Three masses 1 kg, 2 kg and 3 kg are located at the vertices of an equilateral triangle of length 1. Calculate the moment of inertia of the triangle about an axis along the altitude of triangle passing through the 1 kg mass.

Answer

To calculate the moment of inertia of the system of masses located at the vertices of an equilateral triangle about an axis along the altitude passing through the 1 kg mass, we can follow these steps:

### Step 1: Understand the Configuration
We have three masses:
- \( m_1 = 1 \, \text{kg} \) at vertex A
- \( m_2 = 2 \, \text{kg} \) at vertex B
- \( m_3 = 3 \, \text{kg} \) at vertex C

The triangle is equilateral with each side measuring \( 1 \, \text{m} \).

### Step 2: Determine the Position of the Masses
The altitude from vertex A (where the 1 kg mass is located) bisects the base BC. The distance from A to the midpoint of BC can be calculated using the properties of an equilateral triangle.

The height \( h \) of the triangle can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2} \, \text{m} \]

The midpoint of BC is at a distance of \( \frac{1}{2} \, \text{m} \) horizontally from A, and the height \( h \) is \( \frac{\sqrt{3}}{2} \, \text{m} \).

### Step 3: Calculate the Distances from the Axis of Rotation
- For mass \( m_1 \) (1 kg at A), the distance \( r_1 = 0 \) (since it is on the axis).
- For mass \( m_2 \) (2 kg at B), the distance \( r_2 = \frac{\sqrt{3}}{2} \, \text{m} \).
- For mass \( m_3 \) (3 kg at C), the distance \( r_3 = \frac{\sqrt{3}}{2} \, \text{m} \).

### Step 4: Calculate the Moment of Inertia for Each Mass
The moment of inertia \( I \) for a point mass is given by:
\[ I = m \cdot r^2 \]

Calculating for each mass:
- For \( m_1 \):
\[ I_1 = 1 \cdot 0^2 = 0 \, \text{kg m}^2 \]
- For \( m_2 \):
\[ I_2 = 2 \cdot \left(\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2} \, \text{kg m}^2 \]
- For \( m_3 \):
\[ I_3 = 3 \cdot \left(\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{4} = \frac{3}{4} \, \text{kg m}^2 \]

### Step 5: Sum the Moments of Inertia
Now, we add up the moments of inertia:
\[ I_{\text{total}} = I_1 + I_2 + I_3 = 0 + \frac{1}{2} + \frac{3}{4} \]

To add these fractions, we can convert \( \frac{1}{2} \) to \( \frac{2}{4} \):
\[ I_{\text{total}} = 0 + \frac{2}{4} + \frac{3}{4} = \frac{5}{4} \, \text{kg m}^2 \]

### Final Result
Thus, the moment of inertia of the triangle about the axis along the altitude passing through the 1 kg mass is:
\[ I_{\text{total}} = \frac{5}{4} \, \text{kg m}^2 \]