point, or when they are in contact across a smooth surface at one point, the con- tact forces exerted on the bodies by each other contribute nothing to R'AS. The restriction to the case N = 2 and a single point of contact 266 SEC. 4.5.10 [Chapter 4 is removed by observing that all contact forces exerted on the bodies Rjt j = 1, 2, . . . , N, by each other can be grouped into pairs such as the pair Ch C2 considered above. Problem: In Fig. 4.5.10c, B represents a uniform rectangular plate of mass M, which is suspended by two thin rods A and A'f each of mass m. The rods are attached to B and to their support by means of smooth ball and socket connections, b FIG. 4.5.10C During one possible motion of this system the plate rotates about a vertical axis passing through the mass center B* of B, while B* moves upward and downward on this axis. Determine the angular velocity of B at the instant when h = 0 (see Fig. 4.5.10c), assuming that the plate is released from rest when h = a/6. Solution: In Fig. 4.5.lOd, which shows rod A at a typical instant during the motion, n0, ih, n2, n3 are unit vectors, n0 being vertical, ih parallel to A, n2 perpendicular to n0 and nh and n3 per- pendicular to ih and n2. Note that n0 = ?cos >}/ ni + sin ^ n3 (A) (2.2.4,2.2.7) = 4> cos (A) ? ^n2 ? 0 sin ^ n3 (B) Laws of Motion] 267 FIG. 4.5. lOd v" = wAX 5.,, - (2.5.9) V / (B) (C) ni, n2, n3 are principal directions of A for A* (see 3.5.9 and Appendix, Fig. 1), and 4A/A* = 0, The kinetic energy KA of A is given by KA = ma* (D) (4.5.5) (4.5.4) (4.5.5) f2 or Similarly From Problem 4.5.8, (E) (F) (G) 268 SEC. 4.5.10 [Chapter 4 Hence, letting S be the system of rigid bodies A, A', B, Ks = KA + KA' + KB (4.5.3) or Ks sin2 24 (H) (I)(E.F.G.H) O The activity (As) of all gravitational and contact forces acting on the bodies of S is found by considering the five forces shown in Fig. 4.5.lOe, where F and F' represent forces exerted on the rods FIG. 4.5.10e B B* Mg by the support (see Vol. I, Problem 4.8.2a). Note that the veloc- ities (vp) and (vp/) of the points P and P' are equal to zero, so that vp . F + v^ ? P = 0 (J) A5 is given by As = yP . F + v*" . F' + v^ ? (- (That is Consequently, d Pitta2, io , As = 0 ? mgayj/ sin ^ ? Mgh (J) (A,C) (P4.5.8) (K) sin2?+i 12A2) or 0 Laws of Motion] SEC. 4.5.10 269 from which it follows that + g(-ma cos ^ + Mh) = C (L) where C is a constant, whose value is found by noting that, when ft = a/6, tf = 0 = ^ = A = O (M) and (see Fig. 4.5. lOd and recall that A has a length "a") COS \f/\ a- ft 5? ?/& 6 so that and g I ? 7; ma - a i ? 1 6 6 / (L,M,N) sin* if) + ? (N) 24 From Fig. 4.5.lOd, . , b . 6sin ^ = - sin - a Z Differentiating, , . bd 6 /Ox f cos}// = ? cos r W/ Now, when ft = 0, ft attains its minimum value. Hence AU-o = o (R) and, as * = 9 = 0 (S) at this instant, it follows that /i b A\ m ~ (Q,S) ^a ? Thus when ft = 0, ma2 b2d2 , M ,n,n /ma SEC. 4.5.11 [Chapter 4 1 + (m/M) (2.2.4) 1 + (m/M) "|1/2 270 and Finally, Hence (When m/M approaches zero, the present result approaches that of Problem 4.5.8.) 4.5.11 When some (or all) of the rigid bodies of the set S con- sidered in Sec. 4.5.10 are connected to each other by light, helical springs (see Vol. I, Sec. 4.8.5), the forces exerted on the bodies by the springs contribute to R'AS. If x is the deformation of such a spring (x > 0 when the spring is stretched), and f(x) is the func- tion which defines the character of the spring (for example, for a linear spring of modulus ky f(x) = kx), the contribution to R'AS of the forces exerted on the bodies of S by this spring is ,, N dx Proof: Figure 4.5.11a shows two of the bodies of S and the FIG. 4.5.11a Laws of Motion] SEC. 4.5 11 271 forces FA and fB exerted on these bodies by a spring of natural length L. n is a unit vector parallel to the axis of the spring. FA and FB are given by F^ = /(x)n, FB = -/(*)n (A) and if R'vA and R'vB are the velocities of A and B, the contribution of F^ and FB to R'AS is (B) Now (A) R>yA __ R>VB = (2.5.15) (2.4.1) at dx so that But Hence dx /T i \ ^n= -37 n - (L + x) -7T d.5.1) at at _ _ (C) (D) A + B /() ^ (B,C,D) ?t Problem: Referring to Problem 4.5.10, determine the circular b 11 ll p J A" B E ^J P1 A1* FIG. 4.5.11b 272 SEC. 4.5.11 [Chapter 4 frequency of small oscillations of the system about the vertical line passing through B*, assuming that the midpoints A* and A'* of the rods A and A' are connected by a light, helical spring of natural length L, where L = 0.9 b (see Fig. 4.5.11b). Assume that for small deformations the spring is linear and has a spring constant k. Solution: Let S be the set of rigid bodies A, A', B. Then the kinetic energy Ks of S is given by (see Eq. (I), Problem 4.5.10) K* = 2-p sin2 ~ 12A2) (A) At a typical instant during the motion, the distance d between A* and A'* is (see Fig. 4.5.11c) b/2 FIG. 4.5.11C The deformation x at this instant is thus given by x = d-L = 6cos|- 0.96 = b (cos | - 0.9^ (B) and, letting f(x) be the function defining the character of the spring, the activity A5 of the gravitational and contact forces acting on the bodies of S is (see Eq. (K), Problem 4.5.10) Laws of Motion] SEC. 4.5.11 273 dx di dxA s = ?mgayj/ sin \f/ ? Mgh ? f(x) -zr (C) That is 7 a As = -mgaxj/ sin f - Mgh + /(x) - tf sin - (D) (C) Next (see Fig. 4.5.lOd), sin y// = - sin - and 4s are thus given, approximately, by ^ 0(A) 12 (F.E.H) 24 (I) (H) or 4a 4a 4 (H,E) (I) (K) a ? h = a cos yfr (G) Hence for motions during which 6 remains small, (E) 2a and (G) so that and (B) while /(x) ttkx = kb (^0.1 - f) (K) (J) Ks ? H^ (M + 2m) (L) and 274 SEC. 4.5.12 [Chapter 4 or A^ /^/ ?I T? (M ~\~ fit) ? ?"j? I 66 (M-) I 4d 4 _j Now f S - A' (N) at (4.5.10,4.5.11)Hence 24^ ?7 (M + 2m) = - *- (M + m) - ?r?at L 24 J (L,M.N) L4a 4 J or where 2 _ (g/o)(M + m) - O.lfc V " Equation (0) is identical with Eq. (I) of Problem 4.5.9. In accordance with the discussion given there, harmonic oscillations with circular frequency p can occur if and only if g(M + m) - 0.1k > 0 (Q) a and the system is in unstable equilibrium, that is, it tends to "buckle," when 6 = 0 and Eq. (Q) is not satisfied. Equation (Q) may thus be regarded as furnishing a stability criterion. Note that this equation is independent of the kinetic energy Ks] that is, it depends solely on the activity As, given in Eq. (M). 