A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration a. If the coefficient of friction between the rod and the bead is u, and gravity is neglected, then the time after which the bead starts slipping is :

Options

A. infinitesimal
B. `(mg)/(4)`
C. `(mg)/(2)`
D. `mg(1-mu)`

Answer

b

Explanation

Here centripetal force for the bead is provided by friction `F = mu(Fomega)/(t) a_(t)` where `a_(t) `=tangential acceleration and `a_(t) =(upsilon)/(t)`
L = distance of point :. linear speed `upsilon = Lomega`
Hence `F= mum (L omega)/(t). mLomega^(2)= mum (Lomega)/(t)`
`omega=(mu)/(t)`
`t=(mu)/(omega)`But=`alpha.t`
`:.t=(mu)/(alphaxxt)`
or `t^(2)=(mu)/(alpha)`
and `t=sqrt(mu)/(alpha)`
Hence choice is (a)