Center of mass of a nonuniform rod
You will be able to: Use a density function to calculate total mass and center of mass by integration.
How do you average a mass that is spread continuously?
A thin rod becomes progressively heavier toward its right end. Equal lengths no longer have equal mass, so balancing it at the halfway mark will not work.
A useful starting point: Choose a system and locate its center of mass →
Words and symbols before equations
- Linear density λ(x)
- Mass per unit length at position x, in kg/m.
- dm
- A small mass element; for a rod dm = λ(x)dx.
- First moment ∫x dm
- Mass-weighted position accumulation, in kg·m.
- Definite integral
- A continuous sum over stated spatial limits.
What this picture assumes
Thin rod on 0 ≤ x ≤ L with λ(x) = 1 kg/m + cx. The positive baseline prevents a zero-mass rod. The graph gives density; darker segments indicate more mass per length.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- M = 12 kg; first moment = 22.5 kg·m; x_cm = 1.875 m. Uniform density gives L/2; a positive gradient shifts the center right.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Split the rod into short pieces. Each contributes mass λ(x)dx and weighted position xλ(x)dx. In the limit of tiny pieces, M = ∫₀ᴸλ(x)dx and x_cm = (1/M)∫₀ᴸxλ(x)dx. Use the same limits in numerator and denominator.
For λ(x) = cx, c has units kg/m². Integration gives M = cL²/2 and ∫x dm = cL³/3. Their ratio is 2L/3. The density scale c cancels because multiplying every small mass equally does not change its relative weight.
The same idea extends to sheets and solids: dm = σdA for surface density σ in kg/m², or dm = ρdV for volume density ρ in kg/m³. Symmetry can eliminate components before integrating. Define the mass distribution first; do not assume uniformity from the shape alone.
A worked example, step by step
A 3 m rod has λ(x) = (2 kg/m²)x from x = 0 to 3 m. Find its mass and center of mass.
- Use dm = 2x dx with SI coefficient units.
- M = ∫₀³2x dx = [x²]₀³ = 9 kg.
- First moment = ∫₀³2x² dx = [(2/3)x³]₀³ = 18 kg·m.
- x_cm = 18/9 = 2 m, to the heavier side of the midpoint 1.5 m.
Integrating density gives mass; the extra factor x is needed for the center-of-mass numerator.
If every piece becomes twice as massive, does x_cm change?
Compare with an explanation
No. Both numerator and denominator double.
Predict. Change one thing. Explain.
Keep rod length fixed and increase the density gradient while retaining a positive baseline density. Predict how mass and center of mass change. Then compare the uniform case with the increasing-density case.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
M = 12 kg; first moment = 22.5 kg·m; x_cm = 1.875 m. Uniform density gives L/2; a positive gradient shifts the center right.
Thin rod on 0 ≤ x ≤ L with λ(x) = 1 kg/m + cx. The positive baseline prevents a zero-mass rod. The graph gives density; darker segments indicate more mass per length.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus 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 questionA rod extends from 0 to 2 m with λ(x) = (3 kg/m²)x. Determine mass, first moment, center of mass and its location relative to the midpoint.
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Compare with the answer and four-point rubric
- 1 point: M = ∫₀²3x dx = 6 kg.
- 1 point: First moment = ∫₀²3x² dx = 8 kg·m.
- 1 point: x_cm = 8/6 = 4/3 m.
- 1 point: It lies right of the 1 m midpoint because density increases toward the right.
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 write dm = λdx?
Density times a short length gives the mass of that piece.
RECALL 2What units should ∫x dm have?
kg·m.
RECALL 3What must be known before integrating?
The density function, geometry, coordinate origin and integration limits.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Center of mass of a nonuniform rod
- M = ∫λ(x)dx; x_cm = ∫xλ(x)dx / M.
- For λ = cx over 0 ≤ x ≤ L: x_cm = 2L/3.
- Use dm = σdA or ρdV when mass is spread over area or volume.
Remember: Integrating density gives mass; the extra factor x is needed for the center-of-mass numerator.
Conditions: Thin rod on 0 ≤ x ≤ L with λ(x) = 1 kg/m + cx. The positive baseline prevents a zero-mass rod. The graph gives density; darker segments indicate more mass per length.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official 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 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. 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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