Unit 7 · Topic 7.5

7.5 Simple and Physical Pendulums

A physical pendulum is a rigid object swinging about a fixed pivot. For small angles, gravity's restoring torque gives SHM with $T = 2\pi\sqrt{\dfrac{I}{mgd}}$ , where d is the distance from the pivot to the center of mass. A simple pendulum is the special case of a point mass on a string, and a torsion pendulum oscillates by twisting a wire.

Key terms

physical pendulum
simple pendulum
torsion pendulum
small-angle approximation
restoring torque

What changed in the 2024 update

Before the Fall 2024 course update, oscillations were Unit 6. They're now Unit 7, so older videos and practice may call this unit "Unit 6."

The current course now lists the physical pendulum by name. It showed up on older exams too, but now it's spelled out. The torsion pendulum is listed here as well. Practice both, along with springs and simple pendulums.

Physical pendulums

Hang any rigid object from a pivot that isn't at its center of mass, like a ruler from one end or a picture frame from a nail, and it swings. That's a physical pendulum.

When the object is turned through an angle θ from hanging straight down, gravity acting at the center of mass makes a restoring torque about the pivot: $\tau = -mgd\sin\theta$ . Here d is the distance from the pivot to the center of mass.

The small-angle approximation

For small angles, measured in radians, $\sin\theta \approx \theta$ . (At 10°, which is 0.175 rad, the error is about 0.5%.) Then Newton's second law for rotation, $\tau = I\alpha$ , becomes $I\frac{d^2\theta}{dt^2} = -mgd\,\theta \quad\Longrightarrow\quad \frac{d^2\theta}{dt^2} = -\frac{mgd}{I}\,\theta$

This has the SHM form from 7.3, with $\omega^2 = \dfrac{mgd}{I}$ . So $T = 2\pi\sqrt{\frac{I}{mgd}}$ where I is the rotational inertia about the pivot, not about the center of mass.

The simple pendulum as a special case

A simple pendulum is a point mass m on a light string of length ℓ. Its rotational inertia about the pivot is $I = m\ell^2$ , and d = ℓ. Substituting: $T = 2\pi\sqrt{\dfrac{m\ell^2}{mg\ell}} = 2\pi\sqrt{\dfrac{\ell}{g}}$ . The mass cancels.

A real object has a longer period than a point mass at its center of mass would, because its spread-out mass adds to I. For example, a hoop hung on a nail through its rim swings like a simple pendulum with length equal to its diameter.

Torsion pendulums

A torsion pendulum is an object, often a disk, hanging from a wire attached at its center. Twist it through an angle θ and the wire twists back with a torque proportional to the angle: $\tau = -\kappa\theta$ . The torsion constant κ (kappa) is the wire's twisting stiffness, in N·m/rad.

Newton's second law gives $I\dfrac{d^2\theta}{dt^2} = -\kappa\theta$ , so it's SHM with $\omega = \sqrt{\dfrac{\kappa}{I}}$ and $T = 2\pi\sqrt{\dfrac{I}{\kappa}}$ . No small-angle approximation is needed, as long as the wire obeys this rule.

Measuring g with a pendulum

Measure the period for several string lengths. Time 10 or 20 swings and divide, which shrinks the effect of your reaction time. Keep the angles small, and measure ℓ to the center of the bob.

