AP Physics C: Mechanics · Unit 6: Energy and Momentum of Rotating Systems · Lesson 6.3

Deep Dive: Angular Momentum and Angular Impulse

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
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Defining Angular Momentum

Linear momentum, p⃗ = mv⃗, tracks translational motion. Angular momentum, L, is its rotational counterpart. For a rigid system rotating about a fixed axis, it's given by:

L = Iω

But angular momentum is more general than "how fast something spins." Any object has an angular momentum about a chosen point, whether or not it's rotating at all, given by the magnitude of a cross product:

L = rp sinθ
💡Here r is the distance from the reference point to the object, p is the object's (linear) momentum, and θ is the angle between the position vector r⃗ and the momentum vector p⃗. This is exactly the diagram in this unit's header art.

Adjust the distance from the reference point (r), the object's momentum (p), and the angle between them (θ) — the same setup as this unit's header art — and watch the angular momentum respond.

r (m)6
p (kg·m/s)4
θ (deg)60°
reference pointrpθ
L = rp sinθ = (6)(4)sin(60°) = 20.78 kg·m²/s
🔑Angular momentum always depends on which point or axis you measure it about. The same object, at the same instant, has a different L about different reference points — there's no such thing as "the" angular momentum of an object without specifying what it's measured relative to.

Here's the counterintuitive part: because L = rp sinθ only needs an angle between r⃗ and p⃗ — not actual rotation — an object moving in a perfectly straight line can have nonzero, and even constant, angular momentum about a point that isn't on its path.

ExampleGuided Example — Angular Momentum of Straight-Line Motion

A 2 kg object moves in a straight line at a constant 5 m/s, passing 3 m from the origin at its closest approach (i.e., the perpendicular distance from the origin to the line of motion is 3 m). Find its angular momentum about the origin, and explain what happens to that value as the object moves further along the line.

Step 1Set up L = rp sinθ
p = mv = (2)(5) = 10 kg·m/s. r and θ both change as the object moves — but the quantity r sinθ is always the perpendicular distance from the origin to the line of motion.
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Angular Impulse

Where work asked "how much energy does a torque transfer over an angular displacement," angular impulse asks a different question: how much does a torque change a system's motion when it acts over a time interval?

angular impulse = ∫τ dt
🔑Angular impulse points in the same direction as the torque causing it. And just like angular displacement collapsed W = ∫τ dθ down to W = τΔθ for a constant torque, a constant torque collapses angular impulse down to a simple product: angular impulse = τΔt.

Angular impulse also has a graphical meaning: it's the area under a graph of torque plotted against time — the direct rotational analog of finding linear impulse from the area under a force-vs-time graph.

tτArea = angular impulseconstant torque → angular impulse = τΔt

Set a constant torque and the time interval it acts over to see the angular impulse it delivers.

τ (N·m)6
Δt (s)4
Angular impulse = τΔt = (6)(4) = 24.0 kg·m²/s
ExampleGuided Example — Angular Impulse from a Motor

A motor exerts a constant 6 N·m torque on a flywheel for 4 seconds. How much angular impulse does the motor deliver?

Step 1Identify the given quantities
τ = 6 N·m (constant)  ·  Δt = 4 s
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The Rotational Impulse–Momentum Theorem

Now connect the two ideas above. The change in a system's angular momentum, ΔL = L − L₀, is related to the angular impulse delivered to it by a rotational version of the impulse–momentum theorem you met back in Unit 4:

ΔL = ∫τ dt
🔑This isn't a new law — it's Newton's second law in rotational form. For a system with constant rotational inertia:
τnet = dL/dt = I(dω/dt) = Iα

That relationship, τnet = dL/dt, gives you two more ways to read a graph. The slope of a graph of angular momentum vs. time equals the net torque exerted on the system at that instant — and the area under a graph of net torque vs. time equals the angular impulse delivered, which is exactly the system's change in angular momentum.

tLslope = τneta steeper line means a larger net torque acting on the system

Start a system with some initial angular momentum, apply a constant net torque for a while, and see where its angular momentum ends up.

L_i (kg·m²/s)0
τnet (N·m)5
Δt (s)4
Angular impulse = τnetΔt = (5)(4) = 20.0 kg·m²/sLf = Li + angular impulse = 0 + 20.0 = 20.0 kg·m²/s
ExampleGuided Example — Spinning Up a Merry-Go-Round

A merry-go-round starts at rest (Li = 0). A constant net torque of 8 N·m is applied for 5 seconds. Find its final angular momentum, and explain how that value could also be read off a graph.

Step 1Find the angular impulse
angular impulse = τnetΔt = (8)(5) = 40 kg·m²/s
← Back to Lesson 6.3Ready for 6.4? Conservation of Angular Momentum looks at what happens to L when the net external torque is exactly zero.