# Physical Pendulum

> AP Physics C: Mechanics · AP Physics C: Mechanics 2024+
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u6-physical-pendulum/

We derive the physical pendulum period formula, work through common rigid object examples, connect it to the simple pendulum limit, and practice lab-style calculation questions aligned to AP C scoring standards.

**Prerequisites:** [Torque and rotational inertia for rigid bodies](https://www.owlsprep.com/study/ap-physics-c-mech-u5-rotational-inertia/); [Simple harmonic motion differential equation](https://www.owlsprep.com/study/ap-physics-c-mech-u6-simple-harmonic-motion-basics/)

## Learning objectives

- Derive the period formula for a physical pendulum using torque and rotational inertia
- Calculate the period of oscillation for arbitrarily shaped rigid objects
- Relate the physical pendulum to the simple pendulum as a special limiting case
- Solve for unknown quantities like rotational inertia using measured oscillation periods

## Derivation of the Physical Pendulum Period

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.

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

*Starting from:* Net torque on displaced object: $\tau = -mgd\sin\theta$, negative sign indicates restoring direction

1. Apply rotational Newton's second law: $\tau = I_{pivot} \alpha = I_{pivot} \frac{d^2\theta}{dt^2}$
2. Substitute torque expression: $I_{pivot} \frac{d^2\theta}{dt^2} = -mgd \sin\theta$
3. Apply small angle approximation $\sin\theta \approx \theta$ for $\theta < 10^\circ$
4. Rearrange to standard SHM differential form: $\frac{d^2\theta}{dt^2} = - \left(\frac{mgd}{I_{pivot}}\right) \theta$
5. Identify angular frequency $\omega = \sqrt{\frac{mgd}{I_{pivot}}}$, so period $T = \frac{2\pi}{\omega}$

*Conclusion:* Final period formula: $T = 2\pi \sqrt{\frac{I_{pivot}}{mgd}}$

**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. $$T = 2\pi \sqrt{\frac{0.667}{(2)(9.8)(0.5)}} = 1.64 \text{ s}$$

**Check your understanding**

Test your understanding of the derivation

1. What approximation is required to reduce the torque equation to SHM form?

   - Large angle approximation
   - Small angle approximation
   - Massless object approximation
   - Zero gravity approximation

   *Why:* $\sin\theta \approx \theta$ 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.

## Special Cases and Simple Pendulum Limit

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 $I_{pivot} = mL^2$ 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, $I_{cm} = \frac{1}{12}mL^2$, so use parallel axis theorem: $I_{pivot} = \frac{1}{12}mL^2 + md^2$
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.

> **Sweet Spot Connection**
>
> This equivalent point mass location is called the center of oscillation, and it is the point where you can strike the rod to produce zero reaction force at the pivot (the sweet spot for a baseball bat).

## Common Rigid Object Problem Setups

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 | $T=2\pi\sqrt{\frac{2L}{3g}}$ |
| Solid uniform disk | Pivoted at edge | $T=2\pi\sqrt{\frac{3R}{2g}}$ |
| Thin hoop | Pivoted at edge | $T=2\pi\sqrt{\frac{2R}{g}}$ |

**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 $I_{cm} = \frac{1}{2}mR^2$
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. $$T = 2\pi \sqrt{\frac{\frac{3}{2}mR^2}{mgR}} = 2\pi \sqrt{\frac{3R}{2g}} = 1.10 \text{ s}$$

## Experimental Physical Pendulum Analysis

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.

**Exam command terms**

Common exam command terms for this topic:

- **Derive** — You must show every step from torque Newton's law to the final period formula, no jumps allowed

- **Describe a procedure** — You must specify measuring multiple periods to reduce timing uncertainty, and state the small angle requirement

- **Calculate rotational inertia** — Rearrange the period formula algebraically first before substituting measured values

**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 $I_{pivot}$
2. $$I_{pivot} = \frac{T^2 mgd}{4\pi^2}$$
3. Substitute measured values, assuming object mass is 1.5kg
4. $$I_{pivot} = \frac{(2.2)^2 (1.5)(9.8)(0.3)}{4\pi^2} = 0.54 \text{ kg m}^2$$

## Common pitfalls

- **Wrong:** Using the simple pendulum point mass formula $T=2\pi\sqrt{L/g}$ for a pivoted uniform rod
  - Why it fails: 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: Calculate the full rotational inertia of the rod about the pivot using the parallel axis theorem, then use the general physical pendulum formula
- **Wrong:** Using the total length of the object instead of the distance from pivot to center of mass for d
  - Why it fails: Restoring torque only depends on the lever arm to the center of mass, not the full dimension of the rigid object
  - Correct: Explicitly mark d on your diagram as the distance between pivot and CM before substituting any values
- **Wrong:** Forgetting to state the small angle approximation during derivation
  - Why it fails: AP graders explicitly award 1 point for noting $\sin\theta \approx \theta$, and you will lose points if you omit it
  - Correct: Write the small angle approximation step clearly immediately after writing the full torque equation
- **Wrong:** Using rotational inertia about the center of mass directly in the period formula without parallel axis theorem
  - Why it fails: The period formula requires rotational inertia about the pivot point, not the object's CM
  - Correct: Always apply $I_{pivot} = I_{cm} + md^2$ before plugging inertia values into the period equation
- **Wrong:** Assuming period always increases as you move the pivot further away from the CM
  - Why it fails: There is a minimum period at the radius of gyration distance from the CM, after which period starts increasing again
  - Correct: Take derivative dT/dd and set to zero to find the minimum period location for any rigid object

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| General Period | $T = 2\pi \sqrt{\frac{I_{pivot}}{mgd}}$ | Small angles only, d = pivot to CM distance |
| Simple Pendulum Limit | $T = 2\pi \sqrt{\frac{L}{g}}$ | Point mass, massless string |
| Rod pivoted at end | $T = 2\pi \sqrt{\frac{2L}{3g}}$ | L = total rod length |
| Disk pivoted at edge | $T = 2\pi \sqrt{\frac{3R}{2g}}$ | R = disk radius |
| Hoop pivoted at edge | $T = 2\pi \sqrt{\frac{2R}{g}}$ | R = hoop radius |

## 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.

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