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Let's consider a rigid body in 3D space, calculate the moment of inertia tensor, and find the three principal moments of inertia.

I can somewhat understand when only one of them is zero. If it is many identical mass points, it means all these points lie in the plane normal to the axis vector of the zero principal moment of inertia. This is just a picture for a special case. What restriction would a zero impose to a general rigid body?

Another natural further question would be what having two zero principal moments of inertia means.

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A point in space has three zero MMOI components.

A line in space has one zero MMOI component along the direction of the line. Any body that exists along a line and has mass distribution offset of the line would fall under this category.

But you cannot have two zero MMOI components because of the perpendicular axis theorem that states that one component of MMOI must be the sum of the two perpendicular components of MMOI. So to have one non-zero component, you must have have at least one other non-zero components. Note that the principal components must be non-negative.

You can visually check this by laying out the three principal MMOI components along the x-axis of a tensor diagram. This is similar to a 3D version of Mohr's circle for the stress tensor. You draw three circles, with the diagonals the three pairs of principal values.

fig1

Now rotate about the 1 direction by an angle $\theta$, and finding the new diagonal terms $I_{22}$ and $I_{33}$ as well as the off-diagonal term $I_{23}$ in the following fashion:

fig2

Rotate the diagonal line between 2 and 3 by an angle $2\theta$ and where the ends end up are the values of the rotated tensor as indicated above.

What this means is that if two principal values are zero, then the third must be zero also due to the way the circle nest within each other.

But the $I_1$ component can be zero, as the whole diagram can slide left and right without violating any rules.

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  • $\begingroup$ Isn't that the theorem only for a plane lamina? $\endgroup$ – xiaohuamao Feb 12 at 21:29
  • $\begingroup$ It applies here also as both are 3×3 symmetric tensors. See edits for a connection to Mohr's circle diagrams. $\endgroup$ – John Alexiou Feb 12 at 21:36

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