An annular ring with inner and outer radii `R_1` and `R_2` is rolling wihtout slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, `F_1/F_2` is

Options

A. `R_(1)/R_(2)`
B. 1
C. `(R_(1)/R_(2))^(2)`
D. `R_(2)/R_(1)`

Explanation

To solve the problem, we need to find the ratio of the forces experienced by two particles located on the inner and outer parts of an annular ring that is rolling without slipping with a uniform angular speed. Let's denote the inner radius as \( R_1 \) and the outer radius as \( R_2 \).

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Particles:**
- The forces acting on the particles at the inner radius \( R_1 \) and outer radius \( R_2 \) can be expressed using the formula for centripetal force. The centripetal force \( F \) acting on a particle in circular motion is given by:
\[ F = m \omega^2 r \]
where \( m \) is the mass of the particle, \( \omega \) is the angular speed, and \( r \) is the radius at which the particle is located.

2. **Calculating the Force on the Inner Particle:**
- For the particle located at the inner radius \( R_1 \):
\[ F_1 = m \omega^2 R_1 \]

3. **Calculating the Force on the Outer Particle:**
- For the particle located at the outer radius \( R_2 \):
\[ F_2 = m \omega^2 R_2 \]

4. **Finding the Ratio of the Forces:**
- Now, we can find the ratio of the forces \( F_1 \) and \( F_2 \):
\[ \frac{F_1}{F_2} = \frac{m \omega^2 R_1}{m \omega^2 R_2} \]
- The mass \( m \) and the angular speed \( \omega \) are common in both the numerator and the denominator, so they cancel out:
\[ \frac{F_1}{F_2} = \frac{R_1}{R_2} \]

5. **Conclusion:**
- Therefore, the ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring is:
\[ \frac{F_1}{F_2} = \frac{R_1}{R_2} \]

### Final Answer:
The ratio of the forces \( \frac{F_1}{F_2} \) is \( \frac{R_1}{R_2} \).

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