Moment of Inertia Calculator – How Hard Is It to Spin?

Same mass, same size — very different spin

Every formula above is a shape factor times mass times size squared. The factor is the whole story: it says how the mass is spread out relative to the axis.

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Understand the result

Same mass, same size, very different spin

Every formula on this page is a shape factor times mass times size squared, and the factor is the whole story: it says how the mass is spread out relative to the axis. The table below puts every shape on the same footing — a mass of 2 kg and a size of 0.5 m for the round shapes, a length of 2 m for the rods — so the factors can be compared directly. Every figure was run through the calculator above, starting from its default solid sphere and switching shapes or values one at a time.

The first two rows show the simplest behaviour: doubling the mass doubles the moment of inertia and nothing else. The default sphere at 2 kg gives 0.20 kg·m², and at 4 kg it gives 0.40 kg·m², while the radius of gyration stays at 0.3162 m, because the radius of gyration depends only on how the mass is distributed, not on how much of it there is. Mass enters the formula once, linearly, so it is the least surprising input on the page: twice the mass is twice as hard to spin, whatever the shape, and any result that does not scale that way has an input error in it.

The hoop row against the sphere row shows why distribution matters more than mass. A hoop has all of its mass at the rim, so its factor is 1, the largest possible for a given radius, and at 2 kg and 0.5 m it gives 0.50 kg·m², two and a half times the sphere’s 0.20 kg·m² for the same mass and the same radius. A solid disc spreads its mass across the whole face, so its factor is 1/2, exactly half the hoop’s, and the same mass and radius give 0.25 kg·m², which the page states directly: a disc is exactly half as much as a hoop for the same mass and radius. The sphere, at 2/5, sits below the disc, because a sphere’s mass is spread through a volume rather than across a face, and more of it sits near the axis on average. The ordering — hoop 1, disc 1/2, sphere 2/5 — is the ordering of how far, on average, the mass sits from the axis, squared.

The rod rows show the largest effect on the page, and it comes from the axis position, not from the shape. A thin rod of 2 kg and 2 m has a moment of inertia of 0.6667 kg·m² about its centre, but 2.6667 kg·m² about one end: exactly four times as much, from the same rod, because the factor quadruples from 1/12 to 1/3. The page’s own example makes the same point with a heavier rod: a thin rod of 3 kg and 2 m has I = 1.00 kg·m² about its centre but I = 4.00 kg·m² about one end, again four times as much. About the end, the far half of the rod sits a long way from the axis, and the squared distance punishes that. The radius of gyration doubles as well, from 0.5774 m about the centre to 1.1547 m about the end, which is the same statement in different units: the whole mass would have to sit twice as far from the axis, concentrated at a point, to be as hard to spin as the end-pivoted rod.

The parallel axis theorem, worked twice

If your axis does not pass through the centre of mass, the parallel axis theorem adds m × d² to the centre value, where d is the distance between the axes. The theorem is the reason the rod’s end value is so much larger than its centre value, and it also handles the case the shape buttons do not: a familiar shape spinning about an unfamiliar axis. The calculator’s parallel-axis option was run for the default sphere to check it: the centre value is 0.20 kg·m², the added term for a 2 kg mass shifted by 1 m is 2 × 1², which is 2.00 kg·m², and the total about the new axis is 2.20 kg·m², with a radius of gyration about the new axis of 1.0488 m. The page’s own guide gives the same example in words: the 2 kg sphere above, moved to an axis 1 m away, gives 0.20 + 2 × 1² = 2.20 kg·m².

The added term grows with the square of the shift, which is why small shifts matter little and large shifts matter enormously. Shifting the default sphere by 0.5 m would add 2 × 0.25, which is 0.50 kg·m², for a total of 0.70 kg·m², while shifting it by 2 m would add 2 × 4, which is 8.00 kg·m², for a total of 8.20 kg·m². The centre value, 0.20, becomes a rounding error beside the shift term once the shift exceeds a metre or so, and that is the practical lesson: for a part mounted far from its spin axis, the mounting distance matters more than the part’s own shape. Engineers use the radius of gyration to compare such parts quickly, because it folds the whole distribution, centre value and shift together, into a single distance: the radius of gyration about the new axis for the shifted sphere is 1.0488 m, which means a point mass of 2 kg at 1.0488 m from the axis would be exactly as hard to spin as the shifted sphere.

