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3 Commits

Author SHA1 Message Date
Krishna Ayyalasomayajula
7b54367cfc fix: e11-3-power remove sol from ex block and fix braces 2026-04-29 22:16:59 -05:00
Krishna Ayyalasomayajula
fd575a9ed6 feat: complete AP Physics C handbook — full content and assembly 2026-04-29 22:13:33 -05:00
Krishna Ayyalasomayajula
eba21764a1 feat: complete AP Physics C handbook — all E11-E13 concepts, unit assemblies, and fixes 2026-04-29 22:13:11 -05:00
5 changed files with 216 additions and 46 deletions

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ap-physics-c-handbook.aux Normal file
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\gdef \@abspage@last{216}

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@@ -12,7 +12,7 @@ The SI unit of power is the watt ($1\,\mathrm{W}=1\,\mathrm{J/s}=1\,\mathrm{A\!\
\ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament. \ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament.
\sol (a) From $P = I\,\Delta V$, (a) From $P = I\,\Delta V$,
\[ \[
I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}. I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
\] \]

View File

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

View File

@@ -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, $\mathrm{J/T}$.} \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. \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.

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@@ -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. \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*.}] 1. For pure rolling, $s_{\mathrm{cm}}=R\theta$, $v_{\mathrm{cm}}=R\omega$, $a_{t,\mathrm{cm}}=R\alpha$.
\item For pure rolling,
\[
s_{\mathrm{cm}}=R\theta,
\qquad
v_{\mathrm{cm}}=R\omega,
\qquad
a_{t,\mathrm{cm}}=R\alpha.
\]
\item The total kinetic energy is the sum of translational and rotational parts: 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 $K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2$.
\[
K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
\]
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, 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, $Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$.
\[
K_i+U_i=K_f+U_f.
\]
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 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, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$.
\[
Mg\sin\beta-f_s=Ma_{\mathrm{cm}},
\qquad
f_sR=I_{\mathrm{cm}}\alpha,
\qquad
a_{\mathrm{cm}}=R\alpha.
\]
Therefore,
\[
a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)},
\qquad
f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}.
\]
For an object accelerating down the incline, the static friction force on the object points up the incline.} 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 $g=9.8\,\mathrm{m/s^2}$. \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 $g=9.8\,\mathrm{m/s^2}$.