Angular Momentum & Its Conservation

Pull your arms in and you spin faster. Nothing outside had to push.

12 minUpdated 21 October 2026angular momentumconservation of angular momentumangular impulsespinning skater

Lesson at a glanceShow the key points

Angular momentum of a particle about a point: l = r × p; magnitude mvr⊥, where r⊥ is the perpendicular distance from the point to the line of motion.
For a rigid body about a fixed axis: L = Iω (10.12).
dL/dt = τ(ext) (10.13): internal torques cancel.
If the net external torque is zero, angular momentum is conserved: I₁ω₁ = I₂ω₂.
Angular impulse ∫τ dt equals the change in angular momentum.
When I changes with L fixed, rotational KE changes: pulling arms in takes work and raises KE.

A figure skater begins a spin with arms wide, then draws them in and blurs into a whirl. No one pushed her. What stayed the same was her angular momentum.

Angular momentum of a particle

Definition · Angular momentum

The angular momentum of a particle about a point O is

\vec l = \vec r \times \vec p

(10.10), where

\vec r

is its position from O and

\vec p = m\vec v

its linear momentum. For a system, add the vectors.

Its size is

l = mvr\sin\theta = mv\,r_\perp \qquad (10.11)

where $r_\perp$ is the perpendicular distance from O to the particle’s line of motion. Even a particle moving in a straight line has angular momentum about a point off that line. As with torque, the angular momentum about an axis is the component of $\vec r\times\vec p$ along it, for any O on the axis.

A rigid body: L = Iω

For a particle going round a circle of radius r, pick O at the centre: r ⊥ v, so $l = mvr = mr^2\omega$ , along the axis. For a rigid body each particle contributes $m_i r_i^2\omega$ , and they share ω:

Angular momentum of a rigid body about its axis(10.12)

L = \sum m_i r_i^2\,\omega = I\omega

Torque changes angular momentum

Differentiate the definition:

\frac{d\vec L}{dt} = \sum\Big(\frac{d\vec r_i}{dt}\times\vec p_i + \vec r_i\times\frac{d\vec p_i}{dt}\Big) = \sum\big(\vec v_i\times m_i\vec v_i + \vec r_i\times\vec F_i\big)

The first term vanishes (a vector crossed with itself) and internal torques cancel in pairs, leaving

Rotational second law, general form(10.13)

\frac{d\vec L}{dt} = \vec\tau_{\text{ext}}

for a rigid body with a fixed axis this is just I dω/dt = Iα = τ

Interactive

Pull your arms in

Seen from above, a skater holds 3 kg weights. No external torque acts, so L = Iω stays fixed. Pull the weights in and I drops, so ω rises. The kinetic energy rises too: the arms did work pulling the weights inward.

I

4.84kg·m²

\omega

1.00rev/s

L = I\omega

30.4kg·m²/s

K = L^2/2I

96J

work done by arms0J

Reach (weights from the axis)

d

0.80m

Worked example

The skater's spin

A skater spins at 1.5 rev/s with moment of inertia 4.0 kg·m². She pulls in her arms, reducing I to 1.6 kg·m². Ignoring friction from the ice, what is her new spin rate, and what happens to her kinetic energy?

Worked example

Dropping a disc onto a spinning disc

A disc with I₁ = 0.020 kg·m² spins at 30 rad/s on a vertical axle. A second disc (I₂ = 0.010 kg·m²), not spinning, is dropped onto the same axle. They slip, then turn together. Find the final ω and the kinetic energy lost.

Angular impulse

Integrating $\tau = dL/dt$ over a time interval gives the rotational version of impulse:

Angular impulse

J = \int_{t_1}^{t_2}\tau\,dt = L_2 - L_1

What to remember

1 $\vec l = \vec r\times\vec p$ ; size $mvr_\perp$ .
2Rigid body about its axis: $L = I\omega$ .
3 $d\vec L/dt = \vec\tau_{\text{ext}}$ ; no external torque ⇒ L conserved.
4Changing I with L fixed changes ω inversely and changes kinetic energy.
5Angular impulse ∫τ dt = change in L.

