Center of Mass

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Systems of Particles and Linear Momentum· 14 min read

1. Core Definition of Center of Mass★☆☆☆☆⏱ 3 min

The center of mass (COM, often called center of gravity for uniform gravitational fields, a common synonym on the AP exam) is the unique point in a system of particles where the entire mass of the system can be considered concentrated for analyzing translational motion. Per the AP Physics C: Mechanics CED, this topic accounts for 1-4% of exam weight, appearing in both multiple-choice and free-response sections, almost always as a foundational step for momentum, rotation, or equilibrium problems.

The core intuition for COM is that it is a mass-weighted average of particle positions: more massive particles pull the COM closer to them, while low-mass particles have little effect. Unlike the geometric center (which only matches COM for uniform-density objects), COM accounts for uneven mass distribution across a system. On the exam, you will typically either calculate the COM position for a given system, or use properties of COM motion to solve for unknown velocities or displacements of system components.

2. Center of Mass for Discrete Particle Systems★★☆☆☆⏱ 4 min

📘 Definition

Center of Mass (Discrete System)

$r_{C M}, m_{i}, r_{i}, M$

For $N$ separate point masses, the COM position vector is the mass-weighted average of all particle positions: $r_{C M} = \frac{\sum _{i = 1}^{N} m _{i} r _{i}}{M}$ , where $M = \sum_{i = 1}^{N} m_{i}$ is total system mass. In Cartesian coordinates, this gives $x_{C M} = \frac{\sum m _{i} x _{i}}{M}, y_{C M} = \frac{\sum m _{i} y _{i}}{M}, z_{C M} = \frac{\sum m _{i} z _{i}}{M}$ .

Example:

For two equal masses separated by distance $d$ , COM lies at the midpoint between them.

A key strategy to simplify calculations is always choosing a coordinate system that eliminates as many terms as possible, such as placing the origin at one of the masses.

📐 Worked Example

Three point masses are arranged on the xy-plane: 2 kg at (0, 0), 1 kg at (3, 0), and 1 kg at (0, 4). What is the distance of the center of mass from the origin?

1
Calculate total system mass:
2
$M = 2 + 1 + 1 = 4 \text{ kg}$
3
Calculate the x-coordinate of COM:
4
$x_{CM} = \frac{(2 \cdot 0) + (1 \cdot 3) + (1 \cdot 0)}{4} = 0.75 \text{ m}$
5
Calculate the y-coordinate of COM:
6
$y_{CM} = \frac{(2 \cdot 0) + (1 \cdot 0) + (1 \cdot 4)}{4} = 1 \text{ m}$
7
Calculate distance from the origin:
8
$d = \sqrt{x_{CM}^2 + y_{CM}^2} = \sqrt{0.75^2 + 1^2} = 1.25 \text{ m}$

Exam tip:

Always place your coordinate origin at one of the masses if possible to eliminate terms from the sum, cutting down on arithmetic errors common on timed MCQ sections.

3. Center of Mass for Continuous Extended Objects★★★☆☆⏱ 4 min

✓ Calculator OK

For extended objects made of continuous matter, we replace the sum over discrete particles with an integral over infinitesimal mass elements $d m$ , each with position $r$ . The general formula becomes:

\vec{r}_{CM} = \frac{1}{M} \int \vec{r} dm

Or in Cartesian coordinates for 2D problems:

x_{CM} = \frac{1}{M} \int x dm, \quad y_{CM} = \frac{1}{M} \int y dm

For uniform-density objects, density $ρ = d m / d V$ is constant, so $ρ$ cancels out, leaving COM dependent only on geometry. The most powerful shortcut for uniform objects is symmetry: if an object has an axis of symmetry, the COM must lie along that axis. For composite objects or objects with cutouts, you can reuse the discrete COM formula: treat each part as a point mass located at its own COM, or use the negative mass method for cutouts (treating the cutout as a negative mass added to the full original object).

📐 Worked Example

A uniform circular plate of radius 2R has a circular hole of radius R cut out of it, such that the hole is tangent to the edge of the original plate. Taking the center of the original plate as the origin, what is the x-coordinate of the COM of the cut plate?

1
Use negative mass: the full original plate has mass proportional to area $(2 R)^{2} = 4 R^{2}$ , with COM at $x = 0$ . The cutout hole has mass proportional to $R^{2}$ , so we treat it as having negative mass $- M_{f u l l} /4$ , with COM at $x = R$ (its center is R from the origin, since it is tangent to the original edge at $x = 2 R$ ).
2
Calculate total mass of the cut plate:
3
$M = M_{full} - \frac{M_{full}}{4} = \frac{3M_{full}}{4}$
4
Substitute into the discrete COM formula:
5
$x_{CM} = \frac{M_{full} \cdot 0 + (-M_{full}/4) \cdot R}{3M_{full}/4} = -\frac{R}{3}$
6
The negative sign indicates the COM is $\frac{R}{3}$ from the origin, opposite the cutout hole.

✓ Quick check

Test your understanding of symmetry and negative mass:

A uniform-density donut (annulus) has outer radius $R$ and inner hole of radius $R /2$ , centered at the origin of the xy-plane. What is the distance between the center of mass of the donut and the origin?
- $0$
- $\frac{R}{2}$
- $\frac{3 R}{4}$
- $\frac{2 R}{3}$
Reveal answer
$0$ —
The donut is uniform and symmetric across all axes through the origin, so COM must lie at the origin. Using negative mass confirms this: the cutout hole is also centered at the origin, so the weighted sum gives $x_{C M} = 0$ .

