Short on time? This page is the recap for the calculus-based course (separate from Physics 1, which does not prove that internal torques cancel or handle continuously changing I). For the derivations, graphs and worked examples, use the full study guide.
Recap
- Total angular momentum = sum of the parts’ angular momenta about one axis (Iω for rigid parts, mvd for point objects), with signs.
- Newton’s third law pairs act along the line joining two particles, so (r₁ − r₂) × F₁₂ = 0: internal torques cancel.
- So dL_sys/dt = τ_ext and ΔL_sys = ∫τ_ext dt. Two interacting parts give each other equal and opposite angular impulses.
- Choose the system so that the net external torque about the axis is zero. Forces through the axis, or parallel to it, have no torque about it.
- Angular momentum is conserved in all interactions. A system’s L changes only by transfer to or from its surroundings.
Key relationships
| Relationship | When to use it |
|---|---|
| L_sys = ΣIᵢωᵢ + Σmᵢvᵢdᵢ | Any system, one axis |
| dL_sys/dt = τ_ext, net | Always |
| L_i = L_f | Net external torque about the axis is zero |
| ΔL_sys = ∫τ_ext dt | External torque acts |
| I₁ω₁ = I₂ω₂ | Shape change, no external torque |
| K = L²/(2I) | Energy at constant L: K ∝ 1/I |
| ω_f = (I_Aω_A + I_Bω_B)/(I_A + I_B) | Two parts end up turning together |
Assumptions behind the numbers
- Internal forces act along the line joining the particles (true for pushes, tensions and gravity).
- Axles frictionless and pivots fixed unless stated.
- Point objects for small masses; every I about the same axis.
Mistakes to avoid
- Constant L means constant ω. Not if I changes.
- Assuming kinetic energy is conserved too.
- Linear momentum in a pivot collision. The pivot force is external.
- Mixing axes in the sum.
- Dropping signs for parts turning the other way.
- Treating one part as a closed system when another part exerts a torque on it.
- “Friction destroys L.” It transfers L to the surroundings.
Quick self-check
- A freely spinning system changes I from 2.0 to 0.50 kg·m². It was turning at 3.0 rad/s. New ω, and factor of K? (12 rad/s; K × 4)
- A disk (I = 0.10 kg·m²) at 12 rad/s is locked to a second disk (I = 0.050 kg·m²) at rest on the same shaft. Final ω? (1.2/0.15 = 8.0 rad/s)
- A 0.020 kg blob of putty moving at 10 m/s strikes a pivoted bar 0.30 m from the pivot, moving perpendicular to the bar. L of the system about the pivot? (0.060 kg·m²/s, before and after)
Next: practice questions.