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0593_C05_fm Page 141 Monday, May 6, 2002 2:15 PM

Planar Motion of Rigid Bodies — Methods of Analysis 141

I

3.67 m

4.098 m Q
Q v

B 2
P P
v
4.95 m
B 3
2.0 m

B 1 R
FIGURE 5.5.7 O 6.098 m 45°
Instant center of zero velocity of B 2 .

Similarly, the distance IQ is 3.67 m, and v is, then,
Q
v = IQ ω νν = (367 )(2 44 νν. ) = 895νν = 6 34 n + 634 n (5.5.22)
Q
.
.
.
.
23 3 3 x y
Then ω becomes:
3
Q
.
ω =− v / QR =− 895 4 95 =− 181 rad sec (5.5.23)
.
.
3
Next, consider an acceleration analysis. Because B has pure rotation, P moves in a circle
1
about O and its acceleration is:
λλ
a = αα 1 × OP + ωω 1 ×(ωω 1 × OP) = 4 n × λλ 1 + − ( 5 n ) ×− ( [ 5 n ) × ]
P
2
2
z
1
z
3
(5.5.24)
= νν − 50 λλ = −8 n − 50 n m sec 2
8
1 1 x y
Because P and Q are both fixed on B , the acceleration of Q may be expressed as:
2
a = a + αα × PQ + ωω ×(ωω × PQ)
Q
P
2 2 2
=−8 n − 50 n + αα 2 n ×(3 0. λλ 2 ) +(2 44. n ) × ( [ 2 44. n ) ×(3 0. λλ 2 )]
y
x
z
2
z
νν
− 8 n − 50 n + 3 0. α 22 − 17 86. λλ 2 (5.5.25)
y
x
y]
)
+
12
n
/
=−8 n − 50 n + 3 0. α 2[ (–12 n ) x ( 3 2/ n − 17 86. ( [ 3 2 n +( ) n y)]
x
y
x
+−58 93 2 6.
=− ( 23 467 1 5. − . α 2 n ) x ( + . α 2 n ) y