A local ice hockey team has asked you to design an apparatus for measuring the s

Question

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 1.90-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity v (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates.

1. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

Explanation / Answer

Angular momentum of the puck just before collision = mvr = 0.163*v*2 = 0.326v .. (1)
After collision, angular momentum = mr^2w + Iw
m = mass of puck, r = 2m, I = moment of inertia of rod = Mr^2/3 = 1.9(2^2)/3 = 2.533 kgm^2
Angular momentum = 0.163*2^2*w = 2.533w = 3.1853w
w = 2*pi/0.736
so, angular momentum = 3.1853*2pi/0.736 = 27.179 = 0.326v (consertvation of angular momentum)
v = 83.3717 m/s

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