Physical Pendulum Calculator
Find the period of a swinging rod, disk, ring or custom body from its moment of inertia, and compare it with a simple pendulum.
🕰️ What is a Physical Pendulum?
A physical pendulum, also called a compound pendulum, is any rigid body that swings under gravity about a fixed horizontal pivot that does not pass through its centre of mass. A swinging door, a meter stick hung from a nail, a leg swinging while walking, a clock pendulum with a heavy bob on a rod and a playground swing with its seat all behave like physical pendulums. Unlike the textbook simple pendulum, where a point mass hangs on a massless string, a physical pendulum has its mass spread out, so how that mass is distributed around the pivot, measured by the moment of inertia I, controls how fast it swings. For small oscillations the period is T = 2π √(I / (m g d)), where m is the mass, g the gravitational acceleration and d the distance from the pivot to the centre of mass.
Finding I is usually the hard part, and this calculator does it for you. For a uniform rod, a solid disk and a thin ring, the moment of inertia about the centre of mass is known: m L² / 12, m R² / 2 and m R². The parallel axis theorem then moves the axis to the pivot, I = Icm + m d². For anything else, choose the custom body and type the Icm that you looked up, measured or computed. The results include the equivalent simple pendulum length Leq = I / (m d), which is the length of a simple pendulum that would swing with exactly the same period, and a direct comparison with a point-mass pendulum of length d. Because the mass is spread out, the extended body is always slower than the point mass, and the ratio shows by how much.
The physical pendulum is a favourite exam and laboratory topic because it shows two surprising facts at once. First, the mass cancels, so only the shape and the pivot position matter. Second, moving the pivot closer to the centre of mass does not keep shortening the period. The chart plots the period against the pivot distance and shows a minimum at d = √(Icm / m), after which the period grows without limit as the pivot approaches the centre. For the energy of a swing, see the kinetic energy of a pendulum calculator, and for the oscillation itself as a wave in time, the harmonic wave equation calculator. Gravity differs between worlds, so a Moon preset is included.
Limits to remember: the formula assumes small angles, a rigid body, a frictionless pivot and no air drag, and it treats the pivot axis as parallel to the axis through the centre of mass. For large swings the true period is longer, and for a body whose shape is not one of the standard solids the custom mode needs a trustworthy Icm.
📐 Formula
📖 How to Use This Calculator
Steps
💡 Example Calculations
Example 1 — A Rod Hung From Its End
A 2 kg uniform rod 1 m long, pivoted at one end, on Earth
Example 2 — A Disk Hung From Its Rim
A 3 kg solid disk of radius 0.3 m, pivoted at its rim, on Earth
Example 3 — A Hoop on a Nail
A 1.5 kg thin ring of radius 0.3 m hung from a nail on its rim, on Earth
Example 4 — A Custom Body
A 1 kg body with Icm = 0.05 kg m², pivoted 0.3 m from its centre of mass
Example 5 — A Rod Pivoted Off the End
The same 2 kg, 1 m rod, pivoted 0.25 m from one end, on Earth
❓ Frequently Asked Questions
🔗 Related Calculators
What is the formula for the period of a physical pendulum?
T = 2 pi sqrt(I / (m g d)), where I is the moment of inertia about the pivot, m the mass, g the gravitational acceleration and d the distance from the pivot to the centre of mass. It holds for small oscillations.
What is the period of a uniform rod pivoted at its end?
I = m L² / 3 and d = L / 2, so T = 2 pi sqrt(2 L / (3 g)). A 1 m rod swings with a period of 1.638 s on Earth, which is about 15 percent longer than a 1 m simple pendulum would need at its centre of mass.
What is the difference between a simple and a physical pendulum?
A simple pendulum has all its mass at the end of a massless string. A physical pendulum is a real extended rigid body, so its moment of inertia matters. Its period is longer than the simple pendulum with the same pivot to centre of mass distance.
What is the equivalent simple pendulum length?
It is L eq = I / (m d), the length of a simple pendulum with the same period. For a rod hung from its end, L eq = 2 L / 3, so a 1 m rod swings like a 0.667 m simple pendulum.
How does the parallel axis theorem apply to a pendulum?
The tables give the moment of inertia I cm about the centre of mass. For a pivot a distance d away, the theorem gives I = I cm + m d². This is the I used in the period formula.
Does the mass of the pendulum affect its period?
No. I is proportional to m, so the mass cancels in I / (m g d). The period depends only on the geometry of the body, the pivot position and g.
What is the period of a disk pivoted at its rim?
I = 3 m R² / 2 and d = R, so T = 2 pi sqrt(3 R / (2 g)). A disk of radius 0.3 m has T = 1.346 s on Earth, and its equivalent simple pendulum length is 1.5 R = 0.45 m.
Where should the pivot be for the shortest period?
At a distance d equal to the radius of gyration about the centre of mass, k = sqrt(I cm / m). The minimum period is 2 pi sqrt(2 k / g). For a rod of length L that is d = L / sqrt(12), about 0.289 L from the centre.
Why can the pivot not be at the centre of mass?
With d = 0, gravity acts through the pivot and gives no torque. There is no restoring force, so the body does not oscillate and the formula would give an infinite period.
Is the formula accurate for large swings?
No. It assumes the angle is small enough that sin(theta) is close to theta. For a 60 degree swing the true period is about 7 percent longer, and it grows without limit as the amplitude approaches 180 degrees.