# Classical mechanics, principal moments of inertia of a rigid body

1. Jul 2, 2011

### fluidistic

1. The problem statement, all variables and given/known data
Determine the principal moments of inertia of a circular cylinder with radius R and height h.

2. Relevant equations
Not sure.

3. The attempt at a solution
This is the first problem of this kind I attempt to solve.
From what I've read on wikipedia, the tensor of inertia can be written as $\mathbf{I} = \begin{bmatrix}I_{11} & I_{12} & I_{13} \\I_{21} & I_{22} & I_{23} \\I_{31} & I_{32} & I_{33}\end{bmatrix}$.
And if I'm lucky, I can find a Cartesian system of coordinates such that $\mathbf{I} = \begin{bmatrix}I_{1} & 0 & 0 \\0 & I_{2} & 0 \\0 & 0 & I_{3}\end{bmatrix}$ where $I_i$ are what I'm looking for. Hence I guess I must approach the problem in finding such a Cartesian system of coordinate. By intuition I'd set it centered over the center of mass of the cylinder. Say the y-axis goes along the height of the cylinder and the x-axis goes along a radius of the cylinder (so that the z-axis also goes along a radius of the cylinder).
But I've no idea how to calculate the $I_i$'s. It likely involve some integral (simple? double? triple?) but I don't know how to set them.
Any explanation/tip would be great.
P.S.:The solution, I think, can be found in wikipedia. I see there that $I_1=\frac{m(3R^2+h^2)}{12}=I_2$. While $I_3=\frac{mR^2}{2}$.

Edit: I found http://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html.
At first glance I don't see how to get rid of the "pi" constant in the denominator of rho, the density. Maybe when evaluating the integral with respect to x.
I've never seen a matrix being integrated before, not sure how to tackle this.

Last edited: Jul 3, 2011
2. Jul 3, 2011

### Matterwave

Do it one at a time. What is the definition of the moment of inertia?

3. Jul 3, 2011

### fluidistic

I'm self studying the upper undergraduate level Classical Mechanics course (I already took the freshman CM course) in order to take a final exam in about 2 weeks. The definition I was introduced in the introductory course was $I= \int _V \rho (r) [d(r)]^2 dV$.
Now I see that there's a moment of inertia tensor and can be written as $\mathbf{I} = \begin{bmatrix}I_{11} & I_{12} & I_{13} \\I_{21} & I_{22} & I_{23} \\I_{31} &I_{32} & I_{33} \end{bmatrix}$.
And if I'm lucky to find the Cartesian system of coordinate that makes this matrix diagonal, the $I_i$'s components are the principal moment of inertia. I guess it means the moment of inertia about the principal axis. I will study more about what are those principal axis.

4. Jul 3, 2011

### I like Serena

If you apply your definition-formula for inertia to the cylinder, you'll get the principal moments of inertia. The rest of the entries in the matrix is zero.
It involves calculating two 3 dimensional integrals, one relative to the axis along the center of the cylinder, and one relative to one of the other axes.

5. Jul 3, 2011

### fluidistic

Thanks for helping me.
Yeah I know I can calculate the moment of inertia with respect to the axis I've specified using the definition I've been introducing in the introductory course. This time I wanted to use the moment of inertia tensor.

For some reason it seems I badly copied and pasted a link, in my first post.
Can someone explain me how to do the triple integral of this matrix? I should set up the triple integral for each entry of the matrix? And I should find that all integrals are worth 0 execpt the ones of the entries on the diagonal?
The integrals seems nasty. Even wolfram alpha returned errors when trying to compute them :/

EDIT: HMM what?! The link doesn't work.
Ok, here's the integral: $\rho \int_{-h/2}^{h/2} \int_{-R}^R \int_{-\sqrt{R^2-y^2}}^\sqrt{R^2-y^2} \begin{bmatrix}y^2+z^2 & -xy & -xz \\ -xy & z^2+y^2 & -yz \\ -xz & -yz & x^2+y ^2 \end{bmatrix} dxdydz$. With $\rho = \frac{m}{\pi R^2 h}$.

Last edited: Jul 3, 2011
6. Jul 3, 2011

### George Jones

Staff Emeritus
Change the xyz coordinates to cylindrical coordinates, and change the region of integration appropriately as well.

7. Jul 3, 2011

### fluidistic

I've edited my post (link wasn't working so I had to write the integral in question). I must triple integrate each components of the matrix? Sorry to ask such a question, I never seen this before.

8. Jul 3, 2011

### George Jones

Staff Emeritus
Yes.

9. Jul 3, 2011

### fluidistic

Ok thanks :D
I realize they chose different name for their axis compared to me, I'll take care of this. (my x-axis isn't their x-axis).

10. Jul 3, 2011

Thanks guys! Actually I've shown so far that $\int _V -xy dV =0$ and I've found $I_3[itex] which matches the result. Will do the rest, seems like I'm doing it right. 11. Jul 3, 2011 ### fluidistic I've finished and found out the result. However I have a question. What if the density of the cylinder varies with the height linearly. For instance [itex]\rho (z)=az$ where z=0 at the base of the cylinder and z=h at the other base.
I think the integral would become $\int _V \rho (z)\begin{bmatrix}y^2+z^2 & -xy & -xz \\ -xy & z^2+y^2 & -yz \\ -xz & -yz & x^2+y ^2 \end{bmatrix} dV$ but my huge doubt is... are the limits of the integrals changed compared to the case of a constant density?
Because for a constant density, regarding the limits of the integrals, I assumed I was over the center of mass of the cylinder. Now where should I set up the center of my coordinate system? In the center of mass (I think so)? Or in the geometric center of the cylinder (I don't think so)?

12. Jul 3, 2011

### I like Serena

In general you want the center for your inertia tensor in the centroid of your body.
That way you can decouple translational and rotational movement without hassle (when using the inertia tensor).

You want your axes to align with the symmetries of the body.
That way usually only your principal moments of inertia will be non-zero.

I haven't done the math for your case where density is linearly dependent on the z coordinate, so I don't know how it will turn out.

13. Jul 3, 2011

### fluidistic

Thanks for the information. :) I might be the odd one out and would find the center of mass of the rigid body and set up my coordinate system there.
In Goldstein's book, I've read that finding the Cartesian system of coordinates such that the tensor moment of inertia is diagonal is simply an eigenvalue problem (solving a cubic polynomial). Once you have the eigenvalues of the matrix, you can diagonalize it. And the values of the entries on the diagonal are precisely the principal moment of inertia of the rigid body. So I guess I better know where they are, say in a non-homogeneous rigid body. Otherwise I don't see how I can set up the limits of the triple integral. I'm guessing that the 3 principal axis cross themselves in the center of mass of the rigid body. Is this true? This question is very important to me in order to understand well.

14. Jul 3, 2011

### SteamKing

Staff Emeritus
For bodies with constant density, the product of inertia terms (the ones not located on the main diagonal of the inertia tensor) are zero if the body has symmetry about the centroidal coordinate axes.

For bodies with or without constant density, the integrals for the various terms in the inertia tensor can be expressed about any convenient axis or coordinate system. Once evaluated, the result can be transferred to the centroid by means of the Parallel Axis Theorem and any rotations required to align the calculations axes with the final axes.