As the object moves from P to P' in time $\Delta t\left(=t^{\prime}-t\right)$, the line turns through an angle $\Delta \theta$ as shown in the figure.
$\Delta \theta$ is called angular distance.
The angular speed $\omega$ as the time rate of change of angular displacement.
$\therefore \omega=\frac{\Delta \theta}{\Delta t}$
Now, if the distance travelled by the object during the time $\Delta t$ is $\Delta \mathrm{S}$, i.e. PP' is $\Delta \mathrm{S}$, then
$\therefore v=\frac{\Delta \mathrm{S}}{\Delta t}$
$\therefore \Delta \mathrm{S}=\mathrm{R} \Delta \theta$
$v=\frac{\mathrm{R} \Delta \theta}{\Delta t}$
$\therefore v=\mathrm{R} \omega$
Centripetal acceleration $a_{\mathrm{C}}$
$a_{c}=\frac{(\mathrm{R} \omega)^{2}}{\mathrm{R}}=\frac{\mathrm{R}^{2} \omega^{2}}{\mathrm{R}}$
$\therefore a_{c}=\mathrm{R} \omega^{2}$