A sphere of mass 10 kg and radius 0.5 m rotates about a tangent. The moment of inertia of the sphere is

Options

A. `5 kg m^(2)`
B. `2.7 kg m^(2)`
C. `3.5 kg m^(2)`
D. `4.5 kg m^(2)`

Answer

To find the moment of inertia of a sphere of mass 10 kg and radius 0.5 m rotating about a tangent, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Moment of Inertia**:
The moment of inertia \( I \) of a solid sphere about its diameter is given by:
\[ I_{\text{diameter}} = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sphere and \( r \) is its radius.

2. **Use the Parallel Axis Theorem**:
To find the moment of inertia about a tangent, we can use the parallel axis theorem:
\[ I_{\text{tangent}} = I_{\text{diameter}} + m d^2 \]
where \( d \) is the distance from the center of the sphere to the tangent point. For a sphere, this distance is equal to the radius \( r \).

3. **Calculate the Moment of Inertia about the Diameter**:
Plug in the values for mass and radius:
\[ I_{\text{diameter}} = \frac{2}{5} \times 10 \, \text{kg} \times (0.5 \, \text{m})^2 \]
\[ = \frac{2}{5} \times 10 \times 0.25 \]
\[ = \frac{2}{5} \times 2 = \frac{4}{5} \, \text{kg m}^2 \]

4. **Calculate the Moment of Inertia about the Tangent**:
Now, apply the parallel axis theorem:
\[ I_{\text{tangent}} = I_{\text{diameter}} + m r^2 \]
\[ = \frac{4}{5} + 10 \times (0.5)^2 \]
\[ = \frac{4}{5} + 10 \times 0.25 \]
\[ = \frac{4}{5} + 2 \]
Convert \( 2 \) to a fraction with a denominator of 5:
\[ = \frac{4}{5} + \frac{10}{5} = \frac{14}{5} \, \text{kg m}^2 \]

5. **Final Calculation**:
\[ I_{\text{tangent}} = \frac{14}{5} = 2.8 \, \text{kg m}^2 \]

### Conclusion:
The moment of inertia of the sphere about the tangent is \( 2.8 \, \text{kg m}^2 \).