Moment of Inertia Calculator

For a solid cylinder rotating about its central axis, I = ½MR², where M is mass in kilograms and R is radius in metres. The output is in kg·m². So a 5 kg solid cylinder with 0.1 m radius has I = 0.5 × 5 × 0.01 = 0.025 kg·m². Notice it doesn't depend on the cylinder's length — only how the mass is distributed perpendicular to the spin axis. The half-factor reflects the integration over the radial mass distribution. This is a workhorse formula for rolling problems, flywheel design, and any time something cylindrical is spinning around its long axis.

Two common cases: a thin rod about its centre gives I = (1/12)ML², while the same rod about one end gives I = (1/3)ML². Here L is the rod's length. So a 2 kg rod 1 m long has I = 1/6 ≈ 0.167 kg·m² about its centre, but 2/3 ≈ 0.667 kg·m² about one end. The end-axis version is exactly four times the centre version, which falls out of the parallel-axis theorem. Choosing the right axis is half the work in rotational dynamics problems. Check which axis the question wants before grabbing a formula.

A thin hollow cylinder (a hoop) rotating about its central axis has I = MR² — all its mass sits at radius R. A thick-walled hollow cylinder with inner radius R and outer radius R has I = ½M(R² + R²). The thin-hoop formula is the limiting case where R ≈ R. So a thin hoop has twice the moment of inertia of a solid disc with the same mass and outer radius — that's why hoops roll down ramps slower than discs of the same mass. Mass distributed further from the axis means more rotational inertia.

When you know the moment of inertia about an axis through the centre of mass, the parallel axis theorem gives the moment about any parallel axis: I = I_cm + Md². Here M is the total mass and d is the distance between the two axes. So a 3 kg disc with I_cm = 0.06 kg·m² rotated about an axis 0.2 m offset has I = 0.06 + 3 × 0.04 = 0.18 kg·m². This is the formula that lets you calculate moments about awkward off-centre axes without re-doing the integral. Indispensable for compound rigid bodies and pendulum problems.

Solid sphere about a diameter: I = (2/5)MR². Hollow thin spherical shell about a diameter: I = (2/3)MR². So a hollow sphere has 5/3 times the moment of inertia of a solid sphere of equal mass and radius — again, because mass on the surface sits further from the axis. A 1 kg solid sphere of 0.1 m radius gives I = 0.4 × 0.01 = 0.004 kg·m². These formulas explain why a hollow ball rolling down a ramp arrives behind a solid one — more rotational inertia means more energy goes into spinning, less into translating.

Moment of inertia carries units of kg·m² in SI: mass times length squared. Some engineering texts use slug·ft² or lb·in², which require careful unit chasing during conversions. Whatever the units, dimensional consistency with the rotational kinetic energy formula KE = ½Iω² (joules) requires I to have those mass-times-length-squared dimensions. So 1 kg·m² is a moderately big moment of inertia — equivalent to a 1 kg point mass on a 1 m string. For most everyday objects, values fall between 10³ and 10² kg·m². Anything wildly outside that range usually signals a unit error.

Moment of inertia depends not just on mass but on how that mass is distributed relative to the rotation axis. Two objects with identical mass can have very different I: a hoop of mass M and radius R has I = MR², while a solid disc of the same mass and radius has only I = ½MR². Mass close to the axis barely contributes; mass far from the axis dominates. That's why figure skaters pull their arms in to spin faster — they decrease I, and angular momentum conservation kicks in. This dependence on geometry is what makes rotational problems tricker than linear ones.