hw5 solutions kinematics of rigid bodies

Introduction to Dynamics (N. Zabaras)

HW5 Solutions Kinematics Of Rigid Bodies

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

March 7, 2016

1

mailto:[email protected]

http://www.zabaras.com/


Compute wB at t = 5 s

considering that gear A

starts form rest and that

aA= ( 3t + 2 ) rad/s2

dd dt

dt

wa w a

(3 2)d t dtw

0 0

(3 2)

t

Ad t dt

w

w

2

51.5 2 47.5 /A t

t t rad sw

A A C Cr rw w

(47.5)(50) (50)Cw

47.5 /C rad sw

B B C Cr rw w

(75) 47.5(50)Bw

31.7 /B rad sw

Problem 1

2


rG = 80 mm

rC= rD = 40 mm

rE = rB = 50mm

rF= 70 mm

G In the reverse gear

transmission shown,

compute wB

G A C Cr rw w

80(40) 40 Cw

80 /C rad sw

80 /D C rad sw w

E E D Dr rw w

(50) 80(40)Ew

64 /E rad sw

64 /F E rad sw w

F F B Br rw w

64(70) 50 Bw

89.6 /B rad sw

Problem 2

3


2(0.4 ) /

? 3

? 3

A

C

t rad s

v when t s

start from rest

s in s

a

(0.4 )(0.3) (0.2) 0.6B Bt ta a

dd dt

dt

wa w a

3

2

0 0

0.6 0.3d t dt t

w

w w

23 : 0.3 3 2.70 /BAt t s rad sw

2.70(0.4) 1.08 /C Bv r m sw

dd dt

dt

w w

3

2 3

0 0

0.3 0.1td t dt

3 : 2.7At t s rad

2.7(0.4) 1.08s r m

Problem 3

4

A A B Br ra a

With the information

shown compute the

speed of C and the

distance it travels at

time t=3 secs.


Rest

aA= 2 rad/s2

uP=?

aP=?

B = 10 revolutions of B

Problem 4

5

The motor shown in the photo is used to turn a wheel and attached

blower contained within the housing. If the pulley A connected to the motor

begins to rotate from rest with a constant angular acceleration of aA= 2

rad/s2, determine the magnitudes of the velocity and acceleration of point P

on the wheel, after the pulley B has turned ten revolutions. Assume

the transmission belt does not slip on the pulley and wheel.


Rest

aA= 2 rad/s2

uP=?

aP=?

B = 10 revolutions of B

2 2( ) ( )

P B B

P P t P n

r

a a a

u w

2 0 2(2)[167.6 0]Aw 25.89 /A rad sw

A A B Br rw w

9.7 /B rad sw

( 9.7 )(0.4) 3.88 /P B Br m su w

2( 0.75)(0.4) 0.3 /t B ta r a m sa

A A B Br ra a

20.75 /B rad sa

2 2 2(9.7) (0.4) 37.636 /n B na r a m sw

2 2 2 2 2( ) ( ) (0.3) (37.636) 37.6 /P P t P na a a m s

A A B Br r [(10)(2 )](0.4)167.6

0.15A rad

Problem 4

6

2 2

0 02 ( )A A Aw w a


Problem 5

Cable C has a constant acceleration

of 9 in/s2 and 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 load B after 2 s, and (c) the

acceleration of the point D on the rim

of the inner pulley at t = 0.

SOLUTION:

• Due to the action of the cable, the

tangential velocity and acceleration

of D are equal to the velocity and

acceleration of C. Calculate the

initial angular velocity and

acceleration.

• Apply the relations for uniformly

accelerated rotation to determine

the velocity and angular position of

the pulley after 2 s.

• Evaluate the initial tangential and

normal acceleration components

of D.

7


Problem 5SOLUTION:

• The tangential velocity and acceleration of D are equal

to the velocity and acceleration of C.

srad4

3

12

sin.12

00

00

00

r

v

rv

vv

D

D

CD

w

w

2srad33

9

sin.9

r

a

ra

aa

tD

tD

CtD

a

a

• Apply the relations for uniformly accelerated rotation to

determine velocity and angular position of pulley after 2 s.

srad10s 2srad3srad4 20 taww

rad 14

s 2srad3s 2srad422

212

21

0

tt aw

revs ofnumber rad 2

rev 1rad 14

N rev23.2N

rad 14in. 5

srad10in. 5

w

ry

rv

B

B

in. 70

sin.50

B

B

y

v

8


Problem 5

• Evaluate the initial tangential and normal

acceleration components of D.

sin.9CtD aa

2220 sin48srad4in. 3 wDnD 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

9


Problem 6

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 rack R and point D 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 point P on the

gear may be written as

Evaluate the velocities of points B

and D.

