How can I determine whether the mass of an object is evenly distributed without doing any permanent damage? Suppose I got all the typical lab equipment. I guess I can calculate its center of mass and compare with experiment result or measure its moment of inertia among other things, but is there a way to be 99.9% sure?
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Perhaps using a hydrostatic balance and Archimedes' Principle? – David H May 17 '13 at 17:10
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2I think this is very difficult if you admit pathological or malicious cases. Pipe-fitters and shipbuilders and the like make heavy use of x-ray and gamma-ray imaging, sonic imaging and other "non-destructive testing" techniques. It's a big field in it's own right. – dmckee --- ex-moderator kitten May 17 '13 at 17:34
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1I'm construing the question to mean that we assume the object is rigid. Otherwise we could shake it, probe it with ultrasound, or find out that it contained gyroscopes or Mexican jumping beans. – May 17 '13 at 21:45
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@BenCrowell Well, ultrasound would detect the difference in may example, but does Alex's "typically lab equipment" include ultrasound? On the other hand, if the ideal object rings nicely when hit by a mallet my construction is probably detectable that way. – dmckee --- ex-moderator kitten May 17 '13 at 22:55
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The non-rigid version is possibly useful in practice, but doesn't seem likely to lend itself to a definite answer that could be given on this site. BTW, it's not obvious to me how to generalize this to relativity, since you basically can't have rigid objects in relativity. – May 17 '13 at 22:58
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2Come to think of it, if you have some perfect examples to use for comparison, the ringing behavior under a well defined impact is powerful test that can be done with quite simple tools. A repeatable drop mallet and a PC with good microphone might be enough... – dmckee --- ex-moderator kitten May 17 '13 at 23:08
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1If you are allowed to go beyond rigid inertia, there are plenty of methods to debunk simple frauds such as dmckee's counter example. You could compare the thermal conductivity between two closely neighboured points to that between two antipodes, or observe its deformation when compressed infinitesimally. Or x-ray the thing, duh. – leftaroundabout May 18 '13 at 09:12
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@leftaroundabout Yes, any kind of tomography would do it. And I think that we can take "typical lab equipment" to include at least an ohm meter and probably enough bits to get a rough thermal conductivity as well. – dmckee --- ex-moderator kitten Jun 05 '13 at 16:59
5 Answers
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Malicious counter example
The desired object is a sphere of radius $R$ and mass $M$ with uniform density $\rho = \frac{M}{V} = \frac{3}{4} \frac{M}{\pi R^3}$ and moment of inertia $I = \frac{2}{5} M R^2 = \frac{8}{15} \rho \pi R^5$.
Now, we design a false object, also spherically symmetric but consisting of three regions of differing density $$ \rho_f(r) = \left\{ \begin{array}{l l} 2\rho\ , & r \in [0,r_1) \\ \frac{1}{2}\rho\ , & r \in [r_1,r_2) \\ 2\rho\ , & r \in [r_2,R) \\ \end{array} \right.$$
We have two constraints (total mass and total moment of inertia) and two unknowns ($r_1$ and $r_2$), so we can find a solution which perfectly mimics our desired object.
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I'm not so sure. You're talking about the moment of inertia about its center. By measuring the moment of inertia about some off-center axis, it seems to me that we could distinguish the uniform sphere from your malicious counterexample. (Let me just do a little calculating here...) – Mike May 17 '13 at 21:34
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1@Mike: By the parallel axis theorem, you can't get any additional information that way. – May 17 '13 at 21:38
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Well, the gravitational effects of the malicious counter example are different :) – Manishearth May 17 '13 at 21:42
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@Ben Crowell. Yep, I agree. I just had to remember what the parallel-axis theorem said. :) +1 to dmckee – Mike May 17 '13 at 21:43
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1@Manishearth. I'll redeem myself by pointing out Birkhoff's theorem, which says that the gravitational effects are the same (outside the body). – Mike May 17 '13 at 21:44
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1@Manishearth: Nope. In Newtonian gravity (as in E&M), Gauss's Law shows you that it just depends on the total mass enclosed (as long as the system is spherically symmetric. Even in GR the gravitation outside a spherically symmetric body is independent of the distribution inside the body. – Mike May 17 '13 at 21:54
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In GR you could spin the sphere. Then Birkhoff wouldn't apply, and I think you'd get geodetic and frame-dragging effects that might be different in the two cases. Even in SR, the rotational dynamics are different, and in any case it's not possible for it to be perfectly rigid as assumed. – May 17 '13 at 22:14
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In less symmetric arrangements the near field gravitational effects could differ, but this being a malicious case, I've made it maximally easy for me and maximally hard for Alex. Bhawhawhawhaha! ::dons turban with point in the middle:: – dmckee --- ex-moderator kitten May 17 '13 at 22:53
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Well, I had thought I could just try multiple ways and any malicious case has to fail one of them...but for this I can't really think of any within Newtonian mechanics. thx – arax May 18 '13 at 04:50
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@AlexSu As I said in the comments on the main question the ringing spectrum would be different when struck. I don't know of a off-the-shelf piece of software to work the problem for you but the problem is not terribly difficult. – dmckee --- ex-moderator kitten May 18 '13 at 05:27
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While you have two equations and two unknowns, the equations aren't linear: the m.o.i. equation will contain $r_1^5$ and $r_2^5$, while the mass equation will contain $r_1^3$ and $r_2^3$. So it's not at all evident that the equations have a solution. – Pulsar Jan 02 '14 at 19:19
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If you have a rigid mass distribution sealed inside a black box, then the only things you can observe about its motion are its velocity vector and its angular velocity vector as functions of time. These can be predicted if you know the total force and total torque that act, plus the mass, center of mass, and moment of inertia tensor. So all that can be determined by any external, mechanical measurements is its mass, its center of mass, the orientation of its principal axes, and the three diagonal elements of the moment of inertia tensor along the principal axes. This is nowhere near enough information to recover the full mass distribution or to determine whether the mass is evenly distributed.
