Obtain the relation between torque of a system of particles and angular moment. 

The total angular moment of a system of particles is the angular moment of individual particles. For a system of $n$ particles,

$\overrightarrow{\mathrm{L}}=\overrightarrow{l_{1}}+\overrightarrow{l_{2}}+\overrightarrow{l_{3}}+\ldots \overrightarrow{l_{n}}$

$=\sum_{i=1}^{n} \overrightarrow{l_{i}} \text { where } i=1,2,3, \ldots n$

and $\overrightarrow{l_{i}}=\overrightarrow{r_{i}} \times \overrightarrow{p_{i}}$

where $\overrightarrow{r_{i}}$ is the position vector of the $i^{\text {th }}$ particle with respect to a given origin and $\vec{p}_{i}$ is the linear momentum of the particle.

The total angular momentum of the system of particles as

$\therefore \overrightarrow{\mathrm{L}}=\sum_{i=1}^{n} \vec{l}_{i}=\sum_{i=1}^{n} \overrightarrow{r_{i}} \times \vec{p}_{i}$

Differentiating w.r.t. time

$\therefore \frac{d \overrightarrow{\mathrm{L}}}{d t}=\sum_{i=1}^{n} \overrightarrow{\tau_{i}} \quad \ldots \text { (1) }\left[\because \overrightarrow{r_{i}} \times \overrightarrow{p_{i}}=\overrightarrow{\tau_{i}}\right]$

$\therefore \vec{\tau}=\sum_{i=1}^{n} \overrightarrow{r_{i}} \times \overrightarrow{\mathrm{F}_{i}} \quad[\because \vec{\tau}=\vec{r} \times \overrightarrow{\mathrm{F}}]$

Here, $\overrightarrow{\mathrm{F}}_{i}$ is the force on the $i^{\text {th }}$ particle is the vector sum of external forces $\mathrm{F}_{\text {iext }}$ acting on the particle and the internal forces $F_{i \text { (int) }}$ exerted on it by the other particles of the system.

$\therefore \overrightarrow{\mathrm{F}}_{i}=\overrightarrow{\mathrm{F}}_{i(\mathrm{ext})}+\overrightarrow{\mathrm{F}}_{i(\mathrm{int})}$

$\therefore \vec{\tau}=\vec{\tau}_{(\mathrm{ext})}+\vec{\tau}_{(\mathrm{int})}$

where $\vec{\tau}_{(\mathrm{ext})}=\sum \overrightarrow{r_{i}} \times \overrightarrow{\mathrm{F}}_{i(\mathrm{ext})}$ and

$\vec{\tau}_{\text {(int) }}=\sum \overrightarrow{r_{i}} \times \overrightarrow{\mathrm{F}_{i}}_{\text {(int) }}$