Angular Momentum Calculator

Enter the moment of inertia I in kg·m² and the angular speed ω in radians per second. The calculator finds angular momentum L = I·ω plus rotational kinetic energy E = ½Iω², revolutions per minute, and the average torque needed to stop the body in one second. For I = 2 kg·m² and ω = 5 rad/s the result is 10 kg·m²/s. The model does not compute L = r×p for a point mass or solve conservation I₁ω₁ = I₂ω₂ as the primary result.

Inputs

Result

Enter the moment of inertia and the angular speed. The angular momentum, the rotational energy, and the L(ω) chart will appear here.

For the moment of inertia of a point mass, use the I = m·r² calculator; for angular speed from the period, ω = 2π/T; and for the rotational energy of the same body, E = ½Iω².

How it works

How L is computed for I = 2 kg·m² and ω = 5 rad/s
I2 kg·m²
× ω5 rad/s
L10 kg·m²/s
I ω L = I·ω kg·m²/s ← kg·m², rad/s
The angular momentum of a body spinning about an axis is the product of the moment of inertia and the angular speed: L = I·ω. Not to be confused with the energy ½Iω².

Angular momentum is the rotational counterpart of linear momentum. For a point mass it is defined as the cross product L = r × p; for a rigid body spinning about a symmetry axis, summing over all points gives L = I·ω, where I is the moment of inertia about that axis and ω the angular speed. The unit is kg·m²/s, equivalently J·s or N·m·s. Angular momentum is conserved when no external torque acts on the body — one of the fundamental conservation laws of physics (Halliday, Resnick, Walker; Young, Freedman).

The calculator works with two numbers: the moment of inertia I in kg·m² and the angular speed ω in rad/s. The result is their product. For 2 kg·m² and 5 rad/s you get L = 10 kg·m²/s. For a bicycle wheel with I = 0.1 kg·m² spinning at ω = 23.8 rad/s (19 mph) the result is 2.38 kg·m²/s, and for a skater with I = 3 kg·m² at two revolutions per second, 37.7 kg·m²/s.

What the result panel shows. Besides the angular momentum in kg·m²/s, the calculator reports the same value in J·s, the revolutions per minute and frequency corresponding to ω, the rotational kinetic energy ½Iω² (25 J in the example), the average torque that would stop the body in one second (M = L/Δt, here 10 N·m), the entered I and ω, and the angular momentum at doubled angular speed — twice as large, because L is linear in ω. The chart draws the straight line L(ω) from ¼ω to 2ω at fixed I.

Limits of the model. The scalar formula L = I·ω applies to rotation about a symmetry axis or another principal axis of the body; for an arbitrary axis the angular momentum is a vector not parallel to ω and requires the inertia tensor. The calculator does not compute the moment of inertia from shape (that is the I = m·r² calculator), does not add the angular momenta of several bodies, and ignores the translational motion of the center of mass.

The unit switch in the header changes the moment of inertia to lb·ft² and the result to lb·ft²/s; the angular speed stays in rad/s, because the radian is shared by both systems. The values in the text stay in SI units as a reference. The moment of inertia and the angular speed must be greater than zero.

Examples on this page: I = 2 kg·m², ω = 5 rad/s → 10 kg·m²/s; a bicycle wheel → 2.38 kg·m²/s; a skater → 37.7 kg·m²/s; an engine flywheel of 0.15 kg·m² at 314 rad/s → 47.1 kg·m²/s; a helicopter rotor of 2,000 kg·m² at 40 rad/s → 80,000 kg·m²/s; the Earth about its own axis → 5.86·10³³ kg·m²/s. The “Fill this example” button enters the same numbers into the fields and refreshes the result panel and the chart.

Formula

L = I · ω

[L] = kg·m² · rad/s = kg·m²/s = J·s = N·m·s; ω = 2πf, E = ½Iω² = L²/(2I), M = dL/dt

Model: angular momentum of a rigid body about a principal axis (L. Euler, equations of rigid-body motion, 1750–1765; for a point L = r × p). The rearrangements I = L/ω and ω = L/I follow from the same formula, but this calculator solves only for L. Order of computation: I·ω in SI units, then ½Iω², revolutions per minute, and the torque L/(1 s).

