Conservation of Angular Momentum
Introduced| English | Meaning |
|---|---|
| angular momentum is conserved/ˈæŋɡjʊlə məʊˈmentəm ɪz kənˈsɜːvd/ | angular momentum is conserved |
The skater's secret
- A spinning skater pulls in her arms and suddenly whirls faster.
- She pushed off nothing — no one gave her a shove.
- Yet her spin rate jumps dramatically.
- When nothing twists a spinning system, a hidden quantity stays fixed.
When the spin is conserved
- If the net external torque is zero, angular momentum is conserved 角动量守恒:
- The total turning stays the same, however the mass rearranges.
- It is the rotational twin of momentum conservation.
Angular momentum is conserved exactly when...
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Zero net external torque means $dL/dt = 0$, so $L$ stays constant.
Shrink I, spin faster
- Because $I\omega$ is fixed, lowering $I$ forces $\omega$ up.
- The skater pulls her arms in, $I$ drops, and $\omega$ rises.
- She stretches back out, $I$ grows, and the spin slows again.
A skater with $I_i = 8\ \text{kg}\cdot\text{m}^2$ spins at $2\ \text{rad/s}$, then pulls in to $I_f = 4\ \text{kg}\cdot\text{m}^2$. Her new angular speed (in rad/s)?
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$I_i\omega_i = I_f\omega_f \Rightarrow 8\times2 = 4\times\omega_f$, so $\omega_f = 4\ \text{rad/s}$.
With angular momentum conserved, decreasing the rotational inertia increases the angular speed.
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$I\omega$ is fixed, so a smaller $I$ forces a larger $\omega$.
From skaters to stars
- A collapsing star spins up as it shrinks — the same rule.
- Two spinning discs that lock together share their total $L$.
- Divers and gymnasts tuck to flip faster, then open to slow down.
Conservation of angular momentum
With no external torque, a spinning skater speeds up by pulling in.
A star's rotational inertia drops to one-quarter of its old value as it collapses. Its spin rate becomes how many times faster?
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$I\omega$ fixed and $I \to I/4$ means $\omega \to 4\omega$ -- four times faster.
A skater has $I_i = 6\ \text{kg}\cdot\text{m}^2$ spinning at $\omega_i = 2\ \text{rad/s}$. She pulls in to $I_f = 2\ \text{kg}\cdot\text{m}^2$.
- $I_i\omega_i = I_f\omega_f \Rightarrow 6 \times 2 = 2 \times \omega_f$.
- So $\omega_f = 6\ \text{rad/s}$ — three times faster.
The skater's own muscles pulling her arms in are internal forces. Can they change her angular momentum?
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Internal forces come in pairs with no net torque; only external torques change $L$.
When the skater pulls her arms in, her rotational kinetic energy stays exactly the same.
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She does work pulling in, so $K_{rot}$ rises even though $L$ is conserved.
The spin is conserved only when the net external torque is zero. Internal forces (the skater's muscles) can't change $L$, but friction, a gravity torque, or an outside push can. Note too: kinetic energy is not conserved here — the skater does work pulling her arms in, so $K_{rot}$ actually rises.
With zero net external torque, angular momentum is conserved: $I_i\omega_i = I_f\omega_f$. Shrinking $I$ forces $\omega$ up — the skater, the collapsing star, the tucking diver. Only external torques can change $L$; internal forces cannot. And note the spin's kinetic energy need not stay the same.
Write from a new spin of 3 rad/s
- Rotational inertia falls from 6 to 2. The initial spin is 1 rad/s.
- No external torque acts. Find the new spin.
An isolated skater's rotational inertia drops from 6 to 2 kg m squared. The spin starts at 1 rad/s. What is the new angular speed?
6 divided by 2 is 3.
What is the new angular speed?
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The product stays 6, so the new speed is 3 rad/s.