# 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?

#### First Step

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.

#### Equation

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, $\omega_f.$

#### Calculations

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

$\omega_f = 6.0\; rad\; s^{-1}$