4.5.12 The law of motion stated in Sec. 4.5.10 is particularly useful when the number N of rigid bodies comprising S is large and a single scalar function of time describes the configuration of S at any instant. Problem: Figure 4.5.12a represents a train S of N gears J?>, j = 1, 2, . . . , 2V, each gear having a radius one-half as large as that of the one on its left, n is a unit vector parallel to the axes of the gears, and T is the torque of a couple applied to R\9 T being given by T = Tn where T is an unspecified function of time L Laws of Motion] SEC. 4.5.12 275 FIG. 4.5.12a Letting Ix be the moment of inertia of R\ about the axis of rotation of Rh determine the scalar angular acceleration a\ of R\ for the n direction (see 2.3.7), assuming that all gears are made of the same material and have the same thickness. Solution: The radius r> of gear Rj is given by Let o)j be the angular speed of Rj for the n direction. Then the requirement that the points of contact Pn and P21 of Ri and R2 (see Fig. 4.5.12a) have the same velocity takes the form = ? r2a>2 so that Similarly and, in general, or (B) (B) 7*3 r3 (A) o>y = (-2)*-1?i (C) The moment of inertia /> of Rj about the axis of rotation of Rj is proportional to r/. Hence, 276 SEC. 4.5.12 and the kinetic energy KRi of Rj is given by [Chapter 4 (4.5.4,4.5.5) * (C,D) or (E) and the kinetic energy Ks of the gear train by N N K* = > KKi = iicoi (4.5.3) y^f (E) But N V*2(1-2''> = 1(1 ? 2~2N) -2;) Hence (1 - 2"^) (F) Of the gravitational and contact forces acting on the bodies of S, the only ones which contribute to ihe activity As (see 4.5.10) are the contact forces exerted on Rt by the couple of torque T. As this couple is equivalent to the simple couple (see Vol. I, Sees. 3.4.2 and 3.5.7) shown in Fig. 4.5.12b, and (G) (2.5.2) (2.5.6) where n' is a unit vector, As is given by FIG. 4.5.12b As (4.5.10) vP" ? (v Laws of Motion] SEC. 4.5.12 277 Next ^- = As dt (4.5.10) Hence (H)(F) or o dt But 77 ?^ (2.3.7) Consequently 3r ai 4(1 - 2 PROBLEM SETS PROBLEM SET 7 (See Sections 1.1.1-1.7.3) (a)* Four rectangular parallelepipeds, Riy i = 1, 2, 3, 4, are arranged as shown in Fig. la. The unit vectors n,y, j = 1, 2, 3, are respectively parallel to the edges of Ri. FIG. la The angles = -2r(s + 2), 6 = 2TZ\ * = 2T(1 - 2z3) For u = 3nn ? 2zni2 + 4zn13, v = 4z2ni3 #1 ft4 Rx & evaluate o|,?j ? v|/=i, u|,= 4 ? v|,-j, u|?.o ? v|.-j, u|,-o ? v|,-*. Results: 2, 2, 0, -3. * Problems containing information relevant to the solution of problems appearing later are marked with an asterisk. 281 282 PROBLEM SET 1 (b) Referring to Problem l(a), determine the measure numbers of the nn components of Ri Rx u|,?o + v|,-j and u|2=0 + v|2=i Results: 2(cos \p cos ? sin \f/ cos 6 sin = 0, ^ = -2radsec"1 For this instant, determine the rin, ni2, and ni3 measure numbers of R'dnl2/dt, first using the definition given in Sec. 1.2.1, then the theorem of Sec. 1.7.1. (i) Referring to Problem l(a), suppose that at a certain in- stant 6 = 0 and R2 is rotating at a rate of 6 revolutions per minute relative to #3, clockwise as seen by an observer to whom n32 appears to point to the right; further, suppose that the rate at which R2 is rotating clockwise relative to 2?3 is decreasing at a rate of 96x rpm/min. Determine the n3i, n32, n33 measure numbers of Rtd2nz3/dt2 and Rxd?n2Z/dt2 for this instant. Results: 0, 192a-2, -144TT2;0, -192TT2, - 144TT2 min"2. PROBLEM SET 2 (See Sections 1.8.1-1.13.2) (a)* A circle of radius r is drawn on a sheet of paper which is then folded to form a cylinder of radius R, a space curve C being thus produced. Letting 7\ and T2 be two tangents to C, each making an angle of sixty degrees with the axis of the cylinder, show that when the angle between T1 and T2 is independent of r/R it is equal to sixty degrees. (b) The tangent T at a point P of a plane curve C makes an 284 PROBLEM SET 2 angle 4> with the tangent To at a point Po of C. Letting s be the arc-length displacement of P relative to Po, show that #= 1 ds p where p is the radius of curvature of C at P. Also, give an example which shows that this relationship is not necessarily valid if C is a space curve. (c)* Referring to Problem 2 (a), and letting r = 3 in. and/2 = 4 in., determine the minimum radius of curvature of C. Result: 2.4 in. (d) Letting r = 3 in. and R = 4 in. in Problem 2(a), find the cosine of the angle between the axis of the cylinder and the bi- normal of C at the points where the radius of curvature of C has a minimum value. Result: dbf (e) Referring to Problem 2(a), and letting r = 3 in. and R = 4 in., determine the torsion of C at the points where the tangent makes an angle of 45 deg with the axis of the cylinder. Result: ?201/584 rad in.-1. (f) At a certain point P of a curve C drawn on the surface of the earth (regarded as a sphere of radius 3960 miles) the principal normal of C makes an angle of sixty degrees with the line joining P to the earth's center. Determine the radius of curvature of C at P. Result: 1980 miles. (g) The radius of curvature at a point P of a curve C is equal to 12 ft. C" is the orthogonal projection of C on a plane which is inclined at sixty degrees to the osculating plane and is normal to the rectifying plane of C at P. If P'is the projection of P (P' is a point of C"), what is the radius of curvature of C" at P'? Answer: 3 ft. PROBLEM SET 3 285 PROBLEM SET 3 (See Sections 2.1.1-2.3.11) (a)* Figure 3a shows a rigid body R which moves in a reference FIG. 3a frame R' in such a way that at a certain time tr the derivatives in R' of the position vectors a, b, c of the points A, B, C relative to a point 0 fixed in R' have the values -12n3, 0, -9ih - 12n2 - 12n3 ft sec-1, ni, n2, n3 are mutually perpendicular unit vectors, and A, B, C are fixed on R. Determine the ni, n2, n3 measure numbers of the angular veloc- ity of R in R' at time t', and explain why it is not necessary to know whether or not ih, n2, n3 are fixed in R, R', or any other refer- ence frame. Result: 4, -3, 0 rad sec"1. (b)* For the purpose of analyzing the motion of an airplane it is sometimes convenient to resolve all vector quantities (for ex- ample, forces) into components parallel to the vectors r, 0, v (see Parts 1.9, 1.10, 1.11 of the