A ball and stick theorem
As in the solution to Problem 8.18, we have
MV₀ = MV + mv,
MV₀d = MVd + Iω, (8.126)
MV₀² = MV² + mv² + Iω².
The speed of the contact point on the stick right after the collision equals the speed of the CM plus the rotational speed relative to the CM. In other words, it equals v + ωd. The desired relative speed is therefore (v + ωd) − V. We can determine the value of this relative speed by solving the above three equations for V, v, and ω. Equivalently, we can just use the results of Problem 8.18. There is, however, a much more appealing method, which is the following.
The first two equations quickly give mvd = Iω. The last equation may then be written as, using Iω² = (Iω)ω = (mvd)ω,
M(V₀ − V)(V₀ + V) = mv(v + ωd). (8.127)
8.9 Solutions 365
If we now write the first equation as
M(V₀ − V) = mv, (8.128)
we can divide Eq. (8.127) by Eq. (8.128) to obtain V₀ + V = v + ωd, or
V₀ = (v + ωd) − V, (8.129)
as we wanted to show. In terms of velocities, the correct statement is that the final relative velocity is the negative of the initial relative velocity. In other words, V₀ − 0 = −[V − (v + ωd)].