Page 361 - Dynamics of Mechanical Systems

0593_C10_fm Page 342 Monday, May 6, 2002 2:57 PM

342 Dynamics of Mechanical Systems

+ ( [ 2 mgh)] 12
/
I mr )(2
v = ω r (10.11.6)
r =
(
Imr r h)
−
+
2
Suppose that W is a uniform circular disk with I then being (1/2)mr . In this case, v
becomes:
[
−
v = r 3 gh ( ) 2 r h ] (10.11.7)
/
3
Observe that the disk could (at least, theoretically) roll over a step whose height h is greater
than r. From a practical perspective, however, the disk will only encounter the nose of the
step if h is less than r.
Observe also that, as with the colliding vehicles, the energy is not conserved when the
wheel impacts the step. From Eq. (10.11.1), the energy loss L is seen to be:

L = ( ) ω 2 +( ) mr ˆ ω 2 −(12 I ) ω 2 −( ) mr ω 2
I ˆ
2
2
12
12
12
−
23
22 2
=−( ) 2 Imrh + 2 m r h m r h   (10.11.8)
12

+
 Imr 2 
For a uniform circular disk, this becomes:
[
L =− mh r −( )] (10.11.9)
h 3

10.12 The Spinning Diagonally Supported Square Plate
For a third example of combined application of the work–energy and momentum conser-
vation principles, consider again the spinning square plate supported by a thin wire along
a diagonal as in Figure 10.12.1 (recall that we studied this problem in Section 9.12). As
before, let the plate have side length a, mass m, and initial angular speed Ω. Let the plate
be arrested along an edge so that it rotates about that edge as in Figure 10.12.2.

Ω

Ω
0
O

Q
Q

a

FIGURE 10.12.1 FIGURE 10.12.2
Suspended spinning square plate. Arrested plate spinning about an edge.