advance mechnics of machines m tech basics

KINEMATICS AND DYNAMICS

OF PLANE MECHANISMS

JEREMY HIRSCHHORN ~

A ssociafe Professor of Mechanical Engineering The University of New South Wales

Sydney, N.S. W., Australia

McGRAW-HILL BOOK COMPANY, INC.

New York San Francisco Toronto London

Original from UNIVERSITY OF MICHIGAN

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CONTENTS

Preface • • • • • • • • • • • • •

Introduction . . • •

Engineering System of Units • • • • •

Chapter 1. Fundamentals of Vector Analysis •

1-1. Scalars and Vectors . . . . . 1-2. Vector Notation; Unit Vector; Representation of Vectors. 1-3. Composition, Subtraction, and Resolution of Vectors . . 1-4. Multiplication of a Vector by a Pure Number or a Scalar . 1-5. The Vectorial, or " Cross," Product of Two Vectors . . 1-6. The Scalar, or " Dot," Product of Two Vectors . . . 1-7. Differentiation of a Vector with Respect to Time . . . 1-8. Problems . . . . . .

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Chapter 2. Kinematics of the Plane Motion of a Particle .

2-1. Definitions . . . 2-2. Cartesian Reference Frame 2-3. Polar Reference Frame. . 2-4. Moving Reference Frame . 2-5. Summary. 2-6. The Radius of Curvature

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2-7. Graphic Differentiation and Integration . .

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2-8. Method of Finite Differences . . . . .

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2-9. Absolute and Relative Motion; Relative Motion of Separate Particles 2-10. Problems. . . . .

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3 3 4 8 8

11 13 14

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15 15 16 18 19 20 23 28 29 30

Chapter 3. Kinematics of the Plane Motion of a Rigid Body . 32

3-1. Definitions and Basic Concepts . • • •

3-2. Types of Plane Motion. . . . •

3-3. The Velocity Pole, or Instant Center of Rotation . •

• 3-4. Determination of Velocities by Means of the Velocity Pole . . . 3-5. Determination of Velocities by Means of Orthogonal Velocity Vectors . 3-6. Po lodes . . . . . . . . . . 3-7. Relative Motion of Physically Connected Particles . 3-8. General Motion as Superposition of Translation and Rotation 3-9. The Velocity Image. . . . . . .

3.:10. The Acceleration Image . 3-11. Graphical Solution of the Velocity and· Acceleration Equations 3-12. The Acceleration Pole, or Instantaneous Center of Accelerations .

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UNIVERSITY OF MICHIGAN

32 34 35 38 38 40 43 44 44 46 47 49

• • • Vlll CONTENTS

3-13. Acceleration of the Velocity Pole Ps . . • •

3-14. Acceleration of the Center of Curvature of the Moving Polode • •

3-15. Problems . . . . . . • •

Chapter 4. Kinematics of Simple Mechanisms . • • • • •

4-1. 4-2. 4-3. 4-4. 4-5. 4-6. 4-7. 4-8 . 4-9.

4-10. 4-11. 4-12. 4-13.

4-14.

Definitions . • •

Degree of Freedom of a Mechanism . • • •

Inversions of a Linkage . . . . Relative Motion of Two Planes; the Relative-velocity Pole . . . . Relative Motion of Three Planes; Kennedy's Theorem . . . . . Velocity Poles in Mechanisms . . . . . Velocity Analysis of Mechanisms by Means of Velocity Poles . . . Velocity Analysis of Mechanisms by Means of Orthogonal Velocities . Velocity Analysis of Mechanisms by Means of Relative Velocities . . Comparison of the Three Methods of Velocity Analysis . . . Acceleration Analysis of Mechanisms by Means of Relative Accelerations Analysis of Mechanisms with Rolling Pairs . . . . . Analysis of Mechanisms with Sliding and Slip-rolling Pairs in Motion; Coriolis Component. . . . . Problems . . .

Chapter 5. Kinematics of the Slider-crank Mechanism • • •

Nomenclature and Conventions . . . Conventional Velocity and Acceleration Diagrams . Simplified Construction of the Velocity Diagram . Klein's Construction of the Acceleration Polygon .

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5-1. 5-2. 5-3. 5-4. 5-5. 5-6. 5-7.

Ritterhaus's Construction of the Acceleration Polygon . . Harmonic Analysis of the Slider Acceleration of In-line Mechanisms . Approximate Expressions for the Slider Displacement, Velocity, and

5-8. 5-9.

5-10.

Acceleration in In-line Mechanisms . . . . . . . Graphical Determination of the Slider Displacement . . The Acceleration-Displacement Curve . . . . . •

Problems. . •

Chapter 6. Kinematics of Complex Mechanisms • • • •

6-1. Complex Mechanisms; Low and High Degree of Complexity . •

6-2. Goodman's Indirect Acceleration Analysis • • • • • •

6-3. Method of Normal Accelerations . • • • • •

6-4. Hall's and Ault's Auxiliary-point Method • • • • •

6-5. Carter's Method . • • • • •

6-6. Comparison of Methods • • • • • • •

6-7. Problems . • • • • • • • • • • • • • •

Chapter 7. Fundamental Principles of Dynamics and Statics. 7-1. Dynamics of a Particle; Laws of Motion . • • • • •

7-2. Dynamics of the Plane Motion of a Rigid Body • • • •

7-3. Equivalent Mass Systems . • • • • • •

7-4. Work, Power, Energy; Conservation of Energy; Virtual Work 7-5. D' Alembert's Principle; Inertia Force • • • •

7-6. Some Simple Problems of Statics. 7-7. Friction . • • • • • • • • •

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52 55 56

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59 61 63 68 69 70 76 80 82 82 83 88

94 100

106

106 107 108 109 112 113

115 116 117 119

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121 122 131 133 140 141 142

145 145 146 152 156 160 162 165

CONTENTS

Chapter 8. Forces in Mechanisms . • • • • • •

8-1. Introduction . • • • •

8-2. Free-body Diagrams • • • • •

8-3. lllustrative Examples • • • •

8-4. Friction in Link Connections •

8-5. Forces in Nonsymmetrical Linkages . •

8-6. Stress Determination in Moving Members • •

8-7. Gyroscopic Effects . • •

8-8. Problems. • • • •

Chapter 9. Dynamic Motion Analysis • • • • • •

9-1. Nomenclature and Conventions • • • •

9-2. The Energy-distribution Method . •

9-3. The Equivalent-mass-and-force Method . • • • •

9-4. The Rate-of-change-of-energy Method •

9-5. Effects of Friction • • • • • • •

9-6. Summary and Conclusions . 9-7. Problems. • • • • • •

Chapter 10. Advanced Kinematics of the Plane Motion . .

10-1. The Inflection Circle; Euler-Savary Equation . 10-2. Analytical and Graphical Determination of d, 10-3. Bobillier's Construction; Collineation Axis . 10-4. Hartmann's Construction . .

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-5. The Inflection Circle for the Relative Motion of Two Moving Planes . 10-6. Application of the Inflection Circle to Kinematic Analysis . 10-7. Polode Curvature (General Case). 10-8. Polode Curvature (Special Case); Hall's Equation . 10-9. Polode Curvature in the Four-bar Mechanism; Coupler Motion .

10-10. Polode Curvature in the Four-bar Mechanism; Relative Motion of the Output and Input Links; Determination of the Output Angular Accelera-tion and Its Rate of Change . .

10-11. Freudenstein's Collineation-axis Theorem; Carter-Hall Circle . 10-12. The Circling-point Curve (General Case) . 10-13. The Circling-point Curve for the Coupler of a Four-bar Mechanism .

Chapter 11. Introduction to Synthesis; Graphical Methods

11-1. The Four-bar Linkage . 11-2. Guiding a Body through Two Distinct Positions . 11-3. Guiding a Body through Three Distinct Positions . 11-4. The Rotocenter Triangle . .

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11-5. Guiding a. Body through Four Distinct Positions; Burmester's Curve 11-6. Function Generation; General Discussion . 11-7. Function Generation; Relative-rotocenter Method. 11-8. Function Generation; Reduction of Point Positions 11-9. Function Generation; Overlay Method . .

11-10. Function Generation; Velocity-pole Method. 11-11. Path Generation; Hrones's and Nelson's Motion Atlas

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11-12. Path Generation; Reduction of Point Positions (Fixed Pivot Coincident with Rotocenter) . . . . . . . .

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167 167 167 168 181 184 188 190 193

199

199 200 205 207 209 213 214

217 217 222 224 225 232 233 235 239 241

243 245 248 251

254 254 262 266 267 277 281 284 289 295 300 304

305

Digitized by Coogle Original from


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11-13. Path Generation; Reduction of Point Positions (Moving Hinge Coinci-dent with Rotocenter) . . . . . . . . . . . . . . 309

11-14. Path Generation; Roberts's Theorem. . . . . . . . . . 314

Chapter 12. Introduction to Synthesis; Analytical Methods . • 319

12-1. Function Generation; Freudenstein's Equation. . . . . . . . 319 12-2. Function Generation; Precision-point Approximation . . . . 321 12-3. Function Generation; Precision-derivative Approximation . . . . 322 12-4. Path Generation. . . . . . . . . . . . . . . . . 323 12-5. Synthesis of Four-bar Mechanisms for Specified Instantaneous Condi-

tions; Method of Components. . . . . . . . . . . . . 324 12-6. Synthesis of Four-bar Mechanisms for Prescribed Extreme Values of the

Angular Velocity of the Driven Link; Method of Components . . . 330

Chapter 13. Introduction to Synthesis; Grapho-analytical Methods . . . . . . . . . . • • . 336

13-1. Function Generation; Precision-derivative Approximation . . . . 336 13-2. Function Generation; Precision-point and Precision-derivative Method 344. 13-3. Path Generation; Coupler Curves with Approximately Circular Elements 346 13-4. Path Generation; Symmetrical Coupler Curves with Approximately

Straight Elements . . . . . . . . . . . . . . . . 349 13-5. Path Generation; Coupler Curves with Extended Rectilinear Portions;

Ball's Point . . . . . . . . . . . . . . . . . . 356

Bibliography

Index. •

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361

INTRODUCTION

The design of a new machine or device for the performance of an operation, or sequence of operations, associated with some particular industrial process, usually involves the following steps:

1. An assessment of the problem 2. A conception of a suitable mechanism in its skeletal form 3. A kinematic analysis, or examination of the mechanism's motion

characteristics from a purely geometrical point of view, which may reveal the need for a modification of the layout

4. A static analysis, or determination of the nature and magnitude of the forces associated with the primary function of the device

5. A choice of suitable materials of construction, based on technological and economic considerations, and a tentative proportioning of the members

6. A dynamic analysis, or determination of the inertia forces and their effects on safety and operational requirements, which may disclose the need for redesign

The chief purpose of this book is to provide the student with the proper tools for carrying out steps 3, 4, and 6 and to give him a basis for a rational approach to some problems of synthesis. It is also hoped that the book will prove a useful source of reference to the practicing

• engmeer. Before proceeding to the detailed investigation of the kinematic and

dynamic behavior of mechanisms, it will be necessary to select a suitable and consistent system of units and it will be advisable to review some fundamental notions, usually discussed in basic courses in mathematics, physics, or general mechanics.

