Calculate the radius of gyration of a sphere (a) about its diameter (b) about any tangent. The radius of sphere is 10 cm.

Answer

To calculate the radius of gyration of a sphere about its diameter and about a tangent, we can follow these steps:

### Given:
- Radius of the sphere, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Part (a): Radius of Gyration about its Diameter

1. **Moment of Inertia about the Diameter**:
The moment of inertia \( I \) of a solid sphere about its diameter is given by the formula:
\[ I = \frac{2}{5} m r^2 \]

2. **Using the Radius of Gyration Formula**:
The radius of gyration \( k \) is related to the moment of inertia by the equation:
\[ I = m k^2 \]
Equating the two expressions for moment of inertia, we have:
\[ \frac{2}{5} m r^2 = m k^2 \]

3. **Canceling Mass \( m \)**:
Since mass \( m \) appears on both sides, we can cancel it out:
\[ \frac{2}{5} r^2 = k^2 \]

4. **Solving for \( k \)**:
Taking the square root of both sides gives:
\[ k = \sqrt{\frac{2}{5} r^2} = r \sqrt{\frac{2}{5}} \]

5. **Substituting the Value of \( r \)**:
Now substituting \( r = 0.1 \, \text{m} \):
\[ k = 0.1 \sqrt{\frac{2}{5}} = 0.1 \times \sqrt{0.4} \approx 0.063 \, \text{m} \]

### Part (b): Radius of Gyration about a Tangent

1. **Moment of Inertia about a Tangent**:
The moment of inertia \( I \) about a tangent to the sphere can be calculated using the parallel axis theorem:
\[ I = I_{cm} + m d^2 \]
where \( d = r \) (the distance from the center of mass to the tangent). Thus:
\[ I = \frac{2}{5} m r^2 + m r^2 = \left(\frac{2}{5} + 1\right) m r^2 = \frac{7}{5} m r^2 \]

2. **Using the Radius of Gyration Formula Again**:
Again, we use the relation \( I = m k^2 \):
\[ \frac{7}{5} m r^2 = m k^2 \]

3. **Canceling Mass \( m \)**:
Canceling \( m \) gives:
\[ \frac{7}{5} r^2 = k^2 \]

4. **Solving for \( k \)**:
Taking the square root of both sides gives:
\[ k = \sqrt{\frac{7}{5} r^2} = r \sqrt{\frac{7}{5}} \]

5. **Substituting the Value of \( r \)**:
Now substituting \( r = 0.1 \, \text{m} \):
\[ k = 0.1 \sqrt{\frac{7}{5}} = 0.1 \times \sqrt{1.4} \approx 0.12 \, \text{m} \]

### Final Answers:
- (a) Radius of gyration about its diameter: \( k \approx 0.063 \, \text{m} \)
- (b) Radius of gyration about a tangent: \( k \approx 0.12 \, \text{m} \)

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