A long, thin, rod of mass M = 0.500kg and length L = 1.00 m is free to pivot abo

Question

A long, thin, rod of mass M = 0.500kg and length L = 1.00 m is free to pivot about a fixed pin located at L/2. The rod is held in a vertical position as shown below. It is struck in an inleastic collision by a ball mass m = 0.300 kg. The ball sticks to the end of the rod. If the ball was moving at an initial speed of v0 = 4m/s, then with what angular velocity (in rad/s) does the rod move the instant after the ball strikes it?

The moment of inertia of a uniform rod rotating around its center of mass is

Explanation / Answer

using conservation of angular momentum

angular momentum before collision = angular momentum after collision

m*vo*r = I*w

I is the moment of inertia of the system = (1/12)*M*L^2 + m*(L/2)^2

I = ((1/12)*0.5*1^2) + (0.3*0.5^2) = 0.1166 kg-m^2

then m*vo*(L/2) = 0.1166*w

then 0.3*4*(0.5) = 0.1166*w

required angular velocity is w = 5.14 rad/s

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