Since T² = (4π²/g)ℓ, a graph of T² against ℓ should be a straight line through the origin with slope 4π²/g. Then g = 4π²/slope.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A swinging rod
A uniform rod 1.0 m long swings from a pivot at one end. Find its period for small swings, and compare it with a simple pendulum 1.0 m long.
Show the solutionHide the solution
1. Step 1: About the end, $I = \tfrac{1}{3}mL^2$ , and the center of mass is d = L/2 from the pivot.
2. Step 2: $T = 2\pi\sqrt{\dfrac{mL^2/3}{mg(L/2)}} = 2\pi\sqrt{\dfrac{2L}{3g}} = 2\pi\sqrt{\dfrac{2(1.0)}{3(9.8)}} \approx 1.64$ s.
3. Step 3: A 1.0 m simple pendulum has $T = 2\pi\sqrt{1.0/9.8} \approx 2.01$ s. The rod swings like a simple pendulum of length ⅔L, because much of its mass is closer to the pivot than the end is.
Answer: About 1.64 s, shorter than the 2.01 s of a 1.0 m simple pendulum
Example 2
Deriving the physical pendulum period
An object of mass m swings from a pivot a distance d from its center of mass, with rotational inertia I about the pivot. Show that for small angles its motion is SHM, and find its period.
Show the solutionHide the solution
1. Step 1: Gravity acts at the center of mass. Its torque about the pivot is $\tau = -mgd\sin\theta$ , with the minus sign because it turns the object back toward θ = 0.
2. Step 2: Newton's second law for rotation: $I\dfrac{d^2\theta}{dt^2} = -mgd\sin\theta$ .
3. Step 3: For small angles, $\sin\theta \approx \theta$ : $\dfrac{d^2\theta}{dt^2} = -\dfrac{mgd}{I}\theta$ . This has the SHM form $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$ .
4. Step 4: So $\omega = \sqrt{\dfrac{mgd}{I}}$ and $T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{I}{mgd}}$ .
Answer: SHM with T = 2π√(I/(mgd))
Example 3Calculator allowed
A torsion pendulum (exam-level)
A uniform disk of mass 2.0 kg and radius 0.10 m hangs from a wire through its center, with torsion constant κ = 0.040 N·m/rad. Find the period. What happens to the period if you swap in a disk of the same mass but twice the radius? What if you double the twist amplitude?
Show the solutionHide the solution
1. Step 1: $I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0)(0.10)^2 = 0.010$ kg·m².
2. Step 2: $T = 2\pi\sqrt{\dfrac{I}{\kappa}} = 2\pi\sqrt{\dfrac{0.010}{0.040}} = 2\pi(0.50) \approx 3.1$ s.
3. Step 3: Doubling R makes I four times as large, so T doubles to about 6.3 s.
4. Step 4: Doubling the amplitude doesn't change the period. The restoring torque is proportional to θ, which is exactly the SHM condition.
Answer: T ≈ 3.1 s; about 6.3 s with twice the radius; no change with twice the amplitude

Common mistakes

Using I about the center of mass in T = 2π√(I/mgd). Use I about the pivot, with the parallel axis theorem if needed.
Using the rod's full length as d. d is the distance from the pivot to the center of mass.
Applying the small-angle result to large swings. The formula is only accurate for small angles.
Thinking a heavier pendulum bob changes the period. For a simple pendulum, the mass cancels.

On the exam

Expect a derivation that starts from τ = Iα, uses sin θ ≈ θ, and ends in the SHM form. Name the small-angle approximation when you use it.
Experimental Design questions often involve pendulums. Describe what you measure and how (time many swings), and which graph gives a straight line (like T² against ℓ).

Connected topics

Videos

Topic 7.5 - Simple and Physical Pendulums
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Modeling pendulums | AP Physics | Khan Academy
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Simple Pendulum - Simple Harmonic Motion Derivation using Calculus
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AP Physics C - Pendulums
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Physics 16.6 Torsion (6 of 14) Torsional Pendulum (Potential Equivalent of SHM)
Michel van BiezenWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 7.5 Simple and Physical Pendulums. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A uniform rod 1.5 m long swings through small angles about a pivot at one end. Its rotational inertia about that end is $\frac{1}{3}ML^2$ . Use g = 10 m/s². What is its period?

Question 2 of 4Calculator allowed

A uniform rod of length L pivoted at one end and a simple pendulum of length L both swing through small angles. How do their periods compare?

Question 3 of 4Calculator allowed

A horizontal disk with rotational inertia 0.020 kg·m² hangs from a wire through its center. When the disk is twisted by an angle θ, the wire exerts a restoring torque $\tau = -\kappa\theta$ , with κ = 0.50 N·m/rad. What is the period of its twisting oscillations?

Question 4 of 4Calculator allowed

A torsion pendulum uses a uniform disk hung from a wire through its center. The disk is replaced by one with the same mass but twice the radius, on the same wire. What happens to the period?

0 of 4 answered