Let the angle made by the radius vector connecting the pellet to center of the ring be  θ, with the vertical downward direction.

OP = R = Radius of the circle
N = Normal reaction
The respective vertical and horizontal equations of forces can be written as:
Mg = Ncosθ ………….. (i)
ml${\mathrm{\omega }}^2$ = Nsinθ ………... (ii)
In ΔOPQ, we have:

$\sin \mathrm{\theta }=\frac{\mathrm{l}}{\mathrm{R}}$

l = R sin θ ……………………… (iii)
Substituting equation (iii) in equation (ii), we get:
$\begin{array}{l}\mathrm{m}(\mathrm{Rsin} \mathrm{\theta }){\mathrm{\omega }}^2=\mathrm{N} \sin \mathrm{\theta }\\ {\mathrm{mR\omega }}^2=\mathrm{N} ...\left(\mathrm{iv}\right)\end{array}$
Substituting equation (iv) in equation (i), we get:
$\begin{array}{l}\mathrm{mg}={\mathrm{mR\omega }}^2\cos \mathrm{\theta }\\ \cos \mathrm{\theta }=\frac{\mathrm{g}}{{\mathrm{R\omega }}^2} ....\left(\mathrm{v}\right)\end{array}$
Since cosθ ≤ 1, the bead will remain at its lowermost point for $\frac{\mathrm{g}}{{\mathrm{R\omega }}^2}\le 1,\text{ i.e., for }\mathrm{\omega }\le \sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$ $\mathrm{For} \mathrm{\omega }=\sqrt{\frac{2\mathrm{g}}{\mathrm{R}}} \mathrm{or}{\mathrm{\omega }}^2=\frac{2\mathrm{g}}{\mathrm{R}} ...\left(\mathrm{iv}\right)$
On equating equations (v) and (vi), we get:

$\begin{array}{l}\frac{2\mathrm{g}}{\mathrm{R}}=\frac{\mathrm{g}}{\mathrm{Rcos} \mathrm{\theta }}\\ \cos \mathrm{\theta }=\frac{1}{2}\\ \therefore \mathrm{\theta }={\cos }^{-1} (0.5)={60}^{\circ }\end{array}$