Ch15-Kinematics of Rigid Bodies

8/14/2019 Ch15-Kinematics of Rigid Bodies

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VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Ninth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2010 The McGraw-Hill Companies, Inc. All rights reserved.

15Kinematics of

Rigid Bodies


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Vector Mechanics for Engineers: DynamicsNinth

Edition

Contents

15 - 2

Introduction

Translation

Rotation About a Fixed Axis: Velocity

Rotation About a Fixed Axis:

Acceleration

Rotation About a Fixed Axis:

Representative Slab

Equations Defining the Rotation of aRigid Body About a Fixed Axis

Sample Problem 5.1

General Plane Motion

Absolute and Relative Velocity in Plane

Motion

Sample Problem 15.2

Sample Problem 15.3

Instantaneous Center of Rotation in

Plane Motion

Sample Problem 15.4

Sample Problem 15.5

Absolute and Relative Acceleration in

Plane Motion

Analysis of Plane Motion in Terms of a

Parameter

Sample Problem 15.6

Sample Problem 15.7

Sample Problem 15.8

Rate of Change With Respect to aRotating Frame

Coriolis Acceleration

Sample Problem 15.9

Sample Problem 15.10

Motion About a Fixed Point

General Motion


Three Dimensional Motion. Coriolis

Acceleration

Frame of Reference in General Motion



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NE


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Edition

Rotation About a Fixed Axis. Velocity

15 - 5

Consider rotation of rigid body about a

fixed axisAA

Velocity vector of the particlePis

tangent to the path with magnitude

dtrdv

dtdsv

sinsinlim

sin

0

rt

rdt

dsv

rBPs

t

locityangular vekk

rdt

rdv

The same result is obtained from

NE


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Edition

Rotation About a Fixed Axis. Acceleration

15 - 6

Differentiating to determine the acceleration,

vr

dt

d

dt

rdr

dt

d

rdt

d

dt

vda

kkk

celerationangular acdt

d

componentonacceleratiradial

componenonacceleratiltangentia

r

r

rra

Acceleration ofPis combination of two

vectors,

NE


7/64 2010 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

Rotation About a Fixed Axis. Representative Slab

15 - 7

Consider the motion of a representative slab in

a plane perpendicular to the axis of rotation.

Velocity of any pointPof the slab,

rv

rkrv

Acceleration of any pointPof the slab,

rrk

rra

2

Resolving the acceleration into tangential and

normal components,

22

rara

rarka

nn

tt

NE




Edition

Equations Defining the Rotation of a Rigid Body About a Fixed Axis

15 - 8

Motion of a rigid body rotating around a fixed axis isoften specified by the type of angular acceleration.

dd

dtd

dtd

ddt

dt

d

2

2

or Recall

Uniform Rotation, = 0:

t 0

Uniformly Accelerated Rotation, = constant:

02

0

2

2

21

00

0

2

tt

t

fNE




Edition

Sample Problem 5.1

15 - 9

Cable Chas a constant acceleration of 9

in/s2and an initial velocity of 12 in/s,

both directed to the right.

Determine (a)the number of revolutions

of the pulley in 2 s, (b) the velocity and

change in position of the loadBafter 2 s,

and (c)the acceleration of the pointDon

the rim of the inner pulley at t= 0.

SOLUTION:

Due to the action of the cable, thetangential velocity and acceleration of

Dare equal to the velocity and

acceleration of C. Calculate the initial

angular velocity and acceleration.

Apply the relations for uniformlyaccelerated rotation to determine the

velocity and angular position of the

pulley after 2 s.

Evaluate the initial tangential and

normal acceleration components ofD.

V t M h i f E i D iNE




Edition

Sample Problem 5.1

15 - 10

SOLUTION:

The tangential velocity and acceleration ofDare equal to the

velocity and acceleration of C.

srad4

3

12

sin.12

00

00

00

r

v

rv

vv

D

D

CD

2srad3

3

9

sin.9

r

a

ra

aa

tD

tD

CtD

Apply the relations for uniformly accelerated rotation to

determine velocity and angular position of pulley after 2 s.

srad10s2srad3srad4 20 t

rad14

s2srad3s2srad4 22

2

12

2

10

tt

revsofnumberrad2

rev1rad14

N rev23.2N

rad14in.5

srad10in.5

ry

rv

B

B

in.70

sin.50

B

B

y

v





Edition

Sample Problem 5.1

15 - 11

Evaluate the initial tangential and normal acceleration

components ofD.

sin.9CtD aa

2220 sin48srad4in.3 DnD ra

22 sin.48sin.9 nDtD aa

Magnitude and direction of the total acceleration,

22

22

489

nDtDD aaa

2sin.8.48Da

9

48

tan

tD

nD

a

a

4.79





Edition

General Plane Motion

15 - 12

General plane motionis neither a translation nor

a rotation.

