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.
Best Answer
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.