A thin smooth straight tube $OA$ is constrained to rotate with constant angular velocity $ω$ about a fixed vertical axis through $O$, and a particle is free to move in the tube. The angle between $OA$ and the upward vertical is a fixed acute angle $α$ and describe a fixed horizontal circle of radius $a$, While the particle is in a state of steady motion the angular velocity of the tube is suddenly reduced to $\dfrac{\omega}{2}$ and is then maintained constant at the new value. Find the time the particle takes to reach O,
My problem is
Is this particle always equilibrium vertically that is $Rsin(\alpha)=mg$ where $R$ is reaction on particle
Edit:
$\underline{Attempt}$
I got $r$ is distance from vertical axis through $O$ which is perpendicular to the this axis then I used velocity and accelerataion in polar coordinates
So $f=ma$ along the $r$
$$-R\cos(\alpha)=m(\ddot{r}- \frac{r{\omega^2}}{4})$$
and $f=ma$ perpendicular to the $r$
$$R\sin({\alpha})=mg$$
so then $-g\cot(\alpha)=\ddot{r}- \dfrac{r{\omega^2}}{4}$
Is that correct?
Analyzing the movement along the rod and defining $\delta(t)$ the distance from $O$, in the absence of friction, the dynamics are given by
$$ m\ddot\delta(t) = \frac 14m\omega^2(\delta(t)\sin\alpha)\sin\alpha-m g\cos\alpha $$
or
$$ \ddot\delta(t) = \frac 14\omega^2\delta(t)\sin^2\alpha-g\cos\alpha $$
Solving this ODE for $\delta(0) = \frac{g\cos\alpha}{\omega^2\sin^2\alpha},\ \ \dot\delta(0)=0$. Note that the rotation will be reduced from $\omega$ to $\omega/2$ so we have
$$ \delta(t) = \frac{g \cot (\alpha ) \csc (\alpha ) \left(4-3 \cosh \left(\frac{1}{2} \omega t \sin (\alpha )\right)\right)}{\omega ^2} $$
solving now $\delta(t^*)=0$ we obtain
$$ t^*=\frac{2\cosh ^{-1}\left(\frac{4}{3}\right) \csc (\alpha )}{\omega } $$