text) associated with the curve C on which some point P of the airplane moves. It then becomes neces- sary to know the angular velocity of a reference frame R in which these unit vectors are fixed, in the reference frame in which C is 286 PROBLEM SET 3 fixed. Letting s be the arc-length displacement of P relative to a point Po fixed on C, and A. and p the torsion and radius of curvature to C at P, determine this angular velocity. Result: (XT + p/p)ds/dt. (c) A point P oscillates on line AB (extended) of the body R of Problem 3(a), in such a way that the displacement x of P rela- tive to A is given by x = 4 + sin b(t - I') ft Determine the magnitude of the first time-derivative in R' of the position vector of P relative to A, for time t1. Result: 13 ft sec"1. (d)* The crank AB of the mechanism shown in Fig. 3d rotates clockwise, performing 15 revolutions per second, thereby causing FIG. 3d the cylinder to oscillate. Determine the absolute value of the angular speed of the piston P for the instants when B is in its leftmost, rightmost, and upmost positions, and state in each case whether the cylinder is rotating clockwise or counterclockwise. Also, find the smallest angle be- tween AB and AD for which the angular velocity of the cylinder is equal to zero. Results: IOTT rad sec""1, c; 30* rad sec"1,cc; 6* rad sec"1, c; 60 deg. PROBLEM SET 3 287 (e) One method for studying in the laboratory certain forces acting on a body B attached to a space vehicle is to simulate a part of B's motion by mounting B in a manner schematically rep- resented by Fig. la, B playing the part of Ri while the earth (that is, the laboratory) is represented by R\. By appropriate choice of the functions ${(), (t), ^(t), any desired angular velocity of B can then be obtained. Letting Qt(0, % = 1, 2, 3, be the n4t measure number of the angular velocity of B in the laboratory, show that the differential equations governing 0, -jr = (Qi sin \p ? Q2 cos \p) cosec 0 d\p -77 = ( ?12i sin }// + fi2 cos \f/) cotan 6 + fi3 (f) At time t' the second time-derivatives in R' of the vectors a, b, c of Problem 3(a) have the values 0, -36n! -48n2, -6n2 ? 75n3 ft sec~2. Determine the magnitude of the angular acceleration of R in R' at time V. Result: 2 rad sec~2. (g) Regarding the earth as a sphere whose angular velocity in a reference frame R is equal to (?r/12)l< rad hr"1, where V is a unit vector parallel to the earth's north-south axis and is fixed in R, determine the magnitude of the angular acceleration in R of a turbine disc which rotates about its own axis at 10,000 revolutions per minute, this axis being normal to the earth's surface at the equator. Result: 0.076 rad sec~2. (h)* Losing altitude at a uniform rate of 300 ft sec"1, the mass center of the propeller of an airplane follows the helical path H shown in Fig. 3h. During this motion the propeller's shaft axis remains tangent to H; the inclination of the wings to the principal 288 PROBLEM SET 3 normal of H is constant; and the propeller rotates clockwise as seen by the pilot, at 1200 revolutions per minute. Vertical - FIG. 3h H ?-25ft Determine the magnitude of the angular acceleration of the propeller. Result: 337 rad sec~2. (i) Referring to Problem 3(d), determine the absolute value of the scalar angular acceleration of the piston for the instant when B is vertically above A, and state whether the rate at which the cylinder is rotating counterclockwise is increasing or decreasing at this instant. Result: 216TT2 rad sec~2, increasing. (j)* Figure 3j represents Peaucellier's straight line mechanism, so called because point R moves on a straight line perpendicular to AB when the driving arm rotates. Use a partly graphical method to determine the angular speeds and scalar angular accelerations (for the k direction) of bars B2) Bzj and B4, for an instant at which the configuration of the mech- anism is that shown in Fig. 3j and Bi has the angular velocity and angular acceleration there indicated. PROBLEM SETS 3 and 4 289 G FIG. 3j Results: -0.075, 0.054, 0.275 rad sec"1; -0.064, -0.030,0.051 rad sec"2. PROBLEM SET 4 (See Sections 2.4.1-2.5.8) (a) A point P oscillates on one of the edges of a rectangular parallelepiped Rf a point P' on the diagonal of one of the faces (see Fig. 4a). The displacement s of P relative to the midpoint 0 of B \ FIG. 4a 290 PROBLEM SET 4 AB, regarded as positive when P lies between 0 and B, is given by 12 s = ? sin wt ft where t is the time in seconds, s', the displacement of P' relative to the midpoint 0' of A 'B', is given by s' = ?; cos 27r? ft 4TT2 and is regarded as positive when P' lies between 0' and #'. Determine the largest value of the magnitude of the accelera- tion of P' relative to P in R during this motion, and find the cosine of the angle between the acceleration of P relative to P' at t = f sec and a unit vector in the direction A'B'. Results: 13 ft see"2, ? ^. (b)* At a certain instant t' the angular velocity (Ru>A) and angular acceleration (RctA) of an aircraft A in a reference frame R (see Fig. 4b) are given by Ro)A = 2ni ? n2 + n3 rad sec~l RCLA = ri! + n2 ? 2n3 rad sec"2 where ni, n2, n3 are mutually perpendicular unit vectors fixed in A. The distance between two points B and C, which are fixed on A, is 50 ft. FIG. 4b Determine the acceleration of B relative to C in R at time t'. Result: -250n2 ft sec~2. PROBLEM SET 4 291 (c) In Fig. 4c, r and 6 are polar coordinates of a point P which moves in a plane, nx and n2 are unit vectors, parallel and perpen- dicular to line OP. FIG. 4C Determine the ih and n2 measure numbers of the acceleration of P, assuming that P moves in such a way that (1)0 remains con- stant; (2) r remains constant; (3) neither 6 nor r remains constant. Explain why the results of (3) are not simply the sums of those of (1) and (2). (d) In Fig. 4d, Q is the point of contact between a sphere S FIG. 4d and a conical shell C in which S moves in such a way that the distance x decreases uniformly at a rate of 4 in. sec"1 and the angle 6 increases uniformly at a rate of 5 rad sec"1. (Point A is fixed on C.) Determine the magnitude of the acceleration of the center of S for an instant at which this point passes through the axis of the cone C. Result: 20 in. sec~2. 