Physical pendulum — compound pendulum

A physical (compound) pendulum is any rigid body that swings about a fixed horizontal axis under gravity — not a point mass on a massless string. Think of a swinging meter stick, a door, or your forearm.

1. Setup — WHAT we are describing

A rigid body of mass $m$ is free to rotate about a fixed horizontal axis $O$ (the pivot). Its center of mass (CM) is a distance $d$ from the pivot. When displaced by angle $\theta$, gravity pulls the CM back toward equilibrium.

• ==Pivot $O$==: fixed axis the body rotates about.
• ==$d$==: distance from pivot to center of mass.
• ==$I$==: moment of inertia about the pivot axis $O$ (not about the CM!).
• ==$\theta$==: angular displacement from vertical (equilibrium).

2. Derivation from scratch — HOW we get the period

Step 1 — Locate the restoring torque. Gravity $mg$ acts downward at the CM, a distance $d$ from $O$. When the body is tilted by $\theta$, the perpendicular lever arm of gravity about $O$ is $d\sin\theta$.

$\tau = -\,m g\, d \sin\theta$

Why this step? The minus sign says the torque opposes the displacement (restoring). The lever arm is $d\sin\theta$ because torque = force × perpendicular distance from the line of action to the axis.

Step 2 — Apply rotational Newton's law.

$I\frac{d^2\theta}{dt^2} = \tau = -mgd\sin\theta$

Why this step? $\tau = I\alpha$ and $\alpha = \ddot\theta$. $I$ is taken about the pivot $O$ because that's the axis of rotation.

Step 3 — Small-angle approximation. For small $\theta$, $\sin\theta \approx \theta$ (radians):

$I\ddot\theta = -mgd\,\theta \quad\Rightarrow\quad \ddot\theta = -\frac{mgd}{I}\,\theta$

Why this step? This now has the unmistakable SHM form $\ddot\theta = -\omega^2\theta$.

Step 4 — Read off $\omega$ and $T$.

$\omega = \sqrt{\frac{mgd}{I}}, \qquad \boxed{T = 2\pi\sqrt{\frac{I}{mgd}}}$

3. Equivalent simple-pendulum length — the 80/20 insight

Compare with the simple pendulum $T = 2\pi\sqrt{L/g}$. They have the same period when

$\frac{L}{g} = \frac{I}{mgd} \;\Rightarrow\; \boxed{L_{\text{eq}} = \frac{I}{md}}$

Using the parallel-axis theorem $I = I_{cm} + md^2 = mk^2 + md^2$ (where $k$ = radius of gyration about CM):

$L_{\text{eq}} = \frac{mk^2 + md^2}{md} = \frac{k^2}{d} + d, \qquad T = 2\pi\sqrt{\frac{k^2+d^2}{gd}}$

4. Minimum period & the center of oscillation

Minimize $T^2 \propto \dfrac{k^2+d^2}{d} = \dfrac{k^2}{d}+d$. Set derivative to zero:

$\frac{d}{dd}\!\left(\frac{k^2}{d}+d\right) = -\frac{k^2}{d^2}+1 = 0 \;\Rightarrow\; \boxed{d = k}$

At $d=k$: $L_{\text{eq}} = k^2/k + k = 2k$, and

$T_{\min} = 2\pi\sqrt{\frac{2k}{g}}.$

The point a distance $L_{\text{eq}}$ from the pivot (along the line through CM) is the center of oscillation. A famous reversibility: pivoting at the center of oscillation gives the same period (basis of Kater's pendulum for measuring $g$).

5. Worked examples

6. Common mistakes (Steel-man + fix)

7. Forecast-then-Verify

8. Active-recall flashcards

Connections

• Simple pendulum — special case where all mass is at distance $L$, $I=mL^2$, $d=L$ ⟹ $T=2\pi\sqrt{L/g}$.
• Moment of inertia and Parallel axis theorem — supply $I=I_{cm}+md^2$.
• Radius of gyration — defines $k$ via $I_{cm}=mk^2$.
• Simple Harmonic Motion — the $\ddot\theta=-\omega^2\theta$ engine behind it.
• Kater's pendulum — uses the reversibility about the center of oscillation to measure $g$.
• Torsional pendulum — rotational oscillation but restored by a wire's torsion, not gravity.

Concept Map

Hinglish (regional understanding)