Locate a center by integrating a nonuniform mass distribution

Define the system before taking an average

A system is the collection of objects or material selected for analysis. Its center of mass depends on that selection and on mass distribution, not on where an observer happens to place the coordinate origin. For discrete particles, xCM=Σmixi/Σmi and yCM=Σmiyi/Σmi. For example, masses 1, 2 and 3 kg at (0,0), (2,0) and (0,2) m have center (2/3,1) m. The mass-weighted center differs from the unweighted average of coordinates.

Replace a sum by differential masses

For a thin rod, dm=lambda(x)dx. Density has units kg/m, so integrating dm gives kilograms. The first moment ∫x dm has units kg m; dividing by total mass returns a coordinate. Both integrals must use the same system and bounds. A density function is not itself the mass of each little segment unless multiplied by its length.

Derive the original graded rod

The workbench uses lambda(x)=lambda 0(1+beta x/L), lambda 0=2 kg/m. Integration from zero to L gives M=lambda 0 L(1+beta/2). The first moment is lambda 0 L²(1/2+beta/3). Thus xCM=L(1/2+beta/3)/(1+beta/2). When beta=0 this reduces to L/2. Multiplying lambda 0 by a common factor cancels from the ratio, but changing the density shape does not.

Check a concrete prediction

At L=3 and beta=2, M=12 kg and the first moment is 21 kg m, so xCM=1.75 m. Density increases toward the right, consistent with a center to the right of the midpoint but still inside the rod. Moving the origin by a constant changes every coordinate and the center by that same amount; it does not move the material physically.

Extend the differential element honestly

For a thin sheet use dm=sigma(x,y)dA, and for a solid use dm=rho(x,y,z)dV. A spherical distribution with rho(r)=rho 0 r/R has differential shell mass 4πr²rho(r)dr, giving M=πrho0 R³, not the integral of rho alone. Spherical symmetry places the center at the sphere’s center despite the radial gradient. This written extension distinguishes one-, two- and three-dimensional density units; the slider model remains a rod.

Transfer and inference limits

For lambda(x)=(3 kg/m²)x on 0 to 4 m, integrate to M=24 kg and first moment 64 kg m, giving 8/3 m. A measured mass and center still do not uniquely determine every detail of lambda(x). Ask for an independent spatial mass constraint before claiming identification. Center-of-mass motion can represent system translation, but does not replace internal deformation or rotation.

Public numerical transfer practice

Try a new practice case

A rod on 0≤x≤4 m has lambda(x)=(3 kg/m²)x. Find its center coordinate in metres.

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.