Page 321 - Dynamics of Mechanical Systems

0593_C09_fm Page 302 Monday, May 6, 2002 2:50 PM

302 Dynamics of Mechanical Systems

FIGURE 9.9.2 FIGURE 9.9.3
Meshing coaxial disks. Wheel rolling over a step.

Because the disks can rotate freely on their common shaft there is no axial momentum
applied to the disks; thus, there is no angular impulse about the shaft axis. This in turn
means that the angular momentum of the disks about the shaft axis is conserved.
Before meshing, the axial angular momentum of the disks is:

A n = I ω n + I ω n (9.9.3)
before A A B B
where n is a unit vector parallel to the shaft axis. After meshing, the angular momentum
of the disks is:

A (
A n = I + )ω n (9.9.4)
I
after B
By equating these expressions we have:

I
ω = (I ω + I ω ) (I + ) (9.9.5)
A A B B A B
Observe the similarity between Eqs. (9.9.2) and (9.9.5). These expressions are useful in
clutch design and in the design of slider mechanisms.

Example 9.9.2: Wheel Rolling Over a Step
As a third example of momentum conservation consider a wheel or disk W rolling in a
straight line and encountering a small step as in Figure 9.9.3. Let the angular speed of W
before striking the step be ω. The objective is to determine the angular speed ω of W just
after impact with the step.
Let W have radius r and mass m, let its axial moment of inertia be I, and let the step
height be h. Then, the angular momentum of W about the step corner O before and after
impact is conserved. Before impact, the angular momentum of W about O is:

(
ω
−
=
A = A + A G O [ I + mv r h)] n (9.9.6)
before WG
where v is the speed of the disk center before impact, and n is an axial unit vector as
shown in Figure 9.9.3. Because W is rolling, v is simply rω. After impact with the step, W
is rotating about O. Its angular momentum after impact is thus:
=
I +
2
ˆ
A = A + A G O [ ω mr ˆ ω n ] (9.9.7)
after WG