Question

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with a constant angular acceleration $\alpha$. If the coefficient of friction between the rod and bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is

Answer 1

Let the angular velocity is $\omega$.

It can be written as,

$\omega =\alpha t$

Since, the required centripetal force at this time is being provided by the friction force.

The required centripetal force is given as,

$F_c=m{\omega }^{2}L$

The limiting value of the frictional force is given as,

$F_f=\mu m\alpha L$

Since, the frictional force should be equal to the required centripetal force.

It can be written as,

$\mu m\alpha L=m{\omega }^{2}L$

$\mu m\alpha L=m{\left(\alpha t\right)}^{2}L$

$t=\sqrt{\frac{\mu }{\alpha }}\sec$

Thus, the time after which the bead starts slipping is $\sqrt{\frac{\mu }{\alpha }}\sec$.