Solve the following :

Question:

A mosquito is sitting on an L.P. record disc rotating on a turn table at 333 per minute. The distance 3 of the mosquito from thecentre of the turn table is $10 \mathrm{~cm}$. Show that the friction coefficient between the record and the mosquito is greater than $\mathrm{It} \pi^{2} / 81$. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

$\omega=33^{\frac{1}{3}} \mathrm{rpm}=\frac{100}{3} \mathrm{rpm}$

$=\left(\frac{100}{3}\right) \times \frac{(2 \pi)}{60}$

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

For mosquito to remain at rest on L.P record.

Frictional force $\geq$ Centrifugal force

$\mu N \geq m R \omega^{2}$

$\mu \mathrm{mg} \geq \mathrm{mR} \omega^{2}$

$\mu \geq \frac{(0.1)}{(10)}\left(\frac{10 \pi}{9}\right)^{2}$

$\mu \geq \frac{\pi^{2}}{81}$

 

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