Three point masses 3kg, 2kg and 1kg are placed at A,B,C of a triangle. AB = 0.3,m AC = 0.5 m and BC = 0.4 m. The M.I of the system about an axis through A and normal to the plane of triangle is

Options

A. `0.43kgm^(2)`
B. `0.25kgm^(2)`
C. `0.18kgm^(2)`
D. `0.07kgm^(2)`

Answer

To find the moment of inertia (M.I) of the system about an axis through point A and normal to the plane of the triangle, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses and Positions**:
- Mass at A (m1) = 3 kg
- Mass at B (m2) = 2 kg
- Mass at C (m3) = 1 kg
- Distances:
- AB = 0.3 m
- AC = 0.5 m
- BC = 0.4 m

2. **Determine the Distances from the Axis**:
- The axis of rotation is through point A.
- Distance from A to B (r2) = AB = 0.3 m
- Distance from A to C (r3) = AC = 0.5 m
- Distance from A to A (r1) = 0 m (since it's the axis itself)

3. **Calculate the Moment of Inertia**:
The moment of inertia (I) about an axis is given by the formula:
\[ I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 \]
Substituting the values:
\[ I = (3 \, \text{kg} \times 0^2) + (2 \, \text{kg} \times (0.3 \, \text{m})^2) + (1 \, \text{kg} \times (0.5 \, \text{m})^2 \]

4. **Calculate Each Term**:
- For mass at A:
\[ 3 \, \text{kg} \times 0^2 = 0 \]
- For mass at B:
\[ 2 \, \text{kg} \times (0.3 \, \text{m})^2 = 2 \times 0.09 = 0.18 \, \text{kg m}^2 \]
- For mass at C:
\[ 1 \, \text{kg} \times (0.5 \, \text{m})^2 = 1 \times 0.25 = 0.25 \, \text{kg m}^2 \]

5. **Sum the Contributions**:
\[ I = 0 + 0.18 + 0.25 = 0.43 \, \text{kg m}^2 \]

### Final Answer:
The moment of inertia of the system about an axis through A and normal to the plane of the triangle is:
\[ I = 0.43 \, \text{kg m}^2 \]