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Angular Momentum Calculator

Result

3.000kg·m²/s

Result: 3.000 kg·m²/s

For a point mass on a circular path, angular momentum is mass times tangential speed times radius: 2 kg at 3 m/s on a 0.5 m string carries 3 kg·m²/s. The speed goes in as metres per second, not rad/s. With no outside torque it stays constant, which is why a skater speeds up when the arms come in.

The numbers at a glance

Held fixed: Mass (kg) 2.000 kg, Tangential speed (m/s) 3.000 m/s.

Radius of the path (m) (m)Result (kg·m²/s)
0.2001.200
0.4002.400
0.500Your value3.000
0.6003.600
0.8004.800
1.0006.000

Worked examples

Case 2
Mass (kg)
0.145kg
Tangential speed (m/s)
40m/s
Radius of the path (m)
0.5m

2.900kg·m²/s

Open with these values
Case 3
Mass (kg)
80kg
Tangential speed (m/s)
6m/s
Radius of the path (m)
1.5m

720.000kg·m²/s

Open with these values

How it's calculated

L = m × v × r

  1. StepEnter the mass in kilograms.
  2. StepEnter the speed along the circle in metres per second — if you have ω, use v = ω × r.
  3. StepEnter the radius of the circular path in metres.
  4. ResultRead the angular momentum in kg·m²/s.

Reference table

Mass, speed, radiusExampleAngular momentum
1, 1, 1The unit case1
0.145, 40, 0.5A baseball on a 0.5 m arc2.9
2, 3, 0.5A ball on a 0.5 m string3
5, 10, 2A weight swung on a 2 m rope100
80, 6, 1.5A cyclist leaning into a bend720
1000, 20, 0.3A heavy flywheel rim6000

Questions

How do I calculate angular momentum?

For a point mass on a circle, multiply mass by tangential speed by radius: L = m × v × r. Kilograms, metres per second and metres give kg·m²/s. A 2 kg ball at 3 m/s on a 0.5 m string has 3 kg·m²/s.

Which speed do I enter?

The tangential speed, in metres per second — how fast the object actually travels along its circular path. If you know the angular velocity ω in rad/s, multiply it by the radius first.

Why does the radius change the answer?

The radius is the lever arm of the motion. The same mass at the same speed carries more angular momentum the farther it is from the axis, because L grows in direct proportion to r.

Is angular momentum conserved?

Yes, as long as no external torque acts on the system its total angular momentum stays constant. That is why a spinning skater speeds up when pulling their arms in: a smaller radius forces a higher speed so that m × v × r is preserved.

Does this work for a wheel or a disc?

Not directly — this is the point-mass formula. An extended body uses L = I × ω with its moment of inertia I, so work out I first and multiply by the angular velocity.

Sources and last check

  1. openstax.org

Information, not professional advice.