A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is `k`. If radius of the ball be `R`, then the fraction of total energy associated with its rotation will be.

Options

A. `(k^(2) + R^(2))/R^(2)`
B. `k^(2)/R^(2)`
C. `k^(2)/(k^(2) + R^(2))`
D. `R^(2)/(k^(2) + R^(2))`

Explanation

To solve the problem, we need to find the fraction of the total energy associated with the rotation of a ball that rolls without slipping. Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Define Variables**:
- Let the mass of the ball be \( M \).
- Let the radius of the ball be \( R \).
- Let the radius of gyration of the ball about an axis through its center of mass be \( k \).
- Let the velocity of the ball be \( V \).

2. **Moment of Inertia**:
- The moment of inertia \( I \) of the ball about its center of mass is given by:
\[ I = M k^2 \]

3. **Relation between Linear Velocity and Angular Velocity**:
- For a ball rolling without slipping, the relationship between linear velocity \( V \) and angular velocity \( \omega \) is:
\[ V = R \omega \quad \Rightarrow \quad \omega = \frac{V}{R} \]

4. **Kinetic Energies**:
- The translational kinetic energy (TKE) of the ball is:
\[ \text{TKE} = \frac{1}{2} M V^2 \]
- The rotational kinetic energy (RKE) of the ball is:
\[ \text{RKE} = \frac{1}{2} I \omega^2 = \frac{1}{2} (M k^2) \left(\frac{V}{R}\right)^2 = \frac{1}{2} M k^2 \frac{V^2}{R^2} \]

5. **Total Energy**:
- The total energy \( E \) of the ball is the sum of translational and rotational kinetic energies:
\[ E = \text{TKE} + \text{RKE} = \frac{1}{2} M V^2 + \frac{1}{2} M k^2 \frac{V^2}{R^2} \]
- Factoring out \( \frac{1}{2} M V^2 \):
\[ E = \frac{1}{2} M V^2 \left(1 + \frac{k^2}{R^2}\right) \]

6. **Fraction of Rotational Kinetic Energy**:
- The fraction of the total energy that is associated with the rotational kinetic energy is given by:
\[ \text{Fraction} = \frac{\text{RKE}}{E} = \frac{\frac{1}{2} M k^2 \frac{V^2}{R^2}}{\frac{1}{2} M V^2 \left(1 + \frac{k^2}{R^2}\right)} \]
- Simplifying this expression:
\[ \text{Fraction} = \frac{k^2 / R^2}{1 + k^2 / R^2} = \frac{k^2}{k^2 + R^2} \]

7. **Final Answer**:
- The fraction of total energy associated with its rotation is:
\[ \frac{k^2}{k^2 + R^2} \]