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I have found the tensor of inertia of a rectangle of sides $a$ and $b$ and mass $m$, around its center, to be $$I_{11}=ma^2/12,$$ $$I_{22}=mb^2/12,$$ $$I_{33}=(ma^2 + mb^2)/12.$$ All other elements of that tensor are equal to zero. I would now like to use this result to determine the tensor of inertia of a hollow cube of side a around its center of mass.

I realise I have to use the parallel axis theorem. I also knoww that the correct equation is I$_{11}$=I$_{22}$=I$_{33}$=ma$^2$/12+ma$^2$/12+ma$^2$/6+4(ma$^2$/12 + m(a/2)$^2$)=5/3*ma$^2$ I simply do not understand why this is correct. Could anyone please explain why this is the correct way to calculate the desired tensor of inertia? Also, why would I be summing all the diagonal elements in my tensor for the rectangle?

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The center of mass does not change so the parallel axis theorem is not needed. What you need is one positive mass and one negative mass of smaller size. –  ja72 Mar 26 at 18:42
    
What is the thickness of the cube, and does it have any walls missing? –  ja72 Mar 26 at 18:45
    
Actually, the parallel axis theorem is much needed. This is taken from an exam and the correct solution applies parallel axis theorem to solve this problem. I simply don't understand why the above equation is correct and would appreciate an explanation. –  peripatein Mar 26 at 18:46
    
The hollow cube is decomposed as a solid cube of side $a$ and a "negative" cube of side $a-2t$ inside it. Since both cubes share the same center of mass I do not see why the parallel axis theorem is needed. Maybe you need to include a sketch in order to understand what the 3D geometry lools like, and about which point is the MMOI taken. –  ja72 Mar 26 at 18:53
    
I am unable to add attachments here (do not have sufficient credit). –  peripatein Mar 26 at 18:54

2 Answers 2

up vote 1 down vote accepted

We will work in units where the mass of each face is $1$ and where the length of the side of the cube is $1$.

The contribution to the moment of inertia of each of the top and bottom faces is, using your result for the moment of inertia of a rectangle, $\frac{1}{6}$.

By symmetry, each of the four other faces has the same contribution to the moment of inertia. Let's calculate the contribution of one of them. Let's project the system along the axis about which we are calcuating the moment of inertia. This operation has no effect on the moment of inertia. Our face now becomes a rod. We know the moment of inertia through the center of the rod is $\frac{1}{12}$, but we are calculating the moment of inertia at a distance $\frac{1}{2}$ away from the center of the rod. Using the parallel axis theorem, we get that the moment of inertia is $\frac{1}{12} + (\frac{1}{2})^2 = \frac{4}{12} = \frac{1}{3}$. Then our total moment of inertia is $\frac{1}{6} + \frac{1}{6} + 4*\frac{1}{3}= \frac{5}{3}$. (The first two terms come from the top and bottom face, the last term comes from the four side faces.)

Putting units back in we get the moment of inertia is $I=\frac{5}{3} M a^2$, where $M$ is the mass of a face and $a$ is the side length of the cube (also the side length of a face).

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What I still don't quite understand, is why the parallel axis theorem not applied to the top and bottom faces, AND, whether I MUST use projection (I deem it slightly confusing; any way to avoid it)? –  peripatein Mar 26 at 20:05
    
Furthermore, why, in order to calculate I_11, am I adding all these other elements and not just the ones relevant? I mean, when I first calculated I_11 for the rectangle I added only the y and z components. –  peripatein Mar 26 at 20:19
    
The parallel axis theorem does not need to be applied to the top and bottom faces because the axis of rotation already passes through the center of mass for those faces. You are right about the projection. It is not necessary. –  NowIGetToLearnWhatAHeadIs Mar 26 at 20:20
    
Alright, and what about my second comment above? –  peripatein Mar 26 at 20:22
    
I don't understand your second comment. All of the contributions are relevant because they all concern mass that is not on the x-axis. But maybe it will help if I say there are extra components for the side faces because the center of masses of the side faces do not lie on the x-axis (unlike when you did the rectangle) so we need to add a term from the parallel axis theorem to correct for this. –  NowIGetToLearnWhatAHeadIs Mar 26 at 20:23

So you have 6 flat rectangular sheets that form a cube.

TwoSheet

The inertia matrix about $1$ on the sheet center of mass is

$$ I_{1}^C = \begin{bmatrix} \frac{1}{12} m (a^2+a^2) & 0 & 0 \\ 0 & \frac{1}{12} m a^2 & 0 \\ 0 & 0 & \frac{1}{12} m a^2 \end{bmatrix} $$

To transfer an 3×3 inertia matrix to the origin, the center of mass vector $\vec{c}=(c_x,c_y,c_z)$ is used as

$$ I_{1}^A = I_{1}^C - m [\vec{c}\times][\vec{c}\times] $$

where $[\vec{c}\times]$ is the 3×3 anti-symmetric cross product operator

$$ [\vec{c} \times] = \begin{bmatrix} 0 & -c_z & c_y \\ c_z & 0 & -c_x \\ -c_y & c_x & 0 \end{bmatrix} $$

For the case shown above we have $\vec{c}=(\frac{a}{2},0,0)$ so

$$ I_{1}^A = \begin{bmatrix} \frac{1}{12} m (a^2+a^2) & 0 & 0 \\ 0 & \frac{1}{12} m a^2 & 0 \\ 0 & 0 & \frac{1}{12} m a^2 \end{bmatrix} - m \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -\frac{a}{2}\\ 0 & \frac{a}{2} & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -\frac{a}{2}\\ 0 & \frac{a}{2} & 0 \end{bmatrix}$$ $$ I_{1}^A = \begin{bmatrix} \frac{1}{6} m a^2 & 0 & 0 \\ 0 & \frac{1}{3} m a^2 & 0 \\ 0 & 0 & \frac{1}{3} m a^2 \end{bmatrix}$$

It turns out the MMOI for the symmetric sheet (shown in light blue above) is identical to $I_{1}^A$. With permutation of the axis you get

$$I_A = 2\begin{bmatrix} \frac{1}{6} m a^2 & 0 & 0 \\ 0 & \frac{1}{3} m a^2 & 0 \\ 0 & 0 & \frac{1}{3} m a^2 \end{bmatrix} + 2 \begin{bmatrix} \frac{1}{3} m a^2 & 0 & 0 \\ 0 & \frac{1}{6} m a^2 & 0 \\ 0 & 0 & \frac{1}{3} m a^2 \end{bmatrix} + \\ 2 \begin{bmatrix} \frac{1}{3} m a^2 & 0 & 0 \\ 0 & \frac{1}{3} m a^2 & 0 \\ 0 & 0 & \frac{1}{6} m a^2 \end{bmatrix} = \begin{bmatrix} \frac{5}{3} m a^2 & 0 & 0 \\ 0 & \frac{5}{3} m a^2 & 0 \\ 0 & 0 & \frac{5}{3} m a^2 \end{bmatrix}$$

Which is what you have.

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When it comes to 3D, doing the parallel axis theorem by inspection is very tricky. I prefer the vector algebra method shown above with the $[c\times]$ operator. –  ja72 Mar 26 at 19:39

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