Center of mass


In many of our examples discussing Newton's laws, we ignore the extent of the object and treat it as if it were a "point mass" -- an object that has a mass but no size; or rather, that is so small we don't care about the size. This of course is an oversimplified model. In order to get a sense of what we can keep from that model and what we have to jettison once we go to a more complete model, let's consider the motion of a complex object. What we learn might be a bit of a surprise: that when looked at in a slightly refined way, Newton's second law holds exactly without corrections, for any object of whatever extent, shape, or rigidity! To see this, we'll start with the simplest possible extended model and then move to the general case. (This also is a nice example of "thinking with math" -- how the structure of the equations can lead us to create new ways of thinking about a system that makes qualitative as well as mathematical sense.)

A simple model of an extended object

Two hard spheres connected by a rigid rod

As often is the case in physics, in order to get sense of what's going on in a complex system, it helps to work out in detail the simplest possible case. Since we are trying to go beyond one point mass, the next more complex system is two point masses connected by a (nearly) massless rigid rod as shown in the figure on the right at the top. We choose the rod to be rigid and massless so its motion is completely determined by the motion of the two masses connected to it. We are only introducing one additional complication -- the position of the second mass-- and not lots of them -- the shape of the connecting rod.

We'll assume the only non-touching (long-range) force acting is gravity so the masses each have contact interactions with the rod but not with each other. We'll also assume that the masses might have any other interactions with the rest of the objects in the universe. So the only real model simplification is about the internal structure of the object - not what it's feeling from outside. Our system schema for the object will look something like what we have shown in the figure on the right below.

This then leads us to the set of three free-body diagrams for the parts of the object:

The object A and the rod exert equal and opposite forces on each other by Newton's third law, as do object B and the rod. Objects A and B feel forces from interacting with other objects in the world, including the earth (gravity). We've shown the interactions in arbitrary directions so we can create a result that is true in general and doesn't depend on choices beyond our simple modeling assumptions.

Each object satisfies its own Newton's 2nd law equation. (You will see why we are writing them as $ma = F^{net}$ instead of $a = F^{net}/m$ in a moment.)

$$m_A \overrightarrow{a}_A = \overrightarrow{F}_A ^{ext} +\overrightarrow{F}_{rod \rightarrow A} $$

$$m_B \overrightarrow{a}_B = \overrightarrow{F}_B ^{ext} +\overrightarrow{F}_{rod \rightarrow B} $$

$$m_{rod} \overrightarrow{a}_{rod} = \overrightarrow{F}_{rod} ^{ext} +\overrightarrow{F}_{A \rightarrow {rod}} +\overrightarrow{F}_{B \rightarrow {rod}} $$

Since we are assuming that the rod is massless and has no external forces acting on it, the N2 equation for the rod simplifies to

$$\overrightarrow{F}_{A \rightarrow {rod}} +\overrightarrow{F}_{B \rightarrow {rod}} = 0$$

(Now you can see why we used the "ma" form of N2 instead of the "F/m" form. It's OK to divide by m as long as it's not zero, but since we are taking the rod's mass to be 0, we can't divide by it.)

By Newton's third law this implies

$$\overrightarrow{F}_{{rod} \rightarrow {A}} +\overrightarrow{F}_{{rod} \rightarrow {B}} = 0$$

What we now want to do is see to what extent we can think about the structured object as one combined object, ignoring its structure. When we're doing this, we aren't looking at the forces internal to the object -- only the external ones. To this end, let's combine the equations by adding the three $ma = F^{net}$ equations. We have seen that the rod's N2 equation is just 0=0 so our adding the equations gives

$$m_A \overrightarrow{a}_A + m_B \overrightarrow{a}_B = \overrightarrow{F}_A ^{ext} +\overrightarrow{F}_{rod \rightarrow A} + \overrightarrow{F}_B ^{ext} +\overrightarrow{F}_{rod \rightarrow B} $$

Regrouping, we get

$$m_A \overrightarrow{a}_A + m_B \overrightarrow{a}_B = \overrightarrow{F}_A ^{ext} + \overrightarrow{F}_B ^{ext} +(\overrightarrow{F}_{rod \rightarrow A} +\overrightarrow{F}_{rod \rightarrow B} )$$

