Question

A particle slides on the surface of a fixed smooth sphere starting from the topmost point. Find the angle rotated by the radius through the particle, when it leaves contact with the sphere.

 

Answer 1

Let the velocity be \[\nu\]  when the body leaves the surface. 

From the free-body diagram,

\[\frac{\text{m}\nu^2}{\text{R}} = \text{mg} \cos \theta\]
\[ [\text{ normal reaction }]\]
\[ \nu^2 = \text{Rg} \cos \theta . . . (\text{i})\]

Again, from the work-energy principle,
Change in K.E. = Work done

\[\Rightarrow \frac{1}{2}\text{m}\nu^2 - 0 = \text{mg} \left( \text{R - R} \cos \theta \right)\]

\[ \Rightarrow \nu^2 = 2\text{ gR} \left( 1 - \cos \theta \right) . . . . (ii)\]

From (i) and (ii),

\[\text{Rg} \cos \theta = 2\text{ gR } \left( 1 - \cos \theta \right)\]

\[3\text{ gR } \cos \theta = 2\text{ gR }\]
\[\cos \theta = \frac{2}{3}\]
\[\theta = \cos^{- 1} \left( \frac{2}{3} \right)\]