Find the law of motion of a physical pendulum: a rigid body…

Find the law of motion of a physical pendulum: a rigid body of mass $m$ rotating freely about a fixed horizontal axis $O$ that does not pass through the center of mass $G$. Let $I$ be the moment of inertia about $O$, $l$ the distance from $O$ to $G$, $\varphi$ the angle of deflection from vertical equilibrium, and $g$ the acceleration due to gravity.

Step 1: Equation of motion

Gravity exerts a restoring torque about $O$ equal to $-mgl\sin\varphi$. By Newton's second law for rotation, $I\ddot\varphi$ equals this torque: $I\ddot\varphi = -mgl\sin\varphi.$

Step 2: Linearize for small oscillations

For small $\varphi$, $\sin\varphi\approx\varphi$, so: $I\ddot\varphi + mgl\,\varphi = 0.$

Dividing by $I$ and introducing $\omega^2 = \dfrac{mgl}{I}$ gives the standard harmonic-oscillator equation: $\ddot\varphi + \omega^2\varphi = 0.$

Step 3: Characteristic equation and general solution

\[ \lambda^2+\omega^2=0 \implies \lambda=\pm i\omega. \]

\[ \varphi(t) = C_1\cos\omega t + C_2\sin\omega t = \Phi\cos(\omega t-\varphi_0), \] where $\Phi=\sqrt{C_1^2+C_2^2}$ is the amplitude and $\varphi_0=\arctan(C_2/C_1)$ is the initial phase.

Step 4: Initial conditions

Suppose the pendulum is released from rest at deflection $\varphi_0^*$: $\varphi(0)=\varphi_0^*$, $\dot\varphi(0)=0$. Then: \[C_1 = \varphi_0^*, \qquad C_2\omega = 0 \implies C_2=0.\]

Step 5: Period and reduced length

\[ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{mgl}}. \]

Defining a reduced (equivalent simple-pendulum) length $L$ so that $T=2\pi\sqrt{L/g}$: $L = \frac{I}{ml}.$

Step 6: Special case: mathematical pendulum

For a point mass on a massless rod, $I=ml^2$, so $\omega^2=g/l$ and $T=2\pi\sqrt{l/g}$: \[\ddot\varphi+\frac{g}{l}\varphi=0, \qquad \varphi(t)=\varphi_0^*\cos\!\left(\sqrt{\frac{g}{l}}\,t\right).\]

Step 7: Energy

Mechanical energy is conserved; in the small-angle approximation: \[E = \frac{1}{2}I\dot\varphi^2 + mgl(1-\cos\varphi) \approx \frac{1}{2}I\dot\varphi^2 + \frac{1}{2}mgl\,\varphi^2 = \frac{1}{2}I\omega^2\Phi^2 = \mathrm{const}.\]

Step 8: Verify

With $\varphi(t)=\varphi_0^*\cos\omega t$: $\dot\varphi=-\varphi_0^*\omega\sin\omega t$, $\ddot\varphi=-\varphi_0^*\omega^2\cos\omega t=-\omega^2\varphi$, so $\ddot\varphi+\omega^2\varphi=0\ \checkmark$. Also $\varphi(0)=\varphi_0^*\ \checkmark$ and $\dot\varphi(0)=0\ \checkmark$.

Step 9: Law of motion

The solution is recorded in the result below.