How does the concept of center of mass apply to discrete particle systems with varying masses and motions, especially when dealing with a large number of particles?

Considering the challenge of calculating positions and masses for numerous particles, what is the practical utility and significance of center of mass in such scenarios?


5 Answers 5


The answer is that regardless of the motions, total momentum of the system can be expressed as the motion of the center of mass times the total mass of the system

$$ \vec{p}_{\rm TOTAL} = \sum_i m_i \vec{v}_i =\left(\sum_i m_i \right) \vec{v}_{\rm COM} = m\;\vec{v}_{\rm COM} $$

This comes from the definition of center of mass as

$$ \sum_i m_i \vec{r}_i = m \; \vec{r}_{\rm COM}$$

where $m=\sum_i m_i$, and each particle location is $\vec{r}_i$ and velocity $\vec{v}_i$


First of all for practical reasons : Suppose you somehow have access to the trajectories of a large number of classical interacting particles and you wish to see whether the data physically makes sense and is not affected by measurement mistakes. As the center of mass is subject to conservation laws, you can for example verify that it is not drifting, rotating, or moving depending on your problem.

In computational physics, such conservation laws are often actually used to generate the data, i.e they are used as additional constrains to solve the underlying (classical or quantum) equations, because in practice numerical errors can lead to deviations from these conservation laws, and therefore to unphysical solutions.

  • $\begingroup$ Had a doubt related to this topic, if a system's external force is 0 then that means acceleration is 0, which also suggests centre of mass's acceleration and net force is also 0. But does this mean that the rest of the particles also have zero acceleration? Cant they move because of internal forces? But again it doesnt seem physically possible in a rigid body atleast $\endgroup$
    – sanya
    Commented Dec 13, 2023 at 14:29
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    $\begingroup$ @YoussefMabrouk Thanks buddy, I think that this is the best answer that I could get for my doubt. $\endgroup$ Commented Dec 13, 2023 at 18:37
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    $\begingroup$ @sanya In a simple 2-body system with the particles gravitationally bound, both particles have non-zero acceleration the whole time, even if the system as a whole (its centre of mass) has zero acceleration (relative to some external frame). CoM having zero acceleration doesn't imply that all (or indeed any) of the particles have zero acceleration. $\endgroup$
    – Ben
    Commented Dec 14, 2023 at 3:59
  • $\begingroup$ @sanya In a solid subject to zero external force the particles move do due to internal forces, but this motion is on average zero, i.e they "randomly" fluctuate around the same position. $\endgroup$ Commented Dec 16, 2023 at 6:20
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    $\begingroup$ @Mathologist Feel free to accept the answer :) $\endgroup$ Commented Dec 26, 2023 at 18:28

The motion of the center of mass (COM) is conserved. No internal interactions between the particles can change the motion of the COM, no matter how complicated those interactions might be. You need external interactions with something outside of the particle system to move the COM.

As for its practical use, that naturally varies quite greatly based on what system you're working with and what practical activity you're trying to undertake. But as one example, consider the motion of the planets around the sun. If you take careful measurements, you find it's a rather complex motion with lots of bobbing and weaving. However, if you stop assuming the planets revolve around the sun, and start assuming that they revolve around the COM of the entire solar system (which isn't quite at the center of mass of the sun. Jupiter is pretty big), you find that their motion is a lot simpler.

For another example, consider the simulation of a bunch of particles. Did you simulate it correctly? Verifying a simulation takes a lot of work, but one of the first things you can do is calculate the center of mass and make sure it isn't accelerating. If its accelerating, you probably made an error. You can also take the angular momentum of all of the particles about the COM and make sure it is constant as well.


The center of mass is the mass-weighted-average of position.

If the system of particles is isolated, then center of mass moves with constant velocity.

As an example, consider the demonstration “MIT Physics Demo -- Push Me, Pull You“: https://m.youtube.com/watch?v=amfw2nABke4


As other answers explain, the motion of the centre of mass remains constant (if there is no external force).

In simple terms, if masses move towards each other, collide, stick together, break or explode apart.... the overall motion of the overall system is unaffected by any of that, and that overall motion of the overall system means the motion of the CoM.

So a system with each particle's velocity adjusted by the velocity of the CoM will behave identically, but with a fixed CoM this time.

That means that we can remove one variable, so that the net motion of the system oas a whole is zero. We can easily check our numeric approximations are sensible by confirming that 1 trillion iterations later the CoM is still where it is (and if not, why not, and do it more accurately/fix any errors). We may find quite a few equations simplify as motion - after adjusting for the CoM motion - is around one fixed point. We get some more intuitive idea for many systems of how it behaves, even though of course it's chaotic it could give intiuitions, or suggest bounds or simplifications. We may not have to handle such a large range of numbers in computations.

Overall its useful!


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