Kinematics of Rigid Bodies - الصفحات الشخصية | الجامعة...

CHAPTER 15

Kinematics of

Rigid Bodies

Dr. Mohammad Abuhaiba, PE

Saturday, April 07, 2012

Chapter Outline

15 - 2

1. Introduction

2. Translation

3. Rotation About a Fixed Axis

• Velocity

• Acceleration

• Representative Slab

• Equations Defining Rotation of

a Rigid Body About a Fixed

Axis

4. General Plane Motion

• Absolute and Relative Velocity

• Instantaneous Center of

Rotation

• Absolute and Relative

Acceleration

• Analysis of Plane Motion in

Terms of a Parameter

5. Rate of Change With

Respect to a Rotating Frame

6. Coriolis Acceleration

7. Motion About a Fixed Point

8. General Motion

9. 3-D Motion. Coriolis

Acceleration

10.Frame of Reference in

General Motion



Introduction

15 - 3

• Kinematics of rigid bodies (RB):

relations between time and the positions,

velocities, and accelerations of the

particles forming a rigid body.

• Classification of RB motions:

General motion

Motion about a fixed point

General plane motion

Rotation about a fixed axis

• curvilinear translation

• rectilinear translation

Translation:



Translation

15 - 4

RB in translation:

direction of any straight line inside

body is constant,

all particles forming RB move in

parallel lines.

For any two particles in the body,

ABAB rrr

Differentiating with respect to time,

AB

AABAB

vv

rrrr

All particles have same velocity.

AB

AABAB

aa

rrrr

Differentiating with respect to time again,

All particles have same acceleration.



Rotation About a Fixed Axis

Velocity

15 - 5

Consider rotation of RB about a fixed

axis AA’

Velocity vector of the

particle P is 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




Acceleration

15 - 6

Differentiating to determine the acceleration,

vrdt

d

dt

rdr

dt

d

rdt

d

dt

vda

kkk

celerationangular acdt

d

componenton accelerati radial

componenton accelerati l tangentia

r

r

rra

Acceleration of P is combination of two

vectors,




Representative Slab

15 - 7

Velocity of any point P of the slab,

rv

rkrv

Acceleration of any point P of the slab,

rrk

rra

2

Resolving the acceleration into

tangential and normal components,

22

rara

rarka

nn

tt



Equations Defining Rotation of a RB About a

Fixed Axis

15 - 8

d

d

dt

d

dt

d

ddt

dt

d

2

2

or• Recall

• Uniform Rotation, = 0:

t 0

• Uniformly Accelerated Rotation, = constant:

020

2

221

00

0

2

tt

t



Sample Problem 5.1

15 - 9

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. number of revolutions of pulley in 2 s,

b. velocity and change in position of load B

after 2 s

c. acceleration of point D on the rim of the

inner pulley at t = 0.

SOLUTION:

Due to the action of the cable,

tangential velocity &

acceleration of D are equal to

velocity & acceleration of C.

Calculate initial angular velocity

& acceleration.

Apply relations for uniformly

accelerated rotation to determine

velocity and angular position of

the pulley after 2 s.

Evaluate initial tangential and

normal acceleration components

of D. Dr. Mohammad Abuhaiba, PE


Sample Problem 5.1 15 - 10

SOLUTION:

Tangential velocity and acceleration of D are equal to

velocity and acceleration of C.

srad4

3

12

sin.12

00

00

00

r

v

rv

vv

D

D

CD

2srad33

9

sin.9

r

a

ra

aa

tD

tD

CtD

Apply relations for uniformly accelerated rotation to

determine velocity & angular position of pulley after 2 s.

srad10s 2srad3srad4 20 t

rad 14

s 2srad3s 2srad422

212

21

0

tt

revs ofnumber rad 2

rev 1rad 14

N rev23.2N

rad 14in. 5

srad10in. 5

ry

rv

B

Bin. 70

sin.50

B

B

y

v



Sample Problem 5.1

15 - 11

Evaluate initial tangential and normal acceleration

components of D.

sin.9CtD aa

2220 sin48srad4in. 3 DnD ra

22 sin.48sin.9 nDtD aa

Magnitude and direction of total acceleration,

22

22

489

nDtDD aaa2sin.8.48Da

9

48

tan

tD

nD

a

a 4.79



Home Work Assignment # 15.1

1, 8, 14, 21, 28, 35

Due Sunday 8/4/2012



2 - 12

General Plane Motion

15 - 13

General plane motion is neither a translation nor a rotation.