General plane motion can be considered as the

sumof a translation and rotation.

Displacement of particlesAandBtoA2andB2

can be divided into two parts:

- translation toA2and

- rotation of aboutA2

toB2

1B

1B





Edition

Absolute and Relative Velocity in Plane Motion

15 - 13

Any plane motion can be replaced by a translation of an

arbitrary reference pointAand a simultaneous rotation

aboutA.

ABAB vvv

rvrkv ABABAB

ABAB rkvv





Edition

Absolute and Relative Velocity in Plane Motion

15 - 14

Assuming that the velocity vAof endAis known, wish to determine the

velocity vBof endBand the angular velocity in terms of vA, l, and .

The direction of vB

and vB/A

are known. Complete the velocity diagram.

tan

tan

AB

A

B

vv

v

v

cos

cos

l

v

l

v

v

v

A

A

AB

A


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V t M h i f E i D iNi

Ed




Edition

Sample Problem 15.2

15 - 17

x

y

SOLUTION:

The displacement of the gear center in one revolution isequal to the outer circumference.

ForxA> 0 (moves to right), < 0 (rotates clockwise).

1

22rx

r

xA

A

Differentiate to relate the translational and angular

velocities.

m0.150sm2.1

1

1

rv

rv

A

A

kk srad8

Vector Mechanics for Engineers D namicsNi

Ed



Vector Mechanics for Engineers: Dynamicsinth

dition

Sample Problem 15.2

15 - 18

For any pointPon the gear, APAAPAP rkvvvv

Velocity of the upper rack is equal to

velocity of pointB:

ii

jki

rkvvv ABABR

sm8.0sm2.1

m10.0srad8sm2.1

ivR

sm2

Velocity of the pointD:

iki

rkvv ADAD

m150.0srad8sm2.1

sm697.1

sm2.1sm2.1

D

D

v

jiv

Vector Mechanics for Engineers: DynamicsNi

Ed



Vector Mechanics for Engineers: Dynamicsinth

dition

Sample Problem 15.3

15 - 19

The crankABhas a constant clockwise

angular velocity of 2000 rpm.

For the crank position indicated,

determine (a)the angular velocity of

the connecting rodBD, and (b)the

velocity of the pistonP.

SOLUTION:

Will determine the absolute velocity ofpointDwith

BDBD vvv

The velocity is obtained from the

given crank rotation data.Bv

The directions of the absolute velocity

and the relative velocity are

determined from the problem geometry.

Dv

BDv

The unknowns in the vector expression

are the velocity magnitudes

which may be determined from the

corresponding vector triangle.

BDD vv and

The angular velocity of the connecting

rod is calculated from .BDv


Ed



Vector Mechanics for Engineers: Dynamicsnth

dition

Sample Problem 15.3

15 - 20

SOLUTION:

Will determine the absolute velocity of pointDwith

BDBD vvv

The velocity is obtained from the crank rotation data.Bv

srad4.209in.3

srad4.209rev

rad2

s60

min

min

rev2000

ABB

AB

ABv

The velocity direction is as shown.

The direction of the absolute velocity is horizontal.

The direction of the relative velocity is

perpendicular toBD. Compute the angle between thehorizontal and the connecting rod from the law of sines.

Dv

BDv

95.13in.3

sin

in.8

40sin


Ed




dition

Sample Problem 15.3

15 - 21

Determine the velocity magnitudes

from the vector triangle.

