Explain the angular momentum of a particle and show that it is the moment of linear momentum about the reference point.

(N/A) 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 the figure,$Q$ is a particle of mass $m$,having position vector $\vec{OQ} = \vec{r}$ in a Cartesian coordinate system.
$\vec{v}$ is the linear velocity of the particle. So,its linear momentum is $\vec{p} = m\vec{v}$.
It is not necessary that the particle $Q$ belongs to a rigid body or that it moves along a curved path.
Let the angle between $\vec{r}$ and $\vec{p}$ be $\theta$.
The vector product of $\vec{r}$ and $\vec{p}$ is defined as the angular momentum $\vec{l}$ of the particle with respect to point $O$.
$\therefore \vec{l} = \vec{r} \times \vec{p}$
The unit of angular momentum is $kg \cdot m^2 \cdot s^{-1}$ or $J \cdot s$,and its dimensional formula is $[M^1 L^2 T^{-1}]$.
The magnitude of $\vec{l}$ depends on the selection of the reference point; therefore,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 the right-handed screw rule. Here,$\vec{l}$ is in the $OZ$ direction.
Now,$\vec{l} = \vec{r} \times \vec{p}$.
$\therefore |\vec{l}| = r p \sin \theta = p(r \sin \theta) = p(OR)$.
$\therefore$ Angular momentum of a particle = (magnitude of linear momentum) $\times$ (the perpendicular distance of the line of action of linear momentum from the reference point).