Angular Momentum Calculator

The angular momentum calculator solves the rotational analogue of linear momentum for two common physical regimes: rigid-body rotation around a fixed axis, and a point mass (or orbiting body) at position r moving with velocity v. The first regime uses the compact form L = I · ω, the second the full cross-product form L = m · v · r · sin θ. Both reduce to the same units (kg·m²·s⁻¹, also written J·s) and both obey the same conservation law when no external torque acts on the system.

Angular momentum is one of the most useful quantities in physics because it is conserved in any isolated rotation. That single fact explains why figure skaters spin faster when they pull their arms in, why planets stay in stable orbits around the Sun, why a spinning top resists being tipped over, and why the Earth's axis of rotation has barely changed in millions of years. The calculator is designed to make that conservation check easy: enter any two of the variables in L = I · ω (or any three of the four in L = m·v·r·sin θ) and the missing one is computed.

How to Use the Angular Momentum Calculator

1. Pick a mode: Rigid-body rotation (L = I · ω) or Point mass / orbital (L = m · v · r · sin θ).
2. In rigid-body mode, fill in the moment of inertia I (kg·m²) and the angular velocity. If you do not know ω in rad/s, you can leave it blank and enter either the frequency f in Hz (revolutions per second) or the period T in seconds, the calculator converts using ω = 2πf = 2π / T.
3. In point-mass mode, fill in the mass m (kg), speed v (m/s), radius r (m), and the angle θ (in degrees) between v and r. Use 90° for circular orbits and any value between 0° and 180° for general motion.
4. Click Calculate. The solved value, the formula used, the rotational kinetic energy, the period and RPM, and (for circular orbits) the orbital period are all reported below.
5. For a quick start, click any of the reference-example chips to load verified numbers. The Earth-Sun preset reproduces the well-known L ≈ 2.66 × 10⁴⁰ kg·m²/s. The figure-skater and CD presets exercise the rigid-body mode with realistic numbers.

The Formula

The angular momentum of a particle of mass m at position vector r moving with velocity v is the cross product

L = r × p = m (r × v)

The magnitude of this vector is

L = m · v · r · sin θ

where θ is the angle between r and v. When the particle is in a circular orbit, v is always perpendicular to r, so θ = 90° and sin θ = 1, giving the simple form L = m·v·r.

For a rigid body rotating about a fixed axis with angular velocity ω, the angular momentum about that axis is

L = I · ω

where I is the moment of inertia of the body about the same axis. The moment of inertia depends on the mass distribution: for a solid disk of mass m and radius r, I = ½ m r²; for a solid sphere, I = ⅖ m r²; for a thin rod of length L rotating about its centre, I = 1⁄12 m L². The shape presets in the calculator load these textbook values directly.

The rotational kinetic energy is

KE = ½ · I · ω² = L² / (2I)

and the rotational frequency in Hz is

f = ω / (2π), with the period T = 1 / f = 2π / ω.

Real-World Examples

Example 1, Earth orbiting the Sun. The Earth has mass m ≈ 5.972 × 10²⁴ kg, an orbital speed v ≈ 29 780 m/s, and a mean orbital radius r ≈ 1.496 × 10¹¹ m. The velocity is perpendicular to the radius, so θ = 90°. The angular momentum is

L = (5.972 × 10²⁴) × (29 780) × (1.496 × 10¹¹) ≈ 2.66 × 10⁴⁰ kg·m²/s

This matches the accepted astrophysical value. The orbital period is T = 2πr / v ≈ 1 year, and the rotational kinetic energy associated with the orbital motion is ½ m v² ≈ 2.65 × 10³³ J. Note that this orbital L is many orders of magnitude larger than the spin angular momentum of the Earth about its own axis (about 7 × 10³³ kg·m²/s).

Example 2, A figure skater's spin. A 60 kg skater with arms extended at an effective radius of 0.25 m has I = m r² = 3.75 kg·m² (treating the arms as point masses). Spinning at 2 revolutions per second gives ω = 2π·2 ≈ 12.57 rad/s, so L = 3.75 × 12.57 ≈ 47.1 kg·m²/s. Pulling the arms in to r = 0.10 m reduces I to 0.6 kg·m²; conservation of L then forces ω to rise to about 78.5 rad/s (about 12.5 rev/s). The rotational KE also increases, the extra energy coming from the work done by the skater pulling her arms in against the centrifugal "force" in the rotating frame.

