My text use the following example to explain the center of mass. There are three balls (mass $m$) sitting in the origin, at $x=l$ and $x=2l$, each two mass are connected with a spring of constant $k$. The system can only move along $x$ direction. To find the center of mass, I setup the coordinate system with first ball placed at $x=0$, the second ball placed at $x=l$ and the third ball placed at $x=2l$. Set $x_1$, $x_2$ and $x_3$ to be the offset from the corresponding equilibrium positions. To find the center of mass, I do the following

$$ x_{com} = \frac{mx_1 + m(x_2+l) + m(x_3+2l)}{m+m+m} = l + \frac{x_1+x_2+x_3}{3} $$

The text said since all the ball have the same mass and they separated equally, so the center of the mass will be at the geometrical center of the system, that is,

$$ x_{com} = l $$

But from the math, we have the last term, I know the conclusion of the text is correct but what's the physical point that we have $x_1+x_2+x_3 = 0$?


1 Answer 1


The center of mas is not what you have defined, the center of mass is:

$$ x_{com}= \frac{mx_1 + mx_2 + mx_3 }{m+m+m} = \frac{x_1 + x_2 + x_3 }{3} $$

and if we use $x_1 = 0$, $x_2=l$, $x_3=2l$:

$$ x_{com}= \frac{0 + l + 2l}{3} = l $$

acording with your geometrical aproach.

  • $\begingroup$ I would like to know what's your coordinates $x_1$, $x_2$ and $x_3$? I think you choose your coordinate refer to the origin, right? But my $x_1$, $x_2$ and $x_3$ are the one refer to the initial equilibrium position, so my $x_1$ same as yours, my $x_2+l$ is your $x_2$ and my $x_3+2l$ is your $x_3$. The reason I have to do that is because there is other sub-questions asking the motion of atoms. I am thinking even the objects are in oscillation, the COM should not be changed. So my com should equal yours. But what's wrong with my calculation? Why $x_1+x_2+x_3=0$ in my case? $\endgroup$ Commented Apr 13, 2013 at 14:24
  • $\begingroup$ Well then your $x_{1,2,3} =0$, so if you sum zeros you get zero. $\endgroup$
    – iiqof
    Commented Apr 14, 2013 at 2:20

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