292 PROBLEM SET 4 (e) A point P moves with a constant speed of 12 in. sec"1 on the curve C described in Problem 2(a). If r = 3 in. and R = 4 in., what is the maximum magnitude of the acceleration of P during this motion? Answer: 60 in. sec~2. (f)* In Fig. 4f, A, B, and C represent points fixed on a ship, and ni, n2, n3 are mutually perpendicular unit vectors. At a certain instant the velocities of A, B, C are vA = 4n2 ? 6n3 ft min"1, B 120' D FIG. I201 4f j C vB = 3ni - 6n2 ft min"1, vc = -4n2 + 6n3 ft min""1. Determine the magnitude of the angular velocity of the ship at this instant. Result: 7 rad tor1. (g) Referring to Problem 3(d), let Q be a point on the piston P. For an instant at which B is vertically above A, determine the speed of Q in the cylinder for the direction DB. Result: -12(5)1/2TT R units of length per second. (h) Referring to Problem 3(j), determine the smallest angle between line AB and the acceleration of point 0. Result: 74 deg. (i) A point P moves on a vertical straight line in such a way that its speed v is given as a function of time t (measured in sec- onds) by v = 3*2 _ g ft sec-i If P is moving upward at t = 1 sec, is v the speed of P for the upward or downward direction? PROBLEM SETS 4 and 5 293 At t = 0, P is 2 ft above a certain point Q fixed on the line on which P is moving. Determine the position of P at time t = 3 sec, and state whether P's acceleration is directed upward or down- ward at t = 4 sec. Results: downward, 1 ft below Q, downward. (j) A particle, projected to the left on a horizontal straight line, moves with a time-independent acceleration directed to the right. If it twice passes a point s feet to the left of the point of projection, first at time th next at time t2, what is the magnitude of the acceleration? Answer: 2s/tit2 ft sec~2. (k) Starting from rest, a point moves with constant accelera- tion on a straight line. During the first ten minutes of the motion the distance-average speed of the point is equal to 8 ft sec"1. What is the time-average speed of the point during this time interval? Note: f, the z-average value of a function f(x) in the interval Xi ^ x ^ x2, is defined as Answer: 6 ft sec"1. PROBLEM SET 5 (See Sections 2.5.9-2.5.15) (a) Referring to Problem 4(f), show that the velocity of point D is equal to zero at the instant under consideration. (b) In Problem 3(h) let A be a point of the propeller, one foot from the propeller's shaft axis, as shown in Fig. 5b. FIG. 5b 294 PROBLEM SET 5 Determine the magnitude of the acceleration of A at an instant at which the propeller is horizontal. Result: 16,200 or 18,000 ft see"2. (c)* A circular hoop H of radius R and mass m rolls on a hori- zontal plane P, the normal to H making a constant angle 0 with the normal to P. The center C of H moves with constant speed v on a circle of radius R. FIG. 5d PROBLEM SET 5 295 Determine the magnitude of the acceleration of that point of H which is in contact with P. Result: (v*/R)(l ? cos0)(l + sin2!?)1'2. (d)* A shaft, terminating in a truncated cone C of semi vertical angle 0 (see Fig. 5d), is supported by a thrust bearing consisting of a fixed race R and four identical spheres S of radius r. When the shaft rotates about its axis, S rolls on R at both of its points of contact with R, and C rolls on S. Determine the dimension b such that the contact between C and S is one of pure rolling. Result: r(l ? sin 0)/(cos 0 ? sin 0). (e)* The axes of two shafts S and S' intersect at a point A. Figure 5e illustrates one method of transmitting rotatory motion from S to S' (and vice-versa): Bevel gears B and B', which are, essentially, frusta of cones, are keyed to the shafts, coming into contact with each other along line CD, the angle being chosen such that B and B' roll on each other at all points of contact. FIG. 5e Show that points A and C must coincide, and find 4> as a func- tion of 6 and the angular speed ratio w/?'. Result: = arc cotan[cotan 6 + (w/a/) cosec 0]. 296 PROBLEM SET 5 (f) Figure 5f shows schematically how the driving shaft D of an automobile may be connected to the two halves, Ai and A2, of the rear axle in such a way as to permit the rear wheels to rotate at different angular speeds in the frame F. This is accomplished as follows: Bevel gears (see Problem 5(e)) Bx and B2 are keyed to FIG. 5f Ai and A2, and these engage bevel gears bi and 62, the latter being free to rotate on pins fixed in a casing C. C can revolve about the common axis of Ax and A2, and a bevel gear E, rigidly attached to C, is driven by bevel gear G, which is keyed to D. Letting o> be the angular speed of D in F for the direction n, and <*>i, o>2 the angular speeds of Ax and A2 in F for the direction n', express a> in terms of ?i, a>2, and 0. Result: + w2) cotan 6. (g)* A trailer consists of a rigid frame AB ? CD, to which identical wheels are attached at C and D (see Fig. 5g), the wheels being free to revolve independently about their common axis. Consider a motion during which the wheels roll on a horizontal plane; the frame remains parallel to this plane; and point A moves 297 FIG. 5g with constant speed vona circle, center at 0. Verify that the system approaches a "steady state/' that is, that 6 approaches a limiting value as time t approaches infinity. Determine the limit- ing values of the accelerations (a*, ac, aD) of points B, C, D and of the angular accelerations (ac, ciD) of the wheels at C and D, ex- pressing all results in terms of the mutually perpendicular unit vectors nh n2, n3 shown in Fig. 5g. Results: QB = -(4t^/25d)n2, Qc = -(3v2/25d)n2, QD = -(5i;725d)n2, a? CLD = (h) Referring to Problem 3(b), and considering a motion dur- ing which the orientation of the airplane in reference frame R remains fixed, determine the distance from P to the airplane's instantaneous axis, and find the magnitude of the minimum veloc- ity of any point of the airplane (or airplane extended). Result: p(l + XV)"1, p|Xs|(l + XV)"1/2. (i) Locate an instantaneous center of bar B* of the mechanism described in Problem 3(j), and use it to find the velocity of the midpoint of this bar. Check the result by reference to Sec. 2.5.9. (j)* A circular tube T revolves at a uniform rate of 4 rad sec"1 about an axis A-A which is fixed in a laboratory L and passes through a point JB of T. (B is fixed on A-A.) The