Physical Pendulum

AP Physics C: Mechanics· 12 min read

1. Derivation of the Physical Pendulum Period★★★☆☆⏱ 15 min

Unlike the idealized simple pendulum where all mass is concentrated at a single point, a physical pendulum accounts for distributed mass across the entire rigid object. The derivation starts from Newton's second law for rotation, rather than linear motion.

📘 Definition

Physical Pendulum

Any rigid body free to rotate about a fixed pivot point that is not aligned with its center of mass, producing a gravitational restoring torque when displaced from equilibrium.

🔬 Derivation

Goal:

Derive the SHM period for a physical pendulum under small angle approximation

Starting from:

Net torque on displaced object: $τ = - m g d sin θ$ , negative sign indicates restoring direction

1
Apply rotational Newton's second law: $τ = I_{p i v o t} α = I_{p i v o t} \frac{d ^{2} θ}{d t ^{2}}$
2
Substitute torque expression: $I_{p i v o t} \frac{d ^{2} θ}{d t ^{2}} = - m g d sin θ$
3
Apply small angle approximation $sin θ \approx θ$ for $θ < 1 0^{\circ}$
4
Rearrange to standard SHM differential form: $\frac{d ^{2} θ}{d t ^{2}} = - (\frac{m g d}{I _{p i v o t}}) θ$
5
Identify angular frequency $ω = \frac{m g d}{I _{p i v o t}}$ , so period $T = \frac{2 π}{ω}$

Result:

Final period formula: $T = 2 π \frac{I _{p i v o t}}{m g d}$

📐 Worked Example

A uniform 1m long rod of mass 2kg is pivoted at one end. Calculate its period of small oscillation.

1
First find rotational inertia of the rod about the pivot at its end
2
$I_{pivot} = \frac{1}{3}mL^2 = \frac{1}{3}(2)(1)^2 = 0.667 \text{ kg m}^2$
3
Find distance d from pivot to center of mass, which is at the midpoint of the rod
4
$d = 0.5 \text{ m}$
5
Substitute all values into the physical pendulum period formula
6
$T = 2\pi \sqrt{\frac{0.667}{(2)(9.8)(0.5)}} = 1.64 \text{ s}$

✓ Quick check

Test your understanding of the derivation

What approximation is required to reduce the torque equation to SHM form?
- Large angle approximation
- Small angle approximation
- Massless object approximation
- Zero gravity approximation
Reveal answer
Small angle approximation —
$sin θ \approx θ$ only holds for angular displacements below ~10 degrees, which is required to produce the linear restoring term for SHM.

Exam tip:

You must explicitly state the small angle approximation in your derivation to earn full points on AP FRQs, even if it seems obvious.

2. Special Cases and Simple Pendulum Limit★★☆☆☆⏱ 8 min

The simple pendulum that you learned earlier is just a special limiting case of the physical pendulum. If all mass of the pendulum is concentrated at a single point located a distance L from the pivot, we can substitute $I_{p i v o t} = m L^{2}$ and $d = L$ into the general formula.

T = 2\pi \sqrt{\frac{mL^2}{mgL}} = 2\pi \sqrt{\frac{L}{g}}

📐 Worked Example

Show that a 0.5m long simple pendulum has the same period as a uniform rod pivoted at a point that is not its end.

1
Set simple pendulum period equal to physical pendulum period for the rod
2
$2\pi \sqrt{\frac{0.5}{9.8}} = 2\pi \sqrt{\frac{I_{pivot}}{mgd}}$
3
For a uniform rod, $I_{c m} = \frac{1}{12} m L^{2}$ , so use parallel axis theorem: $I_{p i v o t} = \frac{1}{12} m L^{2} + m d^{2}$
4
Cancel terms and solve for d, you will find d = 0.408m from the rod CM, matching the 0.5m equivalent simple pendulum length.

3. Common Rigid Object Problem Setups★★★☆☆⏱ 10 min

AP exam problems almost always use standard symmetric rigid objects for physical pendulum questions, including uniform rods, solid disks, and thin hoops. You are expected to recall their rotational inertia about their center of mass and apply the parallel axis theorem correctly.