Because the concept of force is of more immediate interest to the engineer than that of mass, force is chosen as one of the three fundamental quantities in the engineering or gravitational system of units, the other two being displacement and time. The fundamental units of measure adopted in this book are, respectively, the pound (lb), the inch (in.), and the second (sec). The reasons for selecting the inch, rather than the foot, as unit of displacement are threefold:

1. Relative displacements of machine parts are generally of the order of a few inches, and sometimes amount to only fractions of an inch.

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2 KINEMATICS AND DYNAMICS OF PLANE MECHANISMS

2. Dimensions of machine elements are usually given in inches. 3. Quantities used in the analysis of stress and strain are based on the

inch, e.g., modulus of elasticity (lb/ in. 2) and moment of inertia of cross section (in. 4). The accompanying table lists the most important quantities and their units of measure.

ENGINEERING SYSTEM OF UNITS

Quantity Symbol

Displacement. . . . . . . . . . . . . . . . . . . s, x Time. . . . . . . . . . . . . . . . . . . . . . . . . . .,. Force . . . . . . . . . . . . . . . . . . . . . . . . . . F, R

Velocity, speed . . ....... .. .. .. .. . • • v, s, :z;

Acceleration . . . . . . . . . . . . . . . . . . . . ~

a, s, .i Angular displacement . . . ...... .. . 8, cp Angular velocity ......... .. . ... . w Angular acceleration ... ...... . . . a Mass . . . . . . . . . . . . . . . . . . . . . . . . . . M Linear momentum .. .. . . ...... .. . mt Torque ... .. .. .... . . .. . . ... . .. . T Moment of inertia. . . . . . . . . . . . . . I Angular momentum . .... .. ... . . . g

Work . .............. .. .. ... .... . w Energy ..... . ... .. ..... .. .. .. . . s Power . . . . . . . . . . . . . . . . . . . . . . . . <P

Unit

• m. sec lb

in./sec in./sec1

rad rad/ sec rad/sec1

lb-sec 2 /in. lb-sec in.-Ibt lb-in.-sec 2

in.-lb-sec lb-in. t lb-in. lb-in./sec

t The reasons for choosing the in.-lb as unit of torque and the lb-in. as unit of work are given in Sees. 1-4 and 1-5.


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CHAPTER 1

FUNDAMENTALS OF VECfOR ANALYSIS

1-1. Scalars and Vectors

Physical quantities are divided into scalars and vectors. Scalars, examples of which are mass, time, and work, are completely defined by magnitude and units of measure. Vectors, such as force, velocity, and acceleration, require, in addition, the specification of direction.

Provided that the physical nature of vector quantities is kept in mind, vector analysis, or mathematical manipulation of vectors, becomes a powerful tool in the investigation of many physical phenomena and helps greatly in their proper understanding.

1-2. Vector Notation; Unit Vector; Representation of Vectors

Vector Notation. Vector quantities will be denoted by bold-faced symbols; their magnitudes, and scalar quantities in general, will be designated by italics:

F, F; r, r; v, v; etc.

Unit Vector. A very useful concept in vector analysis is the unit vector, a directed element of length one, having no physical units. It will be denoted by the symbol i with an appropriate lower-case suffix:

i, denotes a unit vector in the direction s

Representation of Vectors. A vector quantity is depicted conveniently by a directed line segment, or arrow, of a length representing, to some suitable scale, the actual magnitude. In the case of translative or lineal vectors, such as force or velocity, the line vector is drawn parallel to the line of action of the quantity considered and pointing in its direction. In depicting rotational vectors, such as torque or angular velocity, the line vector is usually1 drawn parallel to the axis around which the action takes place, pointing in the direction in which a right-hand screw would

1 When dealing with the balancing of rotating and reciprocating masses, a different convention is adopted.

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advance if turned in the sense of the particular quantity. In the special case of a two-dimensional system, the simple device of a curved arrow is

Lineal vector

,---I-f-Plane of action

......

'------::;"-Sense .of action

Rotational vector

FIG. 1-1

frequently used to indicate the sense as clockwise (cw) or counterclockwise (ccw), and the magnitude is stated separately.

1-3. Composition, Subtraction, and Resolution of Vectors

Resultant. By definition, the resultant of a vector system is a vector obtained by the process of composition, or geometrical addition.

Parallelogram Method of Composition. The resultants of two vectors a and b is constructed by setting off the vectors from a common origin, or pole, and then completing the parallelogram, as shown in Fig. 1-2. The diagonal which originates at the pole represents the resultant in magnitude and direction.

0

FIG. 1-2

The construction may be extended progressively to any number of vectors. In Fig. 1-3, the resultant s of a, b, and c is found by first constructing the resultant s' of a and b and then combining it with c.

Polygon Method of Composition. Examination of Fig. 1-3 reveals that the same result would be obtained by arranging the individual vectors "in order," i.e., tail to tip, and then joining the initial point, or tail, of the first vector with terminal point, or tip, of the last vector. Furthermore, it is evident that the sequence in which the vectors are taken has no effect on the result; i.e., the commutative law is valid in the geometri-


FUNDAME~TALS OF VECTOR A~ALYSIS 5

s'

0

FIG. 1-3

cal addition of vectors, just as it is in the algebraic addition of scalars:

a + b + c = b + a + c, etc.

B

8

• /

• /

• /

0 0

FIG. 1-4

Free, Sliding, and Fixed Vectors. It is important to realize that, although in the mathematical sense two vectors are considered equal if they have the same magnitude and direction, they may produce, or be the result of, different physical effects. (For instance, two equal parallel forces applied in turn to a given rigid body would produce the same acceleration of its center of gravity but different angular accelerations.) For this reason, three types of vector are distinguished in mechanics, viz., the free, the sliding, and the fixed . A free vector is tied neither to a specific point of application nor to a particular line of action. The velocity and acceleration of a rigid body in translation are examples of free vectors, because in this type of motion all particles have the same motion characteristics. A sliding vector is tied only to a specific line of action. The dynamic effect of a force acting on a body is not affected· by

oig1tized by Go gle Original from



a disp]acement of the force along its line of action. Hence, in dynamics, forces are regarded as sliding vectors. Another example of a sliding vector is the following. In a rigid body the distances between the particles do not change. Consequently, collinear particles of such a body have the same velocity component in the direction of the line containing

Ft

' ~,,

•

Fz

them. Thus this velocity component is a sliding vector. A fixed vector is tied to a particular point of application. The velocity of a specific point or particle is an example of a fixed vector. A second example is illustrated in Fig. 1-5.

Although F 1 and F 2 are equal and act along the same line, their static effects differ: Ft causes compression of the upper part of the pin, while F 2 produces tensile stresses in the portion below the collar. Hence, if the distribution of direct stresses due to a given force is to be investigated, the force must be

Fm. 1-5 regarded as a fixed vector. Position of the Resultant of Concurrent Vectors. The

resultant of a system of concurrent vectors passes through the common point of intersection.

Position of the Resultant of Coplanar Nonconcurrent Vectors. The resultant of a system of coplanar nonconcurrent vectors has physical significance only if the vectors concerned are forces or momenta. As indicated in Fig. 1-3, its line of action may be found by a gradual addition of the sliding component vectors. Alternatively, the line of action may be determined from the condition that the torque of the resultant about any point in the plane is equal to the algebraic sum of the torques of the component vectors with respect to the same reference point. Experience shows that the effect of the resultant is equal to the combined effect of the original system. For instance, a body in motion under the action of a two-dimensional force system will acquire a definite acceleration of its center of gravity and a definite angular acceleration. Identical dynamic effects would also be produced by a single force, equal to the resultant, applied along the proper line of action.

Subtraction of Vectors. The subtraction of a vector is equivalent to the addition of its negative, i.e., of a vector having the same magnitude but opposite direction:

d =a- b =a+ (-b)

Resolution of Vectors. The individual vectors which together form the resultant are called its components. Resolution of a given vector is the process of finding its components in specified directions.

A vector can be resolved uniquely into only two related components. If more than two directions are prescribed, the number of possible combinations of components becomes infinite.



FUNDAMENTALS OF VECTOR ANALYSIS 7

In determining the components, the procedure of the parallelogram method is reversed, as shown in Fig. 1-6.

Direction a Direction n

/ Direction t

\ Direction b

\ b

0 0

FIG. 1-6 FIG. 1-7

Of great practical importance are mutually perpendicular components, or projections, of a vector. Projections will be denoted by the same letter as the vector, with appropriate lower-case suffixes, as shown in Fig. 1-7.

Occasionally, as in the kinematic analysis of complex mechanisms, a vector may be specified by two projections in independent directions. The corresponding construction of the vector is shown in Fig. 1-8. The difference between it and the ordinary composition should be noted carefully.

Direction p

•

Direction l I

., 0

.F'IG. 1-8

Analytical Composition of Vectors. The vectors are referred to a cartesian system of coordinates, and their components in the x and y directions are calculated as shown in general terms for the vector q:

and (1-1)

The components Bz and s11 of the resultant are computed by adding algebraically the corresponding component projections:

and

The magnitude of the resultant is given by

3 = (8z2 + 8v2)1

Digitized by Coogle

(1-2)

(1-3)



and its directional angle is calculated from

t Su an 8, = -s~

(1-4)

In order to obtain the correct algebraic signs of the components, the directional angles are measured in the counterclockwise sense from the positive x axis, as indicated in Fig. 1-9.

+.)'

0 +z

FIG. 1-9

+.)'

8 8 ;y

0 s..,

1-4. Multiplication of a Vector by a Pure Number or a Scalar

From a formalistic point of view the product M a is a vector which has the same direction as a and a magnitude },tf times as great, irrespective of whether 1\f is a pure number or a scalar. From the physical point of view, however, there is a considerable difference in the nature of the two resulting vectors, because a pure number does not change the physical character of the original vector, while a scalar quantity does: if, in the product A = lr! a, M is a pure number and a an acceleration, then A is also an acceleration; if, however, M represents mass, then A becomes a force.

It follows, therefore, that a vector quantity may be expressed as the product of its magnitude (a scalar) and the appropriate unit vector; e.g.,

• q = mlr

signifying that the vector q has a magnitude m and acts in the direction r. (It is not necessary to denote the magnitude and direction by the same letter as the vector itself.)

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1-5. The Vectorial, or "Cross," Product of Two Vectors . . . .

The vectorial product of a and b, denoted by a X b, is defined as a rotational vector of magnitude ·(ab sin 8~), normal to the plane of a and




band pointing in the direction determined by the shorter rotation of a (the first vector) to b (the second vector) . The angle (Jab and the sense of rotation are found from an auxiliary diagram in which the vectors are set off from a common origin.

axb

a

FIG. 1-10

a X b = ab( sin Oab)in

Arrow of (Jab indicates sense of rotation

0

--

a

FIG. 1-11

The following properties of the vectorial product may be deduced from the definition:

1. a X b = - (b X a). 2. If a and b are parallel, a X b = 0. 3. If a and b are mutually perpendicular, mag (a X b) = ab. 4. If a X b = a X c, the components of b and cat right angles to a are

equal, b sin Dab = c sin Oac (Fig. 1-11). The vectorial product may be expressed in terms of the cartesian com

ponents of the vectors in the following manner:

a X b = (a:~: + a11) X (bz + b 71)

= (az X hz) + (az X b 71) + (a11 X bz) + (a11 X by)

According to rule 2, the products az X bz and a 11 X b 71 vanish, so that

a X b = (az X b 11) + (a11 X bx)

In Fig. 1-12 the positivez axis is so directed as to represent the rotation +x to +y. Consequently, a rotation +y to +x would be represented by a vector pointing in the direc-tion of the negative z axis. Noting this, and applying rule 3, the expression for a X b is simplified to

z

or a X b = (azb11 - aybz)iz

mag (a X b) = azb11 - a11bz (1-6) 0 b.c

a

An important example of the vectorial product is the torque, or turning moment, of a force about an axis.

