The radius of gyration of a uniform thin hollow spherical shell of mass $3 ext{ kg}$ about a tange

Question

The radius of gyration of a uniform thin hollow spherical shell of mass $3 	ext{ kg}$ about a tangential axis is measured to be $K = 6 	ext{ m}$. The actual geometric radius $R$ of the spherical shell is

Professor PC Thomas Classes · Answer

(c) : The moment of inertia of a hollow sphere (spherical shell) about an axis passing through its center is $I_{cm} = \frac{2}{3}MR^2$. By the parallel axis theorem, the moment of inertia about a tangent is $I_{tangent} = I_{cm} + MR^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$. We also know $I = MK^2$. Equating the two gives $MK^2 = \frac{5}{3}MR^2 \Rightarrow K^2 = \frac{5}{3}R^2 \Rightarrow R = K\sqrt{\frac{3}{5}}$. Substituting $K=6 	ext{ m}$, we get $R = 6\sqrt{\frac{3}{5}} = \frac{6\sqrt{15}}{5} 	ext{ m}$.