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How do I find the above mentioned moment of inertia?

Steps I've tried:

1.) Triple integrations that proved to be to big.

2.) I noticed that the if we split a $2\times 2\times 2$ into individual $1\times1\times1$ components, the body diagonal of the $2\times 2\times 2$ either passes through or is parallel to body diagonals of the $1\times 1\times 1$ cubes.

If the moment of inertia of the $1\times 1\times 1$ about body diagonal be $I$, then the moment if inertia of the $2\times 2\times 2$ about its body diagonal will be $8I$ because it has 8 times as much mass. I though I could get an equation in terms of $I$ by equating $8I$ and moment of inertia of the individual $1\times 1\times 1$ cubes about the body diagonal of the $2\times 2\times 2$ using parallel axis theorem but it turns out to be greater than $8I$. Could someone resolve my mistake?

P.S: I am not familiar with the concept of tensors.

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If the moment of inertia of the 1x1x1 about body diagonal be I, then the moment if inertia of the 2x2x2 about its body diagonal will be 8I because it has 8 times as much mass.

This is the source of your confusion. The moment of inertia of a solid, uniform density cube about any axis that passes through the center of the cube is $\mathrm I = \frac 1 6 {ml}^2$, where $m$ is the mass of the cube and $l$ is the length of any one of the cube's sides. Since the mass of a solid, uniform density cube is given by $m={\rho l}^3$, another way to write the moment of inertia for such a cube is $\mathrm I = \frac 1 6 {\rho l}^5$. This means that the moment of inertia of your 2x2x2 cube will be 32 times that of the moment of inertia of your 1x1x1 cube.

You will get exactly the same result (a factor of 32) if you consider that 2x2x2 cube to consist of eight 1x1x1 cubes and apply the parallel axis theorem.

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  • $\begingroup$ I feel.....stupid. I didn't consider increse in length in the moment of inertia term. So this means my merhod should work right? We've got 32I on the left hand side and 8I+something on the RHS. $\endgroup$ – Skawang Dec 8 '15 at 15:09
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One of the big tricks that you find helpful is that the inertia matrix of a sphere is of the same form of that of a cube. I'll describe what this means.

For a sphere, imagine the moment of inertia (MoI) about three perpendicular axes through the centre of mass. Each MoI will be the same due to symmetry. Now, rotate the three axes you were just using, and the MoI about each axis is still the same:

enter image description here

Now, the trick says you can do the same for a cube! So, you can easily calculate the MoI about an axis going through the centre of mass perpendicular to one of the faces of a cube. So you know the MoI about three perpendicular axes. Now, you can rotate the axes such that one aligns with the diagonal, like such:

enter image description here

And, like a cube, the MoI about the new axes will be the same as the MoI about the old axes, and so you can say: the moment of inertia about a body diagonal of a cube is the same as the moment of inertia about an axis through the centre perpendicular to one of the faces.

The reason this trick works is because a cube and a sphere have inertia matrices of the same form, and performing rotation operations on those matrices causes no change on either matrix, i.e. rotating axes does not change moment of inertia.

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    $\begingroup$ Sorry but I am not familiar with writing vector equations in matrix form. Could you perhaps give the reason for the MOI of cube being same about both axes for cube in simpler words or offer an alternate derivation for the same? $\endgroup$ – Skawang Dec 9 '15 at 17:45
  • $\begingroup$ The inertia matrix is more complex to cover it in detail without getting quite off-topic from the point I'd like to make, but for this question, all you really need to know is that the inertia matrix contains the moments of inertia about a particular set of axes, and you can consider that matrix to fully describe the inertial behaviour of a body when rotating it about those axes. For some bodies, changing the axes of concern yields a different inertia matrix, and so the rotational inertia is different when rotating about those new axes. For a sphere, that is not the case. $\endgroup$ – Involutius Dec 9 '15 at 18:11
  • $\begingroup$ I'm sorry if that isn't clear, and I'll see if I can input an alternative solution, although David's answer seems to do so well. $\endgroup$ – Involutius Dec 9 '15 at 18:13
  • $\begingroup$ It is obvious why rotating all 3 axes of a sphere by 45 degrees leaves the moment of inertia unchanged. It is not obvious why doing the same for a cube should have the same effect. Your explanation is a tautology : 1st paragraph says One of the big tricks ... is that the inertia matrix of a sphere is of the same form of that of a cube then last paragraph says The reason this trick works is because a cube and a sphere have inertia matrices of the same form. $\endgroup$ – sammy gerbil Jul 24 '18 at 11:45

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