As part of a carnival game, a 0.578-kg ball is thrown at a stack of 16.8-cm tall

Question

As part of a carnival game, a 0.578-kg ball is thrown at a stack of 16.8-cm tall, 0.383-kg objects and hits with a perfectly horizontal velocity of 12.8 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 4.60 m/s in the same direction, the topmost object now has an angular velocity of 1.63 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 11.8 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass?

Explanation / Answer

I would say he almost got it right, but you can't just add linear with angular momentum. You need to convert the initial linear momentum into angular momentum around the axis of rotation, then add it up with the other angular momentums:

0.578*12.8*(11.8/100) = 0.578*4.6*(11.8/100) + I * 1.63
I=.3371m/s^2

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