Angular Momentum (L = Iω) Explained — Derivation, Examples & Quizzes

Q: What is angular momentum?

L = I*omega for a rigid body: the rotational analog of linear momentum p = mv. It measures the 'quantity of spin' — big I or big omega means big L.

Q: How is torque related to angular momentum?

Net torque is the rate of change of angular momentum: Sigma-tau = dL/dt, the rotational analog of F = dp/dt. This is the deeper form of Sigma-tau = I*alpha.

Q: What are the units of angular momentum?

kg*m^2/s: (kg*m^2)*(1/s). Equivalently N*m*s (torque times time) — torque applied over time delivers angular momentum.

Where it comes from

Linear momentum p = mv measures “quantity of motion” — what a force must overcome to stop something. Spin needs the same concept: a flywheel and a bicycle wheel can share an ω yet be wildly different to stop. The spin-quantity must involve I:

Before reading on: linear momentum is p = mv. By analogy, angular momentum should pair I (rotational mass) with which rotational velocity — θ, ω, or α? And why not the others?

linear

p = m v

Mass × velocity. Force changes it: F = dp/dt.

angular

L = I ω

Rotational mass × angular velocity. Torque changes it: τ = dL/dt. (θ is position, α is acceleration — neither is a “velocity.”)

And the deeper definition, L = r × p, says: a particle’s angular momentum about a point is its linear momentum levered by its position — magnitude rp sin θ. For a rigid body, every particle’s r × p lines up along the axis and sums to Iω (derivation below).

Derivation

Start with one particle circling at radius r: its L = r × p has magnitude rmv (v perpendicular to r). Write v = rω, sum over the body, and Iω emerges.

L_i

=

r_i × p_i, |L_i| = r_i m_i v_i sin 90°

Step 1 — one particle. Definition: L = r × p. For circular motion v is perpendicular to r, so sin θ = 1 and |L_i| = r_im_iv_i. Direction: along the axis (right-hand rule).

=

m_i r_i² ω

Step 2 — trade v for ω. v_i = r_iω: r_im_i(r_iω) = (m_ir_i²)ω. There’s the familiar mr² again.

L

=

Σ m_i r_i² ω = I ω

Step 3 — sum the body. One shared ω factors out; Σ m_ir_i² = I. Every particle’s L points along the same axis, so magnitudes add. ∎

Στ

=

dL/dt

Bonus — Newton’s 2nd, deeper form. Since L = Iω with fixed I, dL/dt = I dω/dt = Iα = Στ. Torque is to L what force is to p: its rate of change.

Why does L point along the axis, not along the motion? It’s a cross product: r × p is perpendicular to both r and p — and both lie in the rotation plane, so L sticks straight out along the axle. The direction encodes the plane of spinning, which is exactly what conservation laws need to track.

How to use it

The procedure, every time:

Rigid body about a fixed axis: L = Iω. Get I (shape lessons), get ω in rad/s, multiply. Direction: right-hand rule along the axis.
Single particle: L = rp sin θ = rmv sin θ, about your chosen point. (θ between r and v, tail-to-tail.)
Torque problems: Στ = ΔL/Δt — torque applied over time delivers angular momentum. Average torque: τ_avg = (L_f − L_i)/Δt.
Units: kg·m²/s. Equivalently N·m·s — torque × time.

τ avg = ΔL / Δt torque over time delivers angular momentum — the rotational impulse Say it: “the average torque equals the change in angular momentum over the time”

Common mistake: computing L = Iω with ω in rpm or degrees/s. L inherits ω’s units — convert to rad/s first or the number is fiction.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: a spinning disk

I. Solid disk M = 5 kg, R = 0.4 m ⇒ I = ½ × 5 × 0.16 = 0.4 kg·m².
L. ω = 12 rad/s ⇒ L = 0.4 × 12 = 4.8 kg·m²/s, along the axle (right-hand rule).
Read it. That’s the disk’s “quantity of spin” — what a brake torque must drain to stop it.

Common mistake: L = mv = 5 × (0.4×12) = 24 (“linear momentum of the rim”). Extended bodies aren’t points — L = Iω accounts for all the mass, not just the rim.

Your turn — Hoop M = 4 kg, R = 0.5 m, ω = 6 rad/s. L = ?

Answer: 6 kg·m²/s. I = 4 × 0.25 = 1.0; L = 1.0 × 6 = 6 kg·m²/s.