General plane motion can be considered as

the sum of a translation and rotation.

Displacement of particles A and B to A2

and B2 can be divided into two parts:

1. translation to A2 and

2. rotation of about A2 to B2 1B



Absolute and Relative Velocity in

Plane Motion

15 - 14

Any plane motion can be replaced by a translation of an

arbitrary reference point A and a simultaneous rotation about A.

ABAB vvv

rvrkv ABABAB

ABAB rkvv




Plane Motion

15 - 15

Assuming that the velocity vA of end A is known, determine the velocity

vB of end B and 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




Plane Motion

15 - 16

Selecting point B as the reference point and solving for velocity vA of end

A and angular velocity leads to an equivalent velocity triangle.

vA/B has same magnitude but opposite sense of vB/A. Sense of relative

velocity is dependent on choice of reference point.

Angular velocity of rod in its rotation about B is the same as its rotation

about A. Angular velocity is not dependent on choice of reference point.



Sample Problem 15.2

15 - 17

The double gear rolls on the

stationary lower rack: the velocity

of its center is 1.2 m/s. Determine

a. angular velocity of gear

b. velocities of upper rack R and

point D of gear.

SOLUTION:

The displacement of gear center in

one revolution is equal to the outer

circumference. Relate translational

and angular displacements.

Differentiate to relate translational

and angular velocities.

Velocity for any point P on gear

may be written as

Evaluate the velocities of points B

and D.

APAAPAP rkvvvv



Sample Problem 15.2

15 - 18

x

y

SOLUTION:

The displacement of gear center in one revolution is equal to outer

circumference.

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

1

22rx

r

xA

A

Differentiate to relate translational and angular velocities.

m0.150

sm2.1

1

1

r

v

rv

A

A

kk

srad8



Sample Problem 15.2

15 - 19

For any point P on the gear, APAAPAP rkvvvv

Velocity of upper rack is equal to

velocity of point B:

ii

jki

rkvvv ABABR

sm8.0sm2.1

m 10.0srad8sm2.1

ivR

sm2

Velocity of the point D:

iki

rkvv ADAD

m 150.0srad8sm2.1

sm697.1

sm2.1sm2.1

D

D

v

jiv



Sample Problem 15.3

15 - 20

The crank AB has a constant cw

angular velocity of 2000 rpm. For

the crank position indicated,

determine

a. angular velocity of connecting

rod BD

b. velocity of piston P

SOLUTION:

Absolute velocity of point D:

BDBD vvv

Velocity is obtained from the

given crank rotation data. Bv

• Directions of the absolute velocity

and relative velocity are

determined from geometry.

Dv

BDv

• Unknowns in the vector expression

are velocity magnitudes

which may be determined from the

corresponding vector triangle.

BDD vv and

• Angular velocity of connecting rod

is calculated from .BDv



Sample Problem 15.3

15 - 21

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

The velocity direction is as shown.

Direction of absolute velocity is horizontal.

Direction of relative velocity is perpendicular

to BD. Compute angle between horizontal and

connecting rod from the law of sines.

Dv

BDv

95.13in.3

sin

in.8

40sin



Sample Problem 15.3

15 - 22

• Determine 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

sft6.43 DP vv

kBD

srad 0.62



Home Work Assignment 15.2

38, 46, 53, 59, 66, 72

Due Tuesday 10/4/2012



2 - 23

Instantaneous Center of Rotation in

Plane Motion

15 - 24

• Plane motion of all particles in a slab can always be

replaced by the translation of an arbitrary point A

and a rotation about A with an angular velocity that

is independent of choice of A.

• Same translational & rotational velocities at A are

obtained by allowing slab to rotate with same

angular velocity about point C on a perpendicular to

velocity at A.

• Velocity of all other particles in slab are the same as

originally defined since the angular velocity and

translational velocity at A are equivalent.

• As far as velocities are concerned, the slab seems to

rotate about the instantaneous center of rotation C.




