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Q & A: Can you get mass distributions from moments of inertia?

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Most recent answer: 08/21/2012
Q:
If you have an arbitrary blob of masses and you can measure the angular inertia of this mass through any arbitrary axes you want... is there any way that you can use this information- perhaps with the Fourier slice theorem, to reconstruct the distribution of the blob of masses? It feels very much like a yes you can answer but I cannot thing of how. I understand this sounds like a homework question. I promise it isn't. I don't even know what class this would be asked in. It's kind of a potluck.
- Kevin Dawes (age 29)
San Diego, CA
A:
My first reaction was to think "yes", but on second thought the answer is no. Take the moment of inertia tensor around the center of mass. The moment of inertia around some axis not through the center of mass is just given by the moment around the parallel axis through the C.O.M. plus the moment due to the displacement of the C.O.M. from the new axis. That extra moment depends only on the total mass and the displacement. Thus it gives no new information about the mass distribution.

So we're back to just the inertia tensor about the C.O.M. It of course has far too little information to give the whole distribution.

In fact, in retrospect, consider the following two distributions.

A.  a gram uniformly distributed on a spherical shell of radius 1 cm.

B. 1/2 gm at the middle and 1/2 gm in a spherical shell at sqrt(2)cm from the origin.

A and B have the same moments of inertia, as do an obviously infinite set of distributions from the same family.

Mike W.

(published on 08/14/2012)

Follow-Up #1: parallel axis theorem

Q:
Sorry to do this to you, but actually using the parallel axis theorem it still gives me different results for a tangential axis with that set of spheres you described. I'm stumped. 2/5 * (0.5g) * (sqrt(2)cm)^2 + 1 gram * sqrt(2)cm^2 != 2/5 * 1g * 1cm^2 + 1g * 1cm^2. I'm going to stop trying after this. I'm totally lost. Thank you for your incredible generosity and indulging someone who hasn't done math in a long time.
- Kevin Dawes (age 29)
San Diego, CA
A:
I'm having a little trouble following what your two expressions stand for. What I'm claiming is that our mass distributions A and B will give the same moments for any axis, and thus the moments can't help you figure out which distribution you have.

It looks like you're maybe calculating the moments about an axis that just touches the outer shell at sqrt(2) cm from the middle.
For A, the gives (2/3)*1gm*1cm2+1gm*(1/2)cm2=(5/3)gm*cm2.
For B, that gives (2/3)*(1/2)gm*2cm2+1gm*(1/2)cm2=(5/3)gm*cm2.

(I think your 2/5 factor is for solid spheres, not spherical shells.)

Mike W.

(published on 08/19/2012)

Follow-Up #2: You're welcome

Q:
Thank you. My math was just scrambled. I'm really sorry about that and thank you for your time.
- Kevin Dawes (age 29)
San Diego, CA
A:
Seriously, it's fun to have an excuse to think about this stuff.

Mike W.

(published on 08/21/2012)

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