Question

A ferris wheel of radius r speeds up with angular acceleration α starting from rest. part a find an expression for the velocity of a rider after the ferris wheel has rotated through angle δθ.

Answer 1

The expression for the angular acceleration is:
$\alpha = \frac{\Delta \omega}{\Delta t}$
where $\Delta \omega = \omega-\omega _0$ is the variation of the angular velocity, with $\omega _0$ being the starting velocity (which in our problem is zero), and $\Delta t$ being the time interval. So we can write the angular velocity after an angle $\delta \theta$ as
$\omega (\delta \theta) = \alpha \Delta t$
We also know the relationship between tangential velocity, v, and the angular velocity v:
$v=\omega r$
with r being the radius of the wheel. Substituting $\omega$ into the previous equation, we can write an expression for v:
$v(\delta \theta )= \alpha r \Delta t$