normal to the 298 PROBLEM SET 5 plane of T makes an angle of 45 deg with A-A. A particle P, mov- ing with constant speed in the tube, reaches at time t* the point P* shown (in two views) in Fig. 5j, and the distance between B and P is increasing at this instant. FIG. 5j If the acceleration of P in L at time t* is parallel to the plane of T, what is the magnitude of this acceleration? Answer: 2(127)1/2 ft sec"2. (k)* Referring to Problem 4(b), let P and Q be particles mov- ing on line BC with constant speeds (in A) of 40 ft sec"1 and 60 ft sec"1, P proceeding from B toward C, Q from C toward B, PROBLEM SETS 5 and 6 299 If P arrives at C, and Q at B, at time t', what is the magnitude of the acceleration in R of P relative to Q at time t'1 Answer: 50(105)V2 ft sec~2. PROBLEM SET 6 (See Sections 3.1.1-3.2.9) (a)* In Fig. 6a, nh n2, n3 are mutually perpendicular unit vec- tors; point Q lies in the XOY plane; and point P has a strength of 6 ft3. FIG. 6a Find the second moment of P with respect to line OX by each of the three methods discussed in Sec. 3.1.7, and state which of the three is the most convenient. (b) Referring to Problem (6a), find the second moment of P with respect to 0 for the pair of directions ih, n2. (Check by using two methods.) (c) The second moments of a point P with respect to a point Q for three mutually perpendicular directions nt, i == 1, 2, 3, are Vij ,U = 1, 2, 3. Letting <*>?/Q, *?/?, %Q be the second moments of P with re- spect to any three mutually perpendicular lines passing through Q, express 4>%Q + ?b/Q + tf/Q in terms of f/Q, i, j = 1, 2, 3. Result: \{Q + 4>2^Q + $3*/?- 300 PROBLEM SET 6 (d)* The points Pi, P2, Pz shown in Fig. 6d have strengths of 10, 20, 30 in2. 4' FIG. 6d Determine the radius of gyration of this set of points with re- spect to line AB; also, the radius of gyration with respect to the same line of a point which has a strength of 10 + 20 + 30 = 60 in.2 and is situated at the centroid of the set of points Pi, P2, P3. Account for the fact that these two radii of gyration are not equal to each other. Results: 2.21 ft, 1.078 ft. (e)* The points Pi, P2, . . . , P8 shown in Fig. 6e have strengths of 1, 2, 3, 4, 1, 2, 3, 4 tons. Use the most convenient method avail- PROBLEM SET 6 301 able to find the second moment of this set of points with respect to point P2 for the pair of directions P*Pi, P7P3. Result: -4(3)1/2 ton ft2. (f) The points Pit i = 1, 2, . . . , n, of a set S of n points have strengths N{. nh n2, n3 are unit vectors, each being parallel to one of the axes OX, OY, OZ of a rectangular cartesian coordinate sys- tem and pointing in the positive direction of the axis to which it is parallel. The coordinates of P? are a\, y*, Z{. Letting f/? be the second moments of S with respect to 0, express %? and <$>li? in terms of Ni} x?, yif Zi, and n. Results: 1 = 1 (g) Referring to Fig. 6a, let P be centroid of a set S of points whose second moments *g/p, i,.;' = 1, 2, 3, in units of ft3, are given in Table 6. TABLE 1 1 2 3 1 2 V3 3 6 j 2 V3 4 5 3 3 5 6 FIG. 6g Assuming that S has a strength of 2 ft, determine the second moment of S with respect to line OQ. Result: 30 ft8. (h) The second moment of a set S of points with respect to a point 0 for three mutually perpendicular directions n?, i = 1, 2, 3, are ??>, i, j = 1, 2, 3. A line La passes through 0 and is perpen- dicular to n3. Under these circumstances *??, the second moment 302 PROBLEM SET 6 of S with respect to Lo, depends on the orientation of La in the plane which passes through 0 and is perpendicular to n3. Determine the maximum value of oa. Result: r + /2 (i) Determine whether or not the following statement is true: Given any set S of points, a point 0, and a plane passing through 0, it is always possible to find two directions parallel to this plane and at right angles to each other such that the second moment of S with respect to 0 for this pair of directions is equal to zero. (j)* Given a set S of points, a point 0, and two lines Lx and L2 passing through 0 and forming a right angle (see Fig. 6j), let ni and 112 be unit vectors respectively parallel to the two lines. Draw a line PQ parallel to nh and construct a circle passing through two points A and B and having its center C at the mid- FIG. 6J point of line AB, A and B being the points whose position vectors, a and b, relative to P are given by PROBLEM SETS 6 and 7 303 b = k(44ioni ~ 4>U?n2) where A* is a constant of proportionality. (Note that when U? 7* 0 point C is the intersection of lines PQ and AB.) Verify that the following graphical method can now be used to find the second moment (ia?) of S with respect to any line La passing through 0 and coplanar with Lx and L2: A line parallel to La is drawn through B, intersecting the circle at point D. Then kSaL6 = PE where E is the foot of the perpendicular dropped from D on line PQ. Let S be the set of four points A, B} Ph P2 shown in Fig. 6d, these points having strengths of 1, 2, 3, 4 slug, respectively. Deter- mine by graphical methods (1) the second moment of S with re- spect to line AB, (2) the smallest second moment of S with respect to any line passing through point A and lying in the plane of Sf and (3) the angle between line A Pi and the line passing through A and lying in the plane of S with respect to which S has the largest second moment. Results: 40.3 slug ft2, 35.5 slug ft2, 21.6 deg. PROBLEM SET 7 (See Sections 3.3.1-3.5.9) (a) Three points of equal strength iV are situated at the corners of an equilateral triangle whose sides have lengths L. Determine the smallest angle between any side of the triangle and any prin- cipal axis of the set of points for one vertex of the triangle, and find the principal second moments for one vertex. Also, locate all principal axes for the centroid of the set. Results: 0 deg; ATL2/2, 3NL2/2, 2WL2. (b) Four points of equal strength are placed at the points B, Pi, P2, Pz of Fig. 6d. Determine the minimum radius of gyration of this set of points. Result: 1.4. ft. 304 PROBLEM SET 7 (e) Referring to Fig. 6a, suppose that P is the centroid of a set S of points, that ? has a strength of 2 slug, that nh n2, n3 are centroidal principal directions of S, and that the corresponding centroidal principal second moments of S have