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This question already has an answer here:

Consider a man rotating on a frictionless turntable , initially with his hands folded . When he opens his hands the moment of inertia increases and as the angular momentum is conserved the kinetic energy decreases. But energy is also conserved so where does this kinetic energy goes ?

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marked as duplicate by BowlOfRed, Buzz, John Rennie newtonian-mechanics Feb 5 at 8:39

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  • $\begingroup$ Labeling the first moment of inertia and angular speed 1 and the second 2, we get:angular momentum:$$L=I_{1}\omega_{1}=I_{2}\omega_{2}$$ Kinetic energy: $$E_{K}=\frac{1}{2}I_{1}\omega_{1}^{2}=\frac{1}{2}I_{2}\omega_{2}^{2}$$ $\endgroup$ – R. Rankin Feb 4 at 20:21
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Your statement, that the angular momentum is conserved, is correct, because there is no external torque acting on the system.

However, the statement, that the energy of rotation is conserved, is not correct (as you realized yourself). The rotating man feels the centrifugal force acting on him. Moving his hands radially outwards, work is actually done on him (force integrated over the distance of motion). Therefore the energy of rotation decreases by this amount of work. This energy difference is dissipated in the muscles of the man. In this way, total energy conservation is obeyed, as we expect. The missing rotational energy transforms into ‘heat’ in the man’s body.

Conversely, if the man pulls his hands in again, he has to do work (the same as was done on him before) moving his hands against the centrifugal force pulling outwards. This work appears as an increased energy of rotation again.

In a more idealized setting you consider a single mass $m$ initially rotating at radius $r_1$ around the rotation axis. The angular momentum is $$ L_1 = mr_1^2\omega_1 $$ and the kinetic energy of rotation is $$ E_1 = \frac{1}{2}mr_1^2\omega_1^2 .$$ Then we increase the radius of rotation to $r_2>r_1$. Angular momentum conservation requires $$ L_1 = mr_1^2\omega_1 = mr_2^2\omega_2 = L_2 .$$ This gives the new angular velocity $$ \omega_2 = \frac{r_1^2}{r_2^2}\omega_1 $$ and the new kinetic energy of rotation $$ E_2 = \frac{1}{2}mr_2^2\omega_2^2 = \frac{1}{2}mr_1^2\omega_1^2\frac{r_1^2}{r_2^2} = E_1 \frac{r_1^2}{r_2^2} < E_1.$$

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The kinetic energy does not decrease. It is only the angular velocity that decreases. The kinetic energy remains the same.

An analogy is the following: suppose you have a ball of mass $m$ moving at velocity $v$. Now suppose if the ball gains mass- becomes of mass $M$ where $M>m$. Then its velocity will decrease. But that doesn't mean the kinetic energy decreases; it will remain the same.

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  • $\begingroup$ $E_k$ indeed can remain constant, but why does it? I think you should add that to compelte your answer. $\endgroup$ – FGSUZ Feb 4 at 22:44

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