Unidad II Cinematica de Cuerpo Rígido Beer_dinamica_8e_presentaciones_c15

8/11/2019 Unidad II Cinematica de Cuerpo Rgido Beer_dinamica_8e_presentaciones_c15

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VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2007 The McGraw-Hill Companies, Inc. All rights reserved.

15Cinemtica de

Cuerpo rigido


2/64


Vector Mechanics for Engineers: DynamicsEighth

Edition

15 - 2

ContenidosIntroduccin

Translacin

Rotacin alrededor de un eje fijo: Velocidad

Rotacin alrededor de un eje fijo:

Aceleracin

Rotacin alrededor de un eje fijo: Placa

representativa

Ecuaciones que definen la rotacin de uncuerpo rgido alrededor de un eje fijo

Problema resuelto 5.1

Movimiento plano General

Velocidad absoluta y relativa en el

movimiento plano general



Centro de rotacin instantneo en el

movimiento plano



Aceleracin absoluta y relativa en el

movimiento plano

Analisis del movimiento plano en trminos de

un parmetro




Razn de cambio con respecto a un sistema dereferencia en rotacin.

Aceleracin de coriolis



Movimiento alrededor de un punto fijo

Movimiento general.


Movimiento tridimensional. Aceleracin de

Coriolis

Sistema de referencia en movimiento general.



3/64


4/64



Edition

15 - 4

Traslacin Considere un cuerpo rgido en traslacin:

- La direccin de toda lnea recta dentro de

un cuerpo es constante,

- Toda partcula que constituyen el cuerpo

se mueven a lo largo de lneas paralela.

Para dos partculas en el cuerpo,

ABAB rrr

Diferenciando con respecto al tiempo,

AB

AABAB

vv

rrrr

Las partculas tienen la misma velocidad.

AB

AABAB

aa

rrrr

Diferenciando con respecto al tiempo,

Las particulas tienen la misma aceleracin.EE


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Edition

15 - 5

Rotacin Alrededor de un eje fijo. Velocidad

Considere rotacin de un cuerpo rgido

alrededor de un eje fijoAA

Vector Velocidad de la partculaP

tangente a la trayectoria de magnitud

dtrdv

dtdsv

sinsinlim

sin

0

rt

rdt

dsv

rBPs

t

locityangular vekk

rdt

rdv

El mismo resultado se obtiene de


6/64

EE


7/64 2007 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

15 - 7

Rotacin alrededor de un eje fijo. Placa representativa Considere el movimiento de una placa

representativa en un plano perpendicular al eje

de rotacin.

La Velocidad de un puntoPde la placa es,

rv

rkrv

La Acceleration de un puntoPde la placa,

rrk

rra

2

Resolviendo la aceleracin en componentes

tangencial y componentes normal,

22

rara

rarka

nn

tt

EE




Edition

15 - 8

Definiendo la ecuacin de rotacin de un cuerpo rgido

Alrededor de un eje fijo

El Movimiento de un cuerpo rgido rotandoalrededor de un eje fijo esta especificado por el tipo

de aceleracin angular.

dd

dtd

dtd

ddt

dt

d

2

2

or Recordando

Rotacin uniforme, = 0:

t 0

Rotacin uniformemente acelerado, = constant:

02

0

2

2

21

00

0

2

tt

t

V M h i f E i D iEE




Edition

15 - 9

Sample Problem 5.1

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 iEE




Edition

15 - 10

Sample Problem 5.1SOLUTION:

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

15 - 11

Sample Problem 5.1 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

15 - 12

Movimiento Plano General

Movimiento plano generalno es ni rotacin ni

traslacin.

El movimiento plano general puede considerares

como la suma de traslacin y rotacin.

El desplazamiento de partculasAandBaA2y

B2pueden ser divida en dos partes:

- traslacin deA2and

- rotacin de alrededor deA2

aB2

1B

1B





Edition

15 - 13

Velocidad Absoluta y Relativa en el movimiento Plano

Cualquier movimiento plano puede ser reemplazado por una

traslacin mediante el movimiento de un punto de referencia

arbitrario A y una rotacin simultaneo alrededor de A.

