Question

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 angle $\varphi$ with the horizontal. Then:

• A. $\sin \varphi =\cos \theta$
• B. $\sin \varphi =\frac{1}{2}\cos \theta$
• C. $\sin \varphi =\frac{2}{3}\cos \theta$
• D. $\sin \varphi =\frac{3}{4}\cos \theta$

Answer 1

The correct option is A $\sin \varphi =\cos \theta$

When the particle is at $X$:

$N-mg\cos \theta =\frac{m{V}^2}{R}$ (equation $1$)

Since the particle moves along the circle.

When the particle is at $Y$ (Just before it leaves the circular track):

$N-mg\sin \varphi =\frac{m{V}^2}{R}$ (equation $2$)

From $1$ and $2$:

$\cos \theta =\sin \varphi$