An angular impulse $\tau\Delta t$ changes angular momentum. So when no outside torque acts, angular momentum cannot change at all, which explains spinning skaters, divers, and why planets speed up near the Sun.
Key idea
If the net outside torque on a system is zero, its angular momentum stays the same: $$I_1\omega_1 = I_2\omega_2.$$ Pull mass toward the axis (smaller $I$) and the spin speeds up; push it outward and the spin slows.
Why it works
Inside a system, every force has a third-law partner, equal, opposite and along the same line, so their torques cancel. Only outside torques can change the total $L$. A skater on smooth ice feels almost no torque about her vertical axis, so her $L$ is fixed. Her kinetic energy is not: writing $K = \tfrac{1}{2}I\omega^2 = \dfrac{L^2}{2I}$, the same $L$ with a smaller $I$ means a bigger $K$. The extra energy comes from the work her arms do pulling the weights inward, against their tendency to keep moving in a straight line.
Worked examples
Worked example: A skater pulls in her arms
A skater spins at $1.5\ \text{rev/s}$ with $I = 4.5\ \text{kg·m}^2$. She pulls in her arms, reducing $I$ to $1.5\ \text{kg·m}^2$. Find her new spin rate and the change in her kinetic energy.
- $\omega_2 = \dfrac{I_1}{I_2}\omega_1 = \dfrac{4.5}{1.5}(1.5) = 4.5\ \text{rev/s}$ (any unit works for this ratio).
- For energy use rad/s: $\omega_1 = 9.42$ and $\omega_2 = 28.3\ \text{rad/s}$.
- $K_1 = \tfrac{1}{2}(4.5)(9.42)^2 = 200\ \text{J}$; $K_2 = \tfrac{1}{2}(1.5)(28.3)^2 = 600\ \text{J}$. Three times as much: her arms did $400\ \text{J}$ of work.
Answer: $4.5\ \text{rev/s}$; $K$ triples
Worked example: A disk dropped on a turntable
A turntable ($I = 0.020\ \text{kg·m}^2$) spins freely at $45$ rpm. A disk with $I = 0.010\ \text{kg·m}^2$, not spinning, is dropped on it, centered. Find the final rate.
- Friction between them is internal: $I_1\omega_1 = (I_1 + I_2)\omega_f$.
- $\omega_f = \dfrac{0.020}{0.030}(45) = 30$ rpm.
- Like a sticking collision: angular momentum is kept, but kinetic energy drops (to two thirds here).
Answer: $30$ rpm
Worked example: Jumping on a merry-go-round
A $30\ \text{kg}$ child runs at $3.0\ \text{m/s}$ along a line tangent to the edge of a resting merry-go-round (a uniform disk, $120\ \text{kg}$, radius $2.0\ \text{m}$) and jumps on at the rim. Find $\omega$ afterward.
- Before: $$L = mvr_\perp = (30)(3.0)(2.0) = 180\ \text{kg·m}^2\text{/s}.$$
- After: $$I = \tfrac{1}{2}(120)(2.0)^2 + (30)(2.0)^2 = 240 + 120 = 360\ \text{kg·m}^2.$$
- $\omega = \dfrac{180}{360} = 0.50\ \text{rad/s}$.
Answer: $0.50\ \text{rad/s}$
Common mistake
Thinking the skater's kinetic energy stays the same when she pulls her arms in. Angular momentum is what stays the same; kinetic energy grows, because $K = L^2/2I$ and $I$ shrinks. Conservation of one quantity doesn't mean conservation of the other, just as momentum but not kinetic energy survives a sticking collision. Check: pulling arms in takes effort, so energy must be going in.