Theory, derivation, and domain

Definition of angular momentum: from a point to a body

The angular momentum of a point mass about a chosen reference point is the cross product of its position vector and its momentum: L = r × p = r × m·v. Its magnitude is r·m·v·sin φ, where φ is the angle between r and v; for circular motion φ = 90°, so L = m·v·r = m·r²·ω. The vector L is perpendicular to the plane of motion and points along the axis of rotation according to the right-hand rule.

For a rigid body we add the angular momenta of all its points. When the body spins about a symmetry axis, every point contributes mᵢrᵢ²ω in the same direction, and the sum gives L = (Σmᵢrᵢ²)·ω = I·ω. The unit kg·m²/s is identical to J·s — the same unit in which the Planck constant h = 6.626·10⁻³⁴ J·s, the elementary quantum of angular momentum in atomic physics, is expressed.

Historical note: Kepler, Euler, Rankine, and Noether

The conservation of angular momentum is already hidden in Kepler’s second law (1609): the radius vector of a planet sweeps out equal areas in equal times, which is equivalent to the constancy of L = m·r²·ω along the orbit. Newton derived this law in the Principia (1687) for any central force. Leonhard Euler, between 1750 and 1765, formulated the general equations of rigid-body motion, in which the time derivative of angular momentum equals the torque, and introduced the moment of inertia.

The term moment of momentum was popularized by William Rankine in his Manual of Applied Mechanics (1858); the phrase angular momentum appeared somewhat later. A deeper justification of the conservation law was given by Emmy Noether in 1918: every continuous symmetry of the laws of physics corresponds to a conserved quantity, and the conservation of angular momentum follows from the invariance of the laws under rotations of space.

Deriving L = I·ω and the link to energy

Step 1. A point of mass mᵢ at distance rᵢ from the axis moves at speed vᵢ = ω·rᵢ, perpendicular to the radius. Its angular momentum about the axis is Lᵢ = rᵢ·mᵢ·vᵢ = mᵢ·rᵢ²·ω. Step 2. Sum over all points of the body: L = Σmᵢrᵢ²·ω = ω·Σmᵢrᵢ². The sum Σmᵢrᵢ² is the moment of inertia I, so L = I·ω.

Step 3. The rotational kinetic energy is E = ½Iω²; substituting ω = L/I gives E = L²/(2I) — at constant L the energy grows as I shrinks. Step 4. For I = 2 kg·m² and ω = 5 rad/s: L = 10 kg·m²/s, E = ½·2·25 = 25 J. Step 5. Newton’s second law for rotation, M = dL/dt, says that stopping this body in 1 s needs an average torque of 10 N·m, and in 0.1 s, 100 N·m.

Conservation of angular momentum

If the net external torque on a system is zero, its total angular momentum stays constant: I₁ω₁ = I₂ω₂. Hence the skater’s spin: pulling in the arms reduces I from about 3 to 1 kg·m², so ω triples, from 12.6 to 37.7 rad/s. The same mechanism speeds up a diver in the tuck position, a neutron star after the collapse of a stellar core (rotation periods of milliseconds), and a falling cat, which reorients itself without any external torque by counter-rotating the front and rear of its body.

Conservation of angular momentum also explains the stability of a gyroscope and of a bicycle wheel: the vector L tends to keep its direction in space, and an external torque — from the weight, for example — does not topple the axis but makes it precess at angular speed Ω = M/L. In the Earth–Moon system the tides transfer angular momentum from the Earth’s spin to the Moon’s orbit: the day lengthens by about 2 ms per century and the Moon recedes by 3.8 cm per year, while the sum of the two angular momenta stays constant.

