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.
 A: 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
$$\frac{1}{2}M\frac{r_2^2}{r_2^2-r_1^2}r_2^2-\frac{1}{2}M\frac{r_1^2}{r_2^2-r_1^2}r_1^2=\frac{1}{2}M\frac{r_2^4-r_1^4}{r_2^2-r_1^2}=\frac{1}{2}M(r_1^2+r_2^2)$$
A: Without working out all the details of the answer for you, the basic concept is that 
$$I_{total-axis-1} = \sum_j \left(I_{j-axis-1} \right).$$
That is, the moment of inertia of an extended object about a certain axis (e.g., axis-1) is the sum of moments of inertia of pieces of that object about the same axis.  If you want to subdivide a large object into two smaller pieces, this concept holds true. 
You must be careful to use the proper masses, positions (radii) when calculating each moment, but the principle of sums will be you started.
A: A solid disk has mass $m = \rho z\pi r^2$. The total mass of annular ring is found from adding uniform density disk of $r_2$ and subtracting a disk of $r_1$
$$ \left. M = \rho z \pi (r_2^2-r_1^2) \right\} \rho = \frac{M}{z \pi (r_2^2-r_1^2)} $$
The mass moment of inertia of solid disk is $I=\frac{m}{2} r^2 = \frac{1}{2} \rho \pi z r^4$
The total mass moment of inertia is found from adding a uniform disk of $r_2$ and subtracting a disk of $r_1$
$$ I = \frac{1}{2} \rho \pi z (r_2^4-r_1^4) $$
Using the density from above this becomes
$$ I = \frac{m}{2} \frac{ r_2^4-r_1^4}{r_2^2-r_1^2} = \frac{m}{2} \left( r_1^2+r_2^2 \right) $$
A: Moment of inertia of disc is (I=mr^2)/2 now what you are saying is this that it is cut from another disc so it is still a disc.so you can put the value of radius in above equation to get answer.