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FIG. 1-12

Original from


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Complete description of the torque in Fig. 1-13 requires specification of its magnitude,

T = lF = r(sin Br, )F

and of its counterclockwise sense. Both statements are incorporated in the vectorial notation

I I

Sense of torque /l ~I

Torque axis o I

I

FIG. 1-13

r Point of application

ofF

T = r X F (1-7)

It should be noted that if the product is to indicate the sense correctly, r must appear as the first vector. This is the reason for choosing the inch-pound (in.-lb) as unit of torque.

Other physical examples of the vectorial product are the velocity and linear and angular momenta of a particle in a rigid body which rotates about a fixed axis. Velocity:

v=wxr

Linear momentum, defined as the product of mass and velocity:

dmt = dM(w X r)

where dM is the mass of the particle. Angular momentum, defined as the moment of the linear momentum with respect to the axis of rotation:

dd. = r X dmt = r X dM (w X r) = dM r X ( (o) X r)

Digitized by Coogle

I I I

z

r

M888dM

FIG. 1-14



Illustrative Example. A force F, of 20 lb magnitude, is inclined to the x axis at an angle of 120°. Its point of application is defined by the cartesian components of the position vector, rz = 6 in. and r 11 = -2 in. Calculate the torque about the origin of the coordinate system. Make a scale drawing of the configuration, and check the result obtained. Compute, and check by measuring, the angle between the position and force vectors.

Solution. By Eqs. (1-7) and (1-6),

where

Fz = F cos 120° = -10 lb and F11 = F sin 120° = 17.32 lb

Hence T = (6)(17.32)- (-2)(-10) = 83.92 in.-lb

The positive result indicates counterclockwise sense.

+y

0

r

-- - ---------------

FIG. 1-15

I I I I

+.r

In Fig. 1-15 the force arm (perpendicular distance of 0 from the line of action of F) l measures 4.2 in., so that

T = (4.2)(20) = 84 in.-lb

The angle between the position and force vectors may be calculated by using the concept of the vectorial product:

T = rF sin Orl'

With T = 83.92 in.-lb, r = 6.32 in., and F = 20 lb, the angle is computed as Or, = 125°14'. The measured angle is 125°15'.

1-6. The Scalar, or "Dot," Product of Two Vectors

The scalar product of a and b, denoted by a · b, is, by definition, a scalar quantity of magnitude (ab cos Ooo).

a · b = ab cos Ooo (1-8)



As in the case of the vectorial product, the angle Oob is determined by setting off the vectors from a common origin.

The following properties of the scalar product may be deduced from the definition:

1. a· b = b ·a. 2. If a and b are parallel, a · b = ab. 3. If a and b are mutually perpendicular, a · b = 0. 4. If a • b = a • c, the projections of b and c in the direction of a are

equal, b cos Oob = c cos Oac (Fig. 1-16).

c I I I I I I I

I I I I

Bab : a 0

FIG. 1-16

The scalar product may be expressed in terms of the cartesian components of the two vectors in the following manner:

a • b = (a.z + a") • (b.z + b") = (a.z • b.z) + (a.z • b") + (a" · b.z) + (a"· b")

By the use of rules 2 and 3, the foregoing expression is simplified to

(1-9)

An important practical example of the scalar product is the concept of work. If the point of application of a force F experiences a displacement s, then the amount of work done in the process is equal to the product of the force component in the direction of the displacement and the magnitude · of the displacement:

or, in vectorial notation, 'W=F·s (1-10)

By common usage the force appears as the first vector in the product, particularly if elemental displacements are considered:

d'W = F·ds

This circumstance was the reason for choosing the pound-inch (lb-in.) as unit of work.

Digitized by Google Original from


. - -


1-7. Differentiation of a Vector with Respect to Time

A vectorial change may involve a change .bot.h in magnitude and in direction. Such a condition is illustrated in Fig. 1-17, where the line vector 0 ~ A represents a certain vector quantity

• q = m1,.

at the time ,., and 0 ~ A', t he same quantity after a time interval Ar. The vector A ~ A' represents, therefore, the total change Aq which has taken place.

q' =q + ~q

0

FIG. 1-17

Direction of i 1 obtained by turning i,. 90° in the sense of increasing eq

The vector Aq may be resolved into two perpendicular components, viz., Aq,. in the original radial direction rand Aqz in the lateral direction l:

and

If the time interval considered is infinitesimal, the above expressions reduce to

so that

and

dq,. = dm i,. and dqz = m dfJq iz dq = dm i,. + m dfJ 9 iz dq _ dm • + dfJ q • dT - dr lr m dr l z

= mi,. + mwq iz

(1-11)

(1-12)

This first term on the right-hand side of the equation represents the rate of growth of the vector, and the second, the rate at which it changes direction.

The Time Derivative of the Unit Vector. Because the length of the unit vector is constant and has the value one, Eq. (1-12) simplifies to

di,. . dT = Will

oigit1zed by Go gle

(1-13)



Thus the time derivative of a unit vector is another vector, of a length numerically equal to the rotational speed Wi of the unit vector, which leads the unit vector by 90° in the sense of rotation.

1-8. Problems

1. Determine analytically and graphically the resultant of the following force system: F1 - 50 lb at 45°, Ft - 100 lb at 150°, Fa .,. 100 lb at 270°, and F4 =- 80 lb at 300°.

i. The point of application of the above resultant has the coordinates r,. == -5 in. and r11 = 4 in. Determine analytically and graphically the torque of the resultant R with respect to the origin 0.

3. The point of application of the resultant of Probs. 1 and 2 experiences a displacements, the components of which ares,. = 10 in. and 8 11 ... 3 in. Determine the work done by (+ ) or against (-) the resultant.

4. Using the concepts of the cross and dot products, calculate, respectively, the angles between the position vector rand the resultant Rand between the resultant R and the displacements. Check by measurement.


CHAPTER 2

KINEMATICS OF THE PLANE MOTION OF A PARTICLE

This chapter reviews very briefly basic kinematic relationships and describes some simple methods of motion analysis.

2-1. Definitions

Position. Location of the particle within a given reference frame. Path. Locus of successive positions of the particle. Displacement. Change in position. Displacement is a vector quantity

whose magnitude, often called distance, is measured in inches. (The term distance is also used to denote the extent of space between fixed points.)

Velocity. Time rate of change of position, or time rate of displacement. It is a vector quantity whose magnitude, called speed, is measured in inches per second (in./sec).

Acceleration. Time rate of change of velocity. It is a vector quantity whose magnitude is expressed in inches per second per second (in. / sec2).

2-2. Cartesian Reference Frame

This coordinate system consists of two perpendicular axes, fixed in space (Fig. 2-1).

Position. The position of the particle is defined by the x and y projections of the position vector r, with the directions indicated by the algebraic signs + or - .

Displacement. The magnitude and direction of the elemental displacement ds may be expressed in terms of its x and y components, as follows:

ds = [(dx) 2 + (dy)2]i

tan q, = '!:1!. dx

Velocity. The x andy components of the velocity vector are the time rates of the corresponding projections of the elemental displacement:

dx . V:e =-=X dT and

15

Digitized by Coogle

dy . Vu = (I; = y (2-1)

Original from



The speed and the direction of the velocity are given by

v = (vx2 + vll2)t V11 dy

tan q," = - = -Vx dx

Since 4>" = 4>, it follows that the velocity vector is tangent to the path.

y

y

0

FIG. 2-1

Acceleration. The acceleration components ax and a11 are obtained by differentiating the corresponding velocity components with respect to time:

dvx .. ax= - =x

dr and

dv11 •• au = - = y

dr

Alternatively, by combining Eqs. (2-1) and (2-2),

and

The magnitude and direction of the acceleration vector follow from

a = (ax2 + ai)l all tan cf>a =az

2-3. Polar Reference Frame

(2-2)

(2-3)

This reference frame consists of a fixed point, or pole, and a fixed axis originating at the pole (Fig. 2-2). ·

Position. The position of the par·ticle is defined by the radius vector

• r = Tlr

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KINEMATICS OF THE PLANE MOTION OF A PARTICLE 17

i.e., by its distance r from the pole, and the angle 8 which the radius vector makes with the reference axis. 8 is measured from the axis in a counterclockwise sense.

Displacement. Since displacement is the change in position, it may be expressed, in accordance with Eq. (1-11), as

ds = dr = dr ir + r d8 iz

Its magnitude and direction are given by

and

ds = [(dr)2 + (r d8)2]l r d8

cp = 8 + A = 8 + arctan dr

where cp = angle between reference axis and displacement A = angle between position and displacement vectors

0 X

FIG. 2-2

Velocity. Since velocity is the time rate of displacement,

ds dr dr • d8 • v = - = - = - lr + r - lz

dT dT dT dT = iir + r8iz = Vr + Vz (2-4)

where f = Vr is the magnitude of the radial, and rO = vz that of the lateral, velocity component.

The magnitude and direction of the velocity are given by

and

V = (vr2 + vz2)t Vz

cl>v = 8 + Av = 8 + arctan -Vr

Acceleration. The acceleration is obtained by differentiating Eq. (2-4) with respect to time:

dr • + . dir + dr A! + dO . + 8. diz a = - lr r - - Ull r - lz r -dT dT dT dT dT



By observing the rules for differentiating unit vectors, derived in Sec. 1-6, the above expression is reduced to

a = (r - r82)ir + (r8 + 2r8)i, = &r + ar (2-5)

where r - r82 = ar is the magnitude of the radial, and r8 + 2r8 = a1,

that of the lateral, acceleration component. The term 2r8 is known as the Coriolis acceleration. It will be dealt with more fully in Sec. 4-13. The magnitude and direction of the acceleration are given by

and

a = (ar2 + a,2)l a,

tPa = 8 + Xa = 8 + arctan -ar

2-4. Moving Reference Frame

This very important coordinate system consists of two mutually perpendicular axes which move with the particle in such a manner that one remains tangent to the path, pointing in the direction of motion, while the other points away from the center of curvature. (The center of curvature is located at the intersection of two normals to the curve, through points an elemental distance ds apart and straddling the given point. It is the center of the so-called osculation circle, which, since it has three infinitely close points in common with the curve element, approximates more closely to the curve, in the vicinity of the given point, than other tangent circles. The radius of the osculation circle is called radius of curvature; Fig. 2-3.)

c

i,.

•

FIG. 2-3

Displacement. The elemental displacement may be expressed as

ds = ds i, = p dq, i,

where p = radius of curvature of path at point considered dq, = change in direction of motion

Velocity. Since velocity is the time rate of displacement,

Digitized by Coogle

•• • V = S1c = pWplt


(2-6)


Acceleration. By differentiating Eq. (2-6) with respect to time, the following expressions for the acceleration are obtained:

... + . di, ... . . a = Sic s- = Sic - swp1,. dT

whieh, combined with Eq. (2-6), and s = v dvjds, gives

2 d 2 ... v. v. v. + a = 81t - - ln = v d 1t - - ln = a, a,. p 8 p

(2-7)

Alternatively,

a= (. + . )" + di, pwp pw, 1t fJWp dT

(2-8)

In the foregoing equations wp and ap are, respectively, the angular speed and angular acceleration of the radius of curvature.