Plane Motion

15 - 25

• If the velocity at two points A and B are known,

the instantaneous center of rotation lies at

intersection of perpendiculars to velocity vectors

through A and B .

• If velocity vectors at A and B are perpendicular to

line AB, instantaneous center of rotation lies at

intersection of line AB with line joining

extremities of velocity vectors at A and B.

• If velocity magnitudes are equal, instantaneous

center of rotation is at infinity and angular

velocity is zero.




Plane Motion

15 - 26

instantaneous center of rotation lies at intersection of

perpendiculars to velocity vectors through A and B .

cosl

v

AC

v AA

tan

cossin

A

AB

v

l

vlBCv

velocities of all particles on rod are as if they were

rotated about C.

particle at center of rotation has zero velocity. particle coinciding with center of rotation changes

with time and acceleration of particle at instantaneous

center of rotation is not zero.

acceleration of particles in the slab cannot be

determined as if the slab were simply rotating about

C. trace of locus of center of rotation on the body is the

body centrode and in space is the space centrode.



Sample Problem 15.4

15 - 27

The double gear rolls on the

stationary lower rack: the

velocity of its center is 1.2 m/s.

Determine:

a. angular velocity of the gear

b. velocities of the upper rack R

and point D of the gear.

SOLUTION:

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 angular velocity about C

based on the given velocity at A.

Evaluate velocities at B and D based

on their rotation about C.



Sample Problem 15.4

15 - 28

SOLUTION:

point C is in contact with stationary lower rack

and, instantaneously, has zero velocity. It must be

the location of instantaneous center of rotation.

Determine angular velocity about C based on the

given velocity at A.

srad8m 0.15

sm2.1

A

AAA

r

vrv

Evaluate velocities at B and D based on their

rotation about C.

srad8m 25.0 BBR rvv ivR

sm2

srad8m 2121.0

m 2121.02m 15.0

DD

D

rv

r

sm2.12.1

sm697.1

jiv

v

D

D



Sample Problem 15.5

15 - 29

The crank AB has a constant cw

angular velocity of 2000 rpm. For

the crank position indicated,

determine:

a. angular velocity of connecting

rod BD

b. velocity of piston P

SOLUTION:

Determine velocity at B from the

given crank rotation data.

direction of velocity vectors at B

and D are known. The

instantaneous center of rotation is

at intersection of perpendiculars

to velocities through B and D.

Determine angular velocity about

center of rotation based on

velocity at B.

Calculate velocity at D based on

its rotation about instantaneous

center of rotation.



Sample Problem 15.5

15 - 30

From Sample Problem 15.3,

95.13

sin.3.628sin.3.4819.403

BB vjiv

instantaneous center of rotation is at intersection

of perpendiculars to 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 angular velocity about center of

rotation based on velocity at B.

in. 10.14

sin.3.628

BC

v

BCv

BBD

BDB

Calculate the velocity at D based on its rotation

about the instantaneous center of rotation.

srad0.62in. 44.8 BDD CDv

sft6.43sin.523 DP vv

srad0.62BD




74, 80, 88, 95




2 - 31

Absolute and Relative Acceleration

in Plane Motion

15 - 32

• Absolute acceleration of a particle of the slab,

ABAB aaa

• Relative acceleration associated with rotation about A includes

tangential and normal components,

ABa

ABnAB

ABtAB

ra

rka

2

2

ra

ra

nAB

tAB



15 - 33

Given

determine

, and AA va

. and

Ba

tABnABA

ABAB

aaa

aaa

Vector result depends on sense of and the

relative magnitudes of nABA aa and Aa

Must also know angular velocity .



Absolute and Relative Acceleration in

Plane Motion

Absolute and Relative Acceleration

in Plane Motion

15 - 34

x components: cossin0 2 llaA

y components: sincos2 llaB

Solve for aB and .

Write in terms of the two component equations, ABAB aaa



Analysis of Plane Motion in Terms of a

Parameter

15 - 35

In some cases, it is advantageous to determine absolute

velocity and acceleration of a mechanism directly.

sinlxA coslyB

cos

cos

l

l

xv AA

sin

sin

l

l

yv BB

cossin

cossin

2

2

ll

ll

xa AA

sincos

sincos

2

2

ll

ll

ya BB



Sample Problem 15.6

15 - 36

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. angular acceleration of the gear

b. acceleration of points B, C, D

SOLUTION:

• The expression of gear position as

a function of is differentiated

twice to define the relationship

between translational and angular

accelerations.

