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This question already has an answer here:

Suppose a boy is standing on a platform which is free to rotate about an axis passing through its center. The Kinetic energy of the boy and the platform is K. If the boy stretches his hands so that the moment of inertia of the sytem(boy + platform) gets doubled. Then I have to find the new Kinetic energy of the system. As there is no torque so angular momentum remains conserved. Since M.O.I gets doubled so angular velocity gets halved. Therefore the new kinetic energy is:-

$ K'= \frac{1}{2}I'×(w')^2 $

$ K'= \frac{1}{2}×2I×(\frac{w}{2})^2 $

$ K'= \frac{1}{2}×\frac{1}{2}×I×w^2 $

$ K'= K/2 $

Thus the kinetic energy gets halved. So is the law of conservation of energy being violated here? Is the total energy not conserved?

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marked as duplicate by John Rennie, user191954, Kyle Kanos, Jon Custer, M. Enns Oct 28 '18 at 23:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The boy is doing work against his own spin by extending his arms outwards. Taken to the differential limit, we can see that the force extending his arms has a tangential component that decreases angular velocity. Work is therefore done against his own rotational kinetic energy.

A similar example is the commonly-asked textbook problem of a mass tied to a string, moving in circular motion on a frictionless table, around a hole through which the string runs. When the string is pulled inwards, work is done on the system as the act of pulling inwards adds to the tangential velocity, as user ja72 has excellently explained here.

So a radial outward force has a tangential component that decreases angular velocity (and kinetic energy), while a radial inward force has such a component that increases it. Energy is not conserved.

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