3.1.3.2. Physical Pendulum

Table of Contents

Rigid Bodies That Swing

A physical pendulum is any rigid body that swings back and forth about a fixed horizontal axis under the influence of gravity. Unlike a simple pendulum, which is modeled as a point mass on a massless string, a physical pendulum has its mass spread out through space. Because of that, its rotational inertia matters.

Examples include a swinging ruler pivoted at one end, a door moving about its hinges, or a metal plate hung from a support and allowed to oscillate. The restoring effect comes from gravity acting at the center of mass, while the motion itself is rotational about the pivot.

Torque and Equation of Motion

Suppose a rigid body of mass $m$ is pivoted at a point and its center of mass is a distance $d$ from the pivot. If the body is displaced by an angle $\theta$ from its equilibrium position, gravity produces a torque that tends to bring it back.

The magnitude of the torque is

$$
\tau = -mgd \sin\theta
$$

The minus sign shows that the torque is restoring, it acts opposite to the displacement.

Using rotational dynamics,

$$
\tau = I \alpha = I \frac{d^2\theta}{dt^2}
$$

where $I$ is the moment of inertia about the pivot axis. So the equation of motion is

$$
I \frac{d^2\theta}{dt^2} + mgd \sin\theta = 0
$$

This is the exact equation for a physical pendulum.

For a physical pendulum, the rotational equation of motion is
$$
I \frac{d^2\theta}{dt^2} + mgd \sin\theta = 0
$$
where $I$ is the moment of inertia about the pivot, and $d$ is the distance from the pivot to the center of mass.

Small Angle Approximation

For small oscillations, the angle $\theta$ is small enough that

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

Then the equation becomes

$$
I \frac{d^2\theta}{dt^2} + mgd \theta = 0
$$

This has the same form as simple harmonic motion. Therefore a physical pendulum performs SHM for small angular displacements.

We can rewrite it as

$$
\frac{d^2\theta}{dt^2} + \frac{mgd}{I}\theta = 0
$$

So the angular frequency is

$$
\omega = \sqrt{\frac{mgd}{I}}
$$

and the period is

$$
T = 2\pi \sqrt{\frac{I}{mgd}}
$$

For small oscillations of a physical pendulum,
$$
\omega = \sqrt{\frac{mgd}{I}}, \qquad
T = 2\pi \sqrt{\frac{I}{mgd}}
$$
These formulas are valid only when the small angle approximation is reasonable.

Meaning of the Quantities

The period depends on two main ideas. The first is how strongly gravity restores the object, measured by $mgd$. The second is how hard the object is to rotate, measured by $I$.

If the center of mass is farther from the pivot, then $d$ is larger, the restoring torque is stronger, and the pendulum tends to swing faster. If the mass is distributed far from the pivot, then $I$ is larger, the body resists angular acceleration more, and the pendulum swings more slowly.

This is why two objects of the same mass can have different periods if their shapes or pivot positions differ.

Equivalent Simple Pendulum Length

A physical pendulum can be compared to a simple pendulum that has the same period. If $L_{\text{eq}}$ is the equivalent length, then

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

Comparing this with

$$
T = 2\pi \sqrt{\frac{I}{mgd}}
$$

gives

$$
L_{\text{eq}} = \frac{I}{md}
$$

This lets us think of the physical pendulum as if all its mass were concentrated at a special effective distance from the pivot.

The equivalent simple pendulum length is
$$
L_{\text{eq}} = \frac{I}{md}
$$
It gives a simple pendulum with the same small oscillation period as the physical pendulum.

Example, Uniform Rod Pivoted at One End

Consider a uniform rod of length $L$ and mass $m$, pivoted at one end. Its center of mass is at

$$
d = \frac{L}{2}
$$

Its moment of inertia about one end is

$$
I = \frac{1}{3}mL^2
$$

Substitute into the period formula:

$$
T = 2\pi \sqrt{\frac{I}{mgd}}
= 2\pi \sqrt{\frac{\frac{1}{3}mL^2}{mg\frac{L}{2}}}
$$

Simplifying,

$$
T = 2\pi \sqrt{\frac{2L}{3g}}
$$

So a rod pivoted at one end behaves differently from a simple pendulum of length $L$, because its mass is distributed along its length.

Comparison with a Simple Pendulum

The difference between the two models is important:

Feature	Simple Pendulum	Physical Pendulum
Mass distribution	Point mass	Extended rigid body
Support	Massless string or rod	Fixed pivot axis
Important rotational quantity	Usually not explicit	Moment of inertia $I$
Restoring effect	Gravity	Gravity
Small angle period	$T = 2\pi\sqrt{L/g}$	$T = 2\pi\sqrt{I/(mgd)}$

A simple pendulum is often an idealization of a physical pendulum.

Stability of Equilibrium

A physical pendulum has stable equilibrium when its center of mass lies directly below the pivot. If displaced slightly, gravity creates a restoring torque and the body oscillates.

If the center of mass is above the pivot, the equilibrium is unstable. A tiny disturbance makes the object rotate farther away instead of returning.

Geometry of the Motion

The center of mass moves along a circular arc centered on the pivot.

Rendering scientific sketch…

Physical pendulum geometry

The weight $mg$ acts downward at the center of mass. Because this force does not usually pass through the pivot when the body is displaced, it creates the restoring torque.

Role of the Pivot Position

The same object can have different periods depending on where it is pivoted. Moving the pivot changes both $d$ and $I$.

For example, if a rod is pivoted near its center, then $d$ becomes small. That weakens the restoring torque. At the same time, the moment of inertia may also change. The resulting period can become quite large. If the pivot passes exactly through the center of mass, then $d = 0$, the restoring torque disappears, and there is no pendulum oscillation due to gravity.

If the pivot is at the center of mass, then $d = 0$, so gravity produces no restoring torque. In that case the body is not a gravitational pendulum.

Practical Importance

Physical pendulums are useful because real objects are extended bodies, not point masses. They are used in clocks, measuring devices, and experiments for finding the acceleration due to gravity or the moment of inertia of an object.

If the period $T$, mass $m$, and distance $d$ are known, then the moment of inertia can be determined from

$$
I = \frac{mgdT^2}{4\pi^2}
$$

Or if $I$ is known, the setup can be used to estimate $g$.

Final Picture

A physical pendulum is a rigid body swinging about a fixed axis. Its behavior depends on both gravity and how the mass is distributed. For small oscillations, it behaves like simple harmonic motion, but its period is controlled by the moment of inertia and the location of the center of mass relative to the pivot.

Key result for a physical pendulum:
$$
T = 2\pi \sqrt{\frac{I}{mgd}}
$$
This formula applies to small oscillations about a stable equilibrium position.

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