• The acceleration of each point on

gear is obtained by adding

acceleration of the gear center and

relative accelerations with respect

to the center. The latter includes

normal and tangential acceleration

components.



15 - 37

SOLUTION:

The expression of the gear position as a function of is differentiated twice to

define the relationship between the translational and angular accelerations.

11

1

rrv

rx

A

A

srad 8m 0.150

sm2.1

1

r

vA

11 rraA

m 150.0

sm3 2

1

r

aA

kk 2srad20



Sample Problem 15.6

15 - 38

jii

jjki

rrka

aaaaaa

ABABA

nABtABAABAB

222

222

2

sm40.6sm2sm3

m100.0srad8m100.0srad20sm3

222 sm12.8sm40.6m5 BB ajisa

Acceleration of each point is obtained by adding acceleration of gear center and

relative accelerations with respect to the center. The latter includes normal and

tangential acceleration components.



Sample Problem 15.6

15 - 39

jii

jjki

rrkaaaa ACACAACAC

222

222

2

sm60.9sm3sm3


jac

2sm60.9

iji

iiki

rrkaaaa ADADAADAD

222

222

2

sm60.9sm3sm3


222 sm95.12sm3m6.12 DD ajisa



Sample Problem 15.6

Sample Problem 15.7

15 - 40

Crank AG of the engine system has

a constant cw angular velocity of

2000 rpm. For the crank position

shown, determine angular

acceleration of the connecting rod

BD and the acceleration of point D.

SOLUTION:

Angular acceleration of rod BD

and acceleration of point D will be

determined from

nBDtBDBBDBD aaaaaa

Acceleration of B is determined

from given rotation speed of AB.

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 & angular acceleration of

connecting rod.



15 - 41

Acceleration of B is determined from the given

rotation speed of AB.

SOLUTION:

Angular acceleration of connecting rod BD and

acceleration of point D will be determined from

nBDtBDBBDBD aaaaaa

22

1232

AB

sft962,10srad4.209ft

0

constantsrad209.4rpm2000

ABB

AB

ra

jiaB

40sin40cossft962,10 2



Sample Problem 15.7

15 - 42

Directions of accelerations are determined from geometry. nBDtBDD aaa

and,,

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

22

1282 sft2563srad0.62ft BDnBD BDa

jianBD

95.13sin95.13cossft2563 2

BDBDBDtBD BDa 667.0ft128

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

jia BDtBD

05.76cos05.76sin667.0

iaa DD



Sample Problem 15.7

15 - 43

nBDtBDBBDBD aaaaaa

Component equations for acceleration of point D are solved simultaneously.

x components: 95.13sin667.095.13cos256340cos962,10 BDDa

95.13cos667.095.13sin256340sin962,100 BDy components:

ia

k

D

BD

2

2

sft9290

srad9940



Sample Problem 15.7

Sample Problem 15.8

15 - 44

In the position shown, crank AB has

a constant angular velocity 1 = 20

rad/s ccw. Determine angular

velocities and angular accelerations

of 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



15 - 45

SOLUTION:

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 x components:

BDDE 1216017 y components:

kk DEBD

srad29.11srad33.29



Sample Problem 15.8

15 - 46

Angular accelerations are determined by simultaneously solving the component

equations for BDBD aaa

jiji

jijik

rra

DEDE

DE

DDEDDED

217021701717

171729.1117172

2

ji

jirra BABBABB

56003200

14820022

jiji

jijik

rra

DBDB

DB

DBBDDBBDBD

2580320,10123

31233.293122

2

x components: 690,15317 BDDE

y components: 60101217 BDDE

kk DEBD

22 srad809srad645



Sample Problem 15.8


105, 112, 119, 126, 133, 141

Due Sunday 22/4/2012



2 - 47

Frame OXYZ is fixed.

