Conservation of Angular Momentum: Study Guide (Physics C: Mechanics 6.4)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 6.4 called Conservation of Angular Momentum. This guide is the calculus-based one. It proves that internal torques cancel using vector products, and it treats systems whose rotational inertia changes continuously. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Total angular momentum of a system

A system is whatever collection of objects you choose to study. Its total angular momentum about an axis is the sum of the angular momenta of all its parts about that same axis:

L_sys = L₁ + L₂ + L₃ + …

Use L = Iω for rigid parts turning about the axis, and L = r × p (size mvd) for point objects. Signs matter. With counterclockwise positive, a disk turning clockwise contributes a negative L. Two parts can have angular momenta that cancel, giving a system with L_sys = 0 even though both parts are turning.

Why internal torques cancel

From Topic 6.3, the rate of change of each part’s angular momentum equals the net torque on it. Add these for every part of the system:

dL_sys/dt = (sum of all torques on all parts) = τ_external + τ_internal

Now look at one internal pair. Particle 1 (position r₁) and particle 2 (position r₂) push or pull on each other. By Newton’s third law, F₁₂ = −F₂₁, and for forces such as contact pushes, tension and gravity, the pair acts along the line joining the particles. Their torques about O add to

r₁ × F₁₂ + r₂ × F₂₁ = (r₁ − r₂) × F₁₂ = 0

because r₁ − r₂ points along the line joining them, parallel to F₁₂. Every internal pair cancels like this, so τ_internal = 0 and

dL_sys/dt = τ_ext, net

Integrating over time:

ΔL_sys = ∫τ_ext, net dt

Two results follow:

• Equal and opposite angular impulses. During any interaction between two parts, the angular impulse that A gives B is equal in size and opposite in direction to the angular impulse that B gives A. What one part gains, the other loses.
• Only the outside can change the total. Any change in a system’s angular momentum must come from an interaction with something outside it, and the change equals the external angular impulse.

Choosing the system

Angular momentum is conserved in every interaction: it is only ever moved from one object to another. Whether a particular system’s L stays constant depends on where you draw its boundary.

Situation	System	External torque about the axis?	L of the system
Two disks on one frictionless shaft, coupled by a clutch	both disks	none (clutch torque is internal)	constant
Same, but you look at one disk only	disk A	clutch torque from disk B	changes
Clay hits a rod pivoted at one end	clay + rod, about the pivot	pivot force passes through the axis: zero torque (gravity’s angular impulse is negligible during the brief impact)	constant about the pivot
A motor on a fixed stand spins up a fan	fan alone	motor torque	increases
Same	fan + motor + stand + Earth	none	constant (the Earth gains the opposite L)
A wheel slowing on a rubbing axle	wheel alone	friction torque	decreases: L passes to the mount and Earth

Collisions with pivots. In the clay-and-rod case the pivot pushes on the rod during the impact, so the linear momentum of clay + rod is not conserved. But that force acts at the axis, so its torque about the pivot is zero. Use angular momentum about the pivot, not linear momentum.

When the net external torque is not zero, angular momentum is transferred between the system and its surroundings at the rate τ_ext. A braking torque passes the system’s angular momentum to whatever holds the brake.

Changing shape: same L, different ω

A system that is not rigid can change its rotational inertia by moving mass towards or away from the axis. If no external torque acts, L stays constant, so

Iω = constant, and ω = L/I

Pull mass in, I falls and ω rises. Push mass out, ω falls. The kinetic energy changes as well:

K = ½Iω² = L²/(2I)

At constant L, K is inversely proportional to I. Reducing I to one third makes ω three times larger and K three times larger. The extra energy comes from work done by internal forces, for example by a person pulling weights inwards. Angular momentum is conserved; kinetic energy generally is not.

Worked example 1: two disks coupled by a clutch

Question. Take counterclockwise as positive. Disk A (I_A = 0.060 kg·m²) turns at +20 rad/s. Disk B (I_B = 0.030 kg·m²) is on the same frictionless shaft, turning at −10 rad/s. A clutch is engaged. While the disks slip, it exerts a constant torque of size 0.40 N·m on each, until they turn together. (a) Find the common final ω. (b) Find the angular impulse on each disk and the time to lock. (c) Find the kinetic energy lost. (d) After locking, a bearing exerts a friction torque of 0.030 N·m on the pair. How long until they stop, and where does their angular momentum go?

(a) Common final ω. System: both disks. The clutch torques are internal and the shaft is frictionless, so L_sys is constant.

