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Chapter 16 : Planar Kinematics of a

Rigid Body

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Chapter Objectives

To classify the various types of rigid-body planar motion. To investigate rigid-body

translation and show how toanalyze motion about a fixedaxis.

To study planar motion using anabsolute motion analysis.

To provide a relative motionanalysis of velocity andacceleration using a translatingframe of reference.

To show how to find theinstantaneous center of zerovelocity and determine thevelocity of a point on a bodyusing this method.

To provide a relative-motionanalysis of velocity and

acceleration using a rotatingframe of reference.

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Rigid-Bod y Planar Motion translation, rotation about a fixed axis and general plane motion.

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Translat ion rectilinear translation, all the particles of the body travel

along straight-line paths. curvilinear translation, if all the particles of the body

travel along the paths having the same radius of curvature.

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Rotat ion abo ut a Fixed A xis

all of the particles moves along circular paths

all segments in the body undergo the same angular

displacement, angular velocity and angular acceleration.

d d dt d dt d //

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If the angular acceleration is constant, = c , then theseequations can be integrated and become

)(2

21

020

2

200

0

c

c

c

t t

t

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Once the angular motion of the body is known,then the velocity of any particle a distance r fromthe axis of rotation is

The acceleration of the particle has two

components. The tangential component accountsfor the change in the magnitude of the velocity

rv or r v

ra t t or r a

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The normal component accounts for the changein the velocity direction

General Plane Motio n

When a body undergoes general plane motion, itsimultaneously translates and rotates.

ra22

nn or r a

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*16.4 Absolute Motion Analysis

1. A body subjected to general plane motion undergoes asimultaneous translation and rotation.

2. One way to define these motions is to use a rectilinear position coordinate s to locate the point along its pathand an angular position coordinate to specify theorientation of the line.

3. By direct application of the time-differential equations v =ds/dt , a = dv/dt , = d /dt , = d /dt , the motion of thepoint and the angular motion of the line can be related.

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PROCEDURE FOR ANALYSIS

Posi t ion Coo rdina te Equat ion .

Locate point P using a position coordinate s, which ismeasured from a fixed origin and is directed along thestraight-line path of motion of point P .

Measure from a fixed reference line the angular position of a line lying in the body.

From the dimensions of the body, relate s to , s = f( ),using geometry and/or trigonometry.

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Posi t ion Coo rdina te Equat ion .

Take the first derivative of s = f( ) w.r.t time to get arelationship between v and .

Take the second derivative to get a relationship between a and .

In each case the chain rule of calculus must be used when

taking the derivatives of the position coordinate equation.

PROCEDURE FOR ANALYSIS

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Example 16.5

The large window is opened using a hydrauliccylinder AB . If the cylinder extends at a constantrate of 0.5 m/s, determine the angular velocity andangular acceleration of the window at the instant = 30

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Posi t ion Coo rdina te Equ at ion . Theangular motion of the window can be obtainedusing the coordinate , whereas the extension or motion along the hydraulic cylinder is definedusing a coordinate s, which measures the length

from the fixed point A to moving point B. Thesecoordinates can be related using the law of cosines, namely,

cos45

cos)1)(2(2)1()2(2

222

s

s

(1)

Example 16.5

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When = 30

,m s 239.1

Tim e Derivativ es. Takingthe time derivatives of Eq. 1,

srad

v s

dt

d

st

ds s

s

/620.0

)(sin2)(

)sin(402

Since v s = 0.5 m/s, then at = 30

(2)

Example 16.5

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Taking the time derivatives of Eq. 2yields,

2

22

22

/415.0

30sin2)620.0(30cos20)5.0(

)(sin2)(cos2

)(sin2)(cos2

srad

sav

dt

d

dt

d

dt

dv sv

dt

ds

s s

s s

Since a s = dv s /dt = 0, then

Example 16.5

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16.5 Relative Motion Analysis: Velocity

The general plane motion of a rigid body can bedescribed as a combination of translation androtation.

To view these component motions separately ,we use a relative-motion analysis involving twosets of coordinate axes.

The x, y coordinate system is fixed and measuresthe absolute position of two points A and B on thebody.

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The origin of the x, y coordinate system will beattached to the selectedbase point A, which

generally has a known motion.

The axes of this coordinatesystem do not rotate with the

body; rather they will only beallowed to translate withrespect to the fixed frame.


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Position.

The position vector r A specifies the location of thebase point A, and therelative-position vector r B/A locates point B with respectto point A.

By vector addition, the position of B is

A B A B /rrr


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Displacement. During an instant of timedt , point A and B undergodisplacements d r A and d r B.

