content(warnings): Add W17-W28, X8-X12, N7 — E&M misconceptions, cross-refs, notes

concepts/em/u11/e11-1-current-density.tex

@@ -14,6 +14,8 @@ dI=\vec{J}\cdot d\vec{A}.
 \]
 Thus $\vec{J}$ links the local flow of charge to the total current through a cross section.}
 
+\wc{Electrons flow opposite to conventional current}{The \emph{conventional current} direction is defined as the direction positive charges would move (from higher to lower potential). In metal wires, the actual charge carriers are electrons, which move \emph{opposite} to the conventional current direction. This historical convention does not affect circuit analysis results.}
+
 \nt{Conventional current is defined to point in the direction that positive charges would move. Therefore $\vec{J}$ points in the conventional-current direction. In a metal wire, the mobile charge carriers are electrons, so $q=-e$ and the electron drift velocity $\vec{v}_d$ points opposite to $\vec{J}$. If the carriers were positive instead, then $\vec{v}_d$ and $\vec{J}$ would point in the same direction.}
 
 \mprop{Microscopic-macroscopic current relations}{Let $n$ denote the carrier number density, let $q$ denote the charge of each carrier, let $\vec{v}_d$ denote the drift velocity, and let $S$ be a surface with oriented area element $d\vec{A}$. Then the current density is

concepts/em/u11/e11-2-resistance-ohm.tex

@@ -8,6 +8,8 @@ R = \frac{V}{I}.
 \]
 The SI unit of resistance is the \emph{ohm}, denoted $\Omega$, where $1\,\Omega = 1\,\mathrm{V/A}$.}
 
+\wc{Current is not ``used up'' by resistors}{Current $I$ is the same through every element in a series circuit. A resistor does not ``consume'' current --- it drops voltage (potential energy per charge). Think of the circuit as a closed loop: whatever current leaves the battery returns to it. What is ``used up'' is electrical potential energy (converted to heat), not charge.}
+
 \dfn{Resistivity, conductivity}{The \emph{resistivity} $\rho$ of a material is an intrinsic property that quantifies how strongly that material opposes the flow of electric current. The \emph{conductivity} $\sigma$ is the reciprocal of resistivity:
 \[
 \sigma = \frac{1}{\rho}.

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

@@ -49,6 +49,8 @@ The assumed direction of each branch current must be declared before writing equ
 \end{enumerate}
 The number of independent equations needed equals the number of unknown branch currents.}
 
+Kirchhoff's rules generalize the equivalent-resistance method from Section 11.4. For single-battery circuits with pure series/parallel arrangements, equivalent resistance is faster. Kirchhoff's rules become necessary when multiple batteries or loops make reduction impossible.
+
 \ex{Illustrative example}{Consider a junction with three wires meeting. Current $I_1 = 3.0\,\mathrm{A}$ enters the junction and current $I_2 = 1.5\,\mathrm{A}$ leaves. The third branch carries current $I_3$. By the junction rule, $I_1 = I_2 + I_3$, so $I_3 = 1.5\,\mathrm{A}$ leaves the junction.}
 
 \nt{For circuits with a single battery and resistors in simple series or parallel, the equivalent-resistance method is faster. Kirchhoff's rules are needed whenever the circuit cannot be reduced to simple series/parallel combinations --- for instance, when there are two or more batteries arranged in different branches, forming multiple loops.}

concepts/em/u11/e11-6-rc-circuits.tex

@@ -16,6 +16,8 @@ I(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC}.
 \]
 Here $q(0) = 0$ and $I(0) = \mathcal{E}/R$. As $t \to \infty$, $q \to C\mathcal{E}$ and $I \to 0$.}
 
+\nt{The RC time constant $\tau=RC$ has units of seconds: $[\Omega]\cdot[\mathrm{F}]=[\mathrm{V/A}]\cdot[\mathrm{C/V}]=[\mathrm{C/A}]=[\mathrm{C/(C/s)}]=[\mathrm{s}]$. After one time constant during charging, the capacitor reaches $1-e^{-1}\approx 63.2\%$ of its final voltage. After five time constants, it reaches $99.3\%$, which is the practical definition of ``fully charged'' in circuit analysis.}
+
 \pf{Derivation of charging equations}{Apply Kirchhoff's voltage law around the loop. The potential drops across the resistor and capacitor sum to the battery emf:
 \[
 \mathcal{E} - IR - \frac{q}{C} = 0,

concepts/em/u12/e12-1-magnetic-force-charge.tex

@@ -24,6 +24,8 @@ where $\theta$ is the angle between the vectors $\vec{v}$ and $\vec{B}$ measured
 \item \textbf{SI unit of $B$:} The tesla, $\mathrm{T} = \dfrac{\mathrm{N}}{\mathrm{C}\cdot\mathrm{m/s}} = \dfrac{\mathrm{N}}{\mathrm{A}\cdot\mathrm{m}} = \dfrac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}$.
 \end{itemize}}
 
