As mosquito is sitting on an L.P. record disc rotating on a trun tabel at `33 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^2/81`. Take `g=10m/s^2`

Explanation

To solve the problem, we need to show that the friction coefficient (μ) between the LP record and the mosquito is greater than \(\frac{\pi^2}{81}\). Here are the steps to derive this:

### Step 1: Calculate the angular speed (ω)
The turntable rotates at \(33 \frac{1}{3}\) revolutions per minute (rpm).

To convert this to radians per second:
\[ \omega = \text{revolutions per second} \times 2\pi \]
First, convert revolutions per minute to revolutions per second:
\[ \text{Revolutions per second} = \frac{33 \frac{1}{3}}{60} = \frac{100}{3 \times 60} = \frac{100}{180} = \frac{5}{9} \text{ revolutions per second} \]
Now, multiply by \(2\pi\):
\[ \omega = \frac{5}{9} \times 2\pi = \frac{10\pi}{9} \text{ radians per second} \]

### Step 2: Identify the radius (R)
The distance of the mosquito from the center of the turntable is given as 10 cm. We need to convert this to meters:
\[ R = 10 \text{ cm} = 0.1 \text{ m} \]

### Step 3: Relate friction force and centripetal force
The maximum friction force (F_max) that can act on the mosquito is given by:
\[ F_{\text{max}} = \mu m g \]
where:
- \(m\) is the mass of the mosquito,
- \(g\) is the acceleration due to gravity (given as \(10 \text{ m/s}^2\)).

The centripetal force required to keep the mosquito moving in a circle is:
\[ F_c = m \omega^2 R \]
Setting the maximum friction force equal to the centripetal force gives:
\[ \mu m g = m \omega^2 R \]

### Step 4: Cancel mass (m) and rearrange
Since \(m\) appears on both sides, we can cancel it out (assuming \(m \neq 0\)):
\[ \mu g = \omega^2 R \]
Now, we can solve for \(\mu\):
\[ \mu = \frac{\omega^2 R}{g} \]

### Step 5: Substitute the values
Substituting the known values:
- \(\omega = \frac{10\pi}{9}\)
- \(R = 0.1\)
- \(g = 10\)

Calculating \(\omega^2\):
\[ \omega^2 = \left(\frac{10\pi}{9}\right)^2 = \frac{100\pi^2}{81} \]

Now substituting into the equation for \(\mu\):
\[ \mu = \frac{\frac{100\pi^2}{81} \times 0.1}{10} = \frac{100\pi^2}{81 \times 10} = \frac{10\pi^2}{81} \]

### Step 6: Compare with \(\frac{\pi^2}{81}\)
Now we need to show that:
\[ \mu > \frac{\pi^2}{81} \]
Since \(\frac{10\pi^2}{81} > \frac{\pi^2}{81}\) (as \(10 > 1\)), we conclude:
\[ \mu > \frac{\pi^2}{81} \]

### Conclusion
Thus, we have shown that the coefficient of friction between the record and the mosquito is indeed greater than \(\frac{\pi^2}{81}\).

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