Products of Inertia: Zero vs. Non-Zero Explained

[doublemint]

Hello,

I am confused about when a product of inertia is zero and non-zero. Does it depend on the rotation axis and its orientation wrt to the symmetry axis?
For example, you are give a square cube of sides a, a, a. My guess would be that the products of interia are zero if the rotation axis is parallel to one of the axes of symmetry. Then for a rectangular cube of sides a, 2a, 3a, if the rotation axis was diagonal through the centre of the rectangular cube, then the product of inertia is non-zero.

Another point I don't get is that when i calculate the product of inertia for the square cube with a rotation axis along one of the axis of symmetry, i get something non-zero.
Let say the square cube is oriented at the centre with mass m.
\rho\int^{a/2}_{0}\int^{a/2}_{0}xydxdy=\frac{a^{4}\rho}{64}
where \rho=\frac{m}{a^{2}}
Am I getting this right?
Any explanation would be helpful!
Thanks
DoubleMint

 

[tiny-tim]

hello doublemint!

I am confused about when a product of inertia is zero and non-zero. Does it depend on the rotation axis and its orientation wrt to the symmetry axis?
For example, you are give a square cube of sides a, a, a. My guess would be that the products of interia are zero if the rotation axis is parallel to one of the axes of symmetry.

that's right … the products of inertia are the off-diagonal elements of the matrix, and if the coordinate axes are the principal axes of the body, then the matrix is diagonal, and so the products of inertia are zero

Another point I don't get is that when i calculate the product of inertia for the square cube …

with a name like doublemint, you shouldn't do things by halves …

integrate from -a/2 to a/2, and everything will be ok ​