Angular momentum
The idea
Just as impulse and momentum let you skip the messy details of forces during collisions, angular momentum L = Iω lets you skip the details of torques during spins. When the net external torque on a system is zero, L is conserved — no matter what the parts of the system do to each other internally. This single statement explains spinning skaters, gyroscopic stability, and why planets sweep equal areas in equal times.
The power of the law shows up when the moment of inertia changes mid-motion. Pulling mass toward the rotation axis lowers I, and since Iω must stay fixed, ω rises in exact proportion. For a point mass, L = mvr; for extended bodies, L = Iω with the appropriate I about the chosen axis. Always identify the system and the axis before claiming conservation — torque-free about one axis does not mean torque-free about another.
A tempting but wrong follow-up is to assume rotational kinetic energy is conserved too. It is not: a skater pulling in her arms does real work against the effective outward pull, and that work appears as extra kinetic energy. K = L²/(2I), so at fixed L a smaller I means strictly more kinetic energy, not the same.
Worked example
A playground merry-go-round is a uniform disk of mass 120 kg and radius 2.0 m, rotating freely at 1.5 rad/s. A 60 kg student stands at the rim, rotating with it. The student then walks to the exact center. Treating the student as a point mass and ignoring friction, find the new angular speed and compare kinetic energies.
- Compute the initial moment of inertia: the disk contributes MR²/2 = 0.5 × 120 × 2.0² = 240 kg·m², and the student at the rim adds mr² = 60 × 2.0² = 240 kg·m², so I_i = 480 kg·m².
- No external torque acts about the axle (the axle force has zero lever arm and friction is ignored), so angular momentum is conserved: L = I_i ω_i = 480 × 1.5 = 720 kg·m²/s.
- At the center the student sits on the axis, so r = 0 and the student contributes nothing: I_f = 240 kg·m².
- Apply conservation: ω_f = L/I_f = 720/240 = 3.0 rad/s — halving the moment of inertia doubles the angular speed.
- Compare energies: K_i = ½ × 480 × 1.5² = 540 J and K_f = ½ × 240 × 3.0² = 1080 J. The extra 540 J is the work the student does walking inward against the tendency to be flung outward — L is conserved, K is not.
Answer. The merry-go-round speeds up to 3.0 rad/s, and the kinetic energy doubles from 540 J to 1080 J because the student does work walking inward.
Check your understanding
- Why is angular momentum conserved here even though the angular speed clearly changes?
- Where does the extra kinetic energy come from when the student walks inward, and where would it go on the walk back out?
- How would the answer change if the student walked only halfway to the center — set up the new I before computing?
- What external influences in a real playground would make L only approximately conserved, and how would you estimate their effect?
Build the foundations first
Angular momentum builds on these concepts. If any feel shaky, start there.