The big idea
Angular momentum, L = Iω, is the rotational version of linear momentum. When no external torque acts on a system, L stays constant. Change the rotational inertia and the spin rate has to change to compensate.
Key ideas
- L = Iω for a rigid body. For a point mass, L = rmv sin θ.
- No external torque → L is conserved: I₁ω₁ = I₂ω₂.
- Pulling mass inward lowers I, so ω rises.
- Rotational kinetic energy is not conserved when I changes. When a skater pulls in, K goes up, and the extra energy comes from her muscles doing work.
- Angular impulse: τΔt = ΔL.
Equations
| L = Iω | rigid body |
| L = rmv sin θ | point particle |
| I₁ω₁ = I₂ω₂ | conservation (no external torque) |
| τΔt = ΔL | angular impulse |
| K = ½Iω² = L²/(2I) | rotational KE |
Worked example
A skater with I = 3.0 kg·m² spins at 2.0 rev/s. She pulls in her arms and I drops to 1.2 kg·m². Find her new spin rate and how her kinetic energy changes.
- No external torque, so L is conserved: (3.0)(2.0) = (1.2)ω₂.
- ω₂ = 6.0/1.2 = 5.0 rev/s.
- K = L²/(2I). L is the same, and I is 2.5 times smaller, so K is 2.5 times larger.
- The extra kinetic energy comes from the work her muscles do pulling her arms in.
Common mistakes
- Assuming kinetic energy is conserved along with L.
- Mixing rev/s and rad/s inside one equation. They're fine as long as both sides use the same unit.
- Forgetting that a point mass moving in a straight line can still have angular momentum about a point.
Try it yourself
A disk (I = 0.25 kg·m²) spins at 10 rad/s. A 0.50 kg lump of clay sticks to its rim at r = 0.50 m. Find the new ω.
Show the answer
I₂ = 0.25 + (0.50)(0.50)² = 0.375 kg·m²; ω₂ = (0.25)(10)/0.375 ≈ 6.7 rad/s.
Going deeper: AP Physics 1
Exam-level reasoning in algebra: answers in symbols, "what happens if it doubles?" questions, multi-step problems and the conceptual traps the AP Physics 1 exam likes to set.
The rotational dictionary
Every linear idea has a rotational twin: mass m ↔ rotational inertia I, velocity v ↔ ω, force F ↔ torque τ, momentum p = mv ↔ L = Iω, and F = ma ↔ τ = Iα. If you know the linear version, you know the rotational one.
Kinetic energy is NOT conserved when a skater pulls in
L = Iω is conserved, so as I falls, ω rises in proportion. But K = ½Iω² = L²/(2I) increases: the skater's muscles do positive work pulling the arms in. Exam questions love to check whether you know which quantity is conserved.
More worked examples
Example 1. A skater spinning at 2.0 rad/s with I = 4.0 kg·m² pulls her arms in to I = 1.6 kg·m². Find her new angular speed and compare her kinetic energies.
- L conserved: (4.0)(2.0) = 1.6ω → ω = 5.0 rad/s.
- K before = ½(4.0)(2.0)² = 8.0 J. K after = ½(1.6)(5.0)² = 20 J.
- The extra 12 J comes from the work her arms do.
Example 2. A constant 12 N·m torque acts on a wheel (I = 3.0 kg·m²) at rest for 4.0 s. Find its final angular momentum and angular speed.
- Angular impulse: ΔL = τΔt = 12 × 4.0 = 48 kg·m²/s.
- ω = L/I = 48/3.0 = 16 rad/s.
Exam-style practice
1. A lump of clay is dropped onto a spinning turntable and sticks. What happens to ω, and is kinetic energy conserved?
Show the solution
I increases, so ω decreases. KE is not conserved: this is a rotational sticky collision.
2. A hoop and a disk have equal mass and radius. Which is harder to start spinning?
Show the solution
The hoop: all its mass is at the rim, so it has the larger I.
3. A spinning star collapses to half its radius (mass unchanged). By what factor does its spin rate change?
Show the solution
I ∝ R², so I drops by 4 and ω becomes 4×.
Going deeper: AP Physics C
This part goes beyond the algebra course: calculus derivations, harder multi-step problems and exam-style practice, written at the level of the AP Physics C exam.
Angular momentum of a particle
L = r × p, with magnitude mvr⊥, where r⊥ is the perpendicular distance from the pivot to the line of motion. A particle moving in a straight line still has a constant angular momentum about any point not on that line. This is what makes "putty hits a rod" problems work.
Newton's 2nd law for rotation
Στext = dL/dt. With no external torque, L is conserved, even in a collision where kinetic energy is lost. In a linear-to-rotational collision, conserve L about the pivot, because the pivot force exerts no torque about that point.
More worked examples
Example 1. A uniform rod (M = 0.90 kg, L = 1.2 m) hangs from a pivot at its top end. A 0.10 kg ball of putty moving horizontally at 6.0 m/s sticks to the bottom end. Find ω just after the collision.
- L before (about the pivot) = mvL = (0.10)(6.0)(1.2) = 0.72 kg·m²/s.
- I after = ML²/3 + mL² = 0.432 + 0.144 = 0.576 kg·m².
- ω = L/I = 0.72/0.576 = 1.25 rad/s.
Example 2. A 100 kg merry-go-round (a disk, R = 2.5 m) turns at 1.0 rad/s with a 30 kg child at its edge. The child walks to 0.5 m from the centre. Find the new ω.
- I₁ = ½(100)(2.5)² + 30(2.5)² = 312.5 + 187.5 = 500 kg·m².
- I₂ = 312.5 + 30(0.5)² = 320 kg·m².
- ω₂ = I₁ω₁/I₂ = 500/320 = 1.56 rad/s.
Exam-style practice
1. A 2 kg particle moves at 3 m/s along the line y = 4 m (parallel to x). What is its angular momentum about the origin?
Show the solution
L = mvr⊥ = 2(3)(4) = 24 kg·m²/s, and it stays constant.
2. A constant torque of 8 N·m acts for 3 s on a wheel at rest. Find the final L.
Show the solution
ΔL = τΔt = 24 kg·m²/s.
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