+The magnetic force on a single charge (this section) generalizes to the magnetic force on a current-carrying wire (Section 12.3) by summing over all charge carriers. The Biot-Savart law (Section 12.4) gives the reverse: how currents produce the $\vec{B}$ field that exerts these forces.
+
 \pf{Lorentz magnetic force law from cross-product geometry}{The vector cross product $\vec{v}\times\vec{B}$ is defined to have magnitude $vB\sin\theta$ and direction given by the right-hand rule. Multiplying by $q$ scales the magnitude by $|q|$ and reverses direction if $q<0$. Thus
 \[
 \vec{F}_B = q\,(\vec{v}\times\vec{B})
@@ -32,6 +34,8 @@ has magnitude $|q|vB\sin\theta$ and the correct directional behaviour. This is t
 
 \cor{Charge at rest or parallel to field}{When $\vec{v}=\vec{0}$ or when $\vec{v}\parallel\vec{B}$, we have $\sin\theta=0$ and therefore $F_B=0$. The magnetic field exerts no force on a stationary charge or on a charge moving exactly along the field lines.}
 
+\wc{A stationary charge produces zero magnetic field}{A point charge at rest produces only an \emph{electric} field. A magnetic field is produced only by \emph{moving} charges (currents) or changing electric fields (displacement current). A single stationary charge $q$ has $\vec{B}=\vec{0}$ everywhere.}
+
 \mprop{Magnetic force vs.\ electric force}{For the same charge $q$ placed in both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, the total Lorentz force is
 \[
 \vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}.
@@ -49,6 +53,8 @@ P = \vec{F}_B\cdot\vec{v} = q\,(\vec{v}\times\vec{B})\cdot\vec{v} = 0.
 \]
 The magnetic force can change the direction of a particle's velocity but never its kinetic energy. This is the mathematical expression of the scalar triple-product identity $(\vec{a}\times\vec{b})\cdot\vec{a}=0$.}
 
+\wc{Magnetic force does zero work --- always}{Because $\vec{F}_B=q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, the power $P=\vec{F}_B\cdot\vec{v}=0$. The magnetic force can change a particle's direction but never its speed or kinetic energy. Even in complex field configurations, the magnetic force contributes zero to the work integral $\int\vec{F}\cdot d\vec{r}$.}
+
 \ex{Illustrative example}{When a charged particle enters a uniform magnetic field perpendicularly, it follows a circular path. The magnetic force provides the centripetal force:
 \[
 |q|\,v\,B = \frac{m\,v^2}{R} \quad\Rightarrow\quad R = \frac{m\,v}{|q|\,B},

concepts/em/u12/e12-2-particle-motion-in-b.tex

@@ -72,6 +72,8 @@ The sense of the circular rotation follows the same rule as cyclotron motion: co
 
 \nt{If $\alpha=0^\circ$, then $v_{\perp}=0$ and the particle travels in a straight line along $\vec{B}$ (no magnetic force). If $\alpha=90^\circ$, then $v_{\parallel}=0$ and the particle undergoes pure circular motion (no drift along $\vec{B}$). Helical motion interpolates between these two extremes. The pitch increases as $\alpha\to 0^\circ$ and approaches zero as $\alpha\to 90^\circ$.}
 
+\wc{Field lines are a visualization tool, not physical threads}{Magnetic field lines are a mathematical construct to visualize $\vec{B}$. The field is continuous and exists at every point --- field lines are just a convenient drawing. Charges do not follow field lines (except when $\vec{v}\parallel\vec{B}$). The density of lines indicates field strength, but the lines themselves have no physical substance.}
+
 \mprop{Cyclotron and helical motion parameters}{For a particle of mass $m$ and charge $q$ in a uniform magnetic field $\vec{B}$, with velocity $\vec{v}$ at angle $\alpha$ to $\vec{B}$:
 \begin{align}
 R&=\frac{m\,v\,\sin\alpha}{|q|\,B} && \text{(helix radius)} \\

concepts/em/u12/e12-5-ampere.tex

@@ -12,6 +12,8 @@ This subsection states Amp\`ere's law and shows how symmetry can reduce a diffic
 \]
 This law is always true. It becomes a practical method for solving for the magnetic field when the current distribution has enough symmetry that one can choose an Amperian loop for which the magnitude $B=|\vec{B}|$ is constant on each field-contributing part of the loop and the angle between $\vec{B}$ and $d\vec{\ell}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the line integral reduces to algebraic terms such as $B\ell$, $-B\ell$, or $0$. Common useful cases are cylindrical symmetry (long straight wires), planar symmetry (infinite current sheets), and solenoidal symmetry (ideal solenoids). The direction of $\vec{B}$ follows the right-hand rule relative to the enclosed current: if the thumb of your right hand points in the direction of the current, your fingers curl in the direction of the magnetic field circulation.}
 
