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1

Chapter 6 Plane Kinematics of Rigid Bodies

Rigid body: A body (system) is so rigid such that the relative positions of all the mass elements do not change during motion.

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2

= angular velocity = angular acceleration

angular displacement)

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3

v

r

O

rrvrv and

• Velocity of a point P relative to another point O with a fixed distance from P (could be a fixed point in a rigid body.

• Direction defined by Right Hand Rule.

P

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rα)rω(ωa rωrωva

or

z and z where r v

,

tana

O

z

xercise

Prove that the above statement is identical to:

0 and ,0 :oteN

ˆ)2(ˆ)( 2

rrθθrθrrθrra

Acceleration of a point P relative to another point O with a fixed distance from P.

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5

Relative motion of two points of a fixed separation (i.e. on a rigid plane):

A/BBA

A/BBAA/B

rω v v

rω v v v

or

Av

Bv BAr /

Av

Bv

A

B BAr /

Example : Draw a vector diagram to show the relation between A/BBA v vv

and ,

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6

Instantaneous Centre of zero velocityDraw two lines perpendicular to the velocities of points A and B of a rigid body. Their intercept is an instantaneous stationary point (zero velocity).

B

B

A

A

r

v

r

vω

OAv /

OBv /

A

BCAr /

C

CBr /

O

y

x

0 so parallel,in not generalin are which , and lar toperpendicu is

)(0)(0

C/O

B/CA/CC/O

C/OB/CC/OB/CB/CB/CA/OB/C

C/OA/CC/OA/CA/CA/CA/OA/C

B/CB/O

A/CA/O

C/OB/CC/OB/CB/O

C/OA/CC/OA/CA/O

vrrv

vrvrrωr vrvrvrrωr vr

rvrv

vrω vv vvrω vv v

Av

Bv

A

BAr

CBr

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)( // BABABA rraa

])[()( )]()[()(

)()()()( /

BABA

BABA

BBAA

BABA

rrrrrrrr

rrrraaa

Relative acceleration is:

BABA

BABAaaavvv

/

/

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Example : For pure rolling without slipping, derive

Let s = 00’ = r

(c) For the new position C' from C : x = s - rsin and y = r - rcos

when = 0, vc = 0

ccoo ,0 , , avav

xrωxθrxsv 0 ˆˆˆ(a)

xrαxωrva 00 ˆˆ(b)

yθrωxθrωyyxxvθ rωθθry

θrωθθrx

c ˆ)sin(ˆ)cos1(ˆˆsinsin

)cos1()cos1(

θrωθrα θθrωθωryθrωθrα θθrωθωrx

cossincossinsin)cos1(sin)cos1(

2

2

(d)

when = 0, y raC ˆ2

y

sinr

r

O 'O

C

s

s

x

'C

xy

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OPv /

13 msvo

o60P

O

R

o60

ov

OPv /

Pv

P

#1

P

o220

2

sm26.4)60180cos(2

s/m23.0

32.02.02.0

vvvvvv

R

v v

vvv

P/O0P/OP

oP/O

P/OoP

Solution:

Example A wheel has a radius R = 0.3 m. Point P is 0.2 m from the centre O, which is moving with vo = 3 m/s. Find .Pv

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Example : The structure has the following dimensions: rB/A = 0.1 m, rC/D = 0.075 m, h = 0.05 m, X = 0.25 m. CD = 2 rad/s. Find : AB , BC

h

D

X

A

B

Cx

y

rad/s 05.0/m/s 0429.0

rad/s 857.0

175.0075.02 0 05.0

Henceˆ) 175.0075.02(ˆ 05.0]ˆ05.0ˆ175.0[ˆˆ075.02

ˆˆˆ

/

B

ABBAB

BC

BC

BCB

BCBC

BC

B/CBCC/DCD

B/CBCcBB

rvv

v

y x yxzy

rzω yrω rωvxvv