A solid spherical ball rolls on a table. Ratio of its rotational kinetic energy to total kinetic energy is

Options

A. `(1)/(2)`
B. `(1)/(6)`
C. `(7)/(10)`
D. `(2)/(7)`

Answer

To find the ratio of the rotational kinetic energy to the total kinetic energy of a solid spherical ball rolling on a table, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Kinetic Energies**:
- The total kinetic energy (TKE) of the rolling ball is the sum of its translational kinetic energy (TKE_trans) and rotational kinetic energy (TKE_rot).
- The translational kinetic energy is given by:
\[ TKE_{\text{trans}} = \frac{1}{2} mv^2 \]
- The rotational kinetic energy is given by:
\[ TKE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

2. **Moment of Inertia for a Solid Sphere**:
- The moment of inertia \( I \) for a solid sphere about an axis through its center is:
\[ I = \frac{2}{5} m r^2 \]

3. **Relate Linear and Angular Velocity**:
- For a rolling object without slipping, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is:
\[ v = r \omega \]
- Therefore, we can express \( \omega \) in terms of \( v \):
\[ \omega = \frac{v}{r} \]

4. **Substituting \( \omega \) into Rotational Kinetic Energy**:
- Substitute \( \omega \) into the equation for rotational kinetic energy:
\[ TKE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \]
- Simplifying this gives:
\[ TKE_{\text{rot}} = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{5} mv^2 \]

5. **Calculate Total Kinetic Energy**:
- Now, substitute \( TKE_{\text{rot}} \) and \( TKE_{\text{trans}} \) into the total kinetic energy:
\[ TKE_{\text{total}} = TKE_{\text{trans}} + TKE_{\text{rot}} = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]
- To combine these, find a common denominator:
\[ TKE_{\text{total}} = \frac{5}{10} mv^2 + \frac{2}{10} mv^2 = \frac{7}{10} mv^2 \]

6. **Find the Ratio**:
- Now, we can find the ratio of the rotational kinetic energy to the total kinetic energy:
\[ \text{Ratio} = \frac{TKE_{\text{rot}}}{TKE_{\text{total}}} = \frac{\frac{1}{5} mv^2}{\frac{7}{10} mv^2} \]
- The \( mv^2 \) terms cancel out:
\[ \text{Ratio} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{1}{5} \cdot \frac{10}{7} = \frac{2}{7} \]

### Final Answer:
The ratio of the rotational kinetic energy to the total kinetic energy of a solid spherical ball rolling on a table is \( \frac{2}{7} \).