SCIENCE · INVESTIGATE AND EXPLAIN

Derive an axis-dependent inertia from the actual density

Moment of inertia weights each mass element by squared perpendicular distance to a specified axis: I = integral(rperp² dm). For a thin rod along x and a perpendicular axis through x = d, that distance is |x − d|. A formula about the left end cannot be reused about the mass center without accounting for the translation.

Learning outcome: Derive the normalized second moment and compare direct integration with the parallel-axis reconstruction for three axes.

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

For a uniform rod, α = 0, the center is 1 m, left-end inertia is 4 kg m² and center inertia is 1 kg m². Translating that center axis by 0.5 m increases inertia by 3(0.5)² = 0.75 kg m².

Open the experiment

Predict

At α = 2, which perpendicular axis minimizes this rod’s inertia?

Your prediction and reason

Experiment

Derive the normalized second moment and compare direct integration with the parallel-axis reconstruction for three axes.

Derive inertia about a translated axis

Worked teaching model · synthetic values · A 2 m,3 kg ideal rod has normalized density lambda0(1+alpha x/L). Axes are perpendicular to the rod through coordinate d. Supplied rigid mass distribution, no apparatus.

Linear density gradient: 2 ratio Axis coordinate: 1 m

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Axis location and normalized second moment

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

Equivalent numerical coordinates for Axis location and normalized second moment

integrated squared distance

axis coordinate (m)	moment of inertia (kg m²)
0	5
0.025	4.826875
0.05	4.6575
0.075	4.491875
0.1	4.33
0.125	4.171875
0.15	4.0175
0.175	3.866875
0.2	3.72
0.225	3.576875
0.25	3.4375
0.275	3.301875
0.3	3.17
0.325	3.041875
0.35	2.9175
0.375	2.796875
0.4	2.68
0.425	2.566875
0.45	2.4575
0.475	2.351875
0.5	2.25
0.525	2.151875
0.55	2.0575
0.575	1.966875
0.6	1.88
0.625	1.796875
0.65	1.7175
0.675	1.641875
0.7	1.57
0.725	1.501875
0.75	1.4375
0.775	1.376875
0.8	1.32
0.825	1.266875
0.85	1.2175
0.875	1.171875
0.9	1.13
0.925	1.091875
0.95	1.0575
0.975	1.026875
1	1
1.025	0.976875
1.05	0.9575
1.075	0.941875
1.1	0.93
1.125	0.921875
1.15	0.9175
1.175	0.916875
1.2	0.92
1.225	0.926875
1.25	0.9375
1.275	0.951875
1.3	0.97
1.325	0.991875
1.35	1.0175
1.375	1.046875
1.4	1.08
1.425	1.116875
1.45	1.1575
1.475	1.201875
1.5	1.25
1.525	1.301875
1.55	1.3575
1.575	1.416875
1.6	1.48
1.625	1.546875
1.65	1.6175
1.675	1.691875
1.7	1.77
1.725	1.851875
1.75	1.9375
1.775	2.026875
1.8	2.12
1.825	2.216875
1.85	2.3175
1.875	2.421875
1.9	2.53
1.925	2.641875
1.95	2.7575
1.975	2.876875
2	3

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
Center coordinate	1.166667	m
Inertia about left-end axis	5	kg m²
Inertia about mass-center axis	0.9166667	kg m²
Inertia about selected axis	1	kg m²
Parallel-axis reconstruction	1	kg m²

An inertia is a second moment about a specified axis, not merely mass times an arbitrary length squared.

Integrate (x−d)² lambda(x) dx, with lambda normalized to fixed M. Expanding gives I(d)=I(0)−2d M xcm+Md². Completing the square yields Icm+M(d−xcm)², so the minimum lies at the mass center, not necessarily the geometric midpoint.

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

An inertia is a second moment about a specified axis, not merely mass times an arbitrary length squared.

Observation to attach to a capture

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

Understand

Derive an axis-dependent inertia from the actual density

The 2 m, 3 kg rod has density λ₀(1 + αx/L), normalized by λ₀ = M/[L(1 + α/2)]. Its center is L(1/2 + α/3)/(1 + α/2), and integrating x²λ gives left-end inertia ML²(1/3 + α/4)/(1 + α/2).

Expand the translated integrand: (x − d)² = x² − 2dx + d². Integration gives I(d) = I(0) − 2dMxcm + Md². Completing the square gives Icm + M(d − xcm)², deriving the parallel-axis relation rather than treating the axis label as an afterthought.

At α = 2, the center is 7/6 m, left-end inertia is 5 kg m² and center inertia is 11/12 kg m². The axis at the geometric midpoint d = 1 m gives 1 kg m², slightly larger. The smallest inertia among parallel axes passes through the mass center, which need not be the geometric center.

Changed-case check

Predict before reading the comparison. For a uniform rod, α = 0, the center is 1 m, left-end inertia is 4 kg m² and center inertia is 1 kg m². Translating that center axis by 0.5 m increases inertia by 3(0.5)² = 0.75 kg m².

Retained investigation

Derive the normalized second moment and compare direct integration with the parallel-axis reconstruction for three axes. Why does changing an axis not change the rod mass?

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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Why does changing an axis not change the rod mass?

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Previous: Break a density-gradient and origin-offset ambiguity Next: Include the ring area when deriving disk inertia

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