What moment of inertia measures
Moment of inertia is rotational mass. In linear motion, mass tells you how much force it takes to produce a given acceleration; in rotation, moment of inertia tells you how much torque it takes to produce a given angular acceleration, through the rotational form of Newton's second law, τ = Iα.
The definition is a sum: I = Σmiri², where each element of mass is weighted by the square of its distance from the axis. That square is the entire character of the quantity. Mass near the axis contributes almost nothing; mass at the rim contributes enormously. Move half a body's mass from the centre to twice the radius and its moment of inertia does not merely double.
Because the distance is measured from an axis, moment of inertia has no meaning until you name the axis. The same rod has a moment of inertia four times larger about one end than about its centre. The same wheel is easy to spin about its own axle and much harder to tumble end over end. Any statement of the form "the moment of inertia of a cylinder is…" is incomplete unless the axis is given, and that omission is the most common error in the whole topic.
The unit is kg·m², which follows directly from the definition. Note that it is a different quantity from the area moment of inertia used in beam bending, which has units of m⁴ and describes the distribution of cross-sectional area rather than mass. The two share a name and a mathematical form and nothing else.
Standard shapes and the parallel-axis theorem
For a continuous body the sum becomes an integral, I = ∫r² dm, and evaluating it for common shapes gives the table of standard results this calculator uses. Every one of them has the form I = k·mR² with a dimensionless coefficient k that depends only on how the mass is distributed:
A thin hoop has all its mass at radius R, so k = 1 — the maximum possible for a body of that radius. A solid disk spreads mass from the centre outward and gets k = ½. A solid sphere spreads it in three dimensions and gets k = ⅖. A hollow spherical shell puts more of it far from the axis, so k = ⅔, larger than the solid sphere. Reading those four numbers tells you everything about the shapes without any integration.
The parallel-axis theorem (also called the Huygens–Steiner theorem) handles any axis that is not through the centre of mass: I = Icm + md², where d is the perpendicular distance between the two parallel axes. Because md² is never negative, the centre-of-mass axis always gives the smallest moment of inertia of any axis in that direction — a fact worth remembering as a sanity check.
The theorem also generates results you might otherwise have to look up. A rod about its centre is mL²/12. Shift the axis to one end, a distance L/2, and you add m(L/2)² = mL²/4. The total is mL²/12 + 3mL²/12 = mL²/3, which is exactly the tabulated end-axis value. This calculator's third test vector is that derivation, run as a numeric check.
The radius of gyration, k = √(I/m), is the radius at which the entire mass would have to be concentrated as a thin ring to give the same moment of inertia. It is a convenient single length that summarises the mass distribution, and it is how rotating-machinery specifications often express inertia.
Worked example: a steel flywheel
A flywheel is a solid steel disk 400 mm in diameter and 50 mm thick, spinning at 3,000 rpm. Find its inertia and the energy it stores.
- Mass. Volume = πR²t = π × 0.2² × 0.05 = 6.2832 × 10⁻³ m³. Steel at 7,850 kg/m³ gives m = 6.2832 × 10⁻³ × 7850 = 49.32 kg.
- Moment of inertia. A solid disk about its own axis: I = ½mR² = 0.5 × 49.32 × 0.04 = 0.9864 kg·m².
- Radius of gyration. k = √(0.9864 ÷ 49.32) = √0.02 = 0.1414 m, which is R/√2 as it must be for any solid disk.
- Angular velocity. ω = 3000 × 2π ÷ 60 = 314.159 rad/s.
- Stored energy. E = ½Iω² = 0.5 × 0.9864 × 98,696 = 48,676 J, or 13.5 Wh.
- Torque to spin it up. Reaching that speed from rest in 10 s needs α = 314.159 ÷ 10 = 31.42 rad/s², so τ = Iα = 0.9864 × 31.42 = 31.0 N·m — and an average power of 48,676 ÷ 10 = 4.87 kW.
Now redesign it as a rim-weighted wheel: same 49.32 kg, but with the mass concentrated in a thin rim at 0.2 m. That makes it a hoop, whose shape coefficient is 1 rather than the disk’s ½, so I = 49.32 × 0.04 = 1.973 kg·m² and the stored energy doubles to 97,352 J for the same mass and the same speed. This is why real flywheels are rim-heavy: energy storage scales with the coefficient, and the coefficient is a design choice.
