As part of a carnival game, a 0.528-kg bail is thrown at a slack of 20.8-cm tall
ID: 1477298 • Letter: A
Question
As part of a carnival game, a 0.528-kg bail is thrown at a slack of 20.8-cm tall, 0.403-kg objects and hits with a perfectly horizontal velocity of 10.6 m/s. Suppose the bail strikes the topmost object as shown to the right. Immediately after the collision, the bail has a horizontal velocity of 4.10 m/s in the same direction, the topmost object now has an angular velocity of 3.63 rad/s about its center of mass and ail the objects below are undisturbed. If the objects center of mass is located 14.6 cm below the point where the bail 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
Mb=0.528kg, Mo=0.403, v1=10.6m/s, v2=4.1m/s, w=3.63rad/s, xcm= 14.6*10^-2m
M.I of object =m*xcm^2 =0.403*(14.6*10^-2)^2=0.008590348kg.m^2
b)energy conservation: 1/2*Mb*V1^2 =1/2*I*w^2+1/2*Mo*Vcm^2+1/2*Mb*V2^2
=>0.528*10.6^2 = 0.008590348*3.63^2+0.403*Vcm^2+0.528*4.1^2
=>50.33=0.403*Vcm^2 =>Vcm =11.17m/s
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