Conservation of Angular Momentum: Study Guide (Physics 1 6.4)

This guide is for the algebra-based Physics 1 course. It uses algebra and graphs only. It builds directly on Topic 6.3, where you met the rotational impulse–momentum theorem, τ_net Δt = ΔL.

The idea: no outside twist, no change

Topic 6.3 showed that a net torque acting for a time changes angular momentum. Now apply that idea to a system: a set of objects you choose to study together.

Only torques from outside the system can change the system’s total angular momentum. So:

• If the net external torque is zero, the total angular momentum is constant: L_initial = L_final.
• If the net external torque is not zero, angular momentum is transferred between the system and its surroundings. The change in the system’s L equals the angular impulse from outside.

This is the rotational partner of conservation of linear momentum from Topic 4.3. The method is the same: choose the system, check the external torques, then write “before = after”.

Total angular momentum of a system

The total angular momentum of a system about an axis is the sum of the angular momenta of its parts about that same axis. Each part gets a sign for its sense of rotation.

Example: take counterclockwise (seen from above) as +. Two disks turn on one axle. Disk 1 has I = 0.40 kg·m² and ω = +5.0 rad/s. Disk 2 has I = 0.20 kg·m² and ω = −8.0 rad/s.

L_total = (0.40)(+5.0) + (0.20)(−8.0) = 2.0 − 1.6 = +0.40 kg·m²/s

The parts can be rigid bodies (L = Iω) or point objects (L = rmv sin θ). All must be measured about the same axis.

Why internal torques cannot change the total

Suppose part A of a system pushes or rubs on part B. By Newton’s third law, B pushes back on A with an equal and opposite force, along the same line. The two forces have the same lever arm about the axis, so their torques are equal in size and opposite in sign. They act for the same time. So:

angular impulse on B from A = −(angular impulse on A from B)

Whatever angular momentum B gains, A loses. The total does not change. This is why only external torques matter for the system as a whole.

Choosing the system

Whether angular momentum is “conserved” depends on the system you choose. Angular momentum itself is never created or destroyed in any interaction. It only moves from one object to another.

Situation	System	Net external torque?	Is the system’s L constant?
A ring dropped onto a turntable with a frictionless axle	turntable only	yes: friction from the ring	no; it loses L to the ring
Same	turntable + ring	no: ring–turntable friction is internal	yes
A wheel slowed by a brake pad fixed to a frame	wheel only	yes: friction from the pad	no; L is transferred to the frame and Earth
A spinning platform whose masses are pulled inwards by strings through the axis	platform + masses	no: string forces pass through the axis	yes

Choose the system so that the unknown or awkward forces are internal, or act through the axis. Then their torques drop out.

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 the net external torque is zero, L = Iω stays the same, so

I₁ω₁ = I₂ω₂

Move mass inwards: I falls, so ω rises. Move mass outwards: I rises, so ω falls. A diver who tucks into a ball spins faster, and slows the spin again by opening out before entering the water.

Testing conservation with a graph

If L is constant, then ω = L × (1/I). A graph of ω against 1/I should be a straight line through the origin, with slope L. Figure 2 shows the platform of Figure 1 with its masses stopped at four different radii.

Worked example 1: a ring dropped onto a turntable

Question. Take counterclockwise (seen from above) as +. A turntable with I₁ = 0.020 kg·m² spins at +12 rad/s on a frictionless axle. A ring with I₂ = 0.010 kg·m² about the same axis is held just above it, not rotating, and released so that it lands centred. Friction between the two makes them turn together. Find (a) the final angular speed, (b) the angular impulse each exerts on the other and (c) the change in kinetic energy.

1. System: turntable + ring. The friction between them is internal. The axle is frictionless, and gravity and the axle’s push act along or through the axis, so the net external torque is zero. L is constant.
2. Before: L = I₁ω₁ + I₂(0) = 0.020 × 12 = 0.24 kg·m²/s.
3. After: L = (I₁ + I₂)ω = 0.030ω.
4. (a) 0.030ω = 0.24, so ω = 8.0 rad/s.
5. (b) Ring: ΔL = 0.010 × 8.0 − 0 = +0.080 N·m·s. Turntable: ΔL = 0.020 × 8.0 − 0.24 = −0.080 N·m·s. Equal and opposite, as Newton’s third law requires.
6. (c) Before: K = ½ × 0.020 × 12² = 1.44 J. After: K = ½ × 0.030 × 8.0² = 0.96 J. The system loses 0.48 J, a third of its kinetic energy, to thermal energy from the sliding friction.

