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0593_C10_fm Page 341 Monday, May 6, 2002 2:57 PM

Introduction to Energy Methods 341

ω
W h
n
r O

FIGURE 10.11.1
Wheel rolling over a step.

10.11 A Wheel Rolling Over a Step
For a second example illustrating the tandem use of the work–energy principle and the
conservation of momentum principle, consider again the case of the wheel rolling over a
step as in Figure 10.11.1 (recall that we considered this problem in Section 9.9). Let the
wheel W have a radius r, mass m, and axial moment of inertia I. Suppose we are interested
in knowing the speed v of the wheel center required for the wheel to roll over the step.
Recall in Section 9.9 that when the wheel encounters the step its angular momentum
about the corner (or nose) O of the step is conserved. By using the conservation of angular
momentum principle we found that the angular speed ˆ ω of W just after impact is (see
Eq. (9.9.8)):

ˆ ω = [ { Imr (r − h / )] ( I mr 2 ω )} (10.11.1)
+
+

where ω is the angular speed of W before impact and h is the height of the step.
After impact, W rotates about the nose O of the step. For W to roll over the step it must
have enough kinetic energy after impact to overcome the negative work of gravity as it
rises up over the step. The work W of gravity as W rolls completely up the step is:
g
W =− mgh (10.11.2)
g

If W just rolls over the step (that is, if W expends all its kinetic energy after impact in
rolling over the step), it will come to rest at the top of the step and its kinetic energy K f
at that point will be zero. The kinetic energy K just after impact is:
i
+
K = ( ) ω 2 +( ) mr ˆ ω 2 = ( )( I mr ) ω ˆ 2 (10.11.3)
2
2
I ˆ
12
12
12
i
The work–energy principle then leads to:
+
W = K − K or − mgh = −( )( I mr ) ω 2 (10.11.4)
2
ˆ
1 2
0
g f i
ˆ ω
Solving we have:
[ 2 )] / 12
+
ˆ ω= 2mgh /(I mr (10.11.5)
Hence, from Eq. (10.11.1), the speed v of W just before impact for step rollover is: