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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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.
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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.
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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.
| Quantity | Value | Unit |
|---|---|---|
| Supplied center-axis inertia | 1 | kg m² |
| Added shift term | 3 | kg m² |
| Shifted-axis inertia | 4 | kg m² |
Accessible plot coordinates
| parallel-axis offset (m) | rotational inertia (kg m²) |
|---|---|
| 0 | 1 |
| 0.1 | 1.03 |
| 0.2 | 1.12 |
| 0.3 | 1.27 |
| 0.4 | 1.48 |
| 0.5 | 1.75 |
| 0.6 | 2.08 |
| 0.7 | 2.47 |
| 0.8 | 2.92 |
| 0.9 | 3.43 |
| 1 | 4 |
| 1.1 | 4.63 |
| 1.2 | 5.32 |
| 1.3 | 6.07 |
| 1.4 | 6.88 |
| 1.5 | 7.75 |
| 1.6 | 8.68 |
| 1.7 | 9.67 |
| 1.8 | 10.72 |
| 1.9 | 11.83 |
| 2 | 13 |
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.
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Why does the algebraic proof depend on starting at the center of mass rather than any convenient reference point?
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