Perhaps it is a simple question, but I am unable to solve it. Let us suppose that we have a body confined in $\textbf{R}^{3}$ whose mathematical description is given by a bounded and closed domain $\textbf{D}\subset\textbf{R}^{3}$. Moreover, let us also suppose that its mass distribution is given by the function $m:\textbf{D}\rightarrow\textbf{R}_{\geq0}$ such that \begin{align*} \int_{\textbf{D}}m(x,y,z)\mathrm{d}x\mathrm{d}y\mathrm{d}z = M > 0 \end{align*}

My question is: given that its center of mass' trajectory is described by the curve $\alpha:[0,1]\rightarrow\textbf{R}^{3}$, how do we describe the mass distribution $m(x,y,z,t)$ along time?

It is worth mentioning here that we are not considering any kind of rotation, and we are dealing with a rigid body.

  • $\begingroup$ You does not provide enough information for this to be solved. Even if it is a rigid body (it does not deforms in any way), you need to know how it rotates. $\endgroup$ – b.Lorenz Dec 21 '18 at 20:19
  • $\begingroup$ Thank you for the comment. I forgot to mention that I am not considering rotations of any kind. Besides, it is a rigid body. $\endgroup$ – APC89 Dec 21 '18 at 20:20
  • $\begingroup$ Please edit that into question body $\endgroup$ – b.Lorenz Dec 21 '18 at 20:21
  • 2
    $\begingroup$ I think it's really simple if you can set the origin to the center of mass, then it's just a coordinate shift. $\endgroup$ – Jasper Dec 21 '18 at 20:32
  • $\begingroup$ @Jasper That should have been an answer, not a comment. $\endgroup$ – Aaron Stevens Dec 23 '18 at 3:34

If the body is rigid and not rotating, the solution is not that difficult:

If for the sake of simplicity you place the origin in the center of mass of the original mass distribution:

$m_{timedependent}(\mathrm{r},t) = m_{initial}( \mathrm{r} -\alpha(t))$

So basically, when calculating the mass found at a given point, you "transform back", and check which point of the original mass distribution the given point corresponds to.

  • $\begingroup$ @APC89 I must admit that I am confused with this domain D thing. For me it seems much easier to define m(r) for the whole space, or at lest for a domain large enough to enclose the whole trajectory. $\endgroup$ – b.Lorenz Dec 21 '18 at 21:28
  • $\begingroup$ I do not doubt you know what you are saying, but I am a simple minded physics student and you insist using advanced notation to this trivial problem. It seems to me that my answer provides right what you are asking: The mass distribution as a function of time and the original coordinates. $\endgroup$ – b.Lorenz Dec 21 '18 at 21:36

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