For rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $$\frac{x}{5}$$. The value of $$x$$ is _____.

$$I = \frac{2}{3}MR^2$$

$$K_{\text{rot}} = \frac{1}{2}I\omega^2$$

$$K_{\text{rot}} = \frac{1}{2} \left(\frac{2}{3}MR^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{3}Mv^2$$

$$K_{\text{trans}} = \frac{1}{2}Mv^2$$

$$K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \frac{5}{6}Mv^2$$

$$\frac{K_{\text{rot}}}{K_{\text{total}}} = \frac{\frac{1}{3}Mv^2}{\frac{5}{6}Mv^2} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$$

$$\frac{x}{5} = \frac{2}{5} \implies x = 2$$