The Missing Link: Point Particles, Extended Objects, and the Center of Mass

by eloch

I’d like to make one more post about Newton’s Laws of Motion before I temporarily abandon them. There are two closely-related issues that I want to address:

  1. There is a glaring problem with Newton’s Laws as I have portrayed them in my last two posts (Why Are Newton’s Laws of Motion Important? and Variations on Newton’s Laws of Motion).
  2. Many students in introductory physics classes get the impression that physics cannot possibly apply in the real world because we always replace real objects with point particles.

In the process of addressing the problem with Newton’s Laws, I will simultaneously show that the aforementioned physics students could not be further from the truth; Newton’s Laws of Motion do apply to objects in the real world, and this is true because of the fact that we can replace extended objects with point particles (not in spite of it)!

Continue forward to see what allows me to make these claims!

Note: parts of this will get a little technical.

Note: if you come across unfamiliar notation or ideas in this post, it may be in your interest to take a look at my glossary.

What Is The Problem?

Consider Newton’s Second Law. In it, we claim that if we know the mass of an object and the force acting on it, we can use {\bf F} = m {\bf a} to find its acceleration. The issue is that talking about the acceleration of a real-life object makes no sense. For example, consider a spinning wheel. It is well known from introductory physics classes that any point moving in a circle experiences an acceleration towards the center of the circle. (Check out the Wikipedia site for a short explanation with some pictures… I don’t want to go into this here.) So this means, for example, that the top of the wheel accelerates downwards, while the bottom accelerates upwards. At the same time, the center has zero acceleration. Which one of these accelerations is the acceleration of the wheel? I don’t know either! The plain fact is that an extended object doesn’t have just one acceleration, because its different parts can move in different ways.

So When Are Newton’s Laws of Motion Valid?

The preceding discussion should convince you that Newton’s Laws of Motion only make sense for objects that only have one position, velocity, and acceleration. In physics, we call these things “point particles” because they exist only at a single point. Most everyday objects are certainly not point particles! Anything that doesn’t act like a point particle is called an “extended object.” I will spend the rest of this post explaining why and how Newton’s Laws can apply to extended objects.

How Do We Do It?

The key point is to understand that an extended object can be thought of as a collection of point particles. It takes an infinite number of point particles to fill up an extended volume of space, but our mathematics can handle this situation. So our first step will be to analyze the motion of a collection of particles, and then to generalize our results to extended objects.

So let’s return to the system of N particles that I introduced near the end of my last post. We previously derived an important theorem about this system:

\displaystyle \dot{\bf P} = {\bf F}^{\rm ext} \ \ \ \ \ (1)

We also noticed that this equation looks suspiciously like Newton’s Second Law, except that “force on the object” is replaced with “net external force on the entire system” and that “momentum of the object” is replaced with “total momentum of the entire system.” Let’s take this analogy a little further and try to totally identify the system with a single particle.

We have established that a collection of particles does not have a single position. But if we had to pick one position that best describes it, what would we use? A good starting guess would be the center of the system. What do we mean by center? (Note: the process that I explain here is a good example of how physicists quantify and clarify imprecise thoughts.) We could write a formula for the average position of all the particles in the system:

\displaystyle {\bf R}_{center} = \frac{1}{N} \sum_{alpha} {\bf r}_{\alpha} \ \ \ \ \ (2)

But this might not totally agree with our intuition. To see why, let’s do a thought experiment. Imagine two systems: System One consists solely of a bowling ball, while System Two is the same bowling ball in conjunction with an ant sitting ten feet away. If we apply Equation 2 to each system, we see that the center of System One is the center of the bowling ball, while the center of System Two is five feet away. So Equation 2 tells us that by adding a tiny ant to the system, we can change its center by a huge amount. Intuitively, we feel that the center of the system shouldn’t change very much when we make such a tiny addition. So instead, let’s make a “weighted average” of the particle positions, in which big particles contribute more than smaller ones:

\displaystyle {\bf R}_{CM} = \frac{1}{M} \sum_{\alpha = 1}^{N} m_{\alpha} {\bf r}_{\alpha} \ \ \ \ \ (3)

In this formula, {M} stands for the total mass of all the particles in the system. It turns out that this is the best way to characterize the position of a system. (I will explain why very soon.) This position is so important that it gets a special name: the “center of mass.”

Note that we can easily compute the velocity of the center of mass:

\displaystyle {\bf V}_{CM} \equiv \dot{{\bf R}}_{CM} = \frac{d}{dt} \frac{1}{M} \sum_{\alpha = 1}^{N} m_{\alpha} {\bf r}_{\alpha} = \frac{1}{M} \sum_{\alpha = 1}^{N} m_{\alpha} \frac{d}{dt} {\bf r}_{\alpha} = \frac{1}{M} \sum_{\alpha = 1}^{N} m_{\alpha} {\bf v}_{\alpha} \ \ \ \ \ (4)

Now, let’s take another look at the total momentum of the system:

\displaystyle {\bf P} = \sum_{\alpha} {\bf p}_{\alpha} \ \ \ \ \ (5)

If the system truly behaves like a single particle, we should be able to write its total momentum as the product of a mass and a velocity. And now we see why the center of mass is so special: it allows us to do just that!

