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0593_C05_fm Page 155 Monday, May 6, 2002 2:15 PM

Planar Motion of Rigid Bodies — Methods of Analysis 155

n y Y
P
D α
ω

Q
C
r ω 4
FIGURE 5.8.3 O α + ω 4
2
Location of center C of zero acceler- r α ω 2 n X
2
ation for rolling disk of Figure 5.8.2. α + ω 4 x
and
η= ( [ y − )( y ˙˙ − ) −( x − )( x ˙˙ − )] (5.8.40)
1
y ˙˙
x ˙˙
x
y
∆ P Q P Q P Q Q P
where, from Eq. (5.8.23), ∆ is:
 2 − ) 2 
∆= − ( y ˙˙ − ) +( x ˙˙ ˙˙ x (5.8.41)
˙˙ y
  P Q Q P  
From Eqs. (5.8.27) and (5.8.33) and from Figure 5.8.3 we have:

x = 0, y = 2 r
P P
x = 0, y = r
Q Q
x =−2 α y =− rω 2 (5.8.42)
˙˙
˙˙
r ,
P P
x =− α y = 0
˙˙
˙˙
r ,
Q Q
Then, ∆ becomes:

2
∆= − − ( rω 2 − ) +− ( rα + rα ) 2  =− (ω 4 + ) (5.8.43)
α
2
2
2
0
r
   
Hence, ξ and η are:
ξ = −1 [ 0 −(2rr α + 2r α)] = α (5.8.44)
− ) − ( r
2
α
2
4
r 2 ( ω + ) α + ω 4
η = −1 [ (2rr ω − ) − 0 ] = ω 2 (5.8.45)
2
0
− ) − ( r
α
2
4
2
r 2 ( ω + ) α + ω 4
These expressions are identical with those of Eq. (5.8.18).