Check yourself

A diver tucks into a ball in mid-air. Her angular velocity…

A 2 kg ball moves at 3 m/s along a line passing 0.5 m from point O. Its angular momentum about O is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchAngular momentumKhan Academy WatchAngular momentum and its conservationWalter Lewin · MIT 8.01x Lecture 20The spinning-stool demonstrations ReadAngular momentumOpenStax University Physics §11.2 ReadConservation of angular momentumOpenStax University Physics §11.3 ReadAngular momentumHyperPhysics ReadAngular momentumWikipedia

PreviousBodies in Equilibrium Next Rotational Kinetic Energy, Work & Power

Angular Momentum & Its Conservation

Pull your arms in and you spin faster. Nothing outside had to push.

12 minUpdated 21 October 2026angular momentumconservation of angular momentumangular impulsespinning skater

Lesson at a glanceShow the key points

Angular momentum of a particle about a point: l = r × p; magnitude mvr⊥, where r⊥ is the perpendicular distance from the point to the line of motion.
For a rigid body about a fixed axis: L = Iω (10.12).
dL/dt = τ(ext) (10.13): internal torques cancel.
If the net external torque is zero, angular momentum is conserved: I₁ω₁ = I₂ω₂.
Angular impulse ∫τ dt equals the change in angular momentum.
When I changes with L fixed, rotational KE changes: pulling arms in takes work and raises KE.

A figure skater begins a spin with arms wide, then draws them in and blurs into a whirl. No one pushed her. What stayed the same was her angular momentum.

Angular momentum of a particle

Definition · Angular momentum

The angular momentum of a particle about a point O is

\vec l = \vec r \times \vec p

(10.10), where

\vec r

is its position from O and

\vec p = m\vec v

its linear momentum. For a system, add the vectors.

Its size is

l = mvr\sin\theta = mv\,r_\perp \qquad (10.11)

A rigid body: L = Iω

Angular momentum of a rigid body about its axis(10.12)

L = \sum m_i r_i^2\,\omega = I\omega

Torque changes angular momentum

Differentiate the definition:

\frac{d\vec L}{dt} = \sum\Big(\frac{d\vec r_i}{dt}\times\vec p_i + \vec r_i\times\frac{d\vec p_i}{dt}\Big) = \sum\big(\vec v_i\times m_i\vec v_i + \vec r_i\times\vec F_i\big)

The first term vanishes (a vector crossed with itself) and internal torques cancel in pairs, leaving

Rotational second law, general form(10.13)

\frac{d\vec L}{dt} = \vec\tau_{\text{ext}}

for a rigid body with a fixed axis this is just I dω/dt = Iα = τ

Interactive

Pull your arms in

I

4.84kg·m²

\omega

1.00rev/s

L = I\omega

30.4kg·m²/s

K = L^2/2I

96J

work done by arms0J

Reach (weights from the axis)

d

0.80m

Worked example

The skater's spin

Worked example

Dropping a disc onto a spinning disc

Angular impulse

Integrating $\tau = dL/dt$ over a time interval gives the rotational version of impulse:

Angular impulse

J = \int_{t_1}^{t_2}\tau\,dt = L_2 - L_1

What to remember

1 $\vec l = \vec r\times\vec p$ ; size $mvr_\perp$ .
2Rigid body about its axis: $L = I\omega$ .
3 $d\vec L/dt = \vec\tau_{\text{ext}}$ ; no external torque ⇒ L conserved.
4Changing I with L fixed changes ω inversely and changes kinetic energy.
5Angular impulse ∫τ dt = change in L.

Check yourself

A diver tucks into a ball in mid-air. Her angular velocity…

A 2 kg ball moves at 3 m/s along a line passing 0.5 m from point O. Its angular momentum about O is…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousBodies in Equilibrium Next Rotational Kinetic Energy, Work & Power