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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.

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    $\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Jan 6 '16 at 16:08
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    $\begingroup$ I edited the question to clarify what you meant, if it doesn't reflect what you were asking, feel free to change it. And which part has mass $M$? The remaining ring or the cut out (smaller) disc? Please be more specific. $\endgroup$ – Aritra Das Jan 6 '16 at 17:14
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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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  • $\begingroup$ I think the OP is referring to the mass of smaller disk as mass $M$ $\endgroup$ – Oswald Jan 6 '16 at 17:10
  • $\begingroup$ But he said the disk "left". I therefore think he is referring to the "ring" left. $\endgroup$ – velut luna Jan 6 '16 at 17:12
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    $\begingroup$ @Kyson I think too, if he was referring to the smaller disc as $M$, and wanted its moment of inertia, I think the question would be too trivial. $\endgroup$ – Aritra Das Jan 6 '16 at 17:15
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    $\begingroup$ Probably, but the question is very badly posted. $\endgroup$ – Oswald Jan 6 '16 at 17:17
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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.

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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

$$ I = \frac{m}{2} \frac{ r_2^4-r_1^4}{r_2^2-r_1^2} = \frac{m}{2} \left( r_1^2+r_2^2 \right) $$

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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.

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  • $\begingroup$ It is not still a disc, it is an annular ring. $\endgroup$ – Aritra Das Jan 6 '16 at 17:11
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    $\begingroup$ @AritraDas The question is not clear, nothing wrong with this interpretation of the question either. $\endgroup$ – Oswald Jan 6 '16 at 17:15
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    $\begingroup$ @TheGhostOfPerdition I think the OP misused the term disc for annular ring, otherwise the question would be too trivial. $\endgroup$ – Aritra Das Jan 6 '16 at 17:16

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