Moment of Inertia Calculator
Find the moment of inertia for twelve common rigid bodies, including hollow cylinders, cones, tori, and rectangular plates, across their primary and secondary rotation axes. Enter mass and dimensions in metric or imperial units and get all axis results instantly, with a step-by-step worked solution.
Formula
Worked example
A solid cylinder (m = 2 kg, r = 0.5 m, h = 0.3 m): spin axis I = 0.5 x 2 x 0.5^2 = 0.25 kg·m². Tipping axis I = 2 x ((1/12) x 0.3^2 + 0.25 x 0.5^2) = 2 x (0.0075 + 0.0625) = 0.14 kg·m². If the spin axis shifts 0.2 m from the centre, parallel-axis adds 2 x 0.04 = 0.08 kg·m², giving 0.33 kg·m².
What moment of inertia means
Moment of inertia (symbol I) is the rotational equivalent of mass. Just as mass measures how strongly an object resists a change in straight-line velocity, moment of inertia measures how strongly it resists a change in its angular velocity about a chosen axis. Its SI unit is the kilogram metre squared (kg·m²). The fundamental reason it exists is that every element of mass contributes m times the square of its distance from the axis. Because that distance is squared, mass located far from the axis resists spinning up far more than the same mass packed close in. This is why a figure skater spins faster when they pull their arms in: reducing the spread of their mass reduces the moment of inertia, and angular momentum conservation forces their angular velocity up.
Shape-by-shape formulas and what they mean
For an idealised point mass at distance r the whole mass sits at one radius, giving I = m r^2. Spreading mass into a solid disk reduces the effective distance so the coefficient drops to 1/2. A thin hoop (all mass at the rim) keeps the full coefficient of 1 because no mass is closer to the axis than r. A solid sphere integrates to 2/5 m r^2, a hollow spherical shell to 2/3 m r^2 (higher, because its mass can only sit at the outer surface). A solid cylinder spinning about its long axis behaves like a stack of disks: I = 1/2 m r^2. When the same cylinder tips about a perpendicular axis through its centre, both the radius and the height contribute: I = m((1/12)h^2 + (1/4)r^2). A hollow cylinder or pipe adds an inner radius term: I_spin = 1/2 m (r1^2 + r2^2). A cone about its central axis gives I = 3/10 m r^2. A torus (donut) about its symmetry axis gives I = m(R^2 + (3/4)a^2), where R is the major radius and a the minor (tube cross-section) radius.
The parallel-axis theorem
The formulas above all assume the rotation axis passes through the body's centre of mass. When you need the moment of inertia about a different, parallel axis, the parallel-axis theorem applies: I_new = I_cm + m d^2, where d is the perpendicular distance between the two axes. The correction is always positive, so the centre-of-mass axis always gives the minimum moment of inertia for any given direction. Toggle "Apply parallel-axis theorem" in this calculator and enter the shift distance d to see the new value. A common example is a rod rotating about one end: its centre value is (1/12)mL^2 but the end value is (1/3)mL^2, which is exactly (1/12)mL^2 + m(L/2)^2 = (1/12 + 1/4)mL^2 = (1/3)mL^2.
Angular dynamics: torque, kinetic energy, and angular acceleration
Once you know the moment of inertia you can solve the three linked equations of rotational dynamics. Torque (tau) equals I times angular acceleration (alpha): tau = I * alpha. Rotational kinetic energy equals (1/2) I omega^2, where omega is angular velocity in rad/s. Angular momentum is L = I * omega. Enable "Compute angular dynamics" and enter omega and alpha to have the calculator evaluate kinetic energy and torque alongside the inertia. These outputs update live as you adjust the shape, mass, or dimensions, making it easy to see how design changes flow through to forces and energy.
Metric and imperial units
All dimension inputs accept metres, centimetres, millimetres, feet, or inches; the mass input accepts kilograms, grams, or pounds. Every value is converted to SI internally before the formula runs, so the output is always in kg·m². If you need the result in a non-SI unit, note that 1 kg·m^2 = 23.73 lb·ft^2 = 8.851 lb·ft·s^2. The show-your-work steps panel always prints the converted SI values so you can verify the calculation.
