Physics - Rotation Question with Solution | TestHub

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is $$I_{\mathrm{CM}}=\frac{1}{2}M{R}^2$$ Where $M$ is the mass of disc and $R$ its radius According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc at its diameter and normal to the disc is $$\begin{array}{rl}I& =I_{\mathrm{CM}}+M{R}^2\\ & =\frac{1}{2}M{R}^2+M{R}^2\\ & =\frac{3}{2}M{R}^2\end{array}$$