The weak form of Newton’s third law states that the force \(\ff_{12}\) of \(m_2\) on \(m_1\) is equal and opposite to the force \(\ff_{21}\) of \(m_1\) on \(m_2\text{.}\) We see that each internal force appears twice in the system of equations (8.2.1), once with a positive sign and once with a negative sign. Therefore, if we add all of the equations in together, the internal forces will all cancel, leaving:
Notice how surprising equation (8.3.1) is. The right-hand side directs us to add up all of the external forces, each of which acts on a different mass; something you were taught never to do in introductory physics. The left-hand side directs us to add up (the second derivatives of) \(n\) “weighted” position vectors pointing from the origin, each to a different mass.
We can simplify the left-hand side of (8.3.1) if we multiply and divide by the total mass \(M = m_1 + m_2 + . . . + m_n\) and use the linearity of differentiation to “factor out” the derivative operator:
\begin{align}
\sum_{i=1}^n m_i \frac{d^2\rr_i}{dt^2}
\amp= M \sum_{i=1}^n \frac{m_i}{M} \frac{d^2\rr_i}{dt^2}\notag\\
\amp= M \frac{d^2}{dt^2}
\left( \sum_{i=1}^n \frac{m_i}{M} \rr_i \right)\tag{8.3.2}\\
\amp= M \frac{d^2\RRv}{dt^2}\notag
\end{align}
Definition3.Position of the Center of Mass.
We recognize (or define) the quantity in the parentheses on the right-hand side of (8.3.2) as the position vector \(\RRv\) from the origin to the center of mass of the system of particles
Figure4.The position vectors \(\rr_1\) and \(\rr_2\) for masses \(m_1\) and \(m_2\) and the displacement vector \(\rr=\rr_2-\rr_1\) between them.
In Figure 4, you can drag the two masses and the origin anywhere you want. You can also control the masses of the particles with sliders. Play with the controls until you get a sense of the geometric meaning of \(\RRv\text{.}\) Note that this figure artificially confines the two masses to a plane, so you can see more clearly what is going on.
Reflection.
Can you draw the position vector \(\RRv\) as the vector sum of the position of the center of mass \(\RRv\) plus another vector?
\begin{equation}
M \frac{d^2\RRv}{dt^2} = \sum_{i=1}^n\FF_i\tag{8.3.4}
\end{equation}
which has the form of Newton’s 2nd Law for a fictitious particle with mass \(M\) sitting at the center of mass of the system of particles and acted on by all of the external forces from the original system.
Definition5.Momentum of the Center of Mass.
We can define the momentum of the center of mass as the total mass times the time derivative of the position of the center of mass:
\begin{equation}
\PPv= M \frac{d\RRv}{dt}\tag{8.3.5}
\end{equation}
If there are no external forces acting, then the acceleration of the center of mass is zero and the momentum of the center of mass is constant in time (conserved)
\begin{equation}
M \frac{d^2\RRv}{dt^2} = \frac{d\PPv}{dt} = 0\tag{8.3.6}
\end{equation}
Notice that the entire discussion above applies even if all of the internal forces are zero (\(\ff_{ij}=0\)), i.e. none of the particles have any way of knowing that the others are even present. Such particles are called non-interacting. The position of the center of mass of the system will still move according to (8.3.4).