Rate of Change With Respect to a Rotating

Frame

15 - 48

Frame Oxyz rotates

about fixed axis OA

with 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 Oxyz and

OXYZQ

QkQjQiQ zyx

Q

With respect to the rotating Oxyz frame,

kQjQiQQ zyx

With respect to the fixed OXYZ frame,

QQQ OxyzOXYZ



Coriolis Acceleration

15 - 49

Frame OXY is fixed and frame Oxy rotates with

angular velocity .

Position vector for the particle P is the same

in both frames but the rate of change depends on

the choice of frame.

Pr

The absolute velocity of the particle P is

OxyOXYP rrrv

Imagine a rigid slab attached to the rotating frame Oxy or F for short. Let P’ be a point on

the slab which corresponds instantaneously to

position of particle P.

OxyP rv F velocity of P along its path on

the slab 'Pv

absolute velocity of point P’ on the slab

Absolute velocity for particle P may be written as

FPPP vvv



15 - 50

FPP

OxyP

vv

rrv

Absolute acceleration for particle P is

OxyOXYP rdt

drra

OxyOxyP rrrra

2

OxyOxyOxy

OxyOXY

rrrdt

d

rrr

but,

OxyP

P

ra

rra

F

• Utilizing the conceptual point P’ on the slab,

• Absolute acceleration for the particle P becomes

22

2

F

F

F

POxyc

cPP

OxyPPP

vra

aaa

raaa

Coriolis acceleration




15 - 51

Consider a collar P which is made to slide at

constant relative velocity u along rod OB. The

rod is rotating at a constant angular velocity w.

The point A on the rod corresponds to the

instantaneous position of P.

cPAP aaaa

F

Absolute acceleration of the collar is

0 OxyP ra F

uava cPc 22 F

The absolute acceleration consists of the radial

and tangential vectors shown

2rarra AA

where




15 - 52

uvvtt

uvvt

A

A

,at

,at

Change in velocity over t is represented by

sum of three vectors

TTTTRRv

2rarra AA

recall,

is due to change in direction of

velocity of point A on the rod,

AAtt

arrt

vt

TT

2

00limlim

TT

result from combined

effects of relative motion of P and rotation of

the rod

TTRR and

uuu

t

r

tu

t

TT

t

RR

tt

2

limlim00

uava cPc 22 F

recall,




Sample Problem 15.9

15 - 53

Disk D of the Geneva mechanism

rotates with constant ccw angular

velocity D = 10 rad/s. At the instant

when = 150o, determine:

a. angular velocity of disk S

b. velocity of pin P relative to disk S

SOLUTION:

The absolute velocity of point P

may be written as

sPPP vvv

Magnitude and direction of

velocity of pin P are

calculated from radius and

angular velocity of disk D.

Pv

Direction of velocity of point

P’ on S coinciding with P is

perpendicular to radius OP.

Pv

Direction of velocity of P

with respect to S is parallel to

the slot.

sPv

Solve the vector triangle for

angular velocity of S and

relative velocity of P. Dr. Mohammad Abuhaiba, PE


15 - 54

SOLUTION:

Absolute velocity of point P may be written as

sPPP vvv

Magnitude and direction of absolute velocity of pin P

are calculated from radius and angular velocity of

disk D. smm500srad 10mm 50 DP Rv

Direction of velocity of P with respect to S is parallel

to slot. From the law of cosines,

mm 1.37551.030cos2 2222 rRRllRr

From the law of sines,

4.42742.0

30sinsin

30sin

R

sin

r

6.17304.4290

The interior angle of the vector triangle is



Sample Problem 15.9

15 - 55

Direction of velocity of point P’ on S coinciding with P

is perpendicular to radius OP. From the velocity

triangle,

mm 1.37

smm2.151

smm2.1516.17sinsmm500sin

ss

PP

r

vv

ks

srad08.4

6.17cossm500cosPsP vv

jiv sP

4.42sin4.42cossm477

smm 500PvDr. Mohammad Abuhaiba, PE


Sample Problem 15.9

Sample Problem 15.10

15 - 56

In the Geneva mechanism, disk D

rotates with a constant ccw

angular velocity of 10 rad/s. At

the instant when j = 150o,

determine 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 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.



