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

 

When we are giving an angular acceleration to the rod, the bead is also having an instantaneous $a=L\alpha .$ This will happen when a force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to friction between the bead and the rod, this does not happen to the extent to which frictional force is capable of holding the bead. The frictional force here provides the necessary centripetal force. If the instantaneous angular velocity is $\omega ,$ then $mL{\omega }^2=\mu \left(ma\right)=\mu maL\alpha \Rightarrow {\omega }^2=\mu \alpha$

By applying $\omega ={\omega }_0+\alpha t,$ we get $\omega =\alpha t$

$\therefore {\alpha }^2{t}^2=\mu \alpha \Rightarrow t=\sqrt{\frac{\mu }{\alpha }}$