As part of a carnival game, a 0.528-kg ball is thrown at a stack of 17.8-cm tall

Question

As part of a carnival game, a 0.528-kg ball is thrown at a stack of 17.8-cm tall, 0.383-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 4.85 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 12.5 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass?

Explanation / Answer

mass of ball , m = 0.528 Kg

mass of object , M = 0.383 Kg

let the moment of inertia is I

1)

Using conseration of angular momentum

angular momentum of system before collision = angular momentum of system after collision

0.125 * 0.528 * 11.3 = I * 1.63 rad/s + 0.163 * 0.528 * 4.85 m/s

solving for I

I = 0.2015 Kg.m^2

the moment of inertia of ball is 0.2015 Kg.m^3

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