A solid sphere rolls on horizontal surface without slipping. What is the ratio of its rotational to translation kinetic energy.

Options

A. `(2)/(5)`
B. `(5)/(2)`
C. `(3)/(2)`
D. 0

Answer

To find the ratio of the rotational kinetic energy to the translational kinetic energy of a solid sphere rolling on a horizontal surface without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Kinetic Energies**:
- The **translational kinetic energy (TKE)** of the sphere is given by:
\[ TKE = \frac{1}{2} m v^2 \]
- The **rotational kinetic energy (RKE)** of the sphere is given by:
\[ RKE = \frac{1}{2} I \omega^2 \]

2. **Determine the Moment of Inertia**:
- For a solid sphere, the moment of inertia (I) about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]

3. **Relate Linear and Angular Velocity**:
- Since the sphere rolls without slipping, the relationship between the linear velocity (v) and the angular velocity (ω) is:
\[ v = r \omega \]
- This can be rearranged to express ω in terms of v:
\[ \omega = \frac{v}{r} \]

4. **Substitute ω into RKE**:
- Substitute ω into the expression for RKE:
\[ RKE = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v^2}{r^2}\right) \]
- Simplifying this gives:
\[ RKE = \frac{1}{2} \cdot \frac{2}{5} m v^2 = \frac{1}{5} m v^2 \]

5. **Calculate the Ratio of RKE to TKE**:
- Now we can find the ratio of RKE to TKE:
\[ \text{Ratio} = \frac{RKE}{TKE} = \frac{\frac{1}{5} m v^2}{\frac{1}{2} m v^2} \]
- The terms \(m\) and \(v^2\) cancel out:
\[ \text{Ratio} = \frac{\frac{1}{5}}{\frac{1}{2}} = \frac{1}{5} \cdot \frac{2}{1} = \frac{2}{5} \]

6. **Final Answer**:
- Therefore, the ratio of the rotational kinetic energy to the translational kinetic energy of a solid sphere rolling without slipping is:
\[ \text{Ratio} = \frac{2}{5} \]