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 then
($g=10m/s^2$)

Answer 1

$\begin{array}{c}3.5rev/second1rev\to 2\pi rad3.5rev\to 2\pi \times 3.5rad\\ \Rightarrow w=7\pi rad/\sec \mu mg\\ =\frac{m{v}^2}{1.25}\mu mg\\ =\frac{m{\left(rw\right)}^2}{r}\mu mg\\ =mr{w}^2\mu \end{array}$

$\begin{array}{c}=\frac{r{w}^2}{g}\\ =\frac{1.25\times {10}^{-2}\times {\left(7\times \frac{22}{7}\right)}^2}{10}\\ =\frac{1.25\times {10}^{-2}\times {22}^2}{10}\\ =0.6\end{array}$