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PRINCIPLES OF DYNAMICS
by
Donald T. Greenwood
Professor, Department of Aeronautical and Astronautical Engineering University of Michigan
PRENTICE-HALL, INC. Englewood Cliffs, New Jersey
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vi PREFACE
to the derivation of the orbital trajectories, some attention is given to the time of flight, the determination of orbits, and to elementary perturbation theory. ,Beginning with Chapter 6, extensive usage is made of the Lagrangian formulation of the equations of motion. The principle of virtual work is often associated with the study of statics. But it is included here because the ideas of virtual displacement and' virtual work are fundamental in the derivation of Lagrange's equations and in obtaining a clear understanding of generalized forces. The introduction of Lagrange's equations of motion at this point goes very smoothly in the classroom because the students now have a sufficient fundamental background. Furthermore, they have been motivated to find ways of easing the kinematical difficulties in problem formulation. Chapters 7 and 8 present the kinematics and dynamics of rigid bodies with particular emphasis upon rotational motion in three dimensions. In addition to the general analysis of free and forced motions of rigid bodies, special attention is given to the forced motion of axially symmetric bodies using the complex notation method. Matrix notation is introduced and is extensively used in these chapters. Matrix notation is continued in the final chapter which is devoted to vibration theory. The finding of eigenvalues and the diagonalization of matrices, which were previously as~ociated with the problem of obtaining principal axes of inertia, are now extended to the solution for the natural modes of vibration of systems with many degrees of freedom. Other topics such as Rayleigh's principle, the use of symmetry, and the free and forced vibrations of damped systems are also included. The material contained in this text can be covered in about four semester hours at the senior or fifth-year levels. It is my opinion, however, that the discussion of homework problems and illustrative examples should be given, an important place in the overall allotment of class time; hence the course could easily be extended to six semester hours. On the other hand, a course of three semester hours could be arranged by omitting portions of Chapters 5 and 9, and by spending less time on problems. The major portion of this book was written during a sabbatical leave from the University of Michigan. I am particularly appreciative of the aid of Professor H. D. Christensen and others of the Aerospace Engineering faculty at the University of Arizona for providing a place to write and for their helpful discussions. Also, the comments and suggestions of Professor Y. C. Fung of Caltech were of great value during the final preparation of the manuscript. Finally, I wish to thank my wife who typed the manuscript, helped with the proofreading, and provided eJ;1couragement throughout this period.
DONALD T. GREENWOOD.
Ann Arbor, Michigan
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PREFACE
Although there has been a steady inlprovelnent in the depth of most introductory courses in dynamics during recent years, the place of an additional course at the intermediate level relnains an important one. A course at this level is normally taken by first-year graduate students or by undergraduate seniors. The greater Inaturity of these students pennits the presentation of the subject from a nlore advanced viewpoint, with additional Inathematical knowledge assunled, and with the inclusion of illustrative examples having more than the usual cOlnplexity. Through the study of the general theory and its application in these exanlples, and by the solution of a variety of problems of comparable difficulty, Inost students can attain a real competence in dynalnics. This textbook has evolved from a set of notes which accompany the first of three courses forming a sequence in the general area of flight mechanics. The nature of this sequence explains the presence of several rocket and satellite problems among the illustrative examples. Nevertheless, an effort has been lllade to treat the subject of dynalllics in a rather general context with the liberal use of idealizations such as particles, massless rods, uniform disks, and so on, without requiring that the ,configurations approximate practical designs in any particular area of present-day technology. The introductory chapter reviews some of the basic concepts of Newtonian mechanics and gives a short discussion of units and the;ir definitions. There is also a review of. those topics in vector analysis which are most commonly used in dynamics. This is in accord with the general policy of giving brief explanations or summaries of new mathematical topics as they arise. It has been my observation that one of the principal sources of difficulty for students of the vectorial approach to dynamics is one of kinematics. Consequently, the kinematical foundations of particle Illotion are discussed rather thoroughly in Chapter 2. Motion in a plane and also general threedimensional motion are included. Particular attention is given to rotating reference frames and to vector derivatives relative to these frames. With this background in kinematics, a general vectorial development. of the dynamics of a single particle and of systems of particles is given in the next two chapters. Thus we find that most 0f the basic principles of dynamics are developed and applied to a general set of particles before the introduction cf systems with distributed mass. Chapter 5 is concerned with orbital motion. The discussion is almost entirely limited to motion in an inverse-square gravitational field. In addit}on
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viii CONTENTS
3-9. The Simple Pendulum
3-10. Examples
4 DYNAMICS OF A SYSTEM OF PARTICLES
4-1. The Equations of Motion
V _4-.3· Work and Kinetic Energy 4-3. Conservation of Mechanical Energy
v 4-4. Linear Impulse and Momentum
v 4-5. Angular Momentum
v 4-6. Angular Impulse
v 4-7. Collisions
4-8. The Rocket Problem
4-9. Examples
5 ORBITAL MOTION
5-1. Kepler's Laws and Newton's Law of Gravitation
...--- ",-,.,-"- -
5-2. The Two-Body Problem
5-3. The Geometry of Conic Sections
5-4. Orbital Relationships
5-5. Time and Position
5-6. Satellite Orbits About the Earth
--.-.~ ....- '
5-7. Elementary Perturbation Theory
5-8. Examples
6 LAGRANGE'S EQUATIONS
I
6-1. Degrees of Freedom
6-2. Generalized Coordinates
6-3. Constraints
6-4. Virtual Work
Generalized Forces
\\ 6-5. '\ 6-6. D5rivation of Lagrange's Equations
6-7. Lagrange Multipliers
7 BASIC CONCEPTS AND KINEMATICS OF_
RIGID BODY MOTION
7-1. Qegree§,.Qf Freedom ora Rigid Body
7-2. Moments of Inertia
116
120
130
130
132
."
'135
140 142 ",-,..
151
152
161 174
185
185 191 197
200 204 207
213
218
229
229
231
232
235 249
252 267
281
281
283


CONTENTS
1 INTRODUCTORY CONCEPTS
1-1. Elements of Vector Analysis
1-2. Newton's Laws of Motion
1-3. Units
1-4. The Basis of Newtonian Mechanics
,.--. ~"--.',
1-5. D'Alembert's Principle
2 KINEMATICS OF A PARTICLE
\,....",.
2-1. Position, Velocity, and Acceleration of a Point
2-2. Angular Velocity
2-3. Rigid Body Motion About a Fixed Point
\"",._ 2~4. Time Derivative of a Unit Vector
\.-- 2-5. Velocity and Acceleration of a Particle In Several Co
ordinate Systems
2-6. Simple Motions of a Point
2-7. Velocity and Acceleration of a Point in a Rigid Body
2-8. Vector Derivatives in Rotating Systems
2-9. Motion of a Particle in a Moving Coordinate System
2-10. Plane Motion
2-11. Examples
3 DYNAMICS OF A PARTICLE
3-1. Direct Integration of the Equations of Motion
3-2. Wark and Kinetic Energy
3-3. Conservative Systems
\.. 3-4. Potential Energy
'--- 3:::5.. Linear Impulse and Momentum
'- 3-6. Angular Momentum and Angular Impulse
3-7. The Mass-Spring-Damper System
3-8. Coulomb Friction
vii
1
1
12
15
19
25
29
29
31
32
34
35
40 ,
45
47 49
52
57
65
66
76
79
81
89
92
97
111
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7-3. Matrix Notation
7-4. Kinetic Energy
7-5. Translation of Coordinate Axes
7-6. Rotation of Coordinate Axes
\
\ 7-7.
'\ Principal Axes
Displacelnents of a Rigid Body
\
I 7-8.
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7-9. Axis and Angle of Rotation
7-10. Reduction of Forces
7-11. Infinitesinlal Rotations
7-12. Eulerian Angles
7-13. Exanlples of Rigid Body Motion in a Plane
8 DYNAMICS OF A RIGID BODY
8-1. General Equations of Motion
CONTENTS
8-2. Equations of Motion in Terms of Eulerian Angles
8-3. Free Motion of a Rigid Body
\II
8-4. The Poinsot Method
8-5. The Motion of a Top
8-6. Other Methods for Axially Symmetric Bodies 8-7. Examples
9 VIBRATION THEORY
9-1. Review of Systems with One Degree of Freedom
9-2. Equations of Motion
9-3. Free Vibrations of a Conservative System
9-4. The Use of Symmetry
9-5. Forced Vibrations of a Conservative System
9-6. Vibrations with Damping
APPENDICES
A. Inertial Properties of Homogeneous Bodies
B. Answers to Selected Problems
INDEX
ix
288
295
297
298
302
317
320
322
328
332
336
362
362
380
383
391
401
417
426
446
446 450 455 472
484
489
498
502
509
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2 INTRODUCTORY CONCEPTS CHAP. 1
for the most part, although some of the insights and procedures of analytical mechanics will also be used. Because vector operations are so important in the solution of dynamical problems, we shall review briefly a few of the basic vector operations. First, however, let us distinguish among scalars, vectors, and other tensors of higher order. A scalar quantity is expressible as a single, real number. Common examples of scalar quantities are mass, energy, temperature, and time. A quantity having direction as well as magnitude is called a vector. 1 Common vector quantities are force, moment, velocity, and acceleration. If one thinks of a vector quantity existing in a three-dimensional space, the essential characteristics can be expressed geometrically by an arrow or a . directed line segment of proper magnitude and direction in that space. But the vector can be expressed equally well by a group of three real numbers corresponding to the components of the vector with respect to some frame of reference; for example, a set of cartesian axes. If one writes the numbers in a systematic fashion, such as in a column, then one can develop certain conventions which relate the position in the column to a given component of the vector. This concept can be extended readily to mathematical spaces with more than three dimensions. Thus, one can represent a vector in an n-dimensional space by a column of n numbers. So far, we have seen that a scalar can be expressed as a single numt)er and that a vector can be expressed as a column of numbers, that is, as a one-:-dimensional array of numbers. Scalars and vectors are each special cases of tensors. Scalars are classed as zero-order tensors, whereas vectors are first-order tensors. In a similar fashion, a second-order tensor is expressible as a two-dimensional array of numbers; a third-order tensor is expressible as a three-dimensional array of numbers, and so on. Note, however, that an array must also have certain transformation properties to be called a tensor. An example of a second-order tensor is the inertia tensor which expresses the essential features of the distribution of mass in a rigid body, as it affects the rotational motion. We shall have no occasion to use tensors of order higher than two; hence no more than a two-dimensional array of numbers will be needed to express the quantities encountered. This circumstance enables us to use matrix notation, where convenient, rather than the more general but less familiar tensor notation. (Matrix notation will be introduced in Chapter 7 in the study of the rotational motion of rigid bodies.) For the most part, we shall be considering motions which can be described mathematically using a space of no more than three dimensions; that is,
1 In addition, vectors must have certain transformation properties. For examph~, equal vectors must remain equal after a rotation of axes. See Sec. 7-6 for a discussion of these rotation equations.


1
INTRODUCTORY CONCEPTS
The science of mechanics is concerned with the study of the interactions of material bodies. Dynan1ics is that branch of mechanics which consists of the study of the motions of interacting bodies and the description of these motions in terms of postulated laws. In this book we shall concentrate on the dynamical aspects of Nelvtonian or classical nonrelativistic mechanics. By omitting quantum mechanics, we eliminate the study of the interactions of elementary particles on the atonlic or nuclear scale. Further, by omitting relativistic effects, we eliminate from consideration those interactions involving relative speeds approaching the velocity of light, whether they occur on an atomic or on a cosmical scale. Nor shall we consider the very large systems studied by astronomers and cosmologists, involving questions of long-range gravitation and the curvature of space. Nevertheless, over a broad range of system dimensions and velocities, Newtonian mechanics is found to be in excellent agreement with observation. It is remarkable that nearly three centuries ago, Newton, aided by the discoveries of Galileo and other predecessors, was able to state these basic laws of motion and the law of gravitation in essentially the same form as they are used at present. Upon this basis, but using the mathematical and physical discoveries and notational improvements of later investigators, we shall present a modern version of classical dynamics.
1-1. ELEMENTS OF VECTOR ANALYSIS
Scalars, Vectors, and Tensors. Newtonian mechanics is, to a considerable extent, vectorial in nature. lts basic equation relates the applied force and the acceleration (both vector quantities) in terms o'f a scalar constant of proportionality called the mass. In contrast to Newton's vectorial approach, Euler, Lagrange, and Hamilton later emphasized the analytical or algebraic approach in which the differential equations of motion are obtained by performing certain operations on' a scalar function, thereby simplifying the analysis in certain respects. Our approach to the subject will be vectorial
1


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4 , INTRODUCTORY CONCEPTS CHAP. '1
Unit Vectors. If a positive scalar and a vector are multiplied together (in either order), the result is another vector having the same direction, but whose magnitude is multiplied by the scalar factor. Conversely, if a vector is multiplied by a negative scalar, the direction of the resulting vector is reversed, but the magnitude is again multiplied by a factor equal to the magnitude of the scalar. Thus one can always think of a given vector as the product of a scalar magnitude and a vector of unit length which designates its direction. We 'can write
(1-1)
where the scalar factor A specifies the magnitude of A and the unit vector eA shows its direction (Fig. 1-1).
Addition of Vectors. The vectors A and B can be added as shown in Fig. 1-2 to give the resultant vector C. To addB to A, translate B until its origin coincides with the terminus or arrow of A. The vector sum is indicated by the line directed from the origin of A to the arrow of B. It can be seen that
C == A + B == B + A (1-2)
since, for either order of addition, the vector C is the same diagonal of the parallelogram formed by using A and B as sides. This is the parallelogram
A
Fig. 1-1. A vector and its corresP9nding unit vector.
Fig. 1-2. The parallelogram rule of vector addition.
rule of vector addition. Since the order of the addition of two vectors is unimportant, vector addition is said to be commutative. This procedure can be extended to find the sum of more than two vectors. For example, a third vector D can be added to the vector C obtained previously, giving the resultant vector E.From Fig. 1-3, we see that
E == C + D == (A + B) + D (1-3)
E == A + (B + D) (1-4)
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SEC. 1-1 INTRODUCTORY CONCEPTS 3
each matrix or array will have no more than three rows or columns and each vector will have no more than three components. An exception will be found in the study of vibration theory in Chapter 9 where we shall consider eigenvectors in a multidimensional space.
Types of Vectors. Considering the geometrical interpretation of a vector as a directed line segment, it is important to recall that its essential features include l11agnitude and direction, but not location. This is not to iJnply that the location of a vector quantity, such as a force, is irrelevant in a physical sense. The location or point of application can be very iInportant, and this will be reflected in the details of the Inathematical fonnulation; for exaInple, in the evaluation of the coefficients in the equations of Inotion. Nevertheless, the rules for the InatheInatical manipulation of vectors do not involve location; therefore, from the mathelnatical point of view, the only quantities of interest are Inagnitude and direction. But frOITI the physical point of view, vector quantities can be classified into three types, nanlely, free vectors, sliding vectors, and bound vectors. A vector quantity having the previously discussed characteristics of magnitude and direction, but no specified location or point of application, is known as a free vector. An exanlple of a free vector is the translational velocity of a nonrotating body, this vector specifying the velocity of any point in the body. Another example is a force vector when considering its effect upon translational motion. On the other hand, when one considers the effect of a force on the rotational ITIotion of a rigid body, not only the magnitude and direction of the force, but also its line of action is important. In this case, the moment acting on the body depends upon the line of action of the force, but is independent of the precise point of application along that line. A vector of this sort is known as a sliding vector. The third type of vector is the bound vector. In this case, the magnitude, direction, and point of application are specified. An example of a bound vector is a force acting on an elastic body, the elastic deformation being dependent upon the exact location of the force along its line of action. Note again that all mathematical operations with vectors involve only their free vector properties of magnitude and direction.
Equality of Vectors. We shall use boldface type to indicate a vector quantity. For example, A is a vector of magnitude A, where A is a scalar. Two vectors A and B are equal if A and B have the same magnitude and direction, that is, if they are represented by parallel line segments of equal length which 'are directed in the same sense. It can be seen that the translation of either A or B, or both, does not alter the equality since they are considered as free vectors. I


6 INTRODUCTORY CONCEPTS CHAP. 1
A - A = (1 - I)A = 0 (1-8)
In general, the distributive law applies to either the scalar or the vector. Thus,
(n + m)A = nA + mA (1-9)
and
n(A + B) = nA. + nB (1-10)
Components of a Vector. If a given vector A is equal to the sum of several vectors with differing directions, these vectors can be considered as component vectors of A. Since component vectors defined in this way are not unique, it is the usual practice in the case of a three-dimensional space to specify three directions along which the component vectors must lie. These directions are indicated by three linearly independent unit vectors, that is,
a set of unit vectors such that none can be expressed as a linear combination of the others.
Suppose we choose the unit vectors el , e2, and e3 with which to express
the given vector A. Then we can write
(1-11)
where the scalar coefficients A}, A 2, and A3 are now determined uniquely.
At, A 2, and A3 are known as the scalar components, or simply the components, of the vector A in the given directions.
If another vector B is expressed in terms of the same set of unit vectors, for example,
z
y
x
Fig. 1-4. The components of a vector in a cartesian coordinate system.
(1-12)
then the components of the vector sum of A and B are just the sums of the corresponding components.
A+B (AI + Bt)el
+ (A2 + B2)e2
+ (A3 + B3)e3 (1-13)
This result applies, whether or not et, e2, and e3 form an orthogonal triad of unit vectors. Now consider a case where the unit vectors are mutually orthogonal, as in the cartesian coordinate system of Fig. 1-4. The vector A can be expressed in terms of the scalar components Ax, A y , and Az, that is, it can
be resolved as follows:


SEC. 1-1 INTRODUCTORY CONCEPTS 5
A
Fig. 1-3. The polygon rule of vector addition.
From Eqs. (1-3) and (1-4) we obtain
(A + B) + D == A + (B + D) == A + B + D (1-5)
illustrating that vector addition is associative. Because of the commutative and associative properties of vector addition, we can dispense with the parentheses in a series of additions and perform the additions in any order. Furthermore, using the graphical procedure of Fig. 1-3, we see that the resultant vector E is drawn from the origin of the first vector A to the terminus of the last vector D, thus closing the polygon. This generalization of the parallelogram rule is termed the polygon rule of vector addition. A similar procedure applies for the case where all vectors do not lie in the same plane. It is important to note that certain physical quantities that are apparently vectorial in nature do not qualify as true vectors in the sense that the usual rules for vector operations do not apply to them. For example, a finite rotational displacement of a rigid body is not a true vector quantity because the order of successive rotations is important, and therefore it does not follow the commutative property of v~ctor addition. Further discussion of this topic will be found in Chapter 7.
Suhtraction of Vectors. The negative of the vector A is the vector
-A == -Ae.'1 (1-6)
which has the same magnitude as A but is opposite in direction. In other words, the vector -A is equal to the product of the vector A and the scalar -1. Subtracting a vector B from another vector A is equivalent to adding its negative.
A - B == A + (- B) (1-7)
In particular, for the case where A == B, we may use the distributive law for multiplication of a vector and a scalar to obtain


8 INTRODUCTORY CONCEPTS CHAP. 1
Now consider the dot product of two vectors A and B, each of which is
expressed in terms of a given set of unit vectors CI , C2, and Ca. From Eqs.
(1-11) and (1-12), we obtain
A
e B = AIBI + A2B2 + AaBa + (AIB2 + A 2B1)CI e C2
+ (AlBa + AaBl)CI e Ca + (A2Ba + Aa B2)e2 e Ca (1-18)
For the common case where the unit vectors form an orthogonal triad, the terms involving dot products of different unit vectors are all zero. For this case, we see from Eq. (1-18) that
(1-19)
Vector Product. Referring again to Fig. 1-5, we define the vector product
or cross product as follows:
A X B = ABsinOk (1-20)
where k is a unit vector perpendicular to, and out of, the page. In general, the direction of k is found by the right-hand rule, that is, it is perpendicular to the plane of A and B and positive in the direction of advance of a righthand screw as it rotates in the sense that carries the first vector A into the second vector B. The angle of this rotation is O. It is customary, but not
necessary, to limit () to the range 0 < 0 < 'iT. Using the right-hand rule., it can be seen that
A X B = -B X A (1-21)
indicating that the vector product is not commutative.
Plane Q
Fig. 1..7. The distributive law for the vector product.


