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 : 

Rod AB is set in angular motion about A, hence the radius of circular path of the bead is L. Its tangential acceleration is $\mathrm{a}=\mathrm{L} \alpha$ and therefore tangential force acting on the bead of mass m is $=\mathrm{mL} \alpha$ 

Where $\alpha$ is the angular acceleration of the bead.Normal (or radial ) force on the bead

$=mL{\omega }^2(\because v=L\omega )$ Where $\omega$ is the angular velocity of the bead.

Frictional force on the bead $=\mu \times$( tangential force) $=\mu mL\alpha$

The bead just starts slipping over the rod when  Frictional force= Normal force

$=\mu mL\alpha =mL{\omega }^2=\mathrm{mL}(\alpha \mathrm{t}{)}^2\omega =0+\alpha t=\alpha t\mu =\alpha {\mathrm{t}}^2\Rightarrow t=\sqrt{\frac{\mu }{\alpha }}$