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 $\pi \frac{2}{81.}$ Take $g=10m/s^2.$

Answer 1

A mosquito is sitting on an L.P. record disc and rotating on a turn table at $33\frac{1}{2}$ rpm.

$n=33\frac{1}{3}rpm$

$=\frac{100}{3\times 60}rps$

$\therefore \omega =2\pi n=2\pi \times \frac{100}{180}$

$=\frac{10\pi }{9}rad/sec$

$r=10cm=0.1m$

$g=10m/se{c}^2$

$\mu mg\ge mr{\omega }^2$

$\therefore \mu >\frac{r{\omega }^2}{g}\ge \frac{0.1\times (10\pi /9{)}^2}{10}$

$\Rightarrow \mu >\left(\frac{{\pi }^2}{81}\right)$