Physics I: Mechanics › Rotation › full formula sheet

Li = Lf  (no net external torque)

Say it: “with no net external torque, the total angular momentum stays constant”

Conservation of L

The skater’s secret — spin is a conserved currency, and shrinking your I spends it on speed.

Notation on this page: Li, Lf are the system’s total angular momentum before and after; “external” means torques from outside the system. Internal torques cancel out.

Before this lesson: Angular momentum, Torque

Where it comes from

The figure skater is the icon: arms out, lazy spin — arms in, blur. No one pushed her mid-spin (the ice is nearly frictionless), so where did the extra speed come from? It didn’t come from anywhere — it was traded:

Before reading on: L = Iω is fixed. She pulls in, halving her I. What happens to ω — and what happens to her kinetic energy ½Iω²? Predict both.
L fixed
I → I/2  ⇒  ω → 2ω
L = Iω constant: half the I demands double the ω.
K = ½Iω²
½(I/2)(2ω)² = 2 × (½Iω²)
Energy doubled! Her muscles did work pulling in — conservation of L never promised conservation of energy.

That last line is the trap most students walk into: L conserved ≠ K conserved. The work came from her arms — internal effort, invisible to L’s ledger but very visible in the energy bill.

Derivation

Start from Στext = dL/dt (last lesson’s bonus step). If no external torque acts, the derivative is zero — and a zero derivative means a constant.

Στext
=
dL/dt
Step 1 — the dynamical law. Net external torque equals the rate of change of the system’s total L. (Internal torques cancel in pairs — Newton’s 3rd — so only external ones appear.)
Στext = 0
⇒
dL/dt = 0
Step 2 — impose the condition. No net external torque ⇒ L isn’t changing. This is the only assumption — the whole conservation law hangs on these four symbols.
Li
=
Lf
Step 3 — integrate. Zero derivative ⇒ constant: L at the start equals L at the end. ∎
I1ω1
=
I2ω2
Step 4 — rigid-body form. For a spinner changing shape, L = Iω on both sides. Compute I before and after, set equal, solve. (Collisions: same idea with summed L’s.)

Why “external”? The skater’s muscles exert huge internal torques (arms pulling in) — but every internal torque has an equal-opposite partner somewhere in the system, so they sum to zero in Στext. Only outsiders (friction, a shove, gravity about an off-center pivot) can touch the total.

How to use it

The procedure, every time:

  1. Define the system. Draw the boundary: skater alone? skater + weights? merry-go-round + kid?
  2. Audit external torques. Friction at the axle? A push from outside? Gravity (about the CM: no torque)? If Στext = 0 (or negligible over the short event), conservation applies.
  3. Write Li = Lf. For shape-changers: I1ω1 = I2ω2. For collisions/stick-ons: (I1ω1)before = (Itotalω2)after.
  4. Solve. Usually for the final ω.
  5. Never assume K is conserved. Check energy separately — internal work routinely changes it.
I1ω1 = I2ω2the workhorse form — I before and after, about the same axisSay it: “I one omega one equals I two omega two”
Common mistake: applying conservation while an external torque acts — e.g. a merry-go-round with axle friction “conserving” L as it slows. Friction is external: L drains. Conservation needs the audit in step 2, not wishful thinking.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: the skater pulls in

  1. Audit. Skater on near-frictionless ice: no external torque about her vertical axis ✓ — L conserved.
  2. Before. Arms out: I1 = 3.0 kg·m², ω1 = 2 rad/s ⇒ L = 6 kg·m²/s.
  3. After. Arms in: I2 = 1.2 kg·m².
  4. Solve. ω2 = L/I2 = 6/1.2 = 5 rad/s — 2.5× faster.
  5. Energy check. K1 = ½×3×4 = 6 J; K2 = ½×1.2×25 = 15 J. Her arms did 9 J of work.
Common mistake: ω2 = ω1 × (I2/I1) (ratio flipped) — giving 0.8 rad/s, slower. Smaller I → bigger ω: the ratio is I1/I2, inverted.
Your turn — I1 = 2.5, ω1 = 4 rad/s; pulls in to I2 = 1.0. ω2 = ?

Answer: 10 rad/s. L = 2.5 × 4 = 10; ω2 = 10/1.0 = 10 rad/s.

Example 2 — the kid jumps on the merry-go-round

  1. System. Disk + kid. The jump is internal to the system (kid was “outside” but lands without external torque about the axle — brief event, axle friction negligible) ✓.
  2. Before. Disk M = 100 kg, R = 1.5 m: I1 = ½×100×2.25 = 112.5; ω1 = 1.5 ⇒ L = 168.75.
  3. After. Kid m = 30 kg lands at rim: I2 = 112.5 + 30×2.25 = 112.5 + 67.5 = 180 kg·m².
  4. Solve. ω2 = 168.75/180 = 0.9375 ≈ 0.94 rad/s — the ride slows as the kid boards.
Common mistake: forgetting the kid’s own I after landing (using 112.5 for I2). The system grew — recompute I after every mass change.
Your turn — Disk M = 80 kg, R = 2 m, ω1 = 2 rad/s; 40-kg kid jumps on at rim. ω2 = ?

