As part of a carnival game, a 0.553-kg ball is thrown at a stack of 23.3-cm tall

Question

As part of a carnival game, a 0.553-kg ball is thrown at a stack of 23.3-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 11.3 m/s. Suppose the ball strikes the very top of the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 4.85 m/s in the same direction, the topmost object now has an angular velocity of 2.43 rad/s and all the objects below are undisturbed. If the object's center of mass is located 16.3 cm below the point where the ball hits, what is the moment of inertia of the object?

Explanation / Answer

Energy lost by the thrown ball = 0.5m(v^2 -u^2) = 0.5*0.553*(11.3^2 - 4.85^2) = 28.80J
This is being transferred to the hit object.
Hence the energy of the hit object
0.5I*^2 + 0.5mv^2 = 28.8

I*^2 + mv^2 = 57.6

v =r

I*2.43^2 + 0.403*((0.163*2.43)^2) = 57.6

Hence solving this we get I = 9.74 kgm^2

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