Check. The final speed is less than 12 rad/s, as it must be. A common wrong answer is the plain mean of 12 and 0, which is 6.0 rad/s. That ignores the different rotational inertias. This is the rotational version of a perfectly inelastic collision: L is conserved, K is not.

Worked example 2: clay hits a pivoted arm

Question. A thin arm lies on a level, frictionless table and can turn about a fixed vertical pin through one end. Its rotational inertia about the pin is 0.040 kg·m², and it is at rest. A 0.040 kg ball of clay slides across the table at 5.0 m/s, perpendicular to the arm. It hits the arm 0.50 m from the pin and sticks. Find the arm’s angular speed just after the impact.

1. System: arm + clay. During the impact the pin pushes on the arm, an external force. But it acts at the pin, so its torque about the pin is zero. Take angular momentum about the pin.
2. Before: the clay moves perpendicular to the arm, so θ = 90° and L = rmv = 0.50 × 0.040 × 5.0 = 0.10 kg·m²/s. The arm has L = 0.
3. After: the clay is a point object 0.50 m from the pin, so its rotational inertia is mr² = 0.040 × 0.50² = 0.010 kg·m². Total I = 0.040 + 0.010 = 0.050 kg·m².
4. Conservation: 0.050ω = 0.10, so ω = 2.0 rad/s.

Interpretation. The clay now moves at ωr = 2.0 × 0.50 = 1.0 m/s. Linear momentum is not conserved here, because the pin exerts an external force. Kinetic energy falls from 0.50 J to 0.10 J. Only angular momentum about the pin is conserved, because the pin’s torque about itself is zero.

Worked example 3: pulling the masses in

Question. Use Figure 1. The platform alone (hub and rail) has I = 0.60 kg·m². Two 1.5 kg masses, treated as points, sit 0.80 m from the axis, and the system turns at 2.0 rad/s. Strings through the axle pull both masses in to 0.20 m. Find the new angular speed and the change in kinetic energy.

1. I₁ = 0.60 + 2 × 1.5 × 0.80² = 0.60 + 1.92 = 2.52 kg·m².
2. I₂ = 0.60 + 2 × 1.5 × 0.20² = 0.60 + 0.12 = 0.72 kg·m².
3. The string forces point at the axis, so the net external torque is zero: I₁ω₁ = I₂ω₂. ω₂ = 2.52 × 2.0 ÷ 0.72 = 7.0 rad/s.
4. K₁ = ½ × 2.52 × 2.0² = 5.04 J. K₂ = ½ × 0.72 × 7.0² = 17.64 J. Kinetic energy rises by about 13 J (12.6 J).

Where did the energy come from? Whoever pulls the strings does work. The masses move inwards along the force, so the force does positive work on them. Since K = L² / (2I), keeping L fixed while I falls by a factor of 3.5 multiplies K by 3.5.

Common misconceptions

• “Angular momentum is conserved, so ω stays the same.” Only if I stays the same. In a shape change, ω changes in the opposite way to I.
• “Kinetic energy is conserved too.” Rarely. Sticking collisions lose K (Worked examples 1 and 2); pulling mass in gains K (Worked example 3).
• “Linear momentum is conserved in every collision.” Not when a pivot or axle exerts an external force. Use angular momentum about the pivot instead.
• “Forces between parts of the system change its angular momentum.” They transfer it between the parts. Their angular impulses are equal and opposite.
• “A wheel slowed by a brake breaks conservation.” For the wheel alone, an external torque acts. Include the brake, frame and Earth, and the total is conserved.
• Averaging angular speeds. Use I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω, never (ω₁ + ω₂)/2.
• Forgetting the moving object’s rotational inertia after it sticks. Add mr² to the total.
• Mixing axes. Every angular momentum in one equation must be about the same axis.

Where this leads

Topic 6.5 (Rolling) combines translation and rotation: a rolling object has kinetic energy of both kinds and, if it does not slip, a fixed link between v and ω. Read the Topic 6.5 study guide next. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.