Assumptions: principal axis, rigid body, L versus E

The scalar formula L = I·ω is exact when the body spins about one of its principal axes — for example the symmetry axis of a wheel, a cylinder, or a sphere. For rotation about an arbitrary axis the angular momentum is the vector L = ηω, where Î is the inertia tensor, and in general it is not parallel to ω; this shows up as the vibration of an unbalanced wheel and the need to balance rotors. The calculator assumes a principal axis and a rigid body with constant I.

It pays to distinguish angular momentum from rotational energy. Both depend on I and ω, but L = I·ω is linear in ω, has the unit kg·m²/s, and is a vector, while E = ½Iω² is quadratic, has the unit joule, and is a scalar. Two wheels with the same L can have different energies: at fixed L the energy is L²/(2I), so the wheel with the smaller I has more energy. The result panel shows both quantities side by side so that the difference is visible.

Units, orders of magnitude, and applications

The SI unit of angular momentum is kg·m²/s; J·s and N·m·s are dimensionally identical, and US units use lb·ft²/s (1 kg·m²/s = 23.73 lb·ft²/s). The calculator shows kg·m²/s and J·s, and after the unit switch lb·ft²/s. Orders of magnitude: a hard-drive platter about 0.02 kg·m²/s, a bicycle wheel 2–3 kg·m²/s, a skater in a spin 30–40 kg·m²/s, a helicopter rotor of the order of 10⁵ kg·m²/s, the Earth about its own axis 5.9·10³³ kg·m²/s, the Moon in its orbit 2.9·10³⁴ kg·m²/s — five times more than the Earth’s spin.

In engineering the angular momentum of a rotor governs navigation gyroscopes, reaction wheels in satellites (changing their ω turns the satellite without propellant), and the reaction torque in helicopters, which the tail rotor counteracts. In atomic physics angular momentum is quantized in units of ħ = h/2π = 1.055·10⁻³⁴ J·s — the same kg·m²/s this calculator displays, only thirty-four orders of magnitude smaller than in the bicycle-wheel example.

Sources

The definition, derivation, and applications come from mechanics textbooks and historical sources: D. Halliday, R. Resnick, J. Walker, Fundamentals of Physics, chapter on rolling, torque, and angular momentum; H. D. Young, R. A. Freedman, University Physics, chapter “Dynamics of Rotational Motion” (angular momentum, conservation, gyroscopes); R. P. Feynman, R. B. Leighton, M. Sands, The Feynman Lectures on Physics, vol. I, chapters 18–20 (rotation, angular momentum, the gyroscope); L. Euler, Theoria motus corporum solidorum seu rigidorum, 1765; W. J. M. Rankine, A Manual of Applied Mechanics, 1858; E. Noether, “Invariante Variationsprobleme,” Nachrichten der Gesellschaft der Wissenschaften zu Göttingen, 1918; J. Kepler, Astronomia nova, 1609; BIPM, SI Brochure, 9th edition. This page is for educational purposes.

How to use

  1. Enter the moment of inertia I in kg·m², for example 2. For a point mass or hoop compute I = m·r² in the moment of inertia calculator.
  2. Enter the angular speed ω in rad/s, for example 5. Multiply revolutions per minute by 2π/60 ≈ 0.1047; convert a period T with ω = 2π/T.
  3. Click Calculate or wait for the live result. For 2 kg·m² and 5 rad/s the calculator shows 10 kg·m²/s.
  4. Check the panel: revolutions per minute, the rotational energy ½Iω², the torque that stops the body in one second, and the angular momentum at doubled ω.
  5. The unit switch in the header changes I to lb·ft² and the result to lb·ft²/s. The Clear button removes the data and hides the panel.

Symbol glossary

L is the angular momentum, I the moment of inertia, ω the angular speed. For 2 kg·m² and 5 rad/s the result is 10 kg·m²/s.

L
Angular momentum [kg·m²/s = J·s], L = I·ω; a vector along the axis of rotation.
I
Moment of inertia [kg·m²]; Σm·r² about the axis.
ω
Angular speed [rad/s]; 1 rpm = 0.1047 rad/s.
E
Rotational kinetic energy [J], ½Iω² = L²/(2I).
M
Torque [N·m], M = dL/dt; stopping in time Δt needs M = L/Δt.
r × p
Definition for a point: the cross product of position and momentum.
ħ
Reduced Planck constant, 1.055·10⁻³⁴ J·s — the quantum of angular momentum in an atom.

