MCQ
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 centre of the round $($as seen from the round$)$. The speed of the round afterwards is:
  • A
    $2\omega$
  • $\omega$
  • C
    $\frac{\omega}{2}$
  • D
    $0$

Answer

Correct option: B.
$\omega$
As no torque is exerted by the person jumping, radially away from the centre of the round $($as seen from the round$)$, let the total moment of inertia of the system is $2I ($round $+$ Person because the total mass is $2M)$ and the round is revolving with angular speed $\omega$ Since the angular momentum of the person when it jumps off the round is $\text{I}\omega$ the actual momentum of round seen from ground is $2\text{I}\omega-\text{I}\omega=\text{I}\omega$ So we conclude that the angular speed remains same, i.e., $\omega$

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