A body having its centre of mass at the origin has three of its particles at `(a,0,0), (0,a,0),(0,0,a).` The moments of inertia of the body about the X and Y axes are 0.20 kg `-m^2` each. The moment of inertia about the Z-axis

Options

A. is 0.20 `kg-m^2`
B. is 0.40 `kg-m^2`
C. is `0.20 sqrt2` `kg-m^2`
D. cannot be deduced with this information.

Answer

To find the moment of inertia of the body about the Z-axis, we can follow these steps:

### Step 1: Identify the particles and their positions
We have three particles located at:
- Particle 1: \( (a, 0, 0) \)
- Particle 2: \( (0, a, 0) \)
- Particle 3: \( (0, 0, a) \)

### Step 2: Understand the moment of inertia formula
The moment of inertia \( I \) about an axis for a system of particles is given by the formula:
\[ I = \sum m_i r_i^2 \]
where \( m_i \) is the mass of the particle and \( r_i \) is the perpendicular distance from the axis of rotation to the particle.

### Step 3: Calculate the moment of inertia about the X-axis
For the X-axis:
- Particle 1 contributes \( 0 \) because it lies on the X-axis.
- Particle 2 contributes \( m_2 a^2 \) (distance from the X-axis is \( a \)).
- Particle 3 contributes \( m_3 a^2 \) (distance from the X-axis is \( a \)).

Thus, the moment of inertia about the X-axis is:
\[ I_x = 0 + m_2 a^2 + m_3 a^2 = (m_2 + m_3) a^2 \]
Given \( I_x = 0.2 \, \text{kg m}^2 \), we have:
\[ (m_2 + m_3) a^2 = 0.2 \quad \text{(Equation 1)} \]

### Step 4: Calculate the moment of inertia about the Y-axis
For the Y-axis:
- Particle 1 contributes \( m_1 a^2 \) (distance from the Y-axis is \( a \)).
- Particle 2 contributes \( 0 \) because it lies on the Y-axis.
- Particle 3 contributes \( m_3 a^2 \) (distance from the Y-axis is \( a \)).

Thus, the moment of inertia about the Y-axis is:
\[ I_y = m_1 a^2 + 0 + m_3 a^2 = (m_1 + m_3) a^2 \]
Given \( I_y = 0.2 \, \text{kg m}^2 \), we have:
\[ (m_1 + m_3) a^2 = 0.2 \quad \text{(Equation 2)} \]

### Step 5: Calculate the moment of inertia about the Z-axis
For the Z-axis:
- Particle 1 contributes \( m_1 a^2 \) (distance from the Z-axis is \( a \)).
- Particle 2 contributes \( m_2 a^2 \) (distance from the Z-axis is \( a \)).
- Particle 3 contributes \( 0 \) because it lies on the Z-axis.

Thus, the moment of inertia about the Z-axis is:
\[ I_z = m_1 a^2 + m_2 a^2 + 0 = (m_1 + m_2) a^2 \]

### Step 6: Solve for \( I_z \)
To find \( I_z \), we need to express \( m_1 + m_2 \) in terms of known quantities. From Equations 1 and 2, we can express:
- \( m_2 + m_3 = \frac{0.2}{a^2} \)
- \( m_1 + m_3 = \frac{0.2}{a^2} \)

However, we have two equations with three unknowns, which means we cannot uniquely determine \( m_1, m_2, \) and \( m_3 \) without additional information. Thus, we cannot deduce \( I_z \) with the information given.

### Conclusion
The moment of inertia about the Z-axis cannot be determined with the information provided.