Question

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. Find the minimum projection-speed \[\nu_0\] for which the particle reaches the top of the track.

Answer 1

Net force on the particle at A and B \[F = \text{ mg }\sin \theta\] Work done to reach B from A, \[\text{W = FS = mg} \sin \theta \text{ l }\] Again, work done to reach B to C 

\[= \text{mgh}\] 
\[ = \text{mg R} \left( 1 - \cos \theta \right)\]

So, total work done

\[= \text{mgl} \sin \theta + \text{mgR} \left( 1 - \cos \theta \right)\]
\[ = \text{mg}\left[ \text{l} \sin \theta + R \left( 1 - \cos \theta \right) \right]\]

Now, change in K.E. = Total work done

\[\Rightarrow \frac{1}{2}\text{m}\nu_2^2 = mg \left[ \text{l} \sin \theta + R \left( 1 - \cos \theta \right) \right]\]

\[ \Rightarrow \nu_2 = \sqrt{2g \left[ R \left( 1 - \cos \theta \right) + \text{l} \sin \theta \right]}\]