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Suppose we have a spinning top, with angular velocity $\omega$, its speed is null and there is no force applied to it. Let $J$ be its moment of inertia.

We can say $E_k = \frac{1}{2}J\omega^2$ and $L = J\omega$.

Let's say I instantly divide its moment of inertia $J$ by two, so its moment of inertia is now $J' = \frac{J}{2}$.

According to the conservation of angular momentum, its new angular velocity $\omega'$ should verify $\omega' = 2\omega$.

According to the conservation of Kinetic Energy, if we solve $E_k = \frac{1}{2}J\omega^2 = \frac{1}{2}\frac{J}{2}\omega'^2$ for $\omega'$, we get $\omega' = \sqrt{2}\omega$.

Can anyone explain the correct answer?

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    $\begingroup$ Essentially a duplicate of Why does a ballerina speed up when she pulls in her arms? $\endgroup$ – John Rennie Apr 4 '19 at 9:43
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    $\begingroup$ That just tells you that both things can't be conserved. One of them is, and which one depends on the situation. $\endgroup$ – knzhou Apr 4 '19 at 10:18
  • $\begingroup$ I Kinetic Energy is not conserved, then where is the extra energy coming from? $\endgroup$ – Euler Pythagoras Apr 4 '19 at 10:44
  • $\begingroup$ I saw another post talking about this, the energy doesn't stay constant, it doubles, and the work to reduce moment of inertia by two is equal to the initial Kinetic Energy. $\endgroup$ – Euler Pythagoras Apr 4 '19 at 11:10
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    $\begingroup$ Possible duplicate of Why does a ballerina speed up when she pulls in her arms? $\endgroup$ – FGSUZ Apr 7 '19 at 19:12
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It's important to remember that these conservation laws only hold under specific circumstances.

Conservation of angular momentum only holds if the net external torque applied to the system is zero. If the net external torque is nonzero, angular momentum is not conserved.

Conservation of energy only holds if the net external work applied to the system is zero. If the net external work is nonzero, energy is not conserved (conserving energy and conserving kinetic energy are equivalent in this case, since there is nowhere else for energy to go).

The problem comes when you expect either or both of these equations to hold when you 'divide the moment of inertia by 2'. But what is it that you are actually doing? You can't just press a button that magically changes the moment of inertia without doing anything else - $J$ is a function of the mass and its distribution.

In order to accomplish this halving, you'll actually need to move some of the mass around - moving it closer to the center and/or taking some away entirely. The final $J$ and $\omega$ will depend on the exact procedure by which you manipulate the mass.

Most ways to achieve your halving of $J$ will conserve neither angular momentum nor kinetic energy, but you could potentially carefully arrange things in order to conserve one or the other. But it will never be possible to do things such that you conserve both - the reasoning in your question proves that that would give a contradiction!

EDIT - To clarify something discussed in the comments on the question: in the case you might be thinking of, where a dancer moves their arms inwards or outwards, it is $J$ that is conserved. It must be, because there is no external torque. But then $E$ is not conserved - it is higher when the arms are close to the body and lower when they are far from it. So where is this energy coming from/going?

The answer is work done by the dancer. Pulling the arms inwards requires extra effort on the dancer's part - this work is where the extra rotational energy comes from. Similarly, if they let their arms go completely limp, they will feel an (effective) centrifugal force pulling them outwards. The rotational energy lost by the dancer is work done by this (effective) force.

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  • $\begingroup$ Good answer thank you! Can we say the same thing about conservation of energy and conservation of momentum? Because I could have ask the same question if a car's mass was instantly halved then we would get the same absurdity $v' = 2v$ if we apply conservation of momentum and $v' = \sqrt{v}$ if we apply conservation of energy. $\endgroup$ – Euler Pythagoras Apr 8 '19 at 12:56
  • $\begingroup$ Precisely. 'Halving the mass of a car' is not really a thing in the real world. There is mass in the car, and all of it has to go somewhere. Which, if any, of the conservation laws applies depends on exactly what happens. $\endgroup$ – Drubbels Apr 8 '19 at 15:44

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