SCIENCE · INVESTIGATE AND EXPLAIN

Normalize a density before integrating its center

For an extended object, center of mass is a normalized first moment: xcm = integral(x dm)/M. A geometric midpoint works only when the mass distribution has the corresponding symmetry. Here a 2 m rod has total mass 3 kg and density λ(x) = λ₀(1 + αx/L), with α nonnegative.

Learning outcome: Integrate total mass, whole first moment and one segment first moment independently. Check the uniform limit and preserve the coordinate origin.

Starting recap

Integrate signed force and normalized mass moments. Specify the system before conserving momentum, and independently test energy and observation constraints.

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Try

At α = 0, the full center is 1 m. A left quarter has mass 0.75 kg and center 0.25 m, not the full rod midpoint.

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Predict

For α = 2, what mass lies in the left half of this normalized 3 kg rod?

Your prediction and reason

Experiment

Integrate total mass, whole first moment and one segment first moment independently. Check the uniform limit and preserve the coordinate origin.

Normalize and integrate a nonuniform rod

Worked teaching model · synthetic values · Rod length2 m and total mass3 kg; linear density lambda=lambda0(1+alpha x/L) on0≤x≤L. lambda0 is normalized for each alpha to preserve total mass. Compare full and left partial-segment moments about the same origin.

Density gradient coefficient alpha: 2 ratio Retained left segment fraction of length: 0.5 fraction

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Normalized linear mass density

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: lambda0(1+alpha x/L) (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Normalized linear mass density

lambda0(1+alpha x/L)

position (m)	linear density (kg/m)
0	0.75
0.025	0.76875
0.05	0.7875
0.075	0.80625
0.1	0.825
0.125	0.84375
0.15	0.8625
0.175	0.88125
0.2	0.9
0.225	0.91875
0.25	0.9375
0.275	0.95625
0.3	0.975
0.325	0.99375
0.35	1.0125
0.375	1.03125
0.4	1.05
0.425	1.06875
0.45	1.0875
0.475	1.10625
0.5	1.125
0.525	1.14375
0.55	1.1625
0.575	1.18125
0.6	1.2
0.625	1.21875
0.65	1.2375
0.675	1.25625
0.7	1.275
0.725	1.29375
0.75	1.3125
0.775	1.33125
0.8	1.35
0.825	1.36875
0.85	1.3875
0.875	1.40625
0.9	1.425
0.925	1.44375
0.95	1.4625
0.975	1.48125
1	1.5
1.025	1.51875
1.05	1.5375
1.075	1.55625
1.1	1.575
1.125	1.59375
1.15	1.6125
1.175	1.63125
1.2	1.65
1.225	1.66875
1.25	1.6875
1.275	1.70625
1.3	1.725
1.325	1.74375
1.35	1.7625
1.375	1.78125
1.4	1.8
1.425	1.81875
1.45	1.8375
1.475	1.85625
1.5	1.875
1.525	1.89375
1.55	1.9125
1.575	1.93125
1.6	1.95
1.625	1.96875
1.65	1.9875
1.675	2.00625
1.7	2.025
1.725	2.04375
1.75	2.0625
1.775	2.08125
1.8	2.1
1.825	2.11875
1.85	2.1375
1.875	2.15625
1.9	2.175
1.925	2.19375
1.95	2.2125
1.975	2.23125
2	2.25

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Normalized initial linear density	0.75	kg/m
Unchanged full mass	3	kg
Full center coordinate	1.166667	m
Left segment mass	1.125	kg
Left segment center coordinate	0.5555556	m
Left segment length	1	m

A center is a normalized first moment; changing density normalization changes the mass account.

M=integral lambda dx fixes lambda0=M/[L(1+alpha/2)]. The first moment gives xcm=L(1/2+alpha/3)/(1+alpha/2). For the left fraction f, integrate only0to fL and divide by that segment mass, not the full rod mass. Segment coordinates retain the original left-end origin.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

A center is a normalized first moment; changing density normalization changes the mass account.

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Understand

Normalize a density before integrating its center

Integrating the density fixes λ₀ = M/[L(1 + α/2)]. The first moment then gives xcm = L(1/2 + α/3)/(1 + α/2). Increasing the gradient while keeping total mass fixed must change λ₀; holding both λ₀ and M fixed would describe incompatible accounts.

For α = 2, λ₀ = 0.75 kg/m and the full center is 7/6 m from the left end. The left half contains 3(1/2 + α/8)/(1 + α/2) = 1.125 kg, and its own center is 5/9 m. Its center requires division by the segment mass rather than by all 3 kg.

For a general left fraction f, integrate only from 0 to fL. The segment center is L(f²/2 + αf³/3)/(f + αf²/2). Both centers use the same original left-end coordinate; shifting the segment origin would require a separate coordinate transformation.

Changed-case check

Predict before reading the comparison. At α = 0, the full center is 1 m. A left quarter has mass 0.75 kg and center 0.25 m, not the full rod midpoint.

Retained investigation

Integrate total mass, whole first moment and one segment first moment independently. Check the uniform limit and preserve the coordinate origin. Why is the segment mass the correct denominator for its own center?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All packets, derivations and prose are original. These bounded mathematical models are not measured experiments, physical procedures, practical certification or institutional credit.

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Why is the segment mass the correct denominator for its own center?

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