SEC. 1-1 INTRODUCTORY CONCEPTS 7
(1-14)
where i, j, and k are unit vectors in the directions of the positive x, y, and z axes, respectively. From Fig. 1-4, it can be seen that the component vectors Axi, Ayj, and Azk fonn the edges of a rectangular parallelepiped \vhose qiagon~ll. is the vector A. A similar situation occurs for the case of nonorthogonal or skewed unit vectors, except that the parallelepiped is no longer rectangular. Nevertheless, a vector along a diagonal of the parallelepiped has its con1ponents represented by edge lengths. In this geolnetrical construction, we are dealing with free vectors, and it is custolnary to place the origins· of the vector A and the unit vectors at the origin of the coordinate system. It is ilnportant to note that, for an orthogonal coordinate system, the components of a vector are identical with the orthogonal projections of the given vector onto the coordinate axes. For the case of a skewed coordinate system, however, the scalar components are not equal, in general, to the corresponding orthogonal projections. This distinction will be in1portant in the discussions of Chapter 8 concerning the analysis of rigid body rotation by means of Eulerian angles; for, in this case, a skewed system of unit vectors is used.
Scalar Product. Consider the two vectors A and B shown in Fig. 1-5. The scalar product or dot product is
A • B == AB cos (j (1-15)
Since the cosine function is an even function, it can be seen that
(1-16)
A
Fig. 1-5. MultipJication of two
implying that the scalar multiplica- vectors. .1 tion of vectors is commutative. The scalar product can also be considered as the product of the n1agnitude of one vector and the orthogonal projection of the second vector upon
c
Fig. 1-6. The distributive law for the dot product.
it. Now, it can be seen from Fig. 1-6 that the sum of the projections of vectors A and B onto a third vector C is equal to the projection
of A + B onto C. Therefore, noting that the multip1ication of scalars is distributive, we obtain
(A + B) • C == A • C + B • C (1-17)
Thus, the distributive property ap
plies to the scalar product of vectors. I


10 INTRODUCTORY CONCEPTS
BxC
oB
Ax(BxC)
Fig. 1-8. Triple products of vectors.
CHAP. 1
Scalar Triple Product. . The product A • (B X C) is known as the scalar triple product. Looking at Fig. 1-8, we· see that B X C is a vector whose magnitude is equal to the area of a parallelogram having Band C as sides and whose direction is perpendicular to the plane of that parallelogram. Considering this plane to be horizontal for the moment, we note that A • (B X C) is just the area of the base multipli~d by the projection of A onto the vertical; that is, its magnitude is the volume of the parallelepiped having A, B, and C as edges. The sign of A .. (B X C) is positive or negative, depending upon whether or not A and B X C lie on the same side of the plane of Band C. Of course, if A, B, and C lie in the same plane, the product is zero. We can find the scalar triple product in terms of cartesian components by using equations in the form of Eqs. (1-14) and (1-26) to obtain
i jk
A • (B X C) == A· Bx By Bz
Cx Cy Cz
Ax Ay Az
Bx By Bz
Cx Cy Cz
(1-28)
From the rules for determinants, we note that the interchange of any two vectors (that is, any two rows) results in a change in the sign of the product. Furthermore, an even number of such interchanges results in a cyclic permutation of the vectors, but no change in the sign of the product. Therefore, since the dot product is commutative, the dot and the cross in a scalar triple product may be interchanged, or a cyclic permutation of the vectors may occur, without affecting the result.
A • (B X C) == (A X B) • C (1-29)
Also,
A • (B X C) == B • (C X A) == C • (A X B) (1-30)
These results can be interpreted geometrically by noting that the parallelepiped whose edges are formed by the vectors A, B, and C, has a volume that is independent of the order of these vectors. The order, as we have seen, influences the sign of the result but not its magnitude.
Vector Triple Product. It can be seen from Fig. 1-8 that the vector triple product A X (B X C) lies in the plane of Band C. On the other hand, (A X B) X C lies in the plane of A and B and is not, i~ general, equal to


SEC. 1-1 INTRODUCTORY CONCEPTS 9
Another way of visualizing the vector product A X B is to note again' that it must be perpendicular to both A and B. Thus if planes a and b pass through a common origin 0 and are normal to A and B, respectively, then A X B must be directed along the line of intersection of these planes with the sense again being determined by the right-hand rule. The magnitude of A X B is equal to A times the projection of B onto plane a, or conversely, it is equal to B tilnes the projection of A onto plane b. Now let us use this approach to evaluate the vector product A X
(B + C) (Fig. 1-7). First, we note that the vectors A X B, A X C, and
A X (8 + C) must all lie in plane a that is normal to A. Furthermore, it is
apparent that the projection of B + C onto plane a is equal to the vector sum of the projections of Band C onto the same plane. Thus, the parallelogram formed by the vectors A X B and A X C is similar to that formed by the projections of Band C onto a, there being a rotation through 90 degrees in plane a and a multiplication by a scalar factor A. Therefore, we obtain that
A X (B + C) == A X B + A X C (1-22)
showing that the distributive law applies to vector products. Using the distributive law, we can evaluate the vector product A X B' in terms of the cartesian components of each. Thus,
But
and therefore,
iXi==jXj==kXk==O
ixl==-jXi==k
j X k == -k X j == i
k X i == - i X k == j
(1-23)
(1-24)
A X B == (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k (1-25)
This result can be expressed more concisely as the following determinant:
i jk
A X B == Ax Ay Az (1-26)
Bx By Bz
In general, if the sequence eb e2, and e3 forms a right-handed set of mutually orthogonal unit vectors, then the vector product can be expressed in terms of the corresponding components as follows:
e1 e2 e3
A X B ~ Ai A2 A3
Bl B2 B3
(1-27)
!••,I.Iif~,!
~f, \'j\
I
I'
iI
\,IIIt,~I.,


INTRODUCTORY CONCEPTS
.!!-. (A X B) == dA X B + A X dB
du du du
CHAP. '1
(1-36)
As an example of differentiation, consider the vector A expressed In terms of its cartesian components:
A == Axi + Ayj + Azk
For the common case where differentiation is with respect to time, and the unit vectors have a fixed orientation in space, we obtain, using Eqs. (1-14), (1-33), and (1-34), that
dA == dAx i + dAy · + dAz k
dt dt dt J dt (1-37)
In the more general case where A is expressed in terms of the unit vectors eh e2, and ea which may change their orientation in space, we can write
A == AIel + A 2 e2 + Aaea Then, we obtain
dA == dAI e + dA 2 e + dAa e + A del + A de2 + A dea
dt dt I dt 2 dt a I dt 2 dt a dt
Of, expressing the result using a dot over a symbol to indicate its time derivative,
(1-38)
where we note that the time rate of change of a unit vector is always per
~--.-------...-.--.- ... -..... - ..-- ..... -.. " - ' -- ..
pendicular to that unit vector. Equations of this sort will be developed more extensively in the study of kinematics in Chapter 2.
1-2 NEWTON'S LAWS OF MOTION
In his Principia, published in ~687, Sir Isaac Newton stated the laws upon which classical mechanics is based. Using modern terminology, these laws can be stated as follows:
I. Every body continues in its state. of rest, or of uniform motion in a straight line, unless. compelled to change that state by forces acting upon it. II. The time rate of change of linear momentum of a body is proportional to the force acting upon it and occurs in the direction in which the force acts. III. To every action there is an equal and opposite reaction; that is, the mutual forces of two bodies acting upon each other are equal in magnitude and opposite in direction.


SEC. 1-2 INTRODUCTORY CONCEPTS 13
The Laws of Motion for a Particle. An understanding of Newton's laws of motion is most easily' achieved by applying them to the study of the motion of particles, where 'a particle is defined as a mass concentrated at a point. Later, when we consider the case of bodies with a continuous distribution of mass, the generalization of the dynamical methods from the discrete to the continuous case win be seen to be quite straightforward. Now let us state three basic laws applying to the motion of a particle. The first is the law of motion which summarizes Newton's first two laws. It can be expressed by the equation
d
F == k dt (mv) == kma (1-39)
where the product mv is known as the linear nlomentum p, that is,
p == mv (1-40)
and where m is the mass of the particle, a is its acceleration, F is the applied external force, and k is a positive constant whose value depends upon the choice of units. The mass m is considered to be constant., since we are not concerned with relativistic effects and any variable-mass systems are treated as collections of particles. Because of the fundamental nature of Eq. (1-39), the units are chosen such that k == 1. So, with a proper choice of units, Eq. (1-39) simplifies to
F == ma (1-41)
which is the usual statement of the law of motion. The second basic law is the law of action and reaction:
When two particles exert forces on each other, these interaction forces are equal in magnitude, opposite in sense, and directed along the straight line joining the particles.
This is essentially a statement of Newton's third law as it applies to two particles, but the collinearity of the interaction forces has been mentioned specifically. The added requirement of collinearity will be found to be essential for the conservation of angular momentum of an isolated mechanical system and applies to all mechanical or gravitational interaction forces. It does not apply, however, to certain forces between moving, charged particles, a situation which will not concern us in this book. The third basic law is the law of addition offorces:
Two forces P and Q acting simultaneously on a particle are together
equivalent to a single force F == P + Q.
By similar reasoning, we can conclude that the simultaneous action of more than two forces on a particle produces the same motion as a single force equal to their vector sum. Newton stated the law of addition of forces as a corollary to his laws of


14 INTRODUCTORY CONCEPTS CHAP. 1
motion. Note that it also implies that a single force ·may be replaced by its component forces in a dynamical calculation. Thus, if we consider a particle of mass m to be moving with respect to a fixed cartesian system under the action of a force F, we can write
F == Fxi + Fyj + Fzk
and the resulting acceleration components are found from
Fx == mx
Fy == my
Fz == mz
where the total acceleration is
a == xi + yj + zk
(1-42)
(1-43)
(1-44)
In other words, the single vector equation given in Eq. (1-41) is equivalent to the three scalar equations of Eq. (1-43). That these equations can be verified experimentally indicates the actual independence of the component accelerations and also that the mass m is a single scalar quantity.2
Frames of Reference. In our previous discussion we have not concerned ourselves with the question of what is a proper reference frame from which to measure the accelerations to be expected in accordance with the laws of motion. The approach which we shall take is to define an inertial or Newtonian .reference frame to be. any rigid set of coordinate axes such that particle motion relative to these axes is described by Newton's laws of motion. In an attempt to find an example of an inertial frame in the physical world, pne might consider first a system fixed in the earth. Such a choice would be adequate for most cases where the distances traveled are short. relative to the earth's radius and where the velocities are sm~ll compared to the velocity of escape from the earth. If one were to analyze long-range missiles or satellites, however, the earth would be a completely inadequate approximation to an inertial frame. A much better approximation would be a system whose origin is at the center of the earth and whose axes are not rotating with respect to the "fixed" stars. Even here, however, measurements of satellites traveling far from the earth would easily show deviations from Newton's laws due to the neglected gravitational forces exerted by the sun, moon, and planets. Finally, one could choose as a Newtonian reference frame a system at the center of the sun (or, more. exactly, at the center of mass of the solar
2 This, of course, presupposes that the particle velocity. is small compared to the velocity of light and therefore that relativistic effects can be neglected.


SEC. 1-3 INTRODUCTORY CONCEPTS 15
system) and nonrotating with respect to the so-called fixed stars. This would appear to be adequate for the foreseeable future. So let us assume that we have found an inertial reference frame, and therefore that Newton's laws apply for motions relative to this frame. It can be shown that any other reference frame that is not rotating but is translating with a uniform velocity relative to an inertial frame is itself an inertial frame. For example, if system B is translating at a constant velocity Vre1 with respect to an inertial systen1 A, then, denoting the velocity of a particle as viewed by observers on A and B by VA and VB, respectively, we see that
VA == VB + Vrel (1-45)
Differentiating with respect to tilne and noting that the derivative of V)'el 'is zero, we obtain
(1-46)
where aA and an are the accelerations of the particle as viewed from systems A and B, respectively. Now the total force applied to the particle is independent of the Inotion of the observer. So fron1 the Newtonian point of view, observers on systen1s A and B see identical forces, masses, and accelerations, and therefore, Eq. (1-41) is equally valid for each observer. One can sumn1arize by saying that the existence of an inertial frame implies the existence of an infinite number of other inertial frames, all having no rotation rate relative to the fixed stars but translating with constant velocities relative to each other. Thu~, even Newtonian mechanics has no single, preferred fralne of reference.
1-3. UNITS
When one attempts to apply the law of motion as given by Eq. (1-41), one is immediately faced with the problem of choosing a proper set of units with which to express the quantities of interest. As we have seen previously, when we set the proportionality constant k equal to unity in Eq. (1-39), the sizes of the units used to specify force, mass, and acceleration are no longer arbitrary. Furthermore, as we shall see, the selection of units for any two of these. quantities fixes the units to be used in measuring the third quantity since the proportionality constant is assulned to be dimensionless, that is, it is a pure number and has no units associated with it. In considering the problem of units let us first discuss the so-called dimensions associated with each unit.
Dimensions. It can be seen that the units which are used in the measurement 'of physical quantities may differ quantitatively as well as qualitatively. For example, the foot and the inch differ in n1agnitude but are qualitatively
1! I:. :
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16 INTRODUCTORY CONCEPTS CHAP. 1
the same in that both are units of length. On the other hand, the foot and , the second are qualitatively different. Now, the qualitative aspects of a given unit are characterized .by its dimensions. Thus, the foot and the inch both have the dimension of length. Similarly, the hour and the second have the dimension of time. It turns out that all units used in the study of mechanics can be expressed in terms of only three dimensions. By common agreement, the dimensions of length and of time are considered to be fundamental. Thus, a physical quantity such as velocity has the dimensions of length per unit time, written [LT-l], regardless of whether the units chosen are milhr, ftlsec, or even knots. Similarly, acceleration has the dimensions [LT-2]. A characteristic of the equations of physics is that they must exhibit dimensional homogeneity. By this, we mean that any terms which are added or subtracted must have the same dimensions and it also implies, of course, that the expressions on each side of an equality must have the same dimensions. Furthermore, any argument of a transcendental function, such as the trigonometric function, exponential function, Bessel function, and so on, must be dimensionless; that is, all exponents associated with the fundamental dimensions of the argument must be zero. One will sometimes find apparent exceptioD.s to the requirement of dimensional homogeneity, but in all cases some of the coefficients will prove to have unsuspected dimensions or else the equation is an empirical approximation not based on physical law. In .checking for dimensional homogeneity, one should note that the unit of angular displacement, the radian, is dimensionless. If we note that the equations of mechanics are principally of the form of Eq. (1-41), or its integrals with respect to space or time (or the corresponding moments), then the requirement of dimensional homogeneity implies that the fundamental dimensions corresponding to mass and force cannot be chosen independently. If one chooses mass to be the third fundamental dimension, then the dimensions of force are determined, and vice versa. Thus, we see that there are two obvious possibilities for choices of fundamental dimensions: (1) the absolute system in which mass, length, and time are the fundamental dimensions; (2) the gravitational system in which force, length, and time are the fundamental dimensions. Both types of systems are in wide use in this country; the former being used more extensively by physicists and other scientists, the latter by engineers.
Systems of Units. Within either the absolute system or the gravitational system, many sets of fundamental units can be chosen. Absolute systems of units in common use include the egs or centimeter-gram-second system and the mks or meter-kilogram-second system where, of course, the centimeter (or meter) is the fundamental unit of length, the gram (or kilogram) is the fundamental unit of mass, and the second is the fundamental unit of time.


. (~
SEC. 1-3 INTRODUCTORY CONCEPTS 17
On the other hand, the English gravitational system, which is the system that we shall use, employs the foot as the fundamental unit of length, the pound as the fundamental unit of force, and the second as the fundamental unit of time. In this case, the unit of mass, the slug, is a derived unit rather than a fundamental unit. A slug is that mass which is given an acceleration of 1 ftlsec by an applied force of 1 lb. In terms of fundamental units, we see that
1 slug == 1 lb sec2/ft
since, from Eq. (1-41) and the principle of dimensional homogeneity, the unit of mass must equal the unit of force divided by the unit of acceleration. Note particularly that we always use the term pound as a unit of force and never as a unit of mass. A possible source of confusion arises from the fact that the legal standard of mass is based on the absolute system and, in that system, a standard unit of mass is the pound, also called the pound-mass. To avoid confusion, however, we shall use the gravitational system exclusively. Some of the quantities that are most commonly used in dynamics are listed in Table 1-1. The letters F, L, and T refer to the dimensions of force, length, and time, respectively.
TABLE 1-1. GRAVITATIONAL UNITS AND DIMENSIONS
Quantity Gravitational Units Dimensions
Length ft [L] Time sec [Tl Force Ib [F]
Mass Ib sec2/ft (slug) [FT2L-l]
Velocity ft/sec [LT-l]
Acceleration ft/sec2 [LT-2] Energy (work) ft Ib [FL]
Angular Velocity rad/sec [T-l] Moment Ib ft [FL] Moment of Inertia lb ft sec2 [FLT2] Linear Momentum lb sec [FT] Angular Momentum lb ft sec [FLT] Linear Impulse lb sec [FT] Angular Impulse lb ft sec [FLT]
Conversion of Units. When making checks of dimensional homogeneity, and even when performing numerical computations, it is advisable to carry along the units, treating them as algebraic quantities. This algebraic manipulation of units often requires the conversion from one set of units to another set having the same dimensions. Normally, one converts to three basic units, such as pounds, feet, and seconds, rather than carrying along derived units,


18 INTRODUCTORY CONCEPTS CHAP. 1 .
such as slugs, or perhaps several different units of length. In this conversion process, one does not change the lnagnitudes of any of the physical quantities, but only their form of expression. A convenient method of changing units is to multiply by one or more fractions whose magnitude is unity, but in which the numerator and the denominator are expressed in different units. Suppose, for example, that one wishes to convert knots into inches per second. One finds that
1 knot == 1 nautical mi/hr 1 nautical mi == 6080 ft
Therefore,
1 ft == 12 in. 1 hr == 3600 sec
1 knot == (1 Jlaut. mi)( 6080-ft- .)(12 in.)( I-hr)
I-hr l..naut. mt I-ft- 3600 sec
== 20.27 in./sec
Weight and Mass. We have seen that, although we shall use the pound as a unit of .force, one of the units of mass in the absolute system is also known as the pound. This emphasizes the need for a clear distinction between weight and mass since the weight of an object is expressed in units of force. Briefly, the weight of a body is the force with which that body is attracted toward the earth. Its mass is the quantity of matter in the body, irrespective of its location in space. For example, if a given body is moved from a valley to the top of an adjacent mountain, its mass is constant; but its weight is less on the mountain top because the magnitude of the gravitational attraction decreases with increasing elevation above sea level. Another approach is provided by the equation
w == mg (1-47)
where w is the weight, m is the mass, and g is the local acceleration of gr?vity, that is, g is the acceleration that would result if the body were released from rest in a vacuum at that location. Although the interpretation of Eq. (1-47) appears to be straightforward, certain complications arise in the detennination of the acceleration of gravity. First, the acceleration of gravity is measured with respect to a reference frame fixed in the earth. This is not an inertial frame in the strict sense because the earth is rotating in space. Therefore, as we shall show more clearly in Chapter 2, the so-called inertial forces must be included in a calculation of the acceleration of gravity. Thus it turns out that centrifugal as well as gravitational forces enter the probleln. The effect of the centrifugal force is to cause a slight change in the magnitude and direction of the acceleration of gravity, the amount of the change depending upon the latitude. A second reason for the slight variation In the acceleration of gravity
I


