Angular Momentum Calculator

Angular Momentum Calculator Guide: Rigid Bodies, Particles, Moment of Inertia, Angular Velocity, and Conservation

Angular momentum is the rotational counterpart of linear momentum and describes rotational motion relative to a specified point or axis.

This calculator implements two common angular-momentum models: a rigid-body model and a point-particle model.

For a rigid body rotating about an appropriate fixed axis, angular momentum along the rotation axis can be written L = Iω, where I is the moment of inertia about that axis and ω is angular velocity. The current calculator uses this relationship directly.

For a point particle, the general angular momentum is the vector cross product L = r × p, where r is the particle’s position vector relative to the chosen origin and p = mv is its linear momentum. OpenStax gives this as the fundamental particle definition.

The magnitude of particle angular momentum is L = rp sinθ = mvr sinθ, where θ is the angle between the position vector and the particle’s velocity or momentum.

The calculator’s simpler particle equation L = mvr therefore assumes perpendicular geometry, θ = 90°, so sinθ = 1. This is appropriate for a particle whose velocity is tangent to a circular path around the chosen origin.

That assumption should remain visible in the interface because mvr is not the universal particle angular-momentum formula.

Angular momentum depends on the reference point or rotation axis. A moving particle can have nonzero linear momentum and zero angular momentum about one origin if its line of motion passes directly through that origin.

For rigid bodies, moment of inertia plays a role analogous to mass in linear momentum. However, moment of inertia depends on how mass is distributed relative to the rotation axis, so the same object can have different angular momentum at the same angular velocity when evaluated about different axes.

Angular momentum is a vector quantity. For fixed-axis rotation its direction is along the rotation axis according to the right-hand rule. In one-dimensional rotational calculations, positive and negative signs can encode the selected axial direction.

Net external torque determines the rate of change of angular momentum. OpenStax states the rotational relation dL/dt = Στ. If the net external torque is zero, total angular momentum of the system is conserved.

This conservation law explains phenomena such as a figure skater spinning faster when pulling the arms inward. Reducing moment of inertia while angular momentum remains approximately constant requires angular velocity to increase.

The calculator solves the selected angular-momentum relationship but does not automatically calculate arbitrary moment-of-inertia geometry, resolve vector cross products, or determine whether external torque is truly zero.

How to Calculate Angular Momentum for Rigid Bodies and Point Particles

Choose Rigid Body or Particle: Use Rigid Body for a rotating object described by moment of inertia and angular velocity. Use Particle for the perpendicular mvr case.
Choose the variable to solve: Select angular momentum or another supported variable such as I, ω, m, v, or r.
Specify the reference axis or origin: Angular momentum depends on the point or axis about which it is evaluated.
Use the correct moment of inertia: Rigid-body I must correspond to the same rotation axis used for ω and L.
Convert rotational speed to angular velocity: Convert RPM or revolutions per second to rad/s when the rigid-body formula expects ω.
Confirm perpendicular geometry in particle mode: The implemented L = mvr formula assumes the particle velocity is perpendicular to the radius vector.
Preserve direction signs: Positive and negative angular momentum should follow a consistent rotational-axis convention.
Check units: In SI, angular momentum should reduce to kg·m²/s.

Formula and variables

For a rigid body rotating about an appropriate fixed axis, angular momentum is proportional to moment of inertia and angular velocity. For a particle, angular momentum depends on its linear momentum and perpendicular lever arm relative to the chosen origin. The calculator uses the maximum/perpendicular particle case where the radius vector and velocity are perpendicular.

Rigid body: L = Iω; Particle, general: L = |r × p| = mvr sinθ; Calculator particle mode: L = mvr for θ = 90°

L — Angular momentum: Rotational momentum about the specified axis or origin, commonly expressed in kg·m²/s.
I — Moment of inertia: Rotational inertia about the selected axis, commonly expressed in kg·m².
ω — Angular velocity: Rotation rate, commonly expressed in rad/s.
m — Particle mass: Mass of the point particle.
v — Particle speed: Magnitude of the particle’s linear velocity.
r — Radius or position distance: Distance from the chosen origin or axis in the perpendicular particle case.
θ — Angle between r and p: Angle appearing in the general particle formula L = rp sinθ. The current mvr calculator mode assumes θ = 90°.
p — Linear momentum: Particle linear momentum, p = mv.