the values 40, 20, 30 slug ft2. Determine the second moment of S with respect to line OQ. Result: 50 slug ft2. (d)* Verify each of the following statements: A centroidal principal axis is a principal axis for each of its points. If a principal axis for a point other than the centroid passes through the centroid, it is a centroidal principal axis. A line which is a principal axis for two of its points is a cen- troidal principal axis. The three principal axes for any point on a centroidal principal axis are parallel to centroidal principal axes. If two principal second moments for a given point are equal to each other, the second moments with respect to all lines passing through this point and lying in the plane determined by the corre- sponding principal axes are equal to each other. FIG. 7e 2" / YT 4 0 PROBLEM SET 7 305 (c)* Determine the second moment of the figure shown in Fig. 7e with respect to 0 for the pair of directions OX, OF, regarding the figure as (1) a surface possessing no plane portions and (2) a solid. Results: 2TT in.4, TT/2 in.6. (f) Locate one principal axis of the figure shown in Fig. 7e for point 0, and show that the X-axis is not a principal axis of the figure for point 0. (g) Determine the radius of gyration of one face of a 1 in. X 1 in. X 1 in. cube with respect to a space diagonal of the cube. Result: (5/18)1'2 in. (h)* Determine the smallest angle between line OA and any- principal axis for point 0 of the shaded plane surface shown in Fig. 7h. c c 3 } 3" B 2' 21 A FIG. 7h FIG. 7j Result: 30 deg. (i) Determine the polar second moment of the shaded plane surface shown in Fig. 7h, with respect to point 0. Result: 373.4 in.4. (j) Particles of equal mass are placed at the points A, B, C shown in Fig. 7j. Find the minimum radius of gyration of this set of particles. 306 PROBLEM SET 7 Result: (5 - V13)1'2 ft. (k) The assembly shown in Fig. 7k consists of two brass discs (527 lb ft"3) attached to a steel shaft (489 lb ft"3). iw L M/8" / - A ~2 3" 3" 6" vi c" FIG. 7k PROBLEM SETS 7 AND 8 307 Determine the value of the angle 5 for which the shaft axis is a principal axis of this assembly. Result: 15.1 deg. PROBLEM SET 8 (See Sections 4.1.1-4.1.7) (a)* A particle P of mass m is free to move between two paral- lel plates. The plates are attached to each other and can revolve about an axis Y which is fixed in the plates and in a reference frame R, as shown in Fig. 8a. z^ ^R "~^x FIG. 8a Letting X, Y, Z be axes of a rectangular cartesian coordinate system fixed in the plates, and nh n2, n3 unit vectors respectively parallel to X, F, Z, express the inertia force acting on P in R in terms of the angular speed w of the plates in R for the n2 direction and the coordinates x and y of P. Result: m[(xa2 - x)nx - yn2 + (xa + 2c*r)n3]. (b)* In Fig. 8b, ABC represents a uniform right-triangular plate of mass m. The plate is supported as follows: Vertex A is fixed and vertex B is attached to a string which is fastened at Z>, 308 PROBLEM SET 8 the length of the string being such that line AB is horizontal. (Line AD is vertical.) dD FIG. 8b Letting AE be a fixed horizontal line, and BF a line parallel to AD, consider motions during which 0, remain "small." Determine the moment of the system of inertia forces acting on the plate (1) about line AB and (2) about line AD. Results: mb (a .. , b-?), sense AB ? ( a + - 6J, sense AD (c)* The apex A of a solid right-circular cone of mass m is fixed. The cone moves as follows: The angle 6 between the axis of the cone and the vertical remains constant; the plane P deter- mined by the axis of the cone and the vertical passing through A revolves about this vertical at a constant rate of 12 radians per unit of time; and the cone revolves about its axis at a constant rate of w radians per unit of time in a reference frame in which plane P is fixed. PROBLEM SET 8 309 FIG. 8C Letting n be a unit vector perpendicular to plane P, as shown in Fig. 8c, determine the moment of the system of inertia forces acting on the cone, about a line passing through.A and parallel to n. Result: (3/20)mQ[(4A2 - r2)Q cos 6 - 2r2o>] sin $ n. (d) Referring to Problem 3(h), let r, 0, v be a vector tangent, binormal, and principal normal to H at the mass center P* of the propeller P, and let ni, n2, and n3 be mutually perpendicular unit vectors fixed in the propeller, as shown in Fig. 8d. FIG. 8d 310 PROBLEM SET 8 Assuming that nh n2, n3 are principal directions of P for P*, and that 011 tt 0, 022 ^ 033 = ^ Slug ft2 determine the r, jff, y measure numbers of the inertia torque acting on P. Repeat after replacing P with a propeller P' consisting of three equally spaced blades, taking n/P* = 22/P* = i slug ft2, 03?3/jP* ? 1 slug ft2 Results: -1.8 sin 2(9, 172 sin 20, -344 cos2 0 ft lb; 0, 0, -344 ftlb. (e)* Figure 8e represents a reciprocating engine mechanism with a counterweighted crank. P and P' are the mass centers of the connecting rod and the crank, respectively. These parts have masses m and m1', and the radius of gyration of the connecting rod with respect to a line passing through P and parallel to the unit vector n3 is equal to k. The piston has a mass M, and its mass center lies on the axis of the cylinder. FIG. 8e Considering motions of the mechanism during which the crank revolves with uniform angular speed, and letting S be the system of all inertia forces acting on the crank, connecting rod, and piston, determine the values of s' for which the resultant of S is perpen- dicular to n2, and find the value of k for which the n3 resolute of the moment of S about point 0 is equal to zero. PROBLEM SET 8 311 Suggestion: Make use of the fact that R sin 0' = L sin 0 from which it follows by repeated differentiation (and keeping in mind that dO'/dt is constant) that \dt) " \dt) sin0 dt2 Results: s' = (m/m')(R/L)(L -?),* = [s(L - s)]1/2. (f)* A uniform circular disc of radius R and mass m is rigidly attached to a shaft whose axis passes through the center of the disc. The shaft revolves about its axis, which is fixed, with a uni- form angular speed o>. Due to misalignment, the shaft axis makes an angle 0 with the normal to the plane of the disc. Determine the magnitude of the moment of the system of in- ertia forces acting on the disc, about any point. Result: (raflV/8) sin 20. (g) A rigid body B consists of three identical uniform square plates, each of mass m, attached to each other as shown in Fig. 8g. FIG. 8g B revolves uniformly about an axis which passes through the mid- point 0 of one of the plates and is perpendicular to this plate. Letting S be a set of three particles, each having a mass m and 312 PROBLEM SETS 8 and 9 placed at the mass center of one of the plates, determine the values of 0 for which the system of inertia forces acting on B is equivalent to the system of inertia forces acting on S. Result: 0, 90 deg. PROBLEM SET 9 (See Sections 4.2.1-4.2.8) (a) Referring to Problem 8(a), let R be fixed on the earth, with Y horizontal, and suppose that the plates are made to revolve in such a way that a> is a positive constant. Considering the plates to be smooth, obtain differential equa- tions governing x and y, and solve these, letting t = 0 when n3 points vertically upward and assuming that at this instant P is at rest relative to the plates, at x = ?g/2ot2, y = yo. Result: x = ? (g/2o)2)(e-ut + sin at), y = y0. (b) Referring to Problem 8(a), let R be fixed on the earth, with Y vertical, and suppose that the plates are made to revolve in such a way that a? is a positive constant. Considering the plates to be smooth, determine the reaction of the plates on P, assuming that x = xOy dx/dt = 0 at t = 0. Result: The force mco2x0(e~wt ? eut) n3, line of action through P. (c) Figure 9c illustrates schematically a mass-spring system which may be used as a component of an inertial navigation device: X r^ r>* V J \ / v J 1 T T FIG. 9C PROBLEM SET 9 313 A particle P of mass ra, attached to a spring S of modulus k} is free to slide in a horizontal tube T, and the tube is rigidly attached to a vehicle V which can move on a straight track parallel to the axis of the tube. When the displacement x of P from P's equilibrium position (x is regarded as positive when S is stretched) is known as a function of time for some time interval, the displacement of V during this time interval can be found. Determine the displacement of V during the time interval 2T for which the displacement-time curve of P is shown in Fig. 9c, assuming that V is at rest at the beginning of this time interval. Result: (7/6) (kx*/m) T\ to the left. (d) A gun at north latitude 45 deg is fired at a target 50 miles to the east of the gun. If the muzzle velocity has a magnitude of 4400 ft sec"1, and the gun is aimed on the basis of computations leaving the earth's rotation out of account, where does the pro- jectile land in relation to the target? Answer: 859 ft south, 3590 ft east of the target. (e)* Determine the minimum magnitude of the velocity (in the reference frame A described in Sec. 4.2.7) with which a particle must be projected from the north pole in order to strike the equator. Result: 4.47 mile sec"1. (f) A particle is projected from the north pole with the velocity found in Problem 9(e). Find the time of flight and the maximum distance from the center of the earth to the particle. Results: 32.3 min, 4780 miles. (g) A particle is projected from the north pole with a velocity having the same direction as that found in Problem 9(e), but hav- ing a magnitude (2)1/4 times as great. Determine the angle between a line joining the earth's center to the north pole and a line passing through the earth's center and the point at which the particle strikes the earth. Result: 180 deg. (h) An earth satellite whose perigee is 540 miles above the earth's surface traverses its orbit once per day. 314 PROBLEM SETS 9 and 10 Determine the apogee distance and the perigee and apogee speeds of the satellite. Results: 48,000 miles, 22,400 mile hi-1, 2100 mile hi-1. (i) Show that a particle projected from the earth's surface in such a way that it moves on a parabolic trajectory does not come into contact with the earth subsequent to the instant of projection, and determine the magnitude of the velocity of projection. Result: 6.95 mile sec"1. (j) The mass center of a ballistic rocket crosses the earth's equatorial plane at a height of 540 miles above the surface of the earth, moving parallel to the earth's axis at this instant. The tra- jectory intersects the earth's axis at a point 6040 miles above the surface of the earth. Determine the speed of the mass center of the rocket for the instants at which it crosses the equatorial plane and the earth's axis, and find the time which elapses between these instants. Result: 24,700 mile hr"1, 17,600 mile hr"1, 31.6 min. PROBLEM SET 10 (See Sections 4.3.1-4.3.5) (a) The motion described in Problem 8(c) can take place only when a>, 12, r, h, and 0 satisfy a certain equation. Obtain this equa- tion, and use it to find 0 if w = 450 rad sec"1, 12 = 10 rad sec"1, r = 1 in., and h = 3 in. Result: 24 deg. (b) A uniform, sharp-edged, circular disc rolls on a horizontal plane, its center moving on a straight line. Show that the plane of the disc must be vertical, and determine the minimum speed with which the center must move in order that the motion be stable if the disc has a radius of % in. Result: 8.05 in. sec"1. (c) A billiard ball is struck in such a way that its center returns PROBLEM SET 10 315 to its starting point with one-half of its initial speed. During what fraction of the time required for this motion does the ball roll? Answer: %. (d) A uniform solid sphere is placed at a point 0 on a horizontal plane and is set in motion with the velocity v and angular velocity a> (both parallel to the plane) shown in Fig. lOd. \ 3lvl I 3lvl R FIG. lOd Letting d be the distance from point 0 to the point of line L at which the sphere crosses L, and D the maximum distance be- tween line L and the contact point of the sphere and the plane before the sphere crosses L, determine d/D. Result: If. (e)* When the vertex B of the plate described in Problem 8(b) is held fixed and the plate is permitted to perform small harmonic oscillations, these take place with a frequency f\. If 0 is given some small value while B is held fixed, and the plate is then released from rest, 6 again varies harmonically, with a frequency /i, and line AB oscillates with the same frequency. Determine /i//2. Result: ?. (f)* One end of a uniform, 21 ft long boom is supported