Question

A mosquito is sitting on an L.P. record disc rotating on a turn table at $33\frac{1}{3}$ revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π²/81.

Answer 1

$\mathrm{Frequency}\mathrm{of}\mathrm{disc}=n=33\frac{1}{3}\mathrm{rev}/\mathrm{m}=\frac{100}{3\times 60}\mathrm{rev}/\mathrm{s}$
$\mathrm{Angular}\mathrm{velocity}=\mathrm{\omega }=2\mathrm{\pi }n=2\mathrm{\pi }\times \frac{100}{180}=\frac{10\mathrm{\pi }}{9}\mathrm{rad}/\mathrm{s}$
$$\begin{gathered} r=10\mathrm{cm}=0.1\mathrm{m} \\ g=10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

It is given that the mosquito is sitting on the L.P. record disc. Therefore, we have:
Friction force ≥ Centrifugal force on the mosquito
⇒ μmg ≥ mrω²
⇒ μ ≥ rω²/g
$$\begin{gathered} \Rightarrow \mu \ge \mathrm{0.1\times }{\left({\displaystyle \frac{\mathrm{10\pi }}{9}}\right)}^2\frac{1}{10} \\ \Rightarrow \mu \ge \frac{{\mathrm{\pi }}^2}{81} \end{gathered}$$