. A boy of mass mboy=40.0kg stands on the rim of a frictionless merry-go round (
ID: 1460284 • Letter: #
Question
. A boy of mass mboy=40.0kg stands on the rim of a frictionless merry-go round (“MGR” for short) of radius RMGR=1.50m and moment of inertia IMGR=135kgm2 about the central axel. The system (boy and merry-go-round) initially rotates at the rate of 0.300rad/s about a central vertical axel of the merry-go-round. Here, the boy can be treated as a point mass, Iboy=mr2, where r is his distance from the axel. (20pts) a. What is the angular momentum of the system? b. If the boy walks to the center of the merry-go-round, what is the final angular velocity of the system?
Explanation / Answer
a) angular momentum of the system = I*w
= (I_MGR + I_boy)*w
= (135 + 40*1.5^2)*0.3
= 67.5 kg.m^2 <<<<<-----------Answer
b) Apply conservation of momentum
final angular momentum = initial angular mometum
I2*w2 = I1*w1
(I_MGR + I_boy)*w2 = 67.5
(135 + 0)*w2 = 67.5
w2 = 67.5/135
= 0.5 rad/s <<<<<-----------Answer
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