Can you help me to do this: Two frames of references $S$ and $S'$ have a common origin $O$ and $S'$ rotates with constant angular velocity $\omega$ with respect to $S$.
A square hoop $ABCD$ is made of fine smooth wire and has side length $2a$. The hoop is horizontal and rotating with constant angular speed $\omega$ about a vertical axis through $A$. A small bead which can slide on the wire is initially at rest at the midpoint of the side $BC$. Choose axes relative to the hoop and let $y$ be the distance of the bead from the vertex $B$ on the side $BC$. Write down the position vector of the bead in your rotating frame. Show that
$\ddot y-\omega^2 y=0$ using the expression for the acceleration. Hence find the time which the bead takes to reach a vertex $C$. I showed that $\frac=(\frac)'+2\vec\omega\times(\frac)'+\vec\omega\times(\vec\omega\times\vec r)$ where $'$ indicates that it's done in rotating frame. $\vec r$ is position vector of a point $P$ measured from the origin. I got that
$\vec r=r\cos\theta\vec i+y\vec j$
$\vec r'=(\dot rcos\theta-r\dot\theta \sin\theta)\vec i+\dot y\vec j$
$\vec r''=(\ddot rcos\theta-\dot r\dot \theta sin\theta-\dot r\dot\theta\sin\theta-r\ddot\theta\sin\theta-r\dot\theta^2cos\theta)\vec i+\ddot y\vec j$
$\omega\times\vec r'=-\omega\dot y\vec i+(\omega\dot r\cos\theta-\omega r\dot\theta\sin\theta)\vec j$
$\vec\omega\times (\vec\omega\times\vec r)=-\omega^2 r\cos\theta\vec i-\omega^2 y\vec j$ I suppose I have to write Newton's second law now, but I don't know which forces do I have in this motion.