Find the moment of inertia in $kg{m}^2$ of a thin disc of mass m and radius R with a hole of radius R/2 around the Z-axis (see figure ), if m= 6kg, R=2m.

The easiest way is to “fill” the hole, and then subtract the contribution of the hole. The area of the full disc is $\pi R^2.$  The area of the hole is $\frac{\pi R^2}{4}.$ The area of the disc with the hole is $\frac{3\pi R^2}{4}.$ Thus by filling the hole, one increases the mass by a factor: $\frac{\pi R^2}{3\pi R^2/4}=\frac{4}{3}.$ 

The moment of inertia of a full disc with a mass of $\frac{4}{3}m$ around the z-axis is derived from the parallel axis theorem, by: $I_{full}=\frac{1}{2}.\frac{4}{3}m{R}^2+\frac{4}{3}m{\left(\frac{R}{2}\right)}^2=m{R}^2\mathrm{......}\left(i\right)$

Similarly, the mass of the small disc which fills the hole is given by: $m_1=\frac{\frac{\pi m{R}^2}{4}}{\frac{3\pi R^2}{4}}=\frac{1}{3}m\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{.....}\left(ii\right)$

Its moment of inetia around the axis is $I_{hole}=\frac{1}{2}.\frac{1}{3}m{\left(\frac{R}{2}\right)}^2=\frac{1}{24}m{R}^2\mathrm{......}\left(iii\right)$

Thus, we obtain: $I_{total}=I_{full}-I_{hole}=\frac{23}{24}m{R}^2\mathrm{......}\left(iv\right)$