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Topic 12 of 19

Angular Momentum

Angular momentum is the rotational analog of linear momentum. Like linear momentum, it is conserved when the net external torque is zero. This principle explains some of the most striking phenomena in mechanics — from a spinning skater to planetary orbits — and is one of the fundamental conservation laws of nature.

Key Concepts

Angular Momentum (particle)

\vec{L} = \vec{r}\times\vec{p} = m\vec{r}\times\vec{v}

. For a particle moving in a circle:

L = mrv = mr^2\omega

. Units: kg·m²/s.

Angular Momentum (rigid body)

For a rigid body rotating about a fixed axis:

L = I\omega

. This is the most common form used in GP1.

Torque–Angular Momentum Relation

\vec{\tau}_\text{net} = d\vec{L}/dt

. Net torque equals the rate of change of angular momentum — the rotational analog of

\vec{F} = d\vec{p}/dt

Conservation of Angular Momentum

If the net external torque on a system is zero,

\vec{L}

is constant:

I_i\omega_i = I_f\omega_f

. Conservation holds for each component independently.

Ice Skater Effect

When a spinning skater pulls in their arms,

I

decreases, so

\omega

must increase to conserve

L = I\omega

. Kinetic energy increases (work is done by the skater's muscles).

Angular Impulse

Analog of impulse:

\vec{\tau}\Delta t = \Delta\vec{L}

. A torque applied for a short time produces a change in angular momentum.

Key Equations

Angular momentum (rigid body)

L = I\omega

For rotation about a fixed axis.

Angular momentum (particle)

L = mvr\sin\theta = mr^2\omega

θ is the angle between r and v; for circular motion L = mvr.

Conservation of angular momentum

I_i\omega_i = I_f\omega_f \quad (\sum\tau_\text{ext} = 0)

Valid when no net external torque acts.

Torque–angular momentum

\vec{\tau}_\text{net} = \frac{d\vec{L}}{dt}

Rotational analog of Newton's Second Law in terms of momentum.

Worked Example

Spinning Skater

Problem

A skater spins at 2.0 rad/s with arms extended ( $I_i = 4.0$ kg·m²). She pulls in her arms, reducing $I$ to 1.5 kg·m². Find her new angular speed.

Solution

No external torque acts (frictionless ice), so $L$ is conserved:

I_i\omega_i = I_f\omega_f

\omega_f = \frac{I_i\omega_i}{I_f} = \frac{4.0\times 2.0}{1.5} \approx 5.3 \text{ rad/s}

Answer New angular speed ≈ 5.3 rad/s — faster than before.

Practice

Exercises

7 problems

Free preview

2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

An ice skater spins with arms out at ω₁ = 3 rad/s (I₁ = 2 kg·m²). When she pulls her arms in (I₂ = 0.5 kg·m²), angular momentum is conserved. Find her new angular velocity ω₂.

ω₂ = rad/s

Exercise 2 / 7 Free

+10 XP

Disc 1 (I₁=4, ω=6 rad/s) couples to stationary disc 2 (I₂=2). Press Couple! to watch angular momentum conserved. Find the final ω.

ω_f = rad/s

3 of 7

A spinning skater has $I_i = 4 \text{ kg·m}^2$ and $\omega_i = 2 \text{ rad/s}$ . She reduces $I$ to $1 \text{ kg·m}^2$ . What is her new angular velocity?

Unlock Exercise 3