Object	Pivot Location	Period Formula
Uniform thin rod	Pivoted at one end	$T = 2 π \frac{2 L}{3 g}$
Solid uniform disk	Pivoted at edge	$T = 2 π \frac{3 R}{2 g}$
Thin hoop	Pivoted at edge	$T = 2 π \frac{2 R}{g}$

📐 Worked Example

A solid disk of radius 0.2m is pivoted at a point on its outer edge. Calculate its small oscillation period.

1
Rotational inertia of disk about CM is $I_{c m} = \frac{1}{2} m R^{2}$
2
Apply parallel axis theorem for pivot at edge, d = R
3
$I_{pivot} = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$
4
Substitute into physical pendulum formula, cancel mass terms
5
$T = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}} = 1.10 \text{ s}$

4. Experimental Physical Pendulum Analysis★★★★☆⏱ 12 min

Physical pendulum labs are extremely common on the AP Physics C exam, where you are asked to measure the period of an irregular object to calculate its unknown rotational inertia, without needing to integrate over its shape.

📐 Worked Example

You measure the period of an irregular object pivoted 0.3m from its CM to be 2.2s. Calculate its rotational inertia about the pivot.

1
Rearrange the period formula to solve for $I_{p i v o t}$
2
$I_{pivot} = \frac{T^2 mgd}{4\pi^2}$
3
Substitute measured values, assuming object mass is 1.5kg
4
$I_{pivot} = \frac{(2.2)^2 (1.5)(9.8)(0.3)}{4\pi^2} = 0.54 \text{ kg m}^2$

5. Common Pitfalls

Wrong move:

Using the simple pendulum point mass formula $T = 2 π L / g$ for a pivoted uniform rod

Why:

The rod's mass is distributed across its full length, not concentrated at the far end, so the simple pendulum assumption does not hold

Correct move:

Calculate the full rotational inertia of the rod about the pivot using the parallel axis theorem, then use the general physical pendulum formula

Wrong move:

Using the total length of the object instead of the distance from pivot to center of mass for d

Why:

Restoring torque only depends on the lever arm to the center of mass, not the full dimension of the rigid object

Correct move:

Explicitly mark d on your diagram as the distance between pivot and CM before substituting any values

Wrong move:

Forgetting to state the small angle approximation during derivation

Why:

AP graders explicitly award 1 point for noting $sin θ \approx θ$ , and you will lose points if you omit it

Correct move:

Write the small angle approximation step clearly immediately after writing the full torque equation

Wrong move:

Using rotational inertia about the center of mass directly in the period formula without parallel axis theorem

Why:

The period formula requires rotational inertia about the pivot point, not the object's CM

Correct move:

Always apply $I_{p i v o t} = I_{c m} + m d^{2}$ before plugging inertia values into the period equation

Wrong move:

Assuming period always increases as you move the pivot further away from the CM

Why:

There is a minimum period at the radius of gyration distance from the CM, after which period starts increasing again

Correct move:

Take derivative dT/dd and set to zero to find the minimum period location for any rigid object

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
General Period	$T = 2 π \frac{I _{p i v o t}}{m g d}$	Small angles only, d = pivot to CM distance
Simple Pendulum Limit	$T = 2 π \frac{L}{g}$	Point mass, massless string
Rod pivoted at end	$T = 2 π \frac{2 L}{3 g}$	L = total rod length
Disk pivoted at edge	$T = 2 π \frac{3 R}{2 g}$	R = disk radius
Hoop pivoted at edge	$T = 2 π \frac{2 R}{g}$	R = hoop radius

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · Set 1 FRQ Q2
Physical pendulum lab analysis
2019 · FRQ Q3
Pivoted rod oscillation calculation
2017 · FRQ Q1
Irregular object period derivation

What's Next

Mastering physical pendulum connects rotational motion concepts to simple harmonic oscillation, a frequent cross-topic FRQ that makes up 10-15% of your AP Physics C Mechanics exam score. You will next apply this framework to analyze torsional pendulums, which use restoring torque from twisted wires instead of gravitational force, and practice full lab design questions that ask you to derive rotational inertia of unknown objects from period measurements. These skills will also carry over to your study of wave motion, where SHM of individual particles forms the foundation for traveling and standing wave behavior. Confirm you can independently derive the full period formula without referencing notes before moving on.