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 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

Answer: (C)

$sin\varphi =\frac{2}{3}cos\theta$

In the ground frame of reference, we can see that the necessary centripetal force for the circular motion of the particle shall be provided by $N-mg\sin \varphi$, where $N$ is the normal force on the particle. As the particle keeps sliding, the component $mg\sin \varphi$ keeps decreasing until at some point it becomes zero where the particle flies off.
Hence at the point Y, if $v$ is the speed of the particle
$m{v}^2/r=mg\sin \varphi$ --- eq. 1
Also, since the normal force is always acting normal to the velocity of the particle, it does not do any work against it. And there is no work done by friction since the surface is smooth. Which means the kinetic energy of the particle at Y should be equal to the potential energy which it lost, which is equal to mg*(height of X - height of Y).
From geometry,
height of $X=r\cos \theta$
height of $Y=r\sin \varphi$
$m{v}^2/2=mgr(\cos \theta -\sin \varphi )$ --- eq. 2
Rewriting eq. 2 as $m{v}^2/r=2mg(\cos \theta -\sin \varphi )$ --- eq. 3
Solving eq. 1 and eq. 3, we get
$2mg(\cos \theta -\sin \varphi )=mg\left(\sin \varphi \right)$
$⟹3\sin \varphi =2\cos \theta$