Figure 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. a.Find the minimum projected speed `v_0` for which the particle reaches the top of the track. b. Assuming that the projection speed is `2v_0` and that the block does not lose contact with the track before reading its top, find the force acting on it when it reaches the top. c. Assuming that the projection speed is only slightly greater than `v_0` where will the block lose contact with the track?

Answer

(a) `sqrt(2g[R(1-cos theta)+lsintheta])`
(b) `6mg(1-cos theta +(l)/(R)sin theta)`
`(c )` The radius through the particle makes an angle `cos^(-1)(2//3)`with the vertical

Explanation

`(a) (1)/(2) mv_(0)^(2)=mglsin theta + mg R (1-cos theta )`
`v_(0)=sqrt(2gR(1-cos theta)+2gl sin theta )`

`(b) C.O.E.`
`=(1)/(2) m(2v_(0))^(2)=mgl sin theta - mgR (1-cos theta )=(1)/(2) mv^(2)`
`=2mv_(0)^(2)-mgl sin theta -mgR(1-cos theta )=(1)/(2) mv^(2)`
`=4mgR(1-cos theta ) +4mgl sin theta - mel sin theta -mgR (1-cos theta)=(1)/(2) mv^(2)`
`=6mgR (1-cos theta)+6mg l sin theta =mv^(2)`
`N=6mg (1- cos theta )+6 mg`` (l)/(R) sin theta `
`=6mg[(1-cos theta )+(l)/(R) sin theta ]`

`mg cos theta =(mv^(2))/(R)`
`=(1)/(2) mv^(2)=(1)/(2) mgR cos theta `
`=mgR (1-cos theta )=(1)/(2) mgR cos theta `
`cos theta =(2)/(3) tehta =cos ^(-1) ((2)/(3))`.