Question

A particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu =\frac{1}{\sqrt{3}}$. The maximum height up to which the particle can remain stationary is

• A. $\frac{\sqrt{3}}{2}$ R
• B. $(1-\frac{\sqrt{3}}{2})R$
• C. $\frac{R}{2}$
• D. $\frac{3R}{8}$

Answer 1

The correct option is B $(1-\frac{\sqrt{3}}{2})R$

Let the particle of mass of $m$ stay at rest on the point $B$ with respect to the point $A$.

The point $B$ is at height $h$.


The horizontal component of the weight of the particle along the radius is $mg\cos \theta$

Hence the normal reaction force by sphere on the particle, $N=mg\cos \theta$

The vertical component of the weight $mg\sin \theta$ tries to move the particle.

The limiting friction $F=\mu N,\mu$ is the coefficient of friction between the particle and the hemisphere.

The condition for staying at the rest is that these two forces have to eliminate each other i.e. $mg\sin \theta =\mu mg\cos \theta$

$\Rightarrow \tan \theta =\mu$

Given that $\mu =\frac{1}{\sqrt{3}}$

$\therefore \tan \theta =\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={30}^o$

Now, from the diagram $\Delta OBC,OC=OB\cos {30}^o$

$\Rightarrow OC=\frac{\sqrt{3}R}{2}$ since, the radius of the sphere, $OB=R$

Now, The height $h=AC=AO-OC$

$\Rightarrow h=R-\frac{\sqrt{3}R}{2}$

$\Rightarrow h=R(1-\frac{\sqrt{3}}{2})$

Hence, the maximum height up to which the particle can remain stationary is $h=R(1-\frac{\sqrt{3}}{2})$

Hence, option (b) is the correct answer.