Particle Angular Momentum
Compute the angular momentum L = m·v·r·sin(θ) of a particle.
L (kg·m²/s)
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Angular momentum: L = r × p
A particle's angular momentum is the vector L = r × p. Here r is the position vector from the axis (m) and p = mv is the linear momentum (kg·m/s). When the motion is circular and perpendicular to r, the magnitude works out to L = m·v·r·sin θ, which collapses to L = mvr once θ = 90°. SI units give L in kg·m²/s, and it stays conserved as long as no external torque acts on the system. Example: a 0.5 kg ball going 10 m/s around a 2 m circle has L = 10 kg·m²/s about the centre.
Applications: orbitals, gyroscopes, pulsars
This one idea covers a lot of ground. It's why a figure skater spins faster when the arms come in (smaller r ⇒ higher ω, since L holds constant). It's behind Kepler's second law, where planets sweep equal areas in equal times. In atoms it's quantised as L = √(ℓ(ℓ+1))·ℏ, which is what defines the s, p, d, f orbitals. It keeps gyroscopes and satellite attitude-control wheels steady, and it's how neutron stars (pulsars) end up spinning hundreds of times a second once the core collapses.
FAQ
Why is L a vector? Because it comes from the cross product r × p. Its direction, found by the right-hand rule, points along the rotation axis, while its magnitude tells you how much rotation there is.
When is L conserved? Any time the net external torque is zero. That's the reason a closed system's total angular momentum never changes.
What about spin? Electrons, protons and neutrons carry intrinsic spin ½ℏ. It's angular momentum with no classical analogue, and it sits at the heart of both fermion statistics and MRI.
What happens if θ = 0? Then sin θ = 0, so L = 0. A particle moving radially, straight toward or away from the axis, has zero angular momentum.
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