Question

A coin is placed on a turntable which is making $40$ revolutions per minute. The distance of the coin from the center of the turntable is $\frac{3}{{\pi }^2}\text{m}$. The minimum friction coefficient between the turntable and the coin is [Take $g=10{\text{m/s}}^2$]

• A. $0.53$
• B. $0.72$
• C. $0.83$
• D. $0.27$

Answer 1

The correct option is A $0.53$

FBD of coin:

The static friction provides the centripetal force for the circular motion
$\therefore f_s=\frac{m{v}^2}{r}$ (or) $f_s=mr{\omega }^2$
Since $f_s\le {\mu }_{s}R$, we get
$mr{\omega }^2\le {\mu }_{s}mg$ ($\because R=mg$)
i.e ${\omega }^2\le \frac{{\mu }_{s}g}{r}$
$\Rightarrow {\mu }_s\ge \frac{r{\omega }^2}{g}$
Given, $\omega =40\text{rev/min}=40\times \frac{2\pi }{60}=\frac{4\pi }{3}\text{rad/s}$
$r=\frac{3}{{\pi }^2}\text{m}$
$\therefore {\mu }_{smin}=\frac{\frac{3}{{\pi }^2}\times {\left(\frac{4\pi }{3}\right)}^2}{10}=\frac{\frac{3}{{\pi }^2}\times \frac{16{\pi }^2}{9}}{10}=\frac{16}{30}=0.53$