# Solve the following :

Question:

Figure (8-E17) shows a smooth track which consists of a straight inclined part of length 1 joining smoothly with the circular part. A particle of mass $m$ is projected up the incline from its bottom. (a) Find the minimum projection-speed v, for which the particle reaches the top of the track. (b) Assuming that the projection-speed is $2 \mathrm{v}_{0}$ 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. (c) Assuming that the projection-speed is only slightly greater than $2 \mathrm{v}_{0}$, where will the block lose contact with the track?

Solution:

(a) $F_{A B}=m g \sin \theta$

$W_{B}=F_{A B} S$

$W_{B}=m g \sin \theta \times 1 \ldots 1$

and

W_{C}=m g R(1-\cos \theta) \ldots 2

and

$\mathrm{W}_{\mathrm{c}}=\mathrm{mgR}(1-\cos \theta) \ldots 2$

Total work done

$W=W_{B}+W_{C}$

and

change in K.E. $=$ Work done

$v_{0}=[\sqrt{2} g(R(1-\cos \theta)+\mid \sin \theta]$

(b) By work energy equation

change in K.E. $=$ Work done

$v_{c}=[\sqrt{6} g(\mid \sin \theta)+R(1-\cos \theta)]$

and

Force,

$F=\left(m v_{c}^{2}\right) / R$

$F=6 m g(l / R \sin \theta+1-\cos \theta)$

(c) $\left(m v^{2}\right) / R=m g \cos \theta$

$v=\sqrt{g R} \cos \theta \ldots .1$

Also,

$v=[\sqrt{2} g R(1-\cos \theta)] \ldots . .2$

equating 1 and 2 we get,

$\theta=\cos ^{-1}\left(\frac{2}{3}\right)$