The theorem also explains the rod’s centre-to-end factor of four without any new formula. A rod about its end is a rod about its centre, shifted by half its length: the centre value is (1/12) × m × L², and the shift term is m × (L/2)², which is (1/4) × m × L². Add them: (1/12 + 1/4) × m × L² is (1/12 + 3/12) × m × L², which is (4/12) × m × L², which is (1/3) × m × L², exactly the end formula. The two rod formulas are therefore not two separate facts to memorise but one formula plus one shift, and any axis position between centre and end follows the same way: the moment of inertia about an axis at distance d from the centre is (1/12) × m × L² + m × d². That single expression covers every pivot point on the rod, and the same pattern covers every shape on the page.

Radius of gyration: one number that compares any two parts

The radius of gyration, k = √(I ÷ m), is the distance from the axis at which you could concentrate the whole mass into a single point and keep the same moment of inertia. For the default 2 kg solid sphere with radius 0.5 m, k = 0.3162 m: a point mass of 2 kg at 0.3162 m from the axis would be exactly as hard to spin as the sphere is. For the hoop at the same mass and radius, k = 0.50 m, the full radius, because all the mass already sits at the rim. For the rod about its centre, k = 0.5774 m, and about its end, k = 1.1547 m, exactly double, matching the factor-of-four change in the moment of inertia, since the radius of gyration is a square root of it. The value of k is that it strips out the mass: two parts with the same k and different masses are the same shape of problem, scaled, and two parts with different k values differ in distribution however their masses compare.

That comparison power is why engineers quote k alongside I rather than instead of it. A flywheel designer choosing between a disc and a hoop of the same mass and radius does not need the full formulas to know the hoop stores more rotational energy at the same speed: its k is 0.50 m against the disc’s lower value, so its moment of inertia is higher, and it is harder to spin up and harder to stop. A designer choosing a pivot position for a rod does not need to recompute the whole formula to know the end pivot is four times as demanding as the centre pivot: the k values of 1.1547 m and 0.5774 m say it directly. And a designer checking a mounted part against its centre value only needs the shift term: if the shift distance exceeds k, the shift term dominates, and the part’s own shape is a detail. The calculator prints k for every result, and reading it beside I turns each answer from a single number into a comparison.

Worked checks: the sphere, three ways

The default sphere is the check to memorise: mass 2 kg, radius 0.5 m, factor 2/5, so I = 0.4 × 2 × 0.25, which is 0.20 kg·m², exactly as the calculator shows, with the formula printed as (2/5) × m × r² and the radius of gyration as 0.3162 m. Double the mass to 4 kg and the answer doubles to 0.40 kg·m², while the radius of gyration stays at 0.3162 m, because mass scales the moment of inertia but not its distribution. Switch the shape to a hoop at the same mass and radius and the factor becomes 1, so I = 1 × 2 × 0.25, which is 0.50 kg·m², with a radius of gyration of 0.50 m, the full radius. Switch to a rod of the same mass and a length of 2 m, pivoted at its centre, and I = (1/12) × 2 × 4, which is 0.6667 kg·m², with a radius of gyration of 0.5774 m; pivoted at its end, I = (1/3) × 2 × 4, which is 2.6667 kg·m², with a radius of gyration of 1.1547 m. Every one of those figures was run through the calculator, and every one follows from factor times mass times size squared, with no other ingredient.

The units check is the last habit worth building. Moment of inertia is measured in kilogram square metres (kg·m²): mass in kilograms times distance in metres, squared. In cgs units it is gram square centimetres, and 1 kg·m² equals 10,000,000 g·cm², as the page’s own guide states. A result in any other unit, or a result whose order of magnitude looks wrong — a flywheel at 0.0002 kg·m², or a pencil at 200 kg·m² — signals a unit slip in the inputs, most often a radius entered in centimetres as if it were metres, which understates the answer by a factor of ten thousand, because the distance is squared. The page’s disclaimer draws the remaining boundary: the calculator uses the standard textbook formulas for ideal rigid bodies with uniform density, spinning about the stated axis, so real objects with uneven density, hollow sections or off-centre axes will differ, and the result should be treated as a solid estimate, not an engineering certification.

