SCIENCE · INVESTIGATE AND EXPLAIN

Infer inertia from parallel-axis evidence without hiding bias

Use the linear relation in squared offset and test its conditioning.

Learning outcome: Compare d 1=.5,1,1.5 at second offset+.1. Derive the error amplification and propose a third-axis prediction before looking at its result. Retain both fitted and proposed held-out evidence.

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

d 1=1,second offset+.1 gives I 1=5,I 2=14.1,M≈3.03333 and Icm≈1.96667. With zero offset the true 3 and 2 are recovered.

Open the experiment

Predict

Does a two-axis exact fit independently establish that the axes and readings are correct?

Your prediction and reason

Experiment

Compare d 1=.5,1,1.5 at second offset+.1. Derive the error amplification and propose a third-axis prediction before looking at its result. Retain both fitted and proposed held-out evidence.

Infer mass and central inertia from parallel axes

Worked teaching model · synthetic values · Synthetic true M=3 kg,Icm=2 kg m²; parallel offsets d1 and d2=2 m. Only second inertia reading is biased by selected offset. Axes and distances are stipulated exact; inference is not an apparatus calibration.

First axis offset d1: 1 m Second inertia reading offset: 0.1 kg m²

Reset keeps your prediction, captures and practice history.

True parallel-axis relation

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

True parallel-axis relation: numerical coordinates

squared offset (m²)	inertia (kg m²)
0	2
0.1	2.3
0.2	2.6
0.3	2.9
0.4	3.2
0.5	3.5
0.6	3.8
0.7	4.1
0.8	4.4
0.9	4.7
1	5
1.1	5.3
1.2	5.6
1.3	5.9
1.4	6.2
1.5	6.5
1.6	6.8
1.7	7.1
1.8	7.4
1.9	7.7
2	8
2.1	8.3
2.2	8.6
2.3	8.9
2.4	9.2
2.5	9.5
2.6	9.8
2.7	10.1
2.8	10.4
2.9	10.7
3	11
3.1	11.3
3.2	11.6
3.3	11.9
3.4	12.2
3.5	12.5
3.6	12.8
3.7	13.1
3.8	13.4
3.9	13.7
4	14

Calculated values under the declared model
Quantity	Value	Unit
First indicated inertia	5	kg m²
Second indicated inertia	14.1	kg m²
Inferred mass	3.03333	kg
Inferred central inertia	1.96667	kg m²
Mass inference bias	0.0333333	kg
Squared-offset separation	3	m²

The parallel-axis relation is linear in squared offset, not offset itself.

Subtract I2−I1=M(d2²−d1²), then recover Icm. A small separation of squared offsets amplifies a fixed inertia-reading difference error. The theorem requires parallel axes and distance from the center axis; it does not apply unchanged to a differently oriented axis.

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

The parallel-axis relation is linear in squared offset, not offset itself.

Observation to attach to a capture

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

Understand

State the geometric theorem precisely

For axes parallel to a center-of-mass axis, I(d)=Icm+Md². The distance d is the perpendicular separation between axes. This does not authorize replacing any two arbitrarily oriented axes by the same formula. Nor can an inertia about an arbitrary noncentral axis simply serve as Icm.

Solve a two-axis inverse problem

Given indicated I 1 and I 2 at known offsets d 1,d 2, subtract to infer M=(I 2−I 1)/(d 2²−d 1²). Then Icm=I 1−Md 1². The algebra requires distinct squared offsets. Two measurements can determine these parameters under the model, but cannot independently validate the model that was imposed.

Introduce one declared reading fault

The synthetic true values are M=3 kg and Icm=2 kg m². The second axis is at 2 m, so its true inertia 14 receives the selected additive offset. The first reading at d 1 is exact by stipulation. At d 1=1 and second offset+.1, inferred M=3+.1/3 and Icm=2−.1/3. Both change even though only one reading was perturbed.

Inspect conditioning rather than extra decimal places

The mass error equals the inertia-difference error divided by d 2²−d 1². As the squared offsets approach each other, the same reading fault produces a larger mass error. Increasing displayed precision does not remove that sensitivity. A third independently measured axis or an independently known mass can provide a held-out check rather than another fitted condition.

Keep physical and evidential boundaries separate

Negative inferred central inertia or mass would contradict the positive-mass model and demand investigation, not a literal negative-inertia interpretation. The current controls keep inferred values positive, so a plausible-looking result can still be biased. Axis placement, orientation, instrument response and rigidity must be independently assessed in any real investigation; none is measured here.

Worked original case

d 1=1,second offset+.1 gives I 1=5,I 2=14.1,M≈3.03333 and Icm≈1.96667. With zero offset the true 3 and 2 are recovered.

Investigation

Compare d 1=.5,1,1.5 at second offset+.1. Derive the error amplification and propose a third-axis prediction before looking at its result. Retain both fitted and proposed held-out evidence.

Public numerical transfer practice

Try a new practice case

For known mass 2 kg and Icm 3 kg m², what is I about a parallel axis 2 m away?

Numerical result in the requested units

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

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

Why is distance-squared separation the relevant conditioning variable rather than raw distance alone?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Normalize a radial density before integrating disk inertia Next: Test support feasibility after integrating a distributed load

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