+\wc{Ampere's law is always true for steady currents}{Like Gauss's law, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}$ holds for \emph{any} steady current distribution and \emph{any} closed path. Symmetry only makes it \emph{useful} for calculating $\vec{B}$. Without symmetry, you know the line integral but cannot extract $\vec{B}$ at each point.}
+
 \nt{Amp\`ere's law is the magnetic analogue of Gauss's law. Gauss's law relates the electric field flux through a closed surface to the enclosed charge, $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$. Amp\`ere's law relates the magnetic field circulation around a closed loop to the enclosed current, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}$. Both are universally valid but are practically useful for finding fields only when the source distribution has high symmetry. The matching of symmetry to geometry is parallel: spherical symmetry $\to$ spherical Gaussian surface, cylindrical symmetry $\to$ circular Amperian loop, planar symmetry $\to$ rectangular Amperian loop.}
 
 \pf{How symmetry reduces the line integral}{Let a long straight wire carry current $I$ along the $+z$ axis. By cylindrical symmetry, the magnetic field circulates around the wire in concentric circles in planes perpendicular to the wire, and its magnitude $B(r)$ depends only on the radial distance $r$ from the wire axis. Choose a circular Amperian loop of radius $r$ centred on the wire. Along this loop, $\vec{B}$ is everywhere tangent to $d\vec{\ell}$, so $\vec{B}\cdot d\vec{\ell}=B(r)\,d\ell$, and $B(r)$ is constant everywhere on the loop. Therefore,

concepts/em/u13/e13-2-faraday.tex

@@ -18,6 +18,8 @@ where $d\vec{A}$ is the oriented area element (direction given by the right-hand
 \]
 The minus sign encodes Lenz's law: the induced current produces a magnetic field that opposes the change in flux that created it.}
 
+\wc{Faraday's law creates EMF, not necessarily current}{A changing magnetic flux induces an \emph{electromotive force} $\mathcal{E}=-d\Phi_B/dt$. Whether current flows depends on whether the loop is conducting and whether there is a complete circuit. A changing flux through an open loop or a broken ring creates EMF (a potential difference) but no current flows.}
+
 \pf{Derivation from Faraday's law}{
 The EMF is the work per unit charge by the induced non-conservative field: $\mathcal{E}=\oint_C\vec{E}\cdot d\vec{\ell}$. When the magnetic flux $\Phi_B=\int_S\vec{B}\cdot d\vec{A}$ through the loop changes in time, a non-conservative electric field is induced with non-zero circulation. Energy conservation requires this circulation to equal the rate of flux change (with the minus sign from Lenz's law):
 \[

concepts/em/u13/e13-3-lenz.tex

@@ -20,10 +20,14 @@ The operational procedure is:
 \item Use the right-hand rule on $\vec{B}_{\text{ind}}$ to find the induced current direction: curl the fingers of your right hand in the current direction; your thumb points along $\vec{B}_{\text{ind}}$.
 \end{enumerate}}
 
+\wc{Lenz's law opposes the \emph{change} in flux, not the field itself}{The induced current creates a field that opposes the \emph{change} in magnetic flux, not the external field itself. If flux is \emph{increasing}, the induced field opposes the external field. If flux is \emph{decreasing}, the induced field is in the \emph{same} direction as the external field (to try to maintain it).}
+
 \nt{The negative sign in Faraday's law \emph{is} Lenz's law written as an equation. If the sign were positive, the induced current would reinforce the flux change, producing more flux in the same direction, which would drive yet more current --- an energy-creating runaway. Lenz's law prevents this by ensuring the induced field opposes the change.}
 
 \thm{Lenz's law (energy-conservation form)}{The direction of induced current in any closed loop is always such that the magnetic force or torque on the loop opposes the motion or change that produced the induction. Equivalently, mechanical work must be done against the magnetic forces to sustain the change in flux; this work is converted to electrical energy (and ultimately to thermal energy in the resistance of the loop).}
 
+Lenz's law is the directional consequence of Faraday's law (Section 13.2) and is deeply connected to energy conservation. The same principle governs motional EMF in Section 13.4 and inductance in Section 13.5.
+
 \pf{Lenz's law from energy conservation}{Suppose a magnet is pushed toward a conducting loop. The induced current creates a magnetic field $\vec{B}_{\text{ind}}$ that opposes the approaching magnet. An external agent must do positive work against the magnetic repulsion to keep the magnet moving. This work supplies the electrical energy dissipated as Joule heating in the loop.
 
