Page 173 - Dynamics of Mechanical Systems

0593_C05_fm Page 154 Monday, May 6, 2002 2:15 PM

154 Dynamics of Mechanical Systems

z (
v =− rω n + ω n × n r y) =−2 rω n (5.8.31)
P
x x
z [
z (
z (
a =− rα n + α n × − n r y) + ω n × ω n × − n r y)] = rω 2 n (5.8.32)
O
x y
z [
z (
z (
a =− rα n + α n × n r y) + ω n × ω n × n r y)]
P
x
(5.8.33)
=−2 rα n − rω 2 n
x y
Observe that the velocity of O is zero, as expected, but the acceleration of O is not zero.
To find the point C with zero acceleration, we can use Eqs. (5.8.11) and (5.8.12): specifically,
for a Cartesian (X–Y) axes system with origin at O, we find the coordinates of C to be:
2
α
x − α˙˙
ω ˙˙ y ω 2 r − ( α ) − () 0 rαω 2
x = x + Q Q =+ =− (5.8.34)
0
C Q α 2 + ω 4 α 2 + ω 4 α 2 + ω 4
and
()
x + ω ˙˙
2
α˙˙ 2 y r − ( α α ) + ω 0 rω 4
y = y + Q Q =+ = (5.8.35)
r
C Q α 2 + ω 4 α 2 + ω 4 α 2 + ω 4
For positive values of ω and α, the position of C is depicted in Figure 5.8.3.
To verify the results, consider calculating the acceleration of C using the expression:
a = a + αα × QC +(ωω × QC) (5.8.36)
Q
C
where QC is:

rαω 2   rω 4  
QC =− n −  r −   n (5.8.37)
α 2 + ω 4 x   α 2 + ω 4   y

Using Eq. (5.8.27), a becomes:
C
rαω 2   rω 4  
a =− rα n − α n +  r −   n
α
C
x α 2 + ω 4 y   α 2 + ω 4   x
(5.8.38)
rαω 2   rω 4  
+ ω 2 n + ω 2  r −   n = 0
α 2 + ω 4 x   α 2 + ω 4   y
Finally, we can check the consistency of Eqs. (5.8.18), (5.8.21), and (5.8.22): from Eqs.
(5.8.21) and (5.8.22), ξ and η are:

ξ= ( [ − )( y ˙˙ y ˙˙ − )( x ˙˙ − )]
1
x ˙˙
∆ x P x Q Q − ) −( y P y Q Q P (5.8.39)
P