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Usually, the spinning chair problem is one of the typical angular momentum problems. And most of them start with a person holding two dumbbells on his two hands and bringing his arms in, then by conservation of angular momentum, the angular speed increases when he brings his arms in, and since the person is exerting force for a certain displacement to bring the dumbbells in, the kinetic energy increases since positive work is done.

But now, if I start the other way round, where the person first has his arms brought in, and after that spread out his arms. Then by conservation of angular momentum, the angular speed decreases since the moment of inertia increases. In this case, it seems that kinetic energy is lost, but I think the person needs to exert force to bring the things out? Then is there any work done by the arms? And also, is there any loss of energy? Or the work done is stored in some other forms?

Thank you for any replies.

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When the person moves out his arms , the work is done by the centrifugal force and no work is done by the arms. When we pull the arms inside our arms do work against centrifugal force.

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Because the person has to exert force to hold the dumbbells against gravity, it would be better to consider this situation a gravity free space. When the person tries to increase the radius he doesn't apply force the outwards rather he decreases the force that he applies inwards ,as long as there is circular motion the person has to supply necessary centripetal force.

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In the non-rotating reference frame, the only force acting on the weights is from the arms (and gravity, but that's not relevant since nothing is moving vertically).

Work done is the dot product of force and displacement (or more usefully, the rate at which work is done is the dot product of force with velocity). There are three scenarios:

  • When holding the weights at a fixed distance, the velocity is perpendicular to the force and so does no work.

  • When pulling the weights in, the velocity has a component in the same direction as the force, which does positive work, increasing the kinetic energy of the weights

  • when letting the weights out, the velocity has a component in the opposite direction to the force from the arms, which therefore does negative work, decreasing the kinetic energy of the weights.

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  • $\begingroup$ OH. I see. Thanks $\endgroup$ – Ho F Nov 19 at 4:12

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