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

The Correct Option is D

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
\(m \omega^{2} r=\mu m g\)
\(\Rightarrow \omega^{2} r=\mu g\)
\(\Rightarrow \mu=\frac{\omega^{2} r}{g}=\frac{(3.5 \times 2 \pi)^{2}\left(1.25 \times 10^{-2}\right)}{10}\)
\(\Rightarrow \mu=0.60\)