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 :

• A. 0.5
• B. 0.3
• C. 0.7
• D. 0.6

Answer 1

The correct option is D 0.6

We know,

$m{\omega }^{2}r=\mu mg$--------(A)

$\omega =2\pi N=2\pi \times 3.5=7\pi$ here N=number of revolution per second

Also given distance of coin from the axis of rotation $r=1.25cm=1.25\times {10}^{-2}$

Putting the above values in A we get,

$\mu =\frac{{\omega }^{2}r}{g}=\frac{49{\pi }^{2}1.25\times {10}^{-3}}{10}=0.60$