1 A crazy calculation
Let’s do the following crazy calculation: imagine we have three particles and they exert arbitrary forces on each other but the system only has two-body (pair-wise) interactions. Let’s calculate the total force for the system as a vector. This is a crazy thing to do because it has no meaning—we may have \(\vec{F}_{12}\) the force exerted on particle 1 by particle 2, and the force \(\vec{F}_{23}\) exerted on particle 2 by particle 3, but adding them together makes no sense as they describe forces on different objects. Nevertheless we can crank the handle and do this mathematically.
For three particles there will be six such forces, and for \(n\) particles there will be \(n(n-1)\) combinations (Note that \(n(n-1)\) is always even). Adding the forces on each particle, particle by particle we have
\[\begin{equation*}\vec{F}_\mathrm{tot} = \vec{F}_{12}+\vec{F}_{13}+\vec{F}_{21}+\vec{F}_{23}+\vec{F}_{31}+\vec{F}_{32}\end{equation*}\]
Since the result is always even we can always group these terms in pairs, and in particular in the following arrangement:
\[\begin{equation*}\vec{F}_\mathrm{tot} = (\vec{F}_{12}+\vec{F}_{21})+(\vec{F}_{23}+\vec{F}_{32})+(\vec{F}_{13}+\vec{F}_{31}).\end{equation*}\]
Each term in a bracket has the force on a particle by some other and vice-versa. Here’s the cool observation: because of Newton’s 3rd law (every action has an equal and opposite reaction) each of those terms is zero and
\(\vec{F}_\mathrm{tot} =0\) !
There is a nice way of interpreting this result. Recall that the force is the change in the momentum of a particle \(\vec{F}=\dot{\vec{p}}\). So we can write
\[\begin{equation*}\vec{F}_\mathrm{tot} = \sum_j\vec{F}_{j} = \frac{d}{dt} \sum_j \vec{p}_j = \dot{\vec{p}}_\mathrm{tot} = 0,\end{equation*}\]
where
\(\vec{F}_{j}\) is the total force on particle
\(j\),
\(\vec{p}_j\) is the net momentum of particle
\(j\) and
\(\vec{p}_\mathrm{tot}\) is the net momentum of the entire system. So when the forces acting on a complicated object are only two-body internal forces, the net momentum of the object is conserved.
N.B. Newton’s third law does not hold at relativistic speeds, so these areguments are only relevant to classical physics.
2 The centre of mass
Imagine we have some complicated object made up of lots of particles which have all sorts of internal two-body interactions as well as external forces. The position of each particle is \(\vec{r}_j\). Adding up all the forces on the object yields
\[\begin{align*}\vec{F}_\mathrm{tot} &= \sum_j \left( \vec{F}_j^\mathrm{ext} + \sum_{k\ne j} \vec{F}_{jk}\right) \\
&= \sum_j \left( \vec{F}_j^\mathrm{ext} + \sum_{k < j}(\vec{F}_{jk}+\vec{F}_{kj})\right) \\
&= \vec{F}_\mathrm{tot}^\mathrm{ext}.\end{align*}\]
This is a trivial extension of the toy calculation we started with. All the inter-particle forces cancel because by Newton’s 3rd law
\(\vec{F}_{jk}=-\vec{F}_{kj}\). Now we have included external forces, the total force vector will just be the total of the external forces. This means
\[\begin{equation*}\vec{F}_\mathrm{tot}^\mathrm{ext} = \frac{d^2}{dt^2} \sum_j m_j r_j.\end{equation*}\]
It would be really quite something if we could reduce this expression to the form
\(F=ma\) that applies for a single particle. Well, to do this is easy! Just multiply and divide by the total mass:
\[\begin{equation*}\vec{F}_\mathrm{tot}^\mathrm{ext} = \frac{(\sum_i m_i)}{(\sum_k m_k)}\frac{d^2}{dt^2} \sum_j m_j \vec{r}_j = (\sum_i m_i) \frac{d^2}{dt^2} \frac{(\sum_j m_j \vec{r}_j) }{(\sum_k m_k) } = M \ddot{\vec{R}}\end{equation*}\]
This is an incredible simplification. It means that it doesn’t matter how complicated the interactions are between particles—the entire object will move like the entire mass
\(M=\sum_i m_i\) is concentrated at a single point
\(\vec{R}\):
\[\begin{equation*}\vec{R} = \frac{\sum_j m_j \vec{r}_j }{\sum_k m_k }\end{equation*}\]
This point has to lie somewhere within the set of particles by the way it’s constructed, but it doesn’t have to correspond to any physical object. This fictitious point is the
centre of mass. It let’s us step away from all the complex dynamics at the level of the particles and describe the entire system simply.
For a continuous object the sum becomes an integral and
\[\begin{equation*}\vec{R} = \frac{1}{M} \int_\mathrm{obj} \!\!\vec{r}\, dm\end{equation*}\]
where
\(\vec{r}\) points to the location of each infinitesimal piece of mass
\(dm\). To actually use this calculation the mass distribution of the object needs to be known.
Example
Say we have a three dimensional object with a constant mass density \(\rho\). In this case the infinitesimal mass element can be written as \(dm=\rho \,dx\,dy\,dz\) and the integral is over the shape of the object:
\[\begin{equation*}\vec{R} = \frac{\rho}{M} \iiint_\mathrm{obj} \!\!\vec{r}\, dx\,dy\,dz\end{equation*}\]
3 Locating the centre of mass—divide and conquer
Because the centre of mass \(\vec{R}\) is just a weighted sum of vectors, we can treat each component along an axis independently and we are free to split up the masses however is convenient. This means that we can make use of any symmetries available to simplify the calculation.
For instance, say an object is made up of two pieces \(A\) and \(B\) that are highly symmetric so that it’s easy to determine where the centre of mass is for each piece by inspection. Denote the masses of each piece as \(M_A\) and \(M_B\) so that \(M=M_A+M_B\), and their centre of masses as \(\vec{R}_A\) and \(\vec{R}_B\). Then
\[\begin{align*}\vec{R} &= \frac{\sum_{i} m_i \vec{r}_i}{\sum_k m_k} = \frac{1}{\sum_k m_k}\left(\sum_{i\in M_A} m_i \vec{r}_i + \sum_{j\in M_B} m_j \vec{r}_j \right) \\
&= \frac{1}{M} \left(\frac{M_A}{M_A}\sum_{i\in M_A} m_i \vec{r}_i + \frac{M_B}{M_B}\sum_{j\in M_B} m_j \vec{r}_j \right) \\
&= \frac{1}{M} (M_A \vec{R}_A+ M_B \vec{R}_B).\end{align*}\]
It should be clear that this procedure can be done for any number of pieces.