\displaystyle M {\bf V}_{CM} = M \frac{1}{M} \sum_{\alpha = 1}^{N} m_{\alpha} {\bf v}_{\alpha} = \sum_{\alpha = 1}^{N} m_{\alpha} {\bf v}_{\alpha} = \sum_{\alpha} {\bf p}_{\alpha} = {\bf P} \ \ \ \ \ (6)

Let’s take a derivative of Equation 6:

\displaystyle \dot{\bf P} = M \dot{\bf V}_{CM} \equiv M {\bf A}_{CM} \ \ \ \ \ (7)

where {{\bf A}_{CM}} is the acceleration of the system’s center of mass. Finally, if we combine Equations 1 and 7, we see that:

\displaystyle {\bf F}^{\rm ext} = \dot{\bf P} = M {\bf A}_{CM} \ \ \ \ \ (8)

This looks a lot like Newton’s Second Law; in fact, it is Newton’s Second Law applied to a point particle with mass {M} that is located at the system’s center of mass. Awesome! Awesome! AWESOME!

Generalization to Extended Objects

So we can apply Newton’s Laws of Motion to a collection of point particles as long as we replace them with a single point particle that has a certain mass and position. What does this mean for a continuous extended object? (I’m going to gloss over this a little bit, but don’t feel like going into details about the integrals right now. I don’t think they would add much to the post.)

Well, we can divide the object into {N} little cube-shaped pieces, each of which has a volume of {dx dy dz}, and treat it as a system of {N} particles. If the object has a position-dependent density {\rho({\bf r})}, the mass of each of these particles is be {\rho dx dy dz}. Now, let’s increase {N} to infinity and shrink {dx}, {dy}, and {dz} to zero. All the sums in the preceding section get replaced by integrals over the volume of the object, but the same conclusions hold. In particular, we see that the center of mass of the object moves as if all the external force were acting directly on it.

Notice that we have not said anything concrete about the motion of any individual parts of the object; the object is allowed to rotate around its center of mass, but its center of mass behaves like a single point-particle.

I hope this doesn’t seem like magic to you. If it does, check out the following video, in which some physicists demonstrate that the center of mass of a tennis racket (and a few other things) moves just like a single particle when you throw it in the air. (Look at this page for an explanation of how point particles travel when they are launched into the air; the picture in the middle called “trajectory” is the most relevant part.)

Conclusions

Strictly speaking, Newton’s Laws can only apply to point particles. However, by breaking an extended object into lots of little pieces, we were able to show that its center of mass behaves like a single particle. Therefore, we see that Newton’s Laws of Motion can apply to the real world because we can replace extended objects with point particles, contrary to the opinions of many physics students.

Further Reading

4 comments on “The Missing Link: Point Particles, Extended Objects, and the Center of Mass”

  1. Bradley says:
    January 29, 2014 at 2:39 pm

    “It takes an infinite number of point particles to fill up an extended volume of space.”

    I’ve been looking for something to cite for this, and your blog is the only thing I’ve found. Is this taken to be obvious? Is it is a theorem? Can you direct me to an academic publication I can cite in support of it?

    Thanks!

    Reply
    • eloch says:
      February 2, 2014 at 12:15 pm

      Hi Bradley,

      Good question! I don’t think it’s obvious but I can give you a simple argument to explain why it should be true. Let’s say we have a region of space with volume V that we want to split apart into lots of smaller regions with volume v. How many small regions do we need? Well, that’s easy to find: we need N = V/v small regions. Point particles have no volume at all (this is part of their definition). So to see how many point particles we need, we can start shrinking v. As we shrink v, N correspondingly becomes larger because N = V/v. If we shrink v all the way to zero, then N has to become infinite to keep their product the same (i.e., so that N*v = V). I think this argument is probably included somewhere in Taylor’s Classical Mechanics, but it likely shows up implicitly during the conversion of a sum to an integral.

      By the way, this is a trick that shows up quite often in physics. For example, an ideal electric dipole is a limiting case of a physical dipole where charge goes to infinity and distance goes to zero in such a way that the dipole moment remains constant. Another good example is the Dirac delta function, which you can think of as a rectangle whose height goes to infinity and width goes to zero in such a way that the area remains constant.

      In a more general sense, you can usually deal with infinities properly by first examining a situation with finite properties and then letting them go to zero or infinity.

      I hope this helps!

      Reply
      • Bradley Rettler says:
        February 17, 2014 at 8:43 pm

        That is a precise statement of the intuitive thought I was going for. Thanks! I couldn’t find a name associated with the site; if you email it to me, I’ll be happy to acknowledge you in my paper.

        -Bradley http://www.bradleyrettler.com

        On Sun, Feb 2, 2014 at 12:15 PM, Physics Soup

      • eloch says:
        February 18, 2014 at 11:14 am

        I’m glad I could help! Thanks for pointing out my missing name; I thought I had it on here somewhere but I guess I didn’t. I updated my “About Physics Soup” page to include my name: https://physicssoup.wordpress.com/about-physics-soup/

        Please let me know when your paper is published, I would be interested to read something involving physics and philosophy!

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