Real-life examples

Example 1

Textbook example: I = {{u:2:inertia:0}}, ω = 5 rad/s, just under 48 rpm. This is the reference value for this page.

L = 10 kg·m²/s

Example 2

A body with I = {{u:0.5:inertia:1}} at ω = 10 rad/s. The moment of inertia is smaller and the speed higher; L is half the value in the first example.

L = 5 kg·m²/s

Example 3

A bicycle wheel with I ≈ {{u:0.1:inertia:1}} at 19 mph, that is ω ≈ 23.8 rad/s. This angular momentum steadies the bike while riding.

L ≈ 2.38 kg·m²/s

Example 4

A figure skater spinning with arms out: I ≈ {{u:3:inertia:0}}, two revolutions per second (ω = 12.57 rad/s). After pulling the arms in, I drops to 1 kg·m² and ω rises to 37.7 rad/s, while L stays the same.

L ≈ 37.7 kg·m²/s

Example 5

An engine flywheel: I ≈ {{u:0.15:inertia:2}} at 3,000 rpm (ω = 314.2 rad/s). Stopping it in one second would take a torque of 47 N·m.

L ≈ 47.1 kg·m²/s

Example 6

The Earth about its own axis: I ≈ 8.04·10³⁷ kg·m², ω = 7.292·10⁻⁵ rad/s. Tidal friction transfers part of this angular momentum to the Moon and lengthens the day by about 2 ms per century.

L ≈ 5.86·10³³ kg·m²/s

Example 7

A hard-drive platter: I ≈ 2.2·10⁻⁵ kg·m² at 7,200 rpm (ω = 754 rad/s). That is why a laptop with a running drive resists slightly when turned quickly.

L ≈ 0.0166 kg·m²/s

Example 8

A toy gyroscope: a rotor with I ≈ 4.5·10⁻⁵ kg·m² spun up to ω = 1,000 rad/s (9,550 rpm). That is enough angular momentum for the gyroscope to balance on the tip of a pencil.

L = 0.045 kg·m²/s

Example 9

The main rotor of a light helicopter: I ≈ {{u:2000:inertia:0}} at ω = 40 rad/s (382 rpm). The reaction torque to this spin is balanced by the tail rotor.

L = 80,000 kg·m²/s

Example 10

A merry-go-round with children: I ≈ {{u:2000:inertia:0}}, one turn every 12.6 s (ω = 0.5 rad/s). The moment of inertia is large, and the angular speed is low.

L = 1,000 kg·m²/s

Example 11

A washing-machine drum in the spin cycle: I ≈ {{u:0.3:inertia:1}} at 1,200 rpm (ω = 125.7 rad/s). This angular momentum shakes the machine when the load is unbalanced.

L ≈ 37.7 kg·m²/s

Example 12

The Moon in its orbit around Earth as a point mass: I = m·r² ≈ 1.086·10⁴⁰ kg·m², ω = 2.66·10⁻⁶ rad/s (period 27.3 days). That is about five times the angular momentum of the Earth’s spin.

L ≈ 2.89·10³⁴ kg·m²/s

Frequently asked questions

What does this calculator compute, and from which formula?

The calculator finds the angular momentum of a body spinning about an axis from L = I·ω: it multiplies the moment of inertia by the angular speed. For 2 kg·m² and 5 rad/s the result is 10 kg·m²/s, that is 10 J·s.

Which units should I use for the inputs?

The moment of inertia in kg·m² and the angular speed in radians per second; the result is in kg·m²/s and J·s. Multiply revolutions per minute by 0.1047, and convert a period T with ω = 2π/T. After switching the header to US units the I field takes lb·ft², and the result is in lb·ft²/s.

How does angular momentum differ from rotational kinetic energy?

