Question 140 of 210

A coin is placed on a disc. The coefficient of friction between the coin and the disc is $\mu$. If the distance of the coin from the center of the disc is $r$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :

A$\dfrac{\mu g}{r}$B$\sqrt{\dfrac{r}{\mu g}}$C$\sqrt{\dfrac{\mu g}{r}}$D$\dfrac{\mu }{\sqrt{rg}}$

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Answer: (C) $\sqrt{\dfrac{\mu g}{r}}$

Normal force acting on the coin is, $N=mg$.
Required friction force to provide centripetal force for the coin is, $f=m{\omega }^{2}r$ and maximum value of static friction force available is, $f=\mu N=\mu mg$.
For limiting condition, we can write
$\Rightarrow \mu mg=mr{\omega }^{2}$
$\Rightarrow \omega =\sqrt{\dfrac{\mu g}{r}}$