\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
\[
\ddot{\theta}+16.3\,\theta=0,
\qquad
T=1.56\,\mathrm{s}.
\]
