A particle moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $R$. The plane of motion is at a depth $d$ below the centre of the hemisphere. The speed of the particle is :-
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Centripetal

Force $=$ mass $\times$ centripetal acceleration

$\mathrm{N} \cos \theta=\mathrm{m} \frac{\mathrm{V}^{2}}{\mathrm{r}}$      $...(i)$

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

$(ii)$ $/(\mathrm{i}),$ tan $\theta=\frac{\mathrm{rg}}{\mathrm{V}^{2}}$

$\Rightarrow V=\sqrt{\frac{r g}{\tan \theta}}$

$V=\sqrt{\frac{g\left(R^{2}-d^{2}\right)}{d}}\left[\begin{array}{c}{r=\sqrt{R^{2}-d^{2}}} \\ {\tan \theta=d / r}\end{array}\right]$

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