Conservation of angular momentum

What just happened?

The skater pulled her arms in — the mass moved closer to the rotation axis, so the moment of inertia dropped. Because nothing braked her motion (no external torque), the angular momentum stayed the same, so the angular speed increased. The pirouette speeds up "on its own".

Conservation of angular momentum:

— when no net external torque acts on the body, its angular momentum $L$ stays constant. I — moment of inertia (kg·m²), ω — angular speed (rad/s).

Why does the skater speed up?

The moment of inertia depends on how far the mass is spread from the axis: $I = m r^2$. Pulling the arms in reduces $r$, so $I$ drops. For the product $I\omega$ to stay constant, ω must increase:

Where else does this apply?

A dancer's and a jumper's pirouette; planets speed up at perihelion as they approach the Sun; a collapsing star shrinks to the size of a city and becomes a neutron star spinning hundreds of times per second — everywhere $L$ is conserved.

Why you need this for the exam:

$L = I\omega = \text{const}$ solves problems about pirouettes, carousels and planets. Trick: if there is no external torque, set $I_1\omega_1 = I_2\omega_2$ equal and compute the unknown speed.

Remember: with no external torque the angular momentum is constant — smaller I means larger ω. That is why the skater spins faster with her arms pulled in: $I\downarrow \Rightarrow \omega\uparrow$.