Audited ·Last updated 27 Jul 2026·3 citations·Tier 2·0 uses

Moment of Inertia Calculator

Free moment of inertia calculator for a rod, disc, cylinder, sphere, or hoop. Pick a shape and get I and radius of gyration instantly.

Moment of Inertia Calculator

Shape and axis
Total mass of the object.
Used only in the two rod modes — the full length of the rod.
Used in the disc, sphere, and hoop modes — the object's radius.
Moment of inertia (I)
0.125
Resistance to angular acceleration about the selected shape's standard axis.
Mass (m)
4 kg
Radius of gyration (k)
0.1768 m
Shape and axis
Solid disc or cylinder, axis through the center, along its length

Background.

This moment of inertia calculator computes I — the rotational analogue of mass — for six standard rigid-body shapes about their conventional rotation axis: a thin rod about its center or about one end, a solid disc or cylinder about its central axis, a solid sphere or a hollow (thin-shell) sphere about a diameter, and a hoop or thin ring about its central axis. Pick a shape, enter mass and the relevant dimension (length for a rod, radius for everything else), and the calculator returns I in kg·m², along with the radius of gyration — the single distance at which all the mass could be concentrated to produce the identical I.

Moment of inertia plays the same role in rotational dynamics that mass plays in linear dynamics: it measures how much an object resists a change in its rotational motion, appearing in the rotational form of Newton's second law, τ = Iα (torque equals moment of inertia times angular acceleration), exactly as F = ma does for straight-line motion. But unlike mass, moment of inertia is not a fixed property of an object alone — it depends critically on how that mass is distributed relative to the axis of rotation. Two objects with identical mass can have wildly different moments of inertia if one concentrates its mass near the axis and the other spreads it far away, because I depends on the square of each mass element's distance from the axis (I = Σmᵢrᵢ²). This calculator's rod-center versus rod-end comparison makes that dependence concrete: an identical rod has four times the moment of inertia about an end axis (I = mL²/3) as it does about a center axis (I = mL²/12) — same object, same mass, same length, four times harder to spin from one end than from the middle.

The six shapes here cover the overwhelming majority of moment-of-inertia problems in introductory and intermediate mechanics: rotating rods and beams, flywheels and discs, spinning spheres (solid, like a bowling ball, or hollow, like a basketball), and hoops or rings (like a bicycle wheel rim, idealized as all mass at one radius). Each has a closed-form result derived from integrating r² over the mass distribution, and all six are standard textbook results confirmed against OpenStax University Physics and MIT's introductory mechanics course materials. Because the calculator is multi-mode, switching shapes never changes which quantities it reports — moment of inertia, mass, radius of gyration, and a plain-language description of what you selected — only the underlying formula and the length-or-radius field that feeds it.

Radius of gyration k, defined by I = mk² (so k = √(I/m)), is a useful companion number because it converts an abstract kg·m² figure into an intuitive length: 'if all this object's mass sat in a thin ring at radius k, it would have the exact same moment of inertia.' For a hoop, k always equals the hoop's actual radius exactly, since every particle of a hoop already sits at that one radius — a clean sanity check built into the math itself. For every other shape, k is smaller than the outer radius, because some mass sits closer to the axis than the boundary.

This calculator is designed to sit alongside Quanta's torque calculator (τ = rF sin θ, which combines with I to predict angular acceleration via τ = Iα) and momentum calculator (whose rotational analogue, angular momentum L = Iω, uses I directly). Moment of inertia is rarely the final answer to a physics problem — it's almost always an ingredient feeding into a torque, energy, or angular-momentum calculation next.

What is moment of inertia calculator?

Moment of inertia (also called rotational inertia), denoted I, is a rigid body's resistance to angular acceleration about a specified axis of rotation. It is the rotational counterpart of mass: just as F = ma relates a net force to the resulting linear acceleration, τ = Iα relates a net torque to the resulting angular acceleration. Formally, I is defined as the sum (or, for a continuous body, the integral) of every mass element multiplied by the square of its perpendicular distance from the axis: I = Σmᵢrᵢ² for discrete masses, or I = ∫r² dm for a continuous body.

