Rotating frame of reference

Question

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 $\frac{d^2\vec r}{dt^2}=(\frac{d^2\vec r}{dt^2})'+2\vec\omega\times(\frac{d\vec r}{dt})'+\vec\omega\times(\vec\omega\times\vec r)$ 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.

Answer 1

I see the problem with your equation now.

When differentiating $\vec r=r\cos\theta\vec i+y\vec j$, you have considered $r$ to be constant, which is wrong.
$r$ is given by $$r=\sqrt{l^2+y^2}$$ where $l$ is the side-length of the square. So $r$ will change with $y$, and you'll have to differentiate $r$ too.
This is where the math gets pretty ugly and dissuading!

To avoid that, what we can do is we can observe that in the rotation frame, the bead will experience an outward centrifugal force. This force will have a component along $BC$. That component can be written as(I'll be borrowing your variables) $$F_{BC}=m\omega^2r\sin\theta$$ $$F_{BC}=m\omega^2 \sqrt{l^2+y^2} \frac{y}{\sqrt{l^2+y^2}}$$

Thus by dividing by $m$ on both sides you get

$$\ddot y=\omega^2y$$

Note that this is the same as applying $\frac{d^2\vec r}{dt^2}=(\frac{d^2\vec r}{dt^2})'+2\vec\omega\times(\frac{d\vec r}{dt})'+\vec\omega\times(\vec\omega\times\vec r)$. It is just that this approach is more problem specific(and a lot easier too!).