Systems and the center of mass

Three moves underlie this topic. First, choose the system: the set of particles or chunks of mass you want to analyze as a group. Second, locate that system's center of mass, a single point whose position is the mass-weighted average of all the parts. Third, when the system is a continuous body rather than a handful of point particles, do that weighted average as an integral over differential mass elements $dm$.

The discrete formula $\vec{r}_{cm} = \sum m_i \vec{r}_i / \sum m_i$ is familiar from Physics 1. The Physics C move is the continuous version $\vec{r}_{cm} = \int \vec{r}\, dm / \int dm$, and the question that arrives with it: what is $dm$? Most of the early trouble is about being able to answer that without flinching, then turning $dm$ into something you can actually integrate, usually via a linear mass density $\lambda = dm/d\ell$.

The center of mass is the point a system moves through as a whole: net external force on a system equals total mass times the acceleration of the COM. If you cannot find the COM cleanly, Newton's second law for the system breaks, and that wrecks Unit 2 through Unit 5. The diagnostic at the end of this topic is built around the small handful of places this goes wrong.

A system is whatever set of objects you draw a boundary around. It is a choice. Take two carts joined by a spring on a frictionless table. You can model them as: (a) two separate systems, each a single cart, with the spring force on each cart treated as an external force; or (b) one system containing both carts and the spring, where the spring force is now internal to the system.

Both choices are legal physics. Which one is useful depends on what you want to compute. The discrete COM has a clean property under choice (b): the COM of the two-cart-plus-spring system is not affected by anything the carts do to each other through the spring. Internal forces cannot accelerate a system's COM. That property, which follows from Newton's third law (Topic 2.3), is why this topic exists in the first place.

You have already used the COM in Physics 1 without naming it. Every time you treated a block as a "point particle," you were locating its COM and tracking that point. T2.1 turns the move into an actual calculation, one that works on extended bodies (rods, disks, plates) whose "point particle" location is not obvious.

The system rule. For the rest of this topic, every problem starts with a sentence of the form: "Treat the following as the system: ..." The set after the colon determines which masses go into the sums and integrals. Change the system, change the sum.

For a system of $N$ point particles with masses $m_1, m_2, \ldots, m_N$ at positions $\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_N$, the center of mass is

$$\vec{r}_{cm} \;=\; \frac{\sum_{i=1}^{N} m_i\, \vec{r}_i}{\sum_{i=1}^{N} m_i} \;=\; \frac{1}{M}\sum_{i=1}^{N} m_i\, \vec{r}_i, \qquad M = \sum_{i=1}^{N} m_i.$$

The denominator is the system's total mass. The numerator is the sum of each particle's mass times its position. The ratio is a mass-weighted average position. Heavier particles count more, lighter ones count less, and the unweighted mean $\sum x_i / N$ matches the COM only when all the masses are equal.

Worked example (1D, three particles)

Three particles sit on the $x$-axis: $m_1 = 3$ kg at $x_1 = 0$, $m_2 = 2$ kg at $x_2 = 5$ m, $m_3 = 5$ kg at $x_3 = 8$ m. Find the $x$-coordinate of the COM.

Total mass: $M = 3 + 2 + 5 = 10$ kg. Weighted sum:

$$\sum m_i x_i \;=\; (3)(0) + (2)(5) + (5)(8) \;=\; 50\ \text{kg}\cdot\text{m}.$$

Divide:

$$x_{cm} \;=\; \frac{50}{10} \;=\; 5\ \text{m}.$$

Two checks. The COM lies between the leftmost and rightmost particle ($0 \le 5 \le 8$), as it must for any weighted average. And the $5$ kg at $x = 8$ pulls the COM to $5$ m, past the unweighted mean of $4.33$ m. The bigger the mass imbalance, the further the COM drifts toward the heavy end.

Why mass weighting and not something else

Mass weighting is not an arbitrary choice. It comes from Newton's second law. The equation $\vec{F}_{net,ext} = M\vec{a}_{cm}$ (Topic 2.5) only works if the COM is defined as the mass-weighted average. Try the unweighted mean instead, or the geometric center, and the equation breaks. So out of all the "averages" you could write down, exactly one matches the physics: this one.

