The diameter of a thin circular disc of mass 2 kg is 0.2 m. Its moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc is

Options

A. `"0.01 kg - m"^(2)`
B. `"0.02 kg - m"^(2)`
C. `"0.03 kg - m"^(2)`
D. `"0.04 kg - m"^(2)`

Explanation

To find the moment of inertia of a thin circular disc about an axis passing through its edge and perpendicular to its plane, we can use the parallel axis theorem. Here’s the step-by-step solution:

### Step 1: Identify the parameters
- Mass of the disc (m) = 2 kg
- Diameter of the disc = 0.2 m
- Radius of the disc (r) = Diameter / 2 = 0.2 m / 2 = 0.1 m

### Step 2: Calculate the moment of inertia about the center
The moment of inertia (I) of a thin circular disc about an axis passing through its center and perpendicular to its plane is given by the formula:
\[ I_{\text{center}} = \frac{1}{2} m r^2 \]

Substituting the values:
\[ I_{\text{center}} = \frac{1}{2} \times 2 \, \text{kg} \times (0.1 \, \text{m})^2 \]
\[ I_{\text{center}} = \frac{1}{2} \times 2 \times 0.01 \]
\[ I_{\text{center}} = 0.01 \, \text{kg m}^2 \]

### Step 3: Apply the parallel axis theorem
The parallel axis theorem states:
\[ I = I_{\text{center}} + m d^2 \]
where \( d \) is the distance from the center of mass to the new axis. In this case, the distance \( d \) is equal to the radius of the disc (0.1 m).

Substituting the values:
\[ I_{\text{edge}} = I_{\text{center}} + m r^2 \]
\[ I_{\text{edge}} = 0.01 \, \text{kg m}^2 + 2 \, \text{kg} \times (0.1 \, \text{m})^2 \]
\[ I_{\text{edge}} = 0.01 \, \text{kg m}^2 + 2 \times 0.01 \]
\[ I_{\text{edge}} = 0.01 \, \text{kg m}^2 + 0.02 \, \text{kg m}^2 \]
\[ I_{\text{edge}} = 0.03 \, \text{kg m}^2 \]

### Conclusion
The moment of inertia of the thin circular disc about an axis passing through the edge and perpendicular to its plane is:
\[ \boxed{0.03 \, \text{kg m}^2} \]

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