A small bead of mass m can moves on a smooth circular wire (radiusR) under the action of a force `F=(km)/r^(2)` directed (r=position of bead from P&K=constant) towards a point P with in the circle at a distance `R/2` from the centre what should be the minimum velocity of bead at the point of the wire nearest the centre of force (P) so that bead will complete the circle

Options

A. `sqrt((3k)/R`
B. `sqrt((8k)/(3R)`
C. `sqrt((6k)/R`
D. none of these

Answer

B

Explanation

value of A at any angular position 'theta' is given by
`f=(km)/r^(2)` here is given by
`r=sqrt((Rcostheta-R//2)^(2)+(Rsintheta)^(2))=sqrt(R^(2)+R^(2)//4-R^(2)costheta)=R/2sqrt(5-4costheta)`
`Also (2sinalpha)/R=sin 0/r i.e sinalpha =sintheta/sqrt(5-4costheta)`
for small angular displacement 'dtheta' work done by this force dw =-frd0cos(9theta-alpha)=-fR sin alpha dtheta
`(4kmR)/(R^(2)(5-4costheta)) sintheta/sqrt(5-4costheta)dtheta`
work done in moving bead from AtoB
`Deltaw=-(4km)/R underset(theta)overset(pi)(int) (sin theta d theta)/(5-4costheta)^(3//2)=-(km)/R underset(theta)overset(pi)(int)(4sin theta d theta)/(5-4costheta)^(3//2)`
`=(km)/R underset(theta) overset(9)(int) (dt)/t^(3//2)=-(km)/R|t^(-1//2)/-(1//2)|_(1)^(9)`
`=+(2km)/R(1/3-1)=-4/3(km)/R`
:.Energy provided at point A most be equal to this work done
` :. 1/2mv_(m)^(2)=4/3(km)/R Rightarrow v_(min)= sqrt((8k)/(3R))`