Answer
The speed of the puck just before it hit the rod was 45.1 m/s
Work Step by Step
We can find the moment of inertia of the rod with the puck in the basket.
$I = \frac{1}{3}M_rR^2+M_pR^2$
$I = \frac{1}{3}(0.800~kg)(2.00~m)^2+(0.163~kg)(2.00~m)^2$
$I = 1.72~kg~m^2$
We can find the angular speed of the rotating rod.
$\omega = \frac{2\pi~rad}{0.736~s}$
$\omega = 8.537~rad/s$
We can find the final angular momentum while the rod is rotating.
$L_2 = I \omega$
$L_2 = (1.72~kg~m^2)(8.537~rad/s)$
$L_2 = 14.7~kg~m^2/s$
We can use conservation of angular momentum to find the speed of the puck just before it hit the rod.
$L_1 = L_2$
$M_pvR = L_2$
$v = \frac{L_2}{M_pR}$
$v = \frac{14.7~kg~m^2/s}{(0.163~kg)(2.00~m)}$
$v = 45.1~m/s$
The speed of the puck just before it hit the rod was 45.1 m/s.
