Question

A disc rotates about its axis of symmetry in a horizontal 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 : $(g = 10 \, m/s^2)$

• A. 0.5
• B. 0.3
• C. 0.7
• D. 0.6

Answer 1

Answer: (D)

0.6

Answer 2

The correct answer is D:0.6
We have
$m \omega^{2} r=\mu m g$
Given: rate of rotation $=3.5\, rev / s$
$\Rightarrow 1$ revolution $=2 \,\pi \,rad$

That is, $3.5$ revolutions $=3.5 \times 2 \,\pi \,rad$
Therefore,
$\omega=3.5 \times 2 \,\pi\, rad / s$
$r=1.25\, cm =1.25 \times 10^{-2}\, m$
Thus, from E (1), we have
$m \omega^{2} r=\mu m g$
$\Rightarrow \omega^{2} r=\mu g$
$\Rightarrow \mu=\frac{\omega^{2} r}{g}=\frac{(3.5 \times 2 \pi)^{2}\left(1.25 \times 10^{-2}\right)}{10}$
$\Rightarrow \mu=0.60$