Question

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 to rotate on a horizontal circular path, about a vertical axis passing through $A$ with constant angular acceleration $\alpha$. If the coefficient of static friction between the rod and the bead is $\mu$ . Then find the time when the bead starts slipping on rod. Neglect the effect of gravity.

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

Answer 1

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

Till, the time when bead starts slipping friction will act at it's limiting value ($f_{max}=\mu N$) to provide the necessary centripetal force for bead.



This is a case of horizontal circular motion. In which the centripetal acceleration ($a_c$) of bead and the tangential acceleration of bead ($a_t$) are shown in X and Y- direction respectively as per the reference axis shown below:


Applying the equation of dynamics for bead in X and Y direction as per FBD:

For bead the normal reaction is acting in tangential direction, to provide the tangential acceleration:

$N=m\times a_t......i$
frictional force acts towards centre of circular path to provide centripetal force:
$f_{max}=mr{\omega }^2$

$\Rightarrow f_{max}=mL{\omega }^2........ii$ ;putting $r=L$

Now,
$f_{max}=\mu N=\mu m{a}_t.........iii$

Also tangential acceleration, $a_t=r\alpha =L\alpha$

$\Rightarrow$Expression for angular speed involving constant angular acceleration $\alpha$
$\therefore \omega ={\omega }_i+\alpha t=0+\alpha t=\alpha t.............iv$

Combining the Eq. ($ii$) ($iii$) and ($iv$):

$mL{\omega }^2=\mu \left(m{a}_t\right)$

$\Rightarrow mL{\omega }^2=\mu mL\alpha$
So, ${\omega }^2=\mu \alpha$
$\Rightarrow (\alpha t{)}^2=\mu \alpha$
$\therefore \alpha t=\sqrt{\mu \alpha }$

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