If the radius of gyration of a solid disc of mass 10 kg about an axis is 0.40 m, then the moment of inertia of the disc about that axis is

Options

A. `1.6 "kg m"^2`
B. `3.2 "kg m"^2`
C. `6.4 "kg m"^2`
D. `9.5 "kg m"^2`

Answer

To find the moment of inertia (I) of a solid disc about a given axis using the radius of gyration (k), we can use the formula:

\[ I = m \cdot k^2 \]

Where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the disc,
- \( k \) is the radius of gyration.

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of the disc, \( m = 10 \, \text{kg} \)
- Radius of gyration, \( k = 0.40 \, \text{m} \)

2. **Substitute the values into the formula**:
\[ I = m \cdot k^2 \]
\[ I = 10 \, \text{kg} \cdot (0.40 \, \text{m})^2 \]

3. **Calculate \( k^2 \)**:
\[ (0.40 \, \text{m})^2 = 0.16 \, \text{m}^2 \]

4. **Multiply the mass by \( k^2 \)**:
\[ I = 10 \, \text{kg} \cdot 0.16 \, \text{m}^2 = 1.6 \, \text{kg} \cdot \text{m}^2 \]

5. **Final Result**:
The moment of inertia of the disc about the given axis is:
\[ I = 1.6 \, \text{kg} \cdot \text{m}^2 \]

### Conclusion:
Thus, the moment of inertia of the solid disc about the specified axis is \( 1.6 \, \text{kg} \cdot \text{m}^2 \).

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