A stone of mass 4kg is whirled in a horizontal circle of radius 1m and makes 2 rev/sec. The moment of inertia of the stone about the axis of rotation is

Options

A. `64 kg xx m^(2)`
B. `4 kg xx m^(2)`
C. `16 kg xx m^(2)`
D. `1 kg xx m^(2)`

Answer

To find the moment of inertia of the stone about the axis of rotation, we can use the formula for the moment of inertia \( I \) of a point mass:

\[ I = m \cdot r^2 \]

where:
- \( m \) is the mass of the object,
- \( r \) is the distance from the axis of rotation.

### Step-by-step Solution:

1. **Identify the mass of the stone**:
The mass \( m \) of the stone is given as 4 kg.

2. **Identify the radius of the circular path**:
The radius \( r \) of the horizontal circle in which the stone is whirled is given as 1 m.

3. **Substitute the values into the moment of inertia formula**:
Using the formula \( I = m \cdot r^2 \):
\[ I = 4 \, \text{kg} \cdot (1 \, \text{m})^2 \]

4. **Calculate \( r^2 \)**:
\[ (1 \, \text{m})^2 = 1 \, \text{m}^2 \]

5. **Multiply the mass by \( r^2 \)**:
\[ I = 4 \, \text{kg} \cdot 1 \, \text{m}^2 = 4 \, \text{kg} \cdot \text{m}^2 \]

6. **Final Result**:
The moment of inertia of the stone about the axis of rotation is:
\[ I = 4 \, \text{kg} \cdot \text{m}^2 \]