Because of that r² weighting, moment of inertia depends strongly on how an object's mass is distributed relative to the axis, not just on the total mass. Moving mass farther from the axis increases I much faster than moving the same mass closer decreases it, since the effect scales with the square of distance. This is why the same rod has a moment of inertia four times larger about an end axis than about a center axis — half the rod's length moves twice as far from an end axis, and squaring that doubling gives the fourfold increase (with an extra contribution from the redistribution of the near half as well, fully accounted for by the exact mL²/3 versus mL²/12 formulas).

Every shape has its own closed-form moment of inertia formula, derived by integrating the r² weighting over that shape's specific mass distribution. This calculator provides the six most commonly needed results — rod (two axis choices), solid disc/cylinder, solid sphere, hollow sphere, and hoop — each about that shape's standard, symmetric axis, which is overwhelmingly the axis used in textbook problems and real engineering calculations (flywheels, wheels, shafts) alike.

How to use this calculator.

  1. Select the shape and rotation axis that matches your object from the dropdown.
  2. Enter the object's mass in kilograms.
  3. If you selected a rod mode, enter the rod's full length. If you selected disc, sphere, or hoop, enter the radius instead.
  4. Read the moment of inertia I, the primary result, in kg·m².
  5. Read the radius of gyration k for an intuitive sense of how far, on average, the mass sits from the axis.
  6. Switch shapes to compare how differently distributed mass of the same total quantity changes I — try rod-center versus rod-end with the same mass and length to see the parallel-axis effect directly.

The formula.

I = m·L² ⁄ 12, m·L² ⁄ 3, m·r² ⁄ 2, ⅖m·r², ⅔m·r², or m·r²

Each of the six formulas comes from integrating r² dm over the shape's mass distribution about its stated axis. For a thin uniform rod of mass m and length L rotating about its center, I = mL²/12; about one end instead, I = mL²/3 — exactly four times larger, because the parallel-axis theorem states I_end = I_center + m·d², where d = L/2 is the distance between the two axes: mL²/12 + m(L/2)² = mL²/12 + mL²/4 = mL²/12 + 3mL²/12 = 4mL²/12 = mL²/3, confirming the two formulas are consistent with each other via that theorem.

For a solid disc or cylinder of mass m and radius r rotating about its central axis (the axis running through the middle, perpendicular to the circular faces), I = mr²/2 — the factor of one-half arises because a solid disc's mass is distributed continuously from the center out to r, and integrating r² over that uniform area gives exactly half of what a hoop with all its mass concentrated at radius r would have. That hoop comparison is direct: for a thin hoop or ring of mass m and radius r, every particle of mass sits at exactly the same distance r from the axis, so I = mr² with no fractional factor at all — the simplest of the six formulas and, not coincidentally, the largest I for a given m and r among shapes with radius r.

A solid sphere of mass m and radius r rotating about any diameter has I = (2/5)mr², smaller than the disc's mr²/2 factor because a sphere's mass is distributed through a three-dimensional volume, with a large fraction concentrated closer to the center than a two-dimensional disc's mass is. A hollow (thin-walled) sphere of the same mass and radius instead has I = (2/3)mr² — larger than the solid sphere, because removing the interior and placing all the mass in a thin shell at radius r pushes the average mass distance closer to r itself, more like the hoop's all-mass-at-r extreme.

Radius of gyration k is defined by the identity I = mk², so k = √(I/m) for any shape once I and m are known. It provides a length-scale interpretation of an otherwise abstract kg·m² number: an object with radius of gyration k has the same rotational inertia as a point mass (or a thin hoop) of the same total mass placed at distance k from the axis. For the hoop specifically, k = r exactly, since I = mr² gives k = √(mr²/m) = r — a useful internal consistency check on the whole calculator.

A worked example.

Example

A solid flywheel disc has a mass of 4 kg and a radius of 0.25 m, and needs to be spun about its central axis. Using the disc formula, I = m·r²/2 = 4 × (0.25)² / 2 = 4 × 0.0625 / 2 = 0.125 kg·m². The radius of gyration is k = √(I/m) = √(0.125/4) = √0.03125 ≈ 0.1768 m — meaning this disc behaves, rotationally, as if all 4 kg of its mass were concentrated in a thin ring about 17.7 cm from the axis, noticeably smaller than the disc's actual 25 cm radius because a solid disc packs proportionally more mass near its center than at its rim. If the same 4 kg and 0.25 m dimensions instead described a thin hoop (imagine a bicycle wheel rim with negligible spoke mass) rather than a solid disc, the moment of inertia would double to I = m·r² = 4 × 0.0625 = 0.25 kg·m², and the radius of gyration would equal the full 0.25 m radius exactly — because every bit of a hoop's mass really is out at the rim, with none of the disc's center-weighted mass distribution to reduce the effective average distance.

shapedisc
mass4
length1
radius0.25

Frequently asked questions.