Moment of inertia of common rigid bodies
| Shape | Axis | Formula | Coefficient k (I = k m dim^2) |
|---|---|---|---|
| Point mass | Perpendicular, at distance r | I = m r^2 | 1 |
| Thin hoop | Spin (z-axis through centre) | I = m r^2 | 1 |
| Thin hoop | Diameter (x or y axis) | I = (1/2) m r^2 | 0.5 |
| Solid disk | Spin (through centre) | I = (1/2) m r^2 | 0.5 |
| Solid disk | Diameter | I = (1/4) m r^2 | 0.25 |
| Solid sphere | Through centre | I = (2/5) m r^2 | 0.4 |
| Hollow sphere (thin shell) | Through centre | I = (2/3) m r^2 | 0.6667 |
| Thin rod | Through centre (perp. to rod) | I = (1/12) m L^2 | 0.0833 |
| Thin rod | Through one end (perp.) | I = (1/3) m L^2 | 0.3333 |
| Solid cylinder | Spin (along height) | I = (1/2) m r^2 | 0.5 |
| Solid cylinder | Tipping (perp. to height) | I = m((1/12)h^2 + (1/4)r^2) | varies |
| Hollow cylinder | Spin (along height) | I = (1/2) m (r1^2+r2^2) | varies |
| Solid cone | Spin (tip to base centre) | I = (3/10) m r^2 | 0.3 |
| Torus | Spin (symmetry axis) | I = m(R^2 + (3/4)a^2) | varies |
| Rectangular plate | Iz (through centre, perp.) | I = (1/12) m (a^2+b^2) | varies |
All formulas assume uniform density and the stated rotation axis. I_cm denotes rotation through the centre of mass.
Frequently asked questions
What are the units of moment of inertia?
The SI unit is the kilogram metre squared (kg·m^2). It combines mass (kg) and squared length (m^2) because every mass element contributes its mass times the square of its distance from the axis. If you work in imperial units, 1 kg·m^2 equals approximately 23.73 lb·ft^2 or 8.851 lb·ft·s^2.
Why does a solid sphere have a smaller coefficient than a hollow sphere?
A solid sphere (I = 2/5 m r^2, coefficient 0.4) spreads its mass from the centre all the way to the outer surface, so much of its mass sits at distances well below r. A thin hollow sphere (I = 2/3 m r^2, coefficient 0.667) concentrates all its mass at the outer radius, so on average each bit of mass is farther from the axis. The farther the mass, the larger the moment of inertia.
How do I use the parallel-axis theorem?
The parallel-axis theorem lets you shift the rotation axis from the centre of mass to any parallel axis at perpendicular distance d away. The new inertia is I_new = I_cm + m d^2. Toggle "Apply parallel-axis theorem" in the calculator, enter d, and the shifted value appears as an extra output. A classic check: a thin rod about its end equals (1/12)mL^2 + m(L/2)^2 = (1/3)mL^2.
What is the difference between the spin axis and the tipping axis for a cylinder?
The spin axis runs along the height of the cylinder through its central axis, like the axle of a wheel. Spinning about this axis gives I = (1/2)m r^2. The tipping axis is perpendicular to the height and passes through the centre of mass, like rolling the cylinder over. This axis gives I = m((1/12)h^2 + (1/4)r^2) because both the height and the radius contribute. A squat, wide cylinder resists tipping more than a tall, narrow one of the same mass.
How does moment of inertia relate to torque and kinetic energy?
Three equations connect them. Torque (N·m) equals I times angular acceleration (alpha in rad/s^2): tau = I * alpha. Rotational kinetic energy (J) equals (1/2) I omega^2, where omega is angular velocity in rad/s. Angular momentum (kg·m^2/s) equals I * omega. Enable "Compute angular dynamics" and supply omega and alpha to compute KE and torque directly from the inertia this calculator finds.
Why is the moment of inertia of a rod about its end four times larger than about its centre?
About its centre the rod gives I = (1/12)mL^2. About one end it gives I = (1/3)mL^2. The ratio 1/3 / (1/12) = 4. This follows from the parallel-axis theorem: shifting the axis by d = L/2 adds m(L/2)^2 = (1/4)mL^2, and (1/12 + 1/4)mL^2 = (1/3)mL^2. The mass near the far end is now much farther from the axis, which drives the large increase.