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A child of mass m is standing, on the pe...

A child of mass `m` is standing, on the periphery of a circular platform of radius `R`, which can rotate about its central axis. The moment of inertia of the platform is `I`. Child jumps off from the platform with a velocity `u` tangentially relative to the platform. Find the angular speed of the platform after the child jumps off.

Text Solution

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Initially the system was at rest, thus the initial angular momentum was zero. As child jumps off from the platform, it gains an angular momentum in the opposite direction. This implies that the platform must also gain the same amount of angular momentum in opposite direction.
When the child jumps off, the platform gains an angular velocity `omega`. Then the net velocity of child with respect to earth is `u - Romega`. As no external torque is present, the net angular momentum of the system must be finally zero, thus
`m(u-Romega)-Iomega`
`omega=(muR)/(mR^(2)+I)`
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