Lesson 14: Conservation of Angular MomentumAP

Angular momentum is momentum's rotational twin, and its conservation law (CED Unit 7) explains phenomena linear momentum cannot touch — skaters spinning up, stars collapsing into pulsars, gyroscopes refusing to fall. For a particle, L = r p sinφ (φ between the radius and the momentum); for a rigid body rotating about a fixed axis, L = Iω. Torque plays the role force played: τ_net = ΔL/Δt. Hence the conservation condition mirrors momentum's: when the net external torque about an axis is zero, angular momentum about that axis is constant — even while I, ω, and the rotational kinetic energy all change inside the system.

The skater effect and its energy secret

A skater pulling in her arms reduces I (mass moves closer to the axis); with no external torque about the vertical axis, L = Iω is fixed, so ω rises in inverse proportion. Rotational kinetic energy K = ½Iω² = L²/(2I) increases as I falls — the extra energy is the work her muscles do pulling the mass inward against its tendency to fly out. L conserved, K not: the same pattern as inelastic collisions, in mirror image.

Rolling: translation and rotation sharing one motion

An object rolling without slipping obeys the constraint v_cm = ωR (and a_cm = αR). Its kinetic energy splits: K = ½mv_cm² + ½Iω². Energy methods then predict race results down a ramp without any force analysis: objects with smaller I/(mR²) (solid sphere 2/5, solid disk 1/2, hoop 1) convert less energy into rotation and arrive with more translational speed — the sphere always beats the disk, which always beats the hoop, independent of mass and radius. Slipping (burnouts, skids) breaks the constraint and the static friction that enforces rolling becomes kinetic.

Systems and centre of mass

For compound systems (child + merry-go-round, satellite + spinning wheel), angular momentum conservation applies to the total about a shared axis, provided external torques about that axis are negligible during the interaction. A person walking on a turntable rotates it the other way; a spacecraft's reaction wheels reorient the craft by exchanging L internally — no propellant, just bookkeeping.

Worked example 1 — the spinning skater

A skater spins at 3.0 rad/s with arms extended, I = 6.0 kg·m². Pulling her arms in reduces I to 2.0 kg·m². Find (a) her new angular velocity, (b) the change in rotational kinetic energy, and account for it. Given: no external torque about the spin axis → L conserved.

L = Iω = 6.0 × 3.0 = 18 kg·m²/s
(a) ω_f = L/I_f = 18/2.0 = 9.0 rad/s
(b) K_i = ½(6.0)(3.0²) = 27 J;  K_f = ½(2.0)(9.0²) = 81 J;  ΔK = +54 J

Check / accounting: the extra 54 J is work done by the skater's muscles pulling her arms inward — internal energy converted to rotational kinetic energy, exactly why K = L²/(2I) grows as I shrinks at fixed L.

Worked example 2 — the ramp race, by energy

A solid disk and a hoop, each of mass m and radius R, are released from rest at height h on a ramp and roll without slipping. Find each one's speed at the bottom and state who wins. Method: mgh = ½mv² + ½Iω² with ω = v/R.

Disk:  I = ½mR² → mgh = ½mv² + ½(½mv²) = (3/4)mv² → v = √(4gh/3) ≈ 1.15√(gh)
Hoop:  I = mR²  → mgh = ½mv² + ½mv² = mv²      → v = √(gh) = 1.00√(gh)

Check: the disk wins — it wastes a smaller share (1/3 of its energy) on rotation than the hoop (1/2). Mass and radius cancel for both, so a toy disk beats a giant hoop every time.

Real-world application: collapsing stars and gyroscopic stability

A rotating gas cloud that collapses from light-years across to a 20 km neutron star shrinks its I by an astronomical factor; conservation of L spins the result up to hundreds of revolutions per second — the pulsars astronomers observe. Engineers use the same stability deliberately: a spinning gyroscope's large L resists reorientation because an external torque can only change L at the rate τ = ΔL/Δt, which is why reaction wheels steer satellites and why a rolling bicycle wheel holds its plane.

Experimental design note

Test angular momentum conservation with a rotating platform: drop a known mass (held at radius r, so its L = mvr is computable) onto a spinning turntable with a marker, measuring ω before and after with a motion sensor or phone slow-motion. Compare I_platform·ω_before with (I_platform + mr²)·ω_after. The honest error discussion: bearing friction applies a small external torque throughout, so L_after falls slightly short — quantifying that shortfall is the mark of an AP-level report.

Common mistakes (from College Board scoring commentary)

• Claiming rotational kinetic energy is conserved when a skater pulls in (it is not; L is).
• Applying L conservation when a clear external torque acts about the axis (e.g., a pendulum's full swing under gravity about the pivot — gravity's torque is not zero).
• In rolling problems, using total K = ½mv² only, forgetting the rotational share.
• Forgetting that v = ωR holds only for rolling without slipping.

