A small block is projected with a speed `v_( 0)` on a horizontal track placed on a sufficiently rough surface which turns into a semi circle ( vertical) of radius R. Find the min value of `V_(o)` so that the body will hit the point A after leaving the track at its highest point. The arrangement is shown in the figure, given that the straight part is rough & the curved path is smooth. The coefficient of friction is `mu`.

Explanation

Let the block escape the point at C with a velocity V horizontally. Since it hits the initial spot A after falling through a height 2R we can write
`(2R)=1/2g t^(2)`, whrer `t=` time of its fall.
`rArr t=2 sqrt(R//g)`
`:.` The distance `AB=2vsqrt(R//g)`
`rArr d=2v_(x) sqrt(R//g)` ...(a)
Work energy theorem is applied to the motion of the body from A to B leads `Delta KE=W_(f)`
`rArr 1/2 mv_(o)^(2)-1/2 mv_(1)^(2)=mu mgd`
`rArr v_(o)=sqrt(v_(1)^(2)+2mu gd)` ...(b)
Energy conservation between B & C yields
`1/2 mv_(1)^(2)-1/2 mv^(2)=mg(2R)`
`rArr v_(1)=sqrt(v^(2)+4gR)` ...(c)
When the block escapes C, its minimum speed v can be given as `(mv^(2))/(R)=mg`
( `:.` the normal contact force `=0`)
`rArr v=sqrt(gR)` ...(d)
(Note that if the particle has speed less than `sqrt(gR)` then it will not reach to the highest point of the curvature.
By using (c) and (d) we obtain
`v_(1)=sqrt(5gR)` ...(e)
using (a) and (d) we obtain
`d=(sqrt(gR))2(sqrt(R/g))=2R` ...(f)
`v_(o)=sqrt(5gR+2 mu g (2R))=sqrt((5+4 mu )gR)`.