Scenario 1: Flywheel Angular Momentum

A flywheel has moment of inertia 4 kg·m² and rotates at 10 rad/s.

Moment of inertia: 4 kg·m²
Angular velocity: 10 rad/s

Use L = Iω.
L = 4 × 10.
L = 40 kg·m²/s.

Result: Angular momentum = 40 kg·m²/s.

The flywheel carries 40 kg·m²/s of angular momentum about the selected rotation axis. The sign or vector direction depends on the rotation direction and chosen axis convention.

Understanding your results

Angular momentum magnitude

This quantifies rotational momentum about the chosen reference axis or origin.

Its value cannot be interpreted independently of that reference.

Positive or negative sign

In a fixed-axis scalar model, the sign indicates which axial direction has been defined as positive.

The physical vector direction follows the right-hand rule.

Rigid-body result

L = Iω applies along the appropriate fixed or symmetry axis under the scalar model used here.

Increasing I or ω increases angular momentum proportionally when the other quantity is fixed.

Particle result

The calculator’s L = mvr form applies when velocity is perpendicular to the radius vector.

For nonperpendicular motion, the general magnitude is mvr sinθ.

Conservation context

If net external torque about the selected origin is zero, total angular momentum is conserved.

Individual components of a system can change while the total remains constant.

Assumptions

Angular momentum is evaluated about a specified origin or axis.
Rigid-body mode uses L = Iω for an appropriate fixed-axis or symmetry-axis rotational model.
The entered moment of inertia corresponds to the same axis as the angular velocity.
Particle mode assumes velocity is perpendicular to the radius vector.
Particle mass is positive.
Radius is nonnegative.
Units are converted into a coherent system before calculation.
A consistent rotational sign convention is used.
Classical mechanics adequately describes the system.

Limitations

The rigid-body scalar equation L = Iω is not a universal vector identity for every three-dimensional rigid-body orientation.
In general three-dimensional rigid-body mechanics, angular momentum and angular velocity need not be parallel, and the inertia tensor may be required.
The calculator’s particle equation L = mvr assumes r and v are perpendicular.
For arbitrary particle motion the correct magnitude is L = mvr sinθ, and the vector relation is L = r × p.
Angular momentum depends on the selected origin or rotation axis.
The calculator does not automatically calculate moment of inertia from arbitrary shape geometry.
The calculator does not determine whether angular momentum is conserved; that requires evaluating net external torque about the relevant origin.
Internal torques can redistribute angular momentum among parts of a system even while total angular momentum is conserved.
Friction or drag do not directly enter L = Iω, but external torques from them can change angular momentum over time.
Relativistic particle momentum requires more advanced expressions when speeds approach the speed of light.
Quantum mechanical angular momentum is governed by different rules and is outside the scope of this classical calculator.

Common mistakes

Using L = mvr when velocity is not perpendicular to radius.
Forgetting the sinθ term in the general particle formula.
Using diameter instead of radius.
Using the wrong moment of inertia for the selected axis.
Mixing RPM directly with I without converting rotational speed appropriately.
Confusing angular momentum with angular velocity.
Confusing angular momentum with torque.
Assuming angular momentum is always conserved.
Ignoring external torque.
Ignoring the reference origin for particle angular momentum.
Treating angular momentum as a scalar in a problem where vector direction matters.
Using mass instead of moment of inertia in the rigid-body model.

Practical use cases

Scenario 2: Rigid rotor

A rotor has I = 6 kg·m² and ω = 20 rad/s.

L = 120 kg·m²/s.

Scenario 3: Solve for moment of inertia

A rotating body has L = 90 kg·m²/s and ω = 15 rad/s.

I = L/ω = 6 kg·m².

Scenario 4: Perpendicular particle motion

A 2 kg particle moves at 5 m/s perpendicular to a radius of 3 m.

L = 2 × 5 × 3 = 30 kg·m²/s.

