Moment Of Inertia Calculator
Rotational stubbornness: your mass and size dressed in seven costumes at once — disc, hoop, sphere, shell, rod — with the parallel-axis theorem proved on your own numbers.
Moment Of Inertia Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Uniform-density ideal shapes
- Axes through the centre unless shifted
In short: 2 kg at r = 0.5 m as a solid disc: I = ½mr² = 0.250000 kg·m² about its axle. The SAME mass and radius as a hoop reads 0.500000 kg·m² — exactly twice, because every gram sits at the full radius instead of averaging half of it. The family card dresses your numbers in all seven costumes, and the parallel-axis card proves the end-pivoted rod: mL²/12 + m(L/2)² = 0.166667 + 0.500000 = 0.666667, derived rather than recited.
Formula
I = Σ m·r² · disc ½mr² · hoop mr² · sphere 2/5·mr² · rod mL²/12 · parallel axis: I = Icm + m·d²
Moment of inertia is mass with a lever arm: every scrap counts twice, once for how much and once for how FAR from the axis it sits (squared). That square is why shape matters as much as size — a hoop doubles a disc of the same mass and radius, and pivoting a rod at its end quadruples-plus its stubbornness versus its middle. The parallel-axis theorem is the shift rule: know the middle reading, pay m·d² to move the axis d away.
Worked Example
- Enter the mass, a radius and a length.
- Read the disc — the reference costume.
- Compare all seven costumes built from your numbers.
- Watch the parallel-axis card prove the rod's end reading from its middle.
Defaults: disc 0.250000; hoop 0.500000 (2×); sphere 0.200000; rod about middle 0.166667, about end 0.666667 — the parallel-axis proof closes it.
Strengths & Limits Of This Model
Where this engine is strong
- Seven costumes from one input set
- Parallel-axis theorem proved live
Where it stops
- No composite bodies
- No principal-axis bookkeeping
Practical Use Cases
Machinery
flywheel and spindle sizing
Teaching
why figure skaters pull arms in
Design
where to put the mass, not just how much
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Moment Of Inertia Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why does the hoop beat the disc at equal mass and radius?
Because r is squared: the disc averages r²/4 over its area (giving the ½ factor), the hoop sits every gram at the full r². Same mass, same radius, double the stubbornness — mass PLACEMENT is the physics, which is why flywheels are rims, not pancakes of uniform fill.
What does the moment of inertia actually control?
How hard the object is to spin up, slow down, or reorient: torque = I × angular acceleration. Double I and every manoeuvre costs double the torque or double the time. The angular-acceleration page consumes this number directly — its motor cards are this page's arithmetic in a gearbox.
Why do figure skaters spin faster when they pull arms in?
Angular momentum I·ω is conserved when no external torque acts: shrink I by pulling mass to the axis and ω must rise in payment. Arms at distance r carry r² weight; tucking them can triple a spin with zero new effort — the square's signature move.
Why is the sphere's factor 2/5 and not something rounder?
Integration over the volume: mass near the axis contributes little (r is small there), mass at the equator a lot, and the average of r² over a solid ball lands at 2/5 of mR². The shell's 2/3 is higher because ALL its mass lives far from the axis — hollow things pay more, same outside.
What is the parallel-axis theorem for?
Reusing homework: every table gives moments about a shape's CENTRE; real axes live elsewhere. The theorem prices the move — add m·d² for the shift d. The card proves it on your rod: middle reading mL²/12 plus m(L/2)² reproduces the end reading mL²/3 exactly.
Which costume is my bicycle wheel?
Nearly a hoop — rim, tyre and spokes put most mass at full radius, so mr² is the honest model. That is why wheel mass feels heavier in acceleration than frame mass: it costs twice the disc-reading per gram, paying the r² tax at full rate.
Why radians and not degrees for what comes next?
The rotational kinematics that consume I (ω = α·t and friends) work only in radians — degrees break the derivative relationships the same way feet break v = at in SI. The angular-velocity page converts your rpm honestly before any arithmetic.
How does this pair with the torque page?
That page prices the twist (force times lever arm); this page prices what the twist has to argue with. τ = I·α joins them: the same motor torque spins the hoop costume half as fast as the disc — which is the whole discipline of flywheel and rotor design in one division.