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 $1.25cm$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : $(g=10m/s^2)$

• A. 0.5
• B. 0.3
• C. 0.7
• D. 0.6

Answer 1

Answer: (D)

We have
$m{\omega }^{2}r=\mu mg$
Given: rate of rotation $=3.5rev/s$
$\Rightarrow 1$ revolution $=2\pi rad$

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