\subsection{Angular Momentum and Angular Impulse}

This subsection connects torque to rotational motion in the same way that linear impulse connects force to linear momentum. The key AP idea is that a net external torque acting for a time interval changes angular momentum about a chosen point.

\dfn{Angular momentum and angular impulse}{Let $O$ denote a chosen reference point. Let a particle of mass $m$ have position vector $\vec{r}$ measured from $O$, velocity $\vec{v}$, and linear momentum $\vec{p}=m\vec{v}$. The angular momentum of the particle about $O$ is
\[
\vec{L}_O=\vec{r}\times \vec{p}.
\]
For a system of particles, the total angular momentum about $O$ is
\[
\vec{L}_O=\sum_i \vec{r}_i\times \vec{p}_i.
\]
Let $\vec{\tau}_{\mathrm{ext},O}(t)$ denote the net external torque about the same point $O$ over a time interval from $t_i$ to $t_f$. The angular impulse about $O$ over that interval is
\[
\vec{J}_{\tau,O}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
Its SI unit is $\mathrm{N\cdot m\cdot s}$, which is equivalent to $\mathrm{kg\cdot m^2/s}$. For a rigid body rotating about a fixed axis with moment of inertia $I$ about that axis and angular velocity $\vec{\omega}$, the angular momentum simplifies to
\[
\vec{L}=I\vec{\omega}.
\]
}

\thm{Torque-angular momentum relation and angular impulse theorem}{Let $O$ denote a chosen reference point. Let $\vec{L}_O(t)$ be the total angular momentum of a particle or system about $O$, and let $\vec{\tau}_{\mathrm{ext},O}(t)$ be the net external torque about $O$. Then
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrating from $t_i$ to $t_f$ gives
\[
\Delta \vec{L}_O=\vec{L}_{O,f}-\vec{L}_{O,i}=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt=\vec{J}_{\tau,O}.
\]
For fixed-axis rotation along a chosen axis with unit vector $\hat{k}$, let $\vec{L}=L\hat{k}$ and let $\vec{\tau}_{\mathrm{ext}}=\tau_{\mathrm{ext}}\hat{k}$. Then
\[
\frac{dL}{dt}=\tau_{\mathrm{ext}},
\qquad
\Delta L=\int_{t_i}^{t_f} \tau_{\mathrm{ext}}\,dt,
\]
and if $\tau_{\mathrm{ext}}$ is constant,
\[
\Delta L=\tau_{\mathrm{ext}}\Delta t.
\]
}

\nt{Angular momentum and torque depend on the chosen point $O$. The same object can have different $\vec{L}_O$ and $\vec{\tau}_{\mathrm{ext},O}$ when a different origin is chosen, so use the same reference point consistently throughout a problem. The direction of $\vec{L}$ and $\vec{\tau}$ is set by the right-hand rule and is perpendicular to the plane of the relevant cross product. In many AP fixed-axis problems, this vector bookkeeping reduces to signed scalars along one axis: counterclockwise may be chosen positive and clockwise negative. The simplification $\vec{L}=I\vec{\omega}$ is valid for rigid rotation about that fixed axis with the stated moment of inertia about the same axis.}

\pf{Short derivation from $\vec{L}=\vec{r}\times \vec{p}$}{For one particle about point $O$,
\[
\vec{L}_O=\vec{r}\times \vec{p}.
\]
Differentiate with respect to time:
\[
\frac{d\vec{L}_O}{dt}=\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r}\times \frac{d\vec{p}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times \vec{F}_{\mathrm{net}}.
\]
Because $\vec{v}\times \vec{v}=\vec{0}$,
\[
\frac{d\vec{L}_O}{dt}=\vec{r}\times \vec{F}_{\mathrm{net}}=\vec{\tau}_{\mathrm{net},O}.
\]
For a system, summing over all particles gives the same form with the net external torque:
\[
\frac{d\vec{L}_O}{dt}=\vec{\tau}_{\mathrm{ext},O}.
\]
Integrating from $t_i$ to $t_f$ yields
\[
\Delta \vec{L}_O=\int_{t_i}^{t_f} \vec{\tau}_{\mathrm{ext},O}\,dt.
\]
For a rigid body rotating about a fixed axis with constant moment of inertia $I$,
\[
\vec{L}=I\vec{\omega},
\]
so along that axis the relation becomes
\[
\vec{\tau}_{\mathrm{ext}}=I\vec{\alpha}.
\]
}

\qs{Worked example}{A flywheel rotates about a frictionless fixed axle through its center. Choose counterclockwise as positive, so the axis direction is $\hat{k}$ by the right-hand rule. The flywheel has moment of inertia $I=0.80\,\mathrm{kg\cdot m^2}$ about the axle. Initially its angular velocity is
\[
\vec{\omega}_i=12\,\hat{k}\,\mathrm{rad/s}.
\]
From $t=0$ to $t=1.5\,\mathrm{s}$, a brake pad exerts a constant external torque
\[
\vec{\tau}_{\mathrm{ext}}=-3.2\,\hat{k}\,\mathrm{N\cdot m}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the angular impulse delivered by the brake,
\item the flywheel's final angular momentum,
\item the flywheel's final angular velocity, and
\item whether the flywheel reverses direction during the interval.
\end{enumerate}}

\sol First compute the initial angular momentum using $\vec{L}=I\vec{\omega}$:
\[
\vec{L}_i=I\vec{\omega}_i=(0.80\,\mathrm{kg\cdot m^2})(12\,\hat{k}\,\mathrm{rad/s})=9.6\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]

For part (a), the torque is constant, so the angular impulse is
\[
\vec{J}_\tau=\vec{\tau}_{\mathrm{ext}}\Delta t.
\]
Thus
\[
\vec{J}_\tau=(-3.2\,\hat{k}\,\mathrm{N\cdot m})(1.5\,\mathrm{s})=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s}.
\]
Using $1\,\mathrm{N\cdot m\cdot s}=1\,\mathrm{kg\cdot m^2/s}$,
\[
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]

For part (b), apply the angular impulse theorem:
\[
\Delta \vec{L}=\vec{J}_\tau=\vec{L}_f-\vec{L}_i.
\]
So
\[
\vec{L}_f=\vec{L}_i+\vec{J}_\tau=(9.6-4.8)\,\hat{k}\,\mathrm{kg\cdot m^2/s}=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}.
\]

For part (c), again use $\vec{L}=I\vec{\omega}$:
\[
\vec{\omega}_f=\frac{\vec{L}_f}{I}=\frac{4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s}}{0.80\,\mathrm{kg\cdot m^2}}=6.0\,\hat{k}\,\mathrm{rad/s}.
\]

For part (d), the final angular velocity still points in the $+\hat{k}$ direction, so the flywheel is still rotating counterclockwise. It slows down, but it does not reverse direction during the $1.5\,\mathrm{s}$ interval.

The results are
\[
\vec{J}_\tau=-4.8\,\hat{k}\,\mathrm{N\cdot m\cdot s},
\qquad
\vec{L}_f=4.8\,\hat{k}\,\mathrm{kg\cdot m^2/s},
\qquad
\vec{\omega}_f=6.0\,\hat{k}\,\mathrm{rad/s}.
\]
This example shows the key AP idea: a torque acting over time changes angular momentum directly, and the sign of the torque determines whether the wheel speeds up or slows down.
