A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively $$(k_{sph} : k_{cyl})$$ is $$2 : \sqrt{x}$$. The value of $$x$$ is _____.

For a solid sphere: $$I = \frac{2}{5}mR^2$$, so $$k_{sph} = R\sqrt{\frac{2}{5}}$$

For a solid cylinder: $$I = \frac{1}{2}mR^2$$, so $$k_{cyl} = R\sqrt{\frac{1}{2}} = \frac{R}{\sqrt{2}}$$

Ratio: $$\frac{k_{sph}}{k_{cyl}} = \frac{R\sqrt{2/5}}{R/\sqrt{2}} = \sqrt{\frac{2}{5}} \times \sqrt{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$

So $$k_{sph} : k_{cyl} = 2 : \sqrt{5}$$

Comparing with $$2 : \sqrt{x}$$, we get $$x = 5$$.

This matches the answer key value of $$\mathbf{5}$$.