Four thin uniform rods each of length L and mass m are joined to form a square . The moment of inertia of square about an axis along its one diagonal is

Options

A. `(mL^(2))/6`
B. `2/3mL^(2)`
C. `(3mL^(2))/4`
D. `(4mL^(2))/3`

Explanation

To find the moment of inertia of a square frame made of four uniform rods about an axis along one of its diagonals, we can follow these steps:

### Step 1: Understand the Configuration
We have a square frame formed by four thin uniform rods, each of length \( L \) and mass \( m \). The square has its diagonals, and we need to find the moment of inertia about one of these diagonals.

### Step 2: Moment of Inertia of a Single Rod
The moment of inertia \( I \) of a single rod about an axis passing through its center and perpendicular to its length is given by the formula:
\[ I_{\text{rod}} = \frac{1}{12} m L^2 \]

### Step 3: Moment of Inertia of the Square Frame
Since the square frame consists of four rods, we can calculate the total moment of inertia about an axis through the center of the square (perpendicular to the plane of the square) by considering the contribution of each rod. However, we will first find the moment of inertia of one rod about the diagonal.

### Step 4: Using the Parallel Axis Theorem
To find the moment of inertia of a rod about an axis that is not through its center, we can use the parallel axis theorem:
\[ I = I_{\text{cm}} + md^2 \]
where \( d \) is the distance from the center of mass of the rod to the new axis.

For a rod lying along one side of the square, the distance \( d \) to the diagonal is \( \frac{L}{2} \). Therefore, for one rod:
\[ I_{\text{diagonal}} = \frac{1}{12} m L^2 + m \left(\frac{L}{2}\right)^2 = \frac{1}{12} m L^2 + m \frac{L^2}{4} = \frac{1}{12} m L^2 + \frac{3}{12} m L^2 = \frac{4}{12} m L^2 = \frac{1}{3} m L^2 \]

### Step 5: Total Moment of Inertia for the Square Frame
Since there are four rods, the total moment of inertia about the diagonal is:
\[ I_{\text{total}} = 4 \times \frac{1}{3} m L^2 = \frac{4}{3} m L^2 \]

### Step 6: Using the Perpendicular Axis Theorem
The perpendicular axis theorem states that for a planar body:
\[ I_z = I_x + I_y \]
where \( I_z \) is the moment of inertia about an axis perpendicular to the plane, and \( I_x \) and \( I_y \) are the moments of inertia about two perpendicular axes in the plane.

Since the square is symmetric, we can say:
\[ I_x = I_y \]
Thus, we have:
\[ I_z = 2I_x \implies I_x = \frac{I_z}{2} \]
Substituting \( I_z = \frac{4}{3} m L^2 \):
\[ I_x = \frac{4}{3} \cdot \frac{1}{2} m L^2 = \frac{2}{3} m L^2 \]

### Final Answer
The moment of inertia of the square about one of its diagonals is:
\[ I = \frac{2}{3} m L^2 \]