A 3.80-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 3.80-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 = 12.0 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 velocity of the rod immediately after the collision?

Explanation / Answer

Angular momentum remians Conserved

Therfore

mvir = mvfr + Iw

mvir = mvfr + ((1/3)ML^2)w

0.25*12*(4/5)*2 = - 0.25*9.50*(4/5)*2 + (1/3)*2*2^2*w

8.6 = (1/3)*2*2^2*w

w = 3.225 rad/sec

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