BDD vv and

BDBD vvv

sin76.05

sin.3.628

50sin95.53sin

BDD vv

sin.9.495

sft6.43sin.4.523

BD

D

v

v

srad0.62

in.8

sin.9.495

l

v

lv

BDBD

BDBD

sft6.43 DP vv

kBD

srad0.62


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Vector Mechanics for Engineers: DynamicsNin

Ed




dition

Instantaneous Center of Rotation in Plane Motion

15 - 23

If the velocity at two pointsAandBare known, the

instantaneous center of rotation lies at the intersectionof the perpendiculars to the velocity vectors throughA

andB.

If the velocity vectors atAandBare perpendicular to

the lineAB, the instantaneous center of rotation lies at

the intersection of the lineABwith the line joining the

extremities of the velocity vectors atAandB.

If the velocity vectors are parallel, the instantaneous

center of rotation is at infinity and the angular velocity

is zero.

If the velocity magnitudes are equal, the instantaneous

center of rotation is at infinity and the angular velocity

is zero.


Ed




dition

Instantaneous Center of Rotation in Plane Motion

15 - 24

The instantaneous center of rotation lies at the intersection of

the perpendiculars to the velocity vectors throughAandB.

cosl

v

AC

v AA

tan

cossin

A

AB

v

l

vlBCv

The velocities of all particles on the rod are as if they were

rotated about C.

The particle at the center of rotation has zero velocity.

The particle coinciding with the center of rotation changes

with time and the acceleration of the particle at the

instantaneous center of rotation is not zero.

The acceleration of the particles in the slab cannot be

determined as if the slab were simply rotating about C.

The trace of the locus of the center of rotation on the body

is the body centrode and in space is the space centrode.


Ed


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ition

Sample Problem 15.4

15 - 25

The double gear rolls on the

stationary lower rack: the velocity

of its center is 1.2 m/s.

Determine (a)the angular velocity

of the gear, and (b)the velocities of

the upper rackRand pointD of the

gear.

SOLUTION:

The point Cis in contact with the stationarylower rack and, instantaneously, has zero

velocity. It must be the location of the

instantaneous center of rotation.

Determine the angular velocity about C

based on the given velocity atA.

Evaluate the velocities atBandDbased on

their rotation about C.


Ed


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ition

Sample Problem 15.4

15 - 26

SOLUTION:

The point Cis in contact with the stationary lower rack

and, instantaneously, has zero velocity. It must be thelocation of the instantaneous center of rotation.

Determine the angular velocity about Cbased on the

given velocity atA.

srad8

m0.15

sm2.1

A

AAA

r

vrv

Evaluate the velocities atBandDbased on their rotation

about C.

srad8m25.0 BBR rvv

ivR

sm2

srad8m2121.0

m2121.02m15.0

DD

D

rv

r

sm2.12.1sm697.1

jiv

v

D

D


Edi


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ition

Sample Problem 15.5

15 - 27

The crankABhas a constant clockwise

angular velocity of 2000 rpm.

For the crank position indicated,

determine (a)the angular velocity of

the connecting rodBD, and (b)the

velocity of the pistonP.

SOLUTION:

Determine the velocity atBfrom thegiven crank rotation data.

The direction of the velocity vectors atB

andDare known. The instantaneous

center of rotation is at the intersection of

the perpendiculars to the velocitiesthroughB andD.

Determine the angular velocity about the

center of rotation based on the velocity

atB.

Calculate the velocity atDbased on its

rotation about the instantaneous center

of rotation.


Edi


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ition

Sample Problem 15.5

15 - 28

SOLUTION:

From Sample Problem 15.3,

95.13

sin.3.628sin.3.4819.403

BB vjiv

The instantaneous center of rotation is at the intersection

of the perpendiculars to the velocities throughB andD.

05.7690

95.5340

D

B

sin50

in.8

95.53sin05.76sin

CDBC

in.44.8in.14.10 CDBC

Determine the angular velocity about the center of

rotation based on the velocity atB.

in.10.14

sin.3.628

BC

v

BCv

BBD

BDB

Calculate the velocity atDbased on its rotation about

the instantaneous center of rotation.

srad0.62in.44.8 BDD CDv

sft6.43sin.523 DP vv

srad0.62BD


Edi


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tion

Absolute and Relative Acceleration in Plane Motion

15 - 29

Absolute acceleration of a particle of the slab,

ABAB aaa

Relative acceleration associated with rotation aboutAincludestangential and normal components,

ABa

ABnAB

ABtAB

ra

rka

2

2

ra

ra

nAB

tAB


Edi


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Vector Mechanics for Engineers: Dynamicsthtion


15 - 30

Givendetermine

,and AA va

.and

Ba

tABnABA

ABAB

aaa

aaa

Vector result depends on sense of and the

relative magnitudes ofnABA

aa andAa

Must also know angular velocity .


