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Question

$\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg} \sin \phi$            $...(i)$

$\mathrm{mg} \mathrm{r} \cos 	heta=\f

Vedclass · Accepted Answer

$\sin \,\phi  = \frac{2}{3}\,\cos \,	heta $

Vedclass · Answer

$\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg} \sin \phi$            $...(i)$

$\mathrm{mg} \mathrm{r} \cos 	heta=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{rg} \sin \phi$

$\frac{\mathrm{v}^{2}}{\mathrm{rg}}=2 \cos 	heta-2 \sin \phi \ldots \ldots(\mathrm{ii})$

$\sin \phi=2 \cos 	heta-2 \sin \phi$

$3 \sin \phi=2 \cos 	heta$

$\sin \phi=\frac{2}{3} \cos \phi$

A particle is released on a vertical smooth semicircular track from point $X$ so that $OX$ makes angle $\theta $ from the vertical ( see figure). The normal reaction of the track on the particle vanishes at point $Y$ where $OY$ makes angle $\phi $ with the horizontal. Then

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