A 3.65-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 3.65-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed v1 = 16.0 m/s to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle before swinging back toward its original position. What is the angular speed of the rod immediately after the collision?

____ rad/s

Explanation / Answer

I = M L^2 / 3 = (3.65 x 2^2) / 3 = 4.87 kg m^2

angular momentum = m (v x r) or I w

distance of collision point from hinged end = 2 - (2 x 1 / 5) = 8/5 m = 1. 6m
Applying angular momentum conservation of the system for the collision about the hinges end.

(0.25 x 16 x 1.6 ) + ( I x 0) = (0.25 x -9.50 x 1.6) + (4.87 w)

6.4 = - 3.8 + 4.87 w

w = 2.09 rad/s

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