AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-4)
Unit 6 · Topic 6.4
6.4 Conservation of Angular Momentum
If no net external torque acts on a system, its total angular momentum stays constant. A spinning skater who pulls in their arms lowers their rotational inertia and spins faster, and in a rotational collision the angular momentum one object loses, another gains. An outside torque changes a system's angular momentum by exactly the angular impulse it delivers.
Key terms
- conservation of angular momentum
- net external torque
- choosing a system
- rotational collision
- changing rotational inertia
Adding up angular momentum
A system's total angular momentum about an axis is the sum of the angular momenta of all its parts about that same axis, with signs for clockwise and counterclockwise. A spinning turntable plus a bug walking on it, or a rod plus a ball that hits it, each have one total L.
When two parts of a system push on each other, Newton's third law means their angular impulses on each other are equal in size and opposite in sense. Whatever angular momentum one part gains, the other loses. So internal interactions can't change the total. Only a torque from outside the system can.
When is angular momentum constant?
If the net external torque on a system is zero, its total angular momentum stays constant: L_i = L_f. If the net external torque isn't zero, angular momentum moves between the system and its surroundings, and the change equals the external angular impulse.
So the answer depends on the system you choose. A child jumps onto a merry-go-round on a frictionless axle. For the merry-go-round alone, angular momentum changes, because the child pushes on it from outside that system. For the system of child plus merry-go-round, it stays constant, because the axle force passes through the axis and gives no torque.
Angular momentum is never created or destroyed. When it seems to vanish, it has gone to something outside your system, often Earth through friction.
Changing shape: the spinning skater
A system that isn't rigid can change its rotational inertia. A skater spinning with arms out has a large I. Pulling the arms in moves mass toward the axis, so I drops. No outside torque acts (the ice is nearly frictionless), so L = Iω stays the same and ω must rise. Halve I and ω doubles.
Kinetic energy doesn't stay the same. K = ½Iω² = L²/(2I), so with L fixed, a smaller I means more kinetic energy. The extra energy comes from the work the skater's muscles do pulling the arms inward. Divers and gymnasts tuck to spin faster and stretch out to slow down for the same reason.
Rotational collisions
When a spinning object grabs or collides with another (a ring dropped onto a spinning turntable, or a ball striking a pivoted rod), the total angular momentum about the axis is conserved if the axle is frictionless. If the objects stick together, kinetic energy is not conserved: some becomes thermal energy, just as in a perfectly inelastic collision in 4.4.
With a fixed pivot, linear momentum usually isn't conserved, because the pivot pushes on the system during the collision. Angular momentum about the pivot still is, because that pivot force acts at the axis and has no lever arm. That's why you choose the pivot as your axis.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A skater pulls in her arms
A skater spins at 4.0 rad/s with her arms out, where her rotational inertia is 3.5 kg·m². She pulls her arms in, reducing it to 1.4 kg·m². Ignore friction. Find her new angular speed and compare her kinetic energy before and after.
Show the solutionHide the solution
- Step 1: No external torque, so I₁ω₁ = I₂ω₂: (3.5)(4.0) = (1.4)ω₂, so ω₂ = 10 rad/s.
- Step 2: K before: ½(3.5)(4.0)² = 28 J. K after: ½(1.4)(10)² = 70 J.
- Step 3: Kinetic energy rose by 42 J. That energy came from her muscles doing work as she pulled her arms in. Angular momentum was conserved; kinetic energy wasn't.
Answer: 10 rad/s; kinetic energy rises from 28 J to 70 J
- Example 2Calculator allowed
A ring dropped onto a turntable
A turntable (I = 0.050 kg·m²) spins freely at 6.0 rad/s. A ring (I = 0.030 kg·m² about the same axis), not spinning, is dropped onto it, centered, and they turn together. Find the final angular speed and the kinetic energy lost.
Show the solutionHide the solution
- Step 1: System: turntable + ring. The only outside torques would come from the axle, which is frictionless, so L is conserved.
- Step 2: (0.050)(6.0) + 0 = (0.050 + 0.030)ω, so ω = 0.30 ÷ 0.080 = 3.75 rad/s.
- Step 3: K before: ½(0.050)(6.0)² = 0.90 J. K after: ½(0.080)(3.75)² ≈ 0.56 J.
- Step 4: Lost: about 0.34 J (37.5%), turned into thermal energy by friction between the ring and turntable as they came to the same speed.
Answer: About 3.8 rad/s; about 0.34 J of kinetic energy lost
- Example 3Calculator allowed
Clay hits a pivoted rod (exam level)
A uniform rod (mass 0.60 kg, length 0.80 m, I = (1/12)ML² about its center) lies on a frictionless horizontal table, pivoted at its center and at rest. A 0.050 kg ball of clay slides at 4.0 m/s perpendicular to the rod, hits one end and sticks. Find the angular speed just afterward. Is linear momentum conserved?
Show the solutionHide the solution
- Step 1: Choose the system rod + clay and the pivot as the axis. The pivot force has no torque about the pivot, so L about the pivot is conserved.
- Step 2: Before: only the clay has L. It moves perpendicular to the rod at distance 0.40 m, so L = mvr = (0.050)(4.0)(0.40) = 0.080 kg·m²/s.
- Step 3: After: I_rod = (1/12)(0.60)(0.80)² = 0.032 kg·m². The clay at the end adds (0.050)(0.40)² = 0.008 kg·m². Total I = 0.040 kg·m².
- Step 4: ω = L/I = 0.080 ÷ 0.040 = 2.0 rad/s.
- Step 5: Linear momentum is not conserved: the pivot pushes on the rod during the hit, so there's an outside force. Only angular momentum about the pivot is conserved.
Answer: 2.0 rad/s; linear momentum isn't conserved because the pivot exerts an external force
Common mistakes
- Assuming kinetic energy is conserved whenever angular momentum is. A skater pulling in their arms gains kinetic energy; sticking collisions lose it.
- Conserving linear momentum in a collision with a pivoted object. The pivot exerts an outside force; use angular momentum about the pivot.
- Calling a system's angular momentum constant without checking for outside torques, such as axle friction or a person pushing off the ground.
- Forgetting to add the new object's rotational inertia (mr²) after it sticks.
On the exam
- Free-response questions often ask you to justify why angular momentum is conserved. Name the system and explain why the net external torque on it is zero.
- Expect bar charts or comparisons of L and K before and after an interaction; L stays the same in an isolated system, while K can rise or fall.
Connected topics
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Check yourself
4 questions on 6.4 Conservation of Angular Momentum. Pick an answer to see if you got it, and why.
A uniform rod of mass 0.60 kg and length 1.0 m lies on a frictionless horizontal table, free to rotate about a fixed vertical pin through its center. Its rotational inertia about the pin is ¹⁄₁₂ML² = 0.050 kg·m². A 0.10 kg ball of putty slides across the table at 5.0 m/s, perpendicular to the rod, and sticks to one end.
Described scenario
What is the angular speed of the rod and putty just after the collision?
For the putty–rod system, which quantity is conserved during the collision?
A student stands at rest on a turntable that is also at rest and can turn on a frictionless axle. The student begins walking around the edge, and the turntable starts rotating the opposite way. Which explanation is best?
A student drops a ring onto a spinning turntable and wants to test whether the angular momentum of the ring–turntable system is conserved. Which set of measurements is needed?
0 of 4 answered