APAAPAP rkvvvv

w

10


Problem 6

x

y

SOLUTION:

• The displacement of the gear center in one

revolution is equal to the outer circumference.

For xA > 0 (moves to right), w < 0 (rotates

clockwise).

1

22rx

r

xA

A

Differentiate to relate the translational and

angular velocities.

m0.150

sm2.1

1

1

r

v

rv

A

A

w

w

kk

srad8ww

11


Problem 6• For any point P on the gear, APAAPAP rkvvvv

w

Velocity of the upper rack is equal

to velocity of point B:

1.2 m s 8rad s 0.10 m

1.2 m s 0.8m s

R B A B Av v v k r

i k j

i i

w

ivR

sm2

Velocity of the point D:

1.2 m s 8rad s 0.150 m

D A D Av v k r

i k i

w

sm697.1

sm2.1sm2.1

D

D

v

jiv

12


Problem 7

The crank AB has a constant

clockwise angular velocity of 2000

rpm.

For the crank position indicated,

determine (a) the angular velocity

of the connecting rod BD, and (b)

the velocity of the piston P.

SOLUTION:

• Will determine the absolute velocity

of point D with

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 .BDv13


Problem 7

SOLUTION:

• Will determine the absolute velocity of point D with

BDBD vvv

• The velocity is obtained from the crank rotation

data. Bv

srad 4.209in.3

srad 4.209rev

rad2

s60

min

min

rev2000

ABB

AB

ABv w

w

The velocity direction is as shown.

• The direction of the absolute velocity is

horizontal. The direction of the relative velocity

is perpendicular to BD. Compute the angle between

the horizontal and the connecting rod from the law of

sines.

Dv

BDv

95.13in.3

sin

in.8

40sin

14


Problem 7

• Determine the velocity magnitudes

from the vector triangle.

BDD vv and

BDBD vvv

sin76.05

sin.3.628

50sin95.53sin

BDDvv

sin.9.495

sft6.43sin.4.523

BD

D

v

v

srad 0.62

in. 8

sin.9.495

l

v

lv

BDBD

BDBD

w

w

sft6.43 DP vv

kBD

srad 0.62w

15


?

?

BC

wheel

w

w

v x rB AB Bw

v ( 30k) x (0.2cos 60 i 0.2sin 60 j)o o

B

v 5.2i 3.0 jB

/v v x rC B BC C Bw

i 5.20i 3.0 j ( k) x (0.2i)C BCu w

i 5.20i (0.2 -3.0) jC BCu w

5.2 /

15 /

C

BC

m s

rad s

u

w

5.252 /

0.1

cwheel rad s

r

uw

i

jk +

Problem 8

16

The bar AB of the linkage has a clockwise

angular velocity of 30 rad/s when = 60°.

Determine the angular velocities of member BC

and the wheel at this instant.


?Cu

Problem 9

17

The cylinder rolls without

slipping between the two moving

plates E and D. Determine the

angular velocity of the cylinder

and the velocity of its center C.

Since no slipping occurs, the

contact points A and B on the

cylinder have the same velocities as

the plates E and D. respectively.

Furthermore, the velocities vA and

vB are parallel, so that by the

proportionality ol' right triangles the

IC is located at a point on line A B.

Assuming this point to be a distance

x from B.


; 0.4B x xu w w

0.25 (0.25 )xw

0.154x m

0.42.6 /

0.154

B rad sx

uw

(0.25 )A m xu w

?Cu

Problem 9

18

/ 2.6(0.154 0.125) 0.075 /C C ICr m su w

The cylinder rolls without

slipping between the two moving

plates E and D. Determine the

angular velocity of the cylinder

and the velocity of its center C.


Problem 10

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 rack R

and point D of the gear.

SOLUTION:

• The point C is in contact with the

stationary lower 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 at A.

• Evaluate the velocities at B and D

based on their rotation about C.

19


Problem 10SOLUTION:

• The point C is in contact with the stationary lower

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 at A.

srad8m 0.15

sm2.1

A

AAA

r

vrv ww

• Evaluate the velocities at B and D based on their

rotation about C.

srad8m 25.0 wBBR rvv

ivR

sm2

srad8m 2121.0

m 2121.02m 15.0

wDD

D

rv

r

sm2.12.1

sm697.1

jiv

v

D

D

20


Problem 11

The crank AB has a constant


rpm.

