Question

A particle is kept at rest at the top of a sphere of diameter $42m$. When disturbed slightly, it slides down. At what height $h$from the bottom, the particle will leave the sphere

• A. $14m$
• B. $16m$
• C. $35m$
• D. $70m$

Answer 1

The correct option is C $35m$

Let $R$ be the radius of the sphere,
$m$ be the mass of the particle sliding over the sphere,
$V$ be the speed of the particle at the instant it leaves of the sphere
$\theta$ be the angle rotated by the particle at the instant it leaves of the sphere

From Free body diagram
using Newtons Law of Motion
we can Write
$\frac{m{V}^2}{R}=mg\cos \theta -N$
For the Particle to leave the sphere,
contact between particle and sphere has to be lost,So normal reaction will become zero.($N=0$ )
$\Rightarrow \frac{m{V}^2}{R}=mg\cos \theta$
Thus speed of the particle, when it leaves the sphere
$\Rightarrow V=\sqrt{gRcos\theta }$
From the above figure height decended by the particle is $AC$
$AC=OA-OC=R-R\cos \theta$
By law of conservation of mechanical energy
Gain in Kinetic Energy = Drop of Potential Energy
$\frac{m{V}^2}{2}=mg\left(AC\right)$
$\frac{m{V}^2}{2}=mgR(1-\cos \theta )$
$mgR\cos \theta =2mgR(1-\cos \theta )$
$cos\theta =\frac{2}{3}$
Height from the bottom will be, $h=R+Rcos\theta$
As $R=21mand\cos \theta =\frac{2}{3}$
$h=21\times \frac{5}{3}=35m$