Angular Momentum of a Particle

Angular Momentum of a Particle

The idea of angular momentum is the rotational analogue of linear momentum. Just as linear momentum ($\mathbf{p} = m\mathbf{v}$) tells us how much "oomph" a moving object has in a straight line, angular momentum tells us how much "rotational oomph" a particle has about a chosen point.

Consider a particle of mass $m$ moving with velocity $\mathbf{v}$. Its linear momentum is $\mathbf{p} = m\mathbf{v}$. Now pick a fixed point $O$ in space. Let $\mathbf{r}$ be the position vector of the particle measured from $O$. The angular momentum $\mathbf{l}$ of the particle about the point $O$ is defined as the cross product of $\mathbf{r}$ and $\mathbf{p}$:

Since $\mathbf{p} = m\mathbf{v}$, we can also write:

The SI unit of angular momentum is $\text{kg m}^2 \text{s}^{-1}$, and its dimensions are $[ML^2T^{-1}]$.

The direction of $\mathbf{l}$ is given by the right-hand rule for the cross product $\mathbf{r} \times \mathbf{p}$. If the particle moves in a straight line that does not pass through $O$, $\mathbf{l}$ is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{p}$.

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Relation Between Torque and Angular Momentum

Just as force causes a change in linear momentum ($\mathbf{F} = d\mathbf{p}/dt$), torque causes a change in angular momentum. Differentiate $\mathbf{l} = \mathbf{r} \times \mathbf{p}$ with respect to time:

Now, $\frac{d\mathbf{r}}{dt} = \mathbf{v}$ and $\mathbf{p} = m\mathbf{v}$, so the first term becomes:

because the cross product of any vector with itself is zero. The second term uses Newton's second law: $\frac{d\mathbf{p}}{dt} = \mathbf{F}$, where $\mathbf{F}$ is the net force on the particle. So:

But $\mathbf{r} \times \mathbf{F}$ is precisely the torque $\boldsymbol{\tau}$ about the same point $O$. Therefore:

This is the rotational analogue of $\mathbf{F} = d\mathbf{p}/dt$. It holds for any fixed point $O$ as long as both torque and angular momentum are measured about that same point.

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Properties of Angular Momentum

The textbook lists three important properties that follow directly from the definition and the torque relation.

Property (I): Angular momentum about a point for motion along a straight line

If a particle moves in a straight line, its angular momentum about any point on that line is zero. About any other point, it is non-zero and constant in magnitude if the speed is constant.

Proof: Let the particle move along a straight line. Choose a point $O$ on that line. Then the position vector $\mathbf{r}$ of the particle is always along the line of motion, and the velocity $\mathbf{v}$ is also along that line. Hence $\mathbf{r}$ and $\mathbf{v}$ are parallel (or anti-parallel). The cross product $\mathbf{r} \times \mathbf{v} = 0$, so $\mathbf{l} = m(\mathbf{r} \times \mathbf{v}) = 0$.

If $O$ is not on the line, then $\mathbf{r}$ and $\mathbf{v}$ are not parallel. However, the magnitude $l = m r v \sin\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{v}$. For uniform motion along a straight line, $r \sin\theta$ is the perpendicular distance from $O$ to the line, which is constant. So $l$ is constant. Since the direction of $\mathbf{l}$ (perpendicular to the plane of $\mathbf{r}$ and $\mathbf{v}$) also remains fixed, $\mathbf{l}$ is constant. Consequently, $d\mathbf{l}/dt = 0$, implying $\boldsymbol{\tau} = 0$, which is consistent because the net force on a particle moving with constant velocity is zero.

Property (II): Angular momentum in circular motion

For a particle moving in a circle of radius $r$ with constant speed $v$, the angular momentum about the centre of the circle is constant in magnitude and direction.

Proof: Take the centre of the circle as the reference point $O$. The position vector $\mathbf{r}$ is always radial outward from $O$, and the velocity $\mathbf{v}$ is tangential. They are perpendicular: $\mathbf{r} \perp \mathbf{v}$. The magnitude of angular momentum is:

Since $r$ and $v$ are constant, $l$ is constant. The direction of $\mathbf{l} = \mathbf{r} \times \mathbf{v}$ is perpendicular to the plane of the circle (by the right-hand rule, it points along the axis of rotation). As the particle moves, $\mathbf{r}$ and $\mathbf{v}$ rotate together, but their cross product always points in the same fixed direction (out of the plane for anticlockwise motion, into the plane for clockwise motion). Hence $\mathbf{l}$ is constant. …