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 α . 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 correct answer is √(μ/α).

To find the time after which the bead starts slipping along the horizontal rod, we analyze the forces acting on the bead of mass $m$ in the horizontal plane. Since gravity is neglected, the only forces acting on the bead are the normal force and the friction force exerted by the rotating rod.

Let the bead be at a distance $L$ from the pivot point $A$. The rod rotates with a constant angular acceleration $\alpha$. Starting from rest, the angular velocity $\omega$ of the rod at any time $t$ is given by:

$\omega =\alpha t$

As the rod rotates, the bead experiences two types of acceleration: tangential acceleration and centripetal acceleration.

1. Tangential Acceleration:

The tangential acceleration keeps the bead speed matching the rotation of the rod at radius $L$. It is given by:

$a_t=L\alpha$

The normal force $N$ exerted by the rod on the bead perpendicular to the rod provides this tangential acceleration:

$N=m{a}_t=mL\alpha$

Since gravity is neglected, there is no vertical normal force, and this horizontal force is the net normal force acting on the bead.

2. Centripetal Acceleration:

The centripetal acceleration is directed along the rod towards the center of rotation $A$. It is given by:

$a_c={\omega }^{2}L={\left(\alpha t\right)}^{2}L={\alpha }^2{t}^{2}L$

The static friction force $f$ acting along the rod towards the center $A$ must provide this centripetal acceleration to prevent the bead from sliding outward:

$f=m{a}_c=mL{\alpha }^2{t}^2$

3. Condition for Slipping:

The maximum available static friction force (limiting friction) is given by:

$f_{\text{max}}=\mu N$

Substituting the expression for the normal force $N$:

$f_{\text{max}}=\mu mL\alpha$

The bead will start slipping when the required centripetal friction force $f$ becomes equal to the limiting static friction force $f_{\text{max}}$:

$f=f_{\text{max}}$

$mL{\alpha }^2{t}^2=\mu mL\alpha$

By simplifying the equation, we can cancel $m$, $L$, and one factor of $\alpha$ from both sides:

$\alpha t^2=\mu$

$t^2=\frac{\mu }{\alpha }$

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

Thus, the time after which the bead starts slipping is:

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