Inertia Tensor of a cylinder at a distance

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Homework Help Overview

The discussion revolves around calculating the inertia tensor for a cylinder, particularly when the rotational axis is offset from the center of mass. Participants are exploring the implications of the parallel axes theorem and how to adjust the inertia tensor accordingly.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how the inertia tensor changes when the rotational axis is moved, questioning whether the matrix remains diagonal and if zeros stay as zeros. They also express uncertainty about calculating the moment of inertia when the axis is the x-axis.
  • Some participants reference the parallel axes theorem as a method to derive the inertia tensor for an offset axis, noting the importance of keeping coordinate systems parallel.
  • One participant shares a specific example involving a given inertia tensor and a distance vector, expressing confusion about applying the formula correctly and questioning their understanding of the calculations involved.

Discussion Status

The discussion is active, with participants sharing their thoughts and questions. While some guidance has been offered regarding the parallel axes theorem, there is no explicit consensus on the calculations or methods being discussed. The original poster's confusion indicates ongoing exploration of the topic.

Contextual Notes

Participants are working within the constraints of homework guidelines, which may limit the depth of assistance they can provide to one another. The original poster's uncertainty about inertia tensors suggests a foundational knowledge gap that is being addressed through discussion.

pinodk
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I have a cylinder, for which i want to find the inertia tensor.
http://www.mip.sdu.dk/~pino/inertiacyl.JPG
Where the rotational axis are either the x (red) or y (green).

I know that the inertia tensor for a cylinder is of the form
http://www.mip.sdu.dk/~pino/inertiamoment-cylinder.jpg
Then I believe that the bottom right element stays the same, since this describes the rotation around the z-axis.
The tricky part for me is the rest of the matrix. I am no expert, and do not understand inertia tensors fully, so I would like some pointers.

My immediate idea is that the matrix should remain in its diagonal form, the zeros will remain zeros, is this correct?

I know that for complex forms i can split up the moments of inertia, so i have the moment of inertia for the blank space d, which is 0. and then i can add the moment of inertia of the cylinder, but how do i calculate this, when the rotational axis is x-axis for example?
 
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Oops, i should have posted this in the homework section,, don't know how to move it, so will repost there.
 
The wikipedia page

http://en.wikipedia.org/wiki/Moment_of_inertia

contains section 'Parallel axes theorem' at the bottom of the page where they show you how to obtain the tensor for any point of rotation that is offset from the center of mass once you know the tensor for origin at the center of mass.

Keep in mind the axes of the two coordinate systems, one at center of mass and one at your point of choice, must remain parallel, hence the name of the theorem.
 
I have looked at the formula listed, but i don't know quite how to use it...
2b8ad3e76e9f3a4df514e2f2289736eb.png


Suppose i have an inertia tensor in the center of mass like this
[tex]\left(\begin{array}{c c c} 1&0&0\\0&1&0\\0&0&1\end{array}\right)[/tex]
The mass m is 2 kg, and the distance vector R is
[tex]\left(\begin{array}{c} 1\\0\\0\end{array}\right)[/tex]

what i thought of doing was
[tex]I+m\cdot R^2=<br /> \left(\begin{array}{c c c} 3&2&2\\0&1&0\\0&0&1\end{array}\right)[/tex]
But when i look at the formula i read it as
the jk'th of I + M*((R dot R when j equals k) -(the j'th of R times the k'th of R))
for example j=1, k=1, i get
1 + 2* (1-1)=1
for j=2, k=3
0+2* (0-0)=0
and so forth. The M*(R... part always yields 0! what am i misunderstanding?
 
Never mind... i got it now... stupid me :frown:
 

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