A coin just remains on a disc rotating at a steady rate of 180r.p.m .A coin is kept at a distance of 2 cm from the axis of rotationn . The coefficient of fricition between the coin and the disc is _____ `[ gg= 9.8 m//s^(2)]`

Options

A. `0.524`
B. `0.624`
C. `0.724`
D. `0.824`

Explanation

To find the coefficient of friction between the coin and the disc, we can follow these steps:

### Step 1: Convert RPM to RPS
Given that the disc rotates at 180 revolutions per minute (RPM), we need to convert this to revolutions per second (RPS).

\[ \text{RPS} = \frac{180 \text{ RPM}}{60} = 3 \text{ RPS} \]

### Step 2: Calculate Angular Velocity (ω)
The angular velocity (ω) in radians per second can be calculated using the formula:

\[ \omega = 2\pi \times \text{RPS} \]

Substituting the value of RPS:

\[ \omega = 2\pi \times 3 = 6\pi \text{ rad/s} \]

### Step 3: Identify the Radius
The distance of the coin from the axis of rotation is given as 2 cm. We need to convert this to meters:

\[ r = 2 \text{ cm} = 0.02 \text{ m} \]

### Step 4: Apply the Centripetal Force Equation
The centripetal force required to keep the coin in circular motion is given by:

\[ F_c = \frac{mv^2}{r} \]

Where \( v \) can be expressed in terms of angular velocity (ω):

\[ v = r\omega \]

Substituting this into the centripetal force equation:

\[ F_c = \frac{m(r\omega)^2}{r} = m r \omega^2 \]

### Step 5: Relate Centripetal Force to Frictional Force
The frictional force (F_f) that provides the necessary centripetal force is given by:

\[ F_f = \mu mg \]

Where \( \mu \) is the coefficient of friction and \( mg \) is the weight of the coin.

### Step 6: Set the Forces Equal
For the coin to remain on the disc without slipping, the frictional force must equal the centripetal force:

\[ \mu mg = m r \omega^2 \]

### Step 7: Cancel Mass (m)
Since mass (m) appears on both sides of the equation, we can cancel it out:

\[ \mu g = r \omega^2 \]

### Step 8: Solve for the Coefficient of Friction (μ)
Rearranging the equation to solve for μ gives:

\[ \mu = \frac{r \omega^2}{g} \]

### Step 9: Substitute Known Values
Now we can substitute the values we have:

- \( r = 0.02 \) m
- \( \omega = 6\pi \) rad/s
- \( g = 9.8 \) m/s²

Calculating \( \omega^2 \):

\[ \omega^2 = (6\pi)^2 = 36\pi^2 \]

Now substituting into the equation for μ:

\[ \mu = \frac{0.02 \times 36\pi^2}{9.8} \]

### Step 10: Calculate the Value
Calculating the value:

\[ \mu = \frac{0.02 \times 36 \times 9.87}{9.8} \approx 0.725 \]

### Conclusion
The coefficient of friction between the coin and the disc is approximately **0.725**.

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