Scenario 5: Solve particle speed

L = 40 kg·m²/s, m = 2 kg, and r = 4 m in perpendicular motion.

v = L/(mr) = 5 m/s.

Scenario 6: Figure skater conservation

If external torque is negligible and moment of inertia decreases, angular velocity increases so that L remains constant.

This illustrates conservation rather than creation of angular momentum.

Planning and decision guide

Linear momentum provides the starting analogy

Linear momentum is p = mv.

Angular momentum describes rotational momentum relative to a point or axis.

Particle angular momentum is a cross product

OpenStax defines particle angular momentum as L = r × p.

Its direction is perpendicular to the plane containing r and p.

Magnitude depends on sinθ

|L| = rp sinθ.

With p = mv, this becomes L = mvr sinθ.

Scenario 7: Perpendicular particle

θ = 90°, so sinθ = 1.

The general formula reduces to L = mvr.

The calculator’s mvr formula is therefore a special case

It represents maximum angular momentum magnitude for fixed m, v, and r.

The page should state the perpendicular assumption explicitly.

Scenario 8: 30-degree particle angle

m = 2 kg, v = 4 m/s, r = 3 m, θ = 30°.

General L = 2 × 4 × 3 × 0.5 = 12 kg·m²/s.

Using mvr without sinθ would incorrectly return 24.

Angular momentum can be zero even for a moving particle

If momentum points directly along the radius vector, θ = 0 or π.

sinθ = 0, so angular momentum about that origin is zero.

Scenario 9: Particle moving directly away from origin

The particle has nonzero linear momentum.

Its angular momentum about that origin is zero because the lever arm is zero.

Angular momentum depends on the origin

Changing the reference point changes r.

The same moving particle can therefore have different angular momentum relative to different origins.

The lever-arm form is often intuitive

Particle angular momentum magnitude can be written L = r_perp p.

r_perp is the perpendicular distance from the origin to the particle momentum line.

Scenario 10: Lever-arm calculation

Linear momentum p = 10 kg·m/s and perpendicular lever arm = 2 m.

L = 20 kg·m²/s.

Rigid-body angular momentum sums particle contributions

A rotating rigid body can be treated as many mass elements.

Under the fixed-axis symmetric formulation, their angular momenta combine to produce L = Iω. OpenStax derives this result explicitly.

Moment of inertia replaces mass in the rotational analogy

p = mv for translation.

L = Iω for the corresponding fixed-axis rigid-body relation.

Moment of inertia depends on mass distribution

I = Σmᵢrᵢ² for discrete point masses about an axis.

Mass farther from the axis contributes disproportionately because distance is squared.

Scenario 11: Point mass inertia

m = 3 kg at r = 2 m.

I = 3 × 2² = 12 kg·m².

Moving mass outward increases L if ω is fixed

If rotational speed remains unchanged while I increases, L = Iω increases.

Maintaining that change requires an external angular impulse or redistribution with another system component.

Moving mass inward increases ω when L is conserved

If external torque is negligible, L remains constant.

Reducing I therefore requires ω to increase.

Scenario 12: Skater model

Initial I = 4 kg·m² and ω = 2 rad/s gives L = 8 kg·m²/s.

If I falls to 2 kg·m², conservation gives ω = 4 rad/s.

Conservation requires zero net external torque

OpenStax states dL/dt = Στ.

If Στ = 0, total angular momentum remains constant.

Angular momentum conservation is axis-dependent

Torque and angular momentum must be evaluated about the same relevant origin or axis.

Zero torque about one point does not automatically imply zero torque about every point.

Torque changes angular momentum

Στ = dL/dt.

A sustained net torque changes either the magnitude, direction, or both components of angular momentum.

Scenario 13: Constant torque

A constant net torque of 5 N·m acts for 4 seconds.

Angular momentum changes by 20 N·m·s = 20 kg·m²/s.

Angular impulse equals change in angular momentum

Integrating torque over time gives angular impulse.

The Angular Impulse Momentum Calculator handles this relationship directly.

The Angular Impulse Calculator handles torque-time area

For constant torque, angular impulse is τΔt.

For variable torque, it is the integral of torque over time.