The component along the t axis is called tangential acceleration. It represents the rate of change of speed. It should be noted that the

• expressiOn

a,= pa

familiar from elementary mechanics, is valid only if p = 0, that is, if p

is either constant (circular motion) or has, at the point considered, an extreme value.

The component along then axis is called normal, or centripetal, acceleration. The second name, meaning "center-seeking," is due to the fact that this component is always directed toward the center of curvature, as evidenced by the minus sign in Eqs. (2-7) and (2-8). It is important to note that the normal acceleration is independent of the rate of change of the radius of curvature. This fact permits the use of the concept of equivalent mechanisms in kinematic analysis, an application of which is illustrated in Sec. 4-13.

The magnitude of the acceleration and its direction relative to the velocity vector are determined from

and

2-5. Summary

a = (a,2 + an2)l a,.

tan '1 =-a,

The important relationshipB derived in the preceding paragraphs are summarized in Table 2-1.



TABLE 2-1. S UMMARY OF KI NEMATIC REL ATIONSHIPS

Coordinate Cartesian Polar Moving

Velocity v., • =X v,. = t • v = 8

(speed) • == r(J VII = y v, =pwp

Acceleration a.,= x a,. = r - r82 a, = v = 8

dv., dv = v.,

dx = v -

ds av = ti a1 = rB + 2t(J = pwp +pap

dv11 v2

= VII a .. = -dy p

2-6. The Radius of Curvature

Figure 2-4 shows the velocity and acceleration vectors of a particle, resolved into their cartesian and moving components. It can be seen that

• an = a Sln 11

Hence • van = va Sin 11

y

a,. t v

At

n

FIG. 2-4

However, by Eq. (1-6),

Hence

and

va sin 11 = mag (v X a) = v:za11 - v11a:z

p = ----v;zall - v11a:z

(v:z2 + v112)f

v:za11 - v11a:z (2-9)

A positive result indicates that the center of curvature lies in the direction obtained by turning· the velocity vector through 90° in the countetclock-

• w1se sense.




In Eq. (2-9), the radius of curvature is expressed in terms of the cartesian velocity and acceleration components. However, since p is a geometrical property of the path, and thus independent of the other motion characteristics of the particle, it may be obtained directly from the path equation, y = f(x) orr = f(O).

The following expressions are known from the study of analytical geometry:

Cartesian coordinates, with y = f(x), y' = *' y" - ~~:

p= [1 + (y')2]f

y"

P 1 d. . h f( ) , dr , d2r o ar coor mates, w1t r = 8 , r = do' r = d()2 :

P = r 2 - rr" + 2(r')2

Illustrative Example 1. A particle moves as follows:

X direction, az = 2 in. /sec2, Vz,O = 0, Xo = 0 y direction, a11 = -3 in./sec2, v11,o = 21 in. /sec, Yo = 0

Determine: (a) The path equation, y = f(x) (b) The acceleration of the particle (c) The velocity at T = 4 sec

(2-10)

(2-11)

(d) The radius of curvature of the path at the point occupied by the particle at T = 4 sec

Solution. (a) The displacement components x andy are first expressed as functions of time. By eliminating T from the resulting equations, the path equation is obtained:

Vz = f az dT + C1

With the boundary condition T = 0, vz = 0,

Vz = 2r X = fvz dr + C2

With the boundary condition T = 0, x = 0,

(a)

X = T 2 (b) Similarly v11 = - 3T + 21 (c) and y = -jr2 + 21r (d)

Elimination ofT from Eqs. (b) and (d) yields the path equation:

y = -!x + 21 Vx



(b) Since both a~ and ~ are constant, a has constant magnitude and direction:

a = (a~2 + ~2} =-= 3.61 in.jsec2

tl>a = arctan~ = arctan ( -1.5} = 303°39' as

(c) From Eqs. (a) and (c), vz.• = 8 in.jsec and v11.• = 9 in.jsec. Hence

v. = 12.04 in./ sec and q,,,. = arctan t = 48°22'

(d) Substitution of the known values into Eq. (2-9) yields

p, = -41.5 in.

The minus sign shows that the velocity vector v, is to be turned in the clockwise sense to indicate the location of the center of curvature. It is left to the reader to verify this result by means of Eq. (2-4) and the path equation established in part a.

ruustrative Example 2. The rigid bar of Fig. 2-5 revolves about its pivot at a constant speed of 1.2 rad/ sec in the counterclockwise sense.

- Guiding groove

0 ~---"---+---

FIG. 2-5

Xu Rod

The bead, free to slide on the bar, is guided by a spiral groove, defined by the equation

(} r = 4.25 + -

11'

where fJ is in radians. Determine, for the instant when the bead is 6 in. from the pivot:

(a) The velocity of the bead (b) The acceleration of the bead

Solution. In order to determine the velocity &nd acceleration components it will be necessary to express r and fJ as functions of time.



In the present case, • 6 = w = 1.2 rad/sec, const

Differentiation with respect to time yields

and integration leads to 8 = 8o + 1.2T

Substitution of Eq. (b) into the path equation yields

T = 4.25 + 8o + 1.2T . 7r

.

leading to • 1.2 t r = , cons 11"

and r = 0

(a)

(b)

(c)

(d)

(a) In polar coordinates, the general expressions for the velocity components are

• Vr = T and

Hence the following values, corresponding to r = 6 in., are obtained in the present case:

1.2. I Vr = In. sec

11" vz = 7.2 in./sec v = 7.21 in./sec

<Pv = 8 + Xv = 8 + arctan~ = 1.7511" rad + arctan 611" = 187°16' Vr

(b) The general equations for the acceleration components are

and al = re + 2r8

From these equations, the following values, corresponding tor = 6 in., are calculated:

ar = -8.64 in./sec2 a,= 0.92 in./sec2 a = 8.69 in./sec2

<l>a = 8 +arctan~= 611" rad +arctan ( -0.1065) = 274°13' ar

2-7. Graphic Differentiation and Integration

In practice, it is often difficult to express the time dependence of the position of a moving point in a form which would lend itself to an analytical determination of its velocity and acceleration. However, it is usually possible to construct, without undue trouble, time graphs of the x and y coordinates. From these curves, the corresponding velocity and acceleration components can be determined either by means of graphic differentiation or by the method of finite differences, described in Sec. 2-8.




The distance s covered by the moving point (measured along the path) can also be represented graphically as a function of time. This curve permits the determination of the speed and the tangential acceleration (v = s, a, = 8).

Graphic Differentiation. Figure 2-6 shows the x displacement of a particle as function of time.

L.,

0 1

Scale factor& Displacement, 1 in. 1::: ks in. Time, 1 in.-= k., sec

2 3

FIO. 2-6

6 'T

Since V:t = x, it follows that, for any given instant, vz is proportional to the slope of the X-T curve at the corresponding point:

(2-12}

where the lengths lz and l" are measured in inches, and the scale factors kz (in./in.) and k" (sec/ in.) ensure dimensional correctness. By repeating this calculation for a sufficient number of points, the velocity curve (Fig. 2-7) may be developed. Similarly, since az = Vz, it follows that, at any given time, az is proportional to the slope of the vz-T curve at the corresponding point:

(2-13)

As in the previous case, l" and l~ are measured in inches. The velocity scale factor k" is expressed in (in.jsec}/ in.

After this computation is repeated for a sufficient number of points, the acceleration curve (Fig. 2-8} may be drawn.



Vs

Scale factors Velocity, 1 in.-iv in./aee ~· Time, 1 in.-=~ sec

5

2 lu

~~

0 1 a 5 T

FIG. 2-7

Equation (2-12) shows that if l,. is made the same for all points of the curve, then the values of Vz are represented directly by the corresponding intercepts lz, which may therefore be used immediately in plotting the velocity diagram. With this constant value of l,. denoted by L,., the velocity scale factor becomes

(in.fsec)/ in.

Similar considerations apply to the determination of the acceleration from the velocity curve; i.e., if z: is made constant L:, then the intercepts l. represent directly the corresponding accelerations to a scale defined by

k 1 k" o=v-k T T

(in./sec2)/in.

Further simplification is achieved if the slope triangles are not drawn on the graphs but are constructed externally, as shown in Figs. 2-6 and 2-7. This procedure has been fo11owed in developing the graphs of Figs. 2-7 and 2-8.

Scale factors

0 1 ~ 3

FIG. 2-8

Digitized by Go gle

.,.

Original from



Dlustrative Example. The application of the general principles, discussed above, to a practical problem is illustrated by Fig. 2-9, which shows a mechanism whose crank q is assumed to rotate at a given constant speed. Required are the motion characteristics of the point B.

The orientation of the superposed coordinate system and the position of its origin are completely arbitrary. For convenience, the origin has been placed at the outermost left-hand position of B, with the axes inclined at 45° to the extreme chord of the path, so that Ymax = Xmax·

10

11

•

8

, , , ,

7A

/Oq

1 2

' .

"

, / / ,

/ , ,' r ,

/ / ,

--------5 ---,,,

4

FIG. 2-9

--_, -,-

----- ,_, 5

The positions of B correspon~ing to the 12 equiangular position~ ?f the crank have been marked on the path. Figure 2-10 shows the x andy components and the path length s as functions of the crank rotation or of time. (Because the crank speed w is constant, the angle turned through by the crank is directly proportional to the time elapsed.) Since the total length of the abscissa represents one crank revolution, i.e., an angle of 21r rad, Ar = 2~/12w sec, which determines the time scale. Successive graphic differentiations of the curves of Fig. 2-10 would yield, respectively, Vz and az, v11 and a11 , v and a,.

Graphic Integration. Occasionally, the inverse problem is encountered in practice; the acceleration-time relationship is known, and the velocity and displacement characteristics are to be determined. Integration of a% yields vz;

Vz - Vz 0 = r T a% dr · lo

which shows that the change in vz from its initial value vz.o is proportional .




to the area . under the az-T curve, up to the ordinate through the point considered; i.a, . -

. ' .. . . (2-14)

Thus, in Fig. 2-8, where Vz,o = 0, the stippled area Aa,3 is a measure of the speed, in the x direction, 3 sec after the start.

Evaluation of areas for a sufficient number of points yields data for the construction of the velocity-time curve.

%, y, s (in.)

I (crank angle) 4 5 6 7 8 9 10 11~12~ ., (time)

ll-r =..!..sec 6r -

FIG. 2-10

A second integration leads to the displacement-time graph :

where A, is the area under the velocity curve.

(2-15)

The areas A a and A, are measured in square inches. They may be evaluated by any one of the standard procedures, such as the use of a planimeter, true graphic integration, Simpson's rule, midordinate method, or simply by drawing the curve on graph paper and counting the squares.



The Velocity-Displacement Curve. Another graph of interest in the kinematic analysis of mechanisms is the velocity-displacement curve, an

example of which is shown in Fig. C1a 2-11.

l, 0 j ..

FIG. 2-11

Since

it follows that, at any given instant, az is proportional to the subnormal la of the velocity-displacement curve at the corresponding point:

l dvz

a = Vz tan E = Vz dx

With the scale factors taken into consideration, the acceleration becomes

(2-16)

where la is measured in inches.

