Moment of Inertia Calculator

Definition

For a single point mass at perpendicular distance r from the axis of rotation, the moment of inertia is

I = m \, r^2

Why r appears squared

A particle twice as far from the axis is four times harder to spin up, because both linear speed and required centripetal-direction momentum grow linearly with r. This is why figure skaters pull in their arms: reducing r is the only way to spin faster, since m can't change.

Common shapes (extended bodies)

For continuous bodies, I is an integral over the volume: I = ∫ r^2 \, dm. Resulting formulas for common geometries are summarized below (where k is the dimensionless shape factor and r is the characteristic radius):

\beginarraylll\textThin rod about center & (1)/(12) m L^2 & \\ \textThin rod about end & (1)/(3) m L^2 & \\ \textSolid disk/cylinder about center & (1)/(2) m r^2 & \\ \textHollow sphere & (2)/(3) m r^2 & \\ \textSolid sphere & (2)/(5) m r^2 & \endarray

Parallel-axis theorem

If you know I about an axis through the center of mass, you can shift to a parallel axis at distance d by adding m d²:

I = I_\mathrmcm + m \, d^2

Units and Newton's second law for rotation

I is measured in \mathrmkg· m^2. It appears in the rotational form of Newton's second law:

\tau_\mathrmnet = I \, α

Standard Shapes and Why the Coefficient Differs

Every common body has I = c\,MR^2 for some coefficient c that depends purely on how the mass is distributed relative to the axis. A thin hoop about its centre has c = 1; a solid disc or cylinder has c = 1/2; a solid sphere has c = 2/5; a thin spherical shell has c = 2/3; a rod about its centre has I = ML^2/12, rising to ML^2/3 about one end.

The pattern is that mass far from the axis counts disproportionately, because the contribution goes as distance *squared*. A hoop keeps all its mass at radius R, so it has the largest coefficient; a solid sphere concentrates much of its mass near the centre and has the smallest. This is exactly why a hoop rolls down a slope more slowly than a solid ball of identical mass and radius: more of the released potential energy goes into rotation rather than translation.

I = ∫ r^2\,dm = c\,MR^2

The Parallel-Axis Theorem

Tables almost always quote the moment of inertia about an axis through the centre of mass. To shift to any parallel axis a distance d away, add Md^2.

This one line handles most real problems. A rod pivoted at its end has I = ML^2/12 + M(L/2)^2 = ML^2/3, recovering the standard result. A compound body — a flywheel with a hub, a pendulum with a rod and a bob — is handled by computing each part about its own centre, shifting each to the common axis, and adding. Note that the correction is always positive: the centroidal axis gives the smallest moment of inertia of any parallel axis, which is why balanced machinery is mounted through its centre of mass.

I = I_\textcm + Md^2

Rotational Energy, Rolling and Flywheels

A rotating body stores K = \tfrac12I\omega^2. A body that both rolls and translates carries both terms, and for rolling without slipping v = \omega R ties them together — which is what lets you predict rolling races without solving any forces.

Flywheels exploit the same relationship in reverse. Because energy goes as \omega^2 but only linearly with I, spinning faster stores far more energy than adding mass, so modern flywheel storage uses light composite rotors at very high speed rather than heavy steel at low speed. The limit is material strength: hoop stress in the rim grows as \rho v^2, so the maximum stored energy per unit mass depends on the ratio of tensile strength to density, not on size.

K_\textrot = \tfrac12I\omega^2, \qquad K_\texttotal = \tfrac12Mv^2 + \tfrac12I\omega^2

Worked Example

Inputs

• mass: 2
• radius: 0.5

Result: momentOfInertia: 0.5 angularEquivalentMass: 0.5

Explanation

A 2 kg point mass placed 0.5 m from a pivot has moment of inertia I = 2 × (0.5)² = 0.5 kg·m². Doubling r to 1.0 m would quadruple I to 2.0 kg·m², making the same torque produce only one-quarter of the angular acceleration.

Second Scenario

Inputs

• mass: 3.5
• radius: 0.5

Result: momentOfInertia: 0.5 angularEquivalentMass: 0.5

Explanation

This scenario uses different inputs (mass = 3.5, radius = 0.5) to show how changing one variable affects the moment of inertia result. Run the calculator above with these values to get the exact updated output with step-by-step work.