Angular momentum is not torque

Angular momentum has SI units kg·m²/s.

Torque has SI units N·m = kg·m²/s².

The time dimension distinguishes them

Torque is angular-momentum change per unit time.

Multiplying torque by time produces angular momentum units.

Scenario 14: Unit check

N·m·s = kg·m²/s² × s.

This simplifies to kg·m²/s.

Angular momentum is also not rotational kinetic energy

Rigid-body rotational kinetic energy is K = 1/2 Iω².

Angular momentum is L = Iω.

The same L can correspond to different rotational energies

If L is fixed, K = L²/(2I).

A smaller moment of inertia at the same angular momentum produces greater rotational kinetic energy.

Scenario 15: Same L, different I

L = 20.

At I = 10, ω = 2.

At I = 5, ω = 4.

The Angular Velocity Calculator handles ω relationships

Angular velocity can be calculated from angle/time, frequency, RPM, or tangential speed relationships.

Angular Momentum Calculator adds the required inertia or particle geometry.

RPM must be converted before L = Iω

ω = RPM × 2π/60.

Using raw RPM directly changes the numerical scale by a large factor.

Scenario 16: Flywheel at 600 RPM

600 RPM = 20π rad/s.

If I = 2 kg·m², L = 40π ≈ 125.664 kg·m²/s.

Angular momentum direction follows the right-hand rule

For fixed-axis rigid rotation, curl the fingers in the rotation direction.

The thumb indicates the angular-momentum vector direction when L and ω are aligned in the scalar model.

Clockwise and counterclockwise can be represented with signs

A scalar one-axis calculation can assign opposite signs to opposite axial directions.

The chosen convention must be consistent.

Scenario 17: Reversing rotation

A rotor changes from +10 to −10 rad/s with constant I.

Its angular momentum also reverses sign.

A general rigid body can have L not parallel to ω

The simple scalar relation works along an appropriate principal or symmetry axis.

For general three-dimensional rotation, L = I_tensor · ω and the inertia tensor is required.

This calculator intentionally stays in the scalar fixed-axis regime

That makes it suitable for flywheels, wheels, disks, shafts, and introductory mechanics.

It should not imply universal applicability to arbitrary tumbling rigid bodies.

Particle orbital motion is a natural mvr application

For circular motion, velocity is perpendicular to radius.

The calculator’s L = mvr formula is therefore directly applicable.

Scenario 18: Circular particle orbit

m = 1 kg, v = 8 m/s, r = 2 m.

L = 16 kg·m²/s.

Planetary angular momentum requires careful modeling

An orbiting body can be approximated as a particle when its translational orbital angular momentum is the quantity of interest.

Its own spin angular momentum is a separate contribution.

Total angular momentum can contain orbital and spin components

A system may possess angular momentum from center-of-mass motion around an origin and from rotation about its own center.

The calculator evaluates one selected model at a time.

Scenario 19: Spinning orbiting body

A satellite can have orbital angular momentum about Earth and spin angular momentum about its own axis.

These are distinct contributions to the system total.

For circular particle motion, L can be expressed with ω

Since v = rω, L = mvr becomes L = mr²ω.

The factor mr² is exactly the point-mass moment of inertia.

This connects the particle and rigid-body formulas

For one point mass I = mr².

Therefore L = Iω reproduces L = mr²ω.

Scenario 20: Point mass using both forms

m = 2 kg, r = 3 m, ω = 4 rad/s.

I = 18 kg·m² and L = 72 kg·m²/s.

Using v = rω = 12 m/s also gives mvr = 72.

Radius must correspond to the selected origin

For particle angular momentum, r is not simply the radius of the object itself.

It is the position distance from the angular-momentum origin.

Diameter should not replace radius

Using diameter doubles r and therefore doubles the mvr result incorrectly.

The geometric input should be the distance from origin to particle.

Zero radius gives zero particle angular momentum

At the chosen origin, r = 0.

The particle angular momentum about that origin is zero regardless of finite p.

Zero velocity gives zero particle angular momentum

If p = mv = 0, r × p is zero.

A stationary particle has zero orbital angular momentum about the chosen origin.