2-8. Method of Finite Differences•

The accuracy of the results obtainable by graphic differentiation depends on the accuracy with which the displacement curve can be constructed, on the choice of scales, and on the skill of the individual in placing tangents. Since errors are usually magnified by successive differentiations, the resulting acceleration-time curve represents, at best, only a close approximation to the true conditions.

The method of finite differences overcomes to a large extent some of the difficulties.

If a function p = p(T) and its derivatives are continuous in the vicinity of a point p~, then, for any value of p close to p~, the function may be represented by Taylor's series. Since this condition is satisfied by the displacements of points in any real mechanism, p = p(T) may be taken to represent the time dependence of the position of any given point. (p = x, x coordinate; p = y, y coordinate; p = s, path lengths.)

Taylor's series, with the higher-order terms neglected, may be written as

(a)

1 B. W. Shaffer and I. Krause, 14The Kinematic Analysii of a Specific Point in a Mechanism by the Use of Finite-difference Expressions," New York University, Research Division, College of Engineering, New York, November, 1957.




or (b)

where P-r and P-r are, respectively, the values of the first and second time derivative of p at the point P-r·

Subtraction of Eq. (b) from (a) yields

(2-17)

and the addition of the two equations gives

(2-18)

Figure 2-12 represents a portion of the X-T curve of a particle. The x components of its velocity and acceleration may be calculated by means of Eqs. (2-17) and (2-18), as follows;

X-r+~-r - X-r-~-r Vz,-r = 2 AT

and x,.+~-r - 2x,. + X-r-~" az,-r = AT2

In the above equations, the ordinates x are measured in inches and AT is expressed in seconds.

Theoretically, the error in the approximation by finite differences will

T T

FIG. 2-12

be smaller, the shorter the time interval AT. On the other hand, the shorter this interval, the greater will be the percentage error in the differences of the measured ordinates. Thus the choice of a suitable AT is a matter of compromise. By retaining the third and fourth derivatives in Taylor's series, theoretically more accurate expressions for P-r

and p,. may be obtained. For practical reasons, however, this refinement is not recommended. Firstly, the resulting expressions are rather cumbersome, and their evaluation becomes very time-consuming. Secondly, the inevitable errors involved in the measurement of the ordinates (five, for each point) make it extremely doubtful whether the results obtained would actually be more accurate than those calculated with the simpler equations.

2-9. Absolute and Relative Motion; Relative Motion of Separate Particles

The motion of a particle or body can be described only in relation to some reference system. ,If the particular reference frame is stationary



relative to the earth, the motion is called absolute. Motion with respect to a moving reference frame is called relative.

Figure 2-13 shows two separate, i.e., physically disconnected, particles A and B within · a stationary reference frame x-y. Their absolute instantaneous positions are defined by the radius vectors r A and r n, while the vector rnA defines the position of B relative to A. The coordinate system ~-71 is a reference frame attached to A and moving with it in translation,

0

i.e., with the axes remaining parallel to y their original attitude.

7J vB The relationship between the three

X

FIG. 2-13

position vectors is given by the equation

(2-19)

0

FIG. 2-14

Differentiation of this equation with respect to time yields the velocity relationship

VB= VA+ VBA (2-20)

which is represented graphically in Fig. 2-14. If the two velocities are set off to scale from a common origin o, so

that o ~a = VA and o ~ b = v8 , then the line vector a~ b represents VnA, the velocity of B relative to A. This is the velocity that B appears to have in the moving system ~-71.

Differentiation of Eq. (2-20) leads to the acceleration relation

(2-21)

which may be represented graphically in a similar manner. The concept of relative motion is a very important one in kinematics.

It will be dealt with extensively in the next chapter. As a mnemotechnic aid in establishing relative-motion equations, it

should be noted that the suffix characters alternate; for example, B, A, BA.

2-10. Problems lr The path of a particle is .defined by . the equations

· · · x = O.Sr' - 0.01T' and y ;. 5 - 0.4-r



(a) Plot the path, y = f(x), for the first seven seconds. (b) Derive expressions for the velocity and acceleration components. (c) Calculate the radius of curvature of the path forT = 2 and 4 sec. 2. For the particle of Pro b. 1: (a) Plot the X-T and y-T graphs for the first 6 sec. (b) Determine, by graphic differentiation, the values of a% and a 11 forT = 0, 1, 2, 3,

4, and 5 sec, and draw the velocity-time curves. Compare the graphically obtained values with those calculated in Prob. 1.

3. For the particle of Pro b. 1: (a) Plot the X-T curve for the first 6 sec. (b) Using the method of finite differences, calculate vz and az forT = 2 and 4 sec,

with 6T = 0.2 sec. (c) Using the graph, Prob. 1a, plot the 8-T curve for the first 5 sec, and determine,

by means of finite differences, the values of v and a, at T = 2 and 4sec. 4. For the particle of Pro b. 1: (a) Plot the v%-x curve for the first 5 sec. (b) Determine graphically the values of a% at T = 2 and 4 sec, and compare them

with the results obtained in Prob. 1. 6. A particle moves along the logarithmic spiral

T = 10e0.02B

with a constant speed of 10 in./sec. (a) Calculate the radius of curvature of the path at the point corresponding to

8 = 50 rad. (b) Determine, in the above position, the magnitude of the acceleration and its

direction relative to the radius vector. 6. The following is an approximate expression for the acceleration of the slider in a

slider-crank mechanism:

a = rw2 (cos wT + ~ cos 2WT) where r = length of crank

n = connecting rod-crank ratio w = crank speed

(a) Derive general expressions for the slider velocity and displacement, based on the initial conditions vo = 0 and xo = 0.

(b) Determine, for the particular case of r = 12 in., w = 20 rad/sec, and n = 4, the time, velocity, and acceleration corresponding to x ::::::1 7.5 in.

(c) Plot the v-x curve, and check graphically the results of part b. (d) Plot the a-T curve and determine, by measuring the appropriate area, the maxi

mum value of the slider velocity. Check the result analytically. 7. The following are particulars of the mechanism shown in Fig. 2-9 : p = 9 in.,

q = 3 in., r = 13.5 in., 8 = 8 in., w = 15 rad/sec, ccw. (a) Plot the 8-T curve for the point B in the vicinity of the position defined by

~ O,OqA :a 90°. (b) Determine in this position, by the method of finite differences, the speed aHd

tangential acceleration of B. (c) Calculate the normal acceleration of B, in the above position, and determine

the direction of the acceleration vector with respect to the fixed link OvO,.

Digitized by Google Original from UNIVERSITY OF MICHIGAN

CHAPTER 3

KINEMATICS OF THE PLANE MOTION

OF A RIGID BODY

3-1. Definitions and Basic Concepts

The present chapter introduces a number of new kinematic concepts. The definitions of the more fundamental ones follow.

Rigid Body. A body in which the distances between the particles do not change. (Although real bodies are not absolutely rigid, the elastic deformations encountered in engineering applications are generally negligibly small from the standpoint of kinematics.)

Plane "A! otion. A type of motion in which all particles move in parallel planes.

Angular Position. In plane motion, angular position is defined as the angle between a line fixed in the reference system and a given line on the body, both lines being parallel to the plane of motion.

Angular Displacement. A change in angular position. It is a vector quantity, and is measured in radians (rad).

1r rad = 180°

The angular-displacement vector has the following peculiarity. Whereas in plane motion the vector sum of a number of finite angular displacements is equal to the actual resultant angular displacement, in threedimensional motion the vector sum of finite angular displacements has no physical meaning. The limitation does not apply to infinitesimal displacements.

Angular Velocity. The time rate of angular displacement. It is a vector whose magnitude, called rotational speed, is measured in radians per second (rad/sec). An alternative unit, frequently used in engineering, is revolutions per minute (rpm).

1r 1 rpm=

30

32

Digitized by Coogle

rad/sec


KINEMATICS OF PLANE MOTION OF RIGID BODY 33

Angular Acceleration. The time rate of the change in angular velocity. It is a vector quantity, and is measured in radians per second per second (rad/sec2).

The assumptions of rigidity and plane motion have important corollaries:

1. Since kinematics is concerned solely with the geometrical aspects of motion, it is sufficient to view, for the purpose of kinematic motion analysis, a single lamina of the body, cut parallel to the plane of motion.

so • p.. , ,, , '

I ' I ' , ~

I I

I

FIG. 3-1

8' ,.,.. .. ----4

A' --- ' ' ' ' ' B'

Moreover, since the actual physical boundaries of the body do not influence the geometry of motion, the lamina may be regarded as large enough in area to embrace any desired point in the plane. Thus, in the kinematic analysis of the plane motion of a rigid body, the terms moving plane, lamina, and body are synonymous.

2. The plane motion of a rigid body is completely described by the motions of any two points. In Fig. 3-1, the positionS' of the pointS is uniquely determined, because A'S' = A oso, B' S' = BoSo, and the

\ \

' \ \ v \

A

\

FIG. 3-2

' vc ' \ \

\

symmetrical position S* is incompatible with the stipulation of plane motion.

3. Rigidity ensures that particles lying in a straight line have equal velocity components (projections) in .the direction of this line (Fig. 3-2).



3-2. Types of Plane Motion

Three types of plane motion may be distinguished: 1. Translation. All points move along identically shaped paths, so

that a given line AB remains parallel to its initial position throughout the motion (Fig. 3-3).

From drA = drB it follows that

Further, since PA = PB and sA = BB,

Hence, in translation, at any given instant, all points of the body have equal velocities and equal accelerations.

FIG. 3-3 FIG. 3-4 --

2. Rotation About a Fixed Axis. In this motion the particles move along circular arcs, concentric with the axis, in a plane perpendicular to it. From the congruence of the triangles OA o Bo and OA 'B', it follows that

~oa = ~ob = ~~

and for an infinitesimal displacement,

so that and

Digitized by Coogle

dOa = dOb = d~ • • • Oa = Ob = ~ = (,) •• •• •• Oa = O, =~=a


(3-1) (3-2)

KINEMATICS 01'' PLANE MOTIO~ OF RIGID BODY 35

Equations (3-1) and (3-2) show that, for the purpose of determining w

and a, ..it is completely immaterial whether the body line considered intersects the axis of rotation or not (Fig. 3-4).

Given w and a, the velocity and acceleration of any point may be computed by means of the equations derived in Chap. 2. Since, in fixed-axis rotation, the center of rotation 0 coincides with the center of curvature of the path of any pointS, rs = ps, where rs is the distance OS, and ps, the radius of curvature. Moreover, since rs does not change,

fs = Ps = 0 and fs = ps = 0

Consequently, both polar and moving coordinates yield the same • expressions:

Vs = wrs vs2

(as)n = wvs = = w2rs rs

(as)t = ars

(3-3)

(3-4)

(3-5)

The velocity vector Vs and the tangential acceleration vector (as)t are perpendicular to the position vector r s = 0--+ S and point in directions governed, respectively, by the sense of w and a. The normal, or centripetal, acceleration component (as)n acts along rs and points toward 0. The foregoing statements are expressed conveniently in vectorial notation:

Vs = w X rs (as)n = w X Vs = w X ((a) X rs) (as)t = a X rs

(3-3a) (3-4a) (3-5a)

3. General Plane Motion. A type of motion which is neither a translation nor a fixed-axis rotation. It will be shown in subsequent sections of this chapter that, as far as velocities alone are concerned, general plane motion may be replaced by an instantaneous rotation about a virtual axis, and that, as regards velocities and accelerations, it is equivalent to a superposition of a translation and a rotation.

