156 lines
4.4 KiB
TeX
156 lines
4.4 KiB
TeX
\subsection{Physical Pendulum and Small-Angle Linearization}
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This subsection models the small oscillations of a rigid body that swings about a fixed pivot under gravity.
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\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
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\[
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d=OC
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\]
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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.
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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.}
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\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.
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Then the exact rotational equation of motion is
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\[
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I\ddot{\theta}=-mgd\sin\theta,
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\]
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or equivalently,
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\[
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I\ddot{\theta}+mgd\sin\theta=0.
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\]
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For small angular displacements, use the linearization $\sin\theta\approx\theta$ to obtain
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\[
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I\ddot{\theta}+mgd\,\theta=0.
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\]
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Therefore the motion is approximately simple harmonic with angular frequency
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\[
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\omega=\sqrt{\frac{mgd}{I}}
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\]
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and period
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\[
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T=2\pi\sqrt{\frac{I}{mgd}}.
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\]
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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 $d=L$.}
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\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
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\[
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\tau=-mgd\sin\theta.
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\]
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For rotation about a fixed axis, Newton's second law for rotation gives
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\[
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\sum \tau=I\ddot{\theta}.
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\]
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Hence,
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\[
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I\ddot{\theta}=-mgd\sin\theta,
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\]
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which is the exact equation.
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If the oscillations are small, then $\sin\theta\approx\theta$, so the equation becomes
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\[
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I\ddot{\theta}+mgd\,\theta=0.
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\]
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Divide by $I$ to get
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\[
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\ddot{\theta}+\frac{mgd}{I}\theta=0.
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\]
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Comparing with the SHM form $q''+\omega^2 q=0$ shows that
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\[
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\omega^2=\frac{mgd}{I},
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\qquad
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\omega=\sqrt{\frac{mgd}{I}}.
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\]
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Therefore,
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\[
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T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{mgd}}.
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\]}
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\ex{Illustrative example}{Show that the simple pendulum is a special case of the physical pendulum formula.
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For a point mass $m$ at distance $L$ from the pivot,
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\[
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I=mL^2,
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\qquad
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d=L.
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\]
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Substitute into the physical-pendulum period formula:
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\[
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T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{mL^2}{mgL}}=2\pi\sqrt{\frac{L}{g}}.
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\]
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This is exactly the small-angle period of a simple pendulum.}
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\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.
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Find:
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\begin{enumerate}[label=(\alph*)]
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\item the pivot-to-center-of-mass distance $d$ and the rod's moment of inertia $I$ about the pivot,
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\item the small-angle differential equation for $\theta(t)$, and
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\item the period of oscillation.
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\end{enumerate}}
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\sol For a uniform rod pivoted about one end, the center of mass is at the midpoint, so
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\[
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d=\frac{L}{2}=\frac{0.90\,\mathrm{m}}{2}=0.45\,\mathrm{m}.
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\]
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The moment of inertia of a uniform rod about one end is
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\[
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I=\frac{1}{3}mL^2.
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\]
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Substitute the given values:
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\[
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I=\frac{1}{3}(1.50)(0.90)^2\,\mathrm{kg\cdot m^2}.
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\]
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Since $(0.90)^2=0.81$,
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\[
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I=\frac{1}{3}(1.50)(0.81)=0.405\,\mathrm{kg\cdot m^2}.
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\]
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For small oscillations, a physical pendulum satisfies
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\[
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I\ddot{\theta}+mgd\,\theta=0.
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\]
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Now compute $mgd$:
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\[
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mgd=(1.50)(9.8)(0.45)=6.615.
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\]
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So the differential equation is
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\[
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0.405\,\ddot{\theta}+6.615\,\theta=0.
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\]
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Divide by $0.405$:
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\[
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\ddot{\theta}+16.3\,\theta=0.
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\]
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Thus,
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\[
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\omega=\sqrt{16.3}=4.04\,\mathrm{rad/s}.
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\]
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The period is
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\[
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T=\frac{2\pi}{\omega}=\frac{2\pi}{4.04}=1.56\,\mathrm{s}.
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\]
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Equivalently, using the period formula directly,
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\[
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T=2\pi\sqrt{\frac{I}{mgd}}=2\pi\sqrt{\frac{0.405}{6.615}}=1.56\,\mathrm{s}.
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\]
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Therefore,
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\[
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d=0.45\,\mathrm{m},
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\qquad
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I=0.405\,\mathrm{kg\cdot m^2},
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\]
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and the small-angle motion is governed by
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\[
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\ddot{\theta}+16.3\,\theta=0,
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\qquad
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T=1.56\,\mathrm{s}.
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\]
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