 If Lenz's law were reversed --- if the induced field \emph{aided} the approaching magnet --- the magnet would accelerate toward the loop without any external work, increasing both the kinetic energy of the magnet and the electrical energy dissipated in the loop, with no energy input. This violates conservation of energy. Therefore, the minus sign in Faraday's law is required by energy conservation. \Qed}

concepts/em/u13/e13-4-motional-emf.tex

@@ -100,6 +100,8 @@ Thus $P_{\text{mech}} = P_{\text{elec}}$, consistent with conservation of energy
 \]
 This is exactly the motional-emf result $\mathcal{E} = B\ell v$. Faraday's law provides the same emf through the ``flux-change'' perspective, while the motional-emf derivation provides it through the ``force-on-charges'' perspective.}
 
+The motional EMF $\mathcal{E}=B\ell v$ is consistent with Faraday's law $\mathcal{E}=-d\Phi_B/dt$ through the flux change perspective. Both derivations (force-on-charges and flux-change) appear in Sections 13.2 and 13.4 respectively and yield identical results.
+
 \ex{Illustrative example}{A metal rod of length $\ell$ rotates with angular speed $\omega$ about one end in a uniform magnetic field $B$ perpendicular to the plane of rotation. Different points on the rod have different speeds $v(r) = r\,\omega$, so the emf must be computed by integration:
 \[
 \mathcal{E} = \int_{0}^{\ell} B\,(r\omega)\,dr = \frac{1}{2}\,B\,\omega\,\ell^{2}.\]}

concepts/em/u13/e13-5-inductance.tex

@@ -12,6 +12,8 @@ Equivalently, from Faraday's law of induction, a changing current induces an EMF
 \]
 where the negative sign reflects Lenz's law: the induced EMF opposes the change in current. The SI unit of inductance is the henry (H), where $1\;\mathrm{H} = 1\;\mathrm{V\!\cdot\!s/A} = 1\;\mathrm{kg\!\cdot\!m^2/(s^2\!\cdot\!A^2)}$.}
 
+\wc{An inductor opposes \emph{change} in current, not current itself}{An ideal inductor with steady (DC) current has zero voltage drop across it ($V=0$ when $dI/dt=0$). It only produces a back-EMF $\mathcal{E}=-L(dI/dt)$ when the current is \emph{changing}. A steady current passes through an ideal inductor as freely as through a wire.}
+
 \nt{Inductance is purely a geometric property. For a fixed geometry (and no ferromagnetic material near the coil), $L$ is constant and independent of the current. The larger the coil, the more turns, and the greater the flux linkage per unit current, the larger the inductance. A coil with $L = 1\;\mathrm{H}$ and $dI/dt = 1\;\mathrm{A/s}$ develops a $1\;\mathrm{V}$ back EMF.}
 
 \thm{Solenoid self-inductance}{For an ideal long solenoid of length $\ell \gg R$, total turns $N$, cross-sectional area $A = \pi R^2$, and vacuum permeability $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$, the magnetic field inside is $B = \mu_0\,n\,I$ where $n = N/\ell$. The flux through each turn is $\Phi_B = B\,A = \mu_0 N I A/\ell$. Therefore

concepts/em/u8/e8-3-electric-field.tex

@@ -8,6 +8,8 @@ This subsection defines the electric field from source charges and shows how it
 \]
 The source charges are the charges that produce the field. The test charge is a charge used only to probe the field at a location. Because the factor of $q_0$ divides out, the field depends on the source-charge configuration and the location $\vec{r}$, not on the particular test charge used to measure it. By convention, the direction of $\vec{E}$ is the direction of the force on a positive test charge. The SI units of electric field are $\mathrm{N/C}$.}
 
+\wc{Electric field fills all space (even if zero)}{The electric field $\vec{E}$ is defined at every point in space, regardless of whether a test charge is present. At points far from all source charges, the field magnitude may be very small or effectively zero, but the field itself extends throughout space. A test charge placed anywhere will experience $\vec{F}=q\vec{E}$ at that location.}
+
 \thm{Point-charge field law and force relation}{Let a point source charge $Q$ be fixed at position vector $\vec{r}_Q$. Let the field point have position vector $\vec{r}$, and define
 \[
 \vec{R}=\vec{r}-\vec{r}_Q,

concepts/em/u8/e8-6-gauss-law.tex

@@ -10,8 +10,14 @@ This subsection states Gauss's law and shows how symmetry can reduce a difficult
 \]
 This law is always true. It becomes a practical method for solving for the electric field when the charge distribution has enough symmetry that one can choose a Gaussian surface for which the magnitude $E=|\vec{E}|$ is constant on each flux-contributing part of the surface and the angle between $\vec{E}$ and $d\vec{A}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the flux integral reduces to algebraic terms such as $EA$, $-EA$, or $0$. Common useful cases are spherical, cylindrical, and planar symmetry.}
 
+Gauss's law is one of Maxwell's four equations (in electrostatic form). Its analogue for magnetism is Gauss's law for magnetic fields, $\oint\vec{B}\cdot d\vec{A}=0$ (Section 12.5), which reflects the absence of magnetic monopoles.
+
 \nt{Gauss's law is always true, but it is not always useful for finding $\vec{E}$. In a general asymmetric charge distribution, knowing only the total flux through a closed surface does not tell you the field at each point on that surface. Also, zero net enclosed charge implies zero \emph{net flux}, not necessarily zero field everywhere. The main strategy is therefore: first identify strong symmetry, then choose a Gaussian surface matched to that symmetry.}
 