I
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1
\
SEC. 1-4 INTRODUCTORY CONCEPTS 19
with latitude is the oblateness of the earth, the polar radius being about 0.3 per cent smaller than the equatorial radius. The two factors previously mentioned result in a variation of the acceleration of gravity at the earth's surface which can be approximated by the expression
g == 32.26 - 0.17 cos 2 () ft/sec2
where g is the local acceleration of gravity and () is the latitude of the point in question. For our purposes, it will generally be sufficient to use the value g == 32.2 ft/sec 70r the acceleration of gravity at the surface of the earth. Furthennore, in any discussion of orbits about the earth, we shall use the symbol go which includes gravitational effects only and refers to a spherical earth.
1-4. THE BASIS OF NEWTONIAN MECHANICS
Now that we have presented the basic laws of ITIotion and have discussed the question of units briefly, let us consider in greater detail the fundamental assumptions of Newtonian mechanics. In particular, let us consider the concepts of space, time, mass, and force from the viewpoint of Newtonian mechanics and indicate experimental procedures for defining the corresponding fundan1entaI units. In the latter process, we shall not be concerned so much with the practical utility of the operational definitions as with their theoretical validity.
Space. Newton conceived of space as being infinite, homogeneous, isotropic, and absolute. The infinite nature of space follows from the implicit
assumption that ordinary, Euclidean geometry applies to it. By homogeneous
~~~~i:~!tcpiCLit~i~. ~el:!ntthat .the l.Qc_a1.J2!92(:x!i(:~_QLsp<,l.<:(:~re .mdep~ndent
~ 0 .."_Q~~JJ.Qn.. or_..dlrec.tJC>.n. In claiming that space is absolute, Newton assumed the existence of a primary inertial frame. fnhis-"'vTew-;-'-ii-was nonrotating relative to the fixed stars and also was fixed relative to the center of the universe which was interpreted as being located at the center of mass of the solar system. Of course, Newton realized that any other coordinate system that is translating uniformly with respect to his primary frame would serve equally well as an inertial frame. On philosophical grounds, however, he chose to give special preference to the inertial frame that is fixed at the "center of the universe." We, too, saw in the discussion of Eq. (1-46) that the law of motion is not changed by the transformation to another system that is translating uniformly : with respect to a given inertial system. But any rotational motion of a ref, erence frame relative to a given inertial frame will produce apparent acceleration terms which change the form of the basic equation of motion. So we


20 INTRODUCTORY CONCEPTS CHAP. 1.
can conclude that Newtonian mechanics requires an absolute reference for rotational motion, but translational motion is relative in that an arbitrary uniform translation may be superimposed. The spatial relationships associated with Newtonian mechanics are measured in units of length. As a standard unit of length, one can choose the distance between two marks on a bar at a standard temperature, or perhaps one could define the standard unit of length in terms of the wavelength of the light corresponding to a given .spectral Iine measured under standard conditions. In either event, the unit is obtained iIi a rather direct fashion and, in this theory, does not depend upon the motion of either the standard or the observer.
Time. Newton conceived of time as an ~b~gJ:t!~--.9.uantity, that is, the
same for all observers and, furthermore, independent of" aIr-objects of the physical world. He considered that a definition of time in terms of natural phenomena such as the rotation of the earth was, at best, an approximation to the uniform flow of "true" time. Later, Mach3 contended that time is merely an abstraction arrived at by changes in the physical world. For example, we say that object A moves uniformly if equal changes in the displacement of A correspond to equal changes in the "position of another system such as the hands of a clock. But one cannot show that the motion of the clock is uniform in itself; a comparison must be made with another system. Thus, time is not ail absolute entity, but is assumed in order to express the interrelationships of the motions of the physical world. In deciding which physical process to use as a basis for the definition of a standard unit of time, we assume the validity of certain physical laws and choose a process that, according to the physical theory, proceeds at a uniform rate, or at a constant frequency, and is relatively easy to measure. For example, according to Newtonian mechanics, the rotation rate of the earth relative to the fixed stars is constant except for a very small retarding effect due to tidal friction. So, for most purposes, the definition of a standard unit of time based upon the rotation rate of the earth, as measured by the interval between successive transits of a given star, would seem to be satisfactory. Also, we shall adopt the Newtonian assumption that time is the same for all observers. Note that this allows for a finite propagation velocity of light if compensation is made for the time of transmission of clock synchronizing signals from one place to another. No relativistic corrections are required if the- relative velocities of all bodies are much smaller than the velocity of light.
3E. Mach, The Science of Mechanics (La Salle, Ill.: The Open Court Publishing Co., 1942). The first edition was published in 1893.
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SEC. 1-4 INTRODUCTORY CONCEPTS 21
Mass. We have seen that reasonably straightforward operational definitions can be given for the fundamental units of length and of time. The third fundamental unit can be chosen as a unit of mass or of force, depending upon whether the absolute or the gravitational system of dimensions is considered to be most appropriate. Let us consider first the choice of mass as a fundamental dimension. Newton defined Inass as "quantity of matter" which was calculated as the product of volume and density. Although this definition may give one an intuitive concept of mass, it is not a satisfactory operational definition but merely replaces one undefined term by another. In an attempt to find a more satisfactory definition of mass, we might take a given body A as a standard. Another body B is assumed to have a mass equal to that of A if bodies A and B balance when tested on a beam balance. In other words, their masses are equal if their weights are equal. If one assumes further that mass is an additive property, that is, that the mass of several bodies acting together is equal to the sum of their individual masses, then one can establish a set of standard masses which are equal to various multiples or fractional parts of the original primary standard. Using the standard masses, one can in theory determine the mass of a given body to an arbitrary precision. ". This relatively convenient procedure of comparing the masses of two bodies by comparing their weights has certain deficiencies when used as a fundamental definition of mass. One difficulty is that it assumes the equiva
lence of inertial mass and s.ravitat(911a!,JJ;lfl§.§.: The mass that we are interested in determlnin-g'~~l's"-fne"inerfial mass m in the equation of motion F == mao On the other hand, the weighing procedure compares the magnitudes of the forces of gravity acting on the bodies. As we shall see in Sec. 5-1, the force of gravity acting on a body is proportional to its mass, but there is no a priori reason for the equality of gravitational mass and inertial mass. This equality must be experimentally determined.4 In comparing the masses of two bodies by comparing their weights, the assumption is made that, during the balancing procedure, the bodies are motionless in an inertial frame. For this case, the upward force exerted on each body by the balance is equal to the 'downward force of gravity on that body. This results in a net force of zero which, in accordance with the law of motion, corresponds to zero acceleration in an inertial frame, in agreement with the initial assumption. Using this method, if one wishes to determine the mass of a moving body,
an additional assumption must be made; namely, that !l1.? mf!§.Lf.J,Lg_QQ1..r.E independent of its motio!1~ In other words, its mass while in 'motion is assumed
---., ........ -
4It is interesting to note that Newton tacitly assumed the equivalence of inertial and gravitational mass. In more recent times, this principle of equivalence has been one of the p~~!ulates of Einstein's general theory of relativity.


2
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as ch( un] Fo] ear effe unl' inte sati
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22 INTRODUCTORY CONCEPTS CHAP. l
to be the same as the mass which was measured while at rest. This assumption of Newtonian mechanics is in agreement with experimental results for cases where the velocity is very small compared with the velocity of light and· relativistic effects are negligible. Now suppose that the mass of a given body is to be determined by weighing, using a calibrated spring scale. The calibration can be accomplished by using a set of standard masses which are obtained in a manner similar to that used previously except that a spring scale is used in place of a beam balance. (Note that in obtaining standard masses, no calibration of the scale is required since weights are merely matched.)
The use of a spring scale to measure the mass of a body relative to the mass of a primary standard will be observed to have all the shortcomings of the beam balance. In addition, the calibration of the scale is accurate only if the weighings occur at a location with an acceleration of gravity equal to the calibration value. This follows from the definition of weight given by Eq. (1-47) and the fact that the spring scale measures weight or force directly rather than matching weights as in the case of the beam balance. In order to avoid the theoretical difficulties involved in determining mass by means of weight measurements, Mach proposed a method of
measuring inertial. mg..§~."..r~tios by measuring accelerations. A dynamically
. . . . ._ _ ..--~.......................H ......
isolated system is assumed, consisting of two mutually interacting masses. Then, making the experimental proposition that the accelerations are opposite in direction and lie along the line connecting their centers, he defines the mass ratio to be the inverse ratio of their acceleration magnitudes. Designating the bodies as A and B, the mass ratio is
(1-48)
where aA and an are the magnitudes of the corresponding accelerations. Then, if mass A is a standard unit, the mass of body B can, in theory, be determined. Mach does not specify the nature of the interaction in his definition. It might be gravitational or perhaps a direct connection as by a rigid rod or elastic rope. Another possibility is a momentary interaction such as occurs in an impact. Besides being a dynamic method of measuring mass, Mach's procedure has the virtue of not specifically assuming the concept of force. Given the definitions of mass, length, and time, the unit of force can then be defined in accordance with the law of motion as the effort required to produce a unit acceleration of a unit mass. Furthermore, the law of action and reaction follows from the law of motion and M·ach's experimental proposition in the definition of a mass ratio. Nevertheless, Mach's procedure has its shortcomings. First is the practical difficulty of finding a system of two interacting masses that is dynami


t
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1
SEC. 1-4 INTRODUCTORY CONCEPTS 23
cally isolated. The earth's gravitational attraction would be an extraneous influence for experiments performed near the earth, although the effect could be minimized by confining the motion to a horizontal plane by constraints with very small friction. Another possibility is to cause the masses to hit each other. As will be seen lnore clearly in Sec. 4-7, the impact will cause very large accelerations of short duration, and during this interval, the effects of other influences will be relatively lninor. But these procedures do not appear to be practical for mass determinations of ordinary objects.
A second objection to Mach's procedure is its tacit assumption that the accelerations are measured relative to an inertial frame of reference. Since the criterion for an inertial fralne is that Newton's law of motion applies in this frame, we see that, again, the definition of mass assumes the validity of a law which presupposes a definition of lnass. A dynamic method of Inass Ineasurement such as Mach proposes would seem to be the least objectionable on theoretical grounds. But, as a practical matter, the measurement of mass by a weighing procedure will undoubtedly continue to be widely used.
Another important point in the discussion of mass is that Newtonian mechanics assumes the principle of conservation of mass. Thus, while a certain quantity of water may be converted to the vapor phase, its mass is not changed in the process. Similarly, if a rocket is fired, the total mass of the rocket body and ejected gases remains constant. This principle enables one to consider a wide variety of physical systems as a group of particles, constant in number, with the mass of each particle remaining the same in spite of changes in its motion or its physical state.
Force. We have seen that if standard units of length, time, and mass are chosen, then the corresp.onding unit of force is determined in accordance with Newton's law of motion. Thus, a unit force applied to a unit mass results in a unit acceleration. Now, suppose that .force, length, and time are chosen as fundamental " dimensions. What is a suitable operational procedure for defining a standard unit of force?
We shall define a unit force in terms of the effort required to produce a certain extension in a given standard spring under standard conditions. Then, noting that springs connected in series transmit equal forces, we can mark the extensions produced by the standard force in each of a set of similar springs, thereby producing a group of standard springs. Next, we observe that if n standard springs are connected in parallel and if each is stretched the standard amount, then the force transmitted by the n springs is n standard force units. So,·' by connecting a given spring in series with various numbers of parallel standard springs, the given spring may be


I
.1
24 INTRODUCTORY CONCEPTS CHAP. :1
calibrated in standard force units and subsequently used for force measurements. This definition of force, which is based upon an elastic deformation, has the advantage of not requiring the measurements to be made in an inertial system if the spring is assumed to have negligible mass. In a practical· case, the so-called inertial forces associated with the mass of the spring must be small compared to the force transmitted by the spring. Nevertheless, there are objections to the choice of force as a fundamental dimension. But before we discuss these objections, let us first note that the forces that concern us in mechanics are of two general types, namely, contact forces and field forces. Contact forces refer to a direct mechanical push or pull which is transmitted by material means. For example, a rod or a rope, or even air or water pressure could be used to apply contact forces to a body at its surface. On the other hand, field forces are associated with action at a distance, such as in the cases of gravitational and electrical forces. Also, field forces are often applied throughout a body rather than at the surface. Now, difficulties arise when one attempts to apply to field fot:ces the concept of a force as a push or pull that is measured by an elastic deflection. For example, if a body is falling freely in a uniform gravitational field, the gravitational force does not produce any elastic deflection and therefore this method cannot be used to measu-re it. Actually, the gravitational force is often found by counteracting it with a contact force such that the acceleration is zero and then measuring this contact force. This is the process of weighing the body. The assumption is made, of course, that the gravitational force is not a function of v~locity or acceleration. More generally, field forces are calculated by observing accelerations and obtaining the corresponding forces from Newton's law of motion. In other words, accelerations are observed and then forces are inferred in accordance with the law of motion. The concept of force as a fundamental quantity in the study of mechanics has been criticized by various scientists and philosophers of science from shortly after Newton's enunciation of the laws of motion until the present time. Briefly, the idea of a force, and a field force in particular, vv'as considered to be an intellectual construction which has no real existence. It is merely another name for the product of mass and acceleration which occurs in the mathematics of solving a problem. Furthermore, the idea of force as a cause of motion should be discarded since the assumed cause and effect relationships cannot be proved. We shall adopt the viewpoint that the existence of contact forces can be detected and measured by springs or by other means of measuring elastic deformations. Field forces will be calculated from observed accelerations, using Newton's laws of motions, or else from rules governing the force that have been established from such observations. For the most part, we
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SEC. 1-5 INTRODUCTORY CONCEPTS 25
shall avoid questions of cause and effect. For example, suppose a stone is tied to the end of a string and is whirled in a circular path. Does the tension in the string cause the stone to follow a circular path, or does the motion in a circular path cause the tension in the string? Conceivably, either view. point could be taken. But generally it is preferable to note that forces and accelerations occur silTIultaneously and neither is specifically cause nor effect. We have seen that the definitions of the fundamental units of ll1echanics depend to some extent upon Newton's laws of motion and these are the laws to be demonstrated. Thus, the logic has a certain circularity. Nevertheless, the validity of the laws of lTIotion has been established for mechanical systems with a wide range of velocities and spatial dimensions. So long as the distances involved are larger than atolTIic dimensions and the velocities are much smaller than the velocity of light, the fundamental laws of Newtonian mechanics apply remarkably well.
1-5. D'ALEMBERT'S PRINCIPLE
Newton's law of motion for a particle is given by Eq. (1-41). Suppose we write it in the form
'F - rna == 0 (1-49)
where F is the sum of the external forces acting on the particle, rn is its mass, and a is the acceleration of the particle relative to an inertial reference frame. Noy.;, if we consider the term -rna to represent another force, known as an inertial force or reversed effective force, then Eq. (1-49) states that the vector sum of all forces, external and inertial, is zero. But this is just the form of the force summation equation or" statics and the methods of analysis for statics problems apply to it, including more advanced methods, such as virtual work (Sec. 6-4). Thus, in a sense, the dynamics problem has been reduced to a statics problem. Briefly, then, d'Alembert's principle states that the laws of static equilibrium apply to a dynamical system if the inertial forces, as well as the actual external forces, are considered as applied forces acting on the system. Of course, one still must solve the differential equations of dynamics rather than the algebraic equations of statics, but the point of view in setting up the equations is similar to that of statics. '. Considerable care must be used in setting up the equations of motion by means of d'Alembert's principle. In particular, the inertial forces should "not be confused with the external forces comprising the total force F that is applied to the particle. In order to keep the distinction clear on force diagrams, we will des"ignate inertial forces by a dashed arrow and external forces by a solid arrow, as shown in Fig. 1-10. By external forces, we mean contact forces and gravitational or other field forces applied to the particle.


l: .1
26 INTRODUCTORY CONCEPTS CHAP. 1
Another approach is to think of the inertial force as a reaction force exerted by the particle on its surroundings in accordance with the law of action and reaction. This viewpoint is convenient when the motion of a particle is given and one desires to calculate the force it exerts on the
F
ma m
- + - - - -o--~.""
Fig. 1-10. Inertial force for an accelerating particle.
remainder of the system. For example, if a particle is whirled in a circular path by means of ~. string attached to a fixed point, the·n the inertial force of the particle is the so-called centrifugal force which is equal to the tensile force in the string. On the other hand, the external force on the particle is the centripetal force of the string acting radially inward toward the fixed point at the center of the circular path. By the law of action and reaction, the centripetal force is equal in magnitude to the centrifugal force but is opposite in direction.
Example 1-1. A particle of mass m is supported by a massless wire of length I that is attached· to a point 0; of a box having a constant acceleration a to the right (Fig. 1-11). Find the angle 0 corresponding to a condition of equilibrium, that is, such that the particle will remain at constant
a•
!
Ito
9 ma ~--Om
m
Img
(0) ( b)
Fig. 1-11. Simple pendulum in an accelerating box.
f) if released from this position. What is the tensile force T in the wire in this case? [The direction of the acceleration due to gravity is shown by the vertical arrow' in Fig. 1-11(a).] . The external forces acting on the particle are the weight rng and the force T due to the wire, as shown in Fig. 1-II(b). In addition, one notes
that the particle has an acceleration a to~the right for constant 0, and therefore the inertial force is of magnitude rna and is directed to the left. Now, according to d'Alembert's principle, the external and inertial
(


I
I
SEC. 1-5 INTRODUCTORY CONCEPTS 27
forces must add vectorially to zero. So, taking horizontal components, we obtain
T sin () - rna == 0
Similarly, taking vertical cOlnponents,
T cos () - rng == 0
Solving these two equations, we obtain
() = tan - 1 ( ; )
and
REFERENCES
1. Housner, G. W., and D. E. Hudson, Applied Mechanics-Dynamics, 2nd ed. Princeton, N.J.: D. Van Nostrand Co., Inc., 1959.
2. Jammer, M., Concepts of Space. Cambridge, Mass.: Harvard University Press, 1954.
3. , Concepts of Force. Cambridge, Mass.: Harvard University Press, 1957.
4. , Concepts of Mass. Cambridge, Mass.: Harvard University Press, 1961.
5. Lass, H., Vector and Tensor Analysis. New York: McGraw-Hill, Inc., 1950.
6. Lindsay, R. B., and H. Margehau, Foundations of Physics, New York: Dover Publications, Inc., 1957.
7. Mach, E., The Science of Mechanics, 6th ed. LaSalle, Ill.: Open Court Publishing Company, 1960.
8. Phillips, H. B., Vector Analysis. New York: John Wiley & Sons, Inc., 1933.
PROBLEMS
1-1. Forces P and Q are applied to a particle of mass m. The magnitudes
