feat: complete AP Physics C handbook — all E11-E13 concepts, unit assemblies, and fixes

concepts/em/u11/e11-5-kirchhoff.tex

@@ -138,7 +138,6 @@ Substitute equation (1) into equation (2):
 \[
 7.0\,I_1 + 3.0\,I_2 = 12.
 \]
-\label{eq:a}
 
 Substitute equation (1) into equation (3):
 \[
@@ -158,9 +157,8 @@ so
 \[
 I_1 = 2.0 - 3.0\,I_2.
 \]
-\label{eq:b}
 
-Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
+Substitute equation (3) into equation (2):
 \[
 7.0\,(2.0 - 3.0\,I_2) + 3.0\,I_2 = 12,
 \]
@@ -177,7 +175,7 @@ Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
 I_2 = \frac{2.0}{18.0} = \frac{1}{9}\,\mathrm{A}.
 \]
 
-From equation (\ref{eq:b}):
+From equation (3):
 \[
 I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}.
 \]

concepts/em/u12/e12-3-force-on-current.tex

@@ -111,9 +111,10 @@ U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}.
 
 \nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.}
 
-\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:
+\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:}
 \[
-\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.}
+\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.
+\]
 
 \qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current.
 

concepts/mechanics/u6/m6-4-rolling.tex

@@ -32,48 +32,13 @@ Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves
 
 \mprop{Rolling kinematics, energy, and incline dynamics}{Consider a rigid body of mass $M$, radius $R$, and moment of inertia $I_{\mathrm{cm}}$ about its center of mass. Let $s_{\mathrm{cm}}$, $v_{\mathrm{cm}}$, and $a_{t,\mathrm{cm}}$ denote the center-of-mass distance, speed, and tangential acceleration, and let $\theta$, $\omega$, and $\alpha$ denote the corresponding angular variables. Let $K_i$ and $U_i$ denote the initial kinetic and potential energies, and let $K_f$ and $U_f$ denote the final kinetic and potential energies.
 
-\begin{enumerate}[label=\textbf{\arabic*.}]
-\item For pure rolling,
-\[
-s_{\mathrm{cm}}=R\theta,
-\qquad
-v_{\mathrm{cm}}=R\omega,
-\qquad
-a_{t,\mathrm{cm}}=R\alpha.
-\]
+1. For pure rolling, $s_{\mathrm{cm}}=R\theta$, $v_{\mathrm{cm}}=R\omega$, $a_{t,\mathrm{cm}}=R\alpha$.
 
-\item The total kinetic energy is the sum of translational and rotational parts:
-\[
-K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
-\]
-Using $v_{\mathrm{cm}}=R\omega$, this may also be written as
-\[
-K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2.
-\]
+2. The total kinetic energy is the sum of translational and rotational parts: $K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$. Using $v_{\mathrm{cm}}=R\omega$, this may also be written as $K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2$.
 
-\item If no nonconservative force removes mechanical energy from the system, then for rolling without slipping,
-\[
-K_i+U_i=K_f+U_f.
-\]
-For a vertical drop of magnitude $h$ from rest,
-\[
-Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
-\]
+3. If no nonconservative force removes mechanical energy from the system, then for rolling without slipping, $K_i+U_i=K_f+U_f$. For a vertical drop of magnitude $h$ from rest, $Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$.
 
-\item For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then
-\[
-Mg\sin\beta-f_s=Ma_{\mathrm{cm}},
-\qquad
-f_sR=I_{\mathrm{cm}}\alpha,
-\qquad
-a_{\mathrm{cm}}=R\alpha.
-\]
-Therefore,
-\[
-a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)},
-\qquad
-f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}.
-\]
+4. For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then $Mg\sin\beta-f_s=Ma_{\mathrm{cm}}$, $f_sR=I_{\mathrm{cm}}\alpha$, $a_{\mathrm{cm}}=R\alpha$. Therefore, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$.
 For an object accelerating down the incline, the static friction force on the object points up the incline.}
 
 \qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.