As part of a carnival game, a 0.578-kg ball is thrown at a stack of 23.8-cm tall
ID: 2259567 • Letter: A
Question
As part of a carnival game, a 0.578-kg ball is thrown at a stack of 23.8-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 10.8 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 4.43 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 16.7 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass?
What is the center of mass velocity of the tall object immediately after it is struck?
Thanks in advance!
(PS, I already got the inertia portion right, so focusing on the center of mass velocity would be OK!)
As part of a carnival game, a 0.578-kg ball is thrown at a stack of 23.8-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 10.8 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 4.43 rad/s about its center of mass and all the objects below are undisturbed. If the object's center of mass is located 16.7 cm below the point where the ball hits, what is the moment of inertia of the object about its center of mass? What is the center of mass velocity of the tall object immediately after it is struck?Explanation / Answer
here initial angular momentum is conserved
L1 = L2
m*v1*r = m*v2*r + I*w
0.578*10.8*0.167 = 0.578*4.6*0.167 + I*4.43
==> I = (0.578*10.8*0.167 - 0.578*4.6*0.167)/4.43
==> I = 0.135 kg.m^2
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