Parallel Axis Theorem Calculator
Apply the parallel axis theorem I = I_cm + Md² to shift a moment of inertia to a parallel axis.
Choose a shape preset or enter I_cm directly.
What the parallel axis theorem does
You know the moment of inertia of an object about an axis through its center of mass. You want the moment of inertia about a different axis, parallel to the first but offset by some distance. The parallel axis theorem (also called Steiner’s theorem, after Jakob Steiner who proved it in 1839) gives you the answer in one step:
I = I_cm + M d²
Where I_cm is the moment of inertia about the center-of-mass axis, M is the total mass, and d is the perpendicular distance between the two parallel axes. The result I is the moment of inertia about the shifted axis. The theorem only works for axes that are parallel; for non-parallel axes you need the full inertia tensor.
Why this matters in real problems
Pure center-of-mass rotation is rare in practice. Real objects rotate about pivots, hinges, bearings, and supports that are NOT at the center of mass. A pendulum swings about its pivot (often at one end of the rod, not the middle). A door swings about its hinge, not its centroid. A flywheel may be mounted on a shaft slightly off-center for engineering reasons. In each case, the relevant moment of inertia is about the actual rotation axis, which requires shifting from the textbook I_cm.
The theorem makes this trivial. You look up I_cm for the basic shape (the well-known formulas in any physics text), measure d, multiply M by d², and add. That is the entire calculation.
Common center-of-mass moments
| Shape | I_cm formula | About which axis |
|---|---|---|
| Solid disk or cylinder | I_cm = ½ M R² | Central axis |
| Hollow thin-walled cylinder | I_cm = M R² | Central axis |
| Solid sphere | I_cm = (2/5) M R² | Any diameter |
| Hollow thin-shell sphere | I_cm = (2/3) M R² | Any diameter |
| Thin rod | I_cm = (1/12) M L² | Through center, perpendicular to length |
| Rectangular plate | I_cm = (1/12) M (a² + b²) | Through center, perpendicular to plate |
| Hollow cylinder (two radii) | I_cm = ½ M (R₁² + R₂²) | Central axis |
These are the seven I_cm formulas you actually need for most undergraduate physics problems. Engineering tables extend the list to I-beams, C-channels, hollow rectangles, and so on.
Sanity check: rod about one end
The known formula for a thin rod rotating about one end is I = (1/3) M L². Let’s verify with the parallel axis theorem starting from I_cm = (1/12) M L² and d = L/2 (the center of mass is at the midpoint, half the rod length from the end):
I = I_cm + M d² = (1/12) M L² + M (L/2)² = (1/12) M L² + (1/4) M L² = (1/12 + 3/12) M L² = (4/12) M L² = (1/3) M L²
Matches. The theorem is consistent with the special-case formulas in every textbook table.
Why d² and not d
Mathematically, the d² scaling falls out of the definition of moment of inertia as the integral of r² dm. When you shift the axis by d, every infinitesimal mass element’s r-vector changes by exactly d (in the shift direction). The cross term integrates to zero by definition of the center of mass (the position vector to the CM is the average of all r-vectors, so by construction it has zero offset from itself), leaving only the constant d² M term.
The intuition: rotational inertia depends on how far mass is from the axis, squared. Shifting the axis effectively “moves” the entire mass distribution by d, and the contribution to I scales as the square of that shift.
A useful corollary: minimum moment of inertia
The parallel axis theorem implies that, among all parallel axes, the one through the center of mass gives the minimum moment of inertia. Any other parallel axis adds M d² (which is non-negative), so it can only be larger. This is why the I_cm value is the natural reference: it is the smallest member of the family of parallel-axis moments.
This corollary has practical consequences. To get an object up to angular speed with the least torque, mount it on its center-of-mass axis. To make a heavy object hard to rotate (like a flywheel meant to store energy), mount it off-center to take advantage of the M d² boost. Both choices use the same theorem.
Where it shows up in engineering
Structural engineering uses parallel axis when computing the moment of inertia of composite beam cross-sections. The beam consists of multiple rectangular regions, each with its own centroidal moment of inertia and offset from the overall centroid. Total I is the sum of each part’s I_cm plus its M d² shift to the section centroid. This is exactly how textbook section-property tables for I-beams are derived.
Robotics uses parallel axis for kinematic chains: each link has its mass and I_cm about its own centroid, but joint dynamics require I about the joint axis, which is offset. Inverse dynamics and torque calculations call the parallel axis theorem on every link, every step.
Sports equipment design uses parallel axis to tune swing weight. A baseball bat or tennis racket rotates about the grip, not the center of mass, so the relevant I depends on both the bat’s intrinsic I_cm and the offset d from grip to mass center. Adjusting d (by adding weight near the head or grip) changes the swing weight without changing total mass.
Limitations
The theorem only handles axis shifts that keep the two axes parallel. If you want to rotate the axis direction (tilt it), you need the full inertia tensor and a similarity transformation, not parallel axis. For 2D problems where the rotation is always perpendicular to the plane, parallel axis covers every case you will encounter.