Example 3, A spinning compact disc. A CD has mass ≈ 15 g, radius ≈ 6 cm, and spins at up to 400 rpm when reading data. Treating it as a solid disk, I = ½ m r² = 0.5 × 0.015 × 0.06² ≈ 2.7 × 10⁻⁵ kg·m². The angular velocity is ω = 400 × 2π / 60 ≈ 41.9 rad/s, so L = I ω ≈ 1.13 × 10⁻³ kg·m²/s and the rotational KE is ½ I ω² ≈ 2.37 × 10⁻² J. This is small compared with a CD player's electrical power, but it is exactly the quantity that must be supplied during spin-up and recovered during spin-down.

Example 4, A satellite in low Earth orbit. A 1 000 kg satellite at 400 km altitude (r = 6.378 × 10⁶ + 4 × 10⁵ ≈ 6.78 × 10⁶ m) has v = √(GM/r) ≈ 7.67 × 10³ m/s. The orbital L is 1 000 × 7.67 × 10³ × 6.78 × 10⁶ ≈ 5.2 × 10¹³ kg·m²/s. The same satellite, despun to a non-rotating state, has L = 0 about its own centre of mass, but its orbital L is unchanged. The two are independent angular momenta about different axes.

Conservation of Angular Momentum

The principle of conservation of angular momentum states that the angular momentum of an isolated system is constant in both magnitude and direction whenever the net external torque is zero. This is the rotational analogue of Newton's first law and is one of the deepest results in classical mechanics. The torque is defined as the time rate of change of angular momentum: τ = dL/dt. If τ = 0, then dL/dt = 0, so L is conserved.

The most important applications include:

• Stable planetary orbits: gravity is a central force, so the torque it exerts on an orbiting body about the centre of force is zero. The orbital L is therefore conserved, which (combined with conservation of energy) gives the three Kepler laws.
• The figure-skater spin: by pulling her arms in, the skater reduces I without any external torque, so L = I ω stays constant, and ω must rise.
• The gyroscope: a spinning rotor has a fixed L direction in inertial space, and any attempt to tilt the rotor applies a torque perpendicular to L. The rotor precesses around the applied torque rather than falling over, which is the basis of inertial navigation.
• Atomic physics and quantum mechanics: in the classical limit, the Bohr model of the hydrogen atom predicts an orbital L = n ℏ; the correct quantum treatment gives L = √(ℓ(ℓ+1)) ℏ. The classical formula in this calculator is the limit ℏ → 0 of the quantum one.
• Black holes: a rotating (Kerr) black hole is fully described by its mass and its angular momentum J, which sets the spin parameter a = J / (M c). The maximum a is the extremal Kerr black hole with a = GM/c.

Rigid-Body Moment of Inertia

The moment of inertia of a body about a given axis is the integral I = ∫ r² dm, where r is the perpendicular distance from the axis to each mass element dm. For a uniform body of total mass M and characteristic size R, I is always of the form I = c · M R², where c is a dimensionless factor that depends on the shape. The shape presets in the calculator cover the most common textbook cases:

For a composite body made of several parts, the moment of inertia about a given axis is the sum of the individual moments: I_total = Σ I_i. For a continuous distribution, integrate I = ∫ r² dm over the body. The parallel-axis theorem relates the moment of inertia about an axis through the centre of mass to the moment of inertia about a parallel axis at distance d: I = I_cm + m d².

Why Conservation Matters in Engineering

In mechanical engineering, the conservation of angular momentum is exploited in flywheels (which store rotational kinetic energy and resist changes in rotation rate), reaction wheels and control moment gyroscopes (used to point satellites and spacecraft), and centrifuge design. In aerospace, the conservation law explains why a spinning bullet or a spinning spacecraft is more stable in flight. In automotive engineering, it explains why a spinning wheel resists being steered (the gyroscopic effect), why a clutch transfers torque smoothly, and why an engine's flywheel keeps it running between power strokes.

