Boneyard Tools

Moment of Inertia Calculator

Choose one of six standard rigid bodies, type its mass in kilograms and the relevant radius or length in metres, and read the moment of inertia in kg x m^2 alongside the exact formula applied. The calculator switches between a radius field (cylinder, hoop, spheres) and a length field (rods) automatically, and results update as you type. Each formula assumes rotation about the standard textbook axis for that shape.

How to use the moment of inertia calculator

  1. Open the 'Shape and axis' menu and pick your body: solid cylinder or disk, hoop, solid sphere, hollow sphere, rod about centre, or rod about end.
  2. Type the mass in kilograms into the 'Mass (m)' field.
  3. Enter 'Radius (r)' in metres for the round shapes, or 'Length (L)' in metres for a rod.
  4. Read the moment of inertia in kg x m^2 in the highlighted box, with the formula shown beneath it.
  5. Click Copy to save the value and the formula together.

Examples

Solid cylinder or disk

shape = solid cylinder, m = 5 kg, r = 0.2 m
I = 1/2 m r^2 = 0.1 kg x m^2

Solid sphere

shape = solid sphere, m = 10 kg, r = 0.3 m
I = 2/5 m r^2 = 0.36 kg x m^2

Rod about its end

shape = rod about end, m = 3 kg, L = 2 m
I = 1/3 m L^2 = 4 kg x m^2

Frequently asked questions

What is moment of inertia?

Moment of inertia is the rotational counterpart of mass: it measures how much an object resists a change in its spin. It grows with both the total mass and, more sharply, with how far that mass sits from the rotation axis, since each bit of mass contributes its mass times the square of its distance.

Which shapes and axes does the calculator support?

Six standard cases: a solid cylinder or disk (I = 1/2 m r^2), a thin hoop or ring (m r^2), a solid sphere (2/5 m r^2), a hollow spherical shell (2/3 m r^2), a thin rod about its centre (1/12 m L^2), and a thin rod about one end (1/3 m L^2). Each uses the common symmetry axis for that body.

Why does a hoop have a larger value than a disk?

All of a hoop's mass sits out at the rim, the farthest point from the axis, so it uses the full I = m r^2. A disk spreads its mass from the centre outward, so the average squared distance is smaller and it gives I = 1/2 m r^2, exactly half the hoop for the same mass and radius.

Why is a rod about its end larger than about its centre?

About the centre the mass is split evenly on both sides and stays relatively close, giving 1/12 m L^2. Pivot at one end and much more of the rod lies far from the axis, giving 1/3 m L^2, which is exactly four times larger for the same rod.

What units should I use?

Enter mass in kilograms and the radius or length in metres, the SI base units. The answer then comes out in kilogram metre squared, written kg x m^2. If you have grams or centimetres, convert them first, since the calculator does not rescale units.

Does it use the thin-rod and thin-hoop idealisations?

Yes. The rod formulas assume a thin rod whose thickness is negligible next to its length, and the hoop assumes all mass at a single radius. For a thick disk, a ring with real width, or a rod with a large cross-section the true value is slightly different, though usually close.

Can I get the moment of inertia about a shifted or parallel axis?

Not directly. These are the standard central or end axes only. To move to a parallel axis a distance d away, apply the parallel-axis theorem yourself: I_new = I + m d^2, using the value this tool gives as the starting I.

What happens if I enter zero or a negative number?

Mass, radius and length must each be greater than zero, so a zero or negative entry returns a short error instead of a result. Non-numeric text is also rejected. Fix the value and the result reappears instantly.

How do I use the result in a physics problem?

The value plugs straight into rotational equations: rotational kinetic energy is 1/2 I omega^2, angular momentum is I omega, and torque equals I times angular acceleration. Keeping I in kg x m^2 and angles in radians keeps those formulas in consistent SI units.

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