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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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2
concepts/em/u11/e11-3-power.tex
View File

@@ -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.
\sol (a) From $P = I\,\Delta V$,
(a) From $P = I\,\Delta V$,
\[
I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
\]

6
concepts/em/u11/e11-5-kirchhoff.tex
View File

@@ -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):
\[
I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}.
\]

5
concepts/em/u12/e12-3-force-on-current.tex
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, $\mathrm{J/T}$.}
\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.

43
concepts/mechanics/u6/m6-4-rolling.tex
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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.
\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 $g=9.8\,\mathrm{m/s^2}$.