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.

What is the standard formula for a solid sphere?

For a solid uniform sphere of mass m and radius r rotating about any diameter: I = (2/5) m r². For a thin spherical shell (hollow sphere) the constant changes to 2/3, giving I = (2/3) m r². The solid sphere's constant is smaller because more of the mass is concentrated near the centre rather than at the rim.

How does the parallel axis theorem work?

The parallel axis theorem states that if you know the moment of inertia I_cm about an axis through the centre of mass, then the moment of inertia about any parallel axis displaced by distance d is I = I_cm + m d². This is extremely useful in engineering: compute I_cm once from the standard formula, then shift the axis to a bearing or pin location by adding m d². The theorem is exact for any rigid body, not an approximation.

What are the units of moment of inertia?

SI units are kg·m² (kilogram square metres). Rotational kinetic energy KE = ½ I ω² is in joules (J) when ω is in rad/s. Angular momentum L = I ω is in kg·m²/s. Torque τ = I α is in newton-metres (N·m). If you have dimensions in cm, convert to metres before calculating — a factor of 10⁻² per length becomes 10⁻⁴ in the final I value.

Why does a skater spin faster when pulling their arms in?

When no external torque acts, angular momentum L = I ω is conserved. Pulling arms inward reduces the effective radius of mass, sharply lowering I. Because L is fixed, ω must increase to compensate — the skater speeds up. The same principle governs collapsing stars (pulsars), gyroscopes, and engineering flywheels. This calculator lets you model this by comparing two configurations: one with arms out (large r) and one with arms in (small r).

What is the difference between axial and transverse moment of inertia for a cylinder?

For a solid cylinder of mass m, radius r, and length L: the axial moment (rotation about the symmetry axis, like a spinning top) is I = (1/2) m r². The transverse moment (rotation about a diameter through the mid-length, like a tumbling baton) is I = (1/12) m (3r² + L²). For a long thin rod the transverse value greatly exceeds the axial value, explaining why a rod is much harder to spin end-over-end than to spin about its own axis.

Moment of Inertia Calculator

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

Shape	Axis	Formula
Solid sphere (mass m, radius r)	Any diameter	I = (2/5) m r²
Thin spherical shell	Any diameter	I = (2/3) m r²
Solid cylinder (radius r, length L)	Symmetry axis	I = (1/2) m r²
Solid cylinder	Transverse mid-length	I = (1/12) m (3r² + L²)
Hollow cylinder (outer R, inner r)	Symmetry axis	I = (1/2) m (R² + r²)
Thin ring / hoop	Symmetry axis	I = m r²
Thin ring / hoop	Diameter	I = (1/2) m r²
Thin rod (length L)	Perp. through centre	I = (1/12) m L²
Thin rod	Perp. through end	I = (1/3) m L²
Rectangular plate (a × b)	x-axis (along a)	I = (1/12) m b²
Rectangular plate (a × b)	Normal to plate	I = (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).

Moment of inertia: I = (1/2) × 50 × 0.4² = 4 kg·m²
Rotational KE: ½ × 4 × 314.16² ≈ 197 392 J ≈ 197 kJ
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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