Visual walkthrough — Physical pendulum — compound pendulum

Before we begin, three plain words we will keep pointing at:

• pivot — the fixed nail the object turns around. We call it $O$.
• centre of mass (CM) — the single balance point where gravity effectively pulls. Its distance from the nail we call $d$.
• ==tilt angle $\theta$== — how far the object has swung away from hanging straight down.

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Step 1 — Hang the body and name the picture

WHAT. A rigid body (imagine a flat paddle) is pinned at a nail $O$. Left to itself it hangs so its CM sits straight below the nail — this is equilibrium. Now we tilt it by an angle $\theta$.

WHY. Every derivation needs a labelled picture first. If we can't point at $d$, $\theta$, and $O$, we can't write a single honest equation about them.

PICTURE. Look at the figure. The teal dashed line is "straight down" (equilibrium). The orange arrow from the nail to the CM makes angle $\theta$ with that line. That single arrow has length $d$.

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Step 2 — Why torque, and where gravity actually pushes

WHAT. Gravity pulls the whole body, but for a body that can only turn about $O$, what matters is not the force but its twisting effect — the torque. We compute the torque of gravity about $O$.

WHY. The body cannot fly sideways; it is nailed. It can only rotate. The law that governs rotation is $\tau =I\alpha$ (rotational Newton), the twin of $F=ma$. So we need the twist, not the raw pull. This is exactly why we reach for torque and not force here.

PICTURE. Gravity $mg$ points straight down from the CM (plum arrow). Torque = force $\times$ perpendicular lever arm — the sideways gap between the line of the force and the nail. That gap is the orange bracket, and geometry says it equals $d\sin \theta$.

$$\tau =\underset{\text{opposes tilt}}{\underbrace{-}}\underset{\text{gravity force}}{\underbrace{mg}}\underset{\text{lever arm}}{\underbrace{d\sin \theta }}$$

The minus sign is not decoration: it says the twist points back toward equilibrium. Tilt right → torque pushes left. A restoring torque.

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Step 3 — Turn the twist into an equation of motion

WHAT. Feed this torque into rotational Newton's law $\tau =I\alpha$.

WHY. $I\alpha$ tells us how the body responds to the twist. $\alpha$ is the angular acceleration — how fast the tilt-rate is changing — which we write $\ddot{\theta }$ (the second rate of change of $\theta$ in time). Setting the response equal to the cause gives the motion.

PICTURE. The figure stacks the "cause" (gravity's restoring torque, orange) against the "response" (the body's resistance to being spun up, teal), meeting at the balance $\tau =I\ddot{\theta }$.

$$\underset{\begin{array}{c}\text{inertia about }O\\ \text{(resistance to spin)}\end{array}}{\underbrace{I}}\underset{\text{angular accel.}}{\underbrace{\ddot{\theta }}}=\underset{\text{restoring torque}}{\underbrace{-mgd\sin \theta }}$$

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Step 4 — The straightening trick: small-angle approximation

WHAT. For small tilts we replace $\sin \theta$ by $\theta$ (with $\theta$ in radians).

WHY. $\sin \theta$ makes the equation nonlinear — no clean formula for the period. But look at the graph: for small $\theta$ the sine curve and the straight line $y=\theta$ lie almost on top of each other. Replacing the curve by the line is what turns a hopeless equation into SHM, the one oscillation we can solve exactly.

PICTURE. The figure overlays $\sin \theta$ (plum curve) and $\theta$ (teal line). They hug near the origin and only split apart past roughly $20^{\circ }$. That hugging region is where our formula is honest.

$$I\ddot{\theta }=-mgd\theta ⟹\ddot{\theta }=-\underset{\text{a positive constant}}{\underbrace{\frac{mgd}{I}}}\theta$$

Every symbol on the right is a fixed number, so the whole fraction is just some positive constant times $\theta$, with a minus sign. Hold that thought.

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Step 5 — Recognise SHM and read off the period

WHAT. Compare $\ddot{\theta }=-\frac{mgd}{I}\theta$ with the master SHM equation $\ddot{\theta }=-{\omega }^2\theta$.

WHY. Any equation of the form "acceleration $=-(\text{constant})\times (\text{position})$" is simple harmonic motion — a pure sinusoidal wobble. The constant is named ${\omega }^2$ because $\omega$ (angular frequency) directly sets the timing. Matching the two forms hands us $\omega$ for free.

PICTURE. The figure shows the tilt $\theta (t)$ tracing a cosine wave in time; one full wave = one period $T$, and steeper restoring (bigger $mgd/I$) squeezes the wave narrower.

Matching term by term:

$${\omega }^2=\frac{mgd}{I}\Rightarrow \omega =\sqrt{\frac{mgd}{I}}\Rightarrow \boxed{T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{mgd}}}$$

Reading the box: a bigger $I$ (mass flung far from the nail) → slower swing → longer $T$. A bigger $mgd$ (heavier, or CM farther out → stronger restoring twist) → faster swing → shorter $T$. The formula's shape is the physics.