+\wc{Gauss's law is always true, not just for symmetric cases}{Gauss's law $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$ holds for \emph{any} charge distribution and \emph{any} closed surface. Symmetry only makes it \emph{useful} for calculating $\vec{E}$. Without symmetry, you know the total flux but cannot solve for the field at each point on the surface.}
+
+\wc{Zero net enclosed charge does not mean zero field}{A Gaussian surface enclosing zero net charge has zero \emph{net flux}, but the electric field on the surface need not be zero. External charges can produce nonzero field on the surface, with field lines entering on one side and leaving on the other, yielding zero net flux. $\oint\vec{E}\cdot d\vec{A}=0$ does \emph{not} imply $\vec{E}=0$.}
+
 \pf{Why symmetry reduces the integral}{Let a point charge $Q$ be at the center of a spherical Gaussian surface of radius $r$. By spherical symmetry, the electric field is radial and has the same magnitude $E(r)$ at every point on the sphere. The outward area element $d\vec{A}$ is also radial, so
 \[
 \vec{E}\cdot d\vec{A}=E(r)\,dA

concepts/mechanics/u2/m2-1-newton-laws.tex

@@ -32,6 +32,8 @@ This is the main working law for AP mechanics.
 These two forces act on different bodies, so they do not cancel on a single free-body diagram.
 \end{enumerate}}
 
+\wc{Net force causes acceleration, not motion}{A body can move at constant velocity with \emph{zero} net force (Newton's law I). Net force determines \emph{how} motion changes, not \emph{if} motion exists. An object with zero net force does not stop --- it continues at whatever velocity it already had.}
+
 \pf{Why the vector law becomes component equations}{Choose Cartesian axes with unit vectors $\hat{\imath}$ and $\hat{\jmath}$. Let the acceleration be
 \[
 \vec{a}=a_x\hat{\imath}+a_y\hat{\jmath},
@@ -56,6 +58,8 @@ In practice, one first draws only the actual forces on the free-body diagram, th
 
 \cor{Zero net force and equilibrium}{If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then Newton II gives $\vec{a}=\vec{0}$. This does \emph{not} mean the velocity must be zero. A body can have zero net force while moving with a nonzero constant velocity. A special case is equilibrium: if a body is initially at rest and $\vec{F}_{\mathrm{ext}}=\vec{0}$, then it remains at rest.}
 
+See also the kinematics in Unit 1 for the constant-velocity case, and Unit 3 for the work-energy connection to zero net force.
+
 \qs{Worked example}{A block of mass $m=5.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field near Earth. Choose the system to be the block.
 
 Find the acceleration of the block and the magnitude of the normal force exerted by the incline on the block.}
@@ -114,6 +118,9 @@ Therefore,
 \[
 N=mg\cos\theta.
 \]
+
+\wc{Normal force is not always equal to weight}{On a horizontal surface with only gravity and normal force, $N=mg$. But on an incline $N=mg\cos\theta$, and if another force pushes on the object, $N$ adjusts accordingly. The normal force is \emph{whatever value is required} to prevent the object from penetrating the surface. Never assume $N=mg$.}
+
 Substitute the stated values:
 \[
 N=(5.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.

concepts/mechanics/u2/m2-7-drag-terminal.tex

@@ -10,6 +10,10 @@ The negative sign means that the drag force always points opposite the velocity.
 
 If an object falls vertically through the fluid and eventually moves with constant velocity, that steady velocity is called the \emph{terminal velocity}. At terminal velocity, the net force is zero, so the acceleration is zero.}
 
+\wc{Terminal velocity is not zero speed}{Terminal velocity is a \emph{nonzero constant speed}: the object is still moving, but its acceleration is zero because drag balances the driving force. The object approaches $v_T$ asymptotically as $t\to\infty$; it never stops.}
+
+\wc{Drag force is not kinetic friction}{Drag $F_D=bv$ or $F_D=\tfrac12 C\rho A v^2$ is velocity-dependent and acts in fluids. Kinetic friction $f_k=\mu_k N$ is approximately velocity-independent and acts at solid-solid contacts. They are physically distinct interactions.}
+
 \thm{Vertical-fall ODE and terminal-speed result}{Choose the vertical axis positive downward. Let $v(t)$ denote the downward velocity of an object of mass $m$ at time $t$, let $g$ denote the gravitational field strength, and let $b>0$ denote the linear-drag coefficient. Then the vertical equation of motion is
 \[
 m\frac{dv}{dt}=mg-bv.
@@ -32,6 +36,8 @@ In particular, if the object is released from rest, then
 v(t)=v_T\left(1-e^{-bt/m}\right).
 \]}
 
