A physical pendulum is an extended rigid body that oscillates about a fixed horizontal pivot under gravity. Unlike a simple pendulum, its mass is distributed throughout the body, so the period depends on its moment of inertia about the pivot.

Let the body have total mass $M$. Let its center of mass be a distance $d$ from the pivot, and let $I_p$ be its moment of inertia about the pivot axis. If $\theta$ is measured from the stable downward equilibrium, gravity produces the torque

$$\tau=-Mgd\sin\theta.$$

The minus sign shows that gravity tends to restore the body toward equilibrium.

Rotational dynamics gives

$$I_p\ddot\theta=-Mgd\sin\theta,$$

so the exact ideal equation is

$$\boxed{\ddot\theta+\frac{Mgd}{I_p}\sin\theta=0}.$$

This equation is nonlinear because of the sine term.

Small-angle period

For sufficiently small oscillations,

$$\sin\theta\approx\theta.$$

The equation becomes

$$\ddot\theta+\frac{Mgd}{I_p}\theta\approx0,$$

which has the harmonic-oscillator form. Therefore

$$\boxed{\omega\approx\sqrt{\frac{Mgd}{I_p}}}$$

and

$$\boxed{T\approx2\pi\sqrt{\frac{I_p}{Mgd}}}.$$

The period depends on both the center-of-mass distance $d$ and the body's distribution of mass through $I_p$.

Worked example

A rigid object has

$$M=1.20,\mathrm{kg},$$

its center of mass is

$$d=0.250,\mathrm m$$

below the pivot, and its moment of inertia about the pivot is

$$I_p=0.180,\mathrm{kg,m^2}.$$

For small oscillations,

$$T\approx2\pi\sqrt{\frac{0.180}{(1.20)(9.81)(0.250)}}.$$

Thus

$$\boxed{T\approx1.55,\mathrm s}.$$

If another object has the same mass and center-of-mass location but a different $I_p$, it generally has a different period. Knowing only where the center of mass lies is not enough: rotational inertia depends on how the entire mass is distributed around the pivot.

The value of $I_p$ can be supplied experimentally, given as a property of the body, or calculated from its geometry using separate moment-of-inertia tools. Those calculations are not part of the physical-pendulum concept itself.

Equivalent simple-pendulum length

A simple pendulum of length $L_{\rm eq}$ has small-angle period

$$T=2\pi\sqrt{\frac{L_{\rm eq}}{g}}.$$

Equating this with the physical-pendulum period gives

$$\boxed{L_{\rm eq}=\frac{I_p}{Md}}.$$

This equivalent length means only that the two pendulums have the same small-oscillation period. The extended rigid body has not become a point mass.

A physical pendulum therefore generalizes the simple pendulum by replacing the single length that characterizes a point mass on a string with the center-of-mass position and rotational inertia of an extended body.