AP Exam Experience

MCQ 1. A diver leaves the board rotating slowly with moment of inertia I and angular velocity ω. Tucking reduces her moment of inertia to I/3. Her new angular velocity is:

• (A) ω/3
• (B) ω
• (C) 3ω
• (D) 9ω

Answer: (C). With negligible external torque (gravity acts at the centre of mass, exerting no torque about it), Iω is constant: (I/3)ω' = Iω gives ω' = 3ω.

MCQ 2. A hoop, a solid disk and a solid sphere of equal mass and radius roll without slipping down the same ramp. The order of arrival at the bottom is:

• (A) hoop, disk, sphere
• (B) sphere, disk, hoop
• (C) all together
• (D) disk, sphere, hoop

Answer: (B). Smaller I/(mR²) leaves more energy for translation: sphere (2/5) < disk (1/2) < hoop (1), so the sphere is fastest.

MCQ 3. A child runs and jumps onto the edge of a stationary merry-go-round, clinging on. For the child + merry-go-round system about the central axis, the quantity conserved in the jump is:

• (A) kinetic energy
• (B) linear momentum
• (C) angular momentum
• (D) mechanical energy

Answer: (C). The axle's external force during the brief jump acts at the axis (no torque about it), so L about that axis is conserved; the clinging is perfectly inelastic, so kinetic energy drops, and the axle impulse spoils linear momentum conservation.

FRQ. A turntable with moment of inertia 3.0 kg·m² rotates freely at 2.0 rad/s. A 1.5 kg block, initially at rest, is gently placed onto the turntable at a distance of 0.80 m from the centre, where it sticks. (a) Calculate the angular velocity of the system after the block lands. (b) Calculate the rotational kinetic energy before and after, and account for any difference. (c) Explain why angular momentum is conserved here even though the block–turntable interaction involves friction. (d) State what would happen to the final angular velocity if the block landed at 0.40 m instead, and justify it using I_block = mr². Scoring guideline. (a) 2 points: I_block = 1.5 × 0.80² = 0.96 kg·m²; L = 3.0 × 2.0 = 6.0 = (3.0 + 0.96)ω → ω ≈ 1.5 rad/s (1 point I and setup, 1 point value). (b) 2 points: K_i = ½(3.0)(2.0²) = 6.0 J; K_f = ½(3.96)(1.5²) ≈ 4.5 J; the ~1.5 J difference is dissipated by the kinetic friction that acts while the block slips to match the surface speed (thermal energy) (1 point values, 1 point accounting). (c) 1 point: friction between block and table is internal to the chosen system, and the axle exerts no torque about the axis, so net external torque is zero and total L is unchanged. (d) 1 point: I_block falls by ×4 to 0.24 kg·m², so ω = 6.0/3.24 ≈ 1.85 rad/s — the final speed is higher because the added inertia is smaller.

Worked example 3 — stepping to the centre

A merry-go-round (I = 200 kg·m² about its axle) rotates at 0.50 rad/s with a 30 kg child standing at its rim, 2.0 m from the centre. The child walks to the centre. Find the new angular velocity. Method: the child at the rim contributes I_child = mr² = 30 × 2.0² = 120 kg·m²; at the centre, essentially zero.

L = (200 + 120) × 0.50 = 160 kg·m²/s
ω_f = L / 200 = 0.80 rad/s

Check: rotational kinetic energy rises from ½(320)(0.25) = 40 J to ½(200)(0.64) = 64 J — the child's muscles (and grip) did the extra 24 J of work against the apparent outward pull while walking in, the skater effect with playground equipment ✓.

Why the Moon is leaving, and other L stories

Earth's spin angular momentum is slowly being transferred, via tidal friction's tiny external torques, to the Moon's orbital angular momentum: our day lengthens by about 2 milliseconds per century and the Moon recedes ~3.8 cm per year — laser reflectors left by Apollo measure it. Nothing in this unit is more instructive: conservation laws with a small print clause ("when net external torque is zero") govern even planet-scale systems once you choose the system (Earth + Moon) honestly and account for the small torque that remains (the Sun's). Choosing the system, and stating the approximation, is the physics.

The yo-yo: Atwood in a mirror

A yo-yo is a falling mass whose string unwinds from an axle — Lesson 6's Atwood machine where the "pulley" is the falling body itself and rotation resists the fall. For a solid-cylinder yo-yo (I = ½mr² about its centre, string at radius r), the equations mg − T = ma and Tr = Iα with a = rα give a = g/(1 + I/(mr²)) = g/(1 + ½) = (2/3)g ≈ 6.7 m/s². The yo-yo falls slower than free fall because gravity's energy splits between translation and spin — the ramp race of Worked example 2 standing vertically. Any "unrolling" problem (toilet roll, spool pulled by its inner hub, cable drum) is this same trio of equations; the spool-pulled-at-an-angle variant, where the rolling direction surprises everyone, is a beloved extension for strong students.