As part of a carnival game, a 0.503-kg ball is thrown at a stack of 16.3-cm tall

Question

As part of a carnival game, a 0.503-kg ball is thrown at a stack of 16.3-cm tall, 0.423-kg objects and hits with a perfectly horizontal velocity of 11.3 m/s. Suppose the ball strikes the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 3.35 m/s in the same direction, the topmost object now has an angular velocity of 4.03 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 11.4 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass?

Explanation / Answer

initial angular about the center of mass Li = mball*vball*r

r = 0.114 m

viball = 11.3 m/s

after hitting

final angular momentum

Lf = mball*vfball*r + (I*w)

from conservation of angular momentum

Lf = Li

(0.503*11.3*0.114) = (0.503*3.35*0.114) + (I*4.03)

I = 0.113 kg m^2 <<<<------answer

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