Two particles of masses `m_(1)` and `m_(2)` are separated by a distance 'd'. Then moment of inertia of the system about an axis passing through centre of mass and perpendicular the line joining them is

Options

A. `((m_(1)m_(2))/(m_(1)+m_(2)))(d^(2))/(2)`
B. `((m_(1)m_(2))/(m_(1)+m_(2)))d`
C. `((m_(1)m_(2))/(m_(1)+m_(2)))d^(2)`
D. `((2m_(1)m_(2))/(m_(1)+m_(2)))d^(2)`

Answer

To find the moment of inertia of a system of two particles with masses \( m_1 \) and \( m_2 \) separated by a distance \( d \) about an axis passing through the center of mass and perpendicular to the line joining them, we can follow these steps:

### Step 1: Find the position of the center of mass (CM)
The position of the center of mass \( R_{cm} \) for two particles can be calculated using the formula:

\[ R_{cm} = \frac{m_1 \cdot r_1 + m_2 \cdot r_2}{m_1 + m_2} \]

Where \( r_1 \) and \( r_2 \) are the distances of \( m_1 \) and \( m_2 \) from the center of mass, respectively.

### Step 2: Determine distances from the center of mass
Let’s denote the distance from \( m_1 \) to the center of mass as \( r_1 \) and from \( m_2 \) as \( r_2 \). The distances can be expressed as:

\[ r_1 = \frac{m_2}{m_1 + m_2} \cdot d \]
\[ r_2 = \frac{m_1}{m_1 + m_2} \cdot d \]

### Step 3: Calculate the moment of inertia (I)
The moment of inertia \( I \) about the center of mass is given by:

\[ I = m_1 \cdot r_1^2 + m_2 \cdot r_2^2 \]

Substituting \( r_1 \) and \( r_2 \) into the equation gives:

\[ I = m_1 \left( \frac{m_2}{m_1 + m_2} \cdot d \right)^2 + m_2 \left( \frac{m_1}{m_1 + m_2} \cdot d \right)^2 \]

### Step 4: Simplify the expression
Expanding the equation:

\[ I = m_1 \cdot \frac{m_2^2}{(m_1 + m_2)^2} \cdot d^2 + m_2 \cdot \frac{m_1^2}{(m_1 + m_2)^2} \cdot d^2 \]

Factoring out \( d^2 \) from both terms:

\[ I = d^2 \left( \frac{m_1 m_2^2}{(m_1 + m_2)^2} + \frac{m_2 m_1^2}{(m_1 + m_2)^2} \right) \]

Combining the terms inside the parentheses:

\[ I = d^2 \cdot \frac{m_1 m_2 (m_2 + m_1)}{(m_1 + m_2)^2} \]

### Step 5: Final expression for moment of inertia
This simplifies to:

\[ I = \frac{m_1 m_2}{m_1 + m_2} \cdot d^2 \]

### Conclusion
Thus, the moment of inertia of the system about the axis passing through the center of mass and perpendicular to the line joining them is:

\[ I = \frac{m_1 m_2}{m_1 + m_2} \cdot d^2 \]

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