content(warnings): Add W12-W16, N5-N6 — rolling, pendulum, and inertia enhancements

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

@@ -17,6 +17,8 @@ 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:

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

@@ -32,6 +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/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,
@@ -43,6 +45,8 @@ For an object accelerating down the incline, the static friction force on the ob
 
 \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/m7-4-simple-pendulum.tex

@@ -35,6 +35,8 @@ 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.