If we consider the generalplane motion by itscomponent parts then theentire body first translatesby an amount d r A so that A,the base point, moves to itsfinal position and point B toB.


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The body is then rotated about A by an amountd so that B undergoes a relative displacementd r B/A and thus moves to its final position B.


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Due to the rotation about A, dr B/A = r B/A d , andthe displacement of B is

A B A Bd d d

/rrrdue to rotation about A

due to translation about A

due to translation and rotation


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Velocity. To determine therelationship between thevelocities of points A and B,

it is necessary to take thetime derivative of the positionequation, or simply divide thedisplacement equation by dt .

This yields,

dt

d

dt

d

dt

d A B A B /rrr


The terms d r B/dt = vB andd r A/dt = v A are measuredfrom the fixed x, y axes andrepresent the absolute

velocities of points A and B,respectively.

The body appears to moveas if it were rotating with an

angular velocity about thez axis passing through A

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vB/A has a magnitude of v B/A = r B/A and adirection which is perpendicular to r B/A.

A B A B /vvv


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The velocity of B is determined by considering theentire body to translate with a velocity of v A, and rotateabout A with an angular velocity .

Vector addition of these two effects, applied to B, yieldsvB.

The relative velocity vB/A represents the effect of circular motion, about A. It can be expressed by thecross product

A B A B // rv

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A B A B /rvv Hence,


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Example 16.9

The bar AB of the linkage has a clockwise angular velocity of 30 rad/s when = 60 . Determine theangular velocities of member BC and the wheel atthis instant.

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Solution (Vector Analysis)Kin emat ics Diagram .

the velocities of point B and C aredefined by the rotation of link AB

and the wheel about their fixedaxes

Example 16.9

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Veloc i ty Equ at ion.

Link AB (rotation about fixed axis):

sm

B AB B

/}0.320.5{)60sin2.060cos2.0()30(

ji jik

rv

Example 16.9

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Link BC (general plane motion):

srad

smv

vv

BC

BC

C

BC C

BC C

BC BC BC

/15

0.32.00

/20.5

)0.32.0(20.5)2.0()(0.320.5

/

jiiik jii

rvv

Example 16.9

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Wheel (rotation about fixed axis):

srad D

D

D

C DC

/52

1.020.5)1.0()(20.5

jk i

rv

Example 16.9

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16.6 Instantaneous Center of ZeroVelocity

The velocity of any point B located on arigid body can be obtained in a verydirect way if one choose the base point

A to be a point that has zero velocity atthe instant considered.

Since v A = 0, therefore vB = x r B/A.

Point A is called the instantaneous

center of zero velocity (IC ) and it lies onthe instantaneous axis of zero velocity .

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This axis is always perpendicular tothe plane of motion and theintersection of the axis with this planedefines the location of the IC.

Since point A is coincident with theIC, then vB = x r B/A and so point B moves momentarily anout the IC in acircular path .

The magnitude of vB is v B = r B/IC . Due to the circular motion, thedirection of vB must always be

perpendicular to r B/IC


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Consider the wheel as shown, if it rolls without slipping , then the point of contact with the groundhas zero velocity .

Hence this point represents the IC for the wheel.


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If it is imagined that the wheel ismomentarily pinned at this point, thevelocities of points B, C, O and so on,can be found using v = r.

The radial distance r B/IC , r C/IC and r O/IC must be determined from the geometryof the wheel.

Location of the IC. To locate the IC , we use the fact thatthe velocity of a point on the body is always perpendicular to the relative-position vector extending from the IC to thepoint. Several possibilities exist:


16 6 I t t C t f Z

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Given the velocity v A of a point A on the body, and the angular velocity of the body . In thiscase, the IC is located along the

line drawn perpendicular to v A at A, such that the distance from A to the IC is r A/IC = v A/ . Note thatthe IC lies up to the right of A since v A must cause a clockwiseangular velocity about the IC .


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Given the line of action of twononparallel velocities v A and vB.Construct at points A and B linesegments that areperpendicular to v

Aand v

B.

Extending these perpendicular to their point of intersection asshown locates the IC at theinstant considered.


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Given the magnitude and direction of two parallel velocitiesv A and vB. Here the location of the IC is determined by proportional

triangles.


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In both cases r A/IC = v A/ and r B/IC =v B/ .

If d is a known distance betweenpoint A and B, then

r A/IC + r B/IC = d for first diagram, andr B/IC - r A/IC = d for second diagram.

