A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centr

Question

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n} \mathrm{~cm}$. The value of $n$ is

Professor PC Thomas Classes · Answer

By the parallel axis theorem, the moment of inertia of the solid sphere about the given axis is $I = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2$, where $R = 10$ cm and $d = 15$ cm. The radius of gyration $K$ is defined such that $I = MK^2$. Therefore, $K^2 = \frac{2}{5}R^2 + d^2 = \frac{2}{5}(10)^2 + (15)^2 = \frac{2}{5}(100) + 225 = 40 + 225 = 265$. Since $K = \sqrt{265}$ cm, the value of $n$ inside the square root is 265.