A ring of mass 10 kg and radius 0.2 m is rotating about its geometrical axis at 20 rev/sec. Its moment of inertia is

Options

A. `"0.2 kg - m"^(2)`
B. `"3.0 kg - m"^(2)`
C. `"0.4 kg - m"^(2)`
D. `"5.0 kg - m"^(2)`

Explanation

To find the moment of inertia of a ring, we can use the formula:

\[ I = m r^2 \]

where:
- \(I\) is the moment of inertia,
- \(m\) is the mass of the ring,
- \(r\) is the radius of the ring.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the ring, \(m = 10 \, \text{kg}\)
- Radius of the ring, \(r = 0.2 \, \text{m}\)

2. **Substitute the values into the moment of inertia formula:**
\[ I = m r^2 \]
\[ I = 10 \, \text{kg} \times (0.2 \, \text{m})^2 \]

3. **Calculate \(r^2\):**
\[ (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

4. **Multiply the mass by \(r^2\):**
\[ I = 10 \, \text{kg} \times 0.04 \, \text{m}^2 = 0.4 \, \text{kg m}^2 \]

5. **Conclusion:**
The moment of inertia of the ring about its geometrical axis is:
\[ I = 0.4 \, \text{kg m}^2 \]

### Final Answer:
The moment of inertia of the ring is \(0.4 \, \text{kg m}^2\).

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