The rotational kinetic energy stored in a flywheel is ½ I ω², which scales as the square of the angular velocity. Doubling the spin rate quadruples the stored energy. This is why high-speed flywheels in modern kinetic-energy recovery systems (KERS) can store tens of kilojoules in a compact package, even though the mass is small.

Common Mistakes

1. Using diameter instead of radius in the moment of inertia. A disk with diameter 10 cm has radius 5 cm. Using r = 10 gives I = 0.5 × m × 100, which is four times too large. Always convert to radius first.
2. Mixing angular velocity units. RPM is not rad/s. Multiply by 2π/60 to convert. The formula L = I ω only works with ω in rad/s.
3. Confusing the angle θ in L = m·v·r·sin θ with the angular position in the orbit. The angle in the formula is the instantaneous angle between the velocity vector and the position vector, not the orbital phase. For a circular orbit, it is always 90°; for a radial trajectory, it is always 0°.
4. Forgetting that L is a vector. For a particle, the direction of L is perpendicular to the plane defined by r and v, given by the right-hand rule. Two particles orbiting in opposite directions in the same plane can have cancelling angular momenta.
5. Treating L and I as if they were the same quantity for non-rigid bodies. For a deformable body, I can change while L is conserved, leading to a change in ω (the skater effect). For a rigid body with fixed mass distribution, I is constant and any change in ω requires an external torque.

Frequently Asked Questions

What are the units of angular momentum? The SI unit is kg·m²·s⁻¹, which is also equivalent to J·s (joule-seconds). The CGS unit is g·cm²·s⁻¹ (also called erg·s in Gaussian units). The Planck constant ℏ ≈ 1.055 × 10⁻³⁴ J·s is the quantum of angular momentum, so macroscopic L values are typically enormous multiples of ℏ.

What is the difference between angular momentum and moment of inertia? Moment of inertia I is a property of the body alone (how the mass is distributed relative to a chosen axis). Angular momentum L = I ω is a property of the body's state of rotation (how fast it is rotating about that axis). A body can have a large I but a small L if ω is small, and vice versa.

Is angular momentum conserved in every situation? No. It is conserved only when the net external torque on the system is zero. Friction in a bearing, air resistance, and contact with a non-rotating surface all apply torques that change L. In an isolated system (no external torques), L is always conserved in both magnitude and direction.

How is angular momentum related to torque? The time rate of change of angular momentum equals the net external torque: τ = dL/dt. This is the rotational analogue of Newton's second law F = dp/dt. Applying a torque over a time Δt changes L by ΔL = τ·Δt, the rotational analogue of impulse.

What is the spin angular momentum of the Earth? The Earth has mass ≈ 5.97 × 10²⁴ kg, mean radius ≈ 6.37 × 10⁶ m, and rotates once per sidereal day (≈ 86 164 s). Treating it as a solid sphere, I = ⅖ m r² ≈ 9.7 × 10³⁷ kg·m² and ω = 2π / 86164 ≈ 7.29 × 10⁻⁵ rad/s, giving L ≈ 7.1 × 10³³ kg·m²/s. This is about 3.7 × 10⁶ times smaller than the orbital L, but it is what makes the day-night cycle and what stabilises the Earth's axial tilt over geological timescales.

Can angular momentum be negative? In scalar form, L is the magnitude of the vector and is always non-negative. In the vector form, L points in a direction given by the right-hand rule; two rotations in opposite senses have opposite signs along a given axis. The calculator returns the magnitude; the sign convention is up to the user.

What is the relationship between angular momentum and the quantum spin of a particle? Elementary particles carry an intrinsic angular momentum called spin, with values of ½ ℏ, ℏ, 3⁄2 ℏ, etc. Unlike classical angular momentum, spin cannot be stopped, an electron always has spin ½ ℏ regardless of its orbital state. Composite particles (protons, neutrons, atoms) also have a total angular momentum that includes both orbital and spin contributions.

Also try these free tools:

• Torque Calculator
• Centripetal Force Calculator
• Moment of Inertia Calculator
• Kepler's Third Law Calculator
• Orbital Velocity Calculator