Figure ( following ) shows a smooth track which consists of a straight inclined part of length l joining smoothly with the circular part. A particle of mass m is projected up the incline from its bottom.Assuming that the projection-speed is only slightly greater than \[\nu_0\] , where will the block lose contact with the track?
Solution
Let the loose contact after making an angle θ.
\[\frac{\text{m} \nu^2}{R} = \text{mg} \cos \theta\]
\[ \Rightarrow \nu^2 = \text{Rg} \cos \theta . . . (\text{i})\]
\[\text{Again,} \frac{1}{2}\text{m}\nu^2 = \text{mg } \left( R - R \cos \theta \right)\]
\[ \Rightarrow \nu^2 = 2gR \left( 1 - \cos \theta \right) . . . (\text{ii})\]
\[\text{From (i) and (ii), }= \cos^{- 1} \left( \frac{2}{3} \right)\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{2}{3} \right)\]
