Two bodies of mass 1 kg and 2 kg are attached to the ends of a 2 metre long weight less rod . This system is rotating about an axis passing through middle point of rod . Calculate M. I of system .

Options

A. `2 "kg m"^(2)`
B. `3 "kg m"^(2)`
C. `4 "kg m"^(2)`
D. `0.5 "kg m"^(2)`

Explanation

To calculate the moment of inertia (M.I) of the system, we will follow these steps:

### Step 1: Understand the System
We have two masses attached to a weightless rod. The masses are:
- Mass \( m_1 = 1 \, \text{kg} \) (at one end of the rod)
- Mass \( m_2 = 2 \, \text{kg} \) (at the other end of the rod)
The length of the rod is \( L = 2 \, \text{m} \), and it rotates about an axis passing through the midpoint of the rod.

### Step 2: Determine Distances from the Axis of Rotation
Since the rod is 2 meters long and the axis of rotation is at the midpoint, the distances of the masses from the axis are:
- Distance of \( m_1 \) from the axis, \( r_1 = 1 \, \text{m} \)
- Distance of \( m_2 \) from the axis, \( r_2 = 1 \, \text{m} \)

### Step 3: Use the Moment of Inertia Formula
The moment of inertia \( I \) for a point mass is given by the formula:
\[ I = m \cdot r^2 \]
where \( m \) is the mass and \( r \) is the distance from the axis of rotation.

### Step 4: Calculate the Moment of Inertia for Each Mass
1. For mass \( m_1 \):
\[ I_1 = m_1 \cdot r_1^2 = 1 \, \text{kg} \cdot (1 \, \text{m})^2 = 1 \, \text{kg} \cdot 1 \, \text{m}^2 = 1 \, \text{kg m}^2 \]

2. For mass \( m_2 \):
\[ I_2 = m_2 \cdot r_2^2 = 2 \, \text{kg} \cdot (1 \, \text{m})^2 = 2 \, \text{kg} \cdot 1 \, \text{m}^2 = 2 \, \text{kg m}^2 \]

### Step 5: Calculate the Total Moment of Inertia
The total moment of inertia \( I \) of the system is the sum of the individual moments of inertia:
\[ I = I_1 + I_2 = 1 \, \text{kg m}^2 + 2 \, \text{kg m}^2 = 3 \, \text{kg m}^2 \]

### Final Answer
The moment of inertia of the system is \( 3 \, \text{kg m}^2 \).

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