# Tangential and Angular Speed

In circular motion, the velocity vector always points in the tangential direction as shown. With the aid of calculus, it is straightforward to find a relationship between the absolute values of the angular and tangential velocities. (Remember they point in different directions!)

We start with the time derivative of the linear displacement, which is the tangential velocity:

 $\rm \mathbf{v = \frac{dx}{dt}}$

Because the displacement is equal to the arc length in circular motion, we have

$\rm \mathbf{v = \frac{ds}{dt}.}$

But the arc length is:

s = r $\theta$,

and so:

$\rm \mathbf{v = \frac{d(r \theta)}{dt} = \theta \frac{dr}{dt} + r \frac{d\theta}{dt}}$

The first term does not contribute to v because r = constant in circular motion. We thus have the relationship between the tangential speed, v, and the angular speed, $\omega$:

$\rm \mathbf{v = r \frac{d\theta}{dt} = r \omega}$

This derivation above really did require calculus and its product rule. Providing the same derivation without the result of calculus is very cumbersome, but only the final result is what is really important here:

v = r$\omega$