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 static friction between the rod and bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is:

• A. $\text{infinitesimal}$
• B. $\frac{\mu }{\sqrt{\alpha }}$
• C. $\frac{1}{\sqrt{\mu \alpha }}$
• D. $\sqrt{\frac{\mu }{\alpha }}$

Answer 1

The correct option is D $\sqrt{\frac{\mu }{\alpha }}$

Tangential force $\left(F_t\right)$ on the bead will be given by the normal reaction $\left(N\right)$

Centripetal force $\left(F_c\right)$ is provided by limiting friction. The bead starts sliding when the centripetal force is just equal to the limiting friction $\left(f_{max}\right)$.


$\therefore$ Applying the Newton's $2^{nd}$ law of motion in tangential and the radial directions:

$F_t=m\times a_t\dots \left(i\right)$

Relation between tangential acceleration and angular acceleration:
$a_t=r\alpha \dots \left(ii\right)$; $r=L$

$\therefore F_t=m\times L\alpha$
$\Rightarrow N=m\times L\alpha \dots \left(iii\right)$

Limiting value of friction
$f_{\text{max}}=\mu N=\mu m\alpha L\dots \left(iv\right)$

Now Angular speed at time $t$ is:
$\omega =\alpha t$, considering ${\omega }_i=0$ at $t=0$

Centripetal force at time $t$ will be
$F_c=m{\omega }^{2}L=m{\alpha }^2{t}^{2}L\dots \left(v\right)$

Equating Eqs. $\left(iv\right)$ and $\left(v\right)$, we get:
$F_c=f_{\text{max}}$
$\Rightarrow m{\alpha }^2{t}^{2}L=\mu m\alpha L$

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

Hence for $t\ge \sqrt{\frac{\mu }{\alpha }},F_c\ge f_{\text{max}}$ i.e., the bead just begins sliding on the rod.