Edit


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15 - 31

xcomponents: cossin0 2 llaA

ycomponents: sincos2

llaB

Solve for aBand .

Write in terms of the two component equations,ABAB aaa

Vector Mechanics for Engineers: DynamicsNint

Edit


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Analysis of Plane Motion in Terms of a Parameter

15 - 32

In some cases, it is advantageous to determine the

absolute velocity and acceleration of a mechanismdirectly.

sinlxA coslyB

cos

cos

l

l

xv AA

sin

sin

l

l

yv BB

cossincossin

2

2

llll

xa AA

sincossincos

2

2

llll

ya BB


Edit


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Sample Problem 15.6

15 - 33

The center of the double gear has a

velocity and acceleration to the right of

1.2 m/s and 3 m/s2, respectively. The

lower rack is stationary.

Determine (a) the angular acceleration

of the gear, and (b)the acceleration of

pointsB, C, andD.

SOLUTION:

The expression of the gear position as a

function of is differentiated twice to

define the relationship between the

translational and angular accelerations.

The acceleration of each point on thegear is obtained by adding the

acceleration of the gear center and the

relative accelerations with respect to the

center. The latter includes normal and

tangential acceleration components.


Edit


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Sample Problem 15.6

15 - 34

SOLUTION:

The expression of the gear position as a function of

is differentiated twice to define the relationshipbetween the translational and angular accelerations.

11

1

rrv

rx

A

A

srad8m0.150sm2.1

1 r

vA

11 rraA

m150.0

sm3 2

1 r

aA

kk 2srad20


Edit


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Sample Problem 15.6

15 - 35

jiijjki

rrka

aaaaaa

ABABA

nABtABAABAB

222

222

2

sm40.6sm2sm3m100.0srad8m100.0srad20sm3

222 sm12.8sm40.6m5 BB ajisa

The acceleration of each point

is obtained by adding the

acceleration of the gear centerand the relative accelerations

with respect to the center.

The latter includes normal and

tangential acceleration

components.


Edit


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Vector Mechanics for Engineers: Dynamicsthion

Sample Problem 15.6

15 - 36

jii

jjki

rrkaaaa ACACAACAC

222

222

2

sm60.9sm3sm3

m150.0srad8m150.0srad20sm3

jac 2sm60.9

iji

iiki

rrkaaaa ADADAADAD

222

222

2

sm60.9sm3sm3

m150.0srad8m150.0srad20sm3

222 sm95.12sm3m6.12 DD

ajisa


Edit


37/64


Vector Mechanics for Engineers: Dynamicshion

Sample Problem 15.7

15 - 37

CrankAGof the engine system has a

constant clockwise angular velocity of

2000 rpm.

For the crank position shown,

determine the angular acceleration of

the connecting rodBDand the

acceleration of pointD.

SOLUTION:

The angular acceleration of theconnecting rodBDand the acceleration

of pointDwill be determined from

nBDtBDBBDBD aaaaaa

The acceleration ofBis determined fromthe given rotation speed ofAB.

The directions of the accelerations

are

determined from the geometry.nBDtBDD

aaa

and,,

Component equations for acceleration

of pointDare solved simultaneously for

acceleration ofDand angular

acceleration of the connecting rod.


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Vector Mechanics for Engineers: Dynamicshion

Sample Problem 15.7

15 - 39

The directions of the accelerations are

determined from the geometry.nBDtBDD

aaa

and,,

From Sample Problem 15.3, BD= 62.0 rad/s, = 13.95o.

221282 sft2563srad0.62ft BDnBD

BDa

jia nBD

95.13sin95.13cossft2563 2

BDBDBDtBD BDa 667.0ft128

The direction of (aD/B)tis known but the sense is not known,

jia BDt

BD

05.76cos05.76sin667.0

iaa DD


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Vector Mechanics for Engineers: DynamicshonSample Problem 15.7

15 - 40

nBDtBDBBDBD aaaaaa

Component equations for acceleration of pointDare solved

simultaneously.

xcomponents:

95.13sin667.095.13cos256340cos962,10 BDDa

95.13cos667.095.13sin256340sin962,100 BD

y components:

iak

D

BD

2

2

sft9290

srad9940


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15 - 41

In the position shown, crankABhas a

constant angular velocity 1= 20 rad/s

counterclockwise.

