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)/(sqrtalpha)`
C. `(1)/(sqrt(mu alpha))`
D. infinitesimal

Answer

A

Explanation

The radius of the circular motion of the bead is r = L . The linear acceleration of the bead is a = `alpha r = alpha L` . If m is the mass of the bead , then force acting on the bead is R = m`alphaL`
`therefore ` Reaction force acting on the bead is R = `m alpha L`,
The bead starts slipping when frictional force between the bead and the rod becomes equal to centrifugal force acting on the bead , i.e.,
`mu R = (mv^(2))/(r)`
or `" " mu m alpha L = mr omega^(2) = m L omega^(2) " " (because v = r omega)`
or `mu alpha = omega^(2) = (alpha t)^(2) " " (because omega = alpha t)`
or `t = sqrt((mu)/(alpha))`