$A$ thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{g / R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega = \sqrt{2g / R}$? Neglect friction.

(D) Let the radius vector joining the bead $P$ with the centre $O$ make an angle $\theta$ with the vertical downward direction.
Let $m$ be the mass of the bead.
The forces acting on the bead are:
$1$. Weight $mg$ acting vertically downwards.
$2$. Normal reaction $N$ from the wire,acting perpendicular to the tangent at $P$.
Resolving the forces:
The vertical component of the normal reaction $N \cos \theta$ balances the weight $mg$,so $N \cos \theta = mg$.
The horizontal component of the normal reaction $N \sin \theta$ provides the necessary centripetal force $m \omega^2 r$,where $r = R \sin \theta$ is the radius of the circular path of the bead.
Thus,$N \sin \theta = m \omega^2 (R \sin \theta)$.
Since $\sin \theta \neq 0$ for $\theta \neq 0$,we get $N = m R \omega^2$.
Substituting $N$ in the vertical equation: $(m R \omega^2) \cos \theta = mg$.
Therefore,$\cos \theta = \frac{g}{R \omega^2}$.
Since $\cos \theta \leq 1$,the bead remains at the lowermost point $(\theta = 0)$ if $\frac{g}{R \omega^2} \geq 1$,which implies $\omega \leq \sqrt{\frac{g}{R}}$.
For $\omega = \sqrt{\frac{2g}{R}}$,we have $\omega^2 = \frac{2g}{R}$.
Substituting this into the expression for $\cos \theta$:
$\cos \theta = \frac{g}{R (2g/R)} = \frac{1}{2}$.
Thus,$\theta = \cos^{-1}(0.5) = 60^{\circ}$.