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Step 6 — Rewrite $I$ so we can actually compute it

WHAT. Split $I$ into "spread about the CM" plus "distance from the CM to the nail" using the Parallel axis theorem.

WHY. We rarely know $I$ about a random nail, but we always know $I_{cm}$ from a table. The parallel-axis theorem carries $I_{cm}$ to the pivot by adding $m{d}^2$. Writing $I_{cm}=m{k}^2$ introduces the Radius of gyration $k$ — the single distance that packages the shape.

PICTURE. Two dots: the CM (with its compact spread $k$) and the nail a distance $d$ away. The theorem is literally "spread about CM" + "shift $m{d}^2$".

$$I=\underset{=m{k}^2}{\underbrace{I_{cm}}}+\underset{\text{shift to pivot}}{\underbrace{m{d}^2}}=m\left(k^2+d^2\right)$$

Substitute into the box — the mass $m$ cancels top and bottom:

$$T=2\pi \sqrt{\frac{m(k^2+d^2)}{mgd}}=2\pi \sqrt{\frac{k^2+d^2}{gd}}\text{and}{L}_{\text{eq}}=\frac{I}{md}=\frac{k^2}{d}+d$$

$L_{\text{eq}}$ is the length of the Simple pendulum that swings in exactly the same time — the whole messy body collapsed into one number.

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Step 7 — Every edge case, so no swing surprises you

WHAT. Push $d$ to its extremes and find the sweet spot in between.

WHY. A formula you trust must survive its limits. We check $d\to 0$, $d\to \infty$, and hunt the minimum. This also completes the "all cases" contract.

PICTURE. $T$ versus $d$: a valley. Left wall ($d\to 0$) shoots to infinity, right side ($d$ large) climbs steadily, and the floor sits at $d=k$.

• Pivot at the CM, $d\to 0$: restoring torque $mgd\sin \theta \to 0$ — nothing pulls it back → $T\to \infty$. The body doesn't oscillate; it just sits there in any orientation.
• Pivot far away, $d\to \infty$: $I\approx m{d}^2$ grows faster than $mgd$, so $T\approx 2\pi \sqrt{d/g}$ keeps rising — it becomes a long, slow simple pendulum.
• The minimum: minimise $\frac{k^2}{d}+d$. Its derivative is $-\frac{k^2}{d^2}+1$; setting it to zero gives $\boxed{d=k}$.

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

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

The point a distance $L_{\text{eq}}$ from the pivot is the centre of oscillation — pivot there and you get the same $T$. That reversibility is the trick behind Kater's pendulum for measuring $g$.

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The one-picture summary

One image, the whole chain: gravity's twist $-mgd\sin \theta$ → rotational law $I\ddot{\theta }$ → straighten with $\sin \theta \approx \theta$ → recognise SHM → read $T=2\pi \sqrt{I/(mgd)}$ → rewrite $I=m(k^2+d^2)$ → the equivalent length $L_{\text{eq}}=k^2/d+d$ with its minimum at $d=k$.

Recall Feynman retelling — the walkthrough in plain words

Nail a paddle to a wall and let it hang; its middle sits straight below the nail. Tip it a little. Gravity, tugging at the middle, tries to swing it back — and how hard it tugs depends on how far sideways the middle has slid, which is $d\sin \theta$. That sideways tug twists the paddle about the nail; the twist is the torque. How the paddle answers a twist is set by its moment of inertia $I$ (how far its mass sits from the nail). Put "twist causes turning" into an equation and you get a wobble law. For tiny tips, "how far it's slid" is basically just the angle itself, and then the paddle rocks like a perfect clock — simple harmonic motion. Reading the timing off gives $T=2\pi \sqrt{I/(mgd)}$. Because $I$ can be split into "how the mass is bunched near its own middle" plus "how far that middle is from the nail," the period is really $2\pi \sqrt{(k^2+d^2)/(gd)}$. Nail it at the exact middle and it won't swing at all (nothing pulls it back); nail it far away and it swings lazily; somewhere in between — at $d=k$ — it swings its very fastest.

Recall Quick self-test

Why does the lever arm use $\sin \theta$ and not $\cos \theta$? ::: Because the sideways gap between gravity's line and the nail is the opposite side of a right triangle with hypotenuse $d$ and angle $\theta$; opposite/hyp $=\sin$. At $\theta =0$ it must vanish, and $\sin 0=0$. What single approximation converts the exact equation into SHM? ::: $\sin \theta \approx \theta$ for small $\theta$ (radians). Where does the mass $m$ go? ::: It cancels — $T$ is independent of mass, like every gravity pendulum. At what $d$ is the period smallest, and what is $L_{\text{eq}}$ there? ::: $d=k$, giving $L_{\text{eq}}=2k$ and $T_{\min }=2\pi \sqrt{2k/g}$.