of these forces are constant but the angle e between their lines of action can be varied, thereby changing the acceleration of the particle. If the maximum possible
acceleration is three times as large as the minimum acceleration, what value of e is required to attain an acceleration which is the mean of these two values? 1-2. In terms of a fixed cartesian system, the position of auton10bile A is
ra == 100ti ft and the position of airplane P is rp == 1000i + 300tj + 2000k ft where the time t is measured in seconds. Find: (a) the velocity of P relative to A and (b) the minimum separation between P and A. 1-3. Given the following triad of unit vectors:


28 INTRODUCTORY CONCEPTS
C1 == II i + 12 j + 13 k
C2 == mlB + m 2i + m3 k
C3 == n1i + n2i + n3 k
CHAP. 1 .
where i, j, k are the usual cartesian unit vectors. (a) Write 3 equations involving the l's, m's, and n's which apply for any unit triad. (b) What additional equation applies if eh e2, and e3 are coplanar? (c) What equations apply if C1, C2, and e3 are mutually orthogonal?
1-4. Consider the vector A == Axi + Ayj + Azk. Find the components AhA2' A3 of the vector A in a skewed coordinate system whose axes have directions specified by the following unit vector triad:
1-5. If units of length, mass, and force were chosen as the fundamental units, what would the dimensions of time and acceleration be? 1-6. Suppose that units of force, length, and time are chosen such that the density of water and the acceleration of gravity are both of unit magnitude. If the pound is taken as the unit of force, what are the sizes of the units of length and time? 1-7. At a certain moment during reentry, a satellite is moving horizontally in the earth's atmosphere. The satellite carries an accelerometer which measures the .contact force acting on a sman mass moving with the satellite. If the accelerometer reading is 64 ft/sec2 in a direction opposite to the velocity of the satellite, and if the acceleration of gravity at that point is 32 ft/sec2, what is the absolute acceleration of the satellite? If the speed of the satel
mm
Fig. PI-8
a
III D
lite is 10,000 ft/sec, what is the angular rate of rotation of its velocity vector? 1-8. The particles at Band C are of equal mass m and are connected by strings to each other and to points A and D as shown, all points remaining in the same horizontal plane. If points A and D move with the same acceleration a along parallel paths, solve for the tensile force in each of the strings. Assume that all points retain their initial relative positions.


2
KINEMATICS OFA PARTICLE
Kinematics is the study of the motions of particles and rigid bodies, disregarding the forces associated with these motions. It is purely mathematical in nature and does not involve any physical laws such as Newton's laws. In this chapter we are concerned primarily with the kinematics of a particle, that is, with the motion of a point. Depending upon the circumstances, we may at times choose to consider the point as being attached to a rigid body and at other times as being the location of an individ ual particle. In any event, we shall be interested in calculating such quantities as the position, velocity, and acceleration vectors of the point. In discussing the motion of a point, one must specify a frame of reference, since the motion will be different, in general, when viewed frOlTI different reference frames. We saw in Sec. 1-2 that Newton's laws of motion apply in a particular set of reference frames known as inertial frames. These frames, nonrotating but translating uniformly relative to one another, thus constitute a special or preferred set in writing the dynamical equations of a particle. From the viewpoint of kinematics, however, there are no preferred frames of reference since physical laws are not involved in the derivation of kinematical equations. Thus, in the strict sense, all motion is considered to be relative and no reference frame is more fundamental or absolute than another. Nevertheless, we shall at times refer to -absolute motions because of the dynamical background, or for ease in exposition.
2-1. POSITION, VELOCITY, AND ACCELERATION OF A POINT
The position of a point P relativ~ to the X YZ reference frame (Fig. 2-1) is given by the vector r drawn from the origin 0 to P. If the point P moves on a curve C, then the velocity v is in the direction of the tangent to the curve at that point and has a magnitUde equal to the speed1 with which it moves along the curve.
lSpeed is a scalar quantity. and is equal to the magnitude of the velocity. Sometimes the term velocity is used loosely in the same sense, the meaning generally being clear from the context.
29


iII!,:iI'
\.,
!.,
30 KINEMATICS OF A PARTICLE CHAP: 2
zz
~----------------y ~----~--------y
XX
Fig. 2-1. Position and velocity vectors
of a particle P as it moves along curve
c.
Fig. 2-2. The hodograph showing velocity and acceleration vectors.
Considering the velocity v as a free vector, we can imagine that it is drawn with its origin at the origin of the coordinate system at each instant of time, as shown in Fig. 2-2. The path C' of the tip of the velocity vector v is then known as the hodograph. Now, in general, the velocity of the tip of a vector drawn from the origin in a given coordinate system is simply the time derivative of the vector in that system. So the velocity of the tip of the vector v along the hodograph is the acceleration a of point P relative to the XYZ system. Hence we summarize by stating that the velocity of point Pis
dr
v ==dt- (2-1)
and the corresponding acceleration is
dv d 2r
a == dt == dt2 (2-2)
Note that a knowledge of the path C does not determine the motion unless the speed along the path is known at all times. Similarly, a knowledge of C' does not determine the motion unless one knows the starting point in space as well as the rate of travel along the hodograph at all times. If
both C; and C' are known and are compatible, then the motion is nearly always' determined, an exception being the case of rectilinear motion at a varying speed. We always assume that C is continuous, corresponding to a finite velocity of the point.


/
SEC. 2-2 KINEMATICS OF A PARTICLE 31
2-2. ANGULAR VELOCITY
In studying the kinematics of a particle, we are primarily concerned with the translation of a point relative to a given reference fra me, that is, with the position vector of the point and with its derivatives with respect to time. On the other hand, the general motion of a rigid body involves changes of orientation as well as changes of location. It is the rate of change of orientation that is expressed by means of the angular velocity vector. Consider the motion of the rigid body shown in Fig. 2-3 during the infinitesimal interval ~t. Anticipating the results of Chasles' theorem (Sec. 7-8), we note thaLJhe_jnfinitesimaL__displacement.. during_this...,intery.aL_can. l:?_e.~Qnsid.ered...as_a.__tr_q.n~latjQggLQi~I?lg.£~m~nt.A..s_.Q{_.alLI?ojJ!!~_in_J.b~b_Qdy
:r!!!~__~_IQt~Ji.Q!H!.L_displacement... ~(J_~ab.o.ut....an..,.axis_thr.o_ugh__th.e..11(JS.g.._.J2..Qjl1L.
.cl~..._fj~.~~..j!1.!he..._bQdy. The order of performing the translation and rotation is ilTIlTIaterial, but Fig. 2-3 shows the case where the translation occurs first. Thus a typical point P moves to pI and the base point A n10ves to A', each undergoing the same displacement ~s. Then the infinitesimal rotation ~(J occurs, moving pI to P" while A' does not move since it is on the axis of rotation. The infinitesimal angular displacement ~(J is a vector whose magnitude
z
J--~r-------~~----y
x
Fig.. 2-3" Infinitesimal displacement of a rigid body.


i.'
32 KINEMATICS OFA PARTICLE CHAP. 2 -~
is equal to the angle of rotation and whose direction is along an axis determined by those points not displaced by the infinitesimal rotation. The sense. is in accordance with the right-hand rule. We define the angular velocity Q) as follows:
1· /lfJ
(jJ == I m 
At-.O Ilt (2-3)
Now, if we should analY-~~tih~__s_ame_infinitesimal-.dis.ple.£~~t_Qf..Jhe.· bo..Q.y__bJJt_choo.se_.a.-.different---base-,..point, ...only.Jhe~.translatioll.aLApart... of .the motion would lt~t.c.hanged;. the rotational displacementwo'JJld be. identical..
tJ:iuswe-~a;;~e~ that the angul~r velocity is a property of the b~y ~~;"~>';h~i~
and is not dependent upon the choice of a base point. Therefore the angular velocity Q) is a free, vector.
N ote 1hat...measurements-"of~the..angular~yelQcitY...Qf"~...Q.Qgy~~,~yj~w.~~Lfr.QllL different reference frcgne~Hl,J..rL~g~11~IJ!1",,.J?Ig9!l~e. different results. There
~~............_,..".......,....__~....~""....~~t._..,~-="t. - .. "' ........ ~""""':~"'''"'::'-''~''''e.:.....~"l.""",,~.!:v-t.o'''''~,:j:''''''''~'''''':;''''''''\'''.. ,., 
fO~Jy.helLKiyjng.Jh.~.~~ngylg.Lyelo~ily',_th~tx~f.t:!renG~_,Jnlm~Ls.houlcLbe_slat.~.Q Qf_ cl~llrly_irrn~li~~l., An inertial frame is usually assumed if no statement is made to the contrary. Also, recall again that angular velocity refers to the motion of a rigid body, or, in essence, to the motion of three rigidly connected points that are not collinear. The term has no unique meaning for the motion of a point or a vector (or a straight line) in three-dimensional space. Finally, the angular velocity vector wil~ u~u~J1Y__flH!J1g~.~J2Qlh.jts.
~itu.d.e.Jl.nd_..dir.es;Ji.QJ1.J;_Qntinuo.u_sJY-._~Jth.J.i.me.:.Qnly. in ...~Q.~_._~LmQl~L.cases .~. is the motion confined to rotation about an axis fixed in space or in the body.
..... _____•__ -.-.~-.---. -~.- ..'... ---...-- ....... ···.··'~··r···. -,- .. ,...... "~~.~ ........-.. ,', -~" '..... - .......,...... _-_..._. _. . . '.. .
z
y
X
Fig. 2-4. Rigid body rotation about a fixed point.
2-3. RIGID BODY MOTION ABOUT A FIXED POINT
As-~~....h~y..~_ s~~.!.'! ,-.__!h~._LQtatiQI1~J
motion relative to a given ~y.~Jem
-doe~-"~·~t--denen(r-uQ·on"the·-ch~ice.of
__..•__ ._0-,.•' .... - ," .• " •• r ......_....._....,. "'..... ........... _....,......<-•••••• '. -"
!he..E_as~_.E.~int. So, in order to concentrate upon the rotational aspects of the motion, (L~s_u.ro~_t!l.'!.tJbebase noint of the rigid body is fixed at the
~.-~ •• - . . . ~" ................ , _... _ .. po~.......--.......... --0 ....... , .... - . _ _.....-.,.. .,,........... " ...... _ ......., ... - . • . • • • . • ~ •• ~., • •
QI!.gi.o~Q_QfJhe.cartes.ian"system.X.yZ, as shown in Fig. 2-4. Let us calculate the velocity v relative to the X YZ system of a point P that is fixed in the bod y. If the X YZ system is fixed in inertial space, then v is the absolute velocity of point P and Q) is the absolute angular velocity of the body.


I
"j""
SEC. 2-3 KINEMATICS OF A PARTICLE 33
In general, the rotation at any instant is taking place about an axis passing through the fixed base point. This axis is known as the instantaneous axis of rotation. Consider first the case where the rotation takes place about an axis that is fixed in the XYZ system. Then lIJ has a fixed direction and
the path of point P is a circle of radius r sin e, as can be seen from Fig. 2-4. The speed with which point P moves along the circle is
S == ror sin e (2-4)
where s is the displacement of P along its path, ro is the magnitude of (J),
r is the constant length of the position vector drawn from 0 to P, and () is
the angle between (J) and r. The velocity of P is of magnitude s and is directed along the tangent to the path. Thus we can write
v==(J)Xr (2-5)
In the general case where co and e may vary with time, the displacement ~s of point P during the infinitesimal interval fl.t is
~s == ror ~t sin e (2-6)
where higher-order terms have been neglected. It is assumed that ~ro and
b..e approach zero as the interval ~t approaches zero, implying that m is finite. Of course, ~r is zero, since point P and the base point 0 are fixed in the body. Dividing Eq. (2-6) by ~t and letting ~s and fl.t approach zero, we again obtain Eq. (2-4). Furthermore, the direction of v is again normal to (J) and r. Hence we see that Eq. (2-5) is valid for the general case of rigid-body rotation about a fixed point. Differentiating Eq. (2-5) with respect to time, we obtain an expression for the acceleration a of point P.
or, noting that t is the velocity v and using Eq. (2-5),
a == (J) X «(J) X r) + mX r (2-7)
where, of course, all vectors are measured relative to the X YZ system. It can be seen that the term (J) X (lIJ X r) in Eq. (2-7) is a vector that is directed radially inward from point P toward the instantaneous axis of rotation and perpendicular to it. It is called the centripetal acceleration. The
term mX r is often called the tangential acceleration. Note, however, that the so-called tangential acceleration is in a direction tangent to the path of P only if OJ is parallel to the'plane of (J) and r. This would occur, for example, if (J) retains a constant direction and changes in magnitude only.


34 KINEMATICS OF A PARTICLE CHAP. -2
2-4. TIME DERIVATIVE OF A UNIT VECTOR
We have obtained the ,derivative with respect to time of the position vector r of a point P in a rigid body that is rotating about a fixed point
o (Fig. 2-4). A similar situation exists in the calculation of the rates of change of unit vectors. As was the case with the position vector to point P, the unit vectors are each of constant length. Furthermore, we have seen that the time derivative of a vector can be interpreted as the velocity of the tip of the vector when the other end is fixed. So let us calculate the velocities of the unit vectors C 1, C2 , and Cg drawn from the origin of the fixed system XYZ and rotating-together as a rigid body with an absolute angular velocity (J) (Fig. 2-5).
z
JIIi::------- Y
x
Fig. 2-5. The rotation of a unit vector triad.
we obtain
C1 C2
e
1 = (01 CO2
10 Similarly,
C1 C2
e
2 = (01 CO2
01 and
C1 C2
e
g = (01 lO2
00
From Eq. (2-5), we see that the velocities of the tips of the unit vectors, that is, their time derivatives, are
e
2 = (J) X C2 (2-8)
e
3 = Q) X C3
For the case where C1, C2, and Cg form an orthogonal 'set of unit vectors, that is, C1 X C2 = C3, and so on, we can perform the vector multiplications using the determinant expression of Eq. (1-27). Writing the angular velocity (J) in the form"""
(J) = (01 C 1 + (02C2 + (OgCg (2-9)
e
3j
~3 = 003 e2 - 002 e3 (2-10)
Cg
(Og = (OlCg - (OgC! (2-11)
0
Cg
lOa = lO2CI - lOl C 2 (2-12)
1


SEC. 2-5 KINEMATICS OF A PARTICLE 35
In the particular case where we consider a rotating triad of cartesian unit vectors i, j, and k, we can make the substitutions el == i, e2 == j, and e3 == k. Noting that
(J) == G>xi + G>y,i + G>zk (2-13)
where OO x , roy, and OOz are the components of (J) in the directions of i, j, and k, respectively, (not along the X, Y, and Z axes), we obtain from Eqs. (2-8), (2-9), (2-10), and (2-11) that
i == (J) X i == OOzj - ooll k
j == (J) X j . OOxk - G>zi (2-14)
k == (J) X k == OO1Ji - OOxi
I t can be seen that, in each case, the time derivative of a unit veQtor lies in a plane perpendicular to the vector, in accordance with the definition of a cross product. Observe that in each case we have calculated the rate of change of a unit vector with respect to a fixed coordinate system but have expressed the result in terms of the unit vectors of the moving system. Since this approach will be used extensively in our later work, the terminology should be made clear. The terms relative to or with respect to a given systeln mean as viewed by an observer fixed in that system and moving with it. On the other hand, the term re.ferred to a certain system means that the vector is expressed in terms of the unit vectors of that system. For example, the absolute acceleration of a certain particle can be given in terms of the unit vectors of a fixed or of a moving coordinate system. In either event, we are considering the same vector but are merely expressing the result in two different forms. On the other hand, the acceleration of the particle relative to a fixed coordinate system and the acceleration relative to a moving co-\r ordinate system would, in general, be quite different vectors.
2-5. VELOCITY AND ACCELERATION OF A PARTICLE IN SEVERAL COORDINATE SYSTEMS
Cartesian Coordinates. Suppose that the position of a particle P relative to the origin 0 of the cartesian system xyz is given by the vector (Fig. 2-6)
r == xi + yj + zk (2-15)
wherei, j, and k are unit vectors fixed in the xyz system. Differentiating Eq. (2-15) with respect to time, we find that the velocity of point P relative to the xyz system is given by
v == t == xi + Yi + ik (2-16) i .
!I[,


"
;II,
i
I'
i;'1::
!I',;I
36 KINEMATICS OF A PARTICLE
z
p
.J-j z
y
I
Fig. 2-6. Cartesian coordinates and unit vectors.
CHAP. 2 .'
z
ez
p t:e~
e,
z y
x
Fig. 2-7. Cylindrical coordinates and unit vectors.
where we note that the time derivatives of the unit vectors i, j, and k are all zero since they are fixed in the given cartesian system. Similarly,- differentiating Eq. (2-16) with respect to time, we find that the acceleration of P relative t6 the xyz system is
a == xi + yj + zk (2-17)
For the common case where the cartesian system is assumed to be fixed in an inertial frame, the preceding expressions refer to the absolute velocity and acceleration.
Cylindrical Coordinates. Now consider the case where the position .of point P relative to the xyz system shown in Fig. 2-7 is expressed in terms of the cylindrical coordinates r, </>' and z. In this case, the position vector of Pis
(2-18)
Note that the 'unit vectors en eq)) and ez form a mutually orthogonal triad whose directions are given by the directions in which P moves for small
increases in r, 1>, and z, respectively. As before, we obtain the velocity by . finding the derivative of r with 'respect to time:
(2-19)
It can be seen that er and eq, change their directions in space as P moves through a general displacement, but ez is always parallel to the z axis. Further, the changes in the directions of Cr and eq, are due solely to changes
in cp, corresponding to rotations about the z axis. Thus the absolute rotation


I
SEC. 2-5 KINEMATICS OF A PARTICLE
rate of the unit vector triad is P( J.. 2,--;0
0) == ,.-.ez
and, in accordance with Eq. (2-8), we find that
er == 0) X er == ¢e1>
e1> == 0) X e1> == -¢er
e
z == 0) X ez == 0
So from Eqs. (2-19) and (2-21), we obtain
37
(2-2@)
(2-21)
v == fer + r¢e1> + ie:: (2-22)
Another differentiation with respect to time results in the acceleration
a == r == rer + (feb + r¢)e1> + zez + fer + r¢e1> + iez
which can be, sinlplified with the aid of Eq. (2-21) to yield
a == (r - rcfo2)e r + (r¢ + 2f¢)e1> + zez (2-23)
SpJzerical Coordinates. The position vector of the point P in terms of spherical coordinates and the corresponding unit vectors is simply
r == re r (2-24)
since, it) this case, the unit vector er is defined to be in the direction of the position vector r (Fig. 2-8). Differentiating Eq. (2-24) with respect to time, we obtain the velocity
In order to ev~luate en eo', and e1>' we note first that changes in the directions of the unit vectors can arise because Qf changes in () or 1>: An increase in () rotates the unit vector triad about an ~xis having the direction e1>, whereas an increase in 1> corresponds to a rotation about the z axis. Thus the total rotation
(2-25)
z
rate of the unit vector triad is Y
0) == Be1> + cfoez (2-26)
where ez is a unit vector in the direction of the positive z axis. Since e1> and ez are perpendicu
lar, it is relatively easy to express ez in terms of the remaining unit vectors:
x
Fig. 2-8.. Sphericiil coordi~tes and unit vectors. I


'II,:!I"'
Ii
38 KINEMATICS OF A PARTICLE CHAP. 2.
Cz = cos f) e, - sin f) Co
Therefore, from Eqs. (2-26) and (2-27), we obtain
(2-27)
(J) = ¢ cos 0 Cr - ¢ sin () Co + f) e4> (2-28)
The unit vectors Cn Co, al1:d c4> are mutually ortho,gonal, so we can use the determinant method of evaluating the cross products involved in obtaining en eo, and c4>. From Eqs. (2-8) and (2-28), we have
Cr
e
r = (I) X er = ¢ cos (j
1
eo
- ¢ sin ()
o
e4>
() (2-29)
o
and, using a similar procedure to evaluate eo and e4>' we can summarize as follows:
eo = -Ocr + ¢ cos 0 c4> (2-30)
e4> = - ¢ sin () Cr - ¢ cos f) Co
Finally, from Eqs. (2-25) and (2-30), the velocity of point P is
v = rer + rOco + r¢ sin f) e4> (2-31)
In order to find the acceleration of P, we differentiate again, obtaining
a = v = rer + (rO + re) Co + (r¢ sin (j + r1> sin f) + rf)¢ cos ()) (p (2-32)
+ rer + riJeo + r¢ sin () e4>
Substituting Eq. (2-30) into (2-32) and collecting terms, the result is
a = (r - r02- r¢2 sin2 0) er + (re + 2fO - r¢2 sin 0 cos f)) eo
+ (r¢ sin f) + 2r¢ sin f) + 2rO¢ cos f)) e4> (2-33)
Tangential and Normal Components. The velocity and acceleration of a point P as it moves on a curved path in space may be expressed in terms of tangential and normal components. Let us assume that the position of P is specified by its distance s along the curve from a given reference point. From Fig. 2-9, it can be seen that, as P moves an infinitesimal distance ds along the curve, the corresponding change in the position vector r is
dr = dset
or
dr
et =d-s (2-34)
where et is a unit vector that is tangent to the path at P and points in the direction of increasing s. The velocity of P is
dr . dr · (2 35)
v = dt = s ds = set 
\
(
\
\


SEC. 2-5
To find the acceleration of P, we differentiate Eq. (2-35) with respect to time, obtaining
(2-36)
Before evaluating et explicitly, we note that, in general, et changes direction continuously with increasing s. The direction of the derivative
KINEMATICS OF A PARTICLE 39
z
det/ds is normal to et, since the y magnitude of et is constan-t. This is known as the normal direction at P and is designated by the unit vector en. The plane of et and en at any x point P is called the osculating plane. Fig. 2-9. Tangential, norn1al, and Therefore, any Inotion of et Inust binormal unit vectors for a curve in take place in the osculating plane at space.
that point, and hence the vector et lies in this plane and points in the ell direction. Its magnitude is equal to its rotation rate (i)b about an axis perpendicular to the osculating plane. Thus
(2-37)
where rob is the rotation rate of et about an axis in the binonnal direction ebo The binormal unit vector eb is perpendicular to the osculating plane and is given by
(2-38)
thus completing the orthogonal triad of unit vectors at P. The angular rate (i)b can be expressed in terms of the radius of curvature
p and the speed s of the point P as it moves alorig the curve.
s
rob == -p (2-39)
Note that the vector pen gives the position of the center of curvature C relative to P. Of course, the position of the center of curvature changes, in general, as P moves along the curve. From Eqs. (2-37) and (2-39), we obtain
.s
et == - en
p (2-40)
and, substituting this expression into Eq. (2-36), we have
'2
.. s
a == set + - en
p (2-41'


40 KINEMATICS OF A PARTICLE CHAP. 2
The acceleration component s is the tangential acceleration and the component $2/P is the normal or centripetal acceleration. The time derivatives of the unit vectors can also be obtained in terms of the angular rotation rate OJ of the unit vector triad. We have
(2-42)
Note that the normal component of OJ is zero. This can be explained by the fact that en was defined to lie in the dIrection of Ct. Any normal component of OJ would result in a binormal component of Ct, in conflict with the original assumption. Of course, (f)t does not influence Ct. From Eqs. (2-8) and (2-42), we find that
Ct = OJ X et = (f)ben
en = OJ X en = - (f)bet + (i)t eb
c
b = OJ X eb = - (i)ten
(2-43)
