Four holes of radius $R$ are cut from a thin square plate of side $4R$ and mass $M$. Determine the moment of inertia of the remaining portion about $Z$-axis.

$M$=Mass of the square plate before the holes were cut.

Mass of each hole $m=\left[\frac{M}{16R^2}\right]{\mathrm{\pi R}}^2=\frac{\mathrm{\pi M}}{16}$.

Therefore, Moment of inertia of the remaining portion,

$$\begin{gathered} I=I_{square}-4I_{hole}=\frac{M}{12}\left(16R^2+16R^2\right)-4\left[\frac{m{R}^2}{2}+m{\left(\sqrt{2}R\right)}^2\right] \\ =\frac{8}{3}M{R}^2-10m{R}^2=\left[\frac{8}{3}-\frac{10\mathrm{\pi }}{16}\right]M{R}^2. \end{gathered}$$