Simple and physical pendulums

Topic 7.5 is about pendulums as examples of simple harmonic motion. A simple pendulum, a small bob on a light string, swings with period $T=2\pi\sqrt{L/g}$. Only the string length $L$ and the gravitational field $g$ affect the period; the bob mass and the swing amplitude (at small angles) do not.

When the swinging object is an extended rigid body, it becomes a physical pendulum, with period $T=2\pi\sqrt{I/(mgd)}$. Here $I$ is the moment of inertia about the pivot and $d$ is the distance from the pivot to the center of mass. The physical pendulum reduces to the simple form when all the mass sits at one distance from the pivot.

The three main mistakes are: misreading what affects the simple-pendulum period, collapsing the physical pendulum to a simple one, and plugging in the wrong distance or the wrong moment of inertia.

For a bob on a light string displaced to a small angle, the restoring component of gravity is proportional to the displacement, which produces SHM. The resulting period is $T=2\pi\sqrt{L/g}$, where $L$ is the string length from the pivot to the bob and $g$ is the local gravitational acceleration.

Two things are absent from this formula: mass and amplitude. The bob mass cancels because gravity supplies both the restoring force and the inertia. The amplitude drops out in the small-angle approximation $\sin\theta\approx\theta$, so a pendulum released from $5^\circ$ has the same period as one released from $10^\circ$.

Because $L$ sits under a square root, the period grows as $\sqrt{L}$, not in direct proportion. Quadrupling the length doubles the period, and to double the period you must multiply the length by four, not by two.

$T=2\pi\sqrt{L/g}$: only length and gravity set the period. Mass and amplitude do not appear.

When a rigid body swings about a fixed pivot, the restoring torque depends on the mass $m$, the gravitational field $g$, and the distance $d$ from the pivot to the center of mass. The moment of inertia $I$ about the pivot sets how hard the body is to rotate. Together these give $T=2\pi\sqrt{I/(mgd)}$.

For a uniform rod of length $L$ pivoted at one end, $I=\tfrac13 mL^2$ and $d=L/2$, so $T=2\pi\sqrt{2L/(3g)}$. For a uniform disk of radius $R$ pivoted at the rim, $I=\tfrac32 mR^2$ and $d=R$. In both cases the mass cancels (since $I$ is proportional to $m$), so the period depends on geometry, not on mass.

$T=2\pi\sqrt{I/(mgd)}$ keeps $I$ about the pivot and $d$ from the pivot to the center of mass. Mass cancels for a fixed shape.

The temptation is to treat every pendulum as simple: pick some length, write $T=2\pi\sqrt{\text{length}/g}$, and move on. For an extended body that gives the wrong answer, because the mass is spread out and the moment of inertia is not simply $md^2$.

For a uniform rod pivoted at one end, the simple model with $L$ as the length gives $T=2\pi\sqrt{L/g}$, but the true period is $T=2\pi\sqrt{2L/(3g)}$, which is shorter. The equivalent length of a physical pendulum, the simple-pendulum length that would match its period, is $L_{\text{eq}}=I/(md)$, which for the end-pivoted rod is $\tfrac23 L$, not $L$.

A physical pendulum matches a simple one of length $I/(md)$, not of length $d$ or the full body length.

Two inputs require care. The distance $d$ runs from the pivot to the center of mass, not to the far end. For an end-pivoted rod of length $L$, $d=L/2$; for a rim-pivoted disk of radius $R$, $d=R$.

The moment of inertia $I$ must be taken about the pivot, not about the center of mass. The parallel-axis theorem converts: $I_{\text{pivot}}=I_{\text{cm}}+md^2$. Using $I_{\text{cm}}$ directly underestimates $I$ and produces a period that is too short.

When the pivot is at the center of mass, $d=0$, so $mgd=0$ and there is no restoring torque: the body does not oscillate at all.

$d$ is pivot-to-center-of-mass. $I$ is about the pivot (use the parallel-axis theorem). Wrong axis or wrong distance changes the answer.

Pendulum problems, simple and physical alike, come undone at three familiar points.

• Misreading the simple-pendulum periodTreating the period as depending on the bob mass or the swing amplitude, or scaling the period directly with length instead of $\sqrt{L}$. Fix: $T=2\pi\sqrt{L/g}$ has no mass or amplitude term, and the square root means quadrupling $L$ only doubles $T$.
• Collapsing the physical pendulum to simpleDropping $I$ and writing $T=2\pi\sqrt{d/g}$ or $T=2\pi\sqrt{L/g}$ for an extended body. Fix: keep the physical-pendulum formula $T=2\pi\sqrt{I/(mgd)}$; the equivalent length is $I/(md)$, not $d$.
• Wrong distance or wrong axisUsing the full length instead of the pivot-to-center distance $d$, or using $I_{\text{cm}}$ instead of $I$ about the pivot. Fix: $d$ runs from pivot to center of mass; $I$ about the pivot via the parallel-axis theorem.

Identify which formula applies (simple or physical), keep $d$ as pivot to center of mass, and find $I$ about the pivot. The whole topic comes down to those three habits.