We have seen that (i)b is equal in magnitude to Ct and is a function of the speed $ and the radius of curvature p. Similarly, (i)t is equal in magnitude to eb and is a function of $ and deb/ds. Thus,
(2-44)
where the magnitude of deb/ds is called the torsion of the curve. It can be seen that (j)tis also the rotation rate of the osculating plane.
2-6. SIMPLE MOTIONS OF A POINT
Circular Motion. Consider the motion of a point P in a circular path of radius a (Fig. 2-10) in the fixed xy plane. The motion will be described in terms of the polar coordin~tes (r, 0) and the corresponding unit vectors. y ' In this case, the coordinate r is the
~--'-----+----- x
Fig. 2-10. Circular motion of a point.
scalar constant a. Thus, we have
and
Using the notation that
00=8
(2-45)
(2-46)
(2-47)
we find that the speed along the path
IS
v = aO = a(i) (2-48)
or


SEC. 2-6 KINEMATICS OF A PARTICLE 41
(2-49)
in agreement with Eq. (2-46), since
I ,. e.
,,{ 0 .-.. a,.UJ. G-
~'- (" - /
ev~ .:: '.J) e ('} (2-50)
The acceleration IS obtained by differentiating Eq. (2-49) with respect to time:
(2-51)
or, noting that
(2-52)
we obtain
2+.
a·== - aro er awee (2-53)
An alternate form is obtained by substituting for w from Eq. (2-48), obtaining
v2 •
a == - - er + vee
a (2-54)
The first term on the right represents the centripetal acceleration; the second term is the tangential acceleration. Note that the same result is obtained by using the tangential and normal unit vectors of Eq. (f-41)
where et == eo, en == - en P == a, and s == v. ~.
Helical Motion. Next consider the velocity and acceleration of a particle moving along a helix which is given in terms of cylindrical coordinates by
r == a,
where k is the tangent of the helix angle and a is constant (Fig. 2-11). Now let
(2-56)
Differentiating the expression for z in Eq. (2-55), we obtain
i == ka¢ == karo (2-57)
and, from Eqs. (2-22), (2-56), and (2-57), we find that the velocity is
v . t == aweq, + karoez (2-58)
The acceleration is found by a similar evaluation of Eq. (2-23), yielding
a == - aro 2er + arocq, + karocz
(2-59)
For the case where the point P
z == ka!fJ (2-55)
z
ta~-1 k
~--~---------------y
x
Fig. 2-11. Helical motion of a point.


·42 KINEMATICS OF A PARTICLE CHAP. 2
moves at uniform speed along the helix, the, angular acceleration cO' is zero and only the radial acceleration remains. ' The radius of curvature at a point on the helix can be calculated by equating the magnitude of the radial acceleration found in Eq. (2-59) to the normal or centripetal acceleration previously found in Eq. (2-41). Noting that the speed is
(2-60)
we find that
S2 == a2 ro2 (1 + k 2) == aro2
pp
from which we obtain
(2-61)
For example, if the helix angle is 45°, the radius of curvature is p == 2a and the center of curvature is diametrically opposite P on the surface of the cylinder r == a. In this case, the locus of the centers of curvature is a helix with the same radius and helix angle.
Harmonic Motion. A common type of particle motion is that in which the particle is attracted toward a fixed point by a force that is directly proportional to the distance of the particle from the point. Consider first the one-dimensional case in which the motion takes place along the x axis of a fixed cartesian system because of an attracting center at x == O. From Newton's law of motion as given by Eq. (1-41) or Eq. (1-43), we recall that the acceleration of the particle P is proportional to the applied force. But we' have assumed that the applied force is proportional to x and is directed toward the origin. So we can write the differential equation
(2-62)
where (02 is a constant. The solution can be written in the form
x == A cos (rot + a) (2-63)
where, in general, the constants A and a are evaluated from the values of
x and x at t == O. If, however, we choose to measure time from the instant when the particle is at the positive extreme of its motion, then a == 0, and
x == A cos rot (2-64)
Motion wherein the position oscillates sinusoidally with time is known as simple harmonic motion (Fig. 2-12). This motion is periodic with a period
T== 27t
(0 (2-65)


...."
SEC. 2-6
where ro is usually measured in radians per second and is known as the circular frequency. 2
Differentiating Eq. (2-64), we obtain expressions for the velocity and acceleration.
x == - Aro sin rot
= Aco cos ( cot + ~) (2-66)
and
KINEMATICS OF A PARTICLE 43
x r-r--+i
II
~~+---~--~--~~---f
Fig. 2-12. Simple harmonic motion.
x == - Aro2 cos rot == Aro2 cos (rot + 7t) (2-67)
It can be seen that each differentiation with respect to time results in a sinusoidal oscillation of the same frequency, but the amplitude is multiplied by ro and the phase angle is advanced by 7t/2 radians. Thus the extreme values of the displacement and acceleration occur at the moment when the velocity is zero, and vice versa. Now consider the case of two-dimensional harmonic motion. Suppose the particle P moves in the x direction in accordance with Eq.. (2-62), as discussed previously. But it also moves independently in the y direction according to the following equation:
Y == - ro2y (2-68)
where the value of ro is assumed to be the same as for the x motion. The solution of this differential equation can be written in the form
y == B cos (rot + {3) (2-69)
where the angle {3 is not zero, in general, since the time reference was chosen such that a == 0 in Eq. (2-63). Using vector notation, we find that the position of P is
r == xi + yj == A cos rot i + B cos (rot + ,8) j (2-70)
where i and j are the usual cartesian unit vectors. Differentiating Eq. (2-70) with respect to time, we see that the velocity is
v == t == - Aro sin rot i - Bro sin (rot + ,8) j (2-71)
and the acceleration is
a == v == - Aro2 cos rot i - Bro2 cos (rot + {3) j (2-72)
The path of the particle P in the xy plane may also be found by first noting from Eq. (2-64) that
2The term frequency refers to the number of cycles per unit time, or T-l. Thus (i) = 27tf In this book, we shall, for convenience, generally use the circular frequency (i) rather than
f and often shall refer to Ct) as the frequency.


44 KINEMATICS OF A PARTICLE CHAP. 2'
cos rot = ~.... (2-73)
Then, using Eqs. (2-69) and (2-73) and the trigonometric identity for cos
(rot + (3), we see that
sin rot = (sin tS) - I (~ cos tS - ~)
Squaring Eqs. (2-73) and (2-74) and adding, we obtain
(sintS)-2 G: + ~: - 2 ~ ~ costS) = 1
which is the equation of an ellipse (Fig. 2-13).
I
104---4- 2 Ay------ll>--l.\
rr------------- -- v
28 ~I~--------~~--------~r_x
L~___.~ IIIIIII
____________ J
Fig. 2-13. Harmonic motion in two dimensions.
(2-74)
(2-75)
'. ~\
We saw that the x and y motions have the same frequency but are independent of each other. The relative phase angle (3 between these motions is determined from the initial conditions and the value of (3 will influence the shape and sense of the motion along the elliptical path. In any event, however, the path is inscribed within the dashed rectangle shown in Fig.
2-13. For example, if fJ = + 1(/2, Eq. (2-75) reduces to
X2 y2 _
A2 + B2 - 1 (2-76)
indicating that the x and y axes are principal axes in this case.
On the other hand, if fJ = 0 or 1(, Eq. '(2-75) reduces to A
x= +-y
-B (2-77)
and the path is a straight line along a diagonal of the rectangle of Fig. 2-13. In this case, we again have simple rectilinear harmonic motion.
For the case where fJ = +1(/2 and A = B, Eq. (2-76) reduces to
-t6t
X2 + y2 = A2 (2-78)
which is the equation of a circle of radius A. Also,


SEC. 2-7 KINEMATICS OF A PARTICLE 45
v2 == X2 + y2 == A2 ro2(sin2rot + cos2rot) == A2ro2 (2-79)
Thus the particle P moves at uniform speed about a circular path. The acceleration is also constant in magnitude and, from Eq. (2-53), is
(2-80)
It is interesting to note that elliptical harmonic Inotion can be considered to be the projection of unifonn circular motion onto a plane that is not parallel to the circle. For the case of simple harmonic motion, the two planes are orthogonal and the path reduces to a line. The velocity and acceleration vectors in the case of elliptical nlotion are found by projecting the corresponding vectors for the case of circular motion. Hence it can be seen that the maxilnum velocity occurs at the ends of the Ininor axis and the maximunl acceleration occurs at the ends of the major axis. In each of these two cases, the vectors at the given points are parallel to the plane onto which they are projected. In case the motions in the x and y directions are of different frequencies, then a different class of curves known as Lissajous figures is generated. Again, the motion relnains within the dashed rectangle of Fig. 2-13 but, unless the frequency ratio is a rational nUlnber, the curve does not close. For the case of rational frequency ratios, the Inotion is periodic and retraces itself with a period equal to the least common multiple of the periods of the two component vibrations. We shall not, however, pursue this topic further.
2-7. VELOCITY AND ACCELERATION OF A POINT IN A RIGID BODY
In Sec. 2-3, we calculated the absolute velocity of a point in a rigid body that is rotating about a fixed base point. Now consider the case shown in Fig. 2-14 where the base point A has a velocity vA relative to the inertial system XYZ. Also, the body is rotating with an angular velocity co relative to this system. Suppose we define the velocity
(2-81)
where v and vA are the absolute velocities of the points P and A, respectively. Then VPA is known as the relative velocity of P with respect
x
z
r--------------------y
Fig.. 2-14. Motion of a point in a rigid body.


46 KINEMATICS OF A PARTICLE CHAP•. 2
to A as viewed by an observer fixed in the XYZ system. Now suppose the observer is translating in an arbitrary fashion but is not rotating relative to X YZ. The velocities of points P and A as viewed by the translating observer would not be the same as in the previous case. The velocity difference would, however, be the same as before because the apparent velocities of the points P and A would each change by the same amount, namely, by the negative of the observer's translational velocity. Thus the velocity of point P relative to point A is identical for any nonrotating observer. On the other hand, it is important to realize that the velocity of P relative to A will be different, in general, when the motion is viewed from various reference frames in relative rotational motion, as we shall demonstrate. Hence a statement of the relative velocity of two points should also specify the reference frame. An inertial or a nonrotating frame is assumed if none is stated explicitly. Returning now to a consideration of the relative velocity vPA as viewed by a nonrotating observer, we note that it can also be considered to be the velocity of P as viewed by an observer on a non rotating system that is translating with A. In this case, the base point A would have no velocity relative to the observer, and therefore we could use the results of Sec. 2-3 which apply to the rotation of a rigid body about a fixed point. Proceeding in this fashion, we obtain from Eq. (2-5) that
Vl'A == CI) X P (2-82)
where p is the position vector of P relative to A and CI) is the absolute angular velocity of the body. (Here we note that if CI) were measured relative to a rotating system, its value would change and so would the value of VP~b in general.) So, from Eqs. (2-81) and (2-82), we find that the absolute velocity of point P is
v == VA + CI) X P (2-83)
The acceleration of P is obtained by differentiating Eq. (2-83) with respect to time:
v == VA + cO X P + CI) X P (2-84)
To evalute p we recall that the time rate of change of a vector is the velocity
of the tip of the vector when the origin remains fixed. Since p is of constant magnitude, we can again use Eq. (2-5), obtaining
p == CI) X P (2-85)
which, of course, is identical with VPti obtained previously. Finally, denoting
the acceleration vA by aA, we obtain from Eqs. (2-84) and (2-85) the absolute
acceleration of P.
a == aA + cO X P + CI) X (CI) X p) (2-86)


SEC. 2-8 KINEMATICS OF A PARTICLE
2-8. VECTOR DERIVATIVES IN ROTATING SYSTEMS
47
Suppose that a vector A is viewed by an observer on a fixed system X YZ (Fig. 2-15) and also by another observer on a rotating system, The rotating systelTI is designated by the unit vector triad e1, e2, and e3 which is rotating with an angular velocity (J) relative to X YZ. Since A is considered to be a free vector during the differentiation process, no generality is lost by taking the point 0 as the common origin of the unit vector triad and also the vector A. Now, as each observer views the vector A, he might choose to express it in terms of the unit vectors of his own system. Thus each observer would give a different set of components. Nevertheless, they would be viewing the same vector and a simple coordinate conversion based upon the relative orientation of the coordinate systems would provide a check of one observation with the other.
z
-,p::;--------y
x
Fig. 2-15. The vector A relative to fixed and rotating reference frames.
On the other hand, if each observer were to calculate the time rate of change of A, the results would, in general, not agree, even after performing the coordinate conversion used previously. To clarify this point, recall from Eq. (1-38) that the absolute rate of change of A, written in terms of the unit vectors of the rotating system, is
A == AIel + A2 e2 + A3 e3 .+ AIel + A 2 e2 + A3 e3 (2-87)
But the rate of change of A, as viewed by an observer in the rotating system,
IS
(2-88)
since the unit vectors are fixed in this system. Now let us consider the last three terms on the right side of Eq. (2-87). Using Eq. (2-8), we see that
AIel + A 2 e2 + A3 e3 == AI(J) X el + A 2 (J) X e2 + A 3 (J) X e3
or, since vector multiplication is distributive,
AIel + A 2 e2 + A 3e3 == (J) X (AIel + A 2 e2 + A 3 e3) == (J)-X A (2-89)
Therefore, from Eqs. (2-87), (2-88), and (2-89), we find that the ab


;,
48 KINEMATICS OF A PARTICLE CHAP.' 2
,solute rate of change of A can be expressed in terms of its value relative to
a rotating system as follows: . (.=U\ :: (d?\ -L ;;:J J( J
c,.(.b-\ 5 '-dt- j b ·
A = (A)r + W X A .~ l ~' (2-90)
where (A)r is the rate of change of A as viewed from the rotating system and W is the absolute angular velocity of the rotating system. It is important to note that A can be any vector whatever. A relatively simple application of Eq. (2-90) would occur for the case where A is the position vector of a given point P relative to the common origin 0 of the fixed and rotating systems. On the other hand, a more complicated situation would arise if, for example, A were the velocity of P relative to another point P', as viewed from a third system that is rotating separately. Thus Eq. (2-90) has a wide application and should be studied carefully. Although we have referred to the coordinate systems in the preceding discussion as fixed or rotating, the derivation of Eq. (2-90) was based upon mathematics rather than upon physical law. Therefore we need not consider either system as being more fundamental than the other. If we call them system A and system B, respectively, we can write
(2-91) ,
where' W BA is the rotation rate of system B as viewed from system A. Since the result must be symmetrical with respect to the two systems, we could also write (2-92)
where we note that W AIJ = -WllA and that A is the same when viewed from either system.
Example 2-1. A turntable
y
rotates with a constant angular velocity W about a perpendicular axis through
o (Fig. 2-16). The position of a point P which is moving relative to the turntable is given by
r=xi+yj
·-------+---x
where the unit vectors i and j are fixed in the turntable. Solve for the absolute velocity and acceleration of P in terms of its motion relative to the turntable.
Fig. 2-16. Motion of a point that is moving on a turntable.
Using Eq. (2-90), we see that the absolute velocity v is given by
v = t '= (t)r + lI) X r (2-93)
But the velocity seen by an observer rotating with the turntable is just


SEC. 2-9 KINEMATICS OF A PARTICLE 49
Also,
ru x r == xru x i + yru x j == - yroi + Xro j
Therefore,
v == (x - yro)i + (P + xro)j (2-94)
where we note that the absolute velocity v is expressed in terms of the unit vectors of the rotating systelll. To find the acceleration, we again apply Eq. (2-90), this time to the absolute velocity vector.
a == v == (v)r + ru X v (2-95)
The term (v)r is found by differentiating Eq. (2-94) with respect to time, assuming that the unit vectors are constant since they are viewed from the rotating systenl.
(V)r == (x - yro)i + tv + xro)j (2-96)
Note that (v)r is not the acceleration of P relative to the rotating system. Rather, it is the time rate of change of the absolute velocity vector as viewed from the rotating system. In genera], the subscript after a differentiated (dotted) vector refers to the coordinate system from which the rate of change is viewed. If the subscript is omitted, a nonrotating reference frame is assumed. Next, using the result given in Eq. (2-94), the term ru X v is evaluated in a straightforward fashion, giving
ru X v == -ro(y + xro)i + ro(x - yro)j (2-97)
Finally, from Eqs. (2-95), (2-96), and (2-97) we obtain that the absolute acceleration is
a == (x - Xro 2 - 2roy)i + (jj - yro 2 + 2rox)j
2-9. MOTION OF A PARTICLE IN A MOVING COORDINATE SYSTEM
..
(2-98)
Now we shall use the general result of Eq. (2-90) to obtain the equations for the absolute velocity and acceleration of a particle P that is in motion relative to a moving coordinate system. In Fig. 2-17, the X YZ system is fixed in an inertial frame and the xyz system translates and rotates relative to it. Suppose that r is the position vector of P and R is the position vector of 0', both relative to point 0 in the fixed XYZ system. Then
r == R + P (2-99)


50 KINEMATICS OF A PARTICLE CHAP. 2
z p. p'
,
w
~-------------------------y
X
Fig. 2-17. The position vectors of a point P relative to a fixed system and a moving system.
where p is the position vector of P relative to 0'. Differentiating with respect to time, we obtain the absolute velocity
v == t == R + P (2-100)
where both derivatives on the right are calculated from the viewpoint of
a fixed observer. Next we express p in terms of its value relative to the rotating xyz system, using Eq. (2-90), that is, .
P == (P)r + (I) X P (2-101)
where (I) is the absolute rotation rate of the xyz system. Then from Eqs. (2-100) and (2-101) we obtain
v == Ii + (P)r + (I) X P (2-102)
In order to explain further the meaning of the terms on the right side of Eq. (2-102), let us define the point P' to be coincident with P at the time
of observation but fixed in the xyz system. We see that Ii is the absolute velocity of 0' and that (j) X P is the velocity of P' relative to 0' as viewed
by a nonrotating observer. Thus Ii + (I) X P represents the absolute velocity of P'. The remaining term (P)r is the velocity of P relati.y.~__~o.~Q!, as view~d by an observer rotating with the xyz system. It lSlrlteresting to note that (P)r can also be interpreted as the velocity of P relative to P', as viewed by a nonrotating observer.3 Therefore the three terms on the right side of
3This result can be seen more clearly if we let p' be a vector drawn from P' to P. At the moment in question, p' = 0, and therefore, from Eq. (2-90), jl' = {P')r. In other words, at this pa rticular moment, the velocity of P relative to P' is the same, whether viewed from a rotating or nonrotating system. Also, of course, the velocity of P relative to any point fixed in the xyz system, whether it be P' or 0', is the same when viewed by an observer moving with the system.


I
i'
SEC. 2-9 KINEMATICS OF A PARTICLE 51
Eq. (2-102) are, respectively, the velocity of 0' relative to 0, the velocity of P relative to P', and the velocity of P' relative to 0', all as viewed by an absolute or nonrotating observer. To obtain the absolute acceleration of P, we find the rate- of change of each of the terms in Eq. (2-102), as viewed by a fixed observer. Thus,
Using Eq. (2-90), we obtain
d · ..
-(R) == R
dt
;t [(.O)T) = (P)T + (l) X (P)T
Also, using Eq. (2-101),
d ( ). ·
dt (I) X P == w X p + w X p
== OJ X P + w X (P)r + W X (w X p)
(2-103)
(2-104)
(2-105)
Finally, adding Eqs. (2-103), (2-104), and (2-105), we obtain an expression for the absolute acceleration o~ P.
a == R + OJ X P + m X (m X p) + (p)r + 2w X (P)r (2-106)
Now let us explain the nature of each of the terms. R is the absolute
acceleration of 0'. The terms OJ X P and (I) X «(I) X p) together represent the acceleration of P' relative to 0', as viewed by a nonrotating observer. The term OJ X P is similar in nature to the tangential acceleration of Eq. (2-7), whereas the term (I) X «(I) X p) represents a centripetal acceleration since it is directed from P toward, and perpendicular to, the axis of rotation "'-through 0'. Thus the first three terms of Eq. (2-106) represent the absolute acceleration of P'. The fourth term (P)r is the acceleration of the point P relative to the xyz system, that is, as viewed by an observer moving with the xyz system. The fifth term 2(1) X (.o)r is known as the CarioUs acceleration. Note that the Coriolis acceleration arises from two sources, namely, Eqs. (2-104) and (2-105). The term in Eq. (2-104) is due to the changing direction in space of the velocity of P relative to the moving system. The term in Eq. (2-105) represents the rate of change of the velocity (I) X P due to'a changing magnitude or direction of the position vector p relative to the
moving system. /"
_/
L-~ote alsQJ.h(!tJbJ~_JastJ~o t~rms of Eq. (2- ~.9?) represent the aC~~!~!.~!!.<?!l ___ . of P relative to P', as viewe-<LbY~~~__ n9nr9t~_ti.Dg_Q.!?serY~r. d.T_hus---we-can---summarize by saying that the fir~!JhIe~_terl}1S---give the acceleration of P' ,whereas the-last two terms give the a9celeration of P relative _!2-..e_:_;_Jh~.-_..__ sum, of course~l~iIlK-L1}_the absQlute ac.cele.J.'-J.JlQ.U__.ofE.. ---.


52 KINEMATICS OF A PARTICLE CHAP. 2
2-10. PLANE MOTION
. If a particle moves so that it remains in a single fixed plane, it is said to move with plane motion. Similarly, plane motion of a rigid body requires that all points of the body move parallel to the same fixed plane. In the latter case, the kinematical aspects of the motion are adequately described in terms of a lamina chosen such that its motion is confined to its own plane. Let us consider, then, the motion of a lamina in its own plane (Fig. 2-18).
Fig. 2-18. The motion of a lamina in its own plane.
Instantaneous Center. In Eq. (2-83), we found that the velocity of a point P in a rigid body is given by
v == VA + (I) X P
where VA is the velocity of the base point A, (I) is the absolute angular velocity of the body, and p is the position vector ofP relative to A. For plane motion, the angular velocity vector (I), if it exists, must be perpendicular to the plane of the motion. Furthermore, we recall that a different base point will, in general, have a different velocity. So if we can find a base point location C whose velocity is zero, we see from Eq. (2-83) that V == (I) X pp (2-107)
where pp is the position vector of P relative to C. Thus the line CP must be perpendicular to V and of length pp such that
v == ropp
Hence, if V and (I) are known, the point C can be determined. Similarly, it can be seen that
VA == (I) X PA.
or, for another point B in the lamina,
(2-108)
(2-109)
VB == (I) X pn (2-110)
Thus one concludes that the instantaneous center C is located at the intersection of two or more lines, each of which is drawn from some point in the lamina in a direction perpendicular to its velocity vector. The point C, whose velocity is instantaneously zero, and about which the lamina appears to be- rotating at that moment, is known as the instan


