Classical mechanics, principal moments of inertia of a rigid body

In summary, the problem involves determining the principal moments of inertia of a circular cylinder with radius R and height h using the moment of inertia tensor. This can be done by setting up and solving two triple integrals, one for each entry on the diagonal of the tensor. The integrals can be simplified by changing the xyz coordinates to cylindrical coordinates and adjusting the region of integration accordingly.
  • #1
fluidistic
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Homework Statement


Determine the principal moments of inertia of a circular cylinder with radius R and height h.


Homework Equations


Not sure.


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 [itex]\mathbf{I} = \begin{bmatrix}I_{11} & I_{12} & I_{13} \\I_{21} & I_{22} & I_{23} \\I_{31} & I_{32} & I_{33}\end{bmatrix}[/itex].
And if I'm lucky, I can find a Cartesian system of coordinates such that [itex]\mathbf{I} = \begin{bmatrix}I_{1} & 0 & 0 \\0 & I_{2} & 0 \\0 & 0 & I_{3}\end{bmatrix}[/itex] where [itex]I_i[/itex] 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 [itex]I_i[/itex]'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 [itex]I_1=\frac{m(3R^2+h^2)}{12}=I_2[/itex]. While [itex]I_3=\frac{mR^2}{2}[/itex].


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.
 
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  • #2
Do it one at a time. What is the definition of the moment of inertia?
 
  • #3
Matterwave said:
Do it one at a time. What is the definition of the moment of inertia?
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 [itex]I= \int _V \rho (r) [d(r)]^2 dV[/itex].
Now I see that there's a moment of inertia tensor and can be written as [itex]\mathbf{I} = \begin{bmatrix}I_{11} & I_{12} & I_{13} \\I_{21} & I_{22} & I_{23} \\I_{31} &I_{32} & I_{33} \end{bmatrix}[/itex].
And if I'm lucky to find the Cartesian system of coordinate that makes this matrix diagonal, the [itex]I_i[/itex]'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
fluidistic said:
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).

fluidistic said:
The definition I was introduced in the introductory course was [itex]I= \int _V \rho (r) [d(r)]^2 dV[/itex].

And if I'm lucky to find the Cartesian system of coordinate that makes this matrix diagonal, the [itex]I_i[/itex]'s components are the principal moment of inertia. I guess it means the moment of inertia about the principal axis.

You appear to have everything your need already.
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
I like Serena said:
You appear to have everything your need already.
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.
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.
Here's the link: http://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html.
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: [itex]\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[/itex]. With [itex]\rho = \frac{m}{\pi R^2 h}[/itex].
 
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  • #6
fluidistic said:
The integrals seems nasty. Even wolfram alpha returned errors when trying to compute them :/

Change the xyz coordinates to cylindrical coordinates, and change the region of integration appropriately as well.
 
  • #7
George Jones said:
Change the xyz coordinates to cylindrical coordinates, and change the region of integration appropriately as well.

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
fluidistic said:
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?

Yes.
 
  • #9
George Jones said:
Yes.

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
Thanks guys! Actually I've shown so far that [itex]\int _V -xy dV =0[/itex] and I've found [itex]I_3[itex] which matches the result. Will do the rest, seems like I'm doing it right.
 
  • #11
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[/itex] where z=0 at the base of the cylinder and z=h at the other base.
I think the integral would become [itex]\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[/itex] 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
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
I like Serena said:
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.
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
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.
 

Related to Classical mechanics, principal moments of inertia of a rigid body

1. What is classical mechanics?

Classical mechanics is a branch of physics that deals with the motion and behavior of physical bodies under the influence of forces. It is based on Newton's laws of motion and is used to explain the motion of macroscopic objects.

2. What is a rigid body?

A rigid body is a theoretical concept used in classical mechanics to describe a body that does not deform or change shape under the influence of external forces. It is assumed to have a fixed shape and size, and all points within the body move in the same way.

3. What are principal moments of inertia?

Principal moments of inertia are values that describe the distribution of mass within a rigid body. They represent the body's resistance to rotational motion around its three principal axes and are used to calculate the body's rotational kinetic energy.

4. How are principal moments of inertia calculated?

The principal moments of inertia for a rigid body can be calculated by using the body's mass distribution and its distance from the three principal axes. The exact equations vary depending on the shape of the body, but they all involve integrals over the body's volume.

5. Why are principal moments of inertia important?

Principal moments of inertia are important because they help us understand and predict the rotational behavior of rigid bodies. They are used in various applications, such as designing machines and vehicles, analyzing the stability of structures, and studying celestial bodies in astronomy.

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