A circular disc is made to rotate in horizontal plane about its centre at the rate of 2 rps. The greatest distance of a coin placed on the disc from its centre so that it does not skid is (`mu` is coefficient of friction)

Options

A. `(mu)/(6.2)m`
B. `6.2 mu cm`
C. `(mu)/(6.2) cm`
D. `(mu g)/(6.2) cm`

Answer

To solve the problem of finding the greatest distance of a coin placed on a rotating disc from its center so that it does not skid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Coin**:
- The coin experiences a centripetal force due to the rotation of the disc, which is provided by the frictional force between the coin and the disc.
- The forces acting on the coin are:
- Centripetal force: \( F_c = m \omega^2 r \)
- Weight of the coin: \( W = mg \)
- Normal force: \( N = mg \)
- Frictional force: \( F_f = \mu N = \mu mg \)

2. **Set Up the Equation for Forces**:
- For the coin to not skid, the frictional force must be equal to the required centripetal force:
\[ F_f = F_c \]
- Thus, we have:
\[ \mu mg = m \omega^2 r \]

3. **Cancel Mass from Both Sides**:
- Since mass \( m \) appears on both sides, we can cancel it:
\[ \mu g = \omega^2 r \]

4. **Solve for Radius \( r \)**:
- Rearranging the equation gives:
\[ r = \frac{\mu g}{\omega^2} \]

5. **Calculate Angular Velocity \( \omega \)**:
- The disc rotates at 2 revolutions per second (rps). To convert this to radians per second:
\[ \omega = 2 \text{ rps} \times 2\pi \text{ radians/revolution} = 4\pi \text{ radians/second} \]

6. **Substitute \( \omega \) into the Radius Equation**:
- Now substitute \( \omega \) into the equation for \( r \):
\[ r = \frac{\mu g}{(4\pi)^2} \]

7. **Calculate \( (4\pi)^2 \)**:
- Calculate \( (4\pi)^2 \):
\[ (4\pi)^2 = 16\pi^2 \]

8. **Substitute \( g \) (approximately 10 m/s²)**:
- Using \( g \approx 10 \text{ m/s}^2 \):
\[ r = \frac{\mu \cdot 10}{16\pi^2} \]

9. **Final Calculation**:
- The value of \( \pi^2 \) is approximately 9.87, so:
\[ r \approx \frac{\mu \cdot 10}{16 \cdot 9.87} \approx \frac{\mu \cdot 10}{157.92} \approx 0.0633\mu \text{ meters} \]
- To convert this to centimeters:
\[ r \approx 6.33\mu \text{ centimeters} \]

### Final Answer:
The greatest distance \( r \) of the coin from the center of the disc so that it does not skid is approximately \( 6.33\mu \) centimeters.