Answer: 1 rad/s. I1 = 0.5 × 80 × 4 = 160; L = 320; I2 = 160 + 40 × 4 = 320; ω2 = 320/320 = 1 rad/s.

Before reading on: a diver goes from layout (I = 12) to tuck (I = 4) mid-flight, starting at 2 rad/s. Predict the tuck spin rate — then decide: did the diver’s kinetic energy change?

Example 3 — the diver’s tuck

  1. Audit. In flight, gravity acts at the CM (no torque about CM), air negligible ✓ — L conserved about the CM.
  2. Before. Layout: I1 = 12, ω1 = 2 ⇒ L = 24.
  3. After. Tuck: I2 = 4 ⇒ ω2 = 24/4 = 6 rad/s — triple the spin.
  4. Energy. K1 = ½×12×4 = 24 J; K2 = ½×4×36 = 72 J. Tripled — the tuck’s muscle work, again.
Common mistake: “L conserved so K conserved” — then ω2 = 2√3 or some energy-based fiction. L and K are separate ledgers; only L is conserved here.
Your turn — I1 = 10, ω1 = 1.5 rad/s; tucks to I2 = 2.5. ω2 = ?

Answer: 6 rad/s. L = 15; ω2 = 15/2.5 = 6 rad/s.

Example 4 — stellar scale: the collapsing star

  1. Setup. A star (uniform sphere, I ∝ MR²) collapses: R1 = 7×10⁸ m, period T1 = 25 days → R2 = 10⁴ m (neutron star). No external torque ✓.
  2. L conservation. I1ω1 = I2ω2 ⇒ ω2/ω1 = (R1/R2)² (M and 2/5 cancel).
  3. Ratio. (7×10⁸/10⁴)² = (7×10⁴)² = 4.9 × 10⁹.
  4. New period. T2 = 25 days / 4.9×10⁹ = 2.16×10⁶ s / 4.9×10⁹ ≈ 4.4 × 10⁻⁴ s ≈ 0.44 ms — a millisecond pulsar. Real ones exist.
Common mistake: ratio = R1/R2 (not squared) — giving 7×10⁴ instead of 4.9×10⁹. I ∝ R², so the spin-up is the squared ratio. The square is doing all the drama.
Your turn — Star R1 = 5×10⁸ m, T1 = 20 days, collapses to R2 = 1.2×10⁴ m. T2 = ?

Answer: ≈ 1.0 ms. Ratio = (5×10⁸/1.2×10⁴)² = (41666.7)² = 1.736×10⁹; T2 = 20 × 86400/1.736×10⁹ = 1.728×10⁶/1.736×10⁹ = 9.95×10⁻⁴ s ≈ 1.0 ms.

Memorization tips

  • Say it aloud: “no net external torque — L stays.” The condition is the law.
  • The workhorse: I1ω1 = I2ω2. Smaller I → bigger ω — the ratio flips.
  • Audit before you conserve: name the system, list external torques. Friction, pushes, off-center gravity — any of them voids the warranty.
  • L ≠ K: conservation of angular momentum says nothing about energy. Internal work (muscles!) routinely pumps K up.
  • Recompute I after: every mass change (kid boards, arms tuck, star collapses) means a new I2. Stale I’s are the #1 arithmetic error.
  • I ∝ R²: collapse/spread problems square the radius ratio. The square does all the drama — pulsars spin in milliseconds because of it.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and conservation of L is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

When is angular momentum conserved?

When the net external torque is zero: Li = Lf. Internal torques (muscles, collisions between parts) can redistribute L but never change the total.

Why does the skater speed up pulling her arms in?

No external torque acts (ice is nearly frictionless), so L = Iω is conserved. Pulling in shrinks I; with L fixed, ω must grow: ω2 = (I1/I2)ω1.

Is kinetic energy conserved too?

No! The skater’s muscles do work pulling in, so her rotational KE actually increases: K goes as 1/I. Conservation of L does not imply conservation of energy.

Do internal torques break conservation?

No — internal torques come in action-reaction pairs that cancel in the total. Only external torques (friction, applied pushes, gravity about a non-CM point) can change the system’s total L.

What is the formula for a shape-changing spinner?

I1ω1 = I2ω2. Compute I before and after the shape change (arms out vs in), set the products equal, and solve for the unknown.

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