Page 273 - Dynamics of Mechanical Systems

0593_C08_fm Page 254 Monday, May 6, 2002 2:45 PM

254 Dynamics of Mechanical Systems

FIGURE 8.8.2 FIGURE 8.8.3
Two loading conditions on a circular disk. Free-body diagram of disk in
Figure 8.8.2a.

Consider two loading conditions on D: First, let D be loaded by a force W applied to
the rim of D as in Figure 8.8.2a. Next, let D be loaded by a weight having mass m attached
by a cable to the rim of D as in Figure 8.8.2b. Let m have the value W/g (that is, the mass
has weight W).
Consider first the loading of Figure 8.8.2a. The force W will cause a clockwise angular
acceleration α as viewed in Figure 8.8.2a. This angular acceleration will in turn induce a
a
counterclockwise inertia torque component when the equivalent inertia force is passed
through O. A free-body diagram is shown in Figure 8.8.3, where I is the axial moment
O
of inertia of D, M is the mass of D, and O and O are horizontal and vertical bearing
x
y
reaction components. By adding forces horizontally and vertically, by setting the results
equal to zero, and by setting moments about O equal to zero, we obtain:
O = 0 (8.8.1)
x
O − mg W = 0 (8.8.2)
−
y
and

I α− W = 0 (8.8.3)
O a r

Next, for the loading of Figure 8.8.2b, the weight will create a tension T in the attachment
cable which in turn will induce a clockwise angular acceleration α of D and a resulting
b
counterclockwise inertia torque. Free-body diagrams for D and the attached weight are
shown in Figure 8.8.4, where the term rα is the magnitude of the acceleration of the
b
weight. By setting the resultant forces and moments about O equal to zero, we obtain:

O = 0 (8.8.4)
x
−
O − Mg T = 0 (8.8.5)
y
I α− Tr − 0 (8.8.6)
O b
mrα+ T W = 0 (8.8.7)
−
b