Zero angular velocity gives zero rigid-body spin angular momentum in this model

L = Iω.

If ω = 0, L = 0 for the modeled rigid-body spin component.

Solving for ω requires nonzero I

ω = L/I.

Division by zero inertia is not a valid ordinary rigid-body calculation.

Solving particle speed requires nonzero m and r

v = L/(mr) in the perpendicular case.

The calculator should reject zero divisors explicitly.

The SI angular-momentum unit is kg·m²/s

From L = Iω, kg·m² × 1/s gives kg·m²/s.

Radians do not add an independent SI dimension.

N·m·s is dimensionally equivalent

Angular impulse has units N·m·s.

These reduce to the same SI dimensions as angular momentum.

Do not label angular momentum as joules

Although some rotational quantities share combinations of kg, m, and s, angular momentum is not energy.

Always retain the `/s` angular-momentum unit structure.

Moment-of-inertia units are kg·m²

Angular velocity adds inverse seconds.

The product therefore yields angular momentum.

Unit conversion should happen before calculation

Convert grams to kilograms, centimeters to meters, and RPM to rad/s if using SI output.

Mixed units can otherwise produce numerically incorrect results.

Scenario 21: Particle with centimeter radius

m = 0.5 kg, v = 4 m/s, r = 50 cm = 0.5 m.

L = 1 kg·m²/s in perpendicular motion.

Angular momentum conservation is a system statement

One object’s angular momentum can change while another part of the system gains the opposite amount.

The conserved quantity is the total for the isolated system.

Scenario 22: Two-body transfer

One rotating component loses 5 kg·m²/s while another gains 5 kg·m²/s.

Total angular momentum can remain unchanged.

External torque determines whether system L is conserved

Internal forces and torques can redistribute angular momentum.

Net external torque controls the time rate of change of total system angular momentum.

A short collision can conserve angular momentum even when energy is not conserved

If external torque is negligible during the impact, angular momentum can be conserved.

An inelastic collision can still lose mechanical kinetic energy.

Scenario 23: Particle sticks to a disk

Initial particle angular momentum can transfer into combined disk-plus-particle rotation.

OpenStax uses this type of example when illustrating angular momentum conservation.

Conservation does not mean I and ω remain individually constant

Only their appropriate product or total vector angular momentum remains constant when the conservation conditions apply.

I and ω can change inversely.

The calculator should distinguish magnitude from vector direction

Most user inputs are scalar magnitudes with signs.

A full three-dimensional vector calculation would require components or cross products.

Particle mode could later add an angle input

A future advanced mode could implement L = mvr sinθ.

The current mvr mode should remain clearly labeled “perpendicular motion.”

A second useful future particle mode is lever arm

L = r_perp p can be easier when the perpendicular distance to the momentum line is known.

This avoids requiring θ explicitly.

The strongest current UI should expose both models

Rigid body: L = Iω.

Point particle: L = mvr, with a visible perpendicular-motion assumption.

Result cards should name the reference

For example: “Angular momentum about the shaft axis.”

This reinforces that angular momentum is not an origin-free property.

Do not over-round angular velocity conversions

RPM-to-rad/s conversions contain π.

Use full internal precision and round only the final angular-momentum result.

The strongest educational output shows the analogy

Linear momentum: p = mv.

Rigid-body angular momentum: L = Iω.

Particle angular momentum: L = r × p.

Frequently asked questions

What is angular momentum?

Angular momentum is a vector measure of rotational momentum about a specified point or axis.

What is the formula for angular momentum?

For a particle, L = r × p. For an appropriate fixed-axis rigid body, L = Iω.

What formula does this calculator use for a rigid body?

L = Iω.

What formula does this calculator use for a particle?

L = mvr, which assumes velocity is perpendicular to the radius vector.

What is the general particle angular momentum formula?

The general vector formula is L = r × p, with magnitude mvr sinθ.

Why is there a sin theta term?

Only the component of momentum perpendicular to the radius contributes to angular momentum about the chosen origin.

When does L = mvr apply exactly?

When the radius vector and velocity are perpendicular, as in ideal circular motion.

What is the SI unit of angular momentum?

kg·m²/s.

