Angular Momentum

Two forms

For a particle:

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

Magnitude: $L = rp\sin\theta = rmv\sin\theta$

where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$.

For a rigid body rotating about a fixed axis:

$L = I\omega$

Particle angular momentum

For a particle at position $\vec{r}$ with momentum $\vec{p} = m\vec{v}$:

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

Direction: Perpendicular to both $\vec{r}$ and $\vec{v}$, by the right-hand rule.

For circular motion: $\vec{r} \perp \vec{v}$ (radius perpendicular to tangential velocity), so $\theta = 90°$:

$L = rmv\sin90° = rmv = mr^2\omega$

This is consistent with $L = I\omega$ for a point mass ($I = mr^2$).

Rigid body angular momentum

For a rigid body rotating about a fixed axis with angular velocity $\omega$:

$L = I\omega$

This follows from summing $L_i = m_i r_i^2 \omega$ over all elements:

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

Units

$[L] = \text{kg} \cdot \text{m}^2/\text{s} = \text{J} \cdot \text{s}$

Relation to torque

$\vec{\tau} = \frac{d\vec{L}}{dt}$

Torque is the rate of change of angular momentum — exactly analogous to $\vec{F} = \frac{d\vec{p}}{dt}$.

Angular momentum of a planet

A planet in orbit has angular momentum $L = mvr\sin\theta$ where $\theta$ is the angle between the radius vector and velocity. For an elliptical orbit, both $v$ and $r$ change, but their product $vr\sin\theta$ remains constant (conservation of angular momentum) — this is Kepler's Second Law.

The complete analogy