Obtain the relation between linear velocity and angular velocity of rotational motion of a body.

(N/A) Consider a rigid body rotating about a fixed axis (say $Z$-axis) with a constant angular velocity $\omega$.
Let a particle $P$ of the rigid body be at a perpendicular distance $r_{\perp}$ from the axis of rotation.
As the body rotates,the particle $P$ moves in a circular path of radius $r_{\perp}$ with its centre $C$ on the axis of rotation.
In a small time interval $\Delta t$,the particle moves through an arc length $\Delta s$ along the circumference of the circle,corresponding to an angular displacement $\Delta \theta$.
The relation between arc length,radius,and angle is given by $\Delta s = r_{\perp} \Delta \theta$.
Dividing both sides by $\Delta t$,we get $\frac{\Delta s}{\Delta t} = r_{\perp} \frac{\Delta \theta}{\Delta t}$.
As $\Delta t \to 0$,the ratio $\frac{\Delta s}{\Delta t}$ becomes the linear speed $v$,and $\frac{\Delta \theta}{\Delta t}$ becomes the angular velocity $\omega$.
Thus,the relation is $v = r_{\perp} \omega$.
In vector form,this is expressed as $\vec{v} = \vec{\omega} \times \vec{r}$,where $\vec{r}$ is the position vector of the particle with respect to any point on the axis of rotation.