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

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2026-05-04 23:50:51 -05:00
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commit 332e838d2d
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@@ -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.