Question

A small block is placed at a top of a sphere. It slides on the smooth surface of the sphere. The angle made by the radius vector of the block with the vertical when the block leaves the sphere is:

• A. ${24}^{\circ }$
• B. ${36}^{\circ }$
• C. ${42}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is C ${42}^{\circ }$

Let the block leave the sphere when it is an angle $\theta$ with respect to the vertical.
The point at which the block leaves the sphere, normal reaction on the block by the sphere is zero.
Resolving weight $mg$ of the block into two components, $mg\cos \theta$ provides the centripetal force.
Thus, $mg\cos \theta =m{a}_c=\frac{m{v}^2}{R}$

$\Rightarrow \cos \theta =\frac{a_c}{g}=\frac{v^2}{Rg}$

By equating the potential energy of the block at the top of the sphere to the KE at the point where it leaves the sphere:

$\frac{1}{2}m{v}^2=mgh$

or, $v^2=2gh$

$\therefore \cos \theta =\frac{2gh}{Rg}=\frac{2h}{R}$

From geometry, $\cos \theta =\frac{R-h}{R}$

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

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

$\Rightarrow \cos \theta =\frac{2}{3}=\cos {42}^{\circ }$

$\Rightarrow \theta ={42}^{\circ }$