+\nt{The linear drag model $F_D=bv$ is valid for low-Reynolds-number flow (small objects, slow speeds, viscous fluids). For most everyday situations (baseballs, skydivers, cars at highway speed), drag is approximately quadratic: $F_D=\tfrac12 C\rho A v^2$, where $C$ is the drag coefficient, $\rho$ is the fluid density, and $A$ is the cross-sectional area. The AP exam typically uses the linear model for analytical tractability, but recognizes the quadratic model qualitatively.}
+
 \pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction:
 \[
 m\frac{dv}{dt}=mg-bv.

concepts/mechanics/u3/m3-1-work.tex

@@ -19,6 +19,8 @@ Therefore the \emph{total work} done by the force as the particle moves along th
 W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds.
 \]}
 
+This definition of work is the foundation for the work-energy theorem (Section 3.2) and the potential-energy formalism (Section 3.3).
+
 \thm{Line-integral form of work and the constant-force special case}{Let a particle move from an initial point to a final point along a path $C$ under a force $\vec{F}$. Then the work done by that force is
 \[
 W=\int_C \vec{F}\cdot d\vec{r}.
@@ -39,6 +41,8 @@ W=F\,\Delta s.
 
 \nt{Work is positive when the force has a component in the same direction as the displacement, negative when that component is opposite the displacement, and zero when the force is perpendicular to the displacement. For a general force, the value of $W=\int_C \vec{F}\cdot d\vec{r}$ can depend on the path $C$, not just on the endpoints. In AP problems, this often appears when the force changes with position or when different paths make the parallel component $F_{\parallel}$ different. Typical zero-work cases include a normal force on motion along a surface or a centripetal force in uniform circular motion, because those forces are perpendicular to the instantaneous displacement.}
 
+\wc{Holding something still does no work}{If you hold a heavy object stationary, you expend biological energy but do \emph{zero mechanical work} because $\Delta\vec{r}=\vec{0}$. Work requires both a force \emph{and} a displacement along that force: $W=\int\vec{F}\cdot d\vec{r}$.}
+
 \pf{Why the line integral gives total work}{Break the path into many small displacement vectors $\Delta \vec{r}_1,\Delta \vec{r}_2,\dots,\Delta \vec{r}_n$. Over each small segment, the work is approximately
 \[
 \Delta W_k\approx \vec{F}_k\cdot \Delta \vec{r}_k.

concepts/mechanics/u3/m3-2-work-energy.tex

@@ -35,6 +35,8 @@ W_{\text{net}}=\Delta K=K_f-K_i=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
 \]
 Thus the net work done on a body equals the change in its kinetic energy.}
 
+\wc{Net work is the algebraic sum}{Net work $W_{\mathrm{net}}=W_1+W_2+\cdots$ includes the \emph{signs} of individual works. If one force does $+120\,\mathrm{J}$ and another does $-24\,\mathrm{J}$, the net work is $+96\,\mathrm{J}$, \emph{not} $144\,\mathrm{J}$.}
+
 \nt{The work-energy theorem is often more efficient than combining Newton's second law with kinematics when the problem asks for a speed after a known displacement or after a known amount of work. It avoids solving for time and often avoids solving for acceleration explicitly. In AP mechanics, \emph{net work} means the algebraic sum of the work done by all forces on the chosen system. A force parallel to the displacement does positive work, a force opposite the displacement does negative work, and a force perpendicular to the displacement does zero work.}
 
 \pf{Short derivation from Newton II}{Let $m$ denote the constant mass of the body, let $\vec{v}$ denote its velocity, and let $d\vec{r}$ denote its infinitesimal displacement. Start with Newton's second law,

concepts/mechanics/u3/m3-3-potential-energy.tex

@@ -17,6 +17,10 @@ Thus the work done by the conservative force is
 W_c=-\Delta U.
 \]}
 
+\nt{The formula $U=mgh$ is a special case valid only near Earth's surface where $g$ is approximately constant. The general gravitational potential energy between two masses $M$ and $m$ separated by distance $r$ is $U=-GMm/r$. When $h\ll R_{\oplus}$, expanding $-GMm/(R_{\oplus}+h)$ to first order in $h/R_{\oplus}$ gives $U\approx- GMm/R_{\oplus} + (GMm/R_{\oplus}^2)h = \text{const} + mgh$, since $g=GM_{\oplus}/R_{\oplus}^2$. Setting $U=0$ at ground level drops the constant term.}
+
+\wc{Potential energy belongs to a system}{Gravitational $U=mgh$ requires both the object \emph{and} the Earth. Spring $U=\tfrac12 kx^2$ requires both the block \emph{and} the spring. An isolated single object cannot have potential energy --- it is stored in the \emph{interaction}.}
+
 \thm{Equivalent conservative-force relations}{Let $\vec{F}_c$ denote a conservative force and let $d\vec{r}$ denote an infinitesimal displacement. Then the following relations hold:
 \[
 \oint \vec{F}_c\cdot d\vec{r}=0,
@@ -146,3 +150,5 @@ W_s=+2.25\,\mathrm{J},
 \qquad
 v_f=3.0\,\mathrm{m/s}.
 \]
+
+The same spring restoring force $-kx$ governs the simple harmonic motion studied in Unit 7 (Sections 7.1--7.4).

