# Disk's moment of inertia about its diameter

1 vote
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I want to point out I worked with polar coordinates:

$\iint_{D} f(x, y)dxdy = \iint_{D'} f(rcos\theta, rsin\theta)rdrd\theta$

$dxdy=dA=r drd\theta$

Let disk rotate about x axis

Being moment of inertia $I = mr^2$, $a$ the radius of the disk and $\gamma=\frac{m}{\pi a^2}$ then we know:

$$I=\int_{0}^{2\pi}\int_{0}^{a} \gamma r^3sin^2\theta drd\theta$$

Actually, this integral gives the correct answer: $\frac{a^2m}{4}$

However, obtaining the desired result does not mean I am right (even the great Euler experienced that).

That is why I want to discuss the method I used. I took as an infinitesimal length $dr$ and as an infinitesimal width $rd\theta$.

Therefore, $dA=r drd\theta$

The distance from the x axis to the infinitesimal 2D fragment is $d=rsin\theta$

Then moment of inertia of the infinitesimal fragment: $I=dm(rsin\theta )^2$

retagged May 24, 2018
Ups some MathJax code lines do not work. I had a look, but did not identify any mistake. What should I do?
I was able to fix it by placing a \ before each _.

1 vote

Yes your method is correct. I think you have made some errors in setting up the integrals, but they come right in the end.

The area of the segment is $dA=r dr d\theta$. The moment of inertia of the segment is $dm(r\sin\theta)^2$ where $dm=M\frac{dA}{\pi a^2}$.

There are several other ways of finding the MI. You can divide the disk into vertical or horizontal strips, then use the MI for a rod. In the latter case (horizontal strips) you need to use the Parallel Axes Theorem.

Without integrating you can use the Perpendicular Axes Theorem, which only applies for plane figures. This relates the MI about the x, y and z axes :
$$I_z=I_x+I_y$$
In this example $I_x=I_y$, and you already know $I_z=\frac12 Ma^2$ for a disk or cylinder.

answered Apr 15, 2018 by (27,556 points)
selected Apr 16, 2018
I think I realized where I got wrong. Then if the disk turned about its y axis instead of its x axis, the moment of inertia of an infinitesimal segment would be
$$I= dm(rcos\theta )^2$$
Right?
Yes that is correct. The distance of the segment from the y axis is now $r\cos\theta$. But you can tell by symmetry that $I_y=I_x$ because if you rotate the axes by $90^{\circ}$ the mass distribution remains the same.