Example 1
I = 2 kg·m², ω = 5 rad/s -> L = 10 kg·m²/s.
Angular momentum calculator
Type I and ω. The calculator computes L = I ω. 2 kg·m² and 5 rad/s is 10 kg·m²/s. 0.5 kg·m² and 10 rad/s is 5. Energy ½Iω² is a different page.
Point I: I = m·r². ω from period: ω = 2π/T. Energy: Eₖ = ½Iω².
Result
Moment of inertia I (kg·m²) and Angular speed ω (rad/s). The result shows up here.
How it works
Angular momentum is L = I ω. At I = 2 kg·m² and ω = 5 rad/s you get L = 10 kg·m²/s. At I = 0.5 and ω = 10 you get 5. At I = 1 and one turn per second, ω = 2π ≈ 6.283 rad/s, L ≈ 6.283 kg·m²/s. That is not energy: ½Iω² is in joules, L in kg·m²/s.
The form has two fields: I in kg·m² and ω in rad/s. The result is in kg·m²/s. rpm does not belong in the ω field. One turn per second is 2π rad/s. The result is rpm beside L: rpm = ω·60/(2π). A comma and a period are the same I: 0,5 and 0.5.
I and ω must be positive. Zero I rejects the model: no body. Zero ω rejects the spinning model. Both fields need a number before 10 kg·m²/s can appear.
I from m and r lives on point-mass I: I = m r². ω from period T lives on angular velocity: ω = 2π/T. Here I and ω are already inputs.
Torque M = F r sinα is twist, not L. The same body can have both M and L, but those are different cards.
Type 2 and 5, click Calculate, and match 10 kg·m²/s. The header symbol does not spin a wheel. Paste the same I and ω into ½Iω² and compare joules with L.
How to use
- In the first field enter I in kg·m², for example 2. I from m and r is first computed on point-mass I.
- In the second field enter ω in rad/s, for example 5. Not rpm: one turn per second is 2π.
- Click Calculate. The calculator multiplies I by ω. 2 and 5 give L = 10 kg·m²/s. rpm sits beside it.
- I and ω must be greater than zero. Both fields need a number.
- For rotational energy, open ½Iω². Torque lives on M = F r sinα.
Formula
L = I·ω
I > 0, ω > 0. Unit of L: kg·m²/s. rpm = ω·60/(2π).
L, I, and ω for angular momentum
L = I·ω. 2 kg·m² and 5 rad/s is 10 kg·m²/s. 0.5 kg·m² and 10 rad/s is 5. Energy ½Iω² is another calculator.
- L
- Angular momentum in kg·m²/s. 2 × 5 = 10. Not a joule and not torque N·m.
- I
- Moment of inertia in kg·m². 2 or 0.5. Here a factor of ω, not ½Iω².
- ω
- Angular velocity in rad/s. 5 or 10. rpm = ω·60/(2π) sits in the formula note.
Real-life examples
Example 2
I = 0.5 kg·m², ω = 10 rad/s -> L = 5 kg·m²/s.
Example 3
I = 1 kg·m², ω = 6.2832 rad/s -> L ≈ 6.283 kg·m²/s.
Example 4
I = 0.1 kg·m², ω = 20 rad/s -> L = 2 kg·m²/s.
Example 5
I = 4 kg·m², ω = 3 rad/s -> L = 12 kg·m²/s.
Example 6
I = 0.02 kg·m², ω = 50 rad/s -> L = 1 kg·m²/s.
Example 7
I = 8 kg·m², ω = 1.5 rad/s -> L = 12 kg·m²/s.
Example 8
I = 1.2 kg·m², ω = 8 rad/s -> L = 9.6 kg·m²/s.
Ways to use this calculator
- You compute 10 kg·m²/s from I = 2 and ω = 5.
- You paste the same I and ω into ½Iω² and compare joules with L.
Frequently asked questions
How much L at I = 2 kg·m² and ω = 5 rad/s?
L = 2 × 5 = 10 kg·m²/s. At I = 0.5 and ω = 10 it is 5.
Which units do I type?
I in kg·m², ω in rad/s. Result in kg·m²/s. Not rpm in the ω field.
Why will I = 0 not run?
No body. I must be positive. Both fields need a number.
Why is L not rotational energy?
Energy is ½Iω², unit joule. L is Iω, unit kg·m²/s.
Where do I get I from m and r?
For a point mass, I = m r² on its own page. Here I is already an input.
Can I type ω in rpm?
No. The field wants rad/s. One turn per second is 2π rad/s. The result is rpm beside L.
Does a comma in 0.5 work?
Yes. 0,5 and 0.5 mean the same I [kg·m²]. Then 0.5 × 10 rad/s is L = 5 kg·m²/s.
How much at I = 1 and one turn per second?
ω = 2π ≈ 6.283 rad/s, so L ≈ 6.283 kg·m²/s. Do not type 1 rpm in the ω field.
Where is torque M?
On M = F r sinα. Here L = Iω, a product of inertia and spin, not a twist.
Where is ω from period T?
On angular velocity, ω = 2π/T. Then come back here with I and ω.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.