Question

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg·m2
What is:
a) The final angular speed?
b) The initial and final kinetic energy?

Answer 1

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular speed

Since the angular momentum must be conserved, we can write

$L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2$

where we have

$I_1 = 2.25 kg m^2$ is the initial moment of inertia

$\omega_1 = 5.00 rad/s$ is the initial angular speed

$I_2 = 2.25 kg m^2$ is the final moment of inertia

$\omega_2$ is the final angular speed

Solving for $\omega_2$, we find

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s$

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

$K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J$

- Final kinetic energy:

$K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J$