As an example, suppose object A is a spherical mass $m$ of uniform density, with radius $a$. Then you can make object B with the same mass distributed uniformly on a hollow shell of radius $b=\sqrt{3/5}a$, so that B has the same moment of inertia as A.
If you want them to look the same visually you can for example create object C by superposing two objects: (1) a uniform sphere like A, but with half the density, and (2) a concentric shell like B but with half the mass per unit area. Then A and C are indistinguishable.
By the shell theorem, A and C are also not distinguishable by their external gravitational fields.
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This is true, but the OP only wants to know if it's evenly distributed or not, which is an easier prblem than "find the distribution". For example, you could measure the moment of inertia around an axis, and compare that to what you'd get if it were homgenously distributed. If there's a discrepancy - you can catch it.
I wonder whether one can find a counterexample of a body that has the full tensorial moment of inertia of a homogeneously distributed one, but is actually not.
– yohBS May 17 '13 at 19:16 -
@yohBS: Thanks for your comment. I've added a specific counterexample to show that uniformity can't be determined. – May 17 '13 at 19:28
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@BenCrowell: Well, in the counterexample the size is different, so it's obvious that the moment of inertia is different. You seem to have gone in the wrong direction -- you have given an object of a different size and distribution with the same I. What we want is two objects that look the same externally and have the same I, but have different distributions of mass. – Manishearth May 17 '13 at 19:51
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@Manishearth: I don't understand why you assert that the moment of inertia is different. The different size is required in order to make the moment of inertia the same. The questions of size or how it "looks externally" is irrelevant. You can for example put objects A and B inside identical cubical boxes, creating objects $A'$ and $B'$. Since the boxes have the same mass and moment of inertia, $A'$ and $B'$ again have the same mass and moment of inertia as one another. I'll edit the answer to include this explanation. – May 17 '13 at 20:11
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@BenCrowell: I didn't assert that I is different. I'm saying that the smaller sphere isn't a counterexample because the question is "can we have two objects of equal size/mass and I but only one is uniform?". And the OP isn't putting things in boxes (which adds nonuniformity), he is just asking if one can determine the uniformity of a given object. Two boxes containing objects are obviously nonuniform. – Manishearth May 17 '13 at 20:17
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The outer shells won't have the same mass and I; one will have to be thicker than the other. – Manishearth May 17 '13 at 20:19
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@Manishearth: The boxes can be made to have the same uniform density as the uniform object. I've edited the answer to show this explicitly. – May 17 '13 at 20:19
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@BenCrowell: Yes, except that now you will end up adding more mass to one box since the object embedded in it is smaller. – Manishearth May 17 '13 at 20:20
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Basically, even if you embed, some other externally visibly parameter will change. We need two externally equivalent objects, where only one has uniform distribution. – Manishearth May 17 '13 at 20:23
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If you cannot tell any external difference between the black box object and a white box object of the same shape whose mass is evenly distributed, or at least no relevant difference, then can't it be concluded that the mass of the black box is evenly distributed? – Kaz May 17 '13 at 23:10
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It has been pointed out that this cannot simply be done by examining the mass distribution (first and second moment of mass). But there is a way to "look inside" most common objects: Take a CT scan. Not sure if you consider that "typical" lab equipment - but it's equipment I have in my lab...
Of course depending on the size of the object and the material composition, it can be quite hard to get a definitive answer - beam hardening effects need to be taken into account, which means you have to know your Xray spectrum. Reconstructing a dense (high Z) object with the correct density everywhere is actually quite hard.
Floris
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If mass is evenly distributed inside the volume occupied by your mass, and you know your mass theoretical density, then if your mass is evenly distributed inside the volume it occupy, it will be exactly equal to the theoretical density, if there is a mismatch between the theoretical density and the measured density, it means that mass is not evenly distributed or that another kind of mass with different density is in your mass volume (like when there are bubbles inside a solid mass) but it may be the case that the bubbles are also evenly distributed then mass is evenly distributed but theoretical density does not match your one element measured density because mass density averages but it is not evenly distributed.
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Tie a rope to the object and attach the other end to a scale. Slowly lower the object into some water, recording the force on the scale and the amount of water displaced at many intervals. Using this data, compute the density of the section of the object that is submerged at all intervals. If the mass of the object is uniformly distributed then the density values you get should not vary.
Thomas
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@Manishearth could you elaborate? Now that I think about it just lowering the object once definitely wont work. – Thomas May 17 '13 at 20:40
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1This doesn't work. The buoyant force only depends on the weight of the displaced water. – May 17 '13 at 21:29