Physics - Rotation Question with Solution | TestHub

Ring has moment of inertia $M{R}^2$ about the symmetric central axis. Using perpendicular axis theorem one can get moment of inertia about the planar diagonal as $\frac{M{R}^2}{2}$ Now, using to the parallel axis theorem, we shift by a distance R to be at the tangential position: $I_t=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$ Given $M{R}^2=4{\mathrm{kgm}}^2$ : $I_t=\frac{3}{2}\times 4{\mathrm{kgm}}^2=6{\mathrm{kgm}}^2$