<s>[INST] moment of inertia when a shape is cut A disk of radius $r_1$ is cut from a disk of radius $r_2$, $(r_2>r_1)$ from the middle of the bigger disk . If the annular ring left has mass $M$ then find the moment of inertia about the axis passing through its centre and perpendicular to its plane.
 [/INST] I suppose your disk has uniform density. Then the mass of the whole disk is
$$M\frac{\pi r_2^2}{\pi (r_2^2-r_1^2)}=M\frac{r_2^2}{r_2^2-r_1^2}$$
and the mass of the smaller disk is
$$M\frac{\pi r_1^2}{\pi (r_2^2-r_1^2)}=M\frac{r_1^2}{r_2^2-r_1^2}$$
The momentum of inertia of the whole disk is
$$\frac{1}{2}M\frac{r_2^2}{r_2^2-r_1^2}r_2^2$$
The moment of inertia of the smaller disk is
$$\frac{1}{2}M\frac{r_1^2}{r_2^2-r_1^2}r_1^2$$
Hence the momentum of inertia of the ring is
$$\frac{1}{2}M\frac{r_2^2}{r_2^2-r_1^2}r_2^2-\frac{1}{2}M\frac{r_1^2}{r_2^2-r_1^2}r_1^2=\frac{1}{2}M\frac{r_2^4-r_1^4}{r_2^2-r_1^2}=\frac{1}{2}M(r_1^2+r_2^2)$$
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