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concepts/mechanics/u6/m6-2-angular-momentum.tex

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+\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.