ABAB vvv

rvrkv ABABAB

ABAB rkvv





Edition

15 - 14

Velocidad Absoluta y Relativa en el movimiento Plano

Asumiendo que la velocidad vAdel extremoAes conocida, se propone determinar

la velocidad vBdel extremoBy la velocidad angular en termino de vA, l, and .

La direccin de vB

y vB/A

son conocidos. Completando el diagrama velocidad.

tan

tan

AB

A

B

vv

v

v

cos

cos

l

v

l

v

v

v

A

A

AB

A

V t M h i f E i D iE

E




Edition

15 - 15

Velocidad Absoluta y Relativa en el Movimiento plano

Seleccionando el puntoBcomo el punto de referencia y resolviendo para la

velocidad vAen el extremoAy la velocidad angular lleva a una velocidad

equivalente en el triangulo.

vA/Btiene la misma magnitud pero sentido opuesto al de vB/A. El sentido de la

velocidad relativa depende del punto de referencia escogido.

La velocidad angular de la varilla en su rotacin alrededor deBes la misma

que su rotacin alrededor deA. La velocidad angular no depende del punto de

referencia escogida.

V t M h i f E i D iEi

Ed




Edition

15 - 16

Sample Problem 15.2

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 displacement of the gear center in

one revolution is equal to the outer

circumference. Relate the translational

and angular displacements. Differentiate

to relate the translational and angular

velocities.

The velocity for any pointPon the gear

may be written as

Evaluate the velocities of pointsB andD.

APAAPAP rkvvvv

V t M h i f E i D iEi

Ed




Edition

15 - 17

Sample Problem 15.2

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: DynamicsEi

Ed



Vector Mechanics for Engineers: Dynamicsighth

dition

15 - 18

Sample Problem 15.2

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


Ed



Vector Mechanics for Engineers: Dynamicsghth

dition

15 - 19

Sample Problem 15.3

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




dition

15 - 20


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

Vector Mechanics for Engineers: DynamicsEig

Ed




dition

15 - 21

Sample Problem 15.3

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


Ed




dition

15 - 22

Centro Instantneo de Rotacin en Movimiento Plano

El movimiento plano de las partculas en una placa

puede ser reemplazada la traslacin de un puntoarbitrario A y su rotacin alrededor deA con una

velocidad angular que es independiente del puntoA.

La traslacin y rotacin de la velocidad deA son obtenida

y que permite que la placa rote con la misma velocidad

angular alrededor del punto C y es perpendicular a la

velocidad deA.

La velocidad de todas las dems partculas de una placa

son las mismas y estn originalmente definida por la

velocidad angular y la velocidad y traslacional deA.

Tal como las velocidades estn involucradas . La placa

parece rotar alrededor de un centro instantneo de

rotacin C.


Ed




dition

15 - 23


Si son conocidas la velocidad de dos partculasAyB,

el centro instantneo de rotacin depende de lainterseccin de las perpendiculares de los vectores

velocidades a travs de las partculas AyB.

Si los vectores velocidadAyBson perpendiculares a la

lneaAB, el centro instantneo de rotacin se encuentra

en la interseccin de la lneaABcon la lnea que se

junta con los extremos de los vectores velocidadAyB.

Si los vectores velocidad son paralelos, el centro

instantneo de rotacin es infinito y la velocidad

angular es cero.

Si las magnitudes de las velocidades son iguales, el

centro instantneo de rotacin es infinito y la velocidad

angular es cero.


Ed




dition

15 - 24


El centro instantneo de rotacin se determina por la

interseccin de la perpendicular de los vectores velocidadAyB

.

cosl

v

AC

v AA

tan

cossin

A

AB

v

l

vlBCv

La velocidad de todas las partculas de la varilla son las

mismas cuando rotan alrededor de C. La partcula con centro de rotacin en la placa la velocidad

es cero.

La partcula coincidente con el centro de rotacin cambian

con el tiempo y la aceleracin en el centro instantneo de

rotacin no es cero.

Las aceleracin de las partculas de la placa no pueden

determinarse como si la placa simplemente rotara en C.

El trazo en lugar del centro de rotacin del cuerpo es el

centrodo y en el espacio el centrodo espacial.


Ed


25/64



ition

15 - 25

Sample Problem 15.4

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


26/64



ition

15 - 26


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


27/64



ition

15 - 27

Sample Problem 15.5

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.