concepts/mechanics/u4/m4-4-collisions.tex

@@ -27,6 +27,12 @@ A collision is \emph{perfectly inelastic} if the objects stick together after th
 \]
 Every perfectly inelastic collision is inelastic.}
 
+\wc{Momentum is conserved in every collision type}{Total momentum of an isolated system is conserved in elastic, inelastic, \emph{and} perfectly inelastic collisions. The word ``elastic'' only tells you whether kinetic energy is also conserved. A common exam error is to drop the momentum equation for inelastic collisions --- \emph{never} do this.}
+
+\wc{``Momentum is lost'' in a collision}{Momentum of the \emph{system} is never lost in any isolated collision. What is ``lost'' in an inelastic collision is \emph{kinetic energy}, which is converted to thermal energy, sound, deformation, etc. Momentum conservation holds regardless of how ``bouncy'' the collision is.}
+
+Collisions connect momentum conservation (Unit 4) with energy conservation (Unit 3): elastic collisions conserve both, inelastic conserve only momentum. See also the coefficient of restitution in Section 4.5 for explosive separation.
+
 \nt{Do not decide whether momentum is conserved by asking whether the collision is elastic. Those are different ideas. Momentum conservation depends on the net external impulse on the chosen system. If the system is isolated during the collision, then total momentum is conserved for elastic, inelastic, and perfectly inelastic collisions alike. Kinetic energy supplies an \emph{extra} condition only in the elastic case. In two-dimensional AP problems, conserve momentum separately in the $x$- and $y$-directions.}
 
 \ex{Illustrative example}{On a frictionless track, cart 1 has mass $m_1=0.40\,\mathrm{kg}$ and initial velocity

concepts/mechanics/u5/m5-3-torque.tex

@@ -30,6 +30,8 @@ For rotation about a chosen fixed axis with unit vector $\hat{k}$,
 \]
 so the scalar $\tau$ is positive or negative according to the declared sign convention.}
 
+Torque about a fixed axis connects directly to angular acceleration through Newton's second law for rotation (Section 5.6), just as net force connects to linear acceleration through Newton's second law (Section 2.1).
+
 \pf{Why $rF\sin\phi$ equals $F\ell$}{From the cross-product magnitude formula,
 \[
 |\vec{\tau}|=|\vec{r}\times \vec{F}|=rF\sin\phi.

concepts/mechanics/u5/m5-4-moment-of-inertia.tex

@@ -16,6 +16,8 @@ I=mr_\perp^2.
 \]
 The SI unit of moment of inertia is $\mathrm{kg\cdot m^2}$. Because the distance to the axis is squared, mass farther from the axis contributes much more strongly to $I$.}
 
+\wc{Moment of inertia does not depend on speed or rotation}{The moment of inertia $I$ is a geometric property of the mass distribution about an axis. It depends only on the object's mass, shape, and axis position --- \emph{not} on angular velocity, angular acceleration, or torque. A spinning top and a stationary top have the same $I$.}
+
 \thm{Key fixed-axis relations and axis dependence}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $\vec{\alpha}=\alpha\hat{k}$ denote its angular acceleration, let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis, and let $I$ denote the moment of inertia about that same axis. Then
 \[
 \tau_{\mathrm{net}}=I\alpha,
@@ -30,6 +32,8 @@ I=I_{\mathrm{cm}}+Md^2.
 \]
 Therefore moment of inertia depends on the chosen axis as well as on the mass distribution. Moving the axis farther from the center of mass increases $I$.}
 
+\nt{The parallel-axis theorem relates the moment of inertia about any axis to the moment of inertia about a parallel axis through the center of mass: $I_{\mathrm{any}}=I_{\mathrm{cm}}+Md^2$, where $d$ is the perpendicular distance between the axes. This is useful for computing $I$ about arbitrary pivot points. The perpendicular-axis theorem ($I_z=I_x+I_y$) applies only to flat planar objects and relates the moment of inertia about an axis perpendicular to the plane to the moments of inertia about two orthogonal in-plane axes.}
+
 \ex{Illustrative example}{A light turntable carries two small clay balls, each of mass $m=0.40\,\mathrm{kg}$. In arrangement A, each ball is at distance $r_A=0.10\,\mathrm{m}$ from the axis. In arrangement B, each ball is at distance $r_B=0.20\,\mathrm{m}$ from the axis.
 
