Definition of Angular Momentum

Angular momentum is one of the most fundamental conserved quantities in physics, describing the 'quantity of rotational motion' possessed by an object or system. Just as linear momentum $p = mv$ captures how much translational motion an object has, angular momentum captures how strongly an object is rotating or orbiting. It is a vector quantity, meaning it has both a magnitude and a direction — the direction being given by the axis around which the rotation occurs. Angular momentum appears everywhere in the physical world: a spinning figure skater pulling in their arms speeds up because their angular momentum is conserved, the Earth maintains its tilt as it orbits the Sun because of angular momentum, and subatomic particles possess intrinsic 'spin' angular momentum that determines their quantum behavior. Understanding angular momentum is essential for analysing everything from gyroscopes and satellites to the collapse of rotating stars into neutron stars. In this article we define angular momentum rigorously for both a single particle and an extended rigid body, derive the key equations, and explore why this quantity is so powerful. The concept connects intimately with torque (the rotational analogue of force) and with the deeper symmetry principle that the laws of physics are the same in all directions — a symmetry that, by Noether's theorem, guarantees the conservation of angular momentum in isolated systems.

Key takeaways

Angular Momentum of a Single Particle
The Perpendicular Distance (Moment Arm)
Angular Momentum of a Rigid Body Rotating About a Fixed Axis
Angular Momentum as a Vector for Rigid Bodies
Connection to Torque and Conservation

Angular Momentum of a Single Particle

Consider a particle of mass $m$ moving with velocity $\mathbf{v}$ at position $\mathbf{r}$ measured from a chosen reference point O. The linear momentum of the particle is $\mathbf{p} = m\mathbf{v}$ . The angular momentum of the particle about O is defined as the cross product: \mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m\mathbf{v} The magnitude is $L = rp\sin\theta = mvr\sin\theta$ , where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{v}$ . The direction of $\mathbf{L}$ is perpendicular to the plane formed by $\mathbf{r}$ and $\mathbf{v}$ , given by the right-hand rule: curl your fingers from $\mathbf{r}$ toward $\mathbf{p}$ , and your thumb points in the direction of $\mathbf{L}$ .

The Perpendicular Distance (Moment Arm)

A useful geometric interpretation: $r\sin\theta$ equals the perpendicular distance $r_\perp$ from the reference point O to the line of the particle's velocity (the 'moment arm'). Therefore $L = p \cdot r_\perp = mv \cdot r_\perp$ . For a planet orbiting the Sun in a circular orbit of radius $r$ , $\theta = 90°$ , so $L = mvr$ . This perpendicular distance interpretation is especially helpful for visualising angular momentum in orbital problems.

Angular Momentum of a Rigid Body Rotating About a Fixed Axis

For a rigid body rotating about a fixed axis with angular velocity $\omega$ , each infinitesimal mass element $dm_i$ at perpendicular distance $r_i$ from the axis has speed $v_i = r_i\omega$ and contributes angular momentum $dL_i = r_i(dm_i \cdot v_i) = r_i^2\,dm_i\,\omega$ . Summing (integrating) over the entire body: L = \left(\sum_i m_i r_i^2\right)\omega = I\omega where $I = \sum_i m_i r_i^2$ is the moment of inertia of the body about that axis. This is the rotational analogue of $p = mv$ , with $I$ replacing $m$ and $\omega$ replacing $v$ .

Angular Momentum as a Vector for Rigid Bodies

In full vector form, for rotation about a principal axis, $\mathbf{L} = I\boldsymbol{\omega}$ , where $\boldsymbol{\omega}$ is the angular velocity vector pointing along the rotation axis (again by the right-hand rule: curl the fingers in the direction of rotation and the thumb gives the direction of $\boldsymbol{\omega}$ ). For general rotation (not about a principal axis) the relationship becomes the tensor equation $\mathbf{L} = \mathbf{I}\,\boldsymbol{\omega}$ , but at high-school level we focus on fixed-axis rotation where $\mathbf{L} = I\boldsymbol{\omega}$ holds simply.

