Question

1. Show that the rate of change of the total angular momentum of a system of particles is equal to the resultant torque exerted by all external forces which act on the system. 

Answer 1

Let the particle of mass m, is moving with velocity $\overrightarrow{v}$ then linear momentum of the particle will be $\overrightarrow{p}$ .and particle is at a distance $\overrightarrow{r}$ from the origin.

Let the angular momentum of the particle is $\overrightarrow{L}.$

So, angular momentum of the particle will be $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

$\Rightarrow \overrightarrow{L}=|\overrightarrow{r}||\overrightarrow{p}|\sin\theta \hat{n}$

where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{p}$ and $\hat{n}$ is the unit vector perpendicular to $\overrightarrow{r}$ and $\overrightarrow{p}$

Now, taking the time derivative of $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

$\Rightarrow \frac{d\overrightarrow{L}}{dt}=\frac{d \overrightarrow{r}}{dt}\times \overrightarrow{p}+\overrightarrow{r}\times \frac{{d\overrightarrow{p}}}{dt}$

from newton's second law of motion,$\Sigma \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$

Hence, $\frac{d\overrightarrow{L}}{dt}=\overrightarrow{r}\times \overrightarrow{F}$

But we know that $\overrightarrow{\tau}= \overrightarrow{r}\times \overrightarrow{F}$

Hence, $\overrightarrow{\tau}=\frac{d\overrightarrow{L}}{dt}$