 For arrangement A,

concepts/mechanics/u5/m5-6-rotation-dynamics.tex

@@ -13,6 +13,8 @@ The net torque about the axis is the sum of the signed torques:
 
 Let $I$ denote the moment of inertia of the body about that same axis, and let $\alpha$ denote the signed angular acceleration. The quantity $I$ measures the rotational inertia of the body: for the same net torque, a larger $I$ gives a smaller $\alpha$. Thus $I$ plays the rotational role that mass plays in translational motion.}
 
+\wc{Heavier objects do not fall faster (in vacuum)}{In the absence of air resistance, all objects fall with the same acceleration $g$, regardless of mass. The gravitational force is larger on a heavier object ($F_g=mg$), but so is the object's inertia ($F=ma$), and the $m$ cancels: $a=g$. Air resistance is what makes feathers fall slower in real life.}
+
 \thm{Newton's second law for fixed-axis rotation}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the moment of inertia about that axis, let $\alpha$ denote the signed angular acceleration, and let $\sum \tau$ denote the net external torque about the axis using the declared sign convention. Then
 \[
 \sum \tau = I\alpha.

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

@@ -18,6 +18,8 @@ These relations are called the \emph{rolling constraint}. They apply only when t
 
 Also, the friction in rolling without slipping is \emph{static} friction. Let $\vec{N}$ denote the normal force, let $N=|\vec{N}|$ denote its magnitude, and let $f_s$ denote the magnitude of the static friction force. Static friction is not automatically equal to $\mu_s N$. Instead, its magnitude is whatever value is required to prevent slipping, provided that value satisfies $f_s\le \mu_s N$. On level ground at constant speed, the needed static friction can even be zero.}
 
+\wc{Static friction does not always oppose motion}{Static friction opposes \emph{impending relative motion at the contact point}, not the motion of the object as a whole. In rolling without slipping down an incline, the object moves downward but static friction points \emph{up} the incline, because without friction the contact point would slide downward. The friction prevents that slip by pulling the contact backward.}
+
 \ex{Illustrative example}{A wheel of radius $R=0.30\,\mathrm{m}$ rolls without slipping on level ground with angular speed $\omega=8.0\,\mathrm{rad/s}$. Find the speed of its center of mass and the speed of the top point of the wheel relative to the ground.
 
 From the rolling constraint,
@@ -41,6 +43,10 @@ Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves
 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.}
 
+\nt{When several objects roll without slipping down the same incline from the same height, the one with the smallest $I_{\mathrm{cm}}/(MR^2)$ ratio reaches the bottom first. Race order: solid sphere ($\tfrac25=0.40$) $>$ solid cylinder ($\tfrac12=0.50$) $>$ hollow cylinder ($\simeq1$) $>$ thin hoop ($1.0$). Mass and radius do not appear in the comparison --- only the shape-dependent coefficient matters.}
+
+\wc{Rolling race order depends on moment of inertia}{Objects with a \emph{smaller} $I/(MR^2)$ ratio accelerate faster and reach the bottom first. For example: solid sphere ($\tfrac25$) beats solid cylinder ($\tfrac12$) beats thin hoop ($1$). The mass and radius cancel out --- only the \emph{shape distribution} matters. More mass near the rim means a larger fraction of energy goes into rotation, leaving less for translation.}
+
 \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 $g=9.8\,\mathrm{m/s^2}$.
 
 Find:

concepts/mechanics/u7/m7-3-shm-energy.tex

@@ -151,3 +151,5 @@ v=1.60\,\mathrm{m/s},
 \qquad
 v_{\max}=2.0\,\mathrm{m/s}\text{ at }x=0.
 \]
+
+\nt{In simple harmonic motion, total mechanical energy $E = K + U$ is constant. At maximum displacement ($x=\pm A$), $K=0$ and $U=E$ (all energy is potential). At equilibrium ($x=0$), $U=0$ and $K=E$ (all energy is kinetic). At $x=\pm A/\sqrt{2}$, $K=U=E/2$. The energy exchange oscillates at twice the frequency of the motion itself.}

concepts/mechanics/u7/m7-4-simple-pendulum.tex

@@ -33,6 +33,10 @@ and small-angle frequency
 f=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}.
 \]}
 
+\nt{The small-angle approximation $\sin\theta\approx\theta$ introduces error that grows with amplitude. For $\theta_{\max}=10^\circ\approx0.17$ rad, the period error is about $0.2\%$. For $\theta_{\max}=30^\circ$, the period error is about $1.7\%$. The exact period is $T_{\mathrm{exact}}=T_0\big(1+\tfrac14\sin^2\frac{\theta_{\max}}{2}+\tfrac{9}{64}\sin^4\frac{\theta_{\max}}{2}+\cdots\big)$, where $T_0=2\pi\sqrt{\ell/g}$. For AP Physics C, the small-angle model is the standard unless stated otherwise.}
+
+\wc{The ``period is independent of amplitude'' is an approximation}{For a simple pendulum, $T=2\pi\sqrt{\ell/g}$ is valid only for \emph{small angles} where $\sin\theta\approx\theta$. At larger amplitudes, the period increases slightly. For a spring-mass system, the period really is amplitude-independent (within the spring's linear range). Do not confuse these two cases.}
+
 \pf{Short derivation from torque and linearization}{About the pivot, the gravitational torque on the bob is restoring, so
 \[
 \tau=-mg\ell\sin\theta.