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

(A) \sin \,\phi = \,\cos \,\phi

(B) \sin \,\phi = \frac{1}{2}\,\cos \,\theta

(D) \sin \,\phi = \frac{3}{4}\,\cos \,\theta

JEE Mains · Physics · STD 11 - Motion in plane

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

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\sin \,\phi = \frac{2}{3}\,\cos \,\theta \)

Detailed explanation

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

\(\mathrm{mg} \mathrm{r} \cos \theta=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{rg} \sin \phi\)

\(\frac{\mathrm{v}^{2}}{\mathrm{rg}}=2 \cos \theta-2 \sin \phi \ldots \ldots(\mathrm{ii})\)

\(\sin \phi=2 \cos \theta-2 \sin \phi\)

\(3 \sin \phi=2 \cos \theta\)

\(\sin \phi=\frac{2}{3} \cos \phi\)

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