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 `alpha.` if the coefficient of friction between the rod and the bead is `mu`, and gravity is neglected, then the time after which the bead starts slipping is

Options

A. `sqrt(mu/alpha)`
B. `mu/(sqrt(alpha))`
C. `1/(sqrt(mu alpha))`
D. Infinitesimal

Answer

A

Explanation

Tangential force `(F_t)` of the bead will be given by the normal reaction (N), while centripetal force `(F_C)` is provided by friction `(f_r)`. The bead starts sliding when the centripetal force is just equal to the limiting friction. Therefore
`F_t = ma = m alpha L = N `
`therefore` limiting value of friction
`(fr)_("max") = muN = mu m alpha L `..... (1)
Angular velocity at time t is `omega := alpha t`
`therefore` Centripetal force at time t will be
`F_c = mLomega^2 = mLalpha^2 t^2` .......... (2)
Equating equations (1) and (2), we get
`t = sqrt((mu)/(alpha))`
For `t gt sqrt((mu)/(alpha)), F_c gt (f_r)_("max")`
İ.e. the bead starts sliding. In the figure Ft is perpendicular to the paper inwards `ox` .