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Can we prove that if, for a system of particles, the centre of mass is at origin then the summation of $xy$ over all the particles of the system is zero, where $x$ and $y$ are the respective coordinates of $i^\text{th}$ particle.

I tried to fix the any one coordinate, say $x$, but the summation of $y$ coordinates of the particles having that x coordinate is not zero, so I was unable to proceed further. Please help.

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No, you can't prove it since it's not generally true. What will be zero is the sum of the coordinates weighted by mass since that's what center of mass is. I.e. $$\Sigma_i m_i \vec{r}_i = 0$$.

EDIT: Since the OP added some extra conditions. Assuming a uniformly distributed density, then: $$\vec{R}_{cm} = \frac{1}{M}\int \rho (\vec{r}) d^3\vec{r} = \frac{\rho}{M}\int_{\rho\neq 0} d^3\vec{r} = 0$$

Then since $\rho/M$ is non-zero, it must be that $$\int_{\rho\neq 0} d^3 \vec{r} = 0$$

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  • $\begingroup$ Sorry i forgot to mention that the system is a continuous rigid uniform body. Can we prove it then? $\endgroup$ – SKTG Sep 27 '19 at 15:55
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No. Take two particles of equal mass at $(1,1)$ and $(-1,-1)$. The summation of $xy$ is $2$, not $0$.

A diagonal rod is a counterexample using a continuous rigid uniform body.

What you are doing (if you properly weight by mass) is calculating one of the off-diagonal components of the moment of inertia tensor $I_{ij}$, in coordinates with origin at the center of mass. In general, all nine components are nonzero.

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  • $\begingroup$ I am extremely sorry. I forgot to mention the most important fact that moment of inertia of the body about x y and z axis is same $\endgroup$ – SKTG Sep 27 '19 at 16:21
  • $\begingroup$ We answer the question you wrote, not the question you meant to write. I suggest asking a new question since it is completely different. $\endgroup$ – G. Smith Sep 27 '19 at 16:26
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If the particles are identical then the sum of their x coordinates will be zero, as will the sum of their y coordinates.

It is easy to prove this one dimension at a time.

If the origin is the centre of gravity then there is zero moment about it due to gravity. The moment is simply the sum over all the particles of the quantity mx, where m is the mass of each particle and x its horizontal distance from the original. For this sum to be zero with a non-zero fixed value for m, the sum over all the particles of the quantity x must be zero.

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  • $\begingroup$ Please use MathJax in answers. $\endgroup$ – G. Smith Sep 27 '19 at 16:37
  • $\begingroup$ Heavens, that sounds a bit technical for someone like me. Is it easy to use? $\endgroup$ – Marco Ocram Sep 27 '19 at 17:05
  • $\begingroup$ Yes, it is. You can learn the basics (fractions, superscripts, subscripts, Greek letters, integrals) in a few minutes and learn more as you need more. Here is a tutoral: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – G. Smith Sep 27 '19 at 17:25
  • $\begingroup$ Your usage here is so simple that all you need to do is put dollar signs around your variables like $x$. $\endgroup$ – G. Smith Sep 27 '19 at 17:26

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