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Angular Momentum of a Point Particle

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Angular momentum does not require rotation

The definition L = r × p applies to any particle with a position and a momentum, whether or not anything is spinning. A particle moving in a perfectly straight line has angular momentum about any origin not on that line, with magnitude L = mvb, where b is the perpendicular distance from the origin to the line of motion. Angular momentum is always defined relative to a chosen origin, and quoting a value without naming the origin is meaningless — the same particle has different L about different points.

Angular momentum and its rate of change
L = r × p |L| = rp sin θ = mvb Στ = dL/dt
For a rigid body rotating about a fixed axis this reduces to the familiar L = Iω. The point-particle form is the more general one and the two must agree wherever both apply.

Why straight-line motion conserves it

A free particle travels in a straight line at constant speed, so its perpendicular distance b from a fixed origin never changes and neither do m or v. Its angular momentum is therefore constant — as it must be, since no force acts and so no torque acts. This example is worth internalizing because it shows conservation of angular momentum operating with nothing rotating at all, and because it is the cleanest demonstration that L depends on the origin: choose the origin on the line of motion and L is zero everywhere instead.

Torque as the rotational Newton's second law

Just as ΣF = dp/dt governs linear motion, Στ = dL/dt governs rotational motion, and both are exact rather than approximations. The consequence is immediate: if the net external torque about a point is zero, the angular momentum about that point is conserved. A skater pulling in her arms reduces I, and with L fixed ω must rise. Kepler's second law — equal areas in equal times — is the same statement for a planet, since the gravitational force is central and therefore exerts no torque about the sun.

Worked example

A 2.0 kg particle moves at a constant 3.0 m/s along a straight line whose perpendicular distance from the origin is 4.0 m. Find its angular momentum about the origin, and state whether it changes with time.

  1. 1.Use the perpendicular-distance form: L = mvb.
  2. 2.L = (2.0)(3.0)(4.0) = 24 kg·m²/s.
  3. 3.Direction: perpendicular to the plane containing the line and the origin, by the right-hand rule.
  4. 4.No force acts, so no torque acts about any point, and by Στ = dL/dt the angular momentum is constant — b, m and v are each unchanging.
Answer: L = 24 kg·m²/s, constant in both magnitude and direction, even though the particle never rotates about anything
Checkpoint

A particle travels in a straight line at constant velocity. Its angular momentum about an origin located ON that line is —

Watch out

Never write an angular momentum or a torque without naming the point it is measured about. Unlike linear momentum, both quantities change value when the reference point changes, and a conservation argument valid about one point can be false about another.

Checkpoint

A planet orbits the sun under gravity alone. Its angular momentum about the sun is conserved because —

Answer the 2 checkpoints as you read.

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