The radius of a sphere is 5 cm. Calculate the radius of gyration about (i) its diameter and (ii) about any tangent.

Answer

To solve the problem of finding the radius of gyration of a sphere about its diameter and about any tangent, we will follow these steps:

### Step 1: Understand the Definitions
The radius of gyration \( k \) is defined as:
\[ k = \sqrt{\frac{I}{m}} \]
where \( I \) is the moment of inertia and \( m \) is the mass of the object.

### Step 2: Calculate the Moment of Inertia about the Diameter
For a solid sphere, the moment of inertia \( I \) about its diameter is given by the formula:
\[ I = \frac{2}{5} m R^2 \]
where \( R \) is the radius of the sphere.

Given that the radius \( R = 5 \) cm, we can substitute this into the formula.

### Step 3: Substitute Values into the Moment of Inertia Formula
Substituting \( R = 5 \) cm into the moment of inertia formula:
\[ I = \frac{2}{5} m (5 \, \text{cm})^2 = \frac{2}{5} m \cdot 25 \, \text{cm}^2 = \frac{50}{5} m = 10 m \, \text{cm}^2 \]

### Step 4: Calculate the Radius of Gyration about the Diameter
Now we can find the radius of gyration \( k \) about the diameter:
\[ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{10 m}{m}} = \sqrt{10} \, \text{cm} \]

### Step 5: Calculate the Moment of Inertia about a Tangent
For a solid sphere, the moment of inertia \( I \) about a tangent can be calculated using the parallel axis theorem:
\[ I_{\text{tangent}} = I_{\text{center}} + m d^2 \]
where \( d \) is the distance from the center of mass to the tangent line. For a sphere, \( d = R \).

Thus, we have:
\[ I_{\text{tangent}} = \frac{2}{5} m R^2 + m R^2 = \frac{2}{5} m R^2 + \frac{5}{5} m R^2 = \frac{7}{5} m R^2 \]

### Step 6: Substitute Values into the Tangent Moment of Inertia Formula
Substituting \( R = 5 \) cm:
\[ I_{\text{tangent}} = \frac{7}{5} m (5 \, \text{cm})^2 = \frac{7}{5} m \cdot 25 \, \text{cm}^2 = \frac{175}{5} m = 35 m \, \text{cm}^2 \]

### Step 7: Calculate the Radius of Gyration about the Tangent
Now we can find the radius of gyration \( k \) about the tangent:
\[ k = \sqrt{\frac{I_{\text{tangent}}}{m}} = \sqrt{\frac{35 m}{m}} = \sqrt{35} \, \text{cm} \]

### Final Results
1. The radius of gyration about the diameter is \( \sqrt{10} \, \text{cm} \).
2. The radius of gyration about any tangent is \( \sqrt{35} \, \text{cm} \).