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What is the center of mass of a regular icosahedron with edge length $2$ centimeters? The object is one mass, but is divided exactly in half and each half is a different material. One material is twice as dense as the other. Densities can be taken to be $1$ g/cm$^3$ and $2$ g/cm$^3$, respectively.

I want to use a circumscribed sphere, which touches all vertices, as a model for this. If that is a valid strategy, how can I calculate center of mass for a sphere which is made equally from two materials? I have learned how to calculate center of mass for a uniform sphere, but I have not been taught how to for a sphere like this.

A-Level Student
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Keith
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  • You must compute the center of mass of the two halves, and then take their weighted average. See here for the half-sphere: https://math.stackexchange.com/questions/495520/center-of-mass-of-semi-sphere – Intelligenti pauca Nov 11 '20 at 23:10
  • In the case of a icosahedron you should also explain how it is divided in half. – Intelligenti pauca Nov 11 '20 at 23:15
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    You asked this question (with one parameter difference) just a few hours ago.. That version was closed for lack-of-context. Re-posting is not the way to circumvent closure (or, in any case, to attract attention to a question). Rather, you should edit the previous version to improve its quality in hopes of attracting re-open votes. (Your previous version has one such vote so far.) Re-posts only add clutter and confusion to the site. – Blue Nov 12 '20 at 00:10
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    To compute CM (center of mass) of your icosahedron. You first decompose the icosahedron into finitely many tetrahedron (each with uniform density), the CM of any tetrahedron $ABCD$ is the average of its vertices (i.e. CM = $\frac{\vec{A}+\vec{B}+\vec{C}+\vec{D}}{4}$ ) and the CM of whole icosahedron is the weighted average of these CM (with weight proportional to masses of the tetrahedron). The actual location of CM depends on how you cut the icosahedron into halves (which you haven't specify). – achille hui Nov 12 '20 at 09:18

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