Going Between Angular and Linear: Now that we see what the rotational kinematic variables are, and how the equations for constant acceleration allows you to use them to find instantaneous values, it's important to see how they can also be used to find their linear counterparts. Some problems may state what the initial and final angular velocities are, and the change in angle during that motion, then ask for the time. In that case you would not have to use any of the linear quantities to solve the problem. Although it then asks what the linear speed was at the end of the motion, you'd have to use the following relationships below.

Change in angular position ==> Distance traveled: $s=\Delta \theta r$

Angular velocity ==> speed: $|\overrightarrow{v}| = v_t = \omega r$

Angular acceleration ==> Tangential component of linear acceleration: $a_t = \alpha r$

The linear coordinate system we use in this case is one with a radial inward direction ($\widehat{r}$), a CCW tangential direction ($\widehat{t}$), and a direction perpendicular to those two ($\widehat{z}$). This means that since the object is always traveling tangent to the circle, the linear velocity has zero radial or z components - the speed of the object is equal to the tangent component of its linear velocity.

$\overrightarrow{v}=\langle v_r, v_t, v_z \rangle = \langle 0, \omega r, 0 \rangle$

For the angular acceleration, using the same $\widehat{r}$,$\widehat{t}$, and $\widehat{z}$, coordinate system, the object has zero z component to its acceleration. If you recall though it does have a radial component equal to the square of its speed divided by the radius of the circle its traversing ($a_r = \frac{v^2}{r}$). The full linear acceleration is thus:

$\overrightarrow{a}=\langle a_r, a_t, a_z \rangle = \langle \frac{v^2}{r}, \alpha r, 0 \rangle$

Now, take a look at the pre-lecture reading and videos below.