Three point masses each of mass m are placed at the corners of an equilateral triangle of side 'a' . Then the moment of inertia of this system about an axis passing along one side of the triangle is

Options

A. `ma^(2)`
B. `3ma^(2)`
C. `(3)/(4)ma^(2)`
D. `(2)/(3)ma^(2)`

Explanation

To find the moment of inertia of three point masses, each of mass \( m \), placed at the corners of an equilateral triangle with side length \( a \), about an axis passing along one side of the triangle, we can follow these steps:

### Step 1: Identify the Configuration
We have three point masses located at the vertices of an equilateral triangle. Let's label the vertices as \( A \), \( B \), and \( C \). The axis of rotation is along side \( AB \).

### Step 2: Determine the Positions of the Masses
- Mass at \( A \): This mass is on the axis of rotation, so its distance from the axis is \( 0 \).
- Mass at \( B \): This mass is also on the axis of rotation, so its distance from the axis is \( 0 \).
- Mass at \( C \): The distance from point \( C \) to the line \( AB \) can be calculated using the height of the triangle.

### Step 3: Calculate the Height of the Triangle
The height \( h \) of an equilateral triangle can be calculated using the formula:
\[ h = \frac{\sqrt{3}}{2} a \]

### Step 4: Calculate the Moment of Inertia for Each Mass
- For mass at \( A \):
\[ I_A = m \cdot 0^2 = 0 \]
- For mass at \( B \):
\[ I_B = m \cdot 0^2 = 0 \]
- For mass at \( C \):
The distance from \( C \) to the line \( AB \) is \( h = \frac{\sqrt{3}}{2} a \):
\[ I_C = m \cdot \left(\frac{\sqrt{3}}{2} a\right)^2 = m \cdot \frac{3}{4} a^2 \]

### Step 5: Sum the Moments of Inertia
The total moment of inertia \( I \) about the axis along side \( AB \) is the sum of the individual moments of inertia:
\[ I = I_A + I_B + I_C = 0 + 0 + m \cdot \frac{3}{4} a^2 = \frac{3}{4} m a^2 \]

### Final Answer
Thus, the moment of inertia of the system about the axis passing along one side of the triangle is:
\[ I = \frac{3}{4} m a^2 \]