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

Last updated date: 13th Jun 2024
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Hint: First find the value of the rotation per unit of time period in seconds and then find the angular velocity of the disc spinning in terms of that revolution per unit of time period. After finding the angular velocity we find the centripetal force of the mosquito i.e.
Centripetal force $=mr{{\omega }^{2}}$
And equate the force by the frictional force of the mosquito sitting on the disc as:
Frictional force of the mosquito $=\mu mg$
where $\omega$ is the angular velocity, $m$ is the mass of the mosquito, $\mu$ is the frictional force, $r$ is the radius of the disc from the where the mosquito is sitting to the center of the disc.

Complete step by step solution:
The disk is moving at a rpm of $33\dfrac{1}{3}$ Now the gravity is given in seconds and not in minutes thereby we need to change the time from minutes to second. To convert the rpm to rps, we divide the rpm value by 60. Hence, the rotation of the record disc per unit of time is:
$n=\dfrac{33\dfrac{1}{3}rpm}{60}$
$\Rightarrow n=\dfrac{100}{180}rps$
Now that we know the rotation per unit time we can find the angular momentum of the disc rotating with the mosquito sitting on it. To find the angular velocity we use the formula as:
$\omega =2\pi n$
Placing the value of n as $\dfrac{100}{180}rps$, we get the angular velocity as:
$\Rightarrow \omega =2\pi n$
$\Rightarrow \omega =2\pi \dfrac{100}{180}$
$\Rightarrow \omega =\dfrac{10}{9}\pi$
Now, the mosquito is sitting $10\text{ }cm$ from the center of the disc and this $10\text{ }cm$ will act like a radius for the mosquito hence, the centripetal force of the mosquito sitting on the disc is denoted by the formula of:
Centripetal force $=mr{{\omega }^{2}}$
And the frictional force acting on the disc due to the mosquito sitting on the disc is given as:
Frictional force of the mosquito $=\mu mg$
Now with frictional force greater than equal to centripetal force, we equate both the forces together we get the friction coefficient as:
$\mu mg\ge mr{{\omega }^{2}}$
$\Rightarrow \mu g\ge r{{\omega }^{2}}$
$\Rightarrow \mu \ge \dfrac{r{{\omega }^{2}}}{g}$
Placing the value of the angular velocity $\omega =\dfrac{10}{9}\pi$ and radius r as $0.1m$ in the above formula we get the value of friction coefficient as:
$\Rightarrow \mu \ge \dfrac{r{{\left( \dfrac{10}{9}\pi \right)}^{2}}}{g}$
$\Rightarrow \mu \ge \dfrac{0.1\times {{\left( \dfrac{10}{9}\pi \right)}^{2}}}{10}$
$\Rightarrow \mu \ge \dfrac{{{\pi }^{2}}}{81}$

Hence, proved that the mosquito is applying a friction coefficient of $\mu \ge \dfrac{{{\pi }^{2}}}{81}$.

Note: Student might go wrong if they take the formula of $\dfrac{m{{v}^{2}}}{r}$ and take v as the rpm or rps value, the rpm or rps is the unit measure of angular velocity but only for a unit period of time hence, the formula $\dfrac{m{{v}^{2}}}{r}$ is not applicable in this scenario whereas the formula for the force of rotation is taken as centripetal force $\left( mr{{\omega }^{2}} \right)$ as the force of the mosquito is acting inside and not on the periphery of the disc.