\subsection{Conservation of Angular Momentum}

This subsection gives the rotational conservation law that parallels conservation of linear momentum. In AP mechanics, the key question is whether the net \emph{external} torque about a chosen origin or axis is zero over the interval of interest.

\dfn{Angular-momentum-isolated system}{Consider a system of particles labeled by an index $i=1,2,\dots,N$. Choose an origin $O$. Let $\vec{r}_i$ denote the position vector of particle $i$ relative to $O$, let $m_i$ denote its mass, let $\vec{v}_i$ denote its velocity, and let $\vec{p}_i=m_i\vec{v}_i$ denote its linear momentum. The total angular momentum of the system about $O$ is
\[
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
\]
Let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about the same origin. The system is said to be \emph{isolated for angular momentum about $O$} over a time interval if the external torque impulse about $O$ is zero:
\[
\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{0}.
\]
In particular, if $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ at every instant in the interval, then the system is angular-momentum-isolated about $O$. For a rigid body rotating about a fixed axis with unit vector $\hat{k}$, one often writes $\vec{L}=L\hat{k}$ and $\vec{\omega}=\omega\hat{k}$, so that in the fixed-axis case $L=I\omega$.}

\thm{Conservation law for angular momentum}{Let $\vec{L}_O$ denote the total angular momentum of a system about a chosen origin $O$, and let $\vec{\tau}_{\mathrm{ext},O}$ denote the net external torque about that same origin. Then
\[
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
Therefore, if the net external torque is zero throughout the interval, or more generally if the external torque impulse is zero, then
\[
\vec{L}_{O,f}=\vec{L}_{O,i},
\]
so total angular momentum about $O$ is conserved. In the common fixed-axis AP case,
\[
L_i=L_f
\qquad\Rightarrow\qquad
I_i\omega_i=I_f\omega_f
\]
when the net external torque about that axis is zero.}

\nt{Angular momentum is conserved only about the origin or axis for which the net external torque is zero, so the choice of origin matters. Internal forces and internal torques can redistribute angular momentum among parts of the system and can change the moment of inertia $I$, but they do not change the system's \emph{total} angular momentum about the chosen origin when $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$. Also, conservation of angular momentum does \emph{not} imply conservation of kinetic energy: for example, a skater can pull in her arms, decrease $I$, increase $\omega$, and increase rotational kinetic energy by doing internal work.}

\pf{Short derivation from $d\vec{L}/dt=\vec{\tau}_{\mathrm{ext}}$}{Start with the angular-momentum form of Newton's second law about the chosen origin $O$:
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrate from $t_i$ to $t_f$:
\[
\int_{t_i}^{t_f} \frac{d\vec{L}_O}{dt}\,dt=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
This gives
\[
\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
If $\vec{\tau}_{\mathrm{ext},O}=\vec{0}$ throughout the interval, or if the integral on the right is zero, then $\vec{L}_{O,f}-\vec{L}_{O,i}=\vec{0}$. Hence
\[
\vec{L}_{O,f}=\vec{L}_{O,i},
\]
which is the conservation of angular momentum. In a fixed-axis problem, this reduces to $I_i\omega_i=I_f\omega_f$.}

\qs{Worked example}{An ice skater spins about a vertical axis with unit vector $\hat{k}$. Assume the net external torque about that axis is negligible. With her arms extended, her moment of inertia is $I_i=3.0\,\mathrm{kg\cdot m^2}$ and her angular velocity is $\vec{\omega}_i=(2.0\,\mathrm{rad/s})\hat{k}$. She then pulls her arms inward so that her final moment of inertia is $I_f=1.2\,\mathrm{kg\cdot m^2}$.

Find:
\begin{enumerate}[label=(\alph*)]
\item the final angular velocity vector $\vec{\omega}_f$,
\item the initial and final angular momentum vectors, and
\item the initial and final rotational kinetic energies.
\end{enumerate}

Explain briefly why the energy result does not contradict conservation of angular momentum.}

\sol Because the net external torque about the vertical axis is negligible, angular momentum about that axis is conserved:
\[
\vec{L}_i=\vec{L}_f.
\]
For fixed-axis rotation, $\vec{L}=I\vec{\omega}$, so
\[
I_i\vec{\omega}_i=I_f\vec{\omega}_f.
\]

For part (a), solve for the final angular velocity:
\[
\vec{\omega}_f=\frac{I_i}{I_f}\vec{\omega}_i
=\frac{3.0}{1.2}(2.0\hat{k})
=(2.5)(2.0\hat{k})
=5.0\hat{k}\,\mathrm{rad/s}.
\]
So,
\[
\boxed{\vec{\omega}_f=(5.0\,\mathrm{rad/s})\hat{k}}.
\]

For part (b), compute the angular momentum before and after:
\[
\vec{L}_i=I_i\vec{\omega}_i=(3.0)(2.0\hat{k})=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
Because angular momentum is conserved,
\[
\vec{L}_f=\vec{L}_i=6.0\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]
Thus,
\[
\boxed{\vec{L}_i=\vec{L}_f=(6.0\,\mathrm{kg\cdot m^2/s})\hat{k}}.
\]

For part (c), use $K_{\mathrm{rot}}=\tfrac12 I\omega^2$.

Initially,
\[
K_{\mathrm{rot},i}=\tfrac12 I_i\omega_i^2
=\tfrac12 (3.0)(2.0)^2
=\tfrac12 (3.0)(4.0)
=6.0\,\mathrm{J}.
\]

Finally,
\[
K_{\mathrm{rot},f}=\tfrac12 I_f\omega_f^2
=\tfrac12 (1.2)(5.0)^2
=\tfrac12 (1.2)(25)
=15\,\mathrm{J}.
\]

So,
\[
\boxed{K_{\mathrm{rot},i}=6.0\,\mathrm{J}},
\qquad
\boxed{K_{\mathrm{rot},f}=15\,\mathrm{J}}.
\]

The rotational kinetic energy increases even though angular momentum stays constant. This does not contradict conservation of angular momentum because the skater does internal work while pulling in her arms. That internal work increases $K_{\mathrm{rot}}$ while the external torque remains negligible, so $\vec{L}$ is still conserved.