How to read the result
Look at the dimensionless coefficient I/(mR²) in the comparison table the calculator prints. That single number tells you where the mass sits, independently of scale. A coefficient near 1 means a rim-loaded body that stores energy well and is slow to accelerate; a coefficient near 0.4 means a compact body that spins up easily. Between two designs of the same mass and radius, the coefficient is the design difference.
Check the radius of gyration against the physical size of the body. It should always be smaller than the largest distance from the axis to any part of the body, and larger than zero for any body with mass off the axis. If it comes out larger than the outer radius, you have used the wrong dimension or the wrong shape.
Watch the energy figure closely for anything that actually spins. Rotational energy goes with the square of speed, so doubling the rpm quadruples the stored energy — and every joule of it has to be dealt with if the rotor fails. That is why containment, burst testing and speed limits dominate flywheel and rotor design far more than the inertia calculation itself.
Finally, remember that these are idealised uniform bodies. A real wheel has a hub, spokes, bolts, a keyway and a rim of varying section. Treat the standard formula as a first estimate — usually good to a few per cent for a genuinely disk-like part, and poor for anything with a strongly non-uniform section. For real components, either sum the contributions part by part, or read the value from CAD, or measure it with a bifilar or torsional pendulum.
Standard mass moments of inertia
| Body and axis | Moment of inertia | Radius of gyration | Coefficient k |
|---|---|---|---|
| Thin hoop or thin-walled tube, own axis | mR² | R | 1 |
| Thin spherical shell, diameter | ⅔mR² | 0.8165R | 0.6667 |
| Solid disk or cylinder, own axis | ½mR² | 0.7071R | 0.5 |
| Solid sphere, diameter | ⅖mR² | 0.6325R | 0.4 |
| Thick-walled tube, own axis | ½m(Ro² + Ri²) | √((Ro²+Ri²)/2) | 0.5–1 |
| Thin rod, perpendicular axis through the end | mL²/3 | 0.5774L | 0.3333 |
| Thin rod, perpendicular axis through the centre | mL²/12 | 0.2887L | 0.0833 |
| Rectangular plate, perpendicular axis at centre | m(a²+b²)/12 | √((a²+b²)/12) | — |
| Solid cylinder, transverse axis at centre | m(3R²+L²)/12 | √((3R²+L²)/12) | — |
Rod coefficients are quoted against L², not R². The thick-walled tube coefficient runs from 0.5 for a solid cylinder up towards 1 as the wall becomes thin.
Mistakes and limits
- Quoting a moment of inertia without naming the axis. The value depends entirely on the axis. A rod about one end has four times the inertia it has about its centre.
- Confusing mass moment of inertia with area moment of inertia. This page gives kg·m² for rotational dynamics. The m⁴ quantity used for beam stiffness is a different thing with a similar name.
- Using diameter where the formula wants radius. Because the dimension is squared, that error inflates the answer by a factor of four.
- Applying the thin-rod formula to a stubby cylinder. The thin-rod result drops the 3R² term. Use the transverse-cylinder option when the radius is not small compared with the length.
- Subtracting md² for an axis inside the body. The parallel-axis term is always added. The centre-of-mass axis already gives the minimum, so no parallel axis can give less.
- Treating a real machined part as a uniform solid. Hubs, webs, bores and bolt circles all change the distribution. Build the value up from parts, or take it from CAD, when it matters.
Where moment of inertia is used
Its primary role is in the rotational form of Newton's second law, τ = Iα, so it sits directly beside the torque calculator and is the rotational counterpart of the mass in the F = ma calculator. Sizing a motor for a machine that has to accelerate a rotating load is exactly this calculation: find I, decide the angular acceleration you need, multiply.
It also appears in rotational kinetic energy, ½Iω², which is the rotational partner of the ½mv² in the kinetic energy calculator. A body that both moves and spins carries both terms, which is why a rolling ball reaches the bottom of a slope slower than a sliding one: some of the released potential energy has gone into spin rather than into forward speed. For a solid sphere rolling without slipping, exactly 2/7 of the kinetic energy is rotational.
And it governs angular momentum, L = Iω, which is conserved when no net torque acts — the reason a skater speeds up on pulling their arms in, reducing I and forcing ω to rise. The linear analogue is in the momentum calculator, and the correspondence between the two sets of equations is exact throughout: force becomes torque, mass becomes moment of inertia, velocity becomes angular velocity, and every result transfers.