Three point masses, each of mass 2kg are placed at the corners of an equilateral triangle of side 2m. What is moment of inertia of this system about an axis along one side of a triangle?

Options

A. `2kg m^(2)`
B. `4 kg m^(2)`
C. `6 kg m^(2)`
D. `8 kg m^(2)`

Explanation

To find the moment of inertia of the system of three point masses about an axis along one side of the triangle, we can follow these steps:

### Step 1: Identify the masses and their positions
We have three point masses, each of mass \( m = 2 \, \text{kg} \), located at the corners of an equilateral triangle with a side length of \( a = 2 \, \text{m} \). Let's label the corners of the triangle as A, B, and C.

### Step 2: Determine the axis of rotation
We need to calculate the moment of inertia about an axis along one side of the triangle. Let's choose the side AB as the axis of rotation.

### Step 3: Calculate the perpendicular distances from the axis
- For mass at point A (which is on the axis), the distance \( r_A = 0 \).
- For mass at point B (which is also on the axis), the distance \( r_B = 0 \).
- For mass at point C, we need to find the perpendicular distance from point C to line AB. The height \( h \) of the equilateral triangle can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \, \text{m} \]
Therefore, the distance \( r_C = h = \sqrt{3} \, \text{m} \).

### Step 4: Use the moment of inertia formula
The moment of inertia \( I \) about the axis is given by the formula:
\[ I = \sum m_i r_i^2 \]
where \( m_i \) is the mass and \( r_i \) is the distance from the axis of rotation.

Substituting the values:
- For mass at A: \( I_A = m_A \cdot r_A^2 = 2 \cdot 0^2 = 0 \)
- For mass at B: \( I_B = m_B \cdot r_B^2 = 2 \cdot 0^2 = 0 \)
- For mass at C: \( I_C = m_C \cdot r_C^2 = 2 \cdot (\sqrt{3})^2 = 2 \cdot 3 = 6 \)

### Step 5: Calculate the total moment of inertia
Now, summing these contributions:
\[ I = I_A + I_B + I_C = 0 + 0 + 6 = 6 \, \text{kg} \cdot \text{m}^2 \]

### Final Answer
The moment of inertia of the system about an axis along one side of the triangle is \( I = 6 \, \text{kg} \cdot \text{m}^2 \).

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