A merry-go-round with, radius 3 m rotates freely with an angular speed of 2? rev

Question

A merry-go-round with, radius 3 m rotates freely with an angular speed of 2? rev/s. Its kinetic energy is 4000 J. A 50 kg person running tangential to the rim of the merry-go-round at 3 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (you may not assume the MGR is a disk) A. What is the Inertia of the MGR? B. What is the final angular speed of the merry-go-round and the person after the person jumps on? *SHOW WORK on how to get answers please*

Explanation / Answer

As we know

Rotational Kinetic Energy = 0.5*I2

Therefore

0.5*I*(2*2)2 = 4000

Therefore

I = 50.66 Kgm2

Apply Law of Conservation of Angular momentum

I + MvR = Ifinalfinal

50.66*12.566 + 50*3*3 = (50.66 + 50*3^2)*final

final = 2.17 rad/sec

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