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 teh force acting on it whenit reaches the top. c. Assuming that teh projection speed is only slightly greater than `v_0` where will the block lose contact with the track?

Answer

A, B, C, D

Explanation

a. Net force on the particle A and B, `F=mgsin theta`
work done to reach B,
`W=FS=mg sintheta l`
Again work done to reach B to C
`=mgh`
`=mghR(1-costheta)`
`So, total work done
`=mg(lsintheta+r(1-costheta)]`

Now, change in K.E.
=work done
`rarr 1/2mv_2^2=mg[lsintheta+R(1-costheta)]`
`rarr v_2=sqrt(2g[R(1-costheta)+lsintheta]`
b. Whenteh block is projected at a speed `2v_0`

Let the velocity of C will be `v_0`.
Applying energyy priciple
`(1/2)mv_0^2 -(1/2)m(2v_0)^2`
=-mg[lsintheta+R(1-costheta)]`
rarr V^2=4v_0^2-2g[lsintheta+R(1-costheta)]`
`=4.2g[lsintheta+R(1-costheta)]- 2g[lsintheta+R(1-costeta)]`
so force acting on teh body
`rarr N=V^2/R=6mg[(l/R)sintheta+1-costheta]`
c. Let the loose contact after making an angle `theta`
`(mv^2)/R=mgcostheta`
`rarr v^2=Rgcostheta` .......i
`Again `1/2 mv^2=mg(R-Rcostheta)`
`rarr v^2=2gR(1-costheta)`.........ii
`Form i and ii `costheta =2/3
`rarr theta=cos^-1(2/3)`