For horizontal forces.

$\mathrm{N} \sin \theta=\mathrm{mx} \omega^{2}$     ......$(i)$

For vertical forces.

$\mathrm{N} \cos \theta=\mathrm{mg}$        .........$(ii)$

$\tan \theta=\frac{x \omega^{2}}{g}$

$\text { But } \quad x=R \sin \theta$

$\therefore $ $\frac{\sin \theta}{\cos \theta}=\frac{R \sin \theta \cdot \omega^{2}}{g}$

${\rm{ or }}\cos \theta  = g/{{\mathop{\rm Rg}\nolimits} ^2}$

${\rm{ or\,\, }}\theta  = {\cos ^{ - 1}}\left( {\frac{g}{{{\mathop{\rm R}\nolimits} {\omega ^2}}}} \right)$