A 2.80 -kg cylindrical rod of length 2.00 m is suspended from a horizontal bar s

Question

A 2.80-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

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

while the rod swings to the right through an angle ? before swinging back toward its original position. What is the angular velocity of the rod immediately after the collision?
rad/s

A 2.80-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 = 16.5 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 velocity of the rod immediately after the collision?

Explanation / Answer

initial angular momentum = final angular momentum

m*vi*(4/5)*L = (1/3)*M*L^2*W - m*vf*(4/5)*L

m*vi*(4/5) = (1/3)*M*L*W - m*vf*(4/5)

0.25*16.5*(4/5) = (1/3)*2.8*2*w - (0.25*9.5*(4/5))

w = 2.785714 rad/s

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