SEC. 2-10 KINEMATICS OF A PARTICLE 53
taneous center of the rotation. It may be located in the lamina or in an imaginary extension of it. The instantaneous center is at a unique point for any given instant in time, provided that the motion is not that of pure translation (00 == 0). In the latter case, the instantaneous center is located at an infinite distance from the point in a direction perpendicular to the velocity. In considering the velocity of various points in the lamina, it should be noted that only certain velocity distributions are possible. In particular, the rigid-body assumption requires that the velocities at all points on any given straight line in the lamina must have equal components along the line. In fact, from Eq. (2-110) we see that the locus of all points having a given speed vn is a circle centered at C and with a radius pJJ == v13/00, the velocity at each point being tangent to the locus. Thus the velocity distribution has a circular symmetry about the instantaneous center. Furthermore, a knowledge of the angular velocity and the location of theinstantaneous center is sufficient to determine the velocity of any point in the lamina, thereby providing a convenient means of calculating velocities for the case of plane lTIotion. It should be emphasized, however, that the acceleration of the instantaneous center is not necessarily ,zero, and therefore the instantaneous center is not particularly useful in performing acceleration calculations. We have seen that the angular velocity (J) is perpendicular to the lamina and is not a function of position. Hence the _angular velocity of any line in the lamina is equal to the angular velocity of the lamina. So if we choose any two points of the lamina, A and P for example, the angular velocity is found by taking the difference of the velocity components at A and P perpendicular to AP and dividing by the distance AP. Using vector notation,
(2-111)
Space and Body Centrodes. In general, the instantaneous center does not retain the same position, either in space or in the lamina. The locus of the instantaneous centers forms a curve in space known as the space centrode. Similarly, the locus of the instantaneous centers relative to the moving lamina is known as the body centrode. -Both curves are in the same plane, of course, but neither curve need be closed. Now .suppose that, at a given moment, the space centrode Sand the body centrode B corresponding to a given motion are as shown in Fig. 2-19.r The . instantaneous center is a point common to both curves at this moment. Also,we note that curve S is fixed in space and
B
5
Fig. 2-19. The space centrode Sand the body centrode B.


54 KINEMATICS OF A PARTICLE CHAP. 2 
curve B is rotating with angular velocity ~. In general, the curves Band S are tangent at C but ha've different curvatures at this- point. This allows for the instantaneous center C to proceed smoothly along curves Band C because of a finite rotation rate ro of the body centrode. But this is just the condition for rolling motion. Hence the body centrode rolls on the space centrode. A further condition for rolling motion is that equal path lengths are traced by point C in curves Band S in any given time interval. If this were not true, slipping would result and pure rolling motion would not occur. Perhaps a source of confusion can be avoided if we emphasize again that, at a given moment, the point known as the instantaneous center has zero velocity. We may speak of the instantaneous center as moving along the curves Band S but, more accurately, a succession of points on these curves are called the instantaneous center at succeeding instants of time. The points on the body centrode are in motion, in general, but when each becomes the instantaneous center of rotation, its velocity is zero at that instant. Each point on the space centrode is at rest at all times. Now let us take a more general view of the instantaneous center. We saw in the case of the lamina of Fig. 2-18 that the instantaneous center C is instantaneously fixed both in the lamina and in space. So if we consider a lamina A moving with plane motion and lamina B fixed in space, then the instantaneous center is fixed in both laminas at that moment. But, as we emphasized earlier, there are no absolute systems from the viewpoint of kinematics. Therefore, in a more general sense, the instantaneous center relative to any two laminas A and B in plane motion is that point C which is instantaneously at rest, as viewed by observers on both A and B. If more than two laminas are in plane motion, then there is, in general, a separate instantaneous center for each pair. Thus, four laminas would have six instantaneous centers. To avoid confusion, however, we shall always assume that the instantaneous center for a given lamina is defined relative to a fixed frame unless a contrary statement is made.
Example 2-2. As an example of plane motion, consider a wheel of radius a which is rolling along a straight line (Fig. 2-20). Assume that the wheel has a uniform angular velocity roo In this case, the in&tantaneous center is at the point of contact C. The space centrode is the straight line on which the wheel is rolling, whereas the body centrode is the circumference of the wheel. Considering the instantaneous center C at the given moment as a base point whose velocity is instantaneously zero, we can use Eq. (2-107) to calculate the velocity of any point on the wheel. The velocity of the center 0' is constant since its distance from C is constant and the rotation rate is assumed to be constant. Its magnitude is
Ij
-l)


SEC. 2-10
vo' = Qro (2-112)
SiiniIarIy, the speed of a general point P is
v = Pl'ro (2-113)
where pp is the distance froln C to P. The path in space of a point P on the circulnference can be shown to be a cycloid. Its speed along its path reaches a maximunl when pp is a maximum, that is, when P is diametrically opposite C and at the top point of the cycloid. At this point,
KINEMATICS OF A PARTICLE 55
c
Fig. 2-20. A wheel rolling on a straight line.
v = 2vol = 2Qro. On the other hand, the velocity of P goes to zero when P coincides with C and pp = O. In tenns of the cycloidal motion, this occurs when P is at the cusp of the cycloid and instantaneously at rest while reversing the direction of its motion. Pf course,_J.h~ acceleration Qf..p .is_not zero when it coin~ides 'Y~Jh~.._.Jn.fact, the acceleration of P is of uniform magnitude Qro2, being at all times directed toward 0' for the case of unfform ro. .
.... ----.-.--.-.-~ .
Example 2-3. A bar of length I is constrained to move in the xy plane such that end A remains on the x axis and the other end B remains on the y aXIS. Assuming that the angular velocity roof the bar is constant, find
y
-----r------------~~------*-----~---x
Fig. 2-21. The motion of a bar such that its ends remain on the cartesian axes.


56 KINEMATICS OF A PARTICLE CHAP. 2.
the space and body centrodes and solve for the motion of-end A as a function of time. Let the position of the bar be specified by the angle f) between the bar and y axis at B, as shown in Fig. 2~21. If we measure time from the moment when point A is at the origin and moving in the positive sense along the x axis, we obtain
f) == rot (2:-114)
The instantaneous center of rotation of the bar is at the intersection of lines drawn perpendicular to the velocity vectors at two points on the bar. Choosing the end points, for example, point A always moves along the x axis and point B along the y axis. Therefore, the instantaneous center C is located at the intersection of the dashed lines of Fig. 2-21, that is, at
x == I sin f)
y == I cos f) (2-115)
indicating that the space centrode is a circle of radius I since the distance
OC is equal to Jx2-+ .,V2 == I. Furthermore, the line OC makes an angle f) with the positive y axis and, from Eq. (2-114), the point C proceed.s around the space centrode at a uniform rate. We have seen that AB and OC are of equal length I and therefore they bisect each other at the center of the rectangle OA CR. Thus the instantaneous center C remains at a constant distance 1/2 from the midpoint of the rod but changes its position relative to the rod. Therefore we conclude that the body centrode is a circle of radius 1/2 and centered at the midpoint of the rod. To find the motion of point A, we note that it moves along the x axis and has the same x coordinate as the instantaneous center. So, ,from Eqs. (2-114) and (2-115), we obtain
x == I sin rot (2-116)
indicating that point A moves with simple harmonic motion along the x axis. Similarly, point B moves with simple harmonic motion along the yaxIS. The instantaneous center C· moves clockwise at the same angular rate as the bar rotates counterclockwise. Therefore point C moves completely around the space centrode during one complete rotation of the bar. Since, however, the instantaneous center must traverse the same distance along the space and body centrodes, it must make two complete circuits of the body centrode for every complete rotation of the bar.
-Ii


SEC. 2-11 KINEMATICS OF A PARTICLE 57
2-11. EXAMPLES
Example 2-4. A particle P moves along a straight radial groove in a circular disk of radius a which is pivoted about a perpendicular axis through its center 0 (Fig. 2-22). The particle moves relative to the disk such that
r == ~ (1 + sin rot)
and the disk rotates according to
ep == epo sin rot
Find the general expression for the absolute acceleration of P. First we shall solve the problem using the general acceleration equation in cylindrical coordinates, Eq. (2-23). Setting i == 0 since the motion is two-dimensional, we obtain the components in' the er and ecp directions as follows:
Fig. 2-22. A particle with radial harmonic motion relative to a harmonically oscillating disk.
. aro2 aro2 cb2 .
aT == r - rep2 - - 2 sin rot - 2 0 (1 + SIn rot) cos2 rot
.. . aro2 cb
acp ~ rep + 2,ep == - 2 0 (1 + sin rot) sin rot + aro2 cbo cos2 rot
(2-117)
Next let us solve the same problem using the general vector expression of Eq. (2-106). Let the rotating coordinate system be fixed in the disk such that the origins 0 and 0' coincide. Then, noting that
p == rer == ~ (1 + sin rot)eT
m == ¢ez == roepo cos rot ez
we evaluate each of the terms.
it == 0
wX p = - aw; 4>0 (1 + sin wt) sin wt e",
2 cb2
ro X (m X p) == - aro2 0 (1 + sin rot) cos2 rot er
•• aro2 •
(P)T == - T SIn rot eT
2ro X (.0) T == aro2 epo cos2 rot e</J
In obtaining the foregoing results, the procedure IS straightforward


58 KINEMATICS OF A PARTICLE CHAP. 2
except for finding the derivatives of p as viewed from the rotating system. Here we differentiate the expression for p, assuming that er is constant since it is fixed in the rotating system. Adding the components in the er and e4> directions, we obtain agreement with the results found previously in Eq. (2-117).
Example 2-5. Suppose that a particle P moves along a line of longitude on a sphere of radius a rotating at a constant angular velocity OJ about the polar axis. If its speed relative to the sphere is v == kt and if the center of the sphere is fixed, find the absolute acceleration of P in terms of the spherical unit vectors en eo, and ecf>' (Fig. 2-23). Let (J (0) == (Jo.
v
Fig. 2-23. The motion of a particle P having a velocity v relative to a sphere.
One approach to the problem is to evaluate the acceleration directly in terms of spherical coordinates, using Eq. (2-33). First we note that r == a and ¢ == 00, where both a and 00 are constants. Also,
. _ v _ kt
{j---
a a (2-118)
Evaluating the spherical components of the acceleration, we obtain
ar == r - r82 - r¢2 sin2 {j k'2 t 2
== - - - - aoo2 sin2 {j
a
ao == rO + 2;8 - r¢2 sin (J cos {j == k - aoo2 sin (J cos ()
acf> == rif; sin () + 2;¢ sin () + 2r8¢ cos () == 200kt cos ()
where we note from Eq. (2-118) that
(2-119)
(2-120)
Now let us solve this problem using Eq. (2-106). The moving coordinate system is assumed to be fixed in the sphere and rotating with it at a constant rate (I). The origin 0' is at the center O. Evaluating the first two terms, we find that
and
ciJxp==O
Before evaluating the next term, we note that
(I) == 00 cos (J er - 00 sin (J eo
p == aer
(2-121)
(2-122)


I
I
SEC. 2-11 KINEMATICS OF A PARTICLE 59
Therefore,
(1) X P == am sin f) eq,
and
(2-123)
It can be seen that the path of P relative to the sphere, that is, the moving system, is just a circular path which is traversed at a constantly increasing speed. From Eq. (2-54), we find that
(2-124)
Also, we see that
so the CorioUs acceleration is
2(1) X (p) r == 2mkt cos (j eq, (2-125)
Adding the tenns given by Eqs. (2-121) to (2-125), we obtain the same resuIts as were obtained previously in Eq. (2-119).
Example 2-6. Find the acceleration of point P on the circumference
of a wheel of radius '2 which is roll~ng on the inside of a fixed circular track
of radius '1 (Fig. 2-24). An arm connecting the fixed point 0 and the wheel hub at 0' moves at a constant angular velocity m. The position of P relative
to the arm is given by the angle cp. Our procedure will be to solve this example using the general vector equation given in Eq. (2-106). The origin of the fixed system is taken at
o. The moving system is fixed in the moving arm and its origin is at the hub 0'. Normal and tangential unit vectors will be used, en having the direction of the line from 0' to 0 and et being perpendicular, as shown. It can be seen that the wheel undergoes plane motion, and since there
Fig. 2-24. A wheel rolling on the inside of a fixed circular track.
is no slipping of the wheel on the track, the instantaneous center of rotation for the wheel is at the contact point C. This enables us to obta~~_~_!:!_~~uation
relating ¢ and~.-,rg_. First consider the velocity -wfih-whlch--ihe hub· moves
al()Iig--Tfs-·cIrc~iar path. The wheel is rotating clockwise at an angular rate ¢ relative to the arm but the arm rotates with an angular velocity 00 in a counterclockwise sense. Therefore the wheel rotates clockwise at an absolute
angular velocity (¢ - (0). Since the wheel is instantaneously rotating about C, the absolute velocity of the hub is


60 KINEMATICS OF A PARTICLE CHAP. 2 .
(2-126)
But the velocity of the hub can also be calculated from the motion of the arm. The distance 00' is (rl - r2) and the arm rotates with an absolute angular velocity co, so
VO' = (rl - r2) coet
From Eqs. (2-126) and (2-127), we obtain
(2-127)
(2-128)
Now let us proceed with the evaluation of the absolute acceleration of P. We find that
(2-129)
which is just the centripetal acceleration of 0' due to its uniform circular
motion about 0, in accordance with Eq. (2-53). The next term, ro X p, is zero because the angular velocity (I) is constant. In order to evaluate the term (I) X «(I) X p), we note first that
where eb is a unit vector pointing out of the plane of the figure in accordance with Eq. (2-38). Also,
Consequently,
(2-130)
which is a centripetal acceleration directed from P toward 0'. The point P, as viewed by an observer on the moving coordinate system,
moves at a constant angular rate ¢ in a circle of radius r 2• Thus we find that the acceleration due to this motion is again centripetal, being directed from P toward 0'.
(ji)r = - r2¢2 (sin cp et + cos cp en)
or, substituting from Eq. (2-128), we obtain
(2-131)
In order to evaluate the Coriolis acceleration term, we first obtain the relative velocity
(P)r = r2¢(cos cp et - sin cp en)
= r1 ro(cos cp et - sin cp en)
where we note again that an observer on the moving system sees point P moving in a circular path with a uniform speed r2¢' Thus we obtain
II
J


SEC. 2-11 KINEMATICS OF A PARTICLE 61
(2-132)
which is directed radially outward in the direction 0'P. Finally, adding the individual acceleration terms given in Eqs. (2-129) to (2-132), we obtain
a = (2rl - '2 - ;~) 002 sin 1> et + [(2rl - r2 - ;!) 002 COS 1>
+ (rl - (2)002] en
(2-133)
The preceding exanlples have illustrated the application of Eq. (2-106) which is the general vector equation for the acceleration of a point in terms of its motion relative to a nl0ving coordinate system. Although this equation is valid for an arbitrary ITIotion of the moving coordinate system, this system should be chosen such that the calculations are made as simple as possible. An unfortunate choice at this point can result in a large increase in the required effort. Roughly speaking, the motion of P relative to the moving system should be of about the same complexity as the absolute motion of 0', provided that the angular velocity (J) is constant or varies in a simple fashion. Also, the choice of unit vectors in expressing the result should be made for convenience. In general, they should form an orthogonal
\1) •
set. ' !.,)
I
REFERENCES \"J v\ , ',\ I( (J
',J 1\\) X :-: )
1. Banach, S., Mechanics, E. J. Scott, trans. Mathematical Monographs, vol. 24, Warsaw, 1951; distributed by Hafner Publishing Co., New York.
2. Halfman, R. L., Dynamics-Particles, Rigid Bodies, and Systems~ Reading, Mass.: Addison-Wesley Publishing Company, ,1962.
3. Kane, T. R., Analytical Elements ofMechanics-Dynamics. New York: Academic Press, Inc., 1961.
4. Shames,!. H., Engineering Mechanics. Englewood Cliffs, N.J.: Prentice-Hall, Inc., 1960.
5. Synge, J. L., and B. A. Griffith, Principles of Mechanics, 3rd ed. New York: McGraw-Hill, Inc' 2 1959,.
6. Yeh, H., and J. I. Abrams, Principles of Mechanics of Solids and FluidsParticle and Rigid-Body Mechanics. New y~)fk: McGraw-Hill, Inc., 1960.
PROBLEMS
2-1 .. A vertical wheel of radius a rons without slipping along a straight horizontal line. If its angular velocity is given by m == at, solve for the acceleration of


'i:
. II"
, Iii
.:1!
62 KINEMATICS OF A PARTICLE
Fig. P2-1
CHAP. 2. ,
a po~nt P on its rim, assuming that P is initially at the highest point of its path. Express the result in terms of the unit vectors er and eo. 2-2. Solve for the hodograph of the two-dimensional harmonic motion given by Eqs. (2-64) and (2-69). Sketch the result. 2-3. A lamina undergoes general motion in its own xy plane. At a given instant, its angular motion is given by (I) == rok, OJ == rok; whereas a base point A in the lamina has an acceleration aA == aAi. Find the position relative to A of a point P whose absolute acceleration is zero. 2-4. A water particle P moves outward along the impeller of a centrifugal pump with a constant tangential velocity of 60 ft/sec relative to the impeller, which is rotating at a uniform rate of 1200 rpm in the direction shown. Find the acceleration of the particle at the point where it leaves the impeller.
pz
y
Fig. P2-4 Fig. P2-5
2-5. A cyclist rides around a circular track (R == 100 ft) such that the point of contact of the wheel on the track moves at a constant speed of 30 ft/sec. He banks his bicycle at 15° with the vertical. Find the acceleration of a tack in the tire (1.25 ft radius) as it passes through the highest point of its path. Use cylindrical unit vectors in expressing the answer. 2-6. An airplane flies with a constant speed v in a level turn to the left at a constant radius R. The propeller is of radius a and rotates about its axis in a clockwise sense (as viewed from the rear) with a constant angular velocity fl.. Find the total acceleration of a point P at the tip of the propel1er, assuming that its axis is always aligned with the flight path. Use cylindrical unit vectors and assume that the velocity of P relative to the airplane is vertically upward at t == O.


II
I!
\.
I
SEC. 2-11 KINEMATICS OF A PARTICLE 63
2-7. A circular disk of radius r2 rolls in its plane on the inside of a fixed circular
cylinder of radius r 1• Find the acceleration of a point P on the wheel at a distance
b from its hub 0'. Assume that ¢ is not constant, where the angle cp is measured between 0'P and the line of centers 0'0.
R
C
Fig.. P2-7 Fig. P2..8
2-8. A wheel of radius a rolls along a general convex curve of varying radius of curvature R such that all motion is confined to a single plane. Tpe contact point 0' traverses the curve at a constant speed Vo. Find the absolute acceleration of a point P on the rim. Express the result in terms of the unit vectors en and Ct2-9. The origin 0' of a moving coordinate system has a constant absolute velocity Vo'; whereas the point P has a constant absolute velocity Vp. If the moving system rotates with a constant angular velocity (1), and p is the position vector of P relative to 0', find: .
2-10.. The points P and P' are each located on the edge of separate circular disks which are in essentially the same plane and rotate at constant but unequal angular
velocities co and co' about a perpendicular axis through the common center o. Assuming that 0 is fixed, solve for the velocity and acceleration of P' relative to P at the moment when P and P' have the least separation. Consider the following cases.: (a) the observer is nonrotating; (b) the observer is rotating with the unprimed system. Express the results In terms of the unit vectors en and et which rotate with the unprimed system.


64 KINEMATICS OF A PARTICLE CHAP. 2
/
Fig. P2-10 Fig. P2-11
2-11. The plane of the windshield of a certain auto, is inclined at an angle a with the vertical. The windshield wiper blade is of length I and oscillates accord
ing to the equation ,y == ,yo sin fJt. Assuming that the auto travels with a constant speed v around a circular path of radius R in a counterclockwise sense, (more exactly, the point 0' traces out a circle of this radius), solve for the acceleration of
the point P at the tip of the wiper. For what value of -t would you expect the
largest force of the blade against the windshield?
Fig. P2-12
2-12. A gyroscope in the form of a wheel of radius a spins with a constant
angular velocity a == n about its axis of symmetry. This axis maintains a constant angle (J with the vertical and pre
cesses at a constant rate ¢. Find the acceleration of a point P on the rim using spherical unit vectors er , eo, e4J. The
angle a is measured in the plane of the rim and specifies the location of P relative to a horizontal diameter through 0'.
"I)


3
DYNAMICS OF A PARTICLE
In Chapter 2, we studied some of the methods of kinematics which can be used to obtain the absolute acceleration of a particle. Then, knowing the acceleration and assuming a knowledge of its mass, one can use Newton's law of motion to determine the total external force acting on the particle. This chapte~ emphasizes the inverse problem, namely, the problem of calculating the nlotion of a particle from a knowledge of the external forces acting upon it. First consider the general case where the force acting on a particle is a function of its position and velocity and also the time. From Eq. (1-41), its differential equation of motion is
mr == F{r, f, t) (3-1)
Assuming that the function F{r, f, t) and the mass m are known, we would like to solve for the position r as a function of time. Unfortunately, an analytic solution of this equation is impossible except in special cases. To see the difficulties more clearly, let us write the vector equation in terms of its cartesian components.
mx == Fx{x, y, z, x, y, i, t)
my == Fy{x, y, z, x, y, i, t) (3-2)
mz == Fz(x, y, z, x, y, i, t)
