Question

A particle of mass $m$ slides from the vertex of the semi-hemisphere, without any initial velocity. At what height from horizontal will the particle leave the sphere?

• A. $R$
• B. $2R$
• C. $\frac{2}{3}R$
• D. $\frac{3}{2}R$

Answer 1

The correct option is C

$\frac{2}{3}R$

Step 1. Given data:

Mass of the particle = $m$

Initial velocity = $0$

Step 2. Explanation:

Let A be the initial velocity and B be the point where the particle leaves the hemisphere.

Now according to the law of conservation of energy.

The energy at point A $\left(E_A\right)$ = Energy at point B $\left(E_B\right)$

$\Rightarrow mgR=mgx+\frac{1}{2}m{v}^2$,

$\Rightarrow mgR=mg.R\cos \theta +\frac{1}{2}m{v}^2,\left[x=R\cos \theta \right]$

$\Rightarrow \frac{1}{2}{v}^2=gR\left(1-\cos \theta \right)$

$v=\sqrt{2gR\left(1-\cos \theta \right)}........\left(1\right)$

Now, when the particle leaves the hemisphere,

As the particle moves down from point A to B, its speed increases and so the normal force gets smaller and smaller until it becomes zero.

At that moment the component of the weight of the particle provides just enough force to produce the centripetal acceleration of the ball.

Normal force $\left(N\right)=0$, Centrifugal force = $\frac{m{v}^2}{R}$

Conditioning for leaving, $mg.\cos \theta =\frac{m{v}^2}{R}$

$\Rightarrow g.\cos \theta =\frac{v^2}{R}$, Putting the value of $v$ from $\left(1\right)$, we get

$$\begin{gathered} g.\cos \theta =\frac{2gR\left(1-\cos \theta \right)}{R} \\ \Rightarrow 3\cos \theta =2 \\ \Rightarrow \cos \theta =\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$

Now, we have $x=R.\cos \theta$. Putting the value of $\cos \theta$, we get

$x=\frac{2}{3}R$

Thus, the correct option is option C.