3-3. The Velocity Pole, or Instant Center of Rotation

Figure 3-;j shows that the finite displacement of the body from the position AoBo to the positionA'B', accomplished by moving A and B along prescribed paths, could also be brought about by a rotation ~a = ~cp,

about an axis through R, the point of intersection of the normal bisectors of A o A' and BoB'. Of course, with the large displacements depicted, the discrepancy between the actual paths and the corresponding circular arcs is rather appreciable. The closest approximation to the true paths is obtained in the limiting case of an infinitesimal displacement, suggested in



Fig. 3-6. Here the virtual axis of rotation passes through P u, the point .of intersection of the two path normals. This point is known as the velocity pole, or instant center of rotation.

As indicated by the double suffix, P12 is a dual point: P 1 is the intersection of the virtual axis with the stationary reference plane 1, and P2, its intersection with the moving plane 2. Pt and P2 may be visualized as instantaneous bearings of a virtual shaft. This mental picture enables one to realize quite readily that, at the instant of their coincidence at

B \ PathofA

il.t/1

FIG. 3-5 FIG. 3-6

P12, Pt and P2 have zero relative velocity. Since Pt is at rest, it follows that P2 is instantaneously at rest also.

Returning to Fig. 3-6, it can be seen that the elemental path segments and the corresponding circular arcs have common tangents. Hence

Consequently, From Eq. (3-1), so that or generally,

TA d8a = ds...t

T A.(Ja = VA

and TB d8b = dsB

and TB(Jb = VB

(Jo = (Jb = W2

and (3-6)

The velocity v s is perpendicular to the instantaneous radius vector rs = P12---+ S, and its direction is determined by the sense of (1)2. In vectorial fonn,

Vs = (1)2 X rs (3-6a)

The results obtained thus far may be summarized as follows: 1. In any plane motion there exists a point which, at a given instant,

has zero velocity. This point coincides with the velocity pole.



2. As far as velocities are concerned, general plane motion may be replaced by a momentary rotation, with the actual angular velocity (I)

about an axis through the instantaneous velocity pole. The equivalence of the two motions does not extend to accelerations

because the velocity pole does not coincide with the centers of curvature of the actual paths of all points, so that in general

For instance, in Fig. 3-7, where A is constrained to move in a circle while B is guided along a rectilinear path, Pu is located at the intersection of the path normals a and b, whereas the centers of curvature of the two paths are, respectively, at 0 and at infinity, along b.

Path of A

-Path normal b

2

0 B Path of B 1

FIG. 3-7

Location of the Velocity Pole in Special Cases. Occasionally, the simple construction of the velocity pole by means of path normals fails, and other considerations must be used to locate its position.

a. Parallel Unequal Velocities. If v A. and Vs are parallel but unequal, either in magnitude or in direction, they are ipso facto perpendicular to AB (otherwise their projections along AB would differ), and consequently the path normals coincide. The velocity pole is best located by means of the construction shown in Fig. 3-8, which is based on Eq. (3-6).

' ' ' ' ' ', '

A

' ' ' va ',,

' -- ' B __ ,, -"4C).. -pll

F1o. 3-8

Digitized by Google

VA ' ' ' \ \ ' \ \ ,p12

B

' \. Vs



b. Parallel Equal Velocities. If v A and VB are parallel and equal in magnitude and in direction, the path normals intersect at infinity. Hence, as far as velocities are concerned, translation may be regarded as a rotation about an axis at infinity, with zero angular speed.

c. Rolling Bodies. Figure 3-9 shows a body 2 which rolls, without slipping on the stationary track 1. The velocity pole coincides with the point of contact because this is the only point on body 2 which has zero velocity at the given instant.

~----"""' v s

4~~~~~~::r~s~~:JS ""--- - --------- ---- - ----

A

FIG. 3-U FIG. 3-10

3-4:. Determination of Velocities by Means of the Velocity Pole

Given the position of the velocity pole and the velocity of one point, the velocity of any other point may be found either by computation or by construction.

By computation, from Eq. (3-6),

Vs VA - =- = W2 rs rA

Thus, with VA, r A, and rs known, both vs and w2 can be calculated. The sense of w2 and the direction of v s are determined by inspection.

By construction, as shown in Fig. 3-10. The procedure is based on Eq. (3-6) and is self-explanatory.

3-6. Determination of Velocities by Means of Orthogonal Velocity Vectors1

An alternative construction, also based on Eq. (3-6), is shown in Fig. 3-11. It may be used to advantage if the velocity pole is located off the drawing paper but the general direction of the desired velocity is known.

The known velocity vector v A is turned through 90° into its path normal a. The turned velocity vector is called the orthogonal velocity of A and is denoted by oVA· A line drawn through the terminal point of oVA, parallel

1 N. Rosenauer and A. H. Willis, "Kinematics of Mechanisms," p. 115, Associated General Publications Pty. Ltd., Sydney, 1953.


KINEMATICS OF PLANE MOTION OF RIGID BODY

r--------1 v8 I

s

A I

I I

I I

FIG. 3-11

I I

I I

I I

39

to AS, cuts off on the ray, or path normal, s the orthogonal velocity vector oVs. Consideration of similar triangles shows that

thus proving the correctness of the construction. The direction of v8 is obtained by turning 0 V s through 90° in the sense 0 V A~ VA·

Illustrative Example 1. In Fig. 3-12 the velocity of A is given. Required are the velocities of B and C. The sketch shows both graphic methods of solution.

------------~

FIG. 3-12

illustrative Example 2. In Fig. 3-13 the velocity of M is given. Required are the angular velocity of the disk and the velocities of A and B.

Solution

VM VM W2 = = 'CW

TM p

and



Figure 3-13a and b shows the two alternative graphic solutions.

1

3-6. Polodes

I I

I VA I

IVM rM I

I I

I I

I I

I I

(a)

B

1

FIG. 3-13

(b) •

...... ...... ... _ .... _

As the motion of the body progresses, the instantaneous center of rotation changes its position within the fixed reference system and in relation to the moving body itself. The traces of the velocity pole in the stationary and moving planes are known as polodes, or centrodes. The locus of P1 is the fixed, or space, polode, and that of P2, the moving, or body, polode. Both curves are of importance in the study of motion and synthesis of mechanisms. Since P. and P2 have no relative velocity at P u, it follows that the relative motion of the two pol odes is pure rolling. Because a finite displacement of a body may be visualized as a succession of infinitesimal rotations about the respective instantaneous velocity poles, it is obvious that the paths of the points in a moving plane can be reproduced in all details by replacing the actual guides, or constraints, with a pair of toothed segments in the shape of the polodes (with the teeth preventing slip), one of which is held stationary while the other is attached to the moving plane. (An application of this device is shown in the next illustrative example.)

The converse is also true. If the motion of a body is produced by the pure rolling of one curve upon another, the two curves are the polodes of the motion. An example is the rolling wheel: the circle representing the wheel is the body polode, and the line representing the ground is the space polode.

Occasionally, when the actual motion constraints are not positive enough, portions of the polodes may be used to assist the body over an awkward in-line, or dead-center, position. For instance, such polode segments would be useful in the mechanism shown in Fig. 3-26 (Sec. 3-16,

Digitized by Google Original from UNIVERSI1Y OF MICHIGAN


Prob. 3), where otherwise, after the in-line position, the motion of the link AB could continue either with crossed or with parallel guiding arms.

Construction of the Fixed Polo de. The fixed pol ode 1r 1 is constructed by locating the velocity pole in various phases of the motion, by means of intersecting path normals.

Construction of the Moving Polode. The moving polode 1r,. may be found by two methods. The first method is illustrated in Fig. 3-14.

Path of B

FIG. 3-14

At the instant when the body A B is in its phase A 11 B11 , the velocity pole is at P~1 • In order to locate the corresponding point P~1 on the body in its initial phase AoBo, the triangle A11B 11Pi1 is referred back, i.e., reerected on the base line A aBo. After this construction is repeated for a sufficient number of phases, the moving polode may be drawn.

(~AoBoP~1 = ~AIIB11Pp; ~AoBoP~ = ~A 1B1P~; etc.)

The second method of construction is that of kinematic inversion. The moving body is considered temporarily fixed, and the reference frame is moved relative to it, with the actual motion constraints retained. The fixed polode for the inversion is the required body polode.

Illustrative Example. Construct the polodes for the link ABC of the ellipse-trammel mechanism of Fig. 3-15a. (The name of this mechanism is derived from the property that, if the pins A and B move along the guiding grooves in the frame, points, such as C, describe true ellipses.)



Fixed Polode. In the phase shown in Fig. 3-15a, the velocity pole Pu lies at the intersection of the path normals a and b. It can be seen that the distance 01P12 = AB. Since this is true for any other position of the link ABC, the fixed polode is a circle of radius AB, centered at 01.

Moving Polode. The construction of the moving polode, by the method of inversion, is shown in Fig. 3-15b. The bar ABC is considered fixed, and the frame movable. In the initial position the velocity pole is at P12.

' I

I •

I ' I ' I o1 -- - - - --o--I '

AI I •

(a) I • I

b

(c)

FIG. 3-15

A •

I (b)

c

p~

As the frame moves around the bar, with the guide grooves sliding on the pins A and B, the velocity pole describes a circle with the center at the midpoint 02 of AB and radius AB/ 2. (The correctness of the statement will be recognized from the fact that the angle A P B equals 90° in all po~Sitions of the frame.)

The two polodes are shown in Fig. 3 .. 15c. If a gear wheel of n external teeth and pitch diameter AB were attached to the arm ABC, and then rolled inside a stationary wheel of 2n internal teeth (and pitch radius A B), points A and B would describe rectilinear paths and C an ellipse. Thus the two circular polodes replace completely the rectilinear guides.


KIXEMATICS OF PLANE MOTIO~ OF RIGID BODY 43

3-7. Relative Motion of Physically Connected Particles

Other than in translation, points or particles of a rigid body in motion have either different velocities or different accelerations or they differ in both aspects. They have therefore relative motion with respect to each other. In the following discussion, equations are derived which relate the motion characteristics of any given point to those of a particular reference point and to the angular velocity and acceleration of the body. These relations form the basis of the most generally applicable method of kinematic and dynamic analysis of mechanisms.

Figure 3-16 shows a body AB in motion, with the angular velocity wand the angular acceleration a assumed to have positive sense. (In mechanics it is conventional to regard the counterclockwise sense as positive.) n denotes the direc-tion from A (the reference point) to B FIG 3-16

(the point whose velocity and accelera-tion are to be determined). The direction t is obtained by turning n through 90° in the positive sense.