As a special case, note that if thebody is translating , v A = v B, then

the IC would be located at infinity,in which case r A/IC = r B/IC . Thisbeing the case, = (v A /r A/IC ) =(v A /r A/IC ) 0, as expected.


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Example 16.11

Block D moves with a speed of 3 m/s. Determinethe angular velocities of links BD and AB, at theinstant shown.

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Solution

As D moves to the right, it causes arm AB to

rotate clockwise about point A. Hence vB isdirected perpendicular to AB .

The instantaneous center of zero velocity for BD is located at the intersection of the line segmentsdrawn perpendicular to vB and vD

Example 16.11

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From the geometry,

mr

mr

IC D

IC B

566.045cos4.0

4.045tan4.0

/

/

Since the magnitude of vD is known, the angular velocity of link BD is

srad r

v

IC D

D BD /30.5

566.03

/

Example 16.11

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The velocity of B is therefore

smr v IC B BD B /12.2)4.0(30.5)( /

From the figure, the angular velocity of AB is

srad r

v

A B

B AB /30.5

4.012.2

/

45

Example 16.11

16 7 Relative Motion Analysis:

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16.7 Relative-Motion Analysis: Acceleration

An equation that relates the accelerations of twopoints on a rigid body subjected to general planemotion,

dt d

dt d

dt d A B A B /vvv

The terms d vB/dt = a B and d v A/dt = a A are

measured from a set of fixed x, y axes andrepresent the absolute accelerations of points B and A.


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The last term represents the acceleration of B w.r.t A as measured by an observer fixed totranslating x, y axes which have their origin at thebase point A.

To this observer, point B appears to move along acircular arc that has a radius of curvature r B/A.

a B/A can be expressed in terms of its tangentialand normal components of motion

n A Bt A B A B )()( // aaaa



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= +



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Since points A and B move along curved paths ,the accelerations of these points will have bothtangential and normal components .

The relative-acceleration components representthe effect of circular motion observed fromtranslating axes having their origin at the base

point A, and can be expressed as ( a B/A)t = x r B/A and ( a B/A)n = - 2r B/A

A B A B A B /2

/ rraa


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Example 16.17

The collar is moving downward with anacceleration of 1 m/s 2. At the instant shown, it hasa speed of 2 m/s which gives links CB and AB anangular velocity AB = CB = 10 rad/s. Determinethe angular accelerations of CB and AB at thisinstant.

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Solution (Vector Analysis)

The kinematic diagrams of both links AB and CD

are as shown. To solve, we will apply theappropriate kinematic equation to each link.

Example 16.17

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A cc elerat ion Equat ion.

Link AB (rotation about to a fixed axis):

jia

j jk a

rra

202.0

)2.0()10()2.0()( 22

AB B

AB B

B AB B AB B

Note that a B has two components since it movesalong a curved path .

Example 16.17

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Link BC (general plane motion):

jii j j ji

ji jik j ji

rraa

20202.02.01202.0

)2.02.0()10()2.02.0()(1202.0

2

/2

/

CBCB AB

CB AB

C BCBC BCBC B

Thus,

202.0120

202.02.0

CB

CB AB

Example 16.17

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22

2

/95/95

/5

srad srad

srad

AB

CB

Solving,

Example 16.17

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Relative Velocity Analysis

General plane motion can also be analyzed using

a relative-motion analysis between two points Aand B.

This method considers the motion in parts; first atranslation of the selected base point A, then arelative rotation of the body about point A,measured from a translating axis.

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The velocities of the teo points A and B are thenrelated using

This equation can be applied in Cartesian vector form, written as

A B A B/

vvv

A B A B /rvv

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In similar manner, for acceleration,

or

Since the relative motion is viewed as circular motion bout the base point, point B will have avelocity vB/A, that is tangent to the circle.

A B A B A B

n A Bt A B A B

/2

/

// )()(

rraa

aaaa

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It also has two components of acceleration,(a B/A)t , and ( a B/A)n.

It is important to also realize that a A and a B mayhave two components if these points move alongcurved paths.

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Ins tantaneou s Center o f Zero Veloc i ty

If the base point A is selected as having zerovelocity, then the relative velocity equationbecomes

In this case, motion appears as if the body isrotating about an instantaneous axis.

A B B /rv

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The instantaneous center of rotation ( IC ) can beestablished provided the directions of the velocitiesof any two points on the body are known.

Since the radial line r will always be perpendicular to each velocity, then the IC is at the point of intersection of these two radial lines.

Its measured location is determined from thegeometry of the body.

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Once it is established, then the velocity of anypoint P on the body can be determined from v = r ,where r extends from IC to point P .

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