But we found that the forces of the rod on the objects are equal and opposite as a result of its zero mass. So the internal forces cancel, giving 

$$m_A \overrightarrow{a}_A + m_B \overrightarrow{a}_B = \overrightarrow{F}_A ^{ext} + \overrightarrow{F}_B ^{ext}$$

Well that's partway there to our goal of treating the complex object as a simple one! The internal forces cancel. But we have two different ma terms. We'd like to have a single N2 equation for our combined object. So let's ask: Can we write the left side as a single $ma$ term? Let's try and see what that would mean. We'll see if we can write the sum of the two terms as a single term for a "combined mass". Let's try to define a combined (center of mass) acceleration by

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

We don't know if an $a_{CM}$ exists or is reasonable, but let's assume there might be something and work with it. Since we know that an acceleration is a second derivative of a position, let's assume that the$a_{CM}$ is the second derivative of a position, $r_{CM}$, and see what that position needs to be to make this work.

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

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

$$\frac{d^2 \overrightarrow{r}_{CM}}{dt^2} = (\frac{m_A}{m_A + m_B}) \frac{d^2 \overrightarrow{r}_A}{dt^2} +  (\frac{m_B}{m_A + m_B}) \frac{d^2 \overrightarrow{r}_B}{dt^2} $$

Since the masses are constant, we can take them inside the differentiations and get

$$\frac{d^2}{dt^2} \overrightarrow{r}_{CM} =  (\frac{m_A}{m_A + m_B}) \frac{d^2}{dt^2} \overrightarrow{r}_A +  (\frac{m_B}{m_A + m_B}) \frac{d^2}{dt^2}\overrightarrow{r}_B $$

From this we can see how we have to choose rCM in order to get a single N2 equation. It suggests that we choose an "average position" as our position for the combined mass

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

Then that combination of positions moves according to Newton's second law:

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

This is a very sensible and satisfying result! The position we need is a weighted average of the two positions, and the weighting is that fraction of the total that each mass is. To summarize:

For our model system of two objects connected by a rod, if we look at the "weighted average" position (weighted by the fraction each object's mass is of the total) then that position moves according to a Newton's second law that uses the total mass and only responds to external forces acting on the system.

In other words

We can treat our complex object as a point mass having the total mass and the average position of the two objects.

For two objects of equal mass, the average position is halfway between them. For other cases, see the examples in the follow-ons.

More complex objects

By looking carefully at our derivation, we can see that the key idea is that the internal forces cancelled when we added up the different ma terms for different parts of our object. Since that is true by N3 for any set of objects, we can expect our result to generalize. Here's the general result:

$$ \overrightarrow{r}_{CM} =  (\frac{m_A}{m_{total}}) \overrightarrow{r}_A +  (\frac{m_B}{m_{total}}) \overrightarrow{r}_B +  (\frac{m_C}{m_{total}}) \overrightarrow{r}_C + ... $$

$$(m_{total}) \overrightarrow{a}_{CM} = \overrightarrow{F}_{A+B+C+...} ^{ext} $$

Since the position we use if the "average" position of the parts of the combined object, we refer to that position as the object's center of mass. In most cases, the center of mass of a complex object is just where you would expect it to be -- at the center of the object. But it the distribution of mass in the object is not uniform, it may be at surprising places. For example, for a ring, the center of mass is at the center of the ring -- where there is no mass! But that's the point that would move under only the external forces. For example, if you threw the ring and spun it, although any point on the ring would follow a crazy looping path, the center of the ring would trace out a smooth parabola! (As long as you can ignore air resistance.) For the details see Center of Mass: General (technical).


This is way more than a mathematical exercise or curiosity! It is an extraordinary powerful result! It say no matter what happens inside an object, the object's average position moves only by external forces. This tells you that you can't reach down, grab your shoes and lift yourself into the air. It also says that a vibrating molecule can't start moving by itself; it needs to use its vibration to "push off" against another object, creating an external force on itself. It say that your car's engine can't make your car go without an interaction with an external object (such as the friction from the ground). 

This is one example of powerful "constraint" principles in physics: principles that are true and constrain a whole system no matter what complexities are occurring on the inside of the system. In this case, our center of mass principle will be translated into a more general and powerful form: the conservation of momentum.

Joe Redish 8/9/15 


Article 353
Last Modified: May 13, 2019