# Center of Mass: General (technical)

#### Prerequisites

In our discussion of the question, "To what extent can we treat a complex object as if it is a point?", we considered the simplest possible extension of a point mass -- two point masses connected by a non-deformable rigid rod. The is "the simplest" to the extent that it only adds one more position -- the position of the second object. What we saw from an analysis of that system is that the "average position" or Center of Mass of the two objects, moves like a point mass that obeys Newton's second law and only feels the external forces on the combined object. By "average", we mean weighted by the fraction of the total mass.

The mathematical result is that the average coordinate

$$ \overrightarrow{r}_{CM} = (\frac{m_A}{m_A + m_B}) \overrightarrow{r}_A + (\frac{m_B}{m_A + m_B}) \overrightarrow{r}_B $$

satisfies a Newton's second law equation:

$$(m_A + m_B) \overrightarrow{a}_{CM} = \overrightarrow{F}_{AB} ^{ext} $$

We got this by looking to see if there was a coordinate for which the equations would work (i.e., look like the equations for a point mass) and we solved for that coordinate and interpreted what it meant. Now, let's see if we can generalize this by taking any arbitrary number of masses, doing the same idea, and seeing if the resulting equations still look like a point mass.

So now, suppose we have a large number of masses: *m*_{1}, *m*_{2}, ... *m*_{N}. We won't bother with rigid rods -- we'll just assume that the masses interact with each other by forces. (We put in the rigid rod in our test case so that the two masses looked like a single extended object. We didn't really need it so we won't use them here.)

Instead of continually writing 1, 2, 3, ...N for all our mass labels, we'll use the more formal (and convenient, once you get use to it) notation of the "summation" -- Σ. Thus, the Newton's second law equation for mass 1 is

$$m_1 \overrightarrow{a}_1 = \overrightarrow{F}_1 ^{ext} +\overrightarrow{F}_{2 \rightarrow 1} +\overrightarrow{F}_{3 \rightarrow 1} + ... +\overrightarrow{F}_{N \rightarrow 1} = \overrightarrow{F}_1 ^{ext} + \sum_{j=2}^N \overrightarrow{F}_{j \rightarrow 1}$$

The $ma$ for mass 1 is = the sum of all external forces from mass 1, plus the force from mass 2, the force from mass 3, the... up to the force on it from mass N. We simplify (really!) the notation for this big sum (it could be over 10^{22} atoms, for example!) by writing the summation marker and introducing an index, j. The sigma symbol is read: "sum over these quantities as j varies from 2 up to N".

For mass 2, the sum will be similar by now we'll have the force from mass 1 acting on mass 2 instead of mass 2 on mass 1. The result looks like this:

$$m_2 \overrightarrow{a}_2 = \overrightarrow{F}_2 ^{ext} +\overrightarrow{F}_{1 \rightarrow 2} +\overrightarrow{F}_{3 \rightarrow 2} + ... +\overrightarrow{F}_{N \rightarrow 2} = \overrightarrow{F}_1 ^{ext} + \sum_{j \neq2}^N \overrightarrow{F}_{j \rightarrow 2}$$

Since we want all the forces from the internal masses that act on mass 2, we can write the index as j → 2 and sum over all values of j (from 1 to N) except for 2, which we write as j ≠ 2.

We write the equation for a general mass in introducing another index label, i. We let this stand for any mass index from 1 to N, just like we did for j. The equation for an arbitrary mass is

$$m_i \overrightarrow{a}_i = \overrightarrow{F}_i ^{ext} + \sum_{j \neq i}^N \overrightarrow{F}_{j \rightarrow i}$$

This reads just like the other equations: The ma of the i-th mass is equal to all the external forces acting on that mass, plus the sum of all the internal forces from all the other masses.

Now, following what we did in the two mass case, let's add up all the ma equations. What we get is what looks like a royal mess! But don't give up yet!

$$\sum_{i=1}^N m_i \overrightarrow{a}_i = \sum_{i=1}^N \overrightarrow{F}_i ^{ext} + \sum_{i=1}^N \sum_{j \neq i}^N \overrightarrow{F}_{j \rightarrow i}$$

Although this looks complicated, if we think about the physical meaning of each term, we can make sense of it. The first term is just the sum over all the $ma$ terms: $m_1a_1 + m_2a_2 + m_3a_3 +... + m_Na_N$. Just what we would expect. The first term on the right of the equals sign is the sum over all the external forces on the different masses, i = 1...N. The last term is the mess — a sum of a sum! But if we remember what we are doing, it makes sense and simplifies. In each of the $ma$ equations, that term was all the forces the internal masses exerted on the mass we were considering. So when we consider all the masses by adding equations, we get all the internal forces of every mass on every other mass. So we might ask: How often does each term $F_{A→B}$ occur for the different A's and B's? The answer can be pretty quickly be seen to be only once! The only place $F_{A→B}$ comes in is in the $ma$ equation for B. But the term $F_{B→A}$ also comes in — in the $ma$ equation for A! By Newton 3, we know these are equal and opposite, so they cancel. And this is true for EVERY A and B! So that ugly double sum all adds up to 0 because of N3. Our result is:

$$\sum_{i=1}^N m_i \overrightarrow{a}_i = \sum_{i=1}^N \overrightarrow{F}_i ^{ext}$$

All our internal forces are gone. All we have to do now is make the ma sum look like a single ma term. If we define a center of mass position analogous to the way we did it for two masses, we would conjecture that the CM position is the weighted sum, each position weighted by the fraction it is of the total mass.

$$ \overrightarrow{r}_{CM} = (\frac{m_1}{m_{T}}) \overrightarrow{r}_1 + (\frac{m_2}{m_{T}}) \overrightarrow{r}_2 + (\frac{m_3}{m_{T}}) \overrightarrow{r}_3 + ... + (\frac{m_N}{m_{T}}) \overrightarrow{r}_N$$

where $m_T$ is the total mass — the sum of the masses of all the individual objects:

Taking the second derivative of each term to make an acceleration and multiplying both sides by $m_T$, we get the sum of all the $ma$'s, just like we need.

$$ \overrightarrow{a}_{CM} = (\frac{m_1}{m_{T}}) \overrightarrow{a}_1 + (\frac{m_2}{m_{T}}) \overrightarrow{a}_2 + (\frac{m_3}{m_{T}}) \overrightarrow{a}_3 + ... + (\frac{m_N}{m_{T}}) \overrightarrow{a}_N$$

Multiplying both sides by $m_T$ we get the sum we need:

$$ m_T \overrightarrow{a}_{CM} = {m_1} \overrightarrow{a}_1 + {m_2} \overrightarrow{a}_2 + {m_3} \overrightarrow{a}_3 + ... + {m_N}\overrightarrow{a}_N = \sum_{i=1}^N m_i \overrightarrow{a}_i$$

Putting this together with our result for the sum of the $ma$'s above, gives our final result: That the center of mass of a collection of objects moves in a way controlled by a SINGLE Newton's second law -- with only the external forces acting on the collection coming in.

$$ m_T \overrightarrow{a}_{CM} = \sum_{i=1}^N \overrightarrow{F}_i^{ext}$$

Note a really striking result. We did not assume any rigid rods, so in fact the set of masses do NOT have to be what looks to us to be a single object! They can in fact be any collection of objects at all, flying every which way! As you can imagine, this surprising but true result is of considerable power and value.

Joe Redish 8/16/15

Last Modified: January 29, 2019