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. Assuming that the projection-speed is \[\nu_0\] and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. 

Answer 1

(b) When the block is projected at a speed: 

Let the velocity at C be \[\nu_0\] .

Applying energy principle,

\[\left( \frac{1}{2} \right) \text{m}\nu_0^2 - \left( \frac{1}{2} \right) \text{m}\left( 2 \nu_0 \right)^2 \]

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

\[ \Rightarrow V^2 = 4 \nu_0^2 - 2g \left[ \text{ l } \sin g \theta + R \left( 1 - \cos \theta \right) \right]\]

\[ = 4 . 2 g \left[ \text{ l } \sin \theta + R \left( 1 - \cos \theta \right) \right] - \]

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

So, force acting on the body,

\[N = \frac{V^2}{R} = 6 \text{ mg } \left[ \left( \frac{l}{R} \right) \sin \theta + 1 - \cos \theta \right]\]