Rotational kinetic energy when moment of inertia is changed

Question

The following quote comes from: https://physics.stackexchange.com/a/93531/155365

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...

However, it only covers a case when work is done to pull arms inwards. It is understandable that the work would be done in such case.

But what is happening in the opposite case when the skater has their arms inwards and lets them go outwards? S/he doesn't do any work but kinetic energy (and total energy) is lost. Why is it lost?

E.g.

I = 2kg * (1m)^2
w = 3.14/s
L = 2kg * m^2 * 3.14/s
Ek = 9.869

After shifting 2kg by 0.1m outwards:
I = 2kg * (1.1m)^2 = 2.42 kg * m^2
L = 2 * 3.14 = 2.42 * w
w = 2.5963
Ek = 8.1566

So, where did 1.7124 J go?

The quoted answer describes the easy case and omits the hard one.

Answer 1

If you consider the kinetic energy of the bracelents separate from the skater you can visualize what is going on.

When near the axis of rotation they are moving fast and when the skater extends their arms some of the (rotational) kinetic energy of the skater is converted to linear kinetic energy for the bracelets.

If the skater has MMOI of $I$ and the bracelets have mass $m$ at a distance $c$, (constant) angular momentum is $L = (I + m c^2) \omega$.

The total kinetic energy is

$$ KE = \frac{1}{2} I \omega^2 + 1/2 m v^2 $$ with $v=\omega c$ the linear velocity of the bracelets.

As a result $$KE = \frac{1}{2} I \omega^2 + 1/2 m (\omega c)^2 = \frac{1}{2} ( I + m c^2) \omega^2 = \frac{1}{2} L \omega $$

Total energy is conseved when you consider $$\Delta \left( \frac{1}{2} L \omega \right) = {\rm Work}$$

It takes work to make it spin faster, and it releases work to slow down. It is converted as heat in the muscles.

Answer 2

It's the same reason that work is done to lift a brick into the air, but energy is lost when dropping it on the ground. There is no mechanism in place to capture the energy of the fall.

In the case of the skater, the mass that is flung outward (the bracelets) are accelerated during the spin (just like the falling brick). If there were an energy capture system, you could hold on to that energy.

But when the brick hits the ground and the bracelets reach the end of the skater's arms, there is an inelastic collision and the energy is lost.

Answer 3

The energy is lost (dissipated) because the ice skater is unable to use/store it. The skater can hold masses on springs and keep arms in the same position. Then change in the rotational kinetic energy is translated into potential energy stored in the springs and then back.