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 µ, and gravity is neglected, then the time after which the bead starts slipping is:

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 $=ma=m\alpha L$

$\therefore$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.
$$\begin{gathered} \mu R=\frac{m{v}^2}{r} \\ \text{⇒} \mu m\alpha L=mL{\omega }^2 \left[\because v=r\omega \right] \\ \Rightarrow \mu \alpha ={\omega }^2=(\alpha t{)}^2 \left[\because \omega =0+\alpha t\right] \\ \Rightarrow \mu \alpha ={\alpha }^2{t}^2 \\ \Rightarrow t=\sqrt{\frac{\mu }{\alpha }} \end{gathered}$$