For the crank position indicated,

determine (a) the angular velocity

of the connecting rod BD, and (b)

the velocity of the piston P.

SOLUTION:

• Determine the velocity at B from the

given crank rotation data.

• The direction of the velocity vectors

at B and D are known. The

instantaneous center of rotation is at

the intersection of the perpendiculars

to the velocities through B and D.

• Determine the angular velocity about

the center of rotation based on the

velocity at B.

• Calculate the velocity at D based on

its rotation about the instantaneous

center of rotation.

21


Problem 11SOLUTION:

• From an earlier Sample Problem,

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

through B and D.

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 at B.

in. 10.14

sin.3.628

BC

v

BCv

BBD

BDB

w

w

• Calculate the velocity at D based on its rotation

about the instantaneous center of rotation.

srad0.62in. 44.8 BDD CDv w

sft6.43sin.523 DP vv

srad0.62BDw

22


Problem 12

Crank AG of the engine system has

a constant clockwise angular

velocity of 2000 rpm.

For the crank position shown,

determine the angular acceleration

of the connecting rod BD and the

acceleration of point D.

SOLUTION:

• The angular acceleration of the crod

BD and the acceleration of point D

will be determined from

nBDtBDBBDBD aaaaaa

• The acceleration of B is determined

from the given rotation speed of AB.

• The directions of the accelerations

are

determined from the geometry.

nBDtBDD aaa

and,,

• Component equations for

acceleration of point D are solved

simultaneously for acceleration of D

and angular acceleration of the

connecting rod.

23


Problem 12

• The acceleration of B is determined from the given

rotation speed of AB.

SOLUTION:

• The angular acceleration of the connecting rod BD

and the acceleration of point D will be determined

from

nBDtBDBBDBD aaaaaa

22

1232

AB

sft962,10srad4.209ft

0

constantsrad209.4rpm2000

ABB

AB

ra w

a

w

jiaB

40sin40cossft962,10 2

24


Problem 12

• The directions of the accelerations

are determined from the geometry.

nBDtBDD aaa

and,,

From an earlier problem, wBD = 62.0 rad/s, = 13.95o.

22

1282 sft2563srad0.62ft BDnBD BDa w

jianBD

95.13sin95.13cossft2563 2

BDBDBDtBD BDa aaa 667.0ft128

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

jia BDtBD

05.76cos05.76sin667.0 a

iaa DD

25


Problem 12

nBDtBDBBDBD aaaaaa

• Component equations for acceleration of point D are

solved simultaneously.

x components:

95.13sin667.095.13cos256340cos962,10 BDDa a

95.13cos667.095.13sin256340sin962,100 BDa

y components:

ia

k

D

BD

2

2

sft9290

srad9940

a

26


Problem 13

In the position shown, crank AB has

a constant angular velocity w1 = 20

rad/s counterclockwise.

Determine the angular velocities

and angular accelerations of the

connecting rod BD and crank DE.

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

27


Problem 13

SOLUTION:

• The angular velocities are determined by

simultaneously solving the component equations for

BDBD vvv

ji

jikrv

DEDE

DEDDED

ww

ww

1717

1717

ji

jikrv BABB

160280

14820

w

ji

jikrv

BDBD

BDBDBDBD

ww

ww

123

312

BDDE ww 328017 x components:

BDDE ww 1216017 y components:

kk DEBD

srad29.11srad33.29 ww

28


Problem 13• The angular accelerations are determined by

simultaneously solving the component equations for

BDBD aaa

jiji

jijik

rra

DEDE

DE

DDEDDED

217021701717

171729.1117172

2

aa

a

wa

ji

jirra BABBABB

56003200

14820022

wa

jiji

jijik

rra

DBDB

DB

DBBDDBBDBD

2580320,10123

31233.293122

2

aa

a

wa

x components: 690,15317 BDDE aa

y components: 60101217 BDDE aa

kk DEBD

22 srad809srad645 aa

29


2

B/A /a a r rB A B Aa w

2cos 45 i sin 45 j 3cos 45 i 3sin 45 j ( k) x(10i) (0.283) (10i)o o o o

B Ba a a

2cos 45 3cos 45 (0.283) (10) [i]o o

Ba

sin 45 3sin 45 (10) [j]o o

Ba a

21.87 /Ba m s

20.344 /rod rad sa

Problem 14

30

The rod AB is confined to

move along the inclined

planes at A and B. If point A

has an acceleration of 3

m/s2 and a velocity of 2 m/s2,

both directed down the plane

at the instant the rod is

horizontal determine the

angular acceleration of the

rod at this instant.

