Question 11.9: A thin bar of length l and massM lies on a frictionless plat...
A thin bar of length l and massM lies on a frictionless plate (the x,y-plane in Fig. 11.16). A hockey puck of mass m and velocity v knocks the bar elastically under 90° at the distance d from the center of gravity. After the collision the puck is at rest.
(a) Determine the motion of the bar.
(b) Calculate the ratio m/M, accounting for the fact that the puck is at rest.
(a) Since the collision is elastic, momentum and energy conservation hold, where momentum conservation refers both to linear and angular momentum. The bar acquires both a translational and a rotational motion from the collision with the puck.
Conservation of the linear momentum immediately leads to
P_{s} = M v_{s} = mv, (11.27)
and the velocity of the center of gravity is
v_{s} = \frac{mv}{M}. (11.28)
Likewise, from the conservation of angular momentum it follows that
L_{s} = \Theta_{s} ω = mvd = D, (11.29)
and for the angular velocity of the bar relative to the center of gravity,
ω = \frac{mvd}{Ml^{2}/12}, (11.30)
where
\Theta_{s} =\int\limits_{0}^{l/2}{\varrho r^{2} dV} = \frac{1}{12}Ml^{2}.Thus, the center of gravity of the bar moves uniformly with v_{s} along the y-axis, while the bar rotates with the angular velocity ω about the center of gravity. Figure 11.17 illustrates several stages of the motion.
(b) The kinetic energy of the bar can be determined by means of the energy conservation law. Before the collision, the kinetic energy of the puck is
T = \frac{1}{2} mv^{2}, (11.31)
while the kinetic energy after the collision consists of two components:
T_{t} = \frac{1}{2}Mv^{2}_{s} “translation energy of the center of gravity”
and
T_{r} = \frac{1}{2} \Theta_{s} ω^{2} “rotation energy about the center of gravity.”
Since the potential energy remains unchanged, it immediately follows that
T = \frac{1}{2} mv^{2} = T_{t} + T_{r} = \frac{1}{2} (Mv^{2}_{s}+\Theta_{s} ω^{2})or
mv^{2} = Mv^{2}_{s}+ \frac{Ml^{2}}{12} ω^{2}. (11.32)
Insertion of (11.28) and (11.30) into (11.32) finally yields
mv^{2} = \frac{m^{2}v^{2}}{M}+ \frac{m^{2}v^{2}d^{2}(12)^{2}}{M^{2}l^{4}} \frac{Ml^{2}}{12},1 = \frac{m}{M}+ 12 \frac{m}{M} \frac{d^{2}}{l^{2}} ,
or
\frac{m}{M}= \frac{1}{1+ 12(d/l)^{2}}for the mass ratio. If the puck kicks the bar at the center of gravity, d = 0, no rotation appears. In order to make the collision elastic, m =M must be satisfied.
If the puck kicks the bar at the point d = l/2, the collision is elastic only ifM = 4m.
In this case the rotation velocity is ω = 6mv/Ml = 6v_{s}/l.
