JEE Mains · Physics · STD 11 - 3.2 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

- A \(\sin \,\phi = \,\cos \,\phi \)
- B \(\sin \,\phi = \frac{1}{2}\,\cos \,\theta \)
- C \(\sin \,\phi = \frac{2}{3}\,\cos \,\theta \)
- D \(\sin \,\phi = \frac{3}{4}\,\cos \,\theta \)
Answer & Solution
Correct Answer
(C) \(\sin \,\phi = \frac{2}{3}\,\cos \,\theta \)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{mg} \sin \phi\) \(...(i)\) \(\mathrm{mg} \mathrm{r} \cos \theta=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{rg} \sin \phi\) \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}=2 \cos \theta-2 \sin \phi \ldots \ldots(\mathrm{ii})\)…
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