Question

State and prove principle of conservation of angular momentum.

Answer 1

Let $\overrightarrow{L}$ be the angular momentum.

It is the vector product of radius vector and linear momentum.$\overrightarrow{r}\times \overrightarrow{p}$

Please note that you should not say it as a product of linear momentum and radius vector, because vector product is not commutative.

Rate of change of $\overrightarrow{L}$ with respect to time $t$ is given by

$\frac{d\overrightarrow{L}}{dt}=\frac{d(\overrightarrow{r}\times \overrightarrow{p})}{dt}$

$=(\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt})+(\overrightarrow{p}\times \frac{d\overrightarrow{r}}{dt})$

$=(\overrightarrow{r}\times \overrightarrow{F})+(\overrightarrow{p}\times \overrightarrow{{}_V})$ $\because$ rate of change of linear momentum is force and rate of change of radius vector is velocity.

$\therefore \frac{d\overrightarrow{L}}{dt}=\overrightarrow{\tau }+m(\overrightarrow{{}_V}\times \overrightarrow{{}_V})$ here $\overrightarrow{r}\times \overrightarrow{F}=$ torque , $\overrightarrow{p}=m\overrightarrow{v}$

$\therefore \frac{d\overrightarrow{L}}{dt}=\overrightarrow{\tau }$ cross product of any vector with itself is zero.

$\therefore$ if $\tau =0$ then $\frac{d\overrightarrow{L}}{dt}$

This means that in the absence of any external torque, the total angular momentum of a rotating body is conserved.

This can be mathematically written as $I_1{\omega }_1=I_2{\omega }_2$ if $\tau =0$

Here $I$ is moment of inertia and $\omega$ is angular velocity.

We know that $I\omega =L$