Center of mass — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Definition of center of mass (COM), coordinate formulas for discrete particle systems, integration methods for continuous objects, symmetry shortcuts, negative mass for cutouts, and motion of the center of mass for AP exam questions.

You should already know: Vector position coordinates, basic single-variable integration, Newton's second law for single objects.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Center of mass?

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, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is almost never a standalone full FRQ question, but rather a foundational step required to solve 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 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 Systems of Particles

For a system of $N$ separate point particles, each with mass $m_i$ and position vector ${\vec{r}}_i=(x_i,y_i,z_i)$ in a chosen coordinate system, the COM position vector ${\vec{r}}_{CM}$ is defined as the mass-weighted average of all particle positions: $${\vec{r}}_{CM}=\frac{{\sum }_{i=1}^N{m}_i{\vec{r}}_i}{M}$$ where $M={\sum }_{i=1}^N{m}_i$ is the total mass of the system. Breaking this into Cartesian coordinates gives separate formulas for each axis: $$x_{CM}=\frac{{\sum }_{i=1}^N{m}_i{x}_i}{M},y_{CM}=\frac{{\sum }_{i=1}^N{m}_i{y}_i}{M},z_{CM}=\frac{{\sum }_{i=1}^N{m}_i{z}_i}{M}$$

This formula behaves as expected: if all mass is concentrated at one particle, $x_{CM}$ equals that particle’s x-coordinate. For two equal masses separated by distance $d$, the COM lies exactly 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: $M=2+1+1=4$ kg.
2. Calculate the x-coordinate of COM: $x_{CM}=\frac{(2\cdot 0)+(1\cdot 3)+(1\cdot 0)}{4}=0.75$ m.
3. Calculate the y-coordinate of COM: $y_{CM}=\frac{(2\cdot 0)+(1\cdot 0)+(1\cdot 4)}{4}=1$ m.
4. Calculate distance from the origin: $d=\sqrt{x_{CM}^2+y_{CM}^2}=\sqrt{0.75^2+1^2}=1.25$ 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 that are common on timed MCQ sections.

3. Center of Mass for Continuous Extended Objects

For extended objects made of continuous matter, we replace the sum over discrete particles with an integral over infinitesimal mass elements $dm$, each with position $\vec{r}$. The general formula becomes: $${\vec{r}}_{CM}=\frac{1}{M}\int \vec{r}dm$$ or in coordinates: $$x_{CM}=\frac{1}{M}\int xdm,y_{CM}=\frac{1}{M}\int ydm$$

For uniform-density objects, density $\rho =dm/dV$ is constant, so $\rho$ cancels out of the numerator and denominator, 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 (multiple uniform parts stuck together) 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 $M_{full}\propto (2R{)}^2=4R^2$, COM at $x=0$. The cutout hole has mass proportional to $R^2$, so $M_{hole}=-M_{full}/4$, and its COM is at $x=R$ (since it is tangent to the edge of the original plate at $x=2R$, so its center is $R$ from the origin).
2. Total mass of the cut plate: $M=M_{full}-M_{full}/4=3M_{full}/4$.
3. Calculate $x_{CM}=\frac{M_{full}\cdot 0+(-M_{full}/4)\cdot R}{3M_{full}/4}=-\frac{R}{3}$. The negative sign indicates the COM is $\frac{R}{3}$ from the origin, opposite the cutout hole.

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, which leaves more time for harder multi-part questions.

4. Motion of the Center of Mass

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 ${\vec{F}}_{net,ext}$ 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, ${\vec{a}}_{CM}=0$, so ${\vec{v}}_{CM}$ 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. 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 $x=0$. Initial COM: $x_{CM,i}=\frac{(50\cdot 0)+(100\cdot 5)}{150}=\frac{10}{3}\approx 3.33$ m.
3. Let $d$ = displacement of the boat relative to the water. Final position of boat COM is $5+d$, final position of person is $10+d$.
4. Set final COM equal to initial COM: $\frac{50(10+d)+100(5+d)}{150}=\frac{10}{3}$. Solving gives $d=-\frac{10}{3}\approx -3.33$ m, meaning the boat moves 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 (and how to avoid them)