The force components Fx, Fy , and Fz are, in general, nonlinear functions of the coordinates, velocities, and time; thus the equations are hopelessly complex from the standpoint of obtaining an analytical solution. Nevertheless, it is the thesis of Newtonian mechanics that a complete knowledge of the external forces acting on a particle determines its motion, provided that the initial values of displacement and velocity are known. For example, using cartesian coordinates, initial values of x, y, z and also X, y, and i would be specified. So a solution to the problem does, in fact, exist. With the aid of modern electronic computers, and using approximate methods, it is possible to obtain solutions to the complete equations that are of sufficient
I· accuracy for engineering purposes.
I.
Any general analytical solution of Eq. (3-2) will contain six arbitrary
6S


66 DYNAMICS OF A PARTICLE CHAP. 3·
coefficients which are evaluated from the six initial conditions. One method of obtaining the general solution is to look for integrals or constants of the . motion, that is, to attempt to find six functions of the form
fk(X, y, Z, x, y, z, t) = ak (k = 1, 2, ... , 6) (3-3)
where the a k are constants. If the functions are all distinct, that is, if none is derivable from the others, then, in principle, they may be solved for the displacement and the velocity of the particle as a function of time and the constants ak' It is usually not possible to obtain all the ak by any direct process. One of the principal topics of advanced classical mechanics, however, is the study of coordinate transformations such that the solution for the constants in terms of the new coordinates is a straightforward process. Sometimes the constants can be given a simple physical interpretation, thereby giving us more insight into the nature of the motion. For example, a constant of the motion might be the total energy or the angular momentum about a given point. Even in cases where we do not solve completely for the motion, a knowledge of some of the constants that are applicable to the given problem may aid in obtaining certain results, such as the limiting values of certain coordinates. In this chapter, we shall discuss some of the simpler methods and principles to be used in solving for the motion of a particle. As will be seen in the following chapters, these principles can be expanded to apply to systems of particles and to rigid-body motion, thereby forming an important part of our treatment of the subject of dynamics.
3-1. DIRECT INTEGRATION OF THE EQUATIONS OF MOTION
Returning now to the general equation of motion as expressed in Eq. (3-1) or Eq. (3-2), let us consider several cases in which direct integration can be used to find the motion of the particle.
Case 1: Constant Acceleration. The simplest case is that in which the external force on the particle is constant in magnitUde and direction. Considering the cartesian components of the motion, we find from Eq. (3-2) that
mx = Fx
my = Fy
mz = Fz
where Fx, Fy , and Fz are each constant.
(3-4)
From Eq. (3-4), we see that the motions in the x, y, and Z directions are


\~
I•
..
SEC. 3-1 DYNAMICS OF A PARTICLE 67
independent. Therefore, let us consider the one-dimensional case of motion parallel to the x axis. Denoting the velocity and acceleration by v and a, respectively, we find that
.. Fx
a = = x = =n-l
and, by direct integration with respect to time, we obtain that
v == Vo + at and
12
X == Xo + Vo t + 2 at
(3-5)
(3-6)
(3-7)
where Xo and Vo are the displacement and velocity at t == O. Of course, similar equations would apply to motion in the direction of the y or z axes. The time required for the particle to attain a given speed v or displacement x is found by solving Eqs. (3-6) and (3-7) for t.
1
t == - (v - vo)
a (3-8)
Also,
(3-9)
An expression relating the speed and displacement can be obtained for this case of constant acceleration by eliminating t between Eqs. (3-8) and (3-9), with the result:
(3-10)
Motion of a Particle in a Uniform Gravitational Field. The motion of a particle in a uniform gravitational field is confined to the plane determined by the initial velocity vector of the particle and the gravitational force. This follows from the fact that the force remains in this plane, and therefore there is no component of acceleration normal to the plane; hence there is no tendency for the particle to leave it. Of course, the velocity vector will change direction in the general case, but it will always lie in the same plane. Let us choose the xy plane as the plane of motion with the y axis directed vertically upward, that is, opposite to the direction ofthe gravitational force. Assume for convenience that the particle is located at
y
L-~------------------~-X
Fig. 3-1. The trajectory of a particle in a uniform gravitational field.


68 DYNAMICS OF A PARTICLE CHAP. 3
the origin 0 at time t = 0 and is moving with velocity Vo at an angle "I above the horizontal, as shown in Fig. 3-1. The initial values of the velocity components are
x(O) = Vo cos "I
yeO) = Vo sin "I
Noting that the acceleration components are
ax = 0
(3-11)
(3-12)
and recalling again that the motion in the x and y directions proceeds independently, we obtain the velocity components from Eq. (3-6).
Vx = Vo cos "I (3-13)
Vy = Vo sin "I - gt (3-14)
Similarly, the displacement components are obtained from Eq. (3-7).
x = vot cos "I
. 12
Y = votslnry - 2 gt
(3-15)
(3-16)
Solving for t from Eq. (3-15) and substituting into Eq. (3-16), we obtain the trajectory
gx2
Y = x tan "I - ~=--....".
2v~ cos2 "I (3-17)
which is the equation of a vertical parabola. The vertex of the parabola occurs at the point of zero slope in this case and is found by differentiating Eq. (3-17) with respect to x and setting the result equal to zero. We obtain
V~ • v~ . 2
Xv = g sIn "I cos "I = 2g sIn "I (3-18)
and, from Eqs. (3-17) and (3-18),
(3-19)
where (xv, Yv) is the vertex location. The time required to attain a given value of x is found from Eq. (3-15).
x
t=--
Vo cos ry (3-20)
where we note that the velocity component in the x direction is constant. Therefore, from Eqs. (3-18) and (3-20), we find that the time to reach the vertex is


I
SEC. 3-1 DYNAMICS OF A PARTICLE 69
Vo •
t == - SIn N
v g I (3-21)
Of course, this value is also obtained from Eq. (3-14) by setting Vy == o. For the case of a trajectory over a flat surface, the range is
V2
R == 2x ==..-2.. sin 2N
v g I (3-22)
and the time of flight is
(3-23)
It can be seen that the maximum range is achieved for "I == 45° and the maximum time of flight for "I == 90°. Now let us calculate the initial flight path angle "I such that the particle passes through a given point P at the coodinates (x, y), as shown in Fig. 3-2.
Noting that sec21' == 1 + tan21', we can write Eq. (3-17) in the forn1
or
y
Envelope of possible trajectories / for a given va
----. ..I...................., ,~
o~--~~--------------------~------x
Fig. 3-2. The required initial flight path angle l' in order that the trajectory pass through a given point P.
gx2
-22 (1 + tan2 'Y) - x tan "I + y == 0
Vo
2v2 2v2y
tan2 l' - _0 tan "I + _0 + 1 == 0 (3-24)
gx gx2
If the point P is within range for a given initial velocity vo, then there are two real roots of this quadratic equation in tan 1', corresponding to the angles 1'1 and "120 Complex roots result for the case where the point is out of range. , The maximum range in a given direction is a point on the dashed envelope of Fig. 3-2 and corresponds to a double root. The value of this double root


70 DYNAMiCS OF A PARTICLE CHAP. 3'
can be found from Eq. (3...:.24) by the method of completing the square, that
IS,
(
V2 )2. 2v2 2v2
tan;y ;.....;0 _0 = tan2 ry - _0 tan ry + _0 Y + 1 = 0
gx gx gx2
from which we obtain that
v2
tanry = _0
gXe
(3-25) ;
(3-26)
where ry is the initial flight path angle such that the trajectory just reaches a point (xe, Ye) on the envelope of possible trajectories for a given Vo• From Eq. (3-25) we obtain the equation of the envelope:
( v~ )2 = 2v~ + 1
gXe gx; Ye
or
2 _ 2v~ ( V~)
Xe - -g- Ye - 2g (3-27)
which is a parabola with the vertex located at x = 0, Y = v~/2g. Assuming that the azimuth angle (that is, the direction of the horizontal velocity component) of the trajectory is arbitrary, any point within the paraboloid formed by rotating the envelope about the vertical line x. = 0 will have at least one trajectory passing through it and will thereby be within the range of a projectile of initial velocity Vo.
Example 3-1. Assuming a given initial speed vo, find the initial flight path angle ry such that maximum range is achieved in a direction 45° above the horizontal. Because the maximum range is obtained at 45° above the horizon, the trajectory must pass through a point Xe = Ye on the envelope. Dividing
Eq. (3-27) by x;, we obtain
1 - 2v~ (Ye V~)
- - gXe Xe - 2gxe
which, by using Eq. (3--26), may be written in the form:
- 2 tan 'Y (1 - ~ tan 'Y) = 1
or
tan2 ry - 2 tan ry - 1 = 0
The only root fitting the requirement that the end point (xe, Ye) lie above the horizontal is
tanry=l+~
or
-\ (


..
SEC. 3-1 DYNAMICS OF A PARTICLE
The slant range in this case is
v2
R == ~ Xe == --.£. (2 - ,v'2)
g
where Xe is evaluated using Eq. (3-26).
71 .
Case 2: F == F(t). Now assume that the external force is a function of time only. Again the general equation of motion can be written in terms of three independent equations giving orthogonal components of the motion. For the case of motion parallel to the x axis, the equation of motion is
lnx == Fx (t) (3-28)
which can be integrated directly to give the velocity
1 ft
v == Vo + - Fx(T) dT
m
0
(3-29)
where Vo is the initial velocity. Another integration results in
x = Xo + vot + ~ f: [ [ Fx(Tt) dTt] dT2 (3-30)
where Xo is the initial displacement.
Example 3-2. Solve for the displacement of a particle which is subject to a force of constant magnitude P which is rotating at a uniform angular rate ro "in the xy plane such that
Fx == P sin rot
Fy == P cos rot
Initially the particle is at the origin and has a velocity Vo in the direction of the positive x axis. First, we can solve for the velocity component parallel to the x axis from Eq. (3-29).
x == Vo + ~ (1 - cos rot)
mro
Another integration results in the displacement
( P) p.
x == 'IJo + - t - --2 sIn rot
. mro mro
in agreement with Eq. (3-30). .
(3-31)
(3-32)
In a similar fashion, one finds that the y components of the velocity and displacement are
. p.
y == - SIn rot
mro
.P
Y == - 2 (1 - cos rot)
mro
(3-33)
(3-34)


.' 72 DYNAMICS OF A PARTICLE CHAP. 3
. It is interesting to note from Eqs. (3-32) and (3-34) that the path of the particle consists of a uniform circular motion plus a uniform translation. To see this more clearly, consider a primed coordinate system translating uniformly relative to the unprimed system such that the particle position in the primed system is
I ( P) p .
x == x - Vo + - t == - - 2 sIn rot
mro mro
P
y' == y == - (1 - cos rot)
mro 2
(3-35)
(3-36)
These equations represent a circular path of radius P/mro 2 centered at x' == 0, y' == P/mro2 as shown in Fig. 3-3. Of course, the primed system is an inertial system since it is translating uniformly relative to the fixed xy system. It can be seen that, even in the case
y'
~-+-~-_ _ _ _ _X'
0' Fig. 3-3. The path of the particle relative to a uniformly translating system.
of general initial conditions, a uniformly translating coordinate system can always be found such that the motion of the particle relative to this system consists of uniform circular motion in a plane.
Case 3: F == F:lx)i + Fy(y)j + Fz(z)k. For this case, we shall again consider only the x component of the motion, noting that similar results apply to the motion in the y and z directions. The differential equation for the motion is
(3-37)
Because the force Fx is a function of the displacement rather than time, it is desirable to integrate with respect to x. This can be accomplished by making the substitution
X==V (3-38)
implying that
I
I


SEC. 3-1 DYNAMICS OF A PARTICLE 73
.. dv dv dx dv
x == dt == dx dt == v dx (3-39)
From Eqs. (3-37) and (3-39), we obtain
dv (
mv -d == Fx x)
x . (3-40)
which can be integrated directly to give
1 IX
2" m(v2 - v~) == Xo Fx(x) dx (3-41)
This result is an application of the principle of work and kinetic energy which will be discussed in Sec. 3-2. Equation (3-41) can be integrated again to give a solution of the form
t == f(x o, vo, x) (3-42)
from which one obtains a solution for x of the form
x == g(xo, vo, t) (3-43)
Example 3-3. A mass m and a linear spring of stiffness k are connected as shown in Fig. 3-4. Assuming one-dimensional motion with no friction, solve for the displacement x as a function of time and the initial conditions.
In this example,
or, from Eq. (3-37),
m
/ / / / / / ~~~~//:
Fig. 3-4. A mass-spring system.
Fx == -kx
mx + kx == 0
Using Eq. (3-41), we obtain
21 m(v2 - v~) == - IX kx dx == -~ k(x2 - x~)
Xo 2
and, solving for the velocit¥, we find that
v= ~~ = [v; - ! (x2 - xnr
or
I
x [ k J-1/
2
t == Xo v~ - m (x2 ~ x~) dx
Evaluating the integral, we obtain
(3-44)
(3-45)
(3-46)
(3-47)


74 DYNAMICS OF A PARTICLE CHAP. 3
t = ,J!H [sin-! x - sin'-! Xo ]
, k· ,.J(m/k)v~ + x~ ,.J(m/k)v~ + x~
or
x = -"/(m/k)v~ + x~ sin (J!!n t + a) (3-48)
where
a = sin- 1 X o
,.J(m/k)v~ '+ x~ (3-49)
This is the solution for the free motion of a mass-spring system with arbitrary initial conditions and represents an example of one-dimensional harmonic motion.
Case 4: F = Fx(x)i + Fy(y)j + Fz(i)k. In the analysis of this case, we shall again consider just the x component of the motion because the form of the equation is such that the three component motions are each independent. The equation of motion can be written as
dv (
m dt = Fx v) (3-50)
from which we obtain
J
v dv
t = m Vo Fx(v) (3-51)
where, in this case, the velocity v is in the x direction. Evaluating the integral, one can solve for velocity as a function of the initial velocity and time:
dx (
v = dt = g va' t) (3-52)
A second integration results in the displacement in the form
x = f(xo, va' t)
Another approach is to write Eq. (3-50) in the form
dv (
mv dx = Fx v)
Integrating this equation results in
J
v vdv
m F ( ) = x - Xo
Vo x v
Now eliminate v from Eqs. (3-51) and (3-55) and one again obtains
x = f(xo, va' t)
as in Eq. (3-53).
(3-53)
(3-54)
(3-55)
Example 3-4. Solve for the motion of a particle of mass m that is moving in a uniform gravitational field and is subject to a linear damping force.
I


SEC. 3-1 DYNAMICS OF A PARTICLE 75
The force of the linear damper always acts in a direction opposite to the velocity vector and its magnitude is directly proportional to the velocity. So if we consider the motion to occur in the xy plane with gravity acting in the direction of the negative y axis, then the total force acting on the particle
IS
F == -ev - mgj
Now the x component of the force is
Fx == x (-ev) == -ex
v
(3-56)
(3-57)
since x/v is the cosine of the angle between the velocity vector and the x axis (Fig. 3-5). Similarly,
Fy == -ey - mg
y
~----------------~~--x
Fig. 3-5. The trajectory of a linearly-damped particle which moves in a uniform gravitational field.
So the differential equations of motion are
mx + ex == 0 and
my + cy == -mg
(3-58)
(3-59)
(3-60)
Consider first the motion in the x direction. Eq. (3-51) can be used directly in this case, resulting in
m JX dx _ m (x)
" t == - c :(;0 X - - c In xo
or
(3-61)
where Xo is the x component of the initial velocity. Another integration results in
(3-62)
where the constant of integration has been evaluated from initial conditions. Now consider the motion· in the y direction. Equation (3-60) can be integrated to obtain


76 DYNAMICS OF A PARTICLE CHAP. 3 .
t = _m fil dy = _m In[Y + (mg/c)]
c Yo Y + (mg/c) c Yo + (mg/c)
or
Y = - n::g + (Yo + n::g) e-ct/m (3-63)
Integrating again, assuming an initial displacement Yo, we obtain
_ mgt + m (. + mg)(l -ct/m) (3-64)
y - Yo - -c- C Yo C - e
From Eqs. (3-62) and (3-64), we note that the displacement in the x direction has a limiting value as t approaches infinity, namely,
mxo
x = Xo + c
On the other hand, the y displacement changes continuously in the negative direction for large t, its limiting velocity being
. mg
y= - c
The physical interpretation of this last result is that the gravitational force
mg and the friction force cy are of equal magnitude and opposite direction in the steady state, resulting in no net external force being applied to the particle. Another point to notice is that the solutions for the motions in both the x and y directions involve a term of the form exp( - ct/m). The exponent contains the ratio
m
r r = c
which has the dimensions of time and is known as the time constant for the system. Now, time constants are often associated with first-order systems, but Eqs. (3-59) and (3-60) are seen to be second-order differential equations. Since they contain no term in x or y, however, they are also first-order in the velocities Vx and Vy as may be observed by writing them in the form:
rrvx + Vx = 0
rrvy + Vy = -rrg
The solutions for the velocities given by Eqs. (3-61) and (3-63) are typical of the form of the response of first-order systems to an initial condition or to a constant forcing term.
3-2. WORK AND KINETIC ENERGY
We now present some general principles of particle mechanics. For the cases where these principles apply, they are directly derivable from Newton's


II
SEC. 3-2 DYNAMICS OF A PARTICLE 77
laws of motion, and thus they contain no new information. Nevertheless, they promote further insight into particle dynamics and, by providing some integrals of the Inotion, aid in the solution of many specific problems. The first of these principles to be presented is the principle of work and kinetic energy: The increase in the kinetic energy of a particle ill going from one point to another is equal to the work done by the external forces acting on the particle as it lnoves over the given interval.
To illustrate the meaning of the principle, as well as to define some of the terms, consider a particle of mass m that Inoves from A to B under the action of an arbitrary external force F, as shown in Fig. 3-6. Starting with
z
ef
m
~----------+------y
x
Fig. 3-6. The path of a particle moving under the action of an external force.
Newton's law of motion in the form
F =mr
let us evaluate the line integral of each side of the equation over. the given
path from A to B. J: F • dr = s: mr •dr (3-65)
where dr is taken in a direction tangent to the curve at each point. We note that
r
CD dr = ~!{(t. t)dt = ~d(V2)
2 dt 2
where it is assumed that the infinitesimal changes in position and velocity occur during the same time interval. Thus, we see that
JB mr •dr = m2 r d(V2) = ~ (~ - ~) (3-66)
A , VA
where vA and VB are the velocities of the particle at points A and B, respec


78 DYNAMICS OF APAR......ICLE CHAP. 3
tively, of the path. From Eqs. (3-65) and (3-66), we obtain the result:
f
BF d 1 2 1 2
ED r = - mVB - - mVA
A 2 2 (3-67)
The integral on the left side of this equation is the work W which is done by the external force F as it moves along the path from A to B.
"fB
W= AFedr (3-68)
The kinetic energy T of a particle relative to an inertial system is
T = ~ mv2 (3-69)
where v is the speed of the particle relative to that system. Therefore, the right side of Eq. (3-67) represents the increase of kinetic energy in going from A to B. Using Eqs. (3-68) and (3-69), we can express the general result of Eq. (3-67) in the form (3-70)
It should be emphasized that calculations of work and kinetic energy are dependent upon which inertial system is used as a reference frame. Nevertheless, even though the individual terms may vary with the reference frame, the general principle of work and kinetic energy is valid in any inertial frame.
Example 3-5. A whirling particle of mass m is pulled slowly by a string toward a fixed center at 0 (Fig. 3-7) in such a manner that the radial component of velocity is small compared to the tangential component. Also,
m
Fig. 3-7. A particle whirling about a fixed point.
'f can be neglected relative to the centripetal acceleration. Using energy methods, find the angular velocity ro as a function of r. Initial conditions are ro(O) = roo and r(O) = ro' In order to use the principle of work and kinetic energy, we must first determine the total external force acting on the particle. The string provides the only means for exerting an external force and therefore this force must be radial in nature. Thus, in accordance with Newton's law of motion, the
-j


SEC. 3-3 DYNAMICS OF A PARTICLE 79
acceleration of the particle is entirely radial. Since;: is assumed to be negligible, the path at any point is approximately circular and the acceleration is just the centripetal acceleration. From Eqs. (1-41) and (2-53), the total external force is the radial force
(3-71)
The work done on the particle in an infinitesimal radial displacement dr is
dW == Fr dr == -mm2r dr (3-72)
But, by the principle of work and kinetic energy, each increment of work done on the particle is accompanied by an equal increase in the kinetic energy. The kinetic energy is
T == ~ mv2 == ~ mr2m2
2 .2
since we neglect the radial velocity. Thus,
dT == mm2 r dr + mr2 m dm
Therefore we obtain
dW== dT
and, from Eqs. (3-72) and (3-74),
- 2m m2 r dr == mr2 m dm
(3-73)
(3-74)
Rearranging and integrating in the interval ro to r, corresponding to the angular rates mo and m, we obtain
or
-2In (;J = In (:J (3-75)
Solving for (i) as a function of r, we find that
6) = (; )\~o (3-76)
This final result can also be written in the form
mr2 co == mr~ (i)0 (3-77)
which is in agreement with the principle of conservation of angular momentum to be developed in Sec. 3-6.
3-3. CONSERVATIVE SYSTEMS
Referring again to Fig. 3-6 let us suppose that the force F acting on the . given particle has the following characteristics: (1) it is a function of position only; (2) the line integral


80 DYNAMICS OF A PARTICLE CHAP. 3
f: F • dr
is a function of the end points only and is independent of the path taken between A and B. Now, from characteristic (1),
f BF • dr == - fA F • dr (3-78)
A B·
for the case where the line integral on the right is taken in the reverse direction along the same path. But, by characteristic (2), it is also true for an integration along any two paths connecting A and B. Therefore,
(3-79)
where the integral is taken around any closed path. A force with the characteristics just cited is said to be a conservative force, that is, it forms a conservative force field. Practically speaking, this means that the force is not dissipative in nature and that any mechanical process taking place under its influence is reversible. The property of reversibility can be clarified as follows: if, at a certain moment, the velocities of all moving parts are reversed, then, following the same physical laws, a reversible mechanical process will retrace its former sequence of positions. and accelerations in reverse order, as though time were running backwards. If the work W, as given by Eq. (3-68), is found to depend only upon the location of the end points, then the integrand must be an exact differential.
F • dr == -dV (3-80)
where the minus sign has been chosen for convenience in the statement of later results. Therefore we find that
W == JE F • dr == - fB dV == VA - VlJ
A A (3-81)
Equation (3-81) states that the decrease in the potential energy V in moving the particle from A to B is equal to the work done on the particle by the conservative force field. Conversely, the increase in potential energy in moving the particle between two points is equal to the work done against the conservative field forces by the particle. The potential energy V is a scalar function of position only, for a given particle. The sum of the potential and kinetic energies is known as the total energy E, and from Eqs. (3-70) and (3-81), we find that this sum is constant for a conservative system.
