Moment of Inertia Calculator

Compute I for 11 shapes, solve rotational KE and angular momentum — with the parallel axis theorem built in.

Free moment of inertia calculator. Covers 11 rigid-body shapes (solid sphere, hollow sphere, solid and hollow cylinder, thin ring, flat disk, rectangular plate, thin rod, point mass) plus the parallel axis theorem. Enter mass and dimensions to get I in kg·m², then derive rotational kinetic energy (½Iω²) and angular momentum (Iω). Runs 100% client-side. It runs free in your browser on Gera Tools, with nothing uploaded.

Last updated Source: Gera Tools

What is the moment of inertia?

The moment of inertia I is the rotational analogue of mass. It measures how strongly a rigid body resists angular acceleration about a given axis. A larger I means more torque is required to achieve the same angular acceleration, following τ = I α (Newton's second law for rotation). Unlike mass, I depends on both the body's mass distribution AND the chosen rotation axis, so the same object can have very different I values about different axes.

The moment of inertia (rotational inertia) is one of the most fundamental quantities in classical mechanics. Just as mass determines how strongly a body resists linear acceleration under a force, the moment of inertia I determines how strongly it resists angular acceleration under a torque. Formally,

τ = I α

where τ is the net torque (N·m), I is the moment of inertia (kg·m²), and α is the angular acceleration (rad/s²). This is Newton’s second law applied to rotation. The moment of inertia appears in rotational kinetic energy, angular momentum, the period of a physical pendulum, and the design of flywheels, gyroscopes, turbines, and vehicle drivetrains.

The defining integral

For a continuous rigid body the moment of inertia about an axis is

I = ∫ r² dm

where r is the perpendicular distance from the mass element dm to the axis. For a discrete collection of point masses this becomes I = Σ mᵢ rᵢ². The integral is evaluated once for each canonical shape and axis combination; the results are the closed-form formulas tabulated below and implemented in this calculator.

Shapes and formulas

ShapeAxisFormula
Solid sphere (mass m, radius r)Any diameterI = (2/5) m r²
Thin spherical shellAny diameterI = (2/3) m r²
Solid cylinder (radius r, length L)Symmetry axisI = (1/2) m r²
Solid cylinderTransverse mid-lengthI = (1/12) m (3r² + L²)
Hollow cylinder (outer R, inner r)Symmetry axisI = (1/2) m (R² + r²)
Thin ring / hoopSymmetry axisI = m r²
Thin ring / hoopDiameterI = (1/2) m r²
Thin rod (length L)Perp. through centreI = (1/12) m L²
Thin rodPerp. through endI = (1/3) m L²
Rectangular plate (a × b)x-axis (along a)I = (1/12) m b²
Rectangular plate (a × b)Normal to plateI = (1/12) m (a² + b²)

The parallel axis theorem extends any of these: I = I_cm + m d², where d is the distance from the centre-of-mass axis to the new parallel axis.

Rotational dynamics tab

Once I is computed, the calculator provides three derived quantities for a given angular velocity ω (rad/s):

  • Rotational kinetic energy: KE = ½ I ω² (joules)
  • Angular momentum: L = I ω (kg·m²/s)
  • Torque for α = 1 rad/s²: τ = I · 1 (N·m) — handy for motor sizing

Worked example — flywheel energy storage

A steel flywheel modelled as a solid disk has mass 50 kg and radius 0.4 m. It spins at 3 000 rpm (≈ 314.16 rad/s).

  1. Moment of inertia: I = (1/2) × 50 × 0.4² = 4 kg·m²
  2. Rotational KE: ½ × 4 × 314.16² ≈ 197 392 J ≈ 197 kJ
  3. Angular momentum: 4 × 314.16 ≈ 1 257 kg·m²/s

To mount the disk on bearings located 0.6 m from the shaft centre (e.g. the combined axis of a larger assembly), apply the parallel axis theorem:

I_new = 4 + 50 × 0.6² = 4 + 18 = 22 kg·m²

The bearing axis sees more than five times the rotational resistance. Entering those values in the “Parallel Axis Theorem” preset confirms the result instantly.

Conservation of angular momentum

Where no external torque acts, L = I ω = constant. This is why a spinning figure skater accelerates when pulling their arms inward (reducing I, so ω increases), why a collapsing stellar core forms a rapidly spinning pulsar, and why a gyroscope resists reorientation. The interplay between shape-dependent I and conserved L is at the heart of rigid-body rotational dynamics.

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