# Help with Conservation of Angular Momentum Question

An ice skater executes a spin about a vertical axis with her feet on a frictionless ice surface. In each hand she holds a small 5kg mass of which are both 1m from the rotation axis and the angular velocity of the skater is 10rad/s. The skater then moves her arms so that both masses are 0.5m from the rotation axis. The skaters own moment of intertia can be taken as being 50kgm^2, independent of her arm position

a)Find the total angular momentum of the skater and the masses both before and after the arm movement. Explain any difference

b) Find the total kinetic energy of the skater and the masses both before and after the arm movement. Explain any difference.

My attempt at part a) was that quite simply plug in the numbers into the equation L=Iw and gather the summation of the 3 objects however I assumed the arms of the skater were two rods with masses at the end and with axis of rotation at the end therefore meaning I use I = 1/3MR^2 however that is not the case, the answer simply uses I = MR^2 which confuses me.

My attempt at part b) was that K=1/2*I*w^2 but I am unable to generate a term of the kinetic energy before and after.

Any help on this would be greatly appreciated. Also any specific topics I could read up on to understand these concepts would be much appreciated.

The angular momentum of the two masses is computed independent of the skater - you were given the total angular momentum of the skater (including arms and hands which are normally considered part of the person) and ONLY have to compute the moment of inertia / angular momentum of the masses. A point mass at the end of a string has $$I=mr^2$$ as you know. The arms of the skater were already accounted for, and the mass of the weights is not distributed along the arms, it is all at the end.

Angular momentum is $I\omega$. You should now be able to compute it from $I_{total}=I_{skater}+I_{masses}$, and $\omega$ is given. It will, of course, not change when the skater pulls in her arms - conservation of angular momentum, and there is no external torque on the skater-plus-masses system.

The moment of inertia of the masses does change when the skater pulls in her arms - you can compute it for the masses, but not for the arms (which are also coming closer). That is a problem with the question - you must assume a massless arm if you want to compute the moment of inertia when the arms are pulled in.

And you need the moment of inertia for the last part, since you can write the angular kinetic energy as

$$KE = \frac12 I \omega^2$$

So it is not enough to know $L$, you actually need to be able to compute the new angular velocity. And for that you must make a simplifying assumption (massless arms).

On that assumption, you can compute the increased kinetic energy from the above (because you know the new angular velocity from the new moment of inertia).

Before:

$$\omega = 10 rad/s\\ I_{skater}=50 kg m^2\\ I_{weights} = 10 kg m^2\\ KE = \frac12 I_{total} \omega^2$$

and you should be able to figure the rest from here...

In the first case it is taken that the masses of the hands are small compare to the masses that holds the hands. Hence we can neglect the masses of two hands and the case becomes two masses of weight 5 kg moving about the rotational axes. That's why we calculate the moment of inertia by

In second case we have to add the moment of inertia skater and the two masses to find the kinetic energy of the total body. And then we can find the kinetic energy by K=1/2*I*w^2

I'll solve it later.