There are four point masses m each on the corners of a square of side length I about one of its diagonals, the moment of inertia of the system is

Options

A. `2ml^(2)`
B. `ml^(2)`
C. `4 ml^(2)`
D. `6 ml^(2)`

Answer

To find the moment of inertia of the system of four point masses located at the corners of a square of side length \( L \) about one of its diagonals, we can follow these steps:

### Step 1: Understand the Configuration
We have a square with four point masses \( m \) at each corner. The square has a side length \( L \). We need to calculate the moment of inertia about one of its diagonals.

### Step 2: Identify the Diagonal
The diagonal of the square divides it into two equal triangles. The length of the diagonal \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{L^2 + L^2} = L\sqrt{2} \]

### Step 3: Determine the Distance from the Diagonal
For the calculation of the moment of inertia, we need to find the perpendicular distance of each mass from the diagonal. The distance from the center of the square to the diagonal is \( \frac{L}{\sqrt{2}} \).

### Step 4: Calculate the Moment of Inertia for Each Mass
The moment of inertia \( I \) for a point mass is given by:
\[ I = m r^2 \]
where \( r \) is the distance from the axis of rotation. For the two masses that are perpendicular to the diagonal, the distance \( r \) is \( \frac{L}{\sqrt{2}} \).

Thus, the moment of inertia for one mass is:
\[ I_1 = m \left(\frac{L}{\sqrt{2}}\right)^2 = m \frac{L^2}{2} \]

### Step 5: Total Moment of Inertia
Since there are two masses contributing to the moment of inertia about the diagonal, the total moment of inertia \( I_{total} \) is:
\[ I_{total} = 2 \times I_1 = 2 \times m \frac{L^2}{2} = m L^2 \]

### Step 6: Conclusion
The moment of inertia of the system about one of its diagonals is:
\[ I = m L^2 \]