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  • A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed \omega . A person of mass M is standing on it.

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed \omega . A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the center of the round (as seen from the round). The speed of the round afterwards is

(a)2\; \omega

(b)\: \omega

(c)\; \frac{\omega }{2}

(d)0

 

Answers (1)

The correct answer is the option (a)\; 2\omega

By the law of conservation of angular momentum,

I_{1}\omega _{1}=I_{2}\omega _{2}

So before the person jumps off the merry-go-round, angular momentum is

(M+M)R^{2}\omega

After the person jumps off it the angular momentum is MR^{2}\omega _{2} where \omega _{2}  is unknown

Equating them together
                                                                 2\; \omega =\omega _{2} 

Hence (a) is the correct option

Posted by

infoexpert23

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