Determine the angular velocities and

angular accelerations of the connecting

rodBDand crankDE.

SOLUTION:

The angular velocities are determined by

simultaneously solving the component

equations for

BDBD vvv

The angular accelerations are determined

by simultaneously solving the component

equations for

BDBD aaa


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15 - 42

SOLUTION:

The angular velocities are determined by simultaneously

solving the component equations for

BDBD vvv

ji

jikrv

DEDE

DEDDED

1717

1717

ji

jikrv BABB

160280

14820

ji

jikrv

BDBD

BDBDBDBD

123

312

BDDE 328017 xcomponents:

BDDE 1216017 ycomponents:

kk DEBD

srad29.11srad33.29


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15 - 43

The angular accelerations are determined by

simultaneously solving the component equations for

BDBD aaa

jiji

jijik

rra

DEDE

DE

DDEDDED

217021701717

171729.111717 2

2

ji

jirra BABBABB

56003200

148200 22

jiji

jijik

rra

DBDB

DB

DBBDDBBDBD

2580320,10123

31233.29312 2

2

xcomponents: 690,15317 BDDE

y components: 60101217 BDDE

kkDEBD

22 srad809srad645


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Vector Mechanics for Engineers: DynamicshonRate of Change With Respect to a Rotating Frame

15 - 44

Frame OXYZis fixed.

Frame Oxyz rotates about

fixed axis OAwith angular

velocity

Vector function varies

in direction and magnitude.

tQ

kQjQiQQ zyxOxyz

With respect to the fixed OXYZ frame,

kQjQiQkQjQiQQ zyxzyxOXYZ

rate of change

with respect to rotating frame.

Oxyzzyx QkQjQiQ

If were fixed within Oxyz then is

equivalent to velocity of a point in a rigid body

attached to Oxyzand

OXYZQ

QkQjQiQ zyx

Q

With respect to the rotating Oxyz frame,

kQjQiQQ zyx

With respect to the fixed OXYZ frame,

QQQ OxyzOXYZ


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Vector Mechanics for Engineers: DynamicshonCoriolis Acceleration

15 - 45

Frame OXYis fixed and frame Oxyrotates with angular

velocity .

Position vector for the particlePis the same in both

frames but the rate of change depends on the choice of

frame.

Pr

The absolute velocity of the particlePis

OxyOXYP rrrv

Imagine a rigid slab attached to the rotating frame Oxyor Ffor short. LetPbe a point on the slab which

corresponds instantaneously to position of particleP.

OxyP rv F velocity ofPalong its path on the slab

'Pv

absolute velocity of pointPon the slab

Absolute velocity for the particle P may be written as

F

PPP

vvv


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Vector Mechanics for Engineers: DynamicshonCoriolis Acceleration

15 - 46

FPP

OxyP

vv

rrv

Absolute acceleration for the particlePis

OxyOXYP r

dt

drra

OxyOxyP rrrra

2

OxyOxyOxy

OxyOXY

rrrdt

d

rrr

but,

OxyP

P

ra

rra

F

Utilizing the conceptual pointPon the slab,

Absolute acceleration for the particlePbecomes

22

2

F

F

F

POxyc

cPP

OxyPPP

vra

aaa

raaa

Coriolis acceleration


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Vector Mechanics for Engineers: DynamicsonCoriolis Acceleration

15 - 47

Consider a collarPwhich is made to slide at constant

relative velocity ualong rod OB. The rod is rotating at

a constant angular velocity . The pointAon the rodcorresponds to the instantaneous position ofP.

cPAP aaaa

F

Absolute acceleration of the collar is

0 OxyP ra

F

uavacPc

22 F

The absolute acceleration consists of the radial and

tangential vectors shown

2rarra AA

where


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ecto ec a cs o g ee s y a csonCoriolis Acceleration

15 - 48

uvvtt

uvvt

A

A

,at

,at

Change in velocity over tis represented by the

sum of three vectors

TTTTRRv

2rarra AA

recall,

is due to change in direction of the velocity of

pointA on the rod,

AAtt arrtvt

TT

2

00 limlim

TT

result from combined effects of

relative motion ofPand rotation of the rod

TTRR and

uuu

t

r

tu

t

TT

t

RR

tt

2

limlim00

uavacPc

22 F

recall,


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g yonSample Problem 15.9

15 - 49

Disk D of the Geneva mechanism rotates

with constant counterclockwise angular

velocity D= 10 rad/s.

