So, not sure if I'm really dumb and not getting something obvious or if there is more complexity here...Let's do 2 objects first. Say C.O.M. is x°.

Then $x° = \dfrac{x_1M_1 +x_2M_2}{M_1+M_2} $

I get the "idea" of this formula–you are in a sense averaging the value of the masses. But is there some blatantly obvious reason for why this is true? It can easily be derived from:

$(x°-x_1)M_1 = (x_2-x°)M_2$

but I don't see why the initial statement ($x° = \dfrac{x_1M_1 +x_2M_2}{M_1+M_2} $) is a priori true. Is there a geometric intuition?

If we are doing a continuous non-constant function:

$\int_{a}^{x°}f(x)(x°-x)dx= \int_{x°}^{b}f(x)(x-x°)dx $

It can be derived that this is equal (if I did it right) to $\dfrac{\int_{a}^{b}xf(x)dx}{\int_{a}^{b}f(x)dx} $

But again, is there a more obvious geometric intuition? Thanks

  • 2
    $\begingroup$ There is nothing to derive: it is a definition. As you say, the intuition the definition is trying to capture is that you are taking a weighted average of positions (not masses), where the weights correspond to mass. $\endgroup$ Mar 15, 2017 at 23:30
  • 1
    $\begingroup$ To the extent there is intuition for the definition, that is a Physics question, not a math one. $\endgroup$
    – Paul
    Mar 15, 2017 at 23:52

2 Answers 2


Notice the definition of center of mass $x_c$ for two particles rearranges to

$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2$$

Let's call the ratio $(\frac{m_1}{m_1+m_2})$ "$W_1$". It can can be interpreted as the percentage of mass that object 1 contributes to the system, generally $(\frac{mass of object 1}{total mass})$. To convert to units of percentage simply multiply the resulting decimal by 100%. Notice that $W_1 \leq 1$, that is object 1 (any object) cannot contribute more that 100% of the mass of the system.

In general a weighted average (arithmetic mean) $x_{av}$ of $n$ components is

$$x_{av} = w_1x_1 + w_2x_2 + ... + w_nx_n = \sum_{i=1}^{n} w_ix_i$$

where $w_i$ is the weight (in the statistical sense) of component $x_i$. For convenience we often wish for the weight factors to be normalized, that is,

$$\sum_{i=1}^{n} W_i = 1$$

Notice here I wrote $W_i$, the normalized weight factors, instead of $w_i$, which are not necessarily normalized.

For your physics application of this math, the mass $m_i$ of object $i$ corresponds to the "crude" weight factor $w_i$. To normalize weight factors (so that none is greater than 100%), we divide the crude factor by the total weight of all components:

$$W_i = \frac{w_i}{\sum_{i=1}^{n} w_i}$$

or, for us, the object mass by the total mass of the system:

$$W_i = \frac{m_i}{\sum_{i=1}^{n} m_i} = \frac{m_i}{M} $$

(where "$M$" is a common abbreviation for the total system mass that is, $M = \sum_{i=1}^{n} m_i$.) This is why our weight factors for calculating center of mass consist of the mass of the object divided by the total mass of the system. For two objects ($n=2$), the weight factor $W_i$ for object $i$ is of course

$$W_i = \frac{m_i}{\sum_{i=1}^{2} m_i} = \frac{m_i}{m_1+m_2} $$


$$x_c = \Big(\frac{m_1}{m_1+m_2}\Big)x_1+\Big(\frac{m_2}{m_1+m_2}\Big)x_2 \\ = \sum_{i=1}^{2} \Big(\frac{m_i}{m_1+m_2}\Big) x_i \\ = \frac{\sum_{i=1}^{2} m_ix_i}{\sum_{i=1}^{2} m_i} \\ = \frac{1}{M} \sum_{i=1}^{2}x_im_i$$

I've written a few forms of $x_c$ above so you can mull over how this expression is constructed. For the general center of mass of a system of $n$ objects, simply replace $2$ with $n$ in the final expression.

Let's explicitly juxtapose our center of mass formula with the weighted average formula to drive the point home:

$$x_{av} = \sum_{i=1}^{n} W_ix_i \\ \\ x_c= \sum_{i=1}^{n} \Big(\frac{m_i}{M}\Big) x_i$$

Thus, the center of mass is the weighted average of position with respect to mass. $$\\ \\$$ P.S. If the system has a continuous distribution of mass rather than $n$ discrete chunks (as for any real object that is not being approximated as a dimensionless point particle), instead of

$$x_c = \frac{1}{M} \sum_{i=1}^{n}x_im_i$$

We can use

$$x_c = \frac{1}{M} \int xdm$$

which arises when we take the limit as $n \to \infty$. (Note this form is not the most practical to calculate so we usually make the substitution $dm = \rho dV$ where $\rho$ is the density of the object, which may vary with position.)

  • 2
    $\begingroup$ This answer gets to the heart of the misconception in the question: the formula is a kind of average of the positions, not an average of the masses. $\endgroup$
    – David K
    Mar 16, 2017 at 13:25

From the point of view of mathematics, there is no further explanation. The formula $$(x°-x_1)M_1 = (x_2-x°)M_2$$ has to be considered as an axiom. The motivation comes from physics, and is related to the following image of masses balancing each other:

enter image description here

  • $\begingroup$ No I understand the $(x°-x_1)M_1$ part...but can you explain the intuition of the weighted average? (see edit) $\endgroup$ Mar 15, 2017 at 23:43
  • $\begingroup$ I am sorry but I don't understand the comment. I saw the edit but I don't see the difference. Please, please, explain in the question marked as EDIT. $\endgroup$
    – zoli
    Mar 15, 2017 at 23:56
  • $\begingroup$ Answers like "Just accept it buddy..." aren't very helpful. By stating "(x°−x1)M1=(x2−x°)M2" is an axiom, you've both made a mistake and obscured what the actual axiom is. The axioms you're referring to are known as "Newton's Law's". From newton's laws we know.. $ \sum_i F_i =0$. If you multiply your "axiom" by $g cos(\theta)$ ($\theta$ being the angle of rotation) $$(x°−x1) g*cos(\theta)*M1=(x2−x°) g*cos(\theta)*M2$$ you get an explanation which is actually helpful and that is that "for the system to be at rest the torque exerted must be balanced" $\endgroup$
    – ThomasTuna
    Nov 9, 2021 at 22:54

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