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I have a a coin infinitely thin, rotating along the diameter. How to derive the formula for it's moment of inertia passing through the diameter.

I was suggested to use the surface density and infinitely small part of the surface area, equidistant from the axis of rotation (marked as $dS$ on the picture).

I've already figured out that:

$$I~=~\int r^2dm~=~\int \rho r^2 dS. $$

And now I'm stuck.

Any help would be much appreciated. Greg


My coin is infinitely thin, so it's two-dimensional object (only $X$ any $Y$ axis). So let's assume that my rotating axis is equal to Y axis. So I have to integrate from $R$ to $-R$ on the $X$ axis. And every $dS$ will have different surface area. But I know, that total area is $S=\pi r^2$. From Pythagorean theorem I know, that $ r^2 + h^2 = R^2 $. And my integrate is $$I ~=~ \int\limits_{-R}^{R} 2\rho r^2 \sqrt {R^2 - r^2} dr $$ But now I'm confused how to solve that - every time I get different solution than in Wolfram Alpha calculator

Ok, I've solved that - final answer is $ I = \frac {1}{4}MR^2 $

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closed as off-topic by David Z Jan 2 at 17:58

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Welcome to Physics Stack Exchange! What do you know about the moment of inertia? What particular aspect of the problem gives you trouble? Please note that simply asking to solve your problem is not welcomed at this site, see our FAQ at – Slaviks Mar 31 '12 at 17:08
I've edited my post already. – gogowitczak Mar 31 '12 at 17:23
Ah!!, it's also rotating about its diameter also! I didn't catch that part also. Silly me, you guys are right its just wrong. – kηives Mar 31 '12 at 19:00
Are you sure that your answer is not $I=\frac{1}{4}MR^2$? – Bernhard Apr 1 '12 at 12:21
@Bernhard - ahh, of course. That was just typo, but hopefully you saw that :) – gogowitczak Apr 1 '12 at 13:31
up vote 3 down vote accepted

Ask yourself the following questions:

  • In what spatial coordinate(s) do you want to integrate?
  • How do you express $dS$ in these spatial coordinates?

An alternative strategy would be to make smart use of the perpendicular axis theorem.

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My coin is infinitely thin, so it's two-dimentional object (only X any Y axis). So let's assume that my rotating axis is equal to Y axis. So I have to integrate from R to -R on the X axis. And every dS will heve different surface area. But I know, that total area is $S=\pi r^2$. And now i'm confused. Shound my integrate look like this? $I = \int\limits_{-R}^{R} \rho r^2 dx$ – gogowitczak Mar 31 '12 at 18:15
Yuor dS is a rectangle. What is the rectangles surface area? Express everything in $r$ and $dr$. – Bernhard Mar 31 '12 at 18:22
$dS = 2r*dr$, am I right? – gogowitczak Mar 31 '12 at 18:42
No, please, look carefully at your picture. At r=R you have a smaller dS than at r=0. So your expression cannot be right. – Bernhard Mar 31 '12 at 18:44
Yeah, you're right. Now i tried $dS = (R-r)dx$, but it's also wrong (I'm getting $I = \frac {2}{3\pi}MR^2$). How should I reduce that $\pi$? – gogowitczak Mar 31 '12 at 18:54

You need to specify the integral to your coordinate system of choice and the problem at hand. This will involve explicitly writing out the infinitesimals for each of the coordinates to figure out the exact form of the integrand, as well as setting the integration limits to completely cover the body.

What does the infinitesimal volume (or in this case, area) element look like? In other words, what is the value of $dS$ expressed in terms of coordinate infinitesimals?

Give this a shot, and post back here if you have a more specific question.

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or see @Bernard 's answer for a more concise version =) – tmac Mar 31 '12 at 17:55

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