Angular momentum L = I·ω is linear in ω, has the unit kg·m²/s, and is a vector along the axis of rotation. Energy E = ½Iω² is quadratic in ω, has the unit joule, and is a scalar. They are linked by E = L²/(2I); the result panel shows both quantities.

Why does a skater speed up when pulling in the arms?

Because without an external torque angular momentum is conserved: I₁ω₁ = I₂ω₂. Reducing the moment of inertia from 3 to 1 kg·m² triples the angular speed, from 12.6 to 37.7 rad/s. The rotational energy triples as well — the muscles supply it.

Where do I get the moment of inertia?

For a point mass or thin hoop I = m·r², for a solid disk ½m·r², for a solid sphere ⅖m·r², for a rod about its center 1/12·m·L². The value for a point mass has its own moment of inertia calculator; for composite bodies the moments of the parts add up.

What does “torque to stop in 1 s” mean?

It is the average torque that would bring the angular momentum to zero within one second: M = L/Δt. It follows from Newton’s second law for rotation, M = dL/dt. For L = 10 kg·m²/s that is 10 N·m; stopping in 0.1 s would need 100 N·m.

Is angular momentum a vector?

Yes. The vector L lies along the axis of rotation, with its sense given by the right-hand rule. The calculator reports the magnitude of that vector for rotation about a principal axis of the body. For an arbitrary axis L is not parallel to ω and the inertia tensor is needed.

Why does a gyroscope not fall over under its own weight?

Because the torque from the weight changes the direction of the vector L, not its length: the axis does not drop but turns horizontally — it precesses — at angular speed Ω = M/L. The larger the rotor’s angular momentum, the slower the precession and the steadier the axis.

Does the formula work for a planet in orbit?

Yes, if the planet is treated as a point: L = m·r²·ω, that is I = m·r². The Moon in its orbit has L ≈ 2.89·10³⁴ kg·m²/s, five times more than the Earth about its own axis. Kepler’s second law is precisely the conservation of this angular momentum.

What are the joule-seconds in the panel?

The same unit as kg·m²/s written differently: 1 J·s = 1 kg·m²/s² · s = 1 kg·m²/s. The Planck constant h = 6.626·10⁻³⁴ J·s, the quantum of angular momentum in atomic physics, is expressed in this unit.

Does the calculator include the translational motion of the body?

No. The result is the angular momentum of the spin about the body’s own axis. A body moving as a whole also has an orbital angular momentum r × p about the reference point; the total angular momentum is the sum of both terms, which this calculator does not compute.

Why is there no result for I = 0 or ω = 0?

A body without a moment of inertia does not exist, and at ω = 0 there is no rotation, so the angular momentum is zero — there is nothing to show. The calculator requires I > 0 and ω > 0, and until then it displays a message instead of a result.

How do I calculate angular momentum?

Use L = I·ω. At I = 2 kg·m² and ω = 5 rad/s you get 10 kg·m²/s. The panel also shows E = ½Iω² = 25 J.

What is L at I = 0.5 kg·m² and ω = 12 rad/s?

6 kg·m²/s. L = 0.5·12 = 6, and rotational energy is ½·0.5·144 = 36 J.

How does L differ from energy ½Iω²?

L grows linearly with ω and has unit kg·m²/s; energy grows with ω squared and is in joules. They link by E = L²/(2I).

Is this the L = mvr formula for a point?

Not as the primary result. Here L = I·ω for a body about an axis. For a point on a circle I = mr², so L = mr²ω = mvr when v = ωr — a separate path.

Does the calculator solve angular-momentum conservation?

No. It computes L from given I and ω. The rule I₁ω₁ = I₂ω₂ (skater) follows from the same L, but the new ω at a changed I must be worked out separately.

The result is the product L = I·ω: for I = 2 kg·m² and ω = 5 rad/s it equals 10 kg·m²/s (237 lb·ft²/s), with a rotational energy ½Iω² of 25 J. For the energy of the same body, use the E = ½Iω² calculator.

Knowledge sources

The definition, unit, history, and derivation below come from recognized physics and metrology sources.

Page updated November 3, 2026.