Just as the moment of a force is the rotational analogue of force, the quantity angular momentum is the rotational analogue of linear momentum.

In figure, $\mathrm{Q}$ is a particle of mass $\mathrm{m}$, having position vector $\overrightarrow{\mathrm{OQ}}=\vec{r}$ in Cartesian co-ordinate system.

$\vec{v}$ is the linear velocity of the particle. So its linear momentum is $\vec{p}=m \vec{v}$.

Here it is not necessary that the particle $\mathrm{Q}$ should be of a rigid body and it should move over a curved path.

Let the angle between $\vec{r}$ and $\vec{p}$ be $\theta$.

The vector product or $\vec{r}$ and $\vec{p}$ is defined as the angular momentum $\vec{l}$ of the particle w.r.t. point $\mathrm{O}$.

$\therefore \quad \vec{l}=\vec{r} \times \vec{p}$

Unit of angular momentum is $k g-m^{2} s^{-1}$ or $J . S$ and dimensional is $\left[M^{1} L^{2} T^{-1}\right]$

The magnitude of $\vec{l}$ depends on the selection of the reference point and so while defining the angular momentum of a particle it is necessary to mention the reference point.

The direction of $\vec{l}$ can be obtained with the help of right handed screw rule. Here $\vec{l}$ is in $\mathrm{OZ}$ direction.

Now $\vec{l}=\vec{r} \times \vec{p}$.

$\therefore|\vec{l}|=r p \sin \theta=p[r \sin \theta]=p[\mathrm{OR}]$

$\therefore$ Angular momentum of a particle $=$ (magnitude of linear momentum) $\times$ (the perpendicular distance) of linear momentum (vector) from the reference point.