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Three pennies each of radius R and mass M attached at their edges. How to find the center of mass of the three pennies?

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    $\begingroup$ Symmetry will tell you where it is. In what form should the answer be? $\endgroup$ – André Nicolas Sep 9 '15 at 20:05
  • $\begingroup$ I want to find the distance of the center of mass from each center of the pennies. $\endgroup$ – Numerical Person Sep 9 '15 at 20:06
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Draw the equilateral triangle $ABC$ with corners the centres of the circles. Let $O$ be the centre of mass.

Now there are various ways to proceed. For example, drop a perpendicular from $O$ to $AB$, meeting $AB$ at $X$. Note that $AX=R$. Then $\frac{R}{OA}=\cos(30^\circ)=\frac{\sqrt{3}}{2}$. It follows that $OA=\frac{2R}{\sqrt{3}}$.

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  • $\begingroup$ in the solution they have written that, the center of each penny to the center of mass is just $R/ \cos 30^0$ which is $\frac{2R}{\sqrt{3}}$. $\endgroup$ – Numerical Person Sep 9 '15 at 20:49
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    $\begingroup$ That is the answer I got. I earlier had a more awkward solution using the Cosine Law. $\endgroup$ – André Nicolas Sep 9 '15 at 20:51
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By symmetry, the center of mass is the barycenter of the triangle with vertex on the centers of the three pennies.

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