The arc length s of a circle is the product of the circle's radius r and the angle $\mathrm{\θ}$ (as illustrated earlier):

$s\=r\cdot \mathrm{theta;}$,

where $\mathrm{\θ}$ is measured in radians.

Taking the derivative with respect to time, you can determine that the speed of an object in uniform circular motion is the product of the radius and the angular velocity:

$\frac{\ⅆs}{\ⅆt}\=r\cdot \frac{DifferentialD;\mathrm{\theta}}{DifferentialD;t}$,

$vequals;r\cdot \mathrm{omega;}$ .

This formula can also be derived in a different way, using vectors. If you take the position vector to be

$r\cdot {\mathit{u}}_{r}$ ,

where ${\mathit{u}}_{r}\=\left(\mathrm{cos}\left(\mathrm{\θ}\right)\,\mathrm{sin}\left(\mathrm{theta;}\right)\right)$, then the (linear) velocity is given by

$\mathit{v}\=\frac{\ⅆ\mathit{r}}{\ⅆt}\=r\cdot {\mathit{u}}_{\mathrm{\theta}}\cdot \frac{\ⅆ\mathrm{\theta}}{\ⅆt}$,

where ${\mathit{u}}_{\mathrm{\theta}}\=\left(-\mathrm{sin}\left(\mathrm{\theta}\right)\,\mathrm{cos}\left(\mathrm{\theta}\right)\right)$. Then the speed is the magnitude of the velocity:

$vequals;\left|\mathit{v}\right|equals;r\cdot \mathrm{omega;}$.

Finally, you can compute the (centripetal) acceleration in the same way:

$\mathit{a}\mathbf{\=}\frac{\mathbf{\ⅆ}\mathit{v}}{\ⅆt}\=-r\cdot {\mathit{u}}_{r}\cdot \frac{\ⅆ\mathrm{\theta}}{\ⅆt}\cdot \mathrm{\ω}\+r\cdot {\mathit{u}}_{\mathrm{\θ}}\cdot \frac{{\ⅆ}^{}\mathrm{\ω}}{\ⅆt}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}equals;\mathbf{}\mathbf{-}\mathit{r}\mathbf{\cdot}{\mathrm{omega;}}^{2}$.

The magnitude of the centripetal acceleration is thus:

$aequals;\left|\mathit{a}\right|equals;r\cdot {\mathrm{omega;}}^{2}equals;v\cdot \mathrm{omega;}equals;\frac{{v}^{2}}{r}$

and the associated centripetal force is:

${F}_{c}equals;\mathrm{ma}equals;\frac{{\mathrm{mv}}^{2}}{r}$.