28/64


Edi


29/64



tion

15 - 29

Aceleracin Absoluta y Relativa en Movimiento Plano

La aceleracin absoluta de una partcula en la placa,

ABAB aaa

La aceleracin relativa asociada con la rotacin alrededor deAincluye las componentes tangencial y normal,

ABa

ABnAB

ABtAB

ra

rka

2

2

ra

ra

nAB

tAB


Edi


30/64


Vector Mechanics for Engineers: Dynamicshthtion

15 - 30


dadodetermine

,and AA va

.and

Ba

tABnABA

ABAB

aaa

aaa

El vector resultante del sentido y de las

magnitudes relativas denABA

aa andAa

Debe conocerse tambin la velocidad angular .


Edit


31/64



15 - 31


xcomponentes: cossin0 2 llaA

ycomponentes: sincos2

llaB

Resolviendo para aB y.

Escribiendo en trminos de ecuacin de dos

componentes,ABAB aaa

Vector Mechanics for Engineers: DynamicsEigh

Edit


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15 - 32

Anlisis de Movimiento Plano en Trminos de un Parametro

En algunos casos, este es una ventaja para determinar

la velocidad y aceleracin absoluta directamente deun mecanismo.

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


33/64



15 - 33

Sample Problem 15.6

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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15 - 34


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


35/64



15 - 35

Sample Problem 15.6

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


36/64


Vector Mechanics for Engineers: Dynamicshthion

15 - 36

Sample Problem 15.6

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



15 - 37

Sample Problem 15.7

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.


Editi


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15 - 38

Sample Problem 15.7

The acceleration ofBis determined from the given rotation

speed ofAB.

SOLUTION:

The angular acceleration of the connecting rodBDand

the acceleration of pointDwill be determined from

nBDtBDBBDBD aaaaaa

221232

AB

sft962,10srad4.209ft

0

constantsrad209.4rpm2000

ABB

AB

ra

jiaB

40sin40cossft962,10 2


Editi


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15 - 39

Sample Problem 15.7

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


Editi


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Vector Mechanics for Engineers: Dynamicsthon

15 - 40

Sample Problem 15.7

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


Editi


41/64



15 - 41

Sample Problem 15.8

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

Vector Mechanics for Engineers: DynamicsEight

Editio


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15 - 42


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


Editio


43/64



15 - 43

Sample Problem 15.8 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


Editio


44/64



15 - 44

Rapidez de cambio con respecto a un Sistema de Rotacion

Sistema fijo OXYZ.

Sistema Oxyz rotando

alrededor del eje fijoOA

con una velocidad angular

La funcin del vector

vara en direccin y

magnitud.

tQ

kQjQiQQ zyxOxyz

Con respecto al sistema fijo OXYZ ,

kQjQiQkQjQiQQ zyxzyxOXYZ

Rapidez de

cambio con respecto al sistema de rotacin. Oxyzzyx QkQjQiQ

Si donde esta fijo dentro de Oxyz

entonces esto es equivalente a la velocidad de unpunto en un cuerpo rgido Oxyzy

OXYZQ

QkQjQiQ zyx

Q

Con respecto al sistema de rotacin Oxyz

kQjQiQQ zyx

Con respecto al sistema fijo OXYZ ,

QQQ OxyzOXYZ


45/64


Editio


46/64



15 - 46

Coriolis Acceleration

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


Editio


47/64


Vector Mechanics for Engineers: Dynamicshon

15 - 47

Coriolis Acceleration 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


Editio


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ecto ec a cs o g ee s y a cshon

15 - 48

Coriolis Acceleration

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 yhon

15 - 49

Sample Problem 15.9

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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50/64


g yhon

15 - 50

Sample Problem 15.9

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


Editio


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g yhon

15 - 51

Sample Problem 15.9

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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Sample Problem 15.10

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.


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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


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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


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Motion About a Fixed Point 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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General Motion 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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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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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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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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Three-Dimensional Motion. Coriolis Acceleration

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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Vector Mechanics for Engineers: DynamicsE

ighth

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dition


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2007 The McGraw-Hill Companies, Inc. All rights reserved. 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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dition


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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


ighth

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dition


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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

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