Example 2 — a particle: L = r × p

Setup. Ball m = 0.5 kg at r = 2 m from origin, moving at v = 3 m/s perpendicular to r.
p. p = mv = 0.5 × 3 = 1.5 kg·m/s.
L. L = rp sin 90° = 2 × 1.5 × 1 = 3 kg·m²/s about the origin.
Notice. The ball isn’t “spinning” — but it has angular momentum about the origin. L = r × p belongs to particles too, not just wheels.

Common mistake: “it’s moving in a straight line, so L = 0.” Angular momentum is about a point: straight-line motion past a point still has r × p ≠ 0 (unless it heads straight at the point).

Your turn — m = 2 kg, r = 1.5 m, v = 4 m/s perpendicular to r. L = ?

Answer: 12 kg·m²/s. p = 8; L = 1.5 × 8 = 12 kg·m²/s.

Before reading on: Earth’s I ≈ 9.8×10³⁷ kg·m², spinning once per day. Is its angular momentum closer to 10²⁹, 10³³, or 10⁴³ kg·m²/s? Estimate the order first.

Example 3 — planetary scale: Earth’s spin

ω. One turn per day: ω = 2π/86400 ≈ 7.27 × 10⁻⁵ rad/s.
L. L = 9.8×10³⁷ × 7.27×10⁻⁵ = 7.1 × 10³³ kg·m²/s.
Read it. 10³³ — a number so large that no earthly torque meaningfully changes it on human timescales. (Tides do, barely — the day lengthens ~2 ms per century.)

Common mistake: ω = 1/86400 (“one per day”) — forgetting the 2π. Angular velocity is radians per second: one revolution = 2π rad.

Your turn — Skater: I = 3 kg·m², ω = 5 rad/s. L = ?

Answer: 15 kg·m²/s. L = 3 × 5 = 15 kg·m²/s. (Next lesson: she pulls her arms in and this L stays put.)

Example 4 — torque delivers L: spinning up a wheel

Setup. Wheel I = 1.2 kg·m², spun from rest to ω = 10 rad/s in Δt = 4 s.
ΔL. L_f − L_i = 1.2 × 10 − 0 = 12 kg·m²/s.
τ_avg. τ = ΔL/Δt = 12/4 = 3 N·m.
The lesson. Torque × time = angular momentum delivered: 3 N·m for 4 s “buys” 12 units of spin. (Check: α = 2.5 rad/s², τ = Iα = 3 ✓.)

Common mistake: dividing by ω instead of Δt (“τ = ΔL/ω”). Torque is L per time — Δt = 4 s is the divisor, not ω = 10.

Your turn — I = 0.8 kg·m², from rest to ω = 20 rad/s in 5 s. τ_avg = ?

Answer: 3.2 N·m. ΔL = 0.8 × 20 = 16; τ = 16/5 = 3.2 N·m.

Memorization tips

Say it aloud: “the angular momentum equals the moment of inertia times the angular velocity.”
Map onto p = mv: I ↔ m, ω ↔ v. Two formulas, one idea: (inertia) × (velocity).
Torque spends L: Στ = dL/dt. Force changes p; torque changes L. The parallel is exact.
Right-hand rule: curl fingers with the spin, thumb points along L. Direction lives on the axis, not in the plane.
Particles have L too: L = rp sin θ about any point — straight-line motion included. Don’t let “not spinning” fool you.
Units tell the story: kg·m²/s = (N·m)·s. Torque applied over time is angular momentum delivered.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and angular momentum is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is angular momentum?

L = I·ω for a rigid body: the rotational analog of linear momentum p = mv. It measures the “quantity of spin” — big I or big ω means big L.

What is L = r × p?

The fundamental definition: angular momentum of a particle is the cross product of its position vector r and its linear momentum p. Its magnitude is r·p·sin θ; for a rigid body this sums to L = Iω.

What direction does angular momentum point?

Along the rotation axis, by the right-hand rule: curl fingers in the spin direction, thumb points along L. Counterclockwise (viewed from above) gives L pointing up/out of the page.

How is torque related to angular momentum?

Net torque is the rate of change of angular momentum: Στ = dL/dt, the rotational analog of F = dp/dt. This is the deeper form of Στ = Iα.

What are the units of angular momentum?

kg·m²/s: (kg·m²)·(1/s). Equivalently N·m·s (torque times time) — torque applied over time delivers angular momentum.

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