1. L_A = 0.060 × 20 = +1.2 kg·m²/s. L_B = 0.030 × (−10) = −0.30 kg·m²/s.
2. L_sys = +0.90 kg·m²/s.
3. ω_f = L_sys/(I_A + I_B) = 0.90/0.090 = +10 rad/s.

(b) Angular impulses and time.

1. ΔL_A = 0.060 × (10 − 20) = −0.60 N·m·s. ΔL_B = 0.030 × (10 − (−10)) = +0.60 N·m·s. Equal and opposite, as Newton’s third law requires.
2. Each torque has size 0.40 N·m, so the time is 0.60/0.40 = 1.5 s.

(c) Kinetic energy. K_i = ½(0.060)(20²) + ½(0.030)(10²) = 12 + 1.5 = 13.5 J. K_f = ½(0.090)(10²) = 4.5 J. 9.0 J is lost, two thirds of the starting energy, turned into thermal energy in the slipping clutch.

(d) After locking. Now the system has an external torque. ΔL = −0.90 kg·m²/s = (−0.030 N·m)Δt, so Δt = 30 s. The angular momentum is not destroyed: it passes through the bearing into the frame and the Earth.

Worked example 2: a bead sliding along a spinning rod

Question. A thin rod spins in a horizontal plane about a vertical axle through its centre. Its rotational inertia about the axle is I_r = 0.040 kg·m². A 0.10 kg bead, treated as a point, can slide along the rod without friction. It starts 0.10 m from the axle, with the system turning at 10 rad/s. The bead then slides outward until it reaches a stop 0.50 m from the axle. The axle is frictionless. (a) Explain why L is conserved. (b) Derive ω as a function of the bead’s distance r, and find ω when it reaches the stop. (c) Find dω/dr and describe the ω–r graph. (d) Compare the rotational kinetic energies, and explain where the difference has gone.

(a) Why L is constant. System: rod + bead. The forces between bead and rod are internal. The axle force acts at the axis, and gravity and the vertical support forces are parallel to the axis, so neither has a torque about it. Net external torque about the axle = 0.

(b) ω(r).

1. I(r) = I_r + mr². At the start, I = 0.040 + 0.10 × 0.10² = 0.041 kg·m², so L = 0.041 × 10 = 0.41 kg·m²/s.
2. ω(r) = L/(I_r + mr²).
3. At the stop: I = 0.040 + 0.10 × 0.50² = 0.065 kg·m², so ω = 0.41/0.065 = 6.3 rad/s.

(c) Shape of the graph.

1. dω/dr = −2Lmr/(I_r + mr²)². It is negative for every r > 0: ω falls continuously as the bead moves out.
2. Near r = 0.10 m the slope is small, because the bead adds little to I there. Further out the mr² term grows, and ω falls faster. The slope is steepest at r = √(I_r/(3m)) ≈ 0.37 m, then eases very slightly as I itself becomes large (Figure 2).

(d) Energy.

1. Start: K = ½ × 0.041 × 10² = 2.05 J. At the stop, before the bead hits it: K_rot = L²/(2I) = 0.41²/(2 × 0.065) = 1.29 J.
2. The rotational kinetic energy has fallen by 0.76 J. With no friction, nothing has been lost: the bead is now also moving outward along the rod. That radial motion carries the missing 0.76 J, a radial speed of √(2 × 0.76/0.10) ≈ 3.9 m/s.

Interpretation. L stayed fixed while ω fell, because I rose. When the bead hits the stop, that radial kinetic energy becomes thermal energy, but L about the axle is still 0.41 kg·m²/s.

Common misconceptions

• “Angular momentum is conserved, so ω cannot change.” L = Iω is fixed; ω changes whenever I does.
• “Kinetic energy is conserved whenever angular momentum is.” Couplings lose kinetic energy; pulling mass in gains it.
• Mixing axes. Every L in the sum must be about the same axis.
• Using linear momentum in a collision with a pivoted body. The pivot force is external; use L about the pivot.
• Forgetting the sign of a part turning the other way. Opposite rotations partly cancel in L_sys.
• “Internal forces can spin a system up from rest.” They can make parts turn, but always in opposite senses, so the total stays zero.
• “Friction destroys angular momentum.” It transfers it to the surroundings. Draw a bigger system to see it conserved.

Where this leads

Topic 6.5, Rolling, combines rotation and translation for wheels and balls that roll without slipping. Topic 6.6, Motion of Orbiting Satellites, uses conservation of angular momentum about the central body: gravity points at the centre, so it exerts no torque there, and a satellite moves faster when it is closer. Next, try the practice questions, then use the revision notes and the checklist. When you are ready, move on to Topic 6.5, Rolling, or return to the course roadmap.