Torque and Angular Momentum

Torque and Angular Momentum

The connection between force and linear motion is straightforward — force causes linear acceleration. For rotational motion, the analogous quantity is torque, which causes angular acceleration. But torque alone doesn't tell the whole story. Just as a moving body carries linear momentum, a rotating body carries angular momentum. The two are linked by a rotational version of Newton's second law.

Defining Torque

Consider a force $\mathbf{F}$ acting on a particle at position $\mathbf{r}$ relative to some origin O. The torque (or moment of force) about O is defined as the vector cross product:

The magnitude of torque is $|\boldsymbol{\tau}| = rF\sin\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$. This magnitude equals the product of the force and the perpendicular distance from the origin to the line of action of the force — that perpendicular distance is called the lever arm or moment arm.

The direction of $\boldsymbol{\tau}$ is given by the right-hand rule: curl the fingers of your right hand from $\mathbf{r}$ toward $\mathbf{F}$, and your thumb points along $\boldsymbol{\tau}$. Torque is perpendicular to both $\mathbf{r}$ and $\mathbf{F}$.

Defining Angular Momentum

For a single particle of mass $m$, moving with velocity $\mathbf{v}$ at position $\mathbf{r}$ relative to an origin, the angular momentum $\mathbf{l}$ about that origin is:

where $\mathbf{p} = m\mathbf{v}$ is the linear momentum. The magnitude is $l = rmv\sin\theta = rp\sin\theta$, with $\theta$ the angle between $\mathbf{r}$ and $\mathbf{p}$.

The SI unit of angular momentum is $\text{kg m}^2/\text{s}$. Its direction, like torque, follows the right-hand rule and is perpendicular to both $\mathbf{r}$ and $\mathbf{p}$.

The Rotational Form of Newton's Second Law

Differentiate angular momentum with respect to time:

Now $\frac{d\mathbf{r}}{dt} = \mathbf{v}$ and $\mathbf{p} = m\mathbf{v}$, so $\frac{d\mathbf{r}}{dt} \times \mathbf{p} = \mathbf{v} \times (m\mathbf{v}) = m(\mathbf{v} \times \mathbf{v}) = 0$ because the cross product of any vector with itself is zero.

The second term: $\mathbf{r} \times \frac{d\mathbf{p}}{dt} = \mathbf{r} \times \mathbf{F}_{\text{net}}$, where $\mathbf{F}_{\text{net}}$ is the net force on the particle. But $\mathbf{r} \times \mathbf{F}_{\text{net}}$ is exactly the net torque $\boldsymbol{\tau}_{\text{net}}$ about the origin.

Therefore:

This is the rotational analogue of $\mathbf{F}_{\text{net}} = d\mathbf{p}/dt$. The net torque on a particle equals the time rate of change of its angular momentum.

Conservation of Angular Momentum

From $\boldsymbol{\tau}_{\text{net}} = d\mathbf{l}/dt$, a direct consequence follows:

When the net external torque on a particle (or system of particles) is zero, its angular momentum is conserved — both magnitude and direction remain constant.

This is the law of conservation of angular momentum, one of the fundamental conservation laws of physics. It holds for any system, from a single particle to a galaxy, provided no external torque acts.

Properties of Angular Momentum

The textbook lists several key properties that follow from the definitions:

Property 1: For a particle moving with constant velocity along a straight line, angular momentum about any point on that line is zero. About any other point, it is constant in magnitude and direction.

Proof: Let the particle move along a straight line with constant velocity $\mathbf{v}$. Its position vector $\mathbf{r}$ from any origin changes with time, but $\mathbf{l} = \mathbf{r} \times m\mathbf{v}$. Since $\mathbf{v}$ is constant, $\mathbf{l}$ changes only if $\mathbf{r}$ changes in a way that alters the cross product. For a point on the line of motion, $\mathbf{r}$ is always parallel to $\mathbf{v}$, so $\mathbf{r} \times \mathbf{v} = 0$. For a point off the line, the perpendicular distance from the origin to the line is constant, and the direction of $\mathbf{l}$ (perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$) is fixed. Hence $\mathbf{l}$ is constant.

Property 2: For a particle in uniform circular motion, angular momentum about the centre of the circle is constant in magnitude and direction.

Proof: For circular motion, $\mathbf{r}$ is always perpendicular to $\mathbf{v}$ (velocity is tangential). So $l = rmv = mr^2\omega$. Since $r$ and $\omega$ are constant, magnitude is constant. The direction of $\mathbf{l}$ is perpendicular to the plane of motion (along the axis of rotation), which is fixed. Hence $\mathbf{l}$ is constant.

Property 3: Angular momentum obeys the superposition principle — for a system of particles, the total angular momentum $\mathbf{L}$ is the vector sum of individual angular momenta:

Property 4: The torque on a system of particles equals the rate of change of its total angular momentum:

where $\boldsymbol{\tau}_{\text{ext}}$ is the net external torque. Internal torques (due to internal forces) cancel in pairs because they are equal, opposite, and act along the same line — their vector sum is zero.

›Proof

For a system of $n$ particles, the total angular momentum is $\mathbf{L} = \sum \mathbf{r}_i \times \mathbf{p}_i$. Differentiate: …