If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation is -
(a) `2/5` (b) `2/7` (c) `3/5` (d) `3/7`

Options

A. `2/5`
B. `2/7`
C. `3/5`
D. `3/7`

Answer

To solve the problem of finding the fraction of total energy associated with rotation for a spherical ball rolling on a table without slipping, we can follow these steps:

### Step-by-Step Solution:

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

2. **Write the Total Kinetic Energy**:
The total kinetic energy \( K \) of the rolling sphere is the sum of its translational kinetic energy \( K_t \) and rotational kinetic energy \( K_r \):
\[ K = K_t + K_r \]
Where:
\[ K_t = \frac{1}{2} mv^2 \quad \text{(translational)} \]
\[ K_r = \frac{1}{2} I \omega^2 \quad \text{(rotational)} \]

3. **Substitute the Moment of Inertia**:
Substitute the moment of inertia into the rotational kinetic energy formula:
\[ K_r = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 \]

4. **Relate Linear and Angular Velocity**:
For a sphere rolling without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = \omega r \quad \Rightarrow \quad \omega = \frac{v}{r} \]

5. **Substitute \( \omega \) in \( K_r \)**:
Replace \( \omega \) in the rotational kinetic energy equation:
\[ K_r = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \frac{v^2}{r^2} \]
Simplifying this gives:
\[ K_r = \frac{1}{5} mv^2 \]

6. **Calculate Total Kinetic Energy**:
Now we can express the total kinetic energy:
\[ K = K_t + K_r = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
To combine these, find a common denominator:
\[ K = \frac{5}{10} mv^2 + \frac{2}{10} mv^2 = \frac{7}{10} mv^2 \]

7. **Find the Fraction of Rotational Energy**:
The fraction of the total energy that is associated with rotation is given by:
\[ \text{Fraction} = \frac{K_r}{K} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} \]
The \( mv^2 \) terms cancel out:
\[ \text{Fraction} = \frac{1/5}{7/10} = \frac{1}{5} \cdot \frac{10}{7} = \frac{2}{7} \]

8. **Final Answer**:
Thus, the fraction of the total energy associated with rotation is:
\[ \frac{2}{7} \]

### Conclusion:
The correct answer is (b) \( \frac{2}{7} \).