0.283 /

?rod

rad s

at horizontal

w

a


2 2(4 / )(0.15 ) 0.6m/soa r rad s ma

2

B/O /a a r rB O B Oa w

2a 0.6i (4k) (0.15i) (6) (0.15i)B

2a { 6i 0.6 j} m/sB

2

A/O /a a r rA O A Oa w

2a 0.6i (4k) (0.15j) (6) (0.15j)A

2a { 1.2i 5.4 j} m/sA

aA=?

aB=?

Problem 15

31


2 2(4 / )(150mm) 600 /sGa r rad s mma

2

B/G /a a r rB G B Ga w

2600j ( 4k) (225j) (3) (225j)Ba

900 2625Ba i j

2 2 2(0.9) (2.625) 2.775m/sBa

1 2.625tan 71.1

0.9

o

?Ba

Problem 16

32

The spool "appears" to be rolling downward without

slipping at point A.

Introduction to Dynamics (N. Zabaras) 2rad/s95ABa

?CBa

?ABa

Problem 17

33

The collar C moves downward with an

acceleration of l m/s2.At the instant

shown, it has a speed of 2 m/s which

gives links CB and AB an angular

velocity wAB=wCB=10 rad/s.

Determine the angular accelerations of

CB and AB at this instant.


2

A B/A /a a r rB AB AB B Aa w

2a 0 ( k) ( 0.2 j) (10) ( 0.2 j)B ABa

a 0.2 i 20 jB ABa

2

B/C /a a r rB C CB CB B Ca w

20.2 i 20j 1j ( k) (0.2i 0.2 j) (10) (0.2i 0.2 j)AB CBa a

0.2 i 20j 1j 0.2 j 0.2 i 20i 20jAB CB CBa a a

0.2 0.2 20AB CBa a

20 1 0.2 20CBa 25rad/sCBa

2rad/s95ABa

?CBa

?ABa

Problem 17

34


ac =?

aBC= ?

Problem 18

35

The crankshaft AB turns with a

clockwise angular acceleration of

20 rad/s2.

Determine the acceleration of the

piston at the instant AB is in the

position shown. At this instant wAB =10

rad/s and wCB= 2.43 rad/s.


/ { 0.25sin 45 i 0.25cos 45 j} { 0.177i 0.177 j} fto o

B Ar

/ {0.75sin13.6 i 0.75cos13.6 j} {0.176i 0.729j} fto o

C Br

2

A B/A /a a r rB AB AB B Aa w

2a 0 ( 20k) ( 0.177i 0.177 j) (10) ( 0.177i 0.177 j)B

2a {21.2i 14.14j} ft/sB

2

C/B /a a r rC B BC BC C Ba w

2j 21.21i 14.14 j ( k) (0.176i 0.729j) (2.43) (0.176i 0.729j)C BCa a

0.17 18.45C BCa a 0 20.17 0.729 BCa

227.7 rad/sBCa 213.6 ft/sCa

ac =?

aBC= ?

Problem 18

36


Problem 19

In the Geneva mechanism, disk D

rotates with a constant counter-


rad/s. At the instant when j =

150o, determine the angular

acceleration of disk S.

SOLUTION:

• The absolute acceleration of the pin P

may be expressed as

csPPP aaaa

• The instantaneous angular velocity of

disk S is determined as discussed in a

sample problem in the lecture notes.

• 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.

37


Problem 19SOLUTION:

• Absolute acceleration of the pin P may be expressed

as csPPP aaaa

• From the sample problem in the corresponding lecture:

jiv

k

sP

S

4.42sin4.42cossmm477

srad08.44.42 w

• Considering each term in the acceleration equation,

jia

Ra

P

DP

30sin30cossmm5000

smm5000srad10mm500

2

222w

jia

jira

jira

aaa

StP

StP

SnP

tPnPP

4.42cos4.42sinmm1.37

4.42cos4.42sin

4.42sin4.42cos2

a

a

w

note: aS may be positive or negative

38


Problem 19

• 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

velocity by 90o in the sense of wS.

ji

ji

jiva sPSc

4.42cos4.42sinsmm3890

4.42cos4.42sinsmm477srad08.42

4.42cos4.42sin2

2

w

• Equating components of the acceleration terms

perpendicular to the slot,

srad233

07.17cos500038901.37

S

S

a

a

kS

srad233a

39

hw5 solutions kinematics of rigid bodies

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