Four particle of mass m each are placed at four corners of a square ABCD of side a. Point O is the centre of the square. Moment of inertia of all four particles about an axis passing through

Options

A. A and B is `2ma^(2)`
B. A and C is `ma^(2)`
C. O and perpendicular to plane of square is `2ma^(2)`
D. O and parallel to CD is `ma^(2)`

Answer

To find the moment of inertia of four particles of mass \( m \) each placed at the corners of a square ABCD of side \( a \) about various axes, we will consider each axis one by one.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a square ABCD with side length \( a \).
- Each corner of the square has a particle of mass \( m \).
- The center of the square is point \( O \).

2. **Moment of Inertia about an Axis through A and B**:
- The axis is along the line joining points A and B.
- For particles at A and B, their distances from the axis are \( r_A = 0 \) and \( r_B = 0 \) respectively.
- For particle C, the distance from the axis is \( r_C = a \) (the side length).
- For particle D, the distance from the axis is also \( r_D = a \).
- The moment of inertia \( I \) is given by:
\[ I = m_A r_A^2 + m_B r_B^2 + m_C r_C^2 + m_D r_D^2 = 0 + 0 + m a^2 + m a^2 = 2m a^2 \]
- Thus, the moment of inertia about the axis through A and B is \( 2m a^2 \).

3. **Moment of Inertia about an Axis through A and C**:
- The axis is along the line joining points A and C.
- For particle A, \( r_A = 0 \).
- For particle B, the distance from the axis is \( r_B = \frac{a}{\sqrt{2}} \) (diagonal distance).
- For particle C, \( r_C = 0 \).
- For particle D, the distance from the axis is also \( r_D = \frac{a}{\sqrt{2}} \).
- The moment of inertia \( I \) is:
\[ I = m_A r_A^2 + m_B r_B^2 + m_C r_C^2 + m_D r_D^2 = 0 + m \left(\frac{a}{\sqrt{2}}\right)^2 + 0 + m \left(\frac{a}{\sqrt{2}}\right)^2 = m \frac{a^2}{2} + m \frac{a^2}{2} = m a^2 \]
- Thus, the moment of inertia about the axis through A and C is \( m a^2 \).

4. **Moment of Inertia about an Axis through O (Center) and Perpendicular to the Plane**:
- The axis is perpendicular to the plane of the square.
- The distance from the center \( O \) to each corner (A, B, C, D) is \( r = \frac{a}{\sqrt{2}} \).
- The moment of inertia \( I \) is:
\[ I = 4 \cdot m \left(\frac{a}{\sqrt{2}}\right)^2 = 4 \cdot m \cdot \frac{a^2}{2} = 2m a^2 \]
- Thus, the moment of inertia about the axis through O and perpendicular to the plane is \( 2m a^2 \).

5. **Moment of Inertia about an Axis Parallel to CD**:
- We can use the perpendicular axis theorem, which states that for a planar object:
\[ I_z = I_x + I_y \]
- Since the square is symmetric, \( I_x = I_y \).
- From previous calculations, we know \( I_z = 2ma^2 \).
- Therefore:
\[ 2ma^2 = 2I_y \implies I_y = ma^2 \]
- Thus, the moment of inertia about the axis parallel to CD is \( ma^2 \).

### Summary of Results:
- Moment of Inertia about axis through A and B: \( 2ma^2 \)
- Moment of Inertia about axis through A and C: \( ma^2 \)
- Moment of Inertia about axis through O and perpendicular to the plane: \( 2ma^2 \)
- Moment of Inertia about axis parallel to CD: \( ma^2 \)