Question

A mosquito is sitting on an L.P. record disc rotating on a turn table at 3313 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

To Prove : The friction coefficient between the record and the mosquito is greater than π2/81.

Explanation:

Step 1:

Given that,

No. of Revolutions made in 1 minutes  $\begin{equation}33 \frac{1}{3}=\frac{100}{3}$

Frequency (f) =$\frac{100}{180} s^{-1}$

$\begin{equation}\text { Frequency }(f)=\frac{5}{9} s^{-1}$

Step 2:

$\\As\ we\ know\ that,\\ω = \pi f$

=2×3.14× $\frac{5}{9}$

=6.28× $\frac{5}{9}$

= $\frac{31.4}{9}$

=3.48

Step 3:

Given that the distance of the mosquito from centre of the turn table is almost 10 cm

Radius = 10 cm = 0.1 m.

Let g = 10 m/s

Case 1:

$\begin{equation}\begin{aligned}&f \geq \frac{m v^{2}}{r}\\&\mu \mathrm{N} \geq \frac{m v^{2}}{r}\\&\mu m g \geq \frac{m v^{2}}{r}\\&\mu g \geq \frac{v^{2}}{r}\\&\mu \geq \frac{v^{2}}{r g}\\&\mu \geq w^{2} r g\\&\mu \geq 3.48^{2} \times 0.1 \times 10\end{aligned}$

$\begin{equation}\mu \geq 12.11$

Case 2:

$\begin{equation}\frac{\pi^{2}}{81}=\frac{3.14^{2}}{81}=\frac{9.85}{81}$

=0.121

Then the friction coefficient (μ) between the record and the mosquito is comparatively greater than π2/81.