Finding the center of mass
You will be able to: Calculate and interpret the center of mass of a small collection of particles.
Why is the balance point closer to the heavier object?
Put a 1 kg bag at the left end of a light 4 m bar and a 3 kg bag at the right end. The balance point is closer to the 3 kg bag: the two positions should not count equally.
A useful starting point: Coordinates and position →
Words and symbols before equations
- Mass m
- Amount of inertia, measured in kilograms (kg).
- Coordinate x or y
- Position relative to a chosen origin, measured in meters (m).
- Center of mass
- Mass-weighted average position; it can lie in empty space.
What this picture assumes
Left mass 1 kg at x=0 m; right mass at x=4 m. Positions are to scale; circle size is schematic. Connecting bar mass is neglected.
Connect the picture to the physics
Multiply each coordinate by its mass, add those products, then divide by total mass. A heavier object contributes more to the average. The symbol Σ means add all the listed terms.
Use the same procedure separately for x and y in two dimensions. Do not replace signed coordinates with distances from the origin. For a symmetric object with uniform density, symmetry locates the center of mass; a ring’s center is in its hole.
The center of mass helps track overall translation even when individual parts move differently. Internal forces cannot change the motion of the center of mass of an isolated, fixed-mass system.
A worked example, step by step
Place 1 kg at x=0 m and 3 kg at x=4 m. Where is their center of mass? Then place a third 2 kg object at (0 m, 3 m). Find the new two-dimensional center.
- Two objects: total mass is 4 kg; weighted position is 1×0 + 3×4 = 12 kg·m.
- Divide: x_cm = 12/4 = 3 m, closer to the 3 kg object.
- With the third object, total mass is 6 kg. x_cm = 12/6 = 2 m; y_cm = (1×0+3×0+2×3)/6 = 1 m.
- The new center is (2 m, 1 m). It need not coincide with any object.
The midpoint works for two equal masses, not for every two-object system.
If every mass doubles, does the center move?
Compare with an explanation
No. Both numerator and denominator double, leaving the weighted average unchanged.
Predict. Change one thing. Explain.
Keep the masses at 0 m and 4 m. Change the right mass. Predict whether the center moves toward it or away. Reset; make the masses equal and interpret the midpoint.
Total mass 4 kg; x_cm = 3 m. The center lies between 0 m and 4 m, closer to the greater mass.
Left mass 1 kg at x=0 m; right mass at x=4 m. Positions are to scale; circle size is schematic. Connecting bar mass is neglected.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTwo particles are 2 kg at (−1 m,0 m) and 4 kg at (2 m,3 m). (a) Find total mass, (b) calculate x_cm, (c) calculate y_cm, and (d) explain which particle the center is closer to.
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Compare with the answer and four-point rubric
- 1 point: Total mass 6 kg.
- 1 point: x_cm = [2(−1)+4(2)]/6 = 1 m.
- 1 point: y_cm = [2(0)+4(3)]/6 = 2 m.
- 1 point: It is closer to the 4 kg particle because that position has greater weight in the average.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why use mass weights?
A larger mass contributes more to the system’s overall position.
RECALL 2How do you find a 2D center?
Calculate weighted x and y coordinates separately.
RECALL 3Must the center be inside material?
No; a ring is a counterexample.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Finding the center of mass
- x_cm = Σ(mᵢxᵢ)/Σmᵢ; y_cm = Σ(mᵢyᵢ)/Σmᵢ.
- Use a common coordinate system. Here the connecting bar has negligible mass.
Remember: The midpoint works for two equal masses, not for every two-object system.
Conditions: Left mass 1 kg at x=0 m; right mass at x=4 m. Positions are to scale; circle size is schematic. Connecting bar mass is neglected.
Refresh Kid · Unit 2 · Objectives 2.1.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.1, objectives 2.1.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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