Why does an identical rod have a different moment of inertia about its end than about its center?
Because moment of inertia depends on the square of each mass element's distance from the axis, and switching from a center axis to an end axis moves every part of the rod farther away — the far end moves from L/2 to L, and even the near end moves from 0 to 0. The parallel-axis theorem quantifies this exactly: I_end = I_center + m·d², where d = L/2 is the distance between the two parallel axes. Plugging in mL²/12 + m(L/2)² simplifies to mL²/3, precisely the end-axis formula — so the fourfold difference isn't a coincidence, it's the parallel-axis theorem doing its job. This calculator's rod-center and rod-end modes let you verify that relationship directly by comparing the two outputs for the same mass and length.
What is the parallel-axis theorem in general?
The parallel-axis theorem states that the moment of inertia about any axis parallel to one through the center of mass equals the center-of-mass moment of inertia plus the total mass times the square of the distance between the two axes: I = I_cm + m·d². It lets you compute I about any off-center parallel axis without re-deriving the integral from scratch, as long as you already know (or can look up) I about the center-of-mass axis. This calculator applies it implicitly in the rod-end formula (built directly into mL²/3), and you can apply it yourself to extend any of the six results here to an off-axis rotation.
Why does a hollow sphere have a larger moment of inertia than a solid sphere of the same mass and radius?
Because a hollow sphere's mass is concentrated entirely at the outer radius r, while a solid sphere's mass fills the whole interior volume, with a large fraction sitting much closer to the center than r. Since I depends on the square of distance from the axis, mass close to the center contributes very little to I, while mass at the full radius contributes the maximum possible amount. Removing the interior mass and redistributing it into a thin shell at radius r therefore increases the average r² weighting, which is exactly why the hollow-sphere coefficient (2/3) exceeds the solid-sphere coefficient (2/5) even though both share the same total mass and outer radius.
What does radius of gyration actually mean physically?
Radius of gyration k is the distance from the axis at which you could concentrate the object's entire mass as a single point (or thin ring) and get the exact same moment of inertia, defined by I = mk² so k = √(I/m). It's a way of converting an abstract rotational-inertia number into an intuitive length. For a hoop, k equals the actual physical radius exactly, because a hoop already behaves like all its mass is concentrated at one radius. For every other shape here — rod, disc, sphere — k is smaller than the object's outer dimension, because some of that shape's mass sits closer to the axis than the boundary, pulling the 'effective' radius inward.
How does moment of inertia connect to torque and rotational kinetic energy?
Moment of inertia is the proportionality constant in the rotational form of Newton's second law, τ = Iα, where τ is net torque and α is angular acceleration — the direct rotational analogue of F = ma. It also appears in rotational kinetic energy, KE_rot = (1/2)Iω², the rotational analogue of (1/2)mv², and in angular momentum, L = Iω, the rotational analogue of linear momentum p = mv. In every one of these relationships, I plays exactly the role mass plays in the corresponding linear-motion equation — which is why moment of inertia is often described as 'rotational mass.'
Does this calculator handle off-axis or non-standard rotation axes?
No — all six formulas here are for each shape's standard, symmetric axis (through the center for a rod, through the geometric center along or through the axis of symmetry for the others), which is the axis used in the overwhelming majority of textbook and engineering problems. For an off-center parallel axis, apply the parallel-axis theorem manually: take this calculator's center-axis result and add m·d², where d is the distance from the center axis to your actual axis. For a genuinely non-parallel or tilted axis, the calculation requires the full moment-of-inertia tensor, which is beyond the scope of this single-axis calculator.
Why are all four calculator outputs finite no matter which shape I pick?
Because the calculator always asks for exactly the two quantities every mode needs — mass, plus either length or radius depending on which shape you picked — and validates that whichever dimension your chosen shape actually uses is a positive number before computing anything. Moment of inertia, mass, and radius of gyration are always well-defined positive numbers for any valid shape and positive dimensions, and the shape description is always a plain-language label of whatever you selected. There is no combination of a valid shape and valid inputs that produces an undefined or infinite result.

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