At the instant when = 150o, determine

(a) the angular velocity of disk S, and (b)

the velocity of pinPrelative to disk S.

SOLUTION:

The absolute velocity of the pointPmay be written as

sPPP vvv

Magnitude and direction of velocity

of pinP are calculated from the

radius and angular velocity of diskD.Pv

Direction of velocity of pointP on

Scoinciding withPis perpendicular to

radius OP.

Pv

Direction of velocity of Pwithrespect to Sis parallel to the slot.

sPv

Solve the vector triangle for the

angular velocity of Sand relative

velocity ofP.


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15 - 50

SOLUTION:

The absolute velocity of the pointPmay be written as

sPPP vvv

Magnitude and direction of absolute velocity of pinP are

calculated from radius and angular velocity of diskD.

smm500srad10mm50 DP Rv

Direction of velocity ofPwith respect to Sis parallel to slot.

From the law of cosines,

mm1.37551.030cos2 2222 rRRllRr

From the law of cosines,

4.42742.0

30sinsin

30sin

R

sin

r

6.17304.4290

The interior angle of the vector triangle is


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15 - 51

Direction of velocity of pointP on Scoinciding withPis

perpendicular to radius OP. From the velocity triangle,

mm1.37

smm2.151

smm2.1516.17sinsmm500sin

ss

PP

r

vv

ks

srad08.4

6.17cossm500cosPsP vv

jiv sP

4.42sin4.42cossm477

smm500Pv


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g ynSample Problem 15.10

15 - 52

In the Geneva mechanism, diskD

rotates with a constant counter-clockwise angular velocity of 10

rad/s. At the instant when j= 150o,

determine angular acceleration of

disk S.

SOLUTION:

The absolute acceleration of the pinPmaybe expressed as

csPPP aaaa

The instantaneous angular velocity of Disk

Sis determined as in Sample Problem 15.9.

The only unknown involved in the

acceleration equation is the instantaneous

angular acceleration of Disk S.

Resolve each acceleration term into the

component parallel to the slot. Solve for

the angular acceleration of Disk S.


Edition


53/64



15 - 53

SOLUTION:

Absolute acceleration of the pinPmay be expressed as

csPPP aaaa

From Sample Problem 15.9.

jivk

sP

S

4.42sin4.42cossmm477

srad08.44.42

Considering each term in the acceleration equation,

jiaRa

P

DP

30sin30cossmm5000

smm5000srad10mm500

2

222

jia

jira

jira

aaa

StP

StP

SnP

tPnPP

4.42cos4.42sinmm1.37

4.42cos4.42sin

4.42sin4.42cos2

note: Smay be positive or negative


Edition


54/64



15 - 54

The relative acceleration must be parallel to

the slot.sPa

sPv

The direction of the Coriolis acceleration is obtained

by rotating the direction of the relative velocityby 90o in the sense of S.

jiji

jiva sPSc

4.42cos4.42sinsmm3890

4.42cos4.42sinsmm477srad08.42

4.42cos4.42sin2

2

Equating components of the acceleration terms

perpendicular to the slot,

srad233

07.17cos500038901.37

S

S

kS

srad233


Edition


55/64


g ynMotion About a Fixed Point

15 - 55

The most general displacement of a rigid body with a

fixed point Ois equivalent to a rotation of the body

about an axis through O. With the instantaneous axis of rotation and angular

velocity the velocity of a particlePof the body is,

rdt

rdv

and the acceleration of the particlePis

.dt

drra

Angular velocities have magnitude and direction and

obey parallelogram law of addition. They are vectors.

As the vector moves within the body and in space,

it generates a body cone and space cone which are

tangent along the instantaneous axis of rotation.

