2.5.5.1. Angular Momentum

Table of Contents

Definition and Physical Meaning

Angular momentum is the rotational counterpart of linear momentum. Linear momentum tells us how hard it is to stop straight line motion. Angular momentum tells us how hard it is to change rotational motion about a point or axis.

For a single particle, angular momentum depends on both its motion and its position relative to a chosen origin. If a particle of momentum $\vec{p}$ is at position $\vec{r}$, then its angular momentum is

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

where $\times$ is the cross product.

This means angular momentum is a vector. Its magnitude is

$$
L = rp\sin\theta
$$

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

If the particle moves directly toward or away from the origin, then $\theta = 0$ or $\pi$, so $\sin\theta = 0$, and the angular momentum is zero. If the motion is perpendicular to the radius, angular momentum is maximum.

For a particle about a chosen origin,
$$
\vec{L} = \vec{r} \times \vec{p}
$$
and
$$
L = rp\sin\theta
$$
Angular momentum depends on the choice of origin.

Direction of Angular Momentum

Because angular momentum is defined by a cross product, its direction is perpendicular to the plane formed by $\vec{r}$ and $\vec{p}$. The direction is found using the right hand rule.

If the particle moves counterclockwise in a plane, the angular momentum points out of the page. If it moves clockwise, the angular momentum points into the page.

This direction matters because rotation is not described fully by a number alone. The axis and sense of rotation are important.

Rendering scientific sketch…

Direction of angular momentum for planar motion

Angular Momentum of a Particle in Circular Motion

A simple and important case is a particle moving in a circle of radius $r$ with speed $v$. Since the momentum is tangent to the circle and the radius points outward, the angle between $\vec{r}$ and $\vec{p}$ is $90^\circ$. Then

$$
L = rp = r(mv) = mvr
$$

This is the angular momentum of a particle about the center of the circle.

If the motion is not exactly circular but the particle still has a perpendicular component of velocity, only that perpendicular part contributes. If $v_\perp$ is the component of velocity perpendicular to $\vec{r}$, then

$$
L = mr v_\perp
$$

Angular Momentum of a Rigid Body About a Fixed Axis

For a rigid body rotating about a fixed axis, each particle in the body has angular momentum. The total angular momentum is the sum over all particles.

In many common cases, such as a rigid body rotating about a symmetry axis, the total angular momentum is

$$
L = I\omega
$$

where $I$ is the moment of inertia about the axis and $\omega$ is the angular velocity.

This looks similar to the linear relation $p = mv$. Here, moment of inertia plays a role similar to mass in rotational motion.

For a rigid body rotating about a fixed axis,
$$
L = I\omega
$$
This form is valid when the body rotates about that axis and $I$ is taken about the same axis.

Comparing Linear and Angular Momentum

It is useful to compare translational and rotational quantities.

Translational motion	Rotational motion
Mass $m$	Moment of inertia $I$
Velocity $v$	Angular velocity $\omega$
Momentum $p = mv$	Angular momentum $L = I\omega$
Force changes momentum	Torque changes angular momentum

This parallel helps build intuition. A large mass gives large linear momentum for a given speed. A large moment of inertia gives large angular momentum for a given angular speed.

Units of Angular Momentum

From $\vec{L} = \vec{r} \times \vec{p}$, the SI units are

$$
\text{m} \cdot \frac{\text{kg}\cdot\text{m}}{\text{s}} = \frac{\text{kg}\cdot\text{m}^2}{\text{s}}
$$

So the SI unit of angular momentum is

$$
\mathrm{kg \cdot m^2/s}
$$

This is the same as the unit of action or torque multiplied by time, but angular momentum is a different physical quantity from torque.

Angular Momentum About Different Origins

Angular momentum always has to be measured about some chosen point. Changing the origin can change both the magnitude and direction of angular momentum.

For example, a particle moving in a straight line may have zero angular momentum about one point, but nonzero angular momentum about another point not on its line of motion.

This is why problems must clearly state the point or axis about which angular momentum is calculated.

Rendering scientific sketch…

Angular momentum depends on the chosen origin

About $O_1$, the perpendicular distance to the line of motion may be simple to identify. About $O_2$, it is different. So the angular momentum is different.

Angular Momentum and the Perpendicular Distance

The magnitude of angular momentum can also be written using the perpendicular distance from the origin to the line of motion. If $b$ is that perpendicular distance, often called the lever arm for momentum, then

$$
L = pb
$$

This form is very useful in geometry based problems. It says that angular momentum depends on how far the motion passes from the chosen point.

If a particle passes closer to the origin, the angular momentum is smaller. If it passes farther away, the angular momentum is larger.

For a particle moving with momentum magnitude $p$, the angular momentum magnitude about an origin can be written as
$$
L = pb
$$
where $b$ is the perpendicular distance from the origin to the line of motion.

Total Angular Momentum of a System

If several particles are present, the total angular momentum about a point is the vector sum of the angular momenta of all particles:

$$
\vec{L}_{\text{total}} = \sum_i \vec{L}_i = \sum_i \vec{r}_i \times \vec{p}_i
$$

Some parts may point in opposite directions and cancel. Others may add together.

For example, two equal masses rotating in opposite directions about the same axis can produce zero total angular momentum if their individual angular momenta are equal in magnitude and opposite in direction.

Relation to Torque

Torque is what changes angular momentum. For a particle or a system, the net external torque is related to the rate of change of angular momentum:

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

This is one of the most important equations in rotational dynamics.

If the net external torque is zero, then the angular momentum does not change.

The rotational form of Newton's second law is
$$
\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}
$$
If
$$
\vec{\tau}_{\text{net}} = 0
$$
then angular momentum is constant.

The full idea of conservation is treated separately, but the connection is already clear here. No net external torque means no change in angular momentum.

Special Case, Particle Moving in a Plane

In many beginner problems, motion happens in the $xy$ plane. Then the angular momentum vector is along the $z$ direction. Instead of writing the full vector, we often use a signed scalar:

$$
L_z = x p_y - y p_x
$$

Positive values usually mean out of the page, and negative values mean into the page.

This is a convenient shortcut for two dimensional motion.

Example

Consider a particle of mass $2\,\mathrm{kg}$ moving at $3\,\mathrm{m/s}$ in a circle of radius $4\,\mathrm{m}$ about the origin. Its angular momentum magnitude is

$$
L = mvr = (2)(3)(4) = 24\,\mathrm{kg\cdot m^2/s}
$$

If the motion is counterclockwise, the direction is out of the page.

If instead the same particle moves in a straight line with momentum magnitude $6\,\mathrm{kg\cdot m/s}$ and the perpendicular distance from the origin to its line of motion is $4\,\mathrm{m}$, then

$$
L = pb = (6)(4) = 24\,\mathrm{kg\cdot m^2/s}
$$

So the same angular momentum magnitude can arise in different kinds of motion.

Common Points to Remember

Angular momentum is not only for spinning objects. Even a single moving particle can have angular momentum if its line of motion does not pass through the chosen origin.

Angular momentum is a vector quantity, so direction is essential.

Angular momentum depends on the reference point.

For rigid bodies rotating about a fixed axis, the simple formula $L = I\omega$ is often the most useful.

Key facts:
$$
\vec{L} = \vec{r} \times \vec{p}
$$
$$
L = mvr \quad \text{for circular motion}
$$
$$
L = I\omega \quad \text{for a rigid body about a fixed axis}
$$
$$
\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}
$$
Always specify the origin or axis when discussing angular momentum.

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