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SCIENCE · INVESTIGATE AND EXPLAIN

Include the ring area when deriving disk inertia

A disk with radial density cannot be treated as a line of equally weighted radii. An annulus has area 2πr dr, so its mass is that area times local surface density. Omitting the area factor changes both total mass and rotational inertia.

Learning outcome: Normalize the density, derive mass and inertia from separate ring integrals, and compare the uniform and graded cases.

Starting recap

Integrate mass moments about named axes, derive torque and angular-momentum constraints, and test orbital energy, angular and numerical consistency against independent evidence.

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Try

At β = 0 and R = 3 m, I = MR²/2 = 18 kg m² and the half-radius region contains 1 kg. Increasing R at fixed mass is not the same operation as adding more material at fixed density.

Open the experiment

Predict

At β = 2, the mass fraction inside half radius is:

Experiment

Normalize the density, derive mass and inertia from separate ring integrals, and compare the uniform and graded cases.

Normalize a radial density before integrating rotation

Worked teaching model · synthetic values · A hypothetical thin disk of fixed mass4 kg has area density sigma0(1+beta r²/R²). Rotation axis is perpendicular through the center. Radial rings have area2pi r dr; no thickness or material-strength model.

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Cumulative mass versus radial fraction

enclosed mass (kg)radius fraction0140

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: normalized ring integral (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Cumulative mass versus radial fraction

normalized ring integral

radius fractionenclosed mass (kg)
00
0.01250.0003125488
0.0250.001250781
0.03750.002816455
0.050.0050125
0.06250.007843018
0.0750.01131328
0.08750.01542974
0.10.0202
0.11250.02563286
0.1250.03173828
0.13750.03852739
0.150.0460125
0.16250.05420708
0.1750.06312578
0.18750.07278442
0.20.0832
0.21250.09439067
0.2250.1063758
0.23750.1191758
0.250.1328125
0.26250.1473086
0.2750.1626883
0.28750.1789766
0.30.1962
0.31250.214386
0.3250.2335633
0.33750.2537618
0.350.2750125
0.36250.2973477
0.3750.3208008
0.38750.3454063
0.40.3712
0.41250.3982188
0.4250.4265008
0.43750.4560852
0.450.4870125
0.46250.5193243
0.4750.5530633
0.48750.5882735
0.50.625
0.51250.6632891
0.5250.7031883
0.53750.7447461
0.550.7880125
0.56250.8330383
0.5750.8798758
0.58750.9285782
0.60.9792
0.61251.031797
0.6251.086426
0.63751.143145
0.651.202013
0.66251.26309
0.6751.326438
0.68751.39212
0.71.4602
0.71251.530742
0.7251.603813
0.73751.679481
0.751.757813
0.76251.838879
0.7751.922751
0.78752.0095
0.82.0992
0.81252.191925
0.8252.287751
0.83752.386754
0.852.489012
0.86252.594606
0.8752.703613
0.88752.816117
0.92.9322
0.91253.051945
0.9253.175438
0.93753.302765
0.953.434012
0.96253.56927
0.9753.708626
0.98753.852172
14

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
QuantityValueUnit
Central area density0.1591549kg/m²
Fixed disk mass4kg
Axial moment of inertia9.333333kg m²
Normalized axial inertia0.5833333ratio
Mass inside half radius0.625kg

The radial area factor and squared lever arm weight mass and inertia differently.

dm=2pi sigma0(1+beta r²/R²)r dr. Normalize M=pi sigma0 R²(1+beta/2); then I=pi sigma0 R⁴(1/2+beta/3). A half-radius region has mass fraction(1/4+beta/32)/(1+beta/2), not automatically one half or one quarter for a nonuniform disk.

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

The radial area factor and squared lever arm weight mass and inertia differently.

dm=2pi sigma0(1+beta r²/R²)r dr. Normalize M=pi sigma0 R²(1+beta/2); then I=pi sigma0 R⁴(1/2+beta/3). A half-radius region has mass fraction(1/4+beta/32)/(1+beta/2), not automatically one half or one quarter for a nonuniform disk.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Include the ring area when deriving disk inertia

A disk with radial density cannot be treated as a line of equally weighted radii. An annulus has area 2πr dr, so its mass is that area times local surface density. Omitting the area factor changes both total mass and rotational inertia.

The hypothetical thin disk has fixed mass 4 kg and density σ₀(1 + βr²/R²). Integrating annuli gives M = πσ₀R²(1 + β/2). Multiplying each annulus by r² before integration gives I = πσ₀R⁴(1/2 + β/3). Therefore I/(MR²) = (1/2 + β/3)/(1 + β/2).

At β = 2 and R = 2 m, the ratio is 7/12 and I = 28/3 kg m². This exceeds the uniform disk value because more of the fixed mass lies farther from the center. It remains below the thin-ring ratio 1; the distribution has not become a ring.

The mass inside half radius is a separate integral: its fraction is (1/4 + β/32)/(1 + β/2). At β = 2 that fraction is 5/32, giving 0.625 kg. Neither a half-radius cut nor a half-area argument alone gives the mass of a nonuniform region. Radius scaling at fixed β changes inertia as R² while the normalized density adjusts to preserve mass.

For an isolated axial redistribution at fixed radius and mass, changing β from 0 to 2 increases inertia by a factor 7/6. With no external axial torque, angular speed becomes 6/7 of its initial value and rotational kinetic energy also becomes 6/7 at fixed angular momentum. The missing rotational energy belongs to internal work or other degrees of freedom; zero external torque is not zero internal work.

Changed-case check

Predict before reading the comparison. At β = 0 and R = 3 m, I = MR²/2 = 18 kg m² and the half-radius region contains 1 kg. Increasing R at fixed mass is not the same operation as adding more material at fixed density.

Retained investigation

Normalize the density, derive mass and inertia from separate ring integrals, and compare the uniform and graded cases. Which assumption must change if radius grows but both density and mass are claimed fixed?

Source and boundary

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

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At β = 2, the mass fraction inside half radius is:

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Which assumption must change if radius grows but both density and mass are claimed fixed?

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