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SCIENCE · INVESTIGATE AND EXPLAIN

Normalize a radial density before integrating disk inertia

Distinguish circumference, area density and rotational distance in a continuous inertia calculation.

Learning outcome: At R=2 compare n=0,2,4. Derive sigma 0 before I, retain normalized shape factors and compare the thin-rim limit. Explain which extra geometric factor appears for a spherical shell.

Starting recap

Declare the fixed axis, integrate the actual mass or load distribution, and test constraint feasibility before interpreting a numerical result.

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Try

M=2,R=2,n=2 gives sigma 0=1/pi kg/m² and I=16/3 kg m². At the same mass/radius, the uniform disk gives 4 and the thin hoop 8.

Open the experiment

Predict

When changing radial density shape at fixed mass, may sigma 0 remain unchanged?

Experiment

At R=2 compare n=0,2,4. Derive sigma 0 before I, retain normalized shape factors and compare the thin-rim limit. Explain which extra geometric factor appears for a spherical shell.

Integrate annular mass with fixed total mass

Worked teaching model · synthetic values · Thin disk mass 2 kg, sigma=sigma0(r/R)^n normalized to the same mass. Axis perpendicular through center; dm=2pi r sigma dr. No axial thickness or actual machining.

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Annular inertia integrand

dI/dr (kg m)radius (m)02160

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Annular inertia integrand: numerical coordinates
radius (m)dI/dr (kg m)
00
0.051.56250e-7
0.15.00000e-6
0.153.79687e-5
0.20.00016
0.250.000488281
0.30.001215
0.350.00262609
0.40.00512
0.450.00922641
0.50.015625
0.550.0251642
0.60.03888
0.650.0580145
0.70.084035
0.750.118652
0.80.16384
0.850.221853
0.90.295245
0.950.38689
10.5
1.050.638141
1.10.805255
1.151.00568
1.21.24416
1.251.52588
1.31.85647
1.352.24202
1.42.68912
1.453.20487
1.53.79688
1.554.4733
1.65.24288
1.656.11491
1.77.09928
1.758.20654
1.89.44784
1.8510.835
1.912.3805
1.9514.0975
216
Calculated values under the declared model
QuantityValueUnit
Normalized edge density sigma00.31831kg/m²
Total fixed mass2kg
Central rotational inertia5.33333kg m²
Shape factor I/(MR²)0.666667ratio
Equal-mass thin-rim inertia8kg m²

Annular mass includes the circumference factor before the additional r² inertia weighting.

Integrate dm=2pi r sigma dr to normalize mass first. Then I=integral r²dm gives MR²(n+2)/(n+4). At n=0 the uniform disk factor is1/2; concentrating mass outward increases it toward the thin-rim value1, not beyond it.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Annular mass includes the circumference factor before the additional r² inertia weighting.

Integrate dm=2pi r sigma dr to normalize mass first. Then I=integral r²dm gives MR²(n+2)/(n+4). At n=0 the uniform disk factor is1/2; concentrating mass outward increases it toward the thin-rim value1, not beyond it.

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

Understand

Choose annuli as the differential elements

A thin annulus at radius r and width dr has area 2pi r dr. With surface density sigma(r), its mass is 2pi r sigma(r)dr. Its contribution to inertia about the perpendicular center axis adds another factor r². Omitting either the circumference factor or the inertia factor changes both units and physical weighting.

Hold total mass fixed explicitly

The family has sigma(r)=sigma 0(r/R)^n and fixed total mass M=2 kg. Integrating mass from 0 toR gives M=2pi sigma 0 R²/(n+2), so sigma 0=M(n+2)/(2piR²). A comparison at different n must recompute this normalization; holding sigma 0 fixed instead would compare different masses.

Integrate the inertia

Now I=2pi sigma 0 integral r³(r/R)^n dr=2pi sigma 0 R⁴/(n+4). Substituting the mass normalization gives I=MR²(n+2)/(n+4). At n=0 this is the uniform-disk factor 1/2. At n=2 it becomes 2/3; larger n puts a greater share of the same mass near the edge.

Check a limiting geometry

For nonnegative n, the shape factor remains below 1 and tends toward 1 as mass concentrates near the rim. A thin hoop of the same mass and radius has I=MR². The limit is a distribution argument, not an instruction to machine an increasingly concentrated material or a guarantee of mechanical strength.

Distinguish a sphere from a disk

Optional geometric extension: a thin spherical shell has inertia(2/3)r² dm about a diameter, not r²dm, because not all shell mass is the same perpendicular distance from that axis. Integrating uniform volume density over shells gives I=(2/5)MR² for a solid sphere. If spherical density instead scales with radius, normalization and shell weighting give(4/9)MR². Importing a disk formula into a sphere would omit geometry.

Worked original case

M=2,R=2,n=2 gives sigma 0=1/pi kg/m² and I=16/3 kg m². At the same mass/radius, the uniform disk gives 4 and the thin hoop 8.

Investigation

At R=2 compare n=0,2,4. Derive sigma 0 before I, retain normalized shape factors and compare the thin-rim limit. Explain which extra geometric factor appears for a spherical shell.

Public numerical transfer practice

Try a new practice case

A uniform thin disk has mass 4 kg and radius 3 m. What is its central perpendicular inertia in kg m²?

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.

Apply

When changing radial density shape at fixed mass, may sigma0 remain unchanged?

Public practice checks a specific claim. It does not establish independent mastery.

Why does a central axis not make every point on a spherical shell equally far from it?

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