Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x_1+x_2+x_3 = 0$?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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? – user1285419 Apr 13 '13 at 14:24
Well then your $x_{1,2,3} =0$, so if you sum zeros you get zero. – Angel Joaniquet Tukiainen Apr 14 '13 at 2:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.