A circular disc is rotating about an axis through its circumference in a vertical plane. If its mass is 0.1 kg and radius 0.5 m, its M.I. In kg. `m^(2)` units about this axis will be

Options

A. 0.0375
B. 0.025
C. 0.5
D. 0.75

Answer

To find the moment of inertia (M.I.) of a circular disc rotating about an axis through its circumference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the disc (m) = 0.1 kg
- Radius of the disc (r) = 0.5 m

2. **Moment of Inertia about the Center of Mass**:
The moment of inertia (I_cm) of a solid disc about an axis through its center is given by the formula:
\[ I_{cm} = \frac{1}{2} m r^2 \]
Substituting the values:
\[ I_{cm} = \frac{1}{2} \times 0.1 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I_{cm} = \frac{1}{2} \times 0.1 \times 0.25 = 0.0125 \, \text{kg m}^2 \]

3. **Using the Parallel Axis Theorem**:
The parallel axis theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is given by:
\[ I = I_{cm} + m d^2 \]
where \(d\) is the distance from the center of mass axis to the new axis. For a disc rotating about an axis through its circumference, \(d = r\).
\[ I = I_{cm} + m r^2 \]
Substituting the values:
\[ I = 0.0125 \, \text{kg m}^2 + 0.1 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ I = 0.0125 \, \text{kg m}^2 + 0.1 \times 0.25 \]
\[ I = 0.0125 \, \text{kg m}^2 + 0.025 \, \text{kg m}^2 \]
\[ I = 0.0375 \, \text{kg m}^2 \]

4. **Final Result**:
The moment of inertia of the circular disc about the axis through its circumference is:
\[ I = 0.0375 \, \text{kg m}^2 \]