Question

A mosquito is sitting on an $L.P.$ record of a gramophone disc rotating on a turn table at $33\frac{1}{3}$ revolution per minute. The distance of the mosquito from the centre of the disc is $10cm$. Show that the friction coefficient between the record and the mosquito is greater than ${\pi }^2/81$. Take $g=10m/s^2$

Answer 1

A mosquito is sitting on an L.P record disc and rotating on a turn table at $33\frac{1}{3}rpm$

$n=33\frac{1}{3}rpm=\frac{100}{3\times 60}rps$

$\therefore \omega =2\pi n=2\pi \times \frac{100}{180}=\frac{10\pi }{9}rad/sec$

$r=10cm=0.1m,g=10m/{sec}^2$

$r=10cm=0.1m,g=10m/{sec}^2$

$\mu mg\ge mr{\omega }^2\Rightarrow \mu =\frac{{\omega }^2}{g}\ge \frac{0.1\times {\left(\frac{10\pi }{9}\right)}^2}{10}\Rightarrow \mu \ge \frac{{\pi }^2}{81}$