checkpoint 1

concepts/mechanics/u7/m7-5-physical-pendulum.tex

@@ -0,0 +1,155 @@
+\subsection{Physical Pendulum and Small-Angle Linearization}
+
+This subsection models the small oscillations of a rigid body that swings about a fixed pivot under gravity.
+
+\dfn{Physical pendulum, pivot-to-CM distance, and angular coordinate}{Let a rigid body of mass $m$ swing in a vertical plane about a fixed pivot point $O$. Let $C$ denote the center of mass of the body, let
+\[
+d=OC
+\]
+denote the distance from the pivot to the center of mass, and let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position.
+
+Such a system is called a \emph{physical pendulum}. Unlike a simple pendulum, the body's mass is distributed throughout the rigid object, so its rotational inertia must be included in the dynamics.}
+
+\thm{Exact torque equation and small-angle SHM model}{Let $m$ denote the mass of the rigid body, let $d$ denote the distance from the pivot to the center of mass, let $I$ denote the moment of inertia of the body about the pivot, let $g$ denote the magnitude of the gravitational field, and let $\theta(t)$ denote the angular displacement from stable equilibrium.
+
+Then the exact rotational equation of motion is
+\[
+I\ddot{\theta}=-mgd\sin\theta,
+\]
+or equivalently,
+\[
+I\ddot{\theta}+mgd\sin\theta=0.
+\]
+
+For small angular displacements, use the linearization $\sin\theta\approx\theta$ to obtain
+\[
+I\ddot{\theta}+mgd\,\theta=0.
+\]
+Therefore the motion is approximately simple harmonic with angular frequency
+\[
+\omega=\sqrt{\frac{mgd}{I}}
+\]
+and period
+\[
+T=2\pi\sqrt{\frac{I}{mgd}}.
+\]
+
+A simple pendulum is the special case in which all the mass is concentrated a distance $L$ from the pivot, so $I=mL^2$ and $d=L$.}
+
+\pf{Short derivation from torque and linearization}{The weight $m\vec{g}$ acts at the center of mass. When the body is displaced by angle $\theta$, the gravitational torque about the pivot is restoring, so
+\[
+\tau=-mgd\sin\theta.
+\]
+For rotation about a fixed axis, Newton's second law for rotation gives
+\[
+\sum \tau=I\ddot{\theta}.
+\]
+Hence,
+\[
+I\ddot{\theta}=-mgd\sin\theta,
+\]
+which is the exact equation.
+
+If the oscillations are small, then $\sin\theta\approx\theta$, so the equation becomes
+\[
+I\ddot{\theta}+mgd\,\theta=0.
+\]
+Divide by $I$ to get
+\[
+\ddot{\theta}+\frac{mgd}{I}\theta=0.
+\]
+Comparing with the SHM form $q''+\omega^2 q=0$ shows that
+\[
+\omega^2=\frac{mgd}{I},
+\qquad
+\omega=\sqrt{\frac{mgd}{I}}.
+\]
+Therefore,
+\[
+T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{mgd}}.
+\]}
+
+\ex{Illustrative example}{Show that the simple pendulum is a special case of the physical pendulum formula.
+
+For a point mass $m$ at distance $L$ from the pivot,
+\[
+I=mL^2,
+\qquad
+d=L.
+\]
+Substitute into the physical-pendulum period formula:
+\[
+T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{mL^2}{mgL}}=2\pi\sqrt{\frac{L}{g}}.
+\]
+This is exactly the small-angle period of a simple pendulum.}
+
+\qs{Worked AP-style problem}{A uniform rod of mass $m=1.50\,\mathrm{kg}$ and length $L=0.90\,\mathrm{m}$ is pivoted about one end and allowed to swing in a vertical plane. Let $\theta(t)$ denote the angular displacement from the stable vertical equilibrium position. Assume the oscillations are small.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the pivot-to-center-of-mass distance $d$ and the rod's moment of inertia $I$ about the pivot,
+\item the small-angle differential equation for $\theta(t)$, and
+\item the period of oscillation.
+\end{enumerate}}
+
+\sol For a uniform rod pivoted about one end, the center of mass is at the midpoint, so
+\[
+d=\frac{L}{2}=\frac{0.90\,\mathrm{m}}{2}=0.45\,\mathrm{m}.
+\]
+
+The moment of inertia of a uniform rod about one end is
+\[
+I=\frac{1}{3}mL^2.
+\]
+Substitute the given values:
+\[
+I=\frac{1}{3}(1.50)(0.90)^2\,\mathrm{kg\cdot m^2}.
+\]
+Since $(0.90)^2=0.81$,
+\[
+I=\frac{1}{3}(1.50)(0.81)=0.405\,\mathrm{kg\cdot m^2}.
+\]
+
+For small oscillations, a physical pendulum satisfies
+\[
+I\ddot{\theta}+mgd\,\theta=0.
+\]
+Now compute $mgd$:
+\[
+mgd=(1.50)(9.8)(0.45)=6.615.
+\]
+So the differential equation is
+\[
+0.405\,\ddot{\theta}+6.615\,\theta=0.
+\]
+Divide by $0.405$:
+\[
+\ddot{\theta}+16.3\,\theta=0.
+\]
+
+Thus,
+\[
+\omega=\sqrt{16.3}=4.04\,\mathrm{rad/s}.
+\]
+The period is
+\[
+T=\frac{2\pi}{\omega}=\frac{2\pi}{4.04}=1.56\,\mathrm{s}.
+\]
+
+Equivalently, using the period formula directly,
+\[
+T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{0.405}{6.615}}=1.56\,\mathrm{s}.
+\]
+
+Therefore,
+\[
+d=0.45\,\mathrm{m},
+\qquad
+I=0.405\,\mathrm{kg\cdot m^2},
+\]
+and the small-angle motion is governed by
+\[
+\ddot{\theta}+16.3\,\theta=0,
+\qquad
+T=1.56\,\mathrm{s}.
+\]