Question

A small body of mass m slides without friction from the top of a hemisphere of radius R. At what height will the body be detached from the centre of the hemisphere?

• A. $h=\frac{R}{2}$
• B. $h=\frac{R}{3}$
• C. $h=\frac{2R}{3}$
• D. $h=\frac{R}{4}$

Answer 1

The correct option is C $h=\frac{2R}{3}$

$\text{Step 1: Write the force equation for the body}$

$\Sigma F_c=m{a}_c$

$\Rightarrow mg\cos \theta -N=\frac{m{v}^2}{R}$

If Body is detached then $N=0$

$\Rightarrow mg\cos \theta =\frac{m{v}^2}{R}$

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

$\text{Step 2: Apply Conservation of mechanical energy}$

Applying conservation of mechanical energy at $1$ & $2$

$K_i+U_i=K_f+U_f$

$\Rightarrow 0+mgR(1-\cos \theta )=\frac{1}{2}m{v}^2+0$

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

$\text{Step 3: Height from which the body is detached}$

From $\left(1\right)$ & $\left(2\right)$

$2-2\cos \theta =\cos \theta \Rightarrow \cos \theta =\frac{2}{3}$

So height $h=R\cos \theta$

$h=R\times \frac{2}{3}\Rightarrow h=\frac{2R}{3}$

Hence option (C) is correct