Angular momentum calculator
L = Iω for a spinning body, a particle, the spinning skater and the effect of a torque.
What angular momentum is
Angular momentum is the rotational counterpart of linear momentum. A rigid body spinning about a fixed axis with angular velocity ω and moment of inertia I has L = I·ω, and a single particle of momentum p at position r from a chosen point has l = r × p, a vector with magnitude r·p·sin θ (OpenStax University Physics, Volume 1, section 11.2). The direction follows the right-hand rule. Its importance comes from one law: the net external torque is the rate of change of angular momentum, dL/dt = Στ, so when the net torque is zero, L stays constant. That explains a skater speeding up in a spin or a gymnast completing two somersaults.
The formulas
Rigid body: L = I·ω in kg·m²/s, with ω in rad/s, and rotational kinetic energy K = ½·I·ω² (section 10.4). Particle: L = m·v·r·sin θ, where θ is the angle between the position vector and the velocity. Conservation: if no net external torque acts, I₁ω₁ = I₂ω₂, so the new angular velocity is ω₂ = ω₁·I₁ ÷ I₂ and the kinetic energy becomes K₂ = K₁·I₁ ÷ I₂ (section 11.3). Torque over time: ΔL = τ·Δt for a constant net torque. The moments of inertia of the standard shapes (disk ½MR², rod about its end ⅓ML², sphere ⅖MR² and so on) are the ones in our moment of inertia calculator.
Worked example: the gymnast’s dismount
| Step | Value |
|---|---|
| Full extension: 80.0 kg as a rod of 1.8 m, I₀ = (1/12) M L² | 21.6 kg·m² |
| Tucked: the same rod 0.9 m long, I = (1/12) M L² | 5.4 kg·m² |
| Starting spin 1.0 rev/s = 2π rad/s: L = I₀ω₀ | 136 kg·m²/s |
| Spin in the tuck: ω = L ÷ I | 4 rev/s |
| Kinetic energy: before → after | 426 J → 1,710 J |
OpenStax Example 11.8 finds 4.0 rev/s in the same way. Because the tuck lasts about half a second (the time to fall from 3.0 m to 1.8 m), he completes about two revolutions in the tuck. The extra 1,280 J of kinetic energy is the work done by his muscles as he pulls his body in.
Common mistakes
The biggest one is entering rpm where the formula needs rad/s; every rpm is 0.1047 rad/s, so 300 rpm is 31.4 rad/s. Another is conserving the wrong quantity: it is the angular momentum that stays constant when the shape changes, not the angular velocity and not the energy. A third is using conservation when something exerts a torque, for example a hand pushing on a spinning wheel or friction in a bearing; then use ΔL = τ·Δt instead. Finally, I and L have to refer to the same axis, and for a single particle the angle θ is measured between the position vector and the velocity, not between the velocity and the axis.
Limits of this method
The shapes assume a uniform density and a body that stays rigid; a person is not a rod, and OpenStax models the gymnast as a rod only to get an estimate. The formula L = I·ω is the one OpenStax gives for a rigid body rotating about a fixed axis, with L pointing along that axis. The conservation tab assumes the net external torque is exactly zero during the change; gravity or air drag acting off the axis violate that, and the answer is then only approximate for a short time interval. Relativistic speeds are not covered.
What are the units of angular momentum, and how do I convert rpm to rad/s?
Angular momentum is moment of inertia times angular velocity, so its SI unit is kg·m²/s (the same as J·s). In US units it is lb·ft²/s: 1 lb·ft² is 0.042140 kg·m². The angular velocity in the formula must be in radians per second: multiply revolutions per minute by 2π ÷ 60 ≈ 0.10472, so 300 rpm is 31.4 rad/s. Using rpm directly is the most common reason for a result that is off by a factor of 9.55.
How do I find the angular momentum of a spinning disk or wheel?
Use L = I·ω with the moment of inertia of the shape: ½MR² for a solid disk or cylinder about its axis, MR² for a thin hoop. A solid disk of 20 lb (9.07 kg) with a radius of 1 ft (0.3048 m) has I = 0.421 kg·m², which is 10 lb·ft². Spinning at 300 rpm it has L = 13.2 kg·m²/s and a rotational kinetic energy of 208 J. The first tab can build I from any of the standard shapes, the same ones as our moment of inertia calculator.
How do I calculate the angular momentum of a particle moving in a straight line?
A particle has angular momentum about any point, l = r × p, with magnitude r·p·sin θ, where θ is the angle between the position vector and the momentum, or equivalently p times the lever arm r⊥ = r sin θ. In OpenStax Example 11.4 a 15.0-kg meteor is at (25, 25) km and moves at 2.0 km/s along −y. The angle between r and p is 135°, so l = 35,400 m × 30,000 kg·m/s × sin 135° = 7.5 × 10⁸ kg·m²/s, along −k̂ as the book finds. It is zero if the particle moves straight toward or away from the origin.
Why does a skater spin faster when she pulls her arms in?
With no net external torque the angular momentum stays constant, I₁ω₁ = I₂ω₂, so a smaller moment of inertia means a larger angular velocity (OpenStax section 11.3). In Example 11.8 a gymnast of 80.0 kg modeled as a rod of 1.8 m (I = 21.6 kg·m²) tucks to half that length (I = 5.4 kg·m²). Her spin rises from 1.0 to 4 rev/s, four times as fast. The kinetic energy rises by the same factor, from 426 J to 1,710 J, because the muscles do work pulling the body in.
Is the kinetic energy conserved along with the angular momentum?
Not in general. Only the total angular momentum is conserved when there is no external torque. In OpenStax Example 11.7 a flywheel spinning at 600 rpm is coupled to a second one at rest whose moment of inertia is three times as large. The pair settles at 150 rpm (15.7 rad/s) and the kinetic energy falls to 0.25 of its starting value, so three quarters of it is lost to friction between the wheels. The calculator shows the energy before and after in the conservation tab.
How does a torque change angular momentum?
The net torque is the rate of change of angular momentum, dL/dt = Στ, which for a constant torque gives ΔL = τ·Δt. In OpenStax Example 11.6 a robot arm with I = 1.67 kg·m² (a 2.0-kg rod 1.0 m long with a 1.0-kg tip) speeds up from rest to 0.1π rad/s in 0.1 s. Its angular momentum becomes 0.524 kg·m²/s, the book writes 0.17π. The torque needed is I·α = 5.24 N·m (the book writes 1.67π), and 5.24 N·m × 0.1 s adds 0.524 kg·m²/s, the same value, as ΔL = τΔt requires. If the torque is zero, L stays the same.
Sources: OpenStax University Physics, Volume 1, sections 10.4 “Moment of Inertia and Rotational Kinetic Energy” (Examples 10.8 and 10.9), 11.2 “Angular Momentum” (Examples 11.4 to 11.6) and 11.3 “Conservation of Angular Momentum” (Examples 11.7 to 11.9). Related: moment of inertia calculator, angular velocity calculator, torque calculator and kinetic energy calculator.