What is I in L = Iω?

I is moment of inertia about the selected rotation axis.

What is omega in L = Iω?

ω is angular velocity, usually expressed in rad/s.

How do I calculate angular momentum from moment of inertia?

Multiply moment of inertia by angular velocity.

How do I calculate moment of inertia from angular momentum?

Use I = L/ω when angular velocity is nonzero.

How do I calculate angular velocity from angular momentum?

Use ω = L/I when moment of inertia is nonzero.

How do I calculate particle angular momentum?

For general motion use L = r × p. For perpendicular motion, magnitude simplifies to L = mvr.

Can a moving particle have zero angular momentum?

Yes. If its momentum points directly toward or away from the chosen origin, the cross product is zero.

Does angular momentum depend on the origin?

Yes for particle orbital angular momentum. Changing the reference point changes the position vector r.

Can angular momentum be negative?

A signed one-axis angular momentum can be negative when it points opposite the chosen positive axis.

What direction does angular momentum point?

Its vector direction follows the right-hand rule. For the fixed-axis rigid-body case, it lies along the rotation axis under the model assumptions.

What is the difference between angular momentum and angular velocity?

Angular velocity measures rotation rate. Angular momentum also depends on rotational inertia or particle position and linear momentum.

What is the difference between angular momentum and torque?

Torque is the rate of change of angular momentum: dL/dt = Στ.

What is conservation of angular momentum?

If net external torque about the chosen reference is zero, total angular momentum remains constant.

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

Pulling the arms inward reduces moment of inertia. If external torque is negligible and angular momentum is conserved, angular velocity increases.

Does angular momentum conservation mean angular velocity is constant?

No. Angular velocity can change when moment of inertia changes while their product remains constant.

Does zero torque mean zero angular momentum?

No. Zero net torque means angular momentum is constant; it can remain nonzero.

Does zero angular velocity mean zero rigid-body angular momentum?

Yes for the modeled spin component L = Iω.

Can the same object have different angular momentum about different axes?

Yes. Moment of inertia and reference geometry depend on the selected axis.

What is moment of inertia?

Moment of inertia measures rotational inertia and depends on both mass and its distribution relative to an axis.

How do I convert RPM before calculating angular momentum?

Use ω = RPM × 2π/60.

Can I use RPM directly in L = Iω?

Not if the formula expects rad/s. Convert RPM first.

Can angular momentum be conserved in a collision?

Yes if net external torque about the chosen reference is negligible during the collision.

Is angular momentum the same as rotational kinetic energy?

No. Angular momentum is L = Iω, while rotational kinetic energy is K = 1/2 Iω².

Can a system have both orbital and spin angular momentum?

Yes. A body can have angular momentum from motion around an external origin and from rotation about its own center.

Can I use this calculator for a planet orbiting the Sun?

The particle mode can represent a simplified perpendicular orbital case, but realistic orbital mechanics may require vector geometry and changing radius or velocity.

Can this calculator handle nonperpendicular particle motion?

The current particle mode does not include an angle input. Use the general formula L = mvr sinθ for nonperpendicular geometry.

Can this calculator calculate angular impulse?

Use the Angular Impulse Calculator for torque integrated over time.

Can this calculator calculate angular velocity?

It can solve ω from L and I in rigid-body mode, while the Angular Velocity Calculator handles broader rotation-rate relationships.

How accurate is an angular momentum calculator?

The formulas can be evaluated accurately from valid inputs. Physical accuracy depends on the reference axis, moment of inertia, particle geometry, vector direction, and whether the chosen model matches the actual system.

Sources and review

Angular Momentum — OpenStax University Physics Volume 1. Accessed 2026-09-01.
Conservation of Angular Momentum — OpenStax University Physics Volume 1. Accessed 2026-09-01.
Angular Momentum — Key Equations — OpenStax University Physics Volume 1. Accessed 2026-09-01.
SI Brochure, 9th Edition — Bureau International des Poids et Mesures. Accessed 2026-09-01.

Reviewed 2026-09-01 by Dr Akawak Ejigu, DBA.

Angular Momentum Conservation

Rigid Body (Spin)

Particle (Orbit)