Question

A point mass situated on the top of a smooth sphere is allowed to slip on the surface of a sphere. If the radius of the sphere is $r$, then the angle which the line, joining the point of leaving the sphere and the centre of the sphere, makes with the vertical is:

• A. zero
• B. $co{s}^{-1}\left(\begin{array}{c}\frac{1}{3}\end{array}\right)$
• C. $si{n}^{-1}\frac{2}{3}$
• D. $co{s}^{-1}\left(\begin{array}{c}\frac{2}{3}\end{array}\right)$

Answer 1

The correct option is D $co{s}^{-1}\left(\begin{array}{c}\frac{2}{3}\end{array}\right)$

From figure, $N=\frac{m{v}^2}{R}-mg\cos \theta ...\left(1\right)$ where $m{v}^2/R$, centrifugal force outwards the centre.

By energy conservation, $\frac{m{v}^2}{2}=mg(R-R\cos \theta )$

or $\frac{m{v}^2}{R}=2mg(1-\cos \theta )....\left(2\right)$

From (1) and (2), $N=2mg(1-\cos \theta )-mg\cos \theta =2mg-3mg\cos \theta$

At break off point , $N=0$ so $2mg=3mg\cos \theta \Rightarrow \theta =co{s}^{-1}\left(\frac{2}{3}\right)$