With the position vector of B relative to A denoted by rBA, the following relation is self-evident:

or fB = fA+ fBA

fB = fA+ rBAin

(3-7) (3-7a)

The velocity equation is obtained by differentiating Eqs. (3-7) and (3-7a) with respect to time, bearing in mind that rBA is constant:

and or

VB = VA+ VBA

VB = V.A + wrBAit

VB = VA+ (w X fRA)

(3-8) (3-8a) (3-8b)

VBA is the velocity B appears to have in a reference system attached to the point A and moving with it in translation. Hence v BA is called the velocity of B relative to A. Equations (3-8) show that the relative velocity of two points on a body is perpendicular to the line connecting them (pointing in a direction determined by the sense of the absolute angular velocity w of the body) and that the relative speed is equal to the product of wand the spacing of the points:

VBA = wrBA (3-9)



Differentiation of Eqs. (3-8) and (3-8a) yields the acceleration relation

and or

&B = &A + &BA

&B = aA - w 2rBAin + arBAi,

&B = aA + (w X VBA) + (a X rB.A)

(3-10) (3-lOa) (3-10b)

The last two equations may also be expressed in the following more general tertns:

(3-10c)

&B...t is the total acceleration of B relative to A. It consists of the two mutually perpendicular components (aB...t)n and (aBA),.

(aB...t)n is the relative normal acceleration. As indicated by -in in Eq. (3-10a), it is directed from B to A. Its magnitude is given by

(3-11)

(aB.A)t is the relative tangential acceleration. rBA, and its direction depends on the sense of a . by

It is perpendicular to Its magnitude is given

(aBA), = arBA.

Hence the magnitude of the total relative acceleration is

aBA = [(aBA)n2 + (aBA)t2]i = rBA(w4 + a 2)i

and its direction with respect to the vector rBA,

-a ,.,. = arctan

2 w

3-8. General Motion as Superposition of Translation and Rotation

(3-12)

(3-13)

(3-14)

· By means of Eqs. (3-8a) and (3-lOa) the motion of any point on the body may be expressed in terms of the velocity and acceleration of an arbitrarily selected reference point A and the rotational velocity and acceleration of the body itself.

It follows, therefore, that the instantaneous motion of the body, as a whole, may be regarded as a superposition of a translation v A and &...t. and a rotation w and a about an axis through A.

In dynamics, for reasons to be discussed in a later chapter, the reference point is usually placed in the center of gravity of the body.

3-9. The Velocity Image

Figure 3-17a shows a body ABCD in motion. It is assumed that VA.,

w, and a are known.



Figure 3-17b, the so-called polar velocity diagram, represents the graphic solution of the velocity equation (3-8). The name of this diagram is due to the fact that all absolute velocity vectors originate at a common pole o. If the absolute velocity of A, v A, is represented to scale by the line vector o ~a, and the relative velocity vector VBA = a~ b (VBA = wrBA) is added to it vectorially, then o ~ b represents the absolute velocity of B, VB.

Equation (3-8) may now be applied in turn to the point pairs C,A and C,B:

Vc = VA+ VcA and Vc =VB+ VcB

VcA, represented by a~ c, is perpendicular to rcA; VcB, represented by b ~ c, is perpendicular to reB; thus point c is easily located by construction. The line vector o ~ c represents the absolute velocity of C.

D

d

I

c

(a) Configuration diagram

d' aD

(c) Polar acceleration diagram

FIG. 3-17

p 0

(b) Polar velocity diagram

o' ., '

Since the corresponding sides are mutually perpendicular, it is obvious that the triangles abc and ABC are similar. The same result is also obtained from Eq. (3-9), which states that

ab ac be AB = AC = BC = w

Digitized by Google

AB = TBA, AC = rcA, etc.



. . As the above argument may be extended to any other point on the ·body, the following laws can be deduced.

In the polar velocity diagram, the terminal points of the absolute velocity vectors form a polygon which is similar to the configuration and out of phase with respect to it by 90° in the sense of (a). This polygon is called the velocity image of the body.

The sides and diagonals of the image polygon represent relative velocities of the corresponding points.

The velocity image of a straight-line segment, such as ABD, is a similar line segment; i.e.,

ad AD --ab AB

The velocity image of a body in translation is a point. (Because VA = VB = vc, etc., points a, b, c, etc., coincide.)

The pole o is the velocity image of the instantaneous center of rotation. Hence the triangles ABP12 and abo are similar.

3-10. The Acceleration Image

Figure 3-17c, the so-called polar acceleration diagram, represents the graphic solution of the acceleration equation (3-10). The absolute acceleration of A, aA, is represented by the line vector o' -+ a'. Added to it geometrically is the vector a'-+ b', which represents aBA, the relative acceleration of B with respect to A. (aBA is known in magnitude and direction because its normal and tangential components are determinable from the data.) The resultant vector o'-+ b' represents the absolute acceleration of B, as.

Equation (3-10) may now be applied in turn to the point pairs C,A and C,B:

and ac =as+ acs

In conformity with the notation used to represent asA, acA would be shown on the diagram as a' -+ c', and ac s as b' -+ c'.

Equation (3-13) shows that

a'b' --AB a' c' b' c' AC = BC = (w4 + a2)!

proving that the triangles a'b'c' and ABC are similar. Hence point c' may be located either by transferring the angles 'Y A and 'Y B to the base line a'b' or by calculating the distances a' c' and b' c', bearing in mind that the circulatory sense a'-+ b'-+ c'-+ a' is the same as A-+ B-+ C-+ A.

Since the same argument may be extended to any other point on the body, the following laws are easily established.



In the polar acceleration diagram, the terminal points of the absolute acceleration vectors form a polygon, known as the acceleration image of the body, which is similar to the configuration and turned relative to it by an angle Jl. = arctan (- a/ w2).

The sides and diagonals of the acceleration-image polygon represent the total relative accelerations of the corresponding points.

The acceleration image of a straight-line segment is a similar line segment; i.e.,

a'd' AD --a'b' AB

The acceleration image of a body in translation is a point.

3-11. Graphical Solution of the Velocity and Acceleration Equations

Velocity Equation

Figure 3-17b illustrated the solution of this equation in the relatively trivial case of fully specified velocities v A and v BA· Of much greater importance in practice is the case where VA is completely defined and only the general direction of motion of another point B is known. The velocity equation may be solved graphically by using the fact that VnA.

is perpendicular to the line AB.

A

\\ Direction of v B

\ \ \B

\ \

\ ~b

"VBA ..... , v \ . ,,

B \ ' \ a \

', ' \ ' 0

F1o. 3-18 .

aA=o'- a' as=o'- b'

asA=o'- b'

FIG. 3-19

a'

o'

The construction shown in Fig. 3-18 yields the magnitudes and specific directions of Vn and VnA, and thus permits the determination of the angular velocity of the body. (w = VnA/TnA = ab/ AB, sense by inspection.)

Acceleration Equation ·

as= aA + aBA



Figure 3-17c illustrated the solution of this equation for fully specified accelerations aA and aBA· In practice, the acceleration equation is encountered much more frequently in a form in which the three vectors appear resolved into their normal and tangential components:

(aB)n + (aB)t = (aA)n + (aA)t + (aBA)n + (aB~)t Because each component has magnitude and direction, 12 variables are involved in the equation. Since the equation can be solved only if 10 of the variables are known, full information regarding the following items must be available:

1. VA, VB, and VBA 2. (aA)c 3. The positions of OA and OB, the centers of curvature of the paths of

A and B. [This particular piece of information is required for the determina tion of the components (aA)n and (aB)n; for example (a.A)n = V.A2/ PA,

directed from A to 0 A·]

The above data, coupled with the fact that (aBA)n = VBA 2 / rBA, directed from B to A, and that the tangential acceleration components are perpendicular to the appropriate normal components (i.e., parallel to the

TABLE 3-1. SoLUTION oF THE AccELERATION EQUATION

Nor mal a cceleration Tangential acceleration T erm

Magnitude Direct ion Vector Magnitude Direction Vector

as = o' -+ b' v s 2 (ob) 2

B to Os o' -+ bo Unknown II VB bo-+ b' --PB PB

&A = o'-+ a' VA 2 (oa) 2

A too~ o' -+ ao Given II VA in a.-+ a' - = PA PA • gtven

sense

&sA = a' -+ b' VsA 2

B to A a'-+ bo Unknown II VsA bo-+ b' TBA

(ab) 2

--AB

NOTES: The symbol II signifies "parallel to." The characters in the vector notation alternate, e.g., o'-+ ao-+ a' and a'-+ bo-+ b'.

corresponding velocities), reduce the number of unknown quantities to two, viz., the magnitudes of (aa)e and (aa.A)e. It should be noted that the as yet undetermined specific directions of these components are not

Original from UNIVERSI1Y OF MICHIGAN


classed as "unknowns." They correspond to the algebraic signs + or -and are yielded automatically by the vector solution. The calculations are best set out in tabular form, as shown in general terms in Table 3-1.

Upon completion of the acceleration diagram (Fig. 3-19), the angular acceleration of the body may be computed. [a = (aBA)t/rBA = bab' I AB, sense by inspection.]

3-12. The Acceleration Pole, or Instantaneous Center of Acceleration

There always exists, on a body in general plane motion, a singular point which, at a given instant, has zero acceleration. This point is known as the acceleration pole, or instantaneous center of acceleration. In the following discussion, the acceleration pole will be denoted by I. The most outstanding characteristic of the acceleration pole is the fact that the acceleration of any point S is proportional to the point's pole distance rsr and that the direction of the acceleration relative to the corresponding radius vector rsr is the same for all points, as shown by the following

• reasoning: as = ar + asr

and, since ar = 0, as = asr

However, by Eq. (3-13),

asr = rsr(w4 + a 2)l = krsr

where k is a constant, and by Eq. (3-14),

-a p. = arctan 2 w

•s

I

independent of position of S

FIG. 3-20

The acceleration image of I obviously coincides with o'. Hence, if the accelerations of any two points, say, A and B, are known, point I may be located on the configuration diagram by constructing the triangle ABI similar to the triangle a'b' i', as shown in Fig. 3-17 a.

In contrast to the velocity pole P 12, the acceleration pole I is only a single point; its projection on the stationary plane has no physical



significance. Another important difference between the characteristics of the two points is the following. Whereas the position of Pais a purely geometrical property of the motion, i.e., determined solely by the. motion constraints, this is not true of the point/. Although the latter is always located on the inflection circle (Chap. 10), which is a geometrical property of the motion, its position on the circle is determined by w and a.

OAA=2 in. AB=1 in. AG=3 in.

(a)

a

I

FIG. 3-21

(c)

(d)

200 in./sec t I

50,000 in./sec2

I I

Illustrative Example. In Fig. 3-21, point A moves with a constant speed of 500 in./sec in a circle of 2 in. radius, while point B moves along a rectilinear path. For the phase shown, determine:

(a} The velocity of G, using the relative-velocity method (b) The angular velocity of the bar AB (c) The location of the velocity pole and va (d) The acceleration of G, using the relative-acceleration method



(e) The angular acceleration of the bar (f) The position of Oa (center of curvature of the path of G) (g) The position of the acceleration pole and aa (h) The acceleration of the point P2

Solution. (a) Since the direction of motion of G is not known at this stage, the equation va = VA + VaA cannot be solved. It becomes necessary to determine, first, the velocity of B, and thus to establish the velocity image of the moving body.

The solution of the equation V B = VA + VBA is shown in Fig. 3-2lb. (vA is specified in magnitude and direction, v n acts along the path of B, and VaA is normal to AB.) The line ab is the velocity image of the line AB on the moving plane. Point g divides ab in the same ratio as G divides AB. The vector o ~ g is the required (absolute) velocity of G. It measures 415 in./ sec.

(b) w = ab/ AB = 390/ 7 = 55.6 rad/sec. Since VaA has the direction a ~ b, w is clockwise.

