The acceleration of the particle in the tangential direction is $\alpha r$ and the acceleration of the particle in the radial direction $\omega^2 r$ . So, the desired angle
$\begin{align*}\theta = \tan^{-1} \left(\frac{\alpha r}{\omega^2 r} \right) = \boxed{\tan^{-1} \left(\frac{\alpha}{\omega^2} \...$
Therefore, answer (E) is correct.

Solution to 2008 Problem 99