A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25\,cm$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $(g\, = 10\,m/s^2)$
JEE MAIN 2018, Difficult
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Using, $\mu mg =\frac{m v^2}{r}=m r \omega^2$
$\omega=2 \pi n=2 \pi \times 3.5=7 \pi rad / \sec$
Radius, $r=1.25 \ cm=1.25 \times 10^{-2} m$
Coefficient of friction, $\mu=$ ?
$\mu m g=\frac{m(r \omega)^2}{r}(v=r \omega)$
$\Rightarrow \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10}$
$=\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6$
$\omega=2 \pi n=2 \pi \times 3.5=7 \pi rad / \sec$
Radius, $r=1.25 \ cm=1.25 \times 10^{-2} m$
Coefficient of friction, $\mu=$ ?
$\mu m g=\frac{m(r \omega)^2}{r}(v=r \omega)$
$\Rightarrow \mu=\frac{r \omega^2}{g}=\frac{1.25 \times 10^{-2} \times\left(7 \times \frac{22}{7}\right)^2}{10}$
$=\frac{1.25 \times 10^{-2} \times 22^2}{10}=0.6$
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