Question 179 of 183

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 the point $Y$ where $OY$ makes an angle $\phi$ with the horizontal. Then

A$sin\phi =\dfrac{2}{3}cos\theta$B$sin\phi =\dfrac{3}{4}cos\theta$C$sin\phi =\dfrac{1}{2}cos\theta$D$sin\phi =cos\theta$

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Answer: (A) $sin\phi =\dfrac{2}{3}cos\theta$

$mg\left(R\cos \theta -R\sin \phi \right)=\dfrac{1}{2}m{V}^{2}.......(1)$
On losing contact $N=0\Rightarrow mgsin\phi =\dfrac{m{V}^{2}}{R}......(2)$



$\Rightarrow mgR\left(cos\theta -sin\phi \right)=\dfrac{1}{2}mgRsin\phi$
$\Rightarrow 2cos\theta =3sin\phi$