I Angular momentum is conserved, but KE is not. How to find out why?

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1. Aug 21, 2016

FG_313

This (photo) is a very typical example of conservation of angular momentum, but my trouble arrises from trying to prove that the difference of energy will have to correspond to work, by calculating the work done by you to alter the moment of inertia. I have spent a lot of time in this, but I still can't understand it satisfactorily. I'll try to explain my reasoning (for the few parts that I managed to do) and post my points of confusion as they arrise, and if there is anything you want to correct or add, it would be certainly very helpfull.

No external torques => L is constant => Iω is constant. Therefore, the angular velocity goes up when we close our arms.
Now, looking at the rotational KE, its clear that is not conserved. That is not itself a problem: you do work in this situation. I woud like to look at it in a more quantitativaly way. Lets approximate the situation in the first place for this:
Two masses M, each with a distance r to the axis (the situation is always symmetrical!). In this case, it can be checked (done it!) that the difference in KE when the masses are at r1 and after at r2 is equal to the work done by "fighting" the centrifugal force on the both ways (by the way, I'm not certain why we don't need to take into account the coriollis force, it is normal to the displacement in the radial direction, so the first thing I thought is that the work has to be zero. However, if I were to apply the "formula" W=∫τdφ when τ is the torque done by fighting the corriollis force, it would lead to apparent contradiction). But this is oversimplificating too much the situation, I'm not satisfied. The following treatment is where I'm having more problems:
Your torso, head, legs and all that will be treated as a cilinder, or any other rigid body with moment of inertia Io, the weights are connected to you by a massless rod and the weights are pontual masses with mass M (each one of them). Let r be the distance between the axis and one of the weights, the total moment of inertia would then be: I=Io+2Mr^2. I want to calculate the work done by bringing the masses from distance r1 to r2, taking the assumptions described, and I'm having trouble with that. Here are some other thoughts on the matter:

If we take into account only the rigid body (Io), he goes from Io*ω1 to Io*ω2, so there has to be a external torque to this system (only the rigid body- No weights!). That would probably be done by the force done by the weights (M), that are a reaction (Newton 3rd law) to the force I have to make to beat the corriolis force. So I guess this torque has to be taken in account when calculating the work done! Than, We cannot ignore the corriolis force when calculating work, like we did in the first case. Furthermore, the thing that makes the first case so easy so easy to work with, is that the centrifugal force has a simple relation to the angular mommentum, which is constant, so the integral of work gives the correct answer. Thank you for the attention!

2. Aug 21, 2016

jbriggs444

The Coriolis force cannot be responsible for applying a torque to the rigid cylinder. The cylinder is not moving radially. The force that is responsible for the torque is the real interaction force between the cylinder and the [rods restraining the] moving masses. Those masses are subject to the Coriolis force. But that Coriolis force is at right angles to the movement of the masses. As you first thought, quite correctly, it does no work.

The real interaction force between the movable masses and the rigid cylinder does do work. It does accelerate the cylinder. But... Newton's third law applies. The partner force does equal negative work on the movable masses, slowing them down. It is a wash -- equal and opposite torques, conserving total angular momentum.

You might gain some insight by considering the trajectories of the movable masses. They are not moving in circles. They are moving in an inward spiral path and are subject to an inward force. That inward force is not at right angles to their path. It tends to speed them up.

3. Aug 22, 2016

A.T.

Start by defining a reference frame, and then stick to that frame. In the inertial frame where angular momentum is conserved, there are no inertial centrifugal and Coriolis forces.

4. Aug 22, 2016

FG_313

I may be wrong, but the work that you do independs on the reference frame picked, and the non-inertial reference frame is much easier in that case. To the response above, I meant that the force that you need to balance out the coriollis effect on the weights would cause a reaction from the weight to your arm, and that would cause the torque, I think.

5. Aug 23, 2016

A.T.

Work is frame dependent.

Which non-inertial frame? You haven't defined one. Note that if your rotating frame has variable ω, you need to consider the Euler force, additionally to centrifugal and Coriolis.

6. Aug 24, 2016

FG_313

Well, I'm clearly not up to this problem yet! How would you find the work?

7. Aug 25, 2016