Three indentical thin rods each of mass m and length L are joined together to form an equilateral triangular frame . The moment of inertia of frame about an axis perpendicular to the palne of frame and apssing through a corner is

Options

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

Explanation

To find the moment of inertia of the equilateral triangular frame made of three identical thin rods about an axis perpendicular to the plane of the frame and passing through one of its corners, we can follow these steps:

### Step 1: Identify the rods and their arrangement
We have three identical rods, each of mass \( m \) and length \( L \), arranged to form an equilateral triangle. Let's label the rods as \( R_1 \), \( R_2 \), and \( R_3 \).

### Step 2: Calculate the moment of inertia of each rod about the axis
1. **Rod \( R_1 \)**: This rod is aligned along one side of the triangle and passes through the axis. The moment of inertia of a rod about an axis through one end perpendicular to its length is given by:
\[ I_1 = \frac{1}{3} m L^2 \]

2. **Rod \( R_2 \)**: This rod is also aligned along one side of the triangle but does not pass through the axis. We need to use the parallel axis theorem to find its moment of inertia about the given axis. The moment of inertia about its center of mass is:
\[ I_{cm} = \frac{1}{3} m L^2 \]
The distance from the center of mass of rod \( R_2 \) to the axis is \( d = \frac{L \sqrt{3}}{2} \) (since it is at a \( 30^\circ \) angle). Therefore, using the parallel axis theorem:
\[ I_2 = I_{cm} + m d^2 = \frac{1}{3} m L^2 + m \left(\frac{L \sqrt{3}}{2}\right)^2 \]
\[ I_2 = \frac{1}{3} m L^2 + m \frac{3L^2}{4} = \frac{1}{3} m L^2 + \frac{3}{4} m L^2 \]
To combine these, we need a common denominator:
\[ I_2 = \frac{4}{12} m L^2 + \frac{9}{12} m L^2 = \frac{13}{12} m L^2 \]

3. **Rod \( R_3 \)**: This rod is similar to rod \( R_2 \) and will have the same moment of inertia about the axis:
\[ I_3 = I_2 = \frac{13}{12} m L^2 \]

### Step 3: Sum the moments of inertia
Now, we can find the total moment of inertia \( I \) of the triangular frame about the given axis:
\[ I = I_1 + I_2 + I_3 \]
Substituting the values we calculated:
\[ I = \frac{1}{3} m L^2 + \frac{13}{12} m L^2 + \frac{13}{12} m L^2 \]
\[ I = \frac{1}{3} m L^2 + \frac{26}{12} m L^2 \]
To combine these, we convert \( \frac{1}{3} \) to a fraction with a denominator of 12:
\[ \frac{1}{3} = \frac{4}{12} \]
So,
\[ I = \frac{4}{12} m L^2 + \frac{26}{12} m L^2 = \frac{30}{12} m L^2 = \frac{5}{2} m L^2 \]

### Final Result
Thus, the moment of inertia of the triangular frame about the specified axis is:
\[ I = \frac{5}{2} m L^2 \]