# What is happening to rotational kinetic energy when moment of inertia is changed?

I know this question is asked here a lot, but I just had to ask this to finalise the concept.

When a system lets say a rod of length $L$ and mass $M$ is rotating with angular speed $omega_1$ its initial angular momentum is $L1 = (1/12)ML^2\omega_1$ and its initial kinetic energy is $KE = (1/24)ML^2{\omega_1}^2$.

Now after some time the rod is folded in half its angular momentum kept conserved i.e. without applying any external force or torque, its new angular velocity becomes $\omega_2 = 4\omega_1$ and its new kinetic energy becomes $KE_2 = (1/6)ML^2{\omega_1}^2$.

This is 4 times the original kinetic energy when no external force works, since the rod is folded, you can even say melted and formed into a smaller and denser rod, it has not undergone compression/expansion of any sort but still there is change in kinetic energy.

The most sense I could make out of this was that all the particles while rotating felt a centripetal force and the particles of half of the rod under this force went in its direction and did some work which appears as the change in kinetic energy. I have written the same and a proof as an answer here.

Now if I am write in my concept where did this energy come from, tension was providing the centripetal force but no work was done against tension as the rod was folded in half not compressed. If I am wrong then where did the energy come from?

Extra: I also tried the analogy of this question in translatory motion, suppose there is body of mass $m$ moving with velocity $v$ suddenly, its mass becomes $m/2$ then its velocity becomes $2v$ and its KE becomes $4KE_1$ there is no need to explain the energy conservation here since mass suddenly does not disappear into thin air, however moment of inertia can be changed and hence the question.

Addendum : Since folding the rod seems to bring about unnecessary questions about ways of folding, you can imagine that if rod was melted and formed into a longer rod all the while the system was rotating and angular momentum conserved, then new length becomes $2L$ new angular velocity becomes ${\omega_1}/4$ and new KE becomes $(1/96)ML^2{\omega_1}^2$. This time the energy becomes ${1/4}^{th}$ of the initial, where did this energy go to ? Certainly movement against centripetal force takes place, but since there is no extension in existing rod, energy can not be stored as spring energy in it, or so I think.

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Well you have done some work folding the rod, as there is an outward reaction force (to balance centripetal acceleration) on any part of the rod, and to fold it you must reduce it's radius, thus do work against this force. Tryi it with a rod of length 4l, with joints at l and 3l (so that the centre of mass doesn't move along rod) and you should hopefully see that this work done by you on the rod balances the excess kinetic energy. – Zephyr Jan 13 '14 at 19:04
How was work done against this forcd which should be along the rod at all times when the rod itself was folded, remeber it is neither being stretched nor compressed, no displacement along the force. – Rijul Gupta Jan 13 '14 at 19:09
the rod will not fold of it's own accord. the easiest way to think of it folding would be little motors at the joints, and it's these motors that would be doing work. (we can also think of us folding it by hand, but then it's harder to picture not disturbing the system) – Zephyr Jan 13 '14 at 19:11
as for no displacement in the direction of the force. imagine putting a dot on each end of the rod, the final result of folding (or melting or whatever) is that the two dots will now be half the distance, therefore one or both dots must have moved in the radial direction. – Zephyr Jan 13 '14 at 19:12

In honor of the Winter Olympics:

Consider an ice skater that has just gone into a spin with arms stretched out. If it helps, the skater is wearing lead bracelets on each wrist. As the skater spins, the angular momentum and angular kinetic energy are constant; friction with the air and with the ice can be eliminated. The skater must exert an inward centripetal force on the bracelets to keep them rotating in the same circle, but this force does no work, since the bracelets are not changing their radius of rotation.

Then the skater pulls in his arms and the bracelets, closer to his axis of rotation.

Several interrelated things happen:

1. since there is no external torque, the angular momentum of the system stays the same;
2. Since the various masses of the skater are all moving towards the axis of rotation, the total moment of inertia of the system decreases;
3. Combining 1 and 2, the angular velocity of the system increases; if the moment is halved, the angular velocity doubles
4. Combining 2 and 3, with some quantitative treatment, the angular kinetic energy increases; if the moment is halved and the angular velocity is doubled, the kinetic energy doubles;
5. The skater does work; his arms are now exerting an inward force through a distance as his arms draw the bracelets inward; the work done can be rigorously shown to be identical to the increase in kinetic energy...
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If initially arms were near and later the skater moves then away, then how will the energy be stored ? Muscular ? – Rijul Gupta Jan 13 '14 at 21:01