ANGULAR MOMENTUM OF A PARTICLE ABOUT A POINT AND ABOUT AN AXIS

Angular momentum: The turning movement of a particle about the axis of rotation is called the angular momentum of the particle and is measured as the product of the linear momentum and the perpendicular distance of its lien of action from the axis of rotation. It is denoted by L.

Therefore, Angular momentum = linear momentum x perpendicular distance from the axis of rotation

In SI, the unit of angular momentum is kg m² s^–1. Its dimensional formula is [M L² T^–1]

Angular Momentum of A Particle About A Point

Suppose a particle A of mass m is moving with linear momentum $\overrightarrow{P}=m\overrightarrow{v}$. Its angular momentum $\overrightarrow{L}$ about point 0 is defined as: $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{P}=\overrightarrow{r}\times \left(m\overrightarrow{v}\right)=m\left(\overrightarrow{r}\times \overrightarrow{v}\right)$

Here, $\overrightarrow{r}$ is the radius vector of particle A about 0 at that instant of time. The magnitude of $\overrightarrow{L}$ is

L = mvr sin$\theta$= mvr_$\perp$

Here, r_$\perp$= r sin$\theta$is the perpendicular distance of line of action of

Velocity $\overrightarrow{v}$ from point O. The direction of $\overrightarrow{L}$ is same as that of $\overrightarrow{r}\times \overrightarrow{v}$.

Illustration 1: A particle of mass m is projected from origin O with speed u at an angle$\theta$with positive x-axis is in vertically upward direction. Find the angular momentum of particle at any time t about O before the particle strikes the ground again.

Solution: $\overrightarrow{L}=m\left(\overrightarrow{r}\times \overrightarrow{v}\right)$

Here, $\overrightarrow{r}\left(t\right)=x\hat{i}+y\hat{j}$

$=\left(u\cos \theta \right)t\hat{i}+\left(ut\sin \theta -\frac{1}{2}g{t}^2\right)\hat{j}$

and $\overrightarrow{v}\left(t\right)=v_x\hat{i}+v_y\hat{j}$= (u cos $\theta$) $\hat{i}$ + (u sin$\theta$– gt) $\hat{j}$

$\therefore$ $\overrightarrow{L}=m\left(\overrightarrow{r}\times \overrightarrow{v}\right)$

$=m\left|\begin{array}{c}\hat{i}\\ \left(u\cos \theta \right)t\\ u\cos \theta \end{array}\begin{array}{c}\hat{j}\\ \left(u\sin \theta \right)t-\frac{1}{2}g{t}^2\\ u\sin \theta -gt\end{array}\begin{array}{c}\hat{k}\\ 0\\ 0\end{array}\right|$= m[(u² sin$\theta$cos $\theta$) t – (u cos $\theta$) gt²

$-\left(u^2\sin \theta \cos \theta \right)t+\frac{1}{2}\left(u\cos \theta \right)g{t}^2]\hat{k}$ $=-\frac{1}{2}m\left(u\cos \theta \right)g{t}^2\hat{k}$

Angular Momentum of A Rigid Body About An Axis

Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen anywhere on the axis. We choose it at the centre of the circle. In this case $\overrightarrow{r}$ and $\overrightarrow{P}$ are perpendicular to each other and $\overrightarrow{r}\times \overrightarrow{P}$ is along the axis. Thus, component of $\overrightarrow{r}\times \overrightarrow{P}$ along the axis is mvr itself. The angular momentum of the whole rigid body about AB is sum of components of all particles, i.e.

$L=\sum _i{m}_i{r}_i{v}_i$

Here, v_i = r_i$\omega$

$\therefore L=\sum _i{m}_i{r}_i^2\omega$ or $L=\omega \sum _i{m}_i{r}_i^2$ or L = I$\omega$ (as $\sum _i{m}_i{r}_i^2=I$)

Relation Between Torque and Angular Momentum

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

$\Rightarrow$ $\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r}\times \frac{d\vec{p}}{dt}=\vec{v}\times \left(m\vec{v}\right)+\vec{r}\times \vec{F}$ = 0 + $\vec{r}\times \vec{F}={\vec{\tau }}_{ext}$ $\Rightarrow$ $\frac{d\vec{L}}{dt}={\vec{\tau }}_{ext}$

This relation is analogous to $\vec{F}=m\vec{a}$, which is applied in rotation.

Conservation of Angular Momentum

We know that $\overrightarrow{\tau }=\frac{\overrightarrow{dL}}{dt}$

When there is no net external torque acting on a particle, then $\frac{d\vec{L}}{dt}=0$. $\Rightarrow$ $\vec{L}$=constant

Therefore, the angular momentum of the particle remains invariant in the absence of any net external torque.

Angular Impulse

The angular impulse of a torque in a given time interval is defined as $\underset{t_1}{\overset{t_2}{\int }}\overrightarrow{\tau }dt$

Hence, $\overrightarrow{\tau }$ is the resultant torque acting on the body. Further, since$\overrightarrow{\tau }=\frac{\overrightarrow{dL}}{dt}\therefore \overrightarrow{\tau }dt=\overrightarrow{dL}$

Or $\underset{t_1}{\overset{t_2}{\int }}\overrightarrow{\tau }dt=angularimpulse={\overrightarrow{L}}_2-{\overrightarrow{L}}_1$

Thus, the angular impulse of the resultant torque is equal to the change in angular momentum. Let us take an example based on the angular impulse.

Illustration 2: The torque $\tau$on a body about a given point is found to be equal to A L where A is constant vector, and L is the angular momentum of the body about that point. From this it follows that

(A) $\frac{dL}{dt}$ is perpendicular to L at all instants of time

(B) the component of L in the direction of A does not change with time.

(C) the magnitude of L does not change with time.

(D) L does not change with time

Solution: (A, B, C) $\vec{\tau }=\frac{d\overrightarrow{L}}{dt}$

Given that, $\vec{\tau }=\vec{A}\times \vec{L}$ $\Rightarrow$ $\frac{d\overrightarrow{L}}{dt}=\vec{A}\times \vec{L}$

From cross–product rule, $\frac{\overrightarrow{dL}}{dt}$ is always perpendicular to the plane containing $\vec{A}$ and $\vec{L}$

By the dot product definition $\vec{L}.\vec{L}=L^2$

Differentiating with respect to time $\vec{L}.\frac{d\vec{L}}{dt}+\vec{L}.\frac{d\vec{L}}{dt}=2L\frac{dL}{dt}$

$\Rightarrow$ $2\vec{L}.\frac{d\vec{L}}{dt}=2L\frac{dL}{dt}$

Since $\frac{d\vec{L}}{dt}$ is perpendicular to $\vec{L}$ $\Rightarrow$ $\frac{dL}{dt}$ = 0 $\Rightarrow$ L = constant