Connection to Torque and Conservation

Taking the time derivative of $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ for a particle: \frac{d\mathbf{L}}{dt} = \frac{d\mathbf{r}}{dt}\times\mathbf{p} + \mathbf{r}\times\frac{d\mathbf{p}}{dt} = \mathbf{v}\times m\mathbf{v} + \mathbf{r}\times\mathbf{F} = \boldsymbol{\tau} because $\mathbf{v}\times\mathbf{v} = \mathbf{0}$ . So $\boldsymbol{\tau} = d\mathbf{L}/dt$ — the net torque equals the rate of change of angular momentum, exactly mirroring Newton's second law $\mathbf{F} = d\mathbf{p}/dt$ . When the net external torque on a system is zero, $d\mathbf{L}/dt = 0$ , meaning $\mathbf{L}$ is constant — the law of conservation of angular momentum.

The Key Equation

$$\mathbf{L} = \mathbf{r} \times \mathbf{p} = I\boldsymbol{\omega}

- **$\mathbf{L}$** — Angular momentum (kg\cdot m²\cdot s⁻¹): Vector quantity describing the rotational motion of a particle or body about a reference point or axis. - **$\mathbf{r}$** — Position vector (m): Vector from the chosen reference point O to the particle's location. - **$\mathbf{p}$** — Linear momentum (kg\cdot m\cdot s⁻¹): Product of the particle's mass and velocity, $\mathbf{p} = m\mathbf{v}$. - **$I$** — Moment of inertia (kg\cdot m²): Measure of a rigid body's resistance to changes in rotational motion; depends on mass distribution relative to the axis. - **$\boldsymbol{\omega}$** — Angular velocity (rad\cdot s⁻¹): Vector describing the rate and axis of rotation of the rigid body. ## Worked Example **Problem.** A solid disc of mass $M = 2.0$ kg and radius $R = 0.30$ m spins about its central axis at $\omega = 40$ rad/s. Simultaneously, a small ball of mass $m = 0.50$ kg moves in a straight horizontal line at $v = 6.0$ m/s past a fixed point O. At the instant shown, the ball is at a perpendicular distance of $r_\perp = 0.80$ m from O. Calculate: (a) the angular momentum of the disc about its axis, and (b) the angular momentum of the ball about O.

$L_{\text{disc}} = 3.6$ kg·m²·s⁻¹; $\quad L_{\text{ball}} = 2.4$ kg·m²·s⁻¹

**Answer.** The disc has angular momentum 3.6 kg\cdot m²\cdot s⁻¹ about its axis; the ball has angular momentum 2.4 kg\cdot m²\cdot s⁻¹ about point O.

Figure

Diagram

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · English

Newton laid the mathematical foundations from which angular momentum is derived. His second law $\mathbf{F} = d\mathbf{p}/dt$ extends directly to the rotational form $\boldsymbol{\tau} = d\mathbf{L}/dt$ , and his law of universal gravitation (a central force directed along $\mathbf{r}$ ) implies zero torque about the Sun — explaining Kepler's second law (equal areas in equal times) as a consequence of angular momentum conservation. Newton reportedly developed his most productive ideas — including calculus, optics, and gravity — during 1665–1666 when Cambridge University was closed due to the Great Plague, a period he called his 'annus mirabilis' (year of wonders).

Moment of Inertia — Moment of inertia $I$ appears directly in the rigid-body formula $L = I\omega$ and determines how angular momentum is distributed in a rotating object.
Conservation of Angular Momentum — When net external torque is zero, $\mathbf{L}$ remains constant — one of the most powerful principles in physics, explaining spinning figure skaters and planetary orbits.
Torque — Torque $\boldsymbol{\tau} = d\mathbf{L}/dt$ is the rotational analogue of force and is the agent that changes angular momentum, directly paralleling $\mathbf{F} = d\mathbf{p}/dt$ .
Linear Momentum — Angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is built from linear momentum $\mathbf{p} = m\mathbf{v}$ ; understanding translational momentum makes the rotational analogy intuitive.
Angular Velocity and Angular Acceleration — Angular velocity $\boldsymbol{\omega}$ appears in $\mathbf{L} = I\boldsymbol{\omega}$ and must be understood to compute angular momentum for rotating rigid bodies.