Systems and center of mass

A system is a collection of objects you choose to analyze together. You make this choice before doing anything else. The system might be a single block on a table. It might be a person, or a person plus their shoes, or a person plus the floor they are walking on. None of these is "right" or "wrong"; each is a valid way to slice up the situation, and each gives a different (but consistent) analysis.

Once you have chosen the system, everything else is the environment. The line between them is the system boundary. Forces between two pieces inside the boundary are internal. Forces from outside the boundary acting on the system are external. This is not bookkeeping. It is the whole point of the topic, because only external forces can change how the system as a whole accelerates.

For the system's overall motion (not how its pieces wiggle relative to each other), you can collapse the entire system to a single point: the center of mass. Where it sits depends on how mass is distributed inside the system, which is what §2 shows you how to compute.

This is the procedure for every COM problem in this unit. Most students who get the wrong answer skipped Step 1 or Step 2.

1. Define the system. Decide which objects are inside the boundary. Everything else is environment. The choice is yours; different choices give different (but valid) analyses.
2. Pick a coordinate origin and axes. Put the origin at one of the masses if you can, since that makes one term in the formula equal to zero. Line the axes up with how the masses are arranged.
3. Apply the mass-weighted formula. Multiply each piece's mass by its position, sum those products, then divide by the total mass:
   $$x_{cm} = \dfrac{\sum m_i x_i}{\sum m_i} = \dfrac{m_1 x_1 + m_2 x_2 + \ldots}{m_1 + m_2 + \ldots}$$
   The result is the COM coordinate along that axis. Keep all signs intact; a piece at $x = -2$ stays at $-2$ in the formula.
4. Repeat once per axis for 2D problems. Run Step 3 for $x$, then again for $y$. The two coordinates together pin the COM to a single point in the plane.

Quick reference card. The $\vec{F}_{ext}$ vs $\vec{F}_{int}$ distinction does most of the work in this topic.

Step 1. A 4 kg mass sits at $x = -2$ m and a 2 kg mass sits at $x = +4$ m. The bar connecting them is massless. The system is the two masses together.

Step 2. The origin is already at $x = 0$, between the two masses. Use it.

Step 3. Apply the mass-weighted formula:

The COM sits at the origin. That might surprise you: the masses are unequal, the positions are unequal, but the COM lands exactly at $x = 0$. No coincidence. The 4 kg mass is half as far from the origin as the 2 kg mass, and exactly twice as heavy. Their contributions to the formula cancel.

Two wrong answers worth flagging.

The midpoint trap. Forgetting the mass weighting gives $(x_1 + x_2)/2 = (-2 + 4)/2 = +1$ m. That puts the COM closer to the 2 kg mass than to the 4 kg mass, which is backwards: the heavier mass should pull the COM toward itself.

The sign-stripped trap. Treating positions as unsigned distances gives $\dfrac{(4)(2) + (2)(4)}{6} = \dfrac{16}{6} \approx +2.67$ m. The mass weighting is right, but dropping the minus sign on $-2$ m kills the cancellation that produced the clean $0$.

Setup. Two skaters stand at rest on frictionless ice and push off each other. Skater A has mass 50 kg; Skater B has mass 80 kg. After the push, A glides west at 4 m/s and B glides east at 2.5 m/s. The system is both skaters together.

What does the COM do?

Start with the principle: $\vec{F}_{ext, net} = M\vec{a}_{cm}$. Vertically, gravity (down) and the normal force from the ice (up) cancel. Horizontally, no external force acts: the skaters push on each other, not on anything outside the system. So $\vec{F}_{ext, net} = 0$, which means $\vec{a}_{cm} = 0$. The system started at rest, so the COM stays at rest. Throughout the push and the glide that follows, the COM does not move.

Verify by computing the COM velocity directly. Take east as positive:

Just as the principle predicted: the two contributions cancel.

Look at what just happened. Both skaters are moving. The pieces of the system are very much in motion. But the COM is not. This is the heart of the topic: internal forces cannot move the COM, no matter how strong they get. They always come in third-law pairs that cancel inside the system. Only an external force can produce $\vec{a}_{cm} \neq 0$.