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I'm looking for some references on a specific moment of inertia, for radial motions of a spherical body. In my calculations, I got this integral : \begin{equation}\tag{1} \bar{I} = \int r^2 \, dm = \int_{\mathcal{V}} \rho(r) \, r^2 \, d^3 x, \end{equation} where $\rho$ is the matter density and $r^2 = x^2 + y^2 + z^2$ defines the usual radial coordinate (the coordinates origin is located at the center of the spherical body). For an uniform mass distribution, this integral is easy to do : \begin{equation}\tag{2} \bar{I} = \frac{3}{5} \; M R^2. \end{equation} Please, don't confuse this with the well known moment of inertia of the sphere, around some rotation axis. This is about radial motions, and not rotation.

I never saw this in any books on mechanics.

Notice that expression (1) above is also half the trace of the inertia tensor : \begin{equation}\tag{3} I_{ij} = \int_{\mathcal{V}} (r^2 \, \delta_{ij} - x_i \, x_j) \, \rho \; d^3 x, \end{equation} Then we have this : \begin{equation}\tag{4} \bar{I} \equiv \frac{1}{2} \; I_{kk}. \end{equation}

I'm not sure the "radial inertia moment" defined by (1) (if it have a proper interpretation) is getting the proper factor.

Any thoughts on this ?

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  • $\begingroup$ Isn't “radial motion” just a plain center of mass motion and the “moment of inertia” is just the mass $M$ of that sphere? $\endgroup$
    – Martin Ueding
    Feb 12, 2017 at 17:53
  • $\begingroup$ @MartinUeding, I'm not sure. Imagine an uniform sphere at the center of your origin system. Give it some radial motion (compression, dilatation, oscillations), without breaking the spherical symetry. I think that integral (1) measures its inertia for such radial motions. $\endgroup$
    – Cham
    Feb 12, 2017 at 17:55
  • $\begingroup$ What physical significance do you think that the physical quantity in eqns (1) or (2) has? Can you write an equation which uses it? eg Something similar to $L = I\omega$ relating angular momentum and angular velocity. Or perhaps this is the question which you are asking us to answer? Just because you can write a mathematical definition does not mean that it has any physical significance. $\endgroup$
    – sammy gerbil
    Feb 14, 2017 at 2:43
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    $\begingroup$ What do you mean by radial motion? Translation and rotation completely describe one object's motion unless you're examining object deformation which is a whole bag of worms. If we have a second object, their motions may be described, for example, by the relative motion of the bodies, their rotations, and the movement of the system. How many bodies, and which way might they be moving? $\endgroup$
    – user121330
    Feb 21, 2017 at 22:50
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    $\begingroup$ I think a little more context could be useful. You mean a uniformly pulsating sphere, as in a pulsating star (not a rigid body), right? In this case, the velocity of a point is $v_i=\dot\epsilon x_i$, and the kinetic energy in a volume $dV$ is $\rho v^2 dV/2$ where $d\bar I=\rho v^2 dV$ is your "pulsatile moment of inertia" of element, where $v^2=\dot\epsilon^2 x_i^2$. If you want to compare this to rotations, remember that $\vec v=\vec\omega\times\vec r$, and you can expand this as $\rho v^2 dV=I_{ij}x_i x_j$. The derivations are similar; you might expect some relation between $\bar I$ and I. $\endgroup$
    – exmachina
    Feb 22, 2017 at 15:12

2 Answers 2

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Of course it is half the trace. Given the three principal components on some coordinate system are $$ \begin{align} I_{xx} & = \int (y^2+z^2) {\rm d}m \\ I_{yy} & = \int (z^2+x^2) {\rm d}m \\ I_{zz} & = \int (x^2+y^2) {\rm d}m \end{align} $$

Add them up to get

$$ I_{xx}+ I_{yy}+I_{zz} = \int (2 x^2+2 y^2+2 z^2) {\rm d}m $$

which is double the value in your definition of

$$ I_{radial} = \int ( x^2+y^2+ z^2) {\rm d}m $$

The bigger question here is, how is the above derived and how is it used? I think the OP needs to provide more details for the question to be answered effectively.

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    $\begingroup$ I found the "radial moment of inertia" $\bar{I}$ by writing down the differential equation of a radially pulsating star. The quantity $\bar{I}$ entered my equation as an inertia factor in front of a second time derivative (from Newton's equation), so I was wondering about its proper physical meaning. $\endgroup$
    – Cham
    Jun 8, 2017 at 15:29
  • $\begingroup$ If there is a physical meaning, then it is not MMOI or something related to it. $\endgroup$
    – John Alexiou
    Jun 9, 2017 at 12:50
  • $\begingroup$ MMOI ? What is that ? $\endgroup$
    – Cham
    Jun 9, 2017 at 13:28
  • $\begingroup$ @Cham - Sorry, Mass Moment of Inertia (MMOI) $\endgroup$
    – John Alexiou
    Jun 9, 2017 at 16:29
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The moment of inertia is the measure of resistance to angular acceleration about an axis. Unless I'm mistaken, what you're after is the modulus of elasticity $E$ (or Poisson's ratio $\nu$) of the object. That dictates the response to radial motion given a uniform pressure field acting on the surface of the sphere.

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  • $\begingroup$ Take note that the integral I defined for radial motions is the trace of the inertia tensor for any rotation. Is the elasticity modulus related to the inertia tensor ? $\endgroup$
    – Cham
    Feb 12, 2017 at 18:50
  • $\begingroup$ No, it's not related to the inertia tensor, but neither is radial motion. The modulus of elasticity is a material property, and it is defined as the ratio of stress to strain for a given material. You simply cannot quantify the resistance to axial (or radial) strain using a moment of inertia tensor. $\endgroup$
    – gdbb89
    Feb 12, 2017 at 18:57
  • $\begingroup$ Why not ? In my calculations, I got that radial integral (which is the trace of the inertia tensor, by some coincidence). So I strongly think that integral has something to say about radial motion "resistence" (i.e. inertia). $\endgroup$
    – Cham
    Feb 12, 2017 at 19:00

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