AP Physics C: Mechanics · Unit 6: Energy and Momentum of Rotating Systems · Lesson 6.4

Deep Dive: Conservation of Angular Momentum

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
6.4.A.16.4.A.26.4.A.2.i6.4.A.2.ivConcept

A System's Total Angular Momentum

The total angular momentum of a system about a rotational axis is just the sum of the angular momenta of its individual parts about that same axis. Nothing new there — it works exactly the way total linear momentum did back in Unit 4.

What makes angular momentum powerful is what happens when the parts of a system interact with each other. Any torque one part exerts on another is matched by an equal-and-opposite torque the second part exerts back on the first — a direct consequence of Newton's third law. Add those two internal torques together and they cancel exactly.

the chosen systeminternal torques — equal & opposite, cancelexternal torque —crosses the boundary,this one counts
🔑Internal torques can shuffle angular momentum between the parts of a system, but they can never change the system's total. Only a torque from outside the system — an external torque — can do that. If the total angular momentum of a system changes at all, that change is exactly equal to the angular impulse delivered by the external torque: ΔL = ∫τext dt.

This is exactly why choosing your system carefully matters so much in these problems: pick a system that includes everything interacting internally, and all those internal torques disappear from your bookkeeping entirely.

6.4.A.2.iiiConcept⚠ Watch Out

Reshaping a System — Without an External Torque

Here's the idea behind every spinning-skater demonstration you've ever seen: a nonrigid system can change its angular speed without any change in its angular momentum, simply by changing shape — moving mass closer to or farther from the rotation axis.

🔑A skater pulling their arms in decreases their rotational inertia I. With no external torque acting (ignoring the small torque from ice friction), their angular momentum L = Iω can't change — so ω has to increase to compensate. Extending the arms does the reverse: I increases, so ω decreases.

This system's angular momentum is locked at L = 24 kg·m²/s — no external torque acts on it. Slide the rotational inertia (think: a skater pulling their arms in or out) and watch ω respond so that L stays exactly the same.

I (kg·m²)6
larger I (arms out) ←→ smaller I (arms in)
L = 24 kg·m²/s (fixed — no external torque)ω = L / I = 24 / 6 = 4.00 rad/sKrot = ½Iω² = 48.0 J — not conserved!
⚠️Notice what the tool above shows: rotational kinetic energy is not constant here, even though angular momentum is. Pulling the arms in takes muscular effort — real work gets done on the system — and that work shows up as extra Krot. Conserving L never implies conserving K.
ExampleGuided Example — The Spinning Skater

A skater spins with a rotational inertia of 4 kg·m² at 3 rad/s, arms extended. She pulls her arms in, reducing her rotational inertia to 1 kg·m². No external torque acts on her. Find her new angular speed.

Step 1Identify the conserved quantity
No external torque acts, so angular momentum is conserved: Li = Lf.
6.4.B.16.4.B.26.4.B.3Math

The Conservation Condition

Put it all together and you get the master rule for this lesson. Angular momentum is conserved in every interaction, everywhere — but whether a particular system's angular momentum stays constant depends entirely on how you draw the boundary of that system:

Στext = 0  ⟹  Li = Lf
💡If the net external torque on your chosen system is zero, its total angular momentum is constant — full stop, no matter how chaotic the internal interactions are. If the net external torque is not zero, angular momentum is transferred between the system and its environment, and Li ≠ Lf.

This is exactly the scenario in this unit's header art: two flywheels, spinning independently, engage a clutch and lock together. No external torque acts on the combined two-flywheel system, so its total angular momentum before engagement exactly equals its total angular momentum after.

Two flywheels — like the ones in this unit's header art — spin independently, then engage a clutch and lock together. No external torque acts on the combined system, so their total angular momentum before matches their total angular momentum after.

I₁ (kg·m²)4
ω₁ (rad/s)8
I₂ (kg·m²)6
ω₂ (rad/s)2
Li = I₁ω₁ + I₂ω₂ = (4)(8) + (6)(2) = 44.0 kg·m²/sωf = Li / (I₁ + I₂) = 44.0 / (4 + 6) = 4.40 rad/sLf = (I₁ + I₂)ωf = 44.0 kg·m²/s — matches Li, as it must
ExampleGuided Example — Engaging Flywheels

A flywheel with I₁ = 3 kg·m² spins at ω₁ = 10 rad/s. It's brought into contact with a second, stationary flywheel with I₂ = 6 kg·m². They stick together and rotate as one. Find their common final angular velocity.

Step 1Identify the system and check for external torque
System = both flywheels together. The only torques involved are the two flywheels acting on each other as they engage — internal torques, which cancel. No external torque acts, so total L is conserved.
← Back to Lesson 6.4Ready for 6.5? Rolling combines translational and rotational motion into a single, tightly constrained relationship: v_cm = ωR.