I find Newton's second law of motion for point particles quite easy to grasp. However, I run into a lot of confusion when I deal a discrete particle/ continuous body system.

In these notes by Jaan Kalda, in page-1, he defines force as the product of inertial mass times the acceleration of the body, and then later in page-2, he defines $ P = \sum m_i v_i$ ,and then says that we can write $ \vec{F} = M \left( \frac{d^2 r_c}{dt^2} \right)$ , finally he equates both of the definitions which results in:

$$ M \frac{d^2 r_c}{dt^2} = \frac{d}{dt}\sum m_i v_i$$

Now, my question is how were we able to equate the two definitions? I tried researching about these results and I found out about Euler's law of motion, does this suggest that we need to include continuum mechanics and calculus of surfaces to properly talk about the all so familiar second law?


2 Answers 2


The steps in the derivation are as follows:

  1. If we have point masses $m_i$ each moving with velocity $v_i$ then the total momentum of the system is $\vec P = \sum m_i \vec v_i$.

  2. We can show that $\frac {d \vec P}{dt} = \vec F$ where $\vec F$ is the net external force on the system (we show this by applying Newton's second law to the individual point masses and then summing; any internal forces cancel out because of Newton's third law).

  3. So $\vec F = \sum m_i \frac {d \vec v_1}{dt} = \sum m_i \frac {d^2\vec r_i}{dt^2} = \frac {d^2}{dt^2} \left( \sum m_i \vec r_i \right)$ where $\vec r_i$ is the position vector of point mass $m_i$.

  4. We introduce the position $r_C$ of the centre of mass of the system, which we define to be the weighted position vector of its point masses i.e $\vec r_C = \frac {\sum m_r \vec r_i}{M}$ where $M= \sum m_i$ is the total mass of the system.

  5. Now we can rewrite the expression for $\vec F$ from (3) above as

$\quad \quad \displaystyle F = \frac {d^2}{dt^2} \left( M \vec r_C \right) = M \frac {d^2 \vec r_C}{dt^2}$

This shows that the motion of the centre of mass of a system of point masses under external forces is the same as if the same external forces were acting on the whole mass of the system concentrated at its centre of mass. Essentially, this is a direct consequence of the fact that Newton's second law is a linear differential equation.

  • $\begingroup$ Hmm this makes the definition of centre of mass very natural. However I haven't seen any book taking this approach. $\endgroup$
    – Babu
    Sep 9, 2020 at 10:58
  • $\begingroup$ @Buraian It's exactly the same derivation as in the Jaan Kalda notes that you linked to, except that Kalda does not motivate the definition of the COM - he just pulls it out of thin air. $\endgroup$
    – gandalf61
    Sep 9, 2020 at 11:11
  • $\begingroup$ The COM has an exact definition. I don't think one can "pull it out of the air" anymore than one can pull $p=mv$ out of the air. $\endgroup$ Sep 9, 2020 at 11:19
  • $\begingroup$ @BioPhysicist Obviously COM has an exact definition, but you can produce that definition ex nihilo as Kalda does or you can explain to students “this is why it is is useful to have a tool in our toolbox that we call centre of mass and which is defined like this”. $\endgroup$
    – gandalf61
    Sep 9, 2020 at 11:54
  • $\begingroup$ What would happen to this derivation if mass changed as a function of time? $\endgroup$
    – Babu
    Sep 10, 2020 at 17:17

The way I understand this is that a model for a body with extension is that of a collection of $N$ point particles. Suppose that these have masses $m_1,\dots,m_N$ and follow trajectories $\vec{r}_1,\dots,\vec{r}_N$. Then we define the center of mass as an "imaginary" particle, whose mass is the total mass of the body $m=\sum_{i=1}^Nm_i$ and whose trajectory is the mass-weighted average of the trajectories $$\vec{r}_c=\frac{1}{m}\sum_{i=1}^Nm_i\vec{r}_i.$$ By taking this weighted average, instead of the simple average $\frac{1}{N}\sum_{i=1}^N\vec{r}_i$ we ensure that the heavier a particle in the body is, the more it contributes to the center of mass. It is useful to check that in the limit all particles have the same mass $m_i=m/N$ the weighted average reduces to the simple average. One should also check that in the limit that a particle $i$ is much more massive than the rest of the particles the center of mass follows the trajectory $\vec{r}_i$.