(3-82)
Equation (3-82) is a mathematical statement of the principle of conservation of mechanical energy. It applies to systems in which the only forces
II \
-I


II,
SEC. 3-4 DYNAMICS OF A PARTICLE 81
that do work on the particle are those arising from a conservative force field. Note that workless forces, such as those due to frictionless, fixed constraints, do not change the applicability of the principle.
3-4. POTENTIAL ENERGY
Let us consider again the potential energy V. We saw that V is a function of position only, for the case of a conservative force field. Therefore, if we express the position of a particle in terms of its cartesian coordinates, we
find that ---------------'- ... --. '-..---------'-.--".,"..-----.----' .... ,..-.._"."., '.
oV oV oV
dV == -dx + -dy + -dz
ox oy OZ
Also,
dr == dxi + dyj + dzk and therefore
F • dr == Fx dx + Fy dy + Fz dz
From Eqs. (3-80), (3-83), and (3-84), we obtain
oV
Fx == ox
F == _oV
y oy
F == _oV
z OZ
(3-83)
(3-84)
(3-85)
since we note that Eq. (3-80) is applicable for an arbitrary infinitesimal displacement; hence the coefficients of dx, dy, and dz in Eq. (3-83) must be equal to the negative of the corresponding coefficients in Eq. (3-84). Thus we can write
F==Fi+FO+Fk== _oVi_oV._oVk
x yJ z ax ay J OZ (3-86)
Now the gradient of the scalar function V is
VV== oV i + oV. + OV k
ax oy J OZ, (3-87)
and therefore we find that the force exerted by a conservative force field acting on the particle is 1
F == -VV (3-88)
This means that the force is in the direction of the largest spatial rate of decrease of V and is equal in magnitude to that rate of decrease.
1The converse is not necessarily true, that is, a force derivable from a potential function V by using Eq. (3-88) may not be conservative, as in the case of a time-varying field.


82 DYNAMICS OF A PARTICLE CHAP. 1
Equation (3-88) is a general vector equ,ation and therefore the gradient need not be expressed in terms of cartesian coordinates. For a general, coordinate system, the component of the force F in the direction of the unit vector el is given by
(3-89)
where Xl is the linear displacement in the direction el at the point under consideration.
Inverse-square Attraction. As an example of a potential energy calculation, consider the case of an inverse-square attraction of a particle of mass m toward a fixed point. The force exerted by the attracting field is entirely radial and is equal to
OV K
F - -- --
T - or - r2 (3-90)
where K is a constant and r is the distance of the particle from the attracting center. Since the force is a function of r only, we can integrate Eq. (3-90) directly to give
K
V= -- + C
r (3-91)
where C is an arbitrary constant of integration. In the usual case of inversesquare attraction, we choose C = 0, implying that the potential energy is always negative and approaches zero as r approaches infinity. In the general case, the fact that the potential energy contains an arbitrary constant would seem to require that all measurable quantities of the motion such as velocity, acceleration, and so on, should be independent of the choice of C, since the motion in a given situation is not arbitrary. That this is actually true is confirmed further by noting that the potential energy enters all computations as a potential energy d(fference, in which case the constant C cancels out. As a result, the choice of C, that is, the choice of the datum or reference point of zero potential energy, is made for convenience in solving the problem at hand.
Gravitational Potential Energy. The most commonly encountered inversesquare force in the study of mechanics is the force of gravitational attraction, and as we have seen, the corresponding gravitational potential must be of the form given by Eq. (3-91). Now let us consider the particular case of gravitational attraction by the earth. Assuming that the mass distribution of the earth is spherically symmetrical about the center, it can be shown (Sec. 5-1) that the attractive force on an external particle is the same as if the entire mass of the earth were concentrated at its center. Hence we see from Eq. (3-90) that the gravitational force on a particle of mass m outside the earth's surface is the radial force
. II


SEC. 3-4 DYNAMICS OF .A PARTICLE 83
Fr = ~ (r > R) (3-92)
where R is the radius of the earth (Fig. 3-8). The constant K can be evaluated
o
Fig. 3-8. The earth and a particle of mass m.
from the knowledge (Sec. 1-3) that, assuming a nonrotating earth, the weight w of a particle is the force of the gravitational attraction of the earth on the particle, measured at the earth's surface. Thus
K
w == -Fr == R2 (3-93)
But, in this instance, the weight is also given by
w == mgo (3-94)
where go is the acceleration of gravity at the surface of a spherical nonrotating earth. From Eqs. (3-93) and (3-94), we obtain that
(3-95)
Therefore, from Eq. (3-92),
(r> R) (3-96)
and, from Eq. (3-91),
(3-97)
But we can also write the potential energy in terms of the height h above the earth's surface. Letting
r==R+h (3-98)
we find from Eq. (3-97) that
V - mgoR
- -1 + (h/R)
For motion near the surface of the earth, that is, for h « R, this equation is approximated by


.I '
84 DYNAMICS OF A PARTICLE CHAP. 3
V~ -mgoR(I- ~)
Now we eliminate the constant term -mgoR by choosing the zero reference: for potential energy at the earth's surface. Then we obtain
V ~ mgoh (h« R) (3-99)
In the calculation of local trajectories near the earth's surface, a reference frame is often chosen that is fixed in the earth and the value of the acceleration ~f gravity includes the effects at that point of the earth's rotation. In this case, the gravitational field is essentially uniform and we use the symbol g for the local acceleration of gravity. Hence we find that
V = mgh (3-100)
Potential Energy of a Linear Spring~ Another commonly encountered form of potential energy is that due to elastic deformation. As an example of elastic potential energy, consider a particle P which is attached by a linear spring of stiffness k to a fixed point 0, as shown in Fig. 3-9. If the elonga
~x
ok P
"...---WV'v,....----- ~ kx ~
dmdm,
Fig. 3-9. The interaction forces of a particle P and a linear spring.
tion x of the spring is measured from its unstressed position, the particle will experience a force
OV
Fx = - - = -kx
ox (3-101)
Direct integration of Eq. (3-101) with respect to x, choosing the zero reference for potential energy at x = 0, results in
V = i kx2 (3-102)
Note again that, in the preceding development, the force Fx is the force exerted by the spring on the particle and not the force of the opposite sign that is applied to the spring by the particle. In other words, the emphasis here is on the potential ability of the spring to do work on its surroundings, and not vice versa.
Example 3-6. A particle of mass m is suspended vertically by a spring of stiffness k in the presence of a uniform gravitational field, the direction of the gravitational force being as shown in Fig. 3-10. If the vertical displacement y of the mass is measured from its position when the spring is unstressed, solve for y as a function of time. The initial conditions are


SEC. 3-4 DYNAMICS OF A PARTICLE 85
Fig. 3-10. A Inass suspended by a linear spring.
yeO) == Yo and yeO) == o. Also find the maximum values of kinetic energy and potential energy in the ensuing 1110tion. The equation of ITIotion can be written with the aid of the free-body diagram shown in Fig. 3-11, where we note that the spring force and the
ky
mg
Fig. 3-11. A free-body diagranl showing the external forces which act on mass m.
gravity force are the only forces acting on the particle. Using Newton's law t.. - of motion we can write
my == -ky - mg
or
my + ky == -mg (3-103)
Rather than solve this equation by direct integration in a fashion similar to Example 3-3, we shall obtain the solution as the sum of two parts; that is,
y == Yt + Ys (3-104)
where the transient solution Yt (also called the complementary function) is the solution to the homogeneous equation
my + ky == 0 (3-105)
and the steady-state solution Ys (also called the particular integral) is a solution which satisfies the complete differential equation.
In general, for this case of an ordinary differential equation with constant coefficients, one assumes that the transient solution contains terms of the form CeAt, where C and A. are constants. 2 The exponential function is chosen, since it retains the same variable part upon differentiation. For the problem at hand, the substitution of y == CeAt into Eq. (3-105) results in .
2This assumption is valid except' for the case of repeated roots, a situation which will not be considered here.


86 DYNAMICS OF A PARTICLE CHAP. 3
(m'A2 + k)CeAt = ,0
Since this result must hold for all values of time and we rule out the trivial case, C = 0, we obtain the so-called characteristic equation
mI·} + k = 0 (3-106)
The roots are imaginary in this case:
A.I,2 = + if!fn (3-107)
where i =.v=I. Thus \ve assume a transient solution of the form
(3-108)
Substituting for the exponential functions in terms of sine and cosine functions, we use the general formula
ei8 = cos f) + i sin f)
to obtain
Yt = (CI + C2) cos f!fn t + i(CI - C2) sin -flfn t
Now if we let
A = C} + C2 and B = iCC} - C2)
we obtain the transient solution in the form
Yt = A cos f!fn t + B sin ,J! t
(3-109)
(3-110)
(3-111)
where A and B are arbitrary constants. In general, there must be as many arbitrary constants as the order of the differential equation, and as we shall see, the constants are ultimately evaluated on the basis of initial conditions. The steady-state solution can be found by several methods. For the common case where the forcing function is represented by a constant or by terms of the form t n (n a positive integer), ekt , sin rot, cos rot, or by sums and products of these terms, the method of undetermined coefficients is applicable. (See a standard textbook on differential equations for details of that method.) Briefly, however, the usual form of the steady-state solution consists of terms of the form of the forcing function plus terms containing any different variable parts obtainable by successive differentiations of the forcing function. In this particular case, the forcing function is the constant term - mg and the steady-state solution is simply
(3-112)
as may be verified by substitution into the differential equation. Therefore the total solution is of the form


.1
I!t
SEC. 3-4 DYNAMICS OF A PARTICLE 87
Y == Yt + Ys == _lng + A cos !k t + B sin !k t
k 'Vm 'Vm (3-113)
Now we find the constants A and B froln the two given initial conditions. Upon evaluating Eq. (3-113) at t == 0, we obtain
yeO) == Yo == A - ~g
Differentiating Eq. (3-113) with respect to time and setting t == 0, we have
Hence
yeO) = 0 = J! B
A - Y + mg
-0 k
B==O
From Eqs. (3-113) and (3-114), the cOInplete solution is
mg ( mg) IF
Y == - T + Yo + T cos Vm t
(3-114)
(3-115)
It can be seen that the solution consists of a sinusoidal oscillation of the mass about the position Y == -lng/k, which is the position of static equilibrium for the particle. The terms transient and steady-state should not be taken literally as describing the nature of certain terms in the solution, although in many instances the transient solution amplitude decreases exponentially with time and the steady-state solution persists with undiminished magnitude. In this example, the transient solution oscillates with constant amplitude. For the case of a system with one or more roots with positive real parts, it would actually show an amplitude which increases with time. On the other hand, the steady-state solution could have a magnitude which increases or decreases with time, depending upon the nature of the forcing function. Now let us calculate the kinetic and potential energies for this system. From Eq. (3-69), the kinetic energy is
(3-116)
The potential energy arises from both the gravitational force and the spring force, since they are conservative in nature. Adding the individual potential energies to obtain the total, we have from Eqs. (3-100) and (3-102),
1
V == mgy + - ky2 (3-117)
2
the reference point of zero potential energy being taken at y == O. We are considering a conservative system and therefore the total energy E is a con


88 DYNAMICS OF A PARTICLE
stant. Evaluating E from its initial value, ":Ie obtain
1
E= T+V= mgyo + 2ky~
CHAP. l
(3-118)
It can be seen that the kinetic energy is maximum when the potential energy is minimum and vice versa. From Eq. (3-116), we see that
(3-119)
and therefore
(3-120)
On the other hand, we see by setting dV/dy = 0 that Vmin occurs at
Y = mg
-T
and is equal to
(3-121)
Therefore,
(3-122)
Thus we see that
(3...: 123)
even though the extreme values of V are not equal to the corresponding extreme values of T. Now let us consider the case where the displacement is measured from its equilibrium position. Calling this vertical displacement z, we can write
Z = y + mkg (3-124)
From Eqs. (3-115) and (3-124), we find that the complete solution can be written in the form
where
Z = Zo cos /!fn t
mg
Zo = Yo + k
(3-125)
(3-126)
Now let us choose the zero reference for potential energy to be at z = 0, that is, we choose a constant such that V = 0 when z = O. Then,
1 1 m 2g2
V = mgz + 2" ky2 - "2 "
1 ( mg)2 1 m2g2
= mgz + "2 k z - T - 2""
,I


SEC. 3-5 DYNAMICS OF A PARTICLE 89
which reduces to
1
V == _kZ2
2 (3-127)
Therefore, if we use the static equilibrium posItion as the zero reference for potential energy, we find that the total energy is
E == _1 lni2 + _1 kZ2 == ~ kz~
2 2 2 (3-128)
Also,
(3-129)
and
(3-130)
In summary, it can be seen that the analysis is simplified by measuring displacements from the position of static equilibrium and setting the potential energy equal to zero at this point. In this case, the total force acting on the particle is - kz and, from Eq. (3-88), we obtain
oV
- -kz
oz 
or
V == ~kZ2
2
in agreement with Eq. (3-127), provided that the reference for zero potential energy is again taken at the static equilibrium position. Note that this expression includes the gravitational as well as the elastic potential energy.
3-5. LINEAR IMPULSE AND MOMENTUM
In Sec. 1-2, we saw that the linear momentum P of a particle is just the product of its mass and velocity.
p == mv (3-131)
Thus Newton's law of motion can be written in the form
F==p (3-132)
where it is assumed that the particle mass is constant and its velocity is measured relative to an inertial frame. I". N ow let us integrate both sides of Eq. (3-132) over the time interval
11 to 12•
(3-133)
where PI and P2 are the values of the linear momentum vector at times 11


90 DYNAMICS OF A PARTICLE CHAP. 3
and t2, respectively. The time integral of the force F is known as the impulse fl)T of the force, that is,
fl)T == Jt
2 F dt
tl
So, from Eqs. (3-133) and (3-134) we obtain of linear impulse' and momentum:
(3-134)
a statement of the principle
The change in the linear momentum of a particle during a given interval is equal to the total impulse of the externalforces acting on the particle over the same interval.
$ == P2 - PI == mV2 - mVl (3-135) .
,.
The length of time over which the integration proceeds does not influence
the result. Of course, we assume that t2 > t 1, but the interval may approach zero. If the duration of the external force approaches zero, but its amplitude becomes very Jarge in such a manner that the time integral of the force remains finite, then the force is known as an impulsive force. The effect of an impulsive force in changing the motion of a particle is expressed entirely by its total impulse $, in accordance with Eq. (3-135). We see that the velocity changes instantaneously, but the position cannot change instantaneously for a finite impulse because the velocity remains finite. It is convenient to express an impulsive force of total impUlse $ occurring at time t == T in the form
F == :17 oCt - T)
where o(t) is the Dirac delta function defined as follows:
such that
{o t * °
B(t) == 00 t == 0
r~ t(t)dt = 1
It can be seen that the total impulse due to F is
r~ F dt = fijT r~ t(t - T) dt = fijT
in agreement with the original assumption.
(3-136)
(3-137)
(3-138)
Equation (3-135) can also be written in terms of its scalar components. Choosing an inertial cartesian system in which to express the motion of the particle, we find that I
~ == Jt
2 Fx dt == m(x)2 - m(x)l
tl .
(3-139)


SEC. 3-5 DYNAMICS OF A PARTICLE 91
where the subscripts 1 and 2 indicate that the evaluations of the given velocity components are to, be made at II and 12, respectively. These equations are obtained directly from Eq. (3-135) or by integrating Eq. (1-43) with respect to time. From Eq. (3-132) we see that if the external force F equals zero, then the linear momentum p is constant. This is the principle of conservation of linear momentum. Similarly, we note from Eq. (3-139) that if any component of F is zero for a certain interval of time, then the corresponding component of the momentunl is conserved during that interval. A comparison of Eq. (3-135), which equates the total impulse to the change of momentum, and Eq. (3-70), which equates the work done on the particle to the change in kinetic energy, reveals some interesting qualitative differences. First, Eq. (3-135) is a vector equation, whereas Eq. (3-70) is a scalar equation. The vector equation is advantageous at times because it gives the direction as well as the lTIagnitude of the velocity. On the other hand, scalar equations are often easier to use. If the force F is given as a function of time, impulse and momentulTI methods are usually called for; but if F is given as a function of position, then work and energy methods can be used to advantage. Another point of interest is that the total impulse fl7, and consequently the change i~ linear momentum, is independent of which inertial frame is chosen for viewing the motion of the particle. This is in contrast to the calculations of work and the change of kinetic energy which, as we have seen, are dependent upon a specific frame of reference.
Example 3-7. A particle m slides along a frictionless wire that is fixed in inertial space, as shown in Fig. 3-12. Assuming that a known external
z
J--~~------y
x
Fig. 3-12. A particle moving over a fixed path under the action of an external force.
force F acts on the system, find a general method for calculating speed changes of the particle.


92 DYNAMICS OF A PARTICLE CHAP. 3
If at each instant we take components of force and acceleration along the path, we find from Eqs. (1-41) and (2-41) that
Fs == ms (3-140)
where Fs is the component, along the tangential direction, of the external
force F, and s is the tangential component of the acceleration. Note that the frictionless constraint forces are normal to the path and do not contribute to Fs. Integrating Eq. (3-140) with, respect to time, we obtain
S
t2 Fs dt == m(s)2 - m(s)l (3-141)
tl
Thus we obtain an essentially one-dimensional or scalar version of the equation of impulse and momentum that applies to motion along a fixed curve In space. Of course, the equation of work and kinetic energy, Eq. (3-67), can be applied directly, resulting in
S: F, ds = ~ m(S)~ - ~ m(s)~ (3-142)
3-6. ANGULAR MOMENTUM AND ANGULAR IMPULSE
Angular Momentum. We have seen that the linear momentum of a particle with respect to a fixed reference frame is the vector mY, where m is the mass of the particle and v is its absolute velocity. Now let us consider the momentum vector as a sliding vector whose line of action passes through the particle, as shown in Fig. 3-13. If r is the position vector of the particle
z
x
Fig. 3-13. A particle moving relative to a fixed point o.
with respect to a fixed reference point 0, then the moment of momentum or angular momentum about 0 is given by


I
I
SEC. 3-6 DYNAMICS OF A PARTICLE 93
H == r X mv (3-143)
Let us differentiate this equation with respect to time and note that
v == r. We obtain
Ho == r X }}1r + t X tnt
and, since the cross product of a vector with itself is zero, this reduces to
H. == r X Inr (3-144)
Now let us consider again Newton's law of motion written in the form
F == mr
and take the cross product of each side with the position vector r. We find that
r X F == r X mr (3-145)
and ilnmediately identify the left side of the eGuation as the nloment M of
the total external force F about the fixed point o.
M==rxF (3-146)
Then, from Eqs. (3-144), (3-145), and (3-146), we obtain
M==H (3-147)
which is a statement of the important principle that the moment about a fixed point of the total external force applied to a particle is equal to the time rate of change of the angular momentum of the particle about the same fixed point. We see from Eq. (3-147) that for the case where the external moment M is zero, the angular momentum H must be constant in magnitude and direction. This is known as the principle of conservation 0.( angular momentunl. The general vector relationship given by Eq. (3-147) can also be written in terms of its components. Thus, choosing a fixed cartesian coordinate system, one obtains the scalar equations:
Mx == Hx
My == Hy
Mz := Hz
(3-148)
When written in this manner, each equation can be interpreted as relating the moment and the rate of change of angular momentum about the corre
sponding fixed axis passing through the point o. So we can see that even though the total angular momentum is not conserved in a given case, one . of the components of M might vanish. This would require the angular momentum about the corresponding axis to be- conserved. •A,.. similar situation occurs when the motion of a particle is confined to a plane (Fig. 3-14). In this case, the angular momentum is essentially scalar


94 DYNAMICS OF A PARTICLE CHAP. 3 .
Fig. 3-14. The motion of a particle in a plane.
in nature since its direction is fixed. Let us assume that the velocity of the particle m has radial and tangential components given by
Vr == f (3-149)
V4> == rro
where ro is the angular velocity of the radius vector as it moves in the plane of the particle motion. The angular momentum is of magnitude
H == mrv4> == mr2 (i) (3-150)
and, in accordance with the right-hand rule, is directed out of the page. We note that H is independent of Vr since the line of action of the corresponding component of linear momentum passes through the reference point O. Now let us assume that the particle is acted upon by radial forces only. Then, regardless of the manner of their variation, the applied moment M will be zero at all times, and therefore the angular momentum will be conserved. From Eq. (3-150), we obtain that
(3-151)
where the subscripts 1 and 2 refer to the values of the variable at the arbitrary times t 1 and t 2, respectively. Returning briefly to a more general discussion of the angular momentum of a particle, let us consider what latitude is permissible in the application of Eq. (3-147). Specifically, in the analysis of the general motion of a single particle, what constitutes a proper choice of (1) a reference frame for calculating the linear momentum vector; (2) a re.ference point O? Eliminating at the outset the trivial case where the reference point 0 is chosen to coincide with the particle, we can state that the reference frame must be inertial and the reference point must be fixed in that frame. These requirements arise because the derivation of Eq. (3-147) is based upon Newton's law of motion and also because t and v are assumed to be equal.
-~ )
I