A merry-go-round, made of a ring-like platform of radius \(R\) and mass \(M,\) is revolving with the angular speed . A person of mass \(M\) is standing on its edge. At one instant, the person jumps off the round such that the final speed of the person is zero. What is the final angular velocity of the merry-go-round?
1. \(2\omega\)
2. \(\omega\)
3. \(\omega/2\)
4. \(0\)
Step 1: Find the angular momentum of the system.
As no external torque acts on the system. angular momentum should be conserved.
Hence I = constant
where I is the moment of inertia of the system and is the angular velocity of the system.
From Eq (i) \(I_{1}\omega _{1}=I_{2}\omega _{2}\)
(where and are angular velocities before and after jumping)
\(\Rightarrow I\omega =\frac{I}{2}\times \omega _{2}\)
(as mass reduced to half, hence, the moment of inertia also to half)
\(\omega_{2} =2\omega \)
Hence, option (1) is the correct answer.
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