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If the centre of mass of a system of particles is at the origin, does this mean that if there is a particle on the positive X-axis, there must be atleast one particle on the negative X-axis?

I feel that this is true, since to get $X_{COM}$ as zero there should be a particle on the negative X -axis.

Am I on the right track?

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    $\begingroup$ Think of a triangle with one vertex on the positive X-axis, symmetrical with respecto the the X-axis. Edit: I'm talking about the axis, not the negative half-plane. $\endgroup$ – anderstood Apr 27 '15 at 15:27
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No, a particle on the positive $x$-axis does not imply a particle on the negative $x$-axis, only that there are particles whose $x$-coordinate is negative.

For example, take three particles with equal mass, and take one positioned on the negative $x$-axis, and the two others symmetrically about the $x$-axis in the $x-y$-plane, so that the sum of their two position vectors is on the $x$-axis with the length of the third one. No particle on the negative $x$-axis, but the center of mass is in the origin.

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By definition, in one dimension the center of mass has position

$$ x_{\mathrm{CM}} = \sum_{i=1}^{n}m_ix_i$$

So if your system has two particles and the center of mass is the origin indeed if $x_1>0$ then $x_2<0$.

In the general case, suppose all particles are to the right of the origin: then your center of mass will be a sum of positive numbers, which cannot be zero. So at least one of the particles must be to the left. Analogously if all particles are to the left your center of mass will be a sum of negative numbers, which cannot be zero. So you need also at least one particle to the right.

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