in a spherical socket. When the boom is not in use, its upper end is attached to a 5 ft cable and rests on a vertical wall, as shown in Fig. lOf. 316 PROBLEM SET 10 FIG. lOf If the cable breaks and the boom begins to slide on the wall, there comes an instant t\ at which the reaction of the wall on the boom vanishes. Determine the relationship between 0i and 6h these being the values of 6 (see Fig. lOf) and 6 at time tx. Result: 8(26)1/20i2. (g)* Referring to Problem 8(e), suppose that n3 is vertical and that k has a value such that the n3 resolute of the moment about point 0 of the system of all contact forces acting on the body con- sisting of the crank, connecting rod, and piston is equal to zero for all motions during which the crank revolves uniformly. Assume that the surface of contact of the cylinder and piston is smooth and that the system of forces exerted on the piston by the contents of the cylinder is equivalent to a force ? piu whose line of action passes through point 0. Letting the reaction of the cylinder on the piston be a couple together with a force R whose line of action passes through the point of intersection of the axes of the cylinder and the crosshead bearing, show that the n3 resolute of the torque of the couple is equal to zero during all motions of the mechanism, and determine the n2 resolute of R for motions during which the crank revolves with con- stant angular speed w, assuming that R/L is so small that the PROBLEM SET 10 317 second and higher powers of R/L are negligible in comparison with unity. Result: p(R/L) sin 6' - (u>2/2)(sm + LM)(R/L)2 sin 20' (h) A uniform circular disc of radius 6 in. and mass 0.5 slug is rigidly attached to the end of a cylindrical shaft whose axis passes through the center of the disc. Due to misalignment, the shaft axis makes an angle of 0.5 deg with the normal to the plane of the disc. The shaft is supported by short bearings, placed as shown in Fig. lOh, and the system of forces exerted on the portion of the shaft shown in this sketch?by a contiguous portion?is equivalent to a couple whose torque is parallel to the axis of the shaft. FIG. lOh Determine the magnitude of each dynamic bearing reaction when the shaft revolves uniformly at 10,000 rpm. Result: 400 lb. (i)* Figure lOi represents a device which can be used to dem- onstrate the stablizing action of a gyroscope. This device consists of a rotor A, a gimbal B, and a frame C. These parts have a com- mon mass center, the point of intersection of the gimbal axis and the rotor axis, and these two axes are perpendicular to each other. The gimbal axis is perpendicular to an edge of C which is sharpened and which can thus serve as a fixed axis of rotation for C when it is placed in contact with a rough support S. 318 PROBLEM SET 10 ii ill 11111111111111111111111111' 111111111111111111 ITT FIG. lOi During one possible motion of this system, A revolves with constant angular speed u in B; B remains at rest relative to C, with A's axis perpendicular to C (as indicated in Fig. lOi); and C, supported by a horizontal surface S, remains at rest in a ver- tical plane. Show that this motion is stable if all bearings are fric- tionless and w exceeds a certain critical value w*, and determine o>*, letting m be the sum of the masses of A, B, and C, J the moment of inertia of A about A's axis, and K the sum of the mo- ments of inertia of A and B about the gimbal axis. Result: (mghKy'2/J. (j) Figure lOj is a schematic representation of a gyroscopic ship's stabilizer consisting of a cylindrical rotor R, a gimbal G, and two electric motors M and M'. M is attached to G and drives R. M', fastened to the ship S, causes G to rotate relative to S. Letting ? be a reference frame fixed relative to the earth, ni, n2, n3 mutually perpendicular unit vectors fixed in G, as shown in Fig. lOj, and tf an instant at which m is parallel to the ship's longitudinal axis, suppose that at tf s show that the angular momentum (R'ARIQ) of R relative to Q in a reference frame R' is parallel to the angular velocity (R' shown in Fig. 12f, where point Q is the foot of the perpendicular dropped on the wall from point P. FIG. 12f Suggestion: Apply D'Alembert's principle (or the angular mo- mentum principle) to find the differential equations governing 0 and during the time interval t\ < t < U. Next, noting that i and and tf> at time th are given by cos"1 ? = 1.33 rad, 4n = 0 use the differential equations and the values of 0i and 6\ found in (1), above, to evaluate the second derivatives, Si and i, of 6 and ^ at time h; then differentiate the differential equations to find the third and higher derivatives of 6 and at time h. As 6 = 7r/2 at the instant t2 at which the boom strikes the ground, the Taylor series | = 0i + ^ (h ? + fj (h ? + ? ? ? can now be used to find (t2 ? h) to any degree of accuracy desired, and the series PROBLEM SET 12 323 then yields the value of , and E represent identical uniform square plates, each of mass m and side L. These are attached to FIG. 12j each other and to a uniform square plate F of mass 5m, side L, by means of smooth hinges. A vertical shaft, to which C is rigidly attached, passes through an opening in F, thus leaving F free to slide on the shaft. Two light, linear springs of length L and mod- ulus k connect the plates as shown in Fig. 12j. One possible motion of this system is described as follows: The shaft is made to revolve with constant angular speed a> about its axis, and 6 remains constant at 45 deg. Determine a>, and find the smallest value of k for which this motion is stable. Results: APPENDIX Centroidal Principal Axes and Squares of Centroidal Principal Radii of Gyration of Curves, Surfaces, and Solids CURVES 327 J \ L/2 P* L/2 ?o Length: L kx2 = 0, k22 - L2/12 FIG. 1. STRAIGHT LINE Length: 2R0 2 sin2 g\ .. R*/, sin_2? FIG. 2. CIRCULAR ARC 328 SURFACES b/2 b/2 p- a/2 a/2 Area: ab kx2 = bJ/12 FIG. 3. RECTANGLE ?-? 2h/3 h/3 f?^?T-~\?? Area: bh/2 ki2 = h?/18, k22 - b?/24 FIG. 4. ISOSCELES TRIANGLE SURFACES 329 Area: 0R2 ,. R2/t . sin 20 16 sin2 0\ . , R2/, sin 26\ 4 V 2d Qfl^ / 4 \ 2d / FIG. 5. CIRCULAR SECTOR Area: 7rab/2 - 64)/36fr2, k22 FIG. 6. SEMI-ELLIPSE a2/4 330 SOLIDS b/2 b/z A ~O C/2 [ c/2 Volume: abc 3/2 12 FIG. 7. RECTANGULAR PARALLELEPIPED -V?