Angular Momentum

A spin bowler's arm sweeps through a wide arc. The cricket ball at the fingertip — 0.7 metres from the shoulder joint — reaches 25 m/s at the instant of release. Linear momentum, p = mv, tells you how hard the ball will hit the stumps. But it says nothing about the arc. Nothing about the shoulder as a pivot, or the arm as a rotating lever.

You already know that torque is the rotational analogue of force. You need a rotational analogue of momentum too — a quantity that captures not just how fast something moves, but how far from the axis it is moving. That quantity is angular momentum. And it unlocks the same kind of reasoning: just as Newton's second law says force equals the rate of change of linear momentum, the rotational version says torque equals the rate of change of angular momentum. Master that single relation and every rotation problem in physics becomes a momentum problem.

The rotational analogue of momentum

The entire structure of rotational mechanics mirrors translational mechanics, quantity by quantity:

Angular momentum sits in the third row: the rotational quantity that plays the same role momentum plays in translation. The fifth row — \tau = dL/dt — is the result you will derive in this article. It is the single most important equation in rotational dynamics.

Angular momentum of a particle

Start with a single particle of mass m at position \vec{r} from a fixed point O, moving with velocity \vec{v}. Its linear momentum is \vec{p} = m\vec{v}.

Reading the definition. The cross product \vec{r} \times \vec{p} produces a vector perpendicular to both \vec{r} and \vec{p}. Its magnitude is:

where \theta is the angle between \vec{r} and \vec{v}. The quantity r\sin\theta is the perpendicular distance from O to the line along which the particle moves — call it r_\perp. So:

Why: only the component of \vec{r} perpendicular to the velocity contributes to angular momentum. If the particle moves directly toward or away from O (\theta = 0° or 180°), then r_\perp = 0 and the angular momentum is zero — there is no rotation about O.

Direction: the right-hand rule

The direction of \vec{L} follows the right-hand rule: curl the fingers of your right hand from \vec{r} toward \vec{p}, and your thumb points along \vec{L}. For a body spinning counterclockwise (viewed from above), \vec{L} points upward along the axis of rotation. For clockwise spin, it points downward.

This vector nature becomes critical in advanced problems — a spinning top that wobbles, a gyroscope that precesses — because the direction of \vec{L} can change even when its magnitude does not. For now, when the axis of rotation is fixed, the direction is simply along the axis.

A straight line has angular momentum too

Here is something that surprises most students: a particle moving in a straight line — not rotating at all — has non-zero angular momentum about any point O that is not on its path.

Picture a cricket ball rolling along the ground at 30 m/s. Choose a point O that is 3 metres to the side of the ball's path (the perpendicular distance). The ball's mass is 0.16 kg. Its angular momentum about O is:

And this angular momentum stays constant as the ball rolls. The perpendicular distance r_\perp does not change (the ball moves parallel to itself, never approaching or receding from O along the perpendicular). No torque acts, so dL/dt = 0, and L remains 14.4 kg·m²/s throughout the motion.

This is not a trick. Angular momentum is always computed about a specific point, and a straight-line path that misses that point creates a non-zero cross product \vec{r} \times \vec{p} at every instant. The angular momentum about a point on the path, by contrast, is zero — because r_\perp = 0.

For circular motion: the clean case

When a particle moves in a circle of radius R about a centre O, the position vector \vec{r} is always perpendicular to the velocity \vec{v} (the velocity is tangent to the circle). So \theta = 90°, \sin\theta = 1, and:

Since v = R\omega (the relation between linear and angular speed for circular motion):

Why: for circular motion, the full radius contributes to angular momentum. The factor mR^2 is the particle's moment of inertia about the centre, so this is already L = I\omega.

Angular momentum of a rigid body

A rigid body is a collection of particles locked together. When the body rotates about a fixed axis with angular velocity \omega, every particle moves in a circle about that axis. The i-th particle, at perpendicular distance r_i from the axis, has angular momentum:

The total angular momentum about the axis is the sum over all particles:

That sum \sum m_i r_i^2 is the moment of inertia I about the axis.

Why: this is the rotational counterpart of p = mv. Replace mass with moment of inertia, velocity with angular velocity, and linear momentum becomes angular momentum. The structure of the analogy is exact.

The SI unit of angular momentum is kg·m²/s (equivalently, J·s or N·m·s). This same unit appears in quantum mechanics as the unit of Planck's reduced constant \hbar — angular momentum is fundamental enough to set the scale of the quantum world.

