4. A boy of mass rusu-40.0kg stands on the rim of a frictionless merry-go round

Question

4. A boy of mass rusu-40.0kg stands on the rim of a frictionless merry-go round ("MGR" for short) of radius RMGR=1.50m and moment of inertia Inte=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, Jo mr, 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=(I_MGR + I_boy)

so,

= (I_MGR + I_boy)*w

put values

= (135 + 40*1.5^2)*0.3

= 67.5 kg.m^2

b) Apply conservation of momentum

final angular momentum = initial angular mometum

I2*w2 = I1*w1

(I_MGR + I_boy)*w2 = 67.5

Put values

(135 + 0)*w2 = 67.5

w2 = 67.5/135

= 0.5 rad/s

I hope help you.

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