Here we consider a “system of interacting particles”. We could define the system to be the thousands of stars orbiting each other in a globular cluster or the metal atoms clinging to each other in the wrench shown below. Or it could refer to almost any organization of matter governed by Newtonian Physics.
Suppose our system has \(N\) particles, which may be exerting forces on each other. Particle \(i\) will have mass \(m_i\) and location \(\vec{r}_i\). You may prefer to keep in mind the star cluster or the wrench for concrete examples of the “system”, so where the word “system” appears below you may replace it with “wrench” or “star cluster”.
Identifying the center of mass of the system (labelled \(G\)) will be critical to the following development. The center of mass is the mass-weighted average location of all the particles. It is located at
\[ \vec{r}_G = \frac{1}{M} \sum m_i \vec{r}_i \]
where \(M\) is the total mass of the all particles in the system. If any of the particles are in motion, the velocity of the system’s center of mass is
\[ \vec{v}_G = \frac{d}{dt} \vec{r}_G = \frac{1}{M} \sum m_i \frac{d}{dt} \vec{r}_i = \frac{1}{M} \sum m_i \vec{v}_i \]
The acceleration of the center of mass can be found by one more time derivative:
\[ \vec{a}_G = \frac{1}{M} \sum m_i \frac{d^2}{dt^2} \vec{r}_i = \frac{1}{M} \sum m_i \vec{a}_i \]
The center of mass of the wrench, for example, would be found on the body, near the heavy end.
We first consider just particle \(i\). It may experience forces from other particles in the system (for example, an atom in the wrench bound to the neighboring atoms). It may also experience forces from outside the system (for example, and atom on the wrench surface located just where an external force happens to be applied).
We may write the net force on particle \(i\) as the sum of
where \(\vec{f}_{ji}\) refers to the internal force by particle \(j\) on particle \(i\).
By Newton’s Second Law, a particle that experiences a net force must accelerate in the direction of the net force. So for particle \(i\),
\[ \left (\sum \vec F \right )_i + \sum_{j \neq i}^{N} \vec{f}_{ji} = m_i \vec{a}_i \]
We next form the equation which is the sum of all these \(\vec{F} = m \vec{a}\) equations for every particle in the system. This is simplified dramatically using Newton’s Third Law, which states that for every internal force \(\vec{f}_{ji}\) there exists another internal force \(\vec{f}_{ij}\) which is equal in magnitude and opposite in direction. Thus in the sum of equations, all the internal forces cancel exactly and only external forces remain.
\[ \sum \vec F_{ext} = \sum_{i=1}^{N} m_i \vec{a}_i \]
The right hand side is more easily understood in terms of the acceleration of the system’s center of mass (see previous section).
\[ \sum \vec F_{ext} = M \vec{a}_G \]
This very important result shows that
An example of the first consequence is the well known impossibility of lifting yourself up by your bootstraps (which would be only internal forces).
As an example of the second consequence, consider two equal and opposite but not collinear forces applied to a wrench initially at rest (such a pair is called a “force couple”). In each of the figures shown, particles of the wrench will accelerate but the center of mass will not move—the wrench will rotate about its center of mass.
Some notes about this figure:
For a complete treatment of the “Dynamics of a System of Particles” I should include next how external forces cause rotation.
I haven’t written that, but the short story is that a net torque about \(G\) produces angular momentum in the system.
For a rigid body in planar motion, the torque causes an angular acceleration (\(\alpha\)) of the body given by
\[ \sum \vec{\tau}_{ext} = I \vec{\alpha} \]
where \(I\) is the rotational inertia of the system about it’s center of mass.
Last modified March 2023 by Archie Paulson (Madison College, WI)