A particle moves in a horizontal circle on smooth inner surface of a hemispherical bowl of

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3-2.Motion in Plane

hard

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 :-

A$\sqrt {\frac{{g({R^2} - {d^2})}}{R}} $

B$\sqrt {\frac{{g({R^2} - {d^2})}}{d}} $

C$\sqrt {\frac{{gR}}{{{R^2} - {d^2}}}} $

D$\sqrt {\frac{{g d^2}}{{{R^2} - {d^2}}}} $

Solution

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

Standard 11

Physics

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easy

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hard

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 :-

Solution

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