• 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 $a_{CM}$ from $F_{net,ext}$. 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. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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? A) $0$ B) $\frac{R}{2}$ C) $\frac{3R}{4}$ D) $\frac{2R}{3}$

Worked Solution: The donut is uniform-density and symmetric along every axis that passes through the origin. For uniform symmetric objects, the COM must lie on all axes of symmetry, so it is located exactly at the origin. We can confirm this with the negative mass method: the donut is a full circle of radius $R$ with a negative mass circle of radius $R/2$ cut out at the origin, so $x_{CM}=\frac{M_R(0)-M_{R/2}(0)}{M_R-M_{R/2}}=0$. The distance from the origin is 0. Correct answer: A.

Question 2 (Free Response)

A non-uniform rod of length $L$ lies along the x-axis from $x=0$ to $x=L$. The linear mass density of the rod is given by $\lambda (x)={\lambda }_0\left(1+\frac{x}{L}\right)$, where ${\lambda }_0$ is a constant. (a) Find the total mass $M$ of the rod in terms of ${\lambda }_0$ and $L$. (b) Find the x-coordinate of the center of mass of the rod in terms of $L$. (c) If a horizontal force $F$ is applied to the end of the rod at $x=L$, what is the acceleration of the center of mass of the rod?

Worked Solution: (a) Total mass is the integral of $dm=\lambda (x)dx$: $$M={\int }_0^L{\lambda }_0\left(1+\frac{x}{L}\right)dx={\lambda }_0{\left[x+\frac{x^2}{2L}\right]}_0^L={\lambda }_0\left(L+\frac{L}{2}\right)=\frac{3}{2}{\lambda }_{0}L$$

(b) Substitute into the COM formula for continuous objects: $$x_{CM}=\frac{1}{M}{\int }_0^{L}x\lambda (x)dx=\frac{2}{3{\lambda }_{0}L}{\int }_0^{L}x{\lambda }_0\left(1+\frac{x}{L}\right)dx=\frac{2}{3L}{\left[\frac{x^2}{2}+\frac{x^3}{3L}\right]}_0^L=\frac{2}{3L}\left(\frac{5L^2}{6}\right)=\frac{5L}{9}$$

(c) By Newton's second law for COM, $M{a}_{CM}=F_{net,ext}$. The only external horizontal force is $F$, so: $$a_{CM}=\frac{F}{M}=\frac{F}{\frac{3}{2}{\lambda }_{0}L}=\frac{2F}{3{\lambda }_{0}L}$$ The location of the applied force does not affect the acceleration of the COM, only the net external force matters.

Question 3 (Application / Real-World Style)

A 75 kg astronaut is doing a spacewalk outside a 12,000 kg spherical space station module. Initially, the astronaut is 10 m from the center of mass of the module, and the entire system is at rest relative to the station. The astronaut then moves to a point 5 m from the COM of the module, towards the module. How far does the module's COM move relative to the original stationary reference frame of the station? What is the direction of motion?

Worked Solution: The system has no net external force, so the system COM remains stationary. Set initial position of the module's COM at $x=0$, so the astronaut starts at $x=10$ m. Initial system COM: $$x_{CM,i}=\frac{(12000\cdot 0)+(75\cdot 10)}{12075}\approx 0.0621\text{ m}$$ Let $d$ = displacement of the module's COM. Final positions: module COM at $d$, astronaut at $5+d$. Set final COM equal to initial COM: $$\frac{12000d+75(d+5)}{12075}=0.0621⟹12075d=375⟹d\approx 0.031\text{ m}$$ The module moves 3.1 cm towards the astronaut's original position, in the direction the astronaut moved. This small displacement makes sense because the module is far more massive than the astronaut, so it only moves a tiny distance to keep the system COM stationary.

7. Quick Reference Cheatsheet

8. 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. Follow-on topics to study next:

• Impulse and momentum
• Conservation of linear momentum
• Torque and rotational statics
• Moment of inertia