Three masses of 2 kg, 3 kg and 4 kg are located at the three vertices of an equilateral triangle of side 1 m. What is the M.I. of the system about an axis along the altitude of the triangle and passing through the 3 kg mass?

Explanation

To find the moment of inertia (M.I.) of the system about an axis along the altitude of the triangle and passing through the 3 kg mass, we can follow these steps:

### Step 1: Understand the Configuration
We have three masses located at the vertices of an equilateral triangle with each side measuring 1 meter. The masses are:
- Mass \( m_1 = 2 \, \text{kg} \) at vertex A
- Mass \( m_2 = 3 \, \text{kg} \) at vertex B (this is where the axis passes through)
- Mass \( m_3 = 4 \, \text{kg} \) at vertex C

### Step 2: Determine the Position of the Masses
In an equilateral triangle, the altitude divides the base into two equal halves. The distance from the vertex B (where the 3 kg mass is located) to the midpoint of the opposite side (which is the base) is \( \frac{1}{2} \) meter.

### Step 3: Calculate the Moment of Inertia
The moment of inertia \( I \) for a system of point masses is given by the formula:
\[ I = \sum m_i r_i^2 \]
where \( m_i \) is the mass and \( r_i \) is the perpendicular distance from the axis of rotation.

1. For the 3 kg mass (at vertex B):
- Distance \( r_2 = 0 \) (since the axis passes through this mass)
- Contribution to M.I: \( I_2 = m_2 \cdot r_2^2 = 3 \cdot 0^2 = 0 \)

2. For the 2 kg mass (at vertex A):
- Distance \( r_1 = \frac{1}{2} \) meter (perpendicular distance from the axis)
- Contribution to M.I:
\[ I_1 = m_1 \cdot r_1^2 = 2 \cdot \left(\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2} \, \text{kg m}^2 \]

3. For the 4 kg mass (at vertex C):
- Distance \( r_3 = \frac{1}{2} \) meter (perpendicular distance from the axis)
- Contribution to M.I:
\[ I_3 = m_3 \cdot r_3^2 = 4 \cdot \left(\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \, \text{kg m}^2 \]

### Step 4: Sum the Contributions
Now, we sum the contributions from all three masses:
\[ I_{\text{net}} = I_1 + I_2 + I_3 = \frac{1}{2} + 0 + 1 = \frac{3}{2} \, \text{kg m}^2 \]

### Final Answer
The moment of inertia of the system about the given axis is:
\[ I_{\text{net}} = \frac{3}{2} \, \text{kg m}^2 \]

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