The reason for defining the center of mass is that it is very useful in describing the motion of a body as a whole. This can be seen from various perspectives. One of them is obtained by first noticing that we can rewrite the definition of the center of mass in the form $$m\vec{r}_c=\sum_{i=1}^N m_i\vec{r}_i.$$ Differentiating with respect to time we obtain $$m\vec{v}_c=\sum_{i=1}^N m_i\vec{v}_i,$$ where $\vec{v}_c=d\vec{r}_c/dt$ is the velocity of the center of mass and $\vec{v}_i=d\vec{r}_i/dt$ is the velocity of particle $i$. The right hand side of this equation has a nice interpretation. Namely, $m_i\vec{v}_i$ is by definition the momentum $\vec{p}_i$ of the particle $i$. If we define the total momentum of the body to be the sum of the momenta of its constituent particles $$\vec{p}=\sum_{i=1}^N\vec{p}_i,$$ we conclude $$m\vec{v}_c=\vec{p}.$$ In other words, the momentum our imaginary particle, the center of mass, is equal to the total momentum of our body.

One can go further and show several other ways in which the motion of the center of mass helps us describe the total motion of the system. Some of them are

  1. The rate of change of momentum of the center of mass (which by the argument above is equal to the rate of change of momentum of the whole body) is equal to the sum of all of the forces acting on each of the particles that make up the body. In fact, one can do better. Let us call internal forces those arising from the interaction of two particle within the body. Further, let us assume these satisfy the weak version of Newton's third law. This states that the force exerted by particle $i$ on particle $j$ is equal in magnitude but opposite in direction to the one exerted by particle $j$ on particle $i$. Then the rate of change of momentum of the center of mass turns out to be equal to the sum of all external forces acting on the body.
  2. There is an analogous story going on with angular momentum. In particular, define the total angular momentum to be the sum of the angular momenta of each of the particles in the body. Let us assume now that the internal forces satisfy the strong version of Newton's third law. This version demands on top of the weak version that the force that particle $i$ exerts on particle $j$ lies on the line joining $i$ and $j$. Then the rate of change of total angular momentum is equal to the sum of the torques on the particles of the body generated by the external forces. Unlike the previous case though, the angular momentum of the center of mass is not equal to the total angular momentum. One has to take into account the angular momentum relative to the center of mass as well.
  3. There are other statements of this form (although even less related to the center of mass) concerning energy. In this sort of statements new concepts, such as center of gravity, appear.

Nice proofs to the statements above can be found in the first chapter of Goldstein's book, Classical Mechanics.

Finally, let me point out that the picture of a body in terms of a discrete set of particles is nothing but a model. One could also build a continuous model by introducing continuous mass densities. Mathematically these models are unified by something called measure theory. A nice reference taking this approach are the notes on Lagrangian mechanics of Andrew Lewis

  • $\begingroup$ what is strong version and weak version of third law? $\endgroup$
    – Babu
    Sep 9, 2020 at 11:32
  • $\begingroup$ Weak version: If a particle $i$ exerts a force on a particle $j$, then particle $j$ exerts a force on $i$ of equal magnitude but opposite direction. $\endgroup$ Sep 9, 2020 at 13:19
  • $\begingroup$ Strong version: If a particle $i$ exerts a force on a particle $j$, then particle $j$ exerts a force on $i$ of equal magnitude but opposite direction. Moreover, both of these forces are parallel to the line joining $i$ and $j$ $\endgroup$ Sep 9, 2020 at 13:21

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