AP Physics C: Mechanics · Unit 7: Oscillations · Lesson 7.5

Deep Dive: Simple and Physical Pendulums

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
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Physical Pendulum Basics

A physical pendulum is a rigid body that undergoes oscillation about a fixed axis — a swinging door, a metronome arm, a rod pivoted at one end. Unlike the idealized simple pendulum, its mass isn't concentrated at a single point, so its period formula has to account for how that mass is actually distributed:

Tphys = 2π√(I / mgd)

Here, I is the rigid body's rotational inertia about the pivot, m is its total mass, g is gravitational acceleration, and d is the distance from the pivot to the body's center of mass. For small amplitudes of motion, this period comes directly from applying Newton's second law in rotational form — which is exactly where the next section picks up.

Adjust a rigid body's rotational inertia, mass, and the distance from its pivot to its center of mass, and watch the period respond.

d (to COM)
I (kg·m²)0.60
m (kg)2.0
d (m)0.40
Tphys = 2π√(I/mgd) = 1.74 s
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Deriving the Restoring Torque

When a physical pendulum is displaced from equilibrium, the gravitational force acting at its center of mass produces a restoring torque:

τ = −mgd sinθ
🔑This is the rotational sibling of Lesson 7.1's restoring-force condition — a torque that always acts to push the system back toward θ = 0, exactly the way F = −kΔx always pushed a linear system back toward its equilibrium position.

As written, τ = −mgd sinθ isn't linear in θ, so it doesn't immediately look like SHM. But for small amplitudes of motion, the small-angle approximation applies:

sinθ ≈ θ

Substituting that in, combined with Newton's second law in rotational form (τ = Iα), gives:

τ = −mgdθ = Iα

Slide the displacement angle and compare the exact restoring torque to the small-angle approximation — notice how well they agree at small angles, and how they drift apart as θ grows.

θ (deg)15°
τ = −mgd sinθ = -1.522 N·m (exact)τ ≈ −mgdθ = -1.539 N·m (small-angle)difference: 1.2%

Rearranging that equation into the form α = d²θ/dt² produces exactly the second-order differential equation that defines SHM — now written in terms of angle instead of linear displacement:

d²θ/dt² = −ω²θ
💡This is the same differential equation from Lesson 7.3, d²x/dt² = −ω²x, just with θ in place of x. Every physical pendulum, at small amplitude, is SHM — for exactly the same mathematical reason a spring-mass system is.
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Simple Pendulum as a Special Case

A simple pendulum — the hanging point mass on a string you met back in Lesson 7.2 — is just a physical pendulum where all the mass is modeled as a single point at distance l from the pivot. Nothing new is needed; the general formula already contains this case.

ExampleGuided Example — Recovering Tp = 2π√(l/g)

Starting from the physical pendulum period formula, show that modeling a simple pendulum as a point mass at distance l reduces it exactly to the formula from Lesson 7.2.

Step 1Start from the physical pendulum formula
Tphys = 2π√(I / mgd).
🔑Notice that mass never mattered — it canceled out algebraically, which is exactly why the simple-pendulum period doesn't depend on mass. That fact wasn't a coincidence back in Lesson 7.2; it falls straight out of the physical-pendulum formula.
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Torsion Pendulum

Not every pendulum swings under gravity. A torsion pendulum is a case of SHM where the restoring torque is proportional to the angular displacement of a rotating system — the classic example being a horizontal disk suspended from a wire attached to its center of mass, which twists back and forth in the horizontal plane as the wire resists being twisted:

Iα = −κΔθ

(This is often written with the symbol k, as in the CED — but it's a torsional constant with units N·m/rad, not the same kind of quantity as a linear spring constant.) The mathematical form is identical to a mass on a spring or a small-angle pendulum: a restoring effect proportional to displacement, producing the exact same SHM differential equation once again.

A horizontal disk hangs from a wire and twists back and forth. Adjust the disk's rotational inertia and the wire's torsion constant to see how the period responds — the same math as a spring, with angle standing in for displacement.

disk twists about the wire
I (kg·m²)0.020
κ (N·m/rad)0.8
T = 2π√(I/κ) = 0.99 s
ExampleGuided Example — A Torsion Balance Disk

A horizontal disk (I = 0.02 kg·m²) hangs from a wire with torsion constant κ = 0.8 N·m/rad — the same basic setup used in a Cavendish-style torsion balance, the classic instrument for measuring the gravitational constant. Find its period of oscillation.

Step 1Start from the torsion-pendulum relation
Iα = −κΔθ leads to the same SHM period formula as a spring: T = 2π√(I/κ).
💡Every restoring mechanism in this unit — a spring's F = −kΔx, a physical pendulum's gravity-driven torque, a torsion pendulum's twisted wire — leads to the exact same differential equation, d²(something)/dt² = −ω²(something). Once you recognize that pattern, you can find the period of almost any oscillating system without starting from scratch.
← Back to Lesson 7.5Course Home →

That's Lesson 7.5, Unit 7, and the entire AP Physics C: Mechanics course, complete — from kinematics and Newton's laws all the way through orbits, rotating systems, and oscillations. Congratulations on making it through all seven units.