To find the ratio of rotational to translational kinetic energy $K_{\text{rot}}/K_{\text{trans}}$ for a rolling sphere, let’s go through the steps:

1. Total Kinetic Energy (K) of a rolling object includes both translational and rotational components: $K = K_{\text{trans}} + K_{\text{rot}}$

2. For a rolling sphere without slipping: $K_{\text{trans}} = \frac{1}{2} M v^2$ $K_{\text{rot}} = \frac{1}{2} I \omega^2$ where $I$ is the moment of inertia of the sphere, $I = \frac{2}{5} M R^2$ for a solid sphere.

3. Relationship Between $v$ and $\omega$: Since the sphere rolls without slipping, $v = \omega R$ Substituting $\omega = \frac{v}{R}$ into $K_{\text{rot}}$: $K_{\text{rot}} = \frac{1}{2} \cdot \frac{2}{5} M R^2 \cdot \left(\frac{v}{R}\right)^2$ Simplifying this gives: $K_{\text{rot}} = \frac{1}{5} M v^2$

4. Translational Kinetic Energy $K_{\text{trans}}$: $K_{\text{trans}} = \frac{1}{2} M v^2$

5. Ratio of $K_{\text{rot}}$ to $K_{\text{trans}}$: $\frac{K_{\text{rot}}}{K_{\text{trans}}} = \frac{\frac{1}{5} M v^2}{\frac{1}{2} M v^2} = \frac{1/5}{1/2} = \frac{2}{5} = 0.400$

Thus, the correct answer is: $\boxed{0.400}$

Would you like more details or have any questions?

---

Related Questions

1. How is the total kinetic energy of rolling objects divided between rotational and translational components?
2. How would the answer change if the object were a hollow sphere instead of a solid sphere?
3. Why does the relationship $v = \omega R$ hold for objects rolling without slipping?
4. What is the moment of inertia of various shapes, such as disks or cylinders, and how does it affect rolling motion?
5. How do you calculate the kinetic energy if the sphere was sliding instead of rolling?

Tip

Remember, for any rolling object, the ratio $\frac{K_{\text{rot}}}{K_{\text{trans}}}$ depends on its moment of inertia and shape.