(c) P12 is located at the intersection of the path normals a and b. va = wP12G = (55.6)(7.48) = 415 in./ sec, which agrees with the previous solution.

(d) As the position of Oa is not known at this stage, the acceleration equation ao = aA + aaA cannot be solved directly. It becomes necessary to determine, first, the acceleration of B and to construct the acceleration image of the moving body.

The numerical solution of the acceleration equation aa = aA + aHA is set out in Table 3.2.

•

TABLE 3-2. AccELERATION EQUATION, SEc. 3-12, ILLUSTRATIVE ExAMPLE

.

Normal acceleration . Term

Magnitude Direction Vector

as = o'-+ b' VB 2

=0 None 01 -+ bo

PB

(PB = co)

&...t = o'-+ a' VA 2

A to OA 01 -+ ao

PA

= 125,000 . .

&s...t = a'-+ b' (ab) 2

B to A a'-+ b11 AB

= 21,700 .

.

Tangential acceleration

Magnitude Direction Vector

Unknown II Vs bo-+ b'

Zero (VA None ao-+ a' const)

. • .

Unknown II VsA b(l-+ b'


.


The acceleration diagram is shown in Fig. 3-21c. The line a'b' is the acceleration image of the line AB. Point g' divides a'b' in the same ratio as point G divides AB. The vector o'--+ g' represents the required (absolute) acceleration of G. It measures 108,000 in./sec2•

(e) a= bab'/AB = 77,700/7 = 11,100 rad/sec2• Since (a8 A}, has the direction ba --+ b', o is counterclockwise.

(f) Figure 3-21c shows the vectors va and aa, with the latter resolved into its normal and tangential components.

From (aa)n = va2/ pa, pa = 4152/91,000 = 1.9 in. The direction of (aa)n indicates that Oa is located below the bar, as shown in Fig. 3-21a.

(g) The acceleration pole is located on the configuration diagram by constructing the triangle ABI similar to the triangle a'b'i'. Hence aa = aar = IG(w4 + a 2)i = (9.4)(11,500) = 108,000 in./sec2, which agrees with the previous solution, and the direction of aa, found by means of the angle E ( E = p. - 180°), is parallel to o' --+ g', as should be.

(h) The acceleration image of P2 is located by constructing the triangle a'b'p~, in Fig. 3-21c, similar to the configuration triangle ABP12• The required acceleration is then represented by the vector o'--+ p~.

3-13. Acceleration of the Velocity Pole P2

Figure 3-22a shows the fixed and moving polodes associated with the motion of plane 2. 01 and 02 are the centers of curvature of the polodes at their point of contact P12, and Pt and P2 are the corresponding radii of curvature.

At the instant of their coincidence at Pn, the points P2 and P1 have zero relative velocity. However, an infinitesimal time interval thereafter, the two points separate. Obviously, the two points have relative acceleration, which, because of the nonslipping, must be directed along the pole normal Pn· Since P1 is stationary, the relative acceleration of P2 with respect to P1 is identical with the absolute acceleration of P2. The objective of the present investigation is the determination of the magnitude and specific direction of ap, .

The following kinematic properties of the velocity pole P2 have already been established:

1. P2 is instantaneously at rest; i.e.,

Vp2 = 0

2. The speed of any point on the body is proportional to its distance from P 2; for example,

and Vs = w2(Pu.S) Vo, = w2(P u02) = W2P2


(a)

KI~EMATICS OF PLANE MOTION OF RIGID BODY 53

The accelerations of 02 and P2 are related by the equation

Resolution of each vector into components parallel, respectively, to the pole normal Pn and the pole tangent Pt yields

along Pn, or and along p,,

(ao,)n = ap, + (ao,P2)n ap, = (ao,),.- (ao,P,)n

(ao,), = (ao,P,)t

(b)

(c)

ap, will be determined by evaluating the two terms on the right-hand side of Eq. (b).

1

"'2 ...

P1 +pz

tp II r2

P, 11"1

Pt

(a)

FIG. 3-22

P2

ip,. p2

pl

(b)

(c)

02

,.1

---....~~ ~~- ,.2

11"1

The first term is the absolute normal acceleration of 02. Because of the rolling motion of the polode, 02 describes a trochoidal path with the instantaneous center of curvature at Ot. The instantaneous radius of curvature of its path is therefore

Po,= Pl + P2



" . . · · Vo 2 P22

(aot)n = 1 = w22 -~-Po, Pl + P2

Consequently, · .

~ . . . . . . . (d)

and the vector is directed from 02 to 01. The second term represents the normal acceleration of 02 relative to P 2•

Consequently, (e)

and the vector is directed from 02 to P 2· Substitution of Eqs. (d) and (e) into (b) and consideration of the direc

tions result in

Similar investigations for the polode combinatiqns of Fig. 3-22b and c lead to the general expression

(3-15)

in which the plus sign applies in the case of a double-convex contact and the minus, in the case of a convexo-concave contact, and where ip" is the unit vector along the pole normal, directed away from the fixed polode. The meaning of "away from" is clarified in Fig: 3-22b and c, where the polodes have purposely been drawn slightly :apart.

The term di will be met again in Chap. 10, which deals with advanced kinematics of the rigid body. It will be shown that di represents the diameter of the so-called inflection circle.

Equation (3-15) reveals the very interesting and important fact that the acceleration of the velocity pole P2 is independent of the angular acceleration «2 of the moving plane. This unique kinematic property forms the basis of Carter's analysis of complex mechanisms, discussed in in Chap. 6.

In the special case of Pl = oo (rectilinear fixed polode), Eq. (3-15) reduces to

rectilinear fixed polode (3-16)

The acceleration of any point S in the moving plane may be expressed now in terms of ap2, w2, and «2 in the following manner:

as = ap2 + asP2 = aP2 + (asp2)n + (asp2)e = w22diip + (w2 X Vs) + (a2 X rs) (3-17)

where r8 = P 12 ~Sis the position vector of S with respect to P12.



3-14. Acceleration of the Center of Curvature of the Moving Polode

The relations derived in this section are very useful in the kinematic analysis of rolling bodies.

The acceleration vector ao2 may be resolved into its normal and tangential components (ao2)n and (ao2)t.

By Eq. (d) of Sec. 3-13, considering the direction:

For double-convex contact: ( ) - 2 P22 • ao, n - - W2 + lp .. Pt P2

For convexo-concave contact: ( ) 2 P22

• ao2 n = W2 lp,. PI- P2

By Eq. (c), (ao,)t = a2(Pu02) = a2p2

(3-18a)

(3-18b)

(3-19)

Equation (3-19) gives the magnitude of the tangential acceleration component. Its direction depends on the sense of a2. Hence, in vectorial notation,

(3-19a)

If the fixed polode is rectilinear (or P1 happens to be a point of inflection), PI = oo, p0 , = oo, and consequently the normal acceleration of 02 vanishes:

rectilinear fixed polode (3-20)

Illustrative Example. Figure 3-23a shows a disk, 8 in. in diameter, which rolls sliplessly on the horizontal ground. Its instantaneous angular velocity and acceleration are, respectively, 2 rad/ sec and 3 rad/ sec2, both clockwise. Draw the velocity and acceleration images of the disk, and hence determine the motion characteristics of A and B.

Solution. The circle and the straight line are the pol odes, P 12 is the velocity pole, and M ( = 02) is the center of curvature of the moving pol ode.

In order to construct the images, the velocities and accelerations of two points are needed. Here, the most convenient points are P 2 and M.

VELOCITY IMAGE (Fig. 3-23b). Vp2

= 0; point p coincides with 0.

VM = w2PuM = 8 in./ sec; VM is represented to scale by the vector o ~ m. The velocity image of the disk periphery is a circle of radius mp and center m. Points a and b are located on this circle by laying off the angles -y A and 'Y n from the line mp in the counterclockwise sense. v A is represented by the vector o ~a; by measurement, VA = 13.8 in./ sec. Similarly, Vn = o ~band VB = 11.3 in./ sec.

ACCELERATION IMAGE (Fig. 3-23c). By Eq. (3-16),



p 0

(a)

(b)

b

a:---.. ___ ...---

(c)

FIG. 3-23

ap, is represented by o'----+ p~. By Eq. (3-20), (aM)n = 0. By Eq. (3-19), (aM)t = a2 P12M = 12 in./sec2•

The total acceleration of Misrepresented by the vector o'----+ m'. The acceleration image of the disk periphery is again a circle. Its radius is m'p~, and its center is at m'. Points a' and b' are located by transferring the appropriate angles. aA, represented by o'----+ a', measures 18.9 in./sec2, and a8 , represented by o'----+ b', measures 30.4 in./sec2 •

3-16. Problems

1. Locate the velocity pole and determine VB, in Fig. 3-24, if (a) w2 = 3 rad/sec, ccw, and (b) ws = 3 rad/sec, cw.

• %·8XlS

AB-4 in. VA =15 in./sec. at 15°

FIG. 3-24 (Prob. 1)


KINEMATICS OF PLANE MOTION OF RIGID BODY

2. Construct the polodes for the body AB in Fig. 3-25.

0AA=0BB=2 in. O..t OB =AB=4 in.

FIG. 3-25 (Prob. 2)

57

B

3. Construct the polodes for the body AB in Fig. 3-26. Assume that the guiding arms remain crossed after passing the dead-center position.

O..t0B=AB=2 in. O..tA= OBB=4 in.

B

A OBB= 1.5 in. AB= 3 in.

--4fl.!:: --Path of A A

FIG. 3-26 (Prob. 3) FIG. 3-27 (Prob. 4)

4. Construct the polodes for the body AB in Fig. 3-27. 5. Point A in Fig. 3-28 has a constant speed of 125 in./sec. Using the method of

relative velocities and accelerations, determine v s, vc, (,)2 and as, ac, a2 . Check vc by means of the velocity pole and ac by means of the acceleration pole.

c • • . •. . . . • • . . . . · .. :: :-· .. _·=.:: :.· • ~-•. ··.·.·:·. ·. : . .·: .· . ·-: ._: : •.· :.::. : : · ' .· . . . . . . . . . . . . . . ·. . . . . . . . . ' . . . . . . . . . . . . : . . . . . . . .

. . . : . . . ·. ' :. .'· : •': :: -~ ·: .. : . .. . • ·. ·-.-~· . -: .

Path of A

B ...---Path of B

I

I • I ' I I

I I

OA0s=4 in. OAA = 1.25 in. OsB=2.5 in.

,, 135° l ~,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,;,/ !

AB=5 in. AC=2 in. BC=4 in.

A OB

FIG. 3-28 (Prob. 5)



6. Construct the velocity and acceleration images of the wheel of Fig. 3-29, and thus determine VA, vs, vc, vn and aA, as, ac, aD. Check the velocities by means of the velocity pole. Locate the acceleration pole, and check the accelerations. Compare the calculated value of the angle p. (or E) with the angle obtained graphically (Fig. 3-21).

D

c

M

A

VM=lOO in./sec at 0° aM""' BOO in./sec2 at 180°

Fw. 3-29 (Prob. 6)

A

MA=2 in.

B

F1o. 3-30 (Prob. 7)

1. The angular velocity of the small wheel in Fig. 3-30 is 10 rad/sec, ccw, const. Draw its velocity and acceleration images, and thereby determine v A and aA.

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