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Let's say we have a solid with uniform mass distribution, but we don't know its shape. However, we know the moment of inertia of the solid with respect to as many axes of your choosing as you want (e.g. you know $I_{Ox}$, $I_{Oy}$ and $I_{Oz}$). Is it possible to deduce the shape of the solid based on the moment of inertia $I_{Ox}$, $I_{Oy}....$? And if so, what would be the minimum number of axes required to do it ?

Do note that the mass is supposed to be evenly distributed, otherwise we could have a cube with the bulk of the mass concentrated in a sphere within, for example, and thus the different moments of inertia would not take into account the "lightweight corners" of the cube.

Then, there is the parallel axis theorem, which reduces the scope of the problem, as we could therefore only choose axes going through the center of mass, for example. Other than that, I'm completely stumped.

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  • $\begingroup$ I think you may run into trouble with "evenly distributed mass." There's all sorts of clever cuts you could make into an object to emulate an unevenly distributed mass from the perspective of some number of cuts. I think you may find that you're trying to reduce a continuous shape into a finite number of measurements, which rarely if ever works (though radars do try to do a similar concept with SAR mapping, using a very very large number of samples) $\endgroup$ – Cort Ammon - Reinstate Monica Jul 28 '16 at 16:27
  • $\begingroup$ Yes, I feel the question is unanswerable altogether, but still I was curious of the thought process behind this problem. I had thought of a sort of "Cantor set" with each point weighted equally as a non-rigorous counter-example. However, I wasn't very clear on the number of measures, but it is not necessarily finite $\endgroup$ – Throwaway Jul 28 '16 at 16:45
  • $\begingroup$ i really dont think this is possible. Thats because, for any axis, the MI = mr^2 (crudely speaking), i can always generate two solids with the same MI along this axis(by changing the mass in one, and by changing the r in another) $\endgroup$ – Lelouch Jul 28 '16 at 16:57
  • $\begingroup$ for visual simplicity i did not use the subscripts and summation. $\endgroup$ – Lelouch Jul 28 '16 at 16:58
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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/48266/2451 $\endgroup$ – Qmechanic Jul 28 '16 at 18:04
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I will only discuss rigid bodies here; I do not understand fluids well and I doubt you were thinking about fluid bodies anyway.

The first thing you need to understand is that the concept of "moment of inertia" can best be understood in terms of the inertia tensor. The word tensor might deter you, but you must face it to understand many concepts in physics. For pedagogical purposes the essential idea of tensors is that what you are interested in should not depend on the coordinate system you use. You suggest computing moments of inertia about many different axes, but this is the same as computing moments of inertia in different coordinate systems. Since the moments of inertia are part of a tensor we do not gain information by computing them in different coordinate systems. A tensor computed in one coordinate system is equivalent to that tensor computed in a different coordinate system.

In particular, the inertia tensor contains only three independent numbers. The easiest way to obtain these three numbers is to compute the moment of inertia about the three principal axes (let me call these the $x$, $y$ and $z$-axes) of the rigid body. Then the inertia tensor is fully specified by the three numbers $I_x$, $I_y$ and $I_z$. If you now compute the moments of inertia about new axes, $x'$, $y'$ and $z'$, I can tell use the new moments of inertia in terms of the principal moments $I_x$, $I_y$ and $I_z$. In other words, computing the moments of inertia in different coordinate systems does not add information.

One interesting consequence is that any rigid body can be modeled as an ellipsoid with uniform density, as long as we only care about rotation and translation. This is because an ellipsoid can be made to have an arbitrary inertia tensor ($I_x$, $I_y$ and $I_z$) by changing the shape of the ellipsoid.

TL;DR The moments of inertia of a uniform density solid are completely determined by only three components, so we cannot determine the shape of a body from moments of inertia alone.

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  • $\begingroup$ Thank you very much for your crystal clear answer. As I thought, there is no general answer. Do you have any resource on the ellipsoid fact at the end (which seems to be the perfect counter-example to the problem) or even on tensor algebra ? $\endgroup$ – Throwaway Jul 28 '16 at 17:20
  • $\begingroup$ Wiki gives the moments of inertia for an ellipsoid under dynamical properties. You can see that by choosing parameters for the ellipsoid you can get any moments of inertia you wish. As for tensors you should be able to find a good discussion in an intermediate mechanics textbook. $\endgroup$ – James Rowland Jul 28 '16 at 17:32
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Cut out a map of your country drawn on a uniform thickness aluminium plate of homogeneous density. Take it to your physics lab and calculate practically the moment of inertia at multiple points in various axes. Do you think with the results alone you can redraw the shape of country i.e the shape of the aluminium object in your hand.

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