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 in angular motion about A with constant angular acceleration $\alpha$. Ifthe 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 :

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

Answer 1

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

Given,

Distance $L$ Acceleration $\alpha$ Coefficent =$\mu$

So, the bead will start sliding when centripetal force is equal to limiting friction.

i. e. $m{w}^{2}L=f\left(max\right)\Rightarrow m{w}^{2}L=uN----\left(1\right)$

We know that,

Here. Normal Reaction is provided by Tangential force.

$F\left(t\right)=mL\alpha =N---\left(2\right)$

So, Substituting value of N in eq. (1),

Now, at time $t$,

$w=0+\alpha t\Rightarrow w=\alpha t$

Thus, $m{w}^{2}L=umL\alpha \Rightarrow m(\alpha t{)}^{2}L=umL\alpha \Rightarrow \alpha t^2=u\Rightarrow t=\sqrt{\frac{u}{\alpha }}$