The point mass and the disc, examined closely

The point mass is the simplest shape on the page and the most useful reference, because every other shape can be described as a point mass that has been spread out. A point mass of 2 kg at 0.5 m from the axis has a moment of inertia of 1 × 2 × 0.25, which is 0.50 kg·m², the same as the hoop, and the equality is not a coincidence: a hoop is a point mass that has been spread around a circle at the same radius, so every bit of it sits at the full distance, and the factor stays 1. The point mass is also the definition of the radius of gyration in disguise: when the page says a point mass of 2 kg at 0.3162 m would be exactly as hard to spin as the default sphere, it is saying that the sphere’s radius of gyration is 0.3162 m, and the sphere behaves, for spinning purposes, like that point mass. Any part can be reduced to its point-mass equivalent this way, which is why the radius of gyration is printed beside every result: it is the part’s behaviour, compressed into one distance.

The solid cylinder or disc sits exactly halfway between the point mass and the sphere in factor terms, and the reason is geometric. A disc’s mass is spread uniformly across a face, so the average of the squared distance from the axis, over the whole face, is half the squared radius: the centre contributes almost nothing, the rim contributes the full radius squared, and the average over the area lands at exactly one half. A sphere’s mass is spread through a volume, so more of it sits near the axis in three dimensions rather than two, and the average of the squared distance lands lower, at two fifths of the squared radius. The factors 1, 1/2 and 2/5 are therefore not arbitrary constants to memorise but averages of squared distance over different geometries: a ring, a face, and a volume. Once that is seen, the ordering can never be confused again, and a new shape’s factor can be guessed by asking how its mass is spread: the more of it near the rim, the closer the factor to 1; the more of it near the axis, the closer to 0.

Where moment of inertia actually decides something

Moment of inertia decides how hard a part is to spin up or slow down, and that single property drives design choices across engineering. A flywheel stores rotational energy in proportion to its moment of inertia times the square of its angular speed, so a flywheel designer who wants more storage without more mass moves the mass outward, toward a hoop, rather than adding mass near the hub, because the squared distance rewards outward mass far more than added mass. A vehicle designer choosing wheels faces the same trade-off in reverse: a heavier rim increases the moment of inertia that the engine must spin up and the brakes must stop, so lightweight rims are not only about total weight but about where the weight sits. A robotics designer sizing a motor for an arm must add the arm’s moment of inertia about the joint, which grows with the square of the arm’s length, so doubling the arm’s reach quadruples the torque needed to accelerate it at the same rate, before any payload is considered.

The same squared-distance rule explains why the parallel axis theorem matters in practice rather than only in textbooks. A part that spins about its own centre is one problem; the same part mounted on a shaft away from its centre is a different problem, and the difference is m × d², which grows fast. A 2 kg part mounted 1 m from its shaft carries an extra 2.00 kg·m² beyond its own centre value, which for the default sphere is ten times the centre value itself. Mounting distance is therefore often the dominant term in a real assembly, and a designer who computes only the centre value will underestimate the motor torque or the braking requirement by an order of magnitude. The calculator’s parallel-axis option exists for exactly that assembly case, and the habit of checking it — centre value, shift term, total — is the habit that separates a textbook answer from an engineering one. The page’s own disclaimer reinforces the boundary: ideal rigid bodies with uniform density are the model, real objects with uneven density, hollow sections or off-centre axes will differ, and the result is a solid estimate, not a certification.

One last check before you trust a result

Before you use any result this calculator shows, run through three checks that take less than a minute. First, confirm the shape and the axis: a rod about its centre and the same rod about its end differ by a factor of four, and a sphere with the parallel-axis option ticked differs from the same sphere without it by the shift term, so the shape buttons and the parallel-axis checkbox are the inputs to verify first. Second, confirm the units: mass in kilograms, distance in metres, and the answer in kilogram square metres, because a radius entered in centimetres understates the answer by a factor of ten thousand, the distance being squared. Third, estimate the answer from the factor: a hoop is mass times radius squared, a disc is half of that, a sphere is two fifths, and a rod about its centre is a twelfth of mass times length squared. A result that matches its factor estimate is worth trusting; one that does not is worth re-entering.

Area moment vs mass moment: same name, different physics

One is measured in m to the fourth and bends beams; the other in kg m squared and spins bodies.

Moment of inertia of composite shapes: add the parts

Split the body, shift each part to the common axis, and sum.

Frequently asked questions

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