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SCIENCE · INVESTIGATE AND EXPLAIN

Move the axis without moving the mass distribution

A body has no single rotational inertia independent of axis. Moving a parallel axis away from its center of mass generally increases the squared distances of its particles. The parallel-axis theorem states I=Icm+Md², where d is the perpendicular separation and Icm is the inertia about the parallel center-of-mass axis. Both the parallel condition and the center-of-mass reference matter.

Learning outcome: For mass 3 kg, compare offsets 0, 0.5, 1 and 2 m. Plot inertia against d² on paper using the ledger, identify the intercept and slope, and distinguish scaling the shift term from scaling the full sum.

Starting recap

Review signed graph slopes and areas, extended-body force diagrams, Newton’s laws and center-of-mass coordinates. State the rotation axis.

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Try

A 3 kg, 2 m rod has supplied center inertia 1 kg m². The parallel axis 1 m away has inertia 1+3×1²=4 kg m². An axis 2 m from the center has 1+3×4=13 kg m². Starting from the first off-center value and adding 3×1² would incorrectly give 7 for the second axis.

Open the experiment

Predict

Will doubling the axis offset double the added inertia? Will it quadruple the entire inertia?

Experiment

For mass 3 kg, compare offsets 0, 0.5, 1 and 2 m. Plot inertia against d² on paper using the ledger, identify the intercept and slope, and distinguish scaling the shift term from scaling the full sum.

Shift a specified axis without changing the body

Worked teaching model · synthetic values · A thin uniform 2 m rod has supplied center-axis inertia ML²/12. Both axes are perpendicular to the rod and parallel to one another. The shifted axis is d from the center-axis. This calculation does not prescribe a physical support.

Reset keeps your prediction, captures and practice history.

02130parallel-axis offset (m)rotational inertia (kg m²)

Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Supplied center-axis inertia1kg m²
Added shift term3kg m²
Shifted-axis inertia4kg m²
Accessible plot coordinates
Primary plotted points
parallel-axis offset (m)rotational inertia (kg m²)
01
0.11.03
0.21.12
0.31.27
0.41.48
0.51.75
0.62.08
0.72.47
0.82.92
0.93.43
14
1.14.63
1.25.32
1.36.07
1.46.88
1.57.75
1.68.68
1.79.67
1.810.72
1.911.83
213

Among these parallel axes, the center-of-mass axis gives the minimum inertia.

I=Icm+Md², with L=2 m and Icm=ML²/12. The theorem requires the reference axis to pass through the center of mass; adding Md² to an arbitrary already-offset inertia generally double-counts incorrectly.

Among these parallel axes, the center-of-mass axis gives the minimum inertia.

I=Icm+Md², with L=2 m and Icm=ML²/12. The theorem requires the reference axis to pass through the center of mass; adding Md² to an arbitrary already-offset inertia generally double-counts incorrectly.

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

Understand

1. Frame the question

A body has no single rotational inertia independent of axis. Moving a parallel axis away from its center of mass generally increases the squared distances of its particles. The parallel-axis theorem states I=Icm+Md², where d is the perpendicular separation and Icm is the inertia about the parallel center-of-mass axis. Both the parallel condition and the center-of-mass reference matter.

2. Build the mechanism

To see why, write a particle’s distance coordinates relative to the center axis as x,y and shift the axis by d along x. The new sum is Σm[(x−d)²+y²]. Expanding produces Icm−2dΣmx+d²Σm. The middle term vanishes because center-of-mass coordinates obey Σmx=0, leaving Icm+Md². No calculus or shape-specific integral is needed for this argument.

The added term is nonnegative, so the center axis gives the minimum inertia among this family of parallel axes. The result does not compare arbitrarily tilted axes. Nor can you generally start with an off-center inertia and simply add M times the distance to another off-center axis; the cross term would not automatically vanish.

For a thin uniform rod of length 2 m, this lesson supplies Icm=ML²/12=M/3. An axis through an end is 1 m from its center, giving Iend=M/3+M=4M/3. The physical body and mass are unchanged; only the axis used to describe its rotational response has changed.

3. Work an example

A 3 kg, 2 m rod has supplied center inertia 1 kg m². The parallel axis 1 m away has inertia 1+3×1²=4 kg m². An axis 2 m from the center has 1+3×4=13 kg m². Starting from the first off-center value and adding 3×1² would incorrectly give 7 for the second axis.

4. Predict and investigate

Predict first: Will doubling the axis offset double the added inertia? Will it quadruple the entire inertia?

For mass 3 kg, compare offsets 0, 0.5, 1 and 2 m. Plot inertia against d² on paper using the ledger, identify the intercept and slope, and distinguish scaling the shift term from scaling the full sum. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

A body has Icm=2 kg m² and M=4 kg. Find inertia about a parallel axis 0.5 m away.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

Why does the algebraic proof depend on starting at the center of mass rather than any convenient reference point?

Retain your first response, an extended-body diagram or labeled graph, units, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic model captures are not a performed physical experiment.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 5 topic mapping: 5.4. Checked 30 October 2026. Planar clockwise/counterclockwise analysis; supplied formulas for extended bodies, no required calculus or three-dimensional torque-vector direction. Framework cross-reference, not AP authorization.

Apply

A body has Icm=1 kg m² and mass 3 kg. What is its inertia about a parallel axis 1 m from its center?

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

Why does the algebraic proof depend on starting at the center of mass rather than any convenient reference point?

Continue

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

Skipping retains experiments and awards no mastery.

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