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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 (angular displacement)
   = angular velocity
     = angular acceleration

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• 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.
 

v r
and

v
v   r   r
P

O

r

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Acceleration of a point P relative to another point O with a
fixed distance from P.
 
   
a  v  ω  r  ω  r or


 
 
a  ω  (ω  r )  α  r
where

   zˆ and
 
   zˆ
zˆ
 
 ,

r
O
an

v
at
Exercise
Prove that the above statement is identical to:

2
a  ( r  r θ ) rˆ  ( r θ  2 rθ ) θˆ
N ote : r  0 , and r  0
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Relative motion of two points of a fixed
separation (i.e. on a rigid plane):



 
v A/B  v A  v B  ω  rA/B
or


 
v A  v B  ω  rA/B
Example : Draw
a vector
diagram to show the relation between
 

v A , v B and v A/B
A

vB

rA / B
B
5

vA



vB
 
  rA / B

vA
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Instantaneous Centre of zero velocity
Draw two lines perpendicular to the velocities of points A and B of a rigid
body. Their intercept is an instantaneous stationary point (zero velocity).
y




 
v A/O  v A/C  v C/O  ω

r

v
 A/C  C/O
v B/O  v B/C  v C/O  ω  rB/C  v C/O
A

rA / C


v A/O  rA/C
v B/O  rB/C

rA/C
rB/C

vA/ O
B
vB / O

rB / C




 v A/O  0  rA/C  ( ω  rA/C )  r A/C
 v A/O  0  rB/C  ( ω  rB/C )  rB/C

vC/O is perpendicu
v C/O  0
ω
vA
rA
C
O
x



 v C/O  r A/C  v C/O
 v C/O  rB/C  v C/O


lar to r A/C and rB/C , which are in general not in parallel,

vB
rB
A

vA

rA

rB
B
vB
C
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so
Relative acceleration is:



v A  vB  vA / B
aA  aB  aA/B



aA/B  aA  aB


 


 


   ( r A )    (  r A )    ( rB )    (  rB )


 
 


   ( r A  rB )    [(   r A )  (  rB )]


 



   ( r A  rB )    (  [ r A  rB ])
 

 


a A  a B    rA / B    (   rA / B )
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Example : For
pure
rolling without slipping, derive

 
vo , ao , vc  0, ac
y
Let s = 00’ = r 

v 0  s xˆ  r θ xˆ  rω xˆ


(b) a 0  v 0  r ω xˆ  rα xˆ
(a)



r
(c) For the new position C' from C :
x = s - rsin and y = r - rcos
s
O
C'
C
x  r θ (1  cos θ )  rω (1  cos θ )
y  r θ sin θ  rω sin θ

v c  x xˆ  y yˆ  rω (1  cos θ ) xˆ  ( rω sin θ ) yˆ

O'
s
y
x
x r sin 
when  = 0, vc = 0
2
(d) x  r ω (1  cos θ )  rω θ sin θ  rα (1  cos θ )  rω sin θ
y  r ω sin θ  rω θ cos θ  rα sin θ  rω 2 cos θ

2
when  = 0, a C  r  yˆ
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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 v P .
Solution:



v P  v o  v P/O
v P/O  0 . 2   0 . 2 
vo
 0 .2 
R
0 .3
o
cos( 180  60 )
v P  v 0  v P/O  2 v 0 v P/O
1
v P  4 . 26 m s
#
2
2
2

vP / O
60
o
P
O
R
 2 m /s

v

vo  3ms
3
P
1
P

v

vP / O
60
o
o
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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




ˆ
v B  v B x  v c  ω BC  rB/C

  ω CD rC/D yˆ  ω BC zˆ  rB/C
  2  0 . 075 yˆ   BC zˆ  [  0 . 175 xˆ  0 . 05 yˆ ]
  0 . 05  BC xˆ  (  2  0 . 075  0 . 175  BC ) yˆ
Hence
v B   0 . 05  BC
0   2  0 . 075  0 . 175  BC
X
B
 BC   0 . 857 rad/s
v B  0 . 0429 m/s
 AB  v B / rB / A   0 . 05 rad/s
y
C
D
x
A
h
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