Exam tip:

For any problem involving a cutout or composite object, use negative mass or composite COM instead of integrating from scratch to save 2-3 minutes on FRQ, leaving more time for harder multi-part questions.

4. Motion of the Center of Mass★★★☆☆⏱ 3 min

A core result from COM analysis is that the COM of a system moves as if all the mass of the system is concentrated at the COM, and all external forces acting on the system act at that point. This gives Newton's second law for the COM:

M \vec{a}_{CM} = \vec{F}_{net, ext}

where $F_{n e t, e x t}$ is the vector sum of only external forces (forces from objects outside the system boundary). For velocity, this relationship becomes:

M \vec{v}_{CM} = \sum_{i} m_i \vec{v}_i = \vec{P}_{total}

The total momentum of the system is always equal to the total mass times the velocity of the COM. If there is no net external force on the system, $a_{C M} = 0$ , so $v_{C M}$ is constant. This is the foundation of momentum conservation for isolated systems, and is often used to solve for unknown displacements of system components when one component moves.

📐 Worked Example

A 50 kg person stands at the left end of a 100 kg, 10 m long uniform boat that is at rest on frictionless water. If the person walks from the left end to the right end of the boat, how far does the boat move relative to the water?

1
The system (person + boat) has no net external force, and is initially at rest, so the COM of the system must remain stationary at its initial position.
2
Set initial coordinates: origin at the initial position of the left end of the boat. Initial boat COM is at $x = 5$ m, person is at $x = 0$ . Calculate initial COM:
3
$x_{CM,i} = \frac{(50 \cdot 0) + (100 \cdot 5)}{150} = \frac{10}{3} \approx 3.33 \text{ m}$
4
Let $d$ = displacement of the boat relative to the water. Final position of boat COM is $5 + d$ , final position of person is $10 + d$ .
5
Set final COM equal to initial COM and solve for $d$ :
6
$\frac{50(10 + d) + 100(5 + d)}{150} = \frac{10}{3} \\ d = -\frac{10}{3} \approx -3.33 \text{ m}$
7
The negative sign means the boat moves approximately 3.33 m to the left, opposite the direction the person walked.

Exam tip:

Always remember that COM motion only depends on external forces—internal forces (like friction between a person and boat) cancel out by Newton's third law and never affect the acceleration of the COM.

5. Common Pitfalls

Wrong move:

Forgetting to use a single global coordinate system for composite objects, using each part's local coordinate system instead of shifting positions.

Why:

Students often calculate COM for each part separately in its own coordinate system, then average those positions without shifting to a shared reference frame.

Correct move:

Define one global coordinate system at the start of the problem, and find the position of each part's COM in that global system before calculating total COM.

Wrong move:

Assuming geometric center equals COM for non-uniform objects.

Why:

Students get used to uniform objects having COM at geometric center, so they generalize this to all objects.

Correct move:

Only use geometric center for COM if the problem explicitly states the object has uniform density; for non-uniform objects, always integrate the given mass distribution.

Wrong move:

Including internal forces when calculating $a_{C M}$ from $F_{n e t, e x t}$ .

Why:

Students confuse internal interactions between system components with external forces from outside the system.

Correct move:

Always draw a clear system boundary, and only include forces from objects outside the boundary when calculating net external force.

Wrong move:

Forgetting to divide the weighted sum by total mass when calculating COM.

Why:

Students rush and accidentally calculate the unweighted sum, or divide by number of particles instead of total mass.

Correct move:

Always calculate total mass $M$ first, and explicitly write the division by $M$ in every step of your calculation.

Wrong move:

Assuming COM must lie inside the object.

Why:

Most common examples (rods, spheres, cubes) have COM inside the object, so students assume this is a general rule.

Correct move:

Check for hollow shapes or cutouts (e.g. donuts, horseshoes); the COM of these objects lies in the empty space, which is a common MCQ trick.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Discrete system COM	$r_{C M} = \frac{\sum m _{i} r _{i}}{M}, M = \sum m_{i}$	Works for any number of point masses, any coordinate system
Continuous object COM	$r_{C M} = \frac{1}{M} \int r d m$	General formula for all extended objects, uniform or non-uniform
Uniform 1D rod	$x_{C M} = L /2$	For a uniform rod of length $L$ from $x = 0$ to $x = L$
Symmetry shortcut	COM lies on all axes of symmetry	Only applies to uniform-density objects
Negative mass method	Treat cutout as negative mass added to full object	Saves time vs full integration for cutout problems
Total momentum relation	$P_{t o t a l} = M v_{C M}$	Holds for all systems, regardless of internal motion
Newton's 2nd law for COM	$M a_{C M} = F_{n e t, e x t}$	Only external forces contribute; internal forces cancel out

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
COM of cutout circular plate
2022 · FRQ
COM motion for person on boat
2021 · MCQ
COM acceleration of non-uniform rod

Going deeper

syllabusOfficial AP Physics C CEDOfficial College Board curriculum document

What's Next

Mastering center of mass is an essential prerequisite for the remaining topics in Unit 4 (Systems of Particles and Linear Momentum) and all of Unit 5 (Rotation). Next, you will use COM properties to analyze momentum conservation for collisions, and extend the concept to find torque and angular momentum for extended rotating bodies. Without correctly finding the COM of a system, you cannot correctly solve collision problems involving multiple objects, nor can you analyze rolling motion or static equilibrium later in the course. Center of mass also directly underpins the concept of center of gravity, which is core to rigid body statics.