For a single rigid extended body, there are no obvious discrete particles to sum over. A meter stick has $\sim 10^{25}$ atoms, and you do not want that sum. Replace it with an integral instead.

What $dm$ means physically

Imagine slicing the body into a large number $N$ of small pieces. The $i$-th piece has mass $\Delta m_i$ and sits at position $\vec{r}_i$. The discrete COM formula gives

$$\vec{r}_{cm} \;\approx\; \frac{1}{M}\sum_{i=1}^{N} \vec{r}_i\, \Delta m_i.$$

As $N$ grows and each $\Delta m_i$ shrinks toward zero, the sum becomes an integral and $\Delta m_i$ becomes the differential mass element $dm$:

$$\boxed{\;\vec{r}_{cm} \;=\; \frac{1}{M}\int \vec{r}\, dm, \qquad M = \int dm.\;}$$

The notation $dm$ stands for "an arbitrarily small piece of the body's mass." It is a piece, not a number: it has no value until you choose how to slice the body and substitute it in terms of something you can integrate.

Turning $dm$ into something you can integrate

The integral as written above is purely formal: you cannot do $\int x\, dm$ until you express $dm$ in terms of the variable you are integrating over. For a 1D rod, that variable is position $x$ along the rod, and the conversion uses the linear mass density $\lambda$:

$$\lambda(x) \;=\; \frac{dm}{dx} \quad\Longleftrightarrow\quad dm \;=\; \lambda(x)\, dx.$$

$\lambda$ has units of kg/m. It tells you how much mass sits per unit length of the rod at position $x$. For a uniform rod, $\lambda$ is a constant. For a non-uniform rod, $\lambda$ is a function of $x$. Either way, once you have $\lambda(x)$, the integral becomes an ordinary calculus problem in $x$.

The substitution rule. Every continuous-COM problem starts the same way: write $\vec{r}_{cm} = (1/M)\int \vec{r}\, dm$, then immediately replace $dm$ with $\lambda\, dx$ (1D), $\sigma\, dA$ (2D), or $\rho\, dV$ (3D). Skipping this step is the dominant error in §6.

Worked example (rod with $\lambda(x) = \alpha x$)

A thin rod of length $L$ lies along the $x$-axis from $x = 0$ to $x = L$. Its linear mass density varies as $\lambda(x) = \alpha x$, where $\alpha$ is a constant with units of kg/m². Find the $x$-coordinate of the rod's COM.

Step 1. Compute the total mass $M$:

$$M \;=\; \int_0^L \lambda(x)\, dx \;=\; \int_0^L \alpha x\, dx \;=\; \alpha\,\frac{x^2}{2}\bigg|_0^L \;=\; \frac{\alpha L^2}{2}.$$

Step 2. Compute the weighted-position integral $\int x\, dm$:

$$\int x\, dm \;=\; \int_0^L x\,\lambda(x)\, dx \;=\; \int_0^L x(\alpha x)\, dx \;=\; \int_0^L \alpha x^2\, dx \;=\; \alpha\,\frac{x^3}{3}\bigg|_0^L \;=\; \frac{\alpha L^3}{3}.$$

Step 3. Divide:

$$x_{cm} \;=\; \frac{1}{M}\int x\, dm \;=\; \frac{\alpha L^3/3}{\alpha L^2/2} \;=\; \frac{2L}{3}.$$

The $\alpha$ canceled out, as it had to: rescaling the density should not move the COM. The result $2L/3$ sits beyond the midpoint, pulled toward the denser end at $x = L$. The uniform-rod result ($x_{cm} = L/2$) is the special case $\lambda = $ constant.

What this example was checking

The three moves to internalize: $x$ stayed inside the integral as the variable (not pulled out as $L/2$); $dm$ was replaced by $\lambda(x)\, dx$ before integrating; the total mass $M$ came from a separate integral of $\lambda$ alone. §6 unpacks the failure modes one by one.

