The point masses of 0.3 kg, 0.2 kg and 0.1 kg are placed at the corner of a right angles `Delta ABC`, as shown in Fig. 8.49. find the moment of inertia of the system (i) about an axis through A and perpendicular to the plane of the diagram and (ii) about an axis along BC.

Answer

To find the moment of inertia of the system about the specified axes, we will follow these steps:

### Given Data:
- Mass at A (0.3 kg)
- Mass at B (0.2 kg)
- Mass at C (0.1 kg)
- Length of AB = 0.3 m
- Length of AC = 0.5 m

### (i) Moment of Inertia about an Axis through A and Perpendicular to the Plane

1. **Identify the distances from the axis**:
- For mass at B (0.2 kg), the distance from A (point A to point B) is AB = 0.3 m.
- For mass at C (0.1 kg), the distance from A (point A to point C) is AC = 0.5 m.

2. **Use the formula for moment of inertia**:
The moment of inertia \( I \) about point A is given by:
\[ I_A = m_B \cdot d_B^2 + m_C \cdot d_C^2 \]
where \( m_B \) and \( m_C \) are the masses at B and C, and \( d_B \) and \( d_C \) are their respective distances from A.

3. **Substitute the values**:
\[ I_A = (0.2 \, \text{kg}) \cdot (0.3 \, \text{m})^2 + (0.1 \, \text{kg}) \cdot (0.5 \, \text{m})^2 \]
\[ I_A = (0.2) \cdot (0.09) + (0.1) \cdot (0.25) \]
\[ I_A = 0.018 \, \text{kg m}^2 + 0.025 \, \text{kg m}^2 \]
\[ I_A = 0.043 \, \text{kg m}^2 \]

### (ii) Moment of Inertia about an Axis along BC

1. **Identify the distances from the axis along BC**:
- For mass at A (0.3 kg), the distance from BC is the perpendicular distance from A to line BC. Since BC is horizontal, we can use the height from A to BC, which is equal to AC = 0.5 m.

2. **Use the formula for moment of inertia**:
The moment of inertia \( I_{BC} \) about the axis along BC is given by:
\[ I_{BC} = m_A \cdot d_A^2 + m_B \cdot d_B^2 \]
where \( m_A \) and \( m_B \) are the masses at A and B, and \( d_A \) is the distance from A to BC, and \( d_B \) is the distance from B to BC (which is 0 since B is on the line).

3. **Substitute the values**:
\[ I_{BC} = (0.3 \, \text{kg}) \cdot (0.5 \, \text{m})^2 + (0.2 \, \text{kg}) \cdot (0 \, \text{m})^2 \]
\[ I_{BC} = (0.3) \cdot (0.25) + (0.2) \cdot (0) \]
\[ I_{BC} = 0.075 \, \text{kg m}^2 + 0 \]
\[ I_{BC} = 0.075 \, \text{kg m}^2 \]

### Final Answers:
- Moment of inertia about an axis through A and perpendicular to the plane: \( I_A = 0.043 \, \text{kg m}^2 \)
- Moment of inertia about an axis along BC: \( I_{BC} = 0.075 \, \text{kg m}^2 \)