Conservation of Angular Momentum

Equivalent forms

L = I \omega = \text{const}

\frac{dL}{dt} = \tau_{ext} = 0

From figure skaters to galaxies, the same one-line invariance governs why spinning systems speed up when they shrink.

Symbol	Name	SI unit	Dimension	Range
$omega_2$	Final angular velocityoutput Spin rate after redistribution	rad/s	T^-1	0 – 100
$I1$	Initial moment of inertia Pre-redistribution rotational inertia	kg·m²	ML^2	0.1 – 10
$omega_1$	Initial angular velocity Pre-redistribution spin rate	rad/s	T^-1	0.1 – 30
$I2$	Final moment of inertia Post-redistribution rotational inertia	kg·m²	ML^2	0.1 – 10

Unit systems

Symbol	SI	CGS	Imperial
$omega_2$	rad/s	rad/s	rad/s
$I1$	kg·m²	g·cm²	slug·ft²
$omega_1$	rad/s	rad/s	rad/s
$I2$	kg·m²	g·cm²	slug·ft²

Where it holds

Isolated system (zero net external torque); rigid or piecewise-rigid bodies. Quantum and relativistic generalizations exist.

Dimensional analysis

I_{2}\omega _{2} \to [ML^{2}][T^{-1}] = [ML^{2}T^{-1}]

I_{1}\omega _{1} \to [ML^{2}T^{-1}]

Both sides

kg\cdot m^{2}/s

.

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Discovery

Leonhard Euler / Pierre-Simon Laplace · 1736

Euler's Mechanica (1736) formulated rotational dynamics; Laplace and Lagrange used L-conservation to derive Kepler's areal law from the inverse-square force.

Try this

A spinning figure skater pulls her arms in — why does she suddenly speed up?

A skater spins at ω₁ = 3 rad/s with arms out (I₁ = 4 kg·m²). She pulls them in (I₂ = 1.2 kg·m²). Find her new spin rate.

Research status: stable

Real-world applications

Figure skater / diver / gymnast technique
Helicopter tail rotors (counter the body's reaction torque)
Pulsar formation from collapsing stars
Spacecraft attitude control via reaction wheels

Common misconceptions

Kinetic energy is conserved too — it ISN'T; pulling arms in adds energy (you do work)
L is a scalar — it's a vector (rotational direction matters)
L is always conserved — only if NET external torque is zero

Experimental verification

Rotating-stool demos with hand weights; pulsars (collapsed stellar cores spinning

\sim 1000\times /s)

; accretion-disk dynamics in astrophysics.

Derivation

From

dL/dt = \tau _ext

,

if \tau _ext = 0

then

L =

constant.

For a system whose moment of inertia changes from

I_{1} to I_{2} via

internal forces,

L = I_{1}\omega _{1} = I_{2}\omega _{2}

.

Limiting cases

I_{2} = I_{1}

⟶

\omega _{2} = \omega _{1}

No mass redistribution → spin unchanged.

I_{2} \to 0

⟶

\omega _{2} \to \infty

Collapsing toward axis: spin rate diverges (pulsar formation).

External torque

\neq 0

⟶ L not conserved

I_{1}\omega _{1} \neq I_{2}\omega _{2} if

friction or external force applies a torque.

What if…

Where does the extra energy come from?

The work done by internal forces pulling mass inward; $KE = \tfrac{1}{2}I\omega ^{2}$ grows as I shrinks.

What if external friction is present?

Apply $\tau$ _friction over time to find L(t); strict conservation breaks down.

Does this work for non-rigid bodies?

Yes — L of any isolated system is conserved regardless of shape changes.

1

Figure skater

Given ·

I1:: 4
omega 1:: 3
I2:: 1.2

Find · omega_2

Steps

$L = I_{1}\omega _{1} = 12 kg\cdot m^{2}/s$
$\omega _{2} = L/I_{2} = 12/1.2$
$\omega _{2} = 10\,\mathrm{rad/s}$ — over $3\times$ faster

Result ·

\omega _{2} = 4\cdot 3/1.2 = 10\,\mathrm{rad/s}

2

Neutron star collapse

Given ·

I1:: 1e+44
omega 1:: 0.000001
I2:: 1e+38

Find · omega_2

Steps

$L = 1e44 \times 1e-6 = 1e38$
$\omega _{2} = 1e38 / 1e38 = 100\,\mathrm{rad/s}$
Stellar-core collapse spins up the remnant pulsar dramatically

Result ·

\omega _{2} = 100\,\mathrm{rad/s} (\sim 16\,\mathrm{Hz})

Angular Momentum→Moment of Inertia (Point Mass)→Parallel Axis Theorem→Rotational Kinetic Energy→