SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish redistribution from a change of rotational axis

Rotational inertia measures how mass distribution affects angular acceleration about a chosen axis. A point contributes mr², using perpendicular distance to that axis. For several points, add those contributions. Two arrangements with equal total mass can have different inertia because squared distances, not mass alone, set the weighting.

Learning outcome: Compare radii0.25,0.5,1 m about the central axis, then shift the axis at fixed radius. Retain the bar, point-pair and axis-shift contributions separately. Explain why a center-of-mass location alone does not determine inertia.

Starting recap

Declare the axis, angular units, recipient and valid contact or material branch. Distinguish supplied shape factors from independently measured inertia.

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Try

At r=0.5 m,d=0, I=2/3+1/2=7/6 kg m². Moving only the points to r=1 gives 8/3 kg m² at the same total mass. Shifting the axis by 0.5 m adds1 kg m² to either arrangement.

Open the experiment

Predict

Moving the same point mass twice as far from an axis multiplies its inertia contribution by:

Your prediction and reason

Experiment

Compare radii0.25,0.5,1 m about the central axis, then shift the axis at fixed radius. Retain the bar, point-pair and axis-shift contributions separately. Explain why a center-of-mass location alone does not determine inertia.

Keep mass fixed while changing its distance from an axis

Worked teaching model · synthetic values · A uniform thin 2 kg bar of length 2 m has supplied central perpendicular inertia ML²/12. Two 1 kg points attach at symmetric ±r along the bar. Total mass 4 kg and center fixed; compare parallel axes at selected offset. No material construction.

Symmetric point-mass radius: 0.75 m Parallel axis offset: 0.5 m

Reset keeps your prediction, captures and practice history.

Inertia versus symmetric point radius

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

Equivalent numerical coordinates for Inertia versus symmetric point radius

bar plus two points

point radius (m)	central inertia (kg m²)
0.25	0.791667
0.26875	0.81112
0.2875	0.831979
0.30625	0.854245
0.325	0.877917
0.34375	0.902995
0.3625	0.929479
0.38125	0.95737
0.4	0.986667
0.41875	1.01737
0.4375	1.04948
0.45625	1.08299
0.475	1.11792
0.49375	1.15424
0.5125	1.19198
0.53125	1.23112
0.55	1.27167
0.56875	1.31362
0.5875	1.35698
0.60625	1.40174
0.625	1.44792
0.64375	1.49549
0.6625	1.54448
0.68125	1.59487
0.7	1.64667
0.71875	1.69987
0.7375	1.75448
0.75625	1.81049
0.775	1.86792
0.79375	1.92674
0.8125	1.98698
0.83125	2.04862
0.85	2.11167
0.86875	2.17612
0.8875	2.24198
0.90625	2.30924
0.925	2.37792
0.94375	2.44799
0.9625	2.51948
0.98125	2.59237
1	2.66667

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
Total unchanged mass	4	kg
Bar central inertia	0.666667	kg m²
Point-pair central inertia	1.125	kg m²
Combined central inertia	1.79167	kg m²
Inertia about shifted parallel axis	2.79167	kg m²
Net torque for alpha 2 rad/s² about shifted axis	5.58333	N m

Inertia depends on squared distances from the selected axis, not just total mass.

Each point contributes m r², so a symmetric pair contributes 2r² in this packet. The supplied uniform-bar result ML²/12 is about its central perpendicular axis. A parallel axis shift adds total M d², including all components. Moving the points preserves the center and total mass; changing the axis is a separate operation.

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

Inertia depends on squared distances from the selected axis, not just total mass.

Observation to attach to a capture

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

Understand

Choose the axis, reference or material model

Build the algebraic relationship

This assembly has a uniform2 kg bar of length 2 m and two1 kg points at symmetric radii±r. The supplied central bar inertia is ML²/12=2/3 kg m². The pair adds2r², giving Icm=2/3+2r². Moving the points preserves both the 4 kg total mass and the center, so this is a clean redistribution comparison.

Check the original numerical packet

A separate parallel axis a distance d from the center has I=Icm+4d². The added term uses the mass of the whole assembly, not just the moving points. It applies to parallel axes with the stated perpendicular separation. Changing axis orientation would require another calculation, not the same translation formula.

Test the boundary and independent evidence

At r=0.75 m,d=0.5 m, central inertia is 43/24≈1.79167 kg m² and shifted inertia 67/24≈2.79167. Net torque for angular acceleration 2 rad/s² about that fixed shifted axis is 67/12≈5.58333 N m. The law τnet=Iα presumes unchanged rigid distribution during the evaluated motion; moving masses while rotating requires an angular-momentum account instead of silently treating I as constant.

Worked original case

At r=0.5 m,d=0, I=2/3+1/2=7/6 kg m². Moving only the points to r=1 gives 8/3 kg m² at the same total mass. Shifting the axis by 0.5 m adds1 kg m² to either arrangement.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

Two1 kg points are at perpendicular distances0.5 m and 1.5 m from a fixed axis. Find their total rotational inertia in kg m².

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. All records are supplied or generated teaching evidence. No rotating equipment, applied loads, material tests or physical assembly is requested. An admissible virtual model state does not certify real equipment safety, supervised practical skill or independent mastery.

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

Which assumptions are hidden if a known inertia formula is applied about a different axis?

Your explanation and limitations

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Previous: Infer a load location while testing the support model Next: Fit rotational inertia without hiding resistance or readout bias

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