# Angular Momentum conserved but not energy?

1. Apr 11, 2011

### say_cheese

New to the forum- looked through the forum, but all are tangential answers.

A skater with mass m, is spinning with arms extended to radius r1, with a tangential speed v1, angular velocity w1; v1=w1*r1, angular momentum mv1r1 =mw1r12; kinetic energy 1/2 m1v12.

Now she pulls in her arms to a radius r2, r2<r1. Angular momentum is conserved. She changes her tangential speed to v2 and angular velocity to w2.
so mv1r1=mv2r2 or v2 = (v1*r1)/r2

But the new kinetic energy
1/2mv22= 1/2mv12*(r1/r2)2

Since r1>r2, kinetic energy has increased by the square of (r1/r2). Where does this extra energy come from? (there should actually be an energy loss in the intermediate stage, where the skater is spiralling in to the new radius). Does the skater have to exert herself?

2. Apr 11, 2011

### Staff: Mentor

Yes. Something similar is often done as a demonstration, with someone sitting on a freely rotating stool while holding weights in his outstretched hands. As he brings the weights in closer to his body, his angular velocity increases, and he has to make an effort to "pull the weights in", that is, he has to do work.

3. Apr 11, 2011

### Staff: Mentor

The skater must do work to bring her arms in. The extra energy comes from her converting chemical energy in her muscles into mechanical energy.
Exactly!

4. Apr 11, 2011

### Staff: Mentor

5. Apr 11, 2011

### say_cheese

Thanks guys/girls!

6. Apr 11, 2011

### say_cheese

Thanks Guys/Girls!

7. Apr 11, 2011

### sophiecentaur

There's always a 'Force timed Distance' somewhere to explain this sort of thing; it's a matter of spotting it.
When you get down to it, it's that or The Twilight Zone.