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UY1: Angular Momentum

Define angular momentum for particles and systems, derive vector τ = d vector L/dt, and apply conservation with worked examples.

  • University Physics Year 1
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Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the core concept page for torque-angular momentum laws in the UY1 mechanics pathway.

Choose origin/axis before doing any algebra

Most angular-momentum errors are setup errors, not calculus errors. Always state the origin/axis explicitly in your first line, then keep torque and angular momentum about the same origin throughout.

If an external impulse acts over a short time, switch to angular impulse form early: Δ vector L = ∫ vector τₑₓₜ dt.

1) At a glance

  • Prerequisite: vectors, cross product, linear momentum, Newton’s second law.
  • Outcomes:
    • Compute vector L = vector r × vector p for a particle.
    • Derive and use vector τₑₓₜ = d vector L/dt.
  • Apply angular-momentum conservation when external torque is zero.
  • Key result: ∑ vector τₑₓₜ = (d vector Lₜₒₜ)/dt
Common traps (origin + axis)
  • Angular momentum depends on the chosen origin (and axis). Always state “about point O” or “about the axle”.
  • Don’t jump to L = Iω unless you truly have a rigid body rotating about a fixed/principal axis.
  • In planar problems, keep direction consistent: CCW about + k hat gives L_z > 0.

2) Setup

Definitions:

  • Torque about origin O: vector τ = vector r × vector F
  • Particle angular momentum about origin O: vector L = vector r × vector p = vector r × m vector v

Sign/direction:

  • Direction from right-hand rule.
  • In planar motion (xy plane), vector L and vector τ are along ±k hat.

Geometry reminder:

  • vector r is the position vector from the chosen origin O to the particle.
  • Only the component of momentum perpendicular to vector r contributes to | vector L|:
    | vector L| = r p sin φ = r_⊥ p
    (where φ is the angle between vector r and vector p).

Unit:

[L] = kg m²s⁻¹ = N m s

3) Core derivation

For one particle:

(d vector L)/dt = (d/dt)(vector r × vector p); = (d vector r)/dt × vector p + vector r × (d vector p)/dt

Now (d vector r)/dt = vector v and vector p = m vector v, so

vector v × vector p = vector v × (m vector v) = vector 0

Hence

(d vector L)/dt = vector r × (d vector p)/dt = vector r × vector F = vector τ

For a system of particles, internal torques cancel in pairs (about the same origin), giving:

∑ vector τₑₓₜ = (d vector Lₜₒₜ)/dt

If ∑ vector τₑₓₜ = 0, then vector Lₜₒₜ is constant.

Rigid-body special case: vector L = I vector ω

For a particle in a circle of radius r with angular speed ω, v = rω and vector r⊥ vector v, so

| vector L| = r(mv) = mr²ω

For a rigid body rotating about a fixed axis that is also a principal/symmetry axis (the common UY1 case), summing over particles gives

vector L = I vector ω

about that axis (the same axis used to define I). This is the bridge to:

Checks

  • Dimension check: τ has unit N m, so τ dt has unit N m s = kg m²s⁻¹, matching angular momentum change.
  • Sign check in planar motion: clockwise vs counterclockwise torque should produce ∓k hat consistently.

4) Worked examples

Example 1: particle moving in a straight line

A particle of mass m moves along +x with speed v at fixed y = y₀. Choose origin at (0,0).

At any instant: vector r = xi hat + y₀j hat and vector p = mvi hat.

vector L = vector r × vector p = (xi hat + y₀j hat) × (mvi hat) = -y₀mvk hat

So | vector L| = mvy₀, constant.

If origin is chosen on the particle’s path (y₀ = 0), then vector L = 0. Both results are correct because origin changed.

Example 2: constant external torque

A wheel has constant net external torque τ₀ about its axle and initial angular momentum L₀ about that axle.

dL/dt = τ₀ ⇒ L(t) = L₀ + τ₀ t

This is rotational analog of linear impulse-momentum.

5) Practice set (with hints + answers)

  1. A particle moves in a circle of radius R at speed v. Find | vector L| about the circle center. Hint: vector r⊥ vector p. Answer: L = mRv.

  2. Net external torque on a system is zero from t = 0 to t = 3 s. If L_z(0) = 4 kg m²s⁻¹, find L_z(3 s). Hint: conservation. Answer: 4 kg m²s⁻¹.

  3. A constant torque 2.0 N m acts for 0.50 s. Find change in angular momentum magnitude. Hint: Δ L = τΔ t. Answer: 1.0 kg m²s⁻¹.

6) Summary and next steps

Angular momentum is origin-dependent but governed by a clean law:

∑ vector τₑₓₜ = (d vector Lₜₒₜ)/dt

This makes torque-time (angular impulse) the driver of angular momentum change and gives immediate conservation when external torque is zero.

Exam-speed checklist:

  1. Declare origin/axis and positive direction.
  2. Determine whether external torque is negligible over the interval.
  3. Use conservation or impulse form before expanding into component algebra.

Next steps: