Calculate the moment of inertia of a ring of mass 2 kg and radius 50 cm about an axis passing through its centre and perpendicular to its plane

Answer

To calculate the moment of inertia of a ring about an axis passing through its center and perpendicular to its plane, 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 mass and radius of the ring:**
- Given mass \( m = 2 \, \text{kg} \)
- Given radius \( r = 50 \, \text{cm} = 0.5 \, \text{m} \) (convert cm to m)

2. **Substitute the values into the formula:**
\[ I = m r^2 = 2 \, \text{kg} \times (0.5 \, \text{m})^2 \]

3. **Calculate \( r^2 \):**
\[ (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \]

4. **Multiply the mass by \( r^2 \):**
\[ I = 2 \, \text{kg} \times 0.25 \, \text{m}^2 = 0.5 \, \text{kg} \cdot \text{m}^2 \]

5. **Final result:**
\[ I = 0.5 \, \text{kg} \cdot \text{m}^2 \]

### Conclusion:
The moment of inertia of the ring about the specified axis is \( 0.5 \, \text{kg} \cdot \text{m}^2 \).