The angular acceleration represents the velocity of

the tip of .


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g ynGeneral Motion

15 - 56

For particlesAandBof a rigid body,

ABAB vvv

ParticleAis fixed within the body and motion of

the body relative toAXYZis the motion of a

body with a fixed point

ABAB rvv

Similarly, the acceleration of the particlePis

ABABAABAB

rra

aaa

Most general motion of a rigid body is equivalent to:

- a translation in which all particles have the same

velocity and acceleration of a reference particleA,and

- of a motion in which particleAis assumed fixed.


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15 - 57

The crane rotates with a constantangular velocity 1= 0.30 rad/s and the

boom is being raised with a constant

angular velocity 2= 0.50 rad/s. The

length of the boom is l= 12 m.

Determine: angular velocity of the boom,

angular acceleration of the boom,

velocity of the boom tip, and

acceleration of the boom tip.

Angular acceleration of the boom,

21

22221

Oxyz

Velocity of boom tip,

rv

Acceleration of boom tip,

vrrra

SOLUTION:

With

Angular velocity of the boom,

21

ji

jirkj

639.10

30sin30cos1250.030.0 21


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15 - 58

jir

kj

639.10

50.030.0 21

SOLUTION:

Angular velocity of the boom,

21 kj

srad50.0srad30.0

Angular acceleration of the boom,

kj

Oxyz

srad50.0srad30.021

22221

i 2srad15.0

Velocity of boom tip,

0639.10

5.03.00

kji

rv

kjiv

sm12.3sm20.5sm54.3


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15 - 59

jir

kj

639.10

50.030.0 21

Acceleration of boom tip,

kjiik

kjikji

a

vrrra

90.050.160.294.090.0

12.320.53

50.030.00

0639.10

0015.0

kjia 222 sm80.1sm50.1sm54.3


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g ynThree-Dimensional Motion. Coriolis Acceleration

15 - 60

With respect to the fixed frame OXYZand rotating

frame Oxyz,QQQ OxyzOXYZ

Consider motion of particlePrelative to a rotatingframe Oxyzor Ffor short. The absolute velocity can

be expressed as

FPP

OxyzP

vv

rrv

The absolute acceleration can be expressed as

onacceleratiCoriolis22

2

F

F

POxyzc

cPp

OxyzOxyzP

vra

aaa

rrrra


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n

Frame of Reference in General Motion

15 - 61

Consider:

- fixed frame OXYZ,

- translating frameAXYZ, and

- translating and rotating frameAxyzor F.

With respect to OXYZandAXYZ,

APAP

APAP

APAP

aaa

vvv

rrr

The velocity and acceleration ofPrelative to

AXYZcan be found in terms of the velocityand acceleration ofPrelative toAxyz.

FPP

AxyzAPAPAP

vv

rrvv

cPP

AxyzAPAxyzAP

APAPAP

aaa

rr

rraa

F

2


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15 - 62

For the disk mounted on the arm, the

indicated angular rotation rates are

constant.

Determine:

the velocity of the pointP,

the acceleration ofP, and

angular velocity and angular

acceleration of the disk.

SOLUTION:

Define a fixed reference frame OXYZat Oand a moving reference frameAxyzor F

attached to the arm atA.

WithPof the moving reference frame

coinciding withP, the velocity of the point

Pis found from

FPPP vvv

The acceleration ofPis found from

cPPP aaaa

F

The angular velocity and angular

acceleration of the disk are

F

FD


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15 - 63

SOLUTION:

Define a fixed reference frame OXYZat Oand a

moving reference frameAxyzor Fattached to thearm atA.

j

jRiLr

1

k

jRr

D

AP

2

F

WithPof the moving reference frame coinciding

withP, the velocity of the pointPis found from

iRjRkrv

kLjRiLjrv

vvv

APDP

P

PPP

22

11

FF

F

kLiRvP

12


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Sample Problem 15.15 The acceleration ofPis found from

cPPP aaaa

F

iLkLjraP 2

111

jRiRk

ra APDDP

2222

FFF

kRiRj

va Pc

2121 22

2

F

kRjRiLaP

2122

21 2

Angular velocity and acceleration of the disk,

FD

kj

21

kjj

F

i

Documents

Ch15-Kinematics of Rigid Bodies