15 - 57

SOLUTION:

Absolute acceleration of the pin P may be

expressed as csPPP aaaa

• From Sample Problem 15.9.

jiv

k

sP

S

4.42sin4.42cossmm477

srad08.44.42

• Considering each term in the acceleration eq,

jia

Ra

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: S may be positive or negative




15 - 58

• The relative acceleration must be parallel to

the slot. sPa

sPv

Direction of Coriolis acceleration is

obtained by rotating direction of the

relative velocity by 90o in the sense of

S.

ji

ji

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





150, 158, 166, 172, 179,




2 - 59

Motion About a Fixed Point

15 - 60

The most general displacement of a RB with a fixed

point O is equivalent to a rotation of body about an

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

velocity the velocity of a particle P of the body is ,

rdt

rdv

acceleration of particle P is

.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.

Angular acceleration represents the velocity of the

tip of

.



General Motion

15 - 61

For particles A and B of a RB,

ABAB vvv

Particle A is fixed within the body and

motion of the body relative to AX’Y’Z’ is the

motion of a body with a fixed point

ABAB rvv

Similarly, acceleration of particle P is

ABABA

ABAB

rra

aaa

Most general motion of a RB is equivalent to:

a translation in which all particles have the same

velocity and acceleration of a reference particle A, and

of a motion in which particle A is assumed fixed.




15 - 62

The crane rotates with a constant

angular 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:

a. angular velocity of boom,

b. angular acceleration of boom,

c. velocity of boom tip

d. acceleration of 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

jir

kj

639.10

30sin30cos12

50.030.0 21



15 - 63

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




15 - 64

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





184, 190, 197, 204, 210

Due Sunday 29/4/2012



2 - 65

3-D Motion. Coriolis Acceleration

15 - 66

With respect to the fixed frame OXYZ and

rotating frame Oxyz,

QQQ OxyzOXYZ

Consider motion of particle P relative to a

rotating frame Oxyz or F for short. Absolute

velocity can be expressed as

FPP

OxyzP

vv

rrv

Absolute acceleration can be expressed as

onaccelerati Coriolis 22

2

F

F

POxyzc

cPp

OxyzOxyzP

vra

aaa

rrrra



Frame of Reference in General

Motion

15 - 67

Consider:

- fixed frame OXYZ,

- translating frame AX’Y’Z’, and

- translating and rotating frame Axyz or F.

• With respect to OXYZ and AX’Y’Z’,

APAP

APAP

APAP

aaa

vvv

rrr

• The velocity and acceleration of P

relative to AX’Y’Z’ can be found in

terms of the velocity and acceleration

of P relative to Axyz.

FPP

AxyzAPAPAP

vv

rrvv

cPP

AxyzAPAxyzAP

APAPAP

aaa

rr

rraa

F

2




15 - 68

For the disk mounted on the arm,

the indicated angular rotation

rates are constant. Determine:

a. velocity of point P

b. acceleration of P

c. angular velocity and angular

acceleration of the disk

SOLUTION:

Define a fixed reference frame

OXYZ at O and a moving reference frame Axyz or F attached to the arm

at A.

With P’ of the moving reference

frame coinciding with P, the

velocity of the point P is found

from FPPP vvv

The acceleration of P is found from

cPPP aaaa

F

The angular velocity and angular

acceleration of the disk are

F

FD



15 - 69

SOLUTION:

Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F

attached to the arm at A.

j

jRiLr

1

k

jRr

D

AP

2

F

With P’ of the moving reference frame

coinciding with P, the velocity of the point P

is found from

iRjRkrv

kLjRiLjrv

vvv

APDP

P

PPP

22

11

FF

F

kLiRvP

12




15 - 70

• The acceleration of P is found from

cPPP aaaa

F

iLkLjraP

2111

jRiRk

ra APDDP

2222

FFF

kRiRj

va Pc

2121 22

2

F

kRjRiLaP

21

22

21 2

• Angular velocity and acceleration of the disk,

FD

kj

21

kjj

211

F

i

21





220, 234, 240

Due Sunday 1/5/2012



2 - 71

Kinematics of Rigid Bodies - الصفحات الشخصية | الجامعة...

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Transcript of Kinematics of Rigid Bodies - الصفحات الشخصية | الجامعة...