Components are independent

In two dimensions, the COM formula splits into two separate, identical-looking scalar problems, one for each component:

$$x_{cm} \;=\; \frac{1}{M}\sum_i m_i x_i, \qquad y_{cm} \;=\; \frac{1}{M}\sum_i m_i y_i.$$

The $x$-component of the COM depends only on the $x$-positions of the particles. The $y$-component depends only on the $y$-positions. There is no cross-coupling: solving for one does not require knowing the other. This is the same components-independent principle that runs through every 2D problem in this course.

Worked example (2D, two particles)

Particle 1 has mass $m_1 = 1$ kg at the origin $(0, 0)$. Particle 2 has mass $m_2 = 3$ kg at the point $(4, 4)$ m. Find the COM.

Total mass: $M = 1 + 3 = 4$ kg. $x$-component:

$$x_{cm} \;=\; \frac{(1)(0) + (3)(4)}{4} \;=\; \frac{12}{4} \;=\; 3\ \text{m}.$$

$y$-component:

$$y_{cm} \;=\; \frac{(1)(0) + (3)(4)}{4} \;=\; \frac{12}{4} \;=\; 3\ \text{m}.$$

COM at $(3, 3)$ m. It sits on the line connecting the two particles (always true for a two-particle system) and is pulled toward the heavier particle, three-quarters of the way from particle 1 to particle 2. The unweighted midpoint would have been $(2, 2)$ m.

Symmetry shortcut

For a uniform body with the relevant symmetry, the COM lies at its symmetry center; more generally, a uniform body's COM sits at its geometric centroid. A uniform rod's COM is at its midpoint, a uniform disk's at the disk's center, a uniform rectangular plate's at the intersection of its diagonals. The reason these work is symmetry: every $dm$ on one side has a mirror $dm$ at the same distance on the other side, and their contributions to the weighted sum cancel.

The shortcut is fast on the exam but has a sharp limit: it requires uniform density. The moment the density varies (as in §4's $\lambda = \alpha x$ rod), the geometric center is no longer the COM. The first pitfall in §6 is exactly this confusion.

Combining sub-systems

Many problems are a uniform body with something extra: a rod with a point mass at one end, a plate with a piece cut out, two uniform pieces joined at an edge. Strategy: break the body into uniform sub-systems, locate each one's COM by symmetry, then treat each sub-system as a point particle at its own COM and use the discrete formula.

Worked example. A uniform rod of mass $M$ and length $L$ lies along the $x$-axis from $x = 0$ to $x = L$. A point particle of mass $M$ is attached to the right end at $x = L$. Find the COM of the rod-plus-point-particle system.

Sub-system A: the rod alone. Mass $M$, COM at $x = L/2$ by symmetry. Sub-system B: the point particle. Mass $M$, located at $x = L$. Apply the discrete formula to the two sub-systems:

$$x_{cm} \;=\; \frac{M\cdot(L/2) + M\cdot L}{M + M} \;=\; \frac{ML/2 + ML}{2M} \;=\; \frac{3ML/2}{2M} \;=\; \frac{3L}{4}.$$

The COM sits three-quarters of the way along the rod, pulled toward the end with the point particle. Notice the move: a continuous rod was replaced by a single point particle of mass $M$ at $x = L/2$, because the rod is uniform and its sub-system COM is known by symmetry. No integral required.

Going beyond 1D rods

For 2D plates, swap $\lambda\, dx$ for $\sigma\, dA$ (with $\sigma$ in kg/m$^2$). For 3D solids, swap it for $\rho\, dV$ (with $\rho$ in kg/m$^3$). Everything else (component independence, the substitution rule, the symmetry shortcut) carries over unchanged. AP Physics C: Mechanics tests 2D and 3D less often than 1D rods, but the technique is identical.

Each center-of-mass error below maps to a code the diagnostic targets: the geometric-center shortcut (PH1), the unweighted average (PH2), and a mishandled $dm$ (PH20).