Biophysics Problem 45

A figure skater is rotating at \(0.64\) revolution per second.

(a) What is her angular velocity in randians per second?
(b) If she now pulls her arms in so that her moment of inertia is reduced by 1/3, what is her new angular velocity?

Part (a) is a simple calculation.  Using \(\omega = 2\; \pi\; f,\) you can calculate the angular velocity.
You should have gotten \(\omega = 4.0 \;rad \;s^{-1}.\)
Now on to part (b).
You should realize that you must use Conservation of Angular Momentum to solve this.  Write out the equation you will use.

The equation you should have is:
\(I_i \;\omega_i = I_f \;\omega_f\)
where subscript '\(\mathrm{i}\)' is used for initial values, and subscript \('f' \) is used for final values.
If her moment of inertia is reduced by 1/3, then her final moment of inertia must be 2/3 of her initial moment of inertia.  (This is the part where most students go astray.  Make sure that this makes sense to you.)
\(I_f = \frac{2}{3}\; I_i\)
A simple substitution should let you calculate the final angular velocity, \(\omega_f.\)

The equation for conservation of angular momentum becomes
\(I_i (4.0 \;rad\; s^{-1}) = \biggl( \frac{2}{3} I_i \biggr) \omega_f\)
From which, you should be able to get 
\(\omega_f = 6.0\; rad\; s^{-1}\)