Explore: how angular momentum depends on radius

For a particle of mass m moving in a circle at angular velocity \omega, the angular momentum is L = m\omega r^2. Notice the r^2 — doubling the radius quadruples the angular momentum. This is why mass far from the axis dominates the angular momentum of a rotating body.

Torque as the rate of change of angular momentum

This is the central result of the article. Just as \vec{F} = d\vec{p}/dt connects force to linear momentum, you can derive that \vec{\tau} = d\vec{L}/dt connects torque to angular momentum.

Here is the derivation for a single particle.

Step 1. Start with the definition of angular momentum.

Why: this is the angular momentum you defined earlier — the quantity whose rate of change you want to find.

Step 2. Differentiate both sides with respect to time, using the product rule for cross products.

Why: the cross product obeys the same product rule as ordinary multiplication, but order matters — \vec{r} stays on the left, \vec{p} on the right.

Step 3. Evaluate the first term. Since \frac{d\vec{r}}{dt} = \vec{v} and \vec{p} = m\vec{v}:

Why: the cross product of any vector with itself is zero. This is the key step — the velocity term drops out entirely, leaving only the force term.

Step 4. Evaluate the second term. By Newton's second law, \frac{d\vec{p}}{dt} = \vec{F}, the net force on the particle. And \vec{r} \times \vec{F} is the definition of torque about O:

Why: this connects force (the linear quantity) to torque (the rotational quantity) through the same position vector \vec{r} that defines angular momentum.

Step 5. Combine the two terms.

Why: torque is what changes angular momentum, at a rate exactly equal to the applied torque. No torque → angular momentum stays constant. A constant torque → angular momentum changes at a steady rate. This is the rotational Newton's second law in its most general form.

For a rigid body rotating about a fixed axis with constant moment of inertia:

This recovers the \tau = I\alpha relation from Newton's Second Law for Rotation — now revealed as a special case of the more general \vec{\tau} = d\vec{L}/dt.

What changes when I is not constant? If the moment of inertia changes with time — say, a figure skater pulling in their arms — then \tau = dL/dt = d(I\omega)/dt does not simplify to I\alpha. You must use the full \tau = dL/dt and track how both I and \omega change together. This is exactly the situation that Conservation of Angular Momentum handles.

Worked examples

Common confusions

• "Angular momentum requires circular motion." No. Any moving particle has angular momentum about any point O that is not on its line of motion. A cricket ball flying in a straight line past the stumps has angular momentum about the base of the stumps. Angular momentum is computed about a reference point, and straight-line paths qualify.

• "L = I\omega works for every situation." Only for a rigid body rotating about a fixed axis. For a single particle moving in a straight line, you must use \vec{L} = \vec{r} \times \vec{p}. The formula L = I\omega is a special case, valid when every part of the body moves in a circle about the same axis.

• "Angular momentum and torque have the same units." They do not. Torque is N·m = kg·m²/s². Angular momentum is kg·m²/s. The difference is one factor of seconds, consistent with \tau = dL/dt — torque is angular momentum per unit time.

• "The direction of \vec{L} is the direction the body moves." No. \vec{L} is perpendicular to the plane of motion, along the axis of rotation. A top spinning on a table spins in the horizontal plane, but \vec{L} points vertically. This is the cross-product structure at work: \vec{r} \times \vec{p} is always perpendicular to both \vec{r} and \vec{p}.

• "If the net torque is zero, the body stops rotating." Exactly backwards. Zero net torque means dL/dt = 0, so angular momentum does not change — the body keeps rotating at the same rate indefinitely. A spinning wheel in space (no friction, no torque) spins forever. This is the rotational analogue of Newton's first law.

Where this leads next

• Conservation of Angular Momentum — when the net external torque is zero, \vec{L} stays constant. This explains figure skaters speeding up as they pull in their arms, collapsing stars spinning faster, and Kepler's second law.
• Torque — the rotational analogue of force, and the quantity that changes angular momentum.
• Newton's Second Law for Rotation — the fixed-axis special case \tau = I\alpha, now understood as a consequence of \tau = dL/dt with constant I.
• Moment of Inertia — the factor that links \omega to L for a rigid body, and the reason mass distribution matters.
• Gyroscopic Precession — what happens when \vec{\tau} is perpendicular to \vec{L}, changing its direction instead of its magnitude.