How it works
T = 2π√( I / (m·g·d) ) f = 1/T ω = 2π/T
A physical pendulum is a rigid body swinging about a fixed pivot. I is the moment of inertia about the pivot (the calculator derives it from each preset via the parallel-axis theorem Ipivot = Icm + m·d², so you never need to look it up), m the mass, d the pivot-to-center-of-mass distance, g gravity. Small-angle approximation.
L_eq = I / (m·d) d = (2π/T)² · I / (m·g) I = m·g·d · (T/2π)²
Frequently asked questions
What is a physical pendulum?
A physical (or compound) pendulum is any rigid body that swings about a fixed pivot that does not pass through its center of mass — a clock pendulum, a swinging rod, a disk on a peg. Unlike the idealized simple pendulum, its mass is distributed, so the period depends on the body's moment of inertia about the pivot: T = 2π√(I / (m·g·d)), where I is the moment of inertia about the pivot, m the mass, d the pivot-to-center-of-mass distance, and g gravity. This calculator evaluates that formula in four modes.
Physical pendulum vs simple pendulum — which calculator do I need?
Use this physical-pendulum calculator when the swinging object has significant size and distributed mass (rod, disk, sphere, real clock pendulum). Use a simple-pendulum calculator when the mass is concentrated at one point on a light string, where T = 2π√(L/g) applies. Our natural-frequency calculator (nrtechstudio.com/tools/natural-frequency/) covers the simple-pendulum case. Boundary: point mass on a massless string → natural-frequency; rigid extended body → this tool.
What do I, m, and d mean in the formula?
I is the moment of inertia about the pivot axis (kg·m²) — not about the center of mass. m is the total mass (kg). d is the distance from the pivot to the body's center of mass (m). g is the local gravitational acceleration (m/s²). All three must be positive; if d were zero (pivot exactly at the center of mass) the formula is undefined because there is no restoring torque.
Worked example: 1 m rod, 1 kg, pivoted at its end?
I about the center of a uniform rod is mL²/12 = 0.0833 kg·m²; the parallel-axis theorem shifts it to the pivot: I = mL²/3 = 0.3333 kg·m². The center of mass is d = L/2 = 0.5 m from the pivot. Then T = 2π√(0.3333 / (1 × 9.80665 × 0.5)) = 1.6382 s. Frequency f = 1/T = 0.6104 Hz, angular frequency ω = 2π/T = 3.8354 rad/s, and the equivalent length is L_eq = I/(m·d) = 0.6667 m. Try these exact inputs in the Period tab to verify.
Why does the calculator use 3/2 mR² for a disk pivoted at its rim, not 1/2 mR²?
Because 1/2 mR² is the disk's moment of inertia about its own center, while the pendulum formula needs I about the pivot. The parallel-axis (Steiner) theorem gives I_pivot = I_cm + m·d². For a disk pivoted at its rim, d = R, so I = 1/2 mR² + mR² = 3/2 mR². Likewise a solid sphere pivoted at its surface uses 7/5 mR² and a thin hoop at its rim uses 2mR². The From Geometry tab derives I about the pivot automatically and shows both values so you can check.
What is the equivalent length of a physical pendulum?
The equivalent length L_eq = I / (m·d) is the length of an ideal simple pendulum that would have exactly the same period. For the 1 m rod pivoted at its end, L_eq = 2L/3 = 0.6667 m — noticeably shorter than the rod itself. The Equivalent Length tab reports L_eq together with T so you can compare a real rigid body directly against the textbook simple-pendulum result.
How can I find my object's center of mass from a measured period?
Measure the small-swing period T, then invert the formula: d = (2π/T)² · I / (m·g). The Inverse tab does this automatically — it is the standard way to calibrate an unknown COM position or to back out a moment of inertia from timing data. Check: for the worked rod example (T = 1.6382 s, I = 0.3333 kg·m², m = 1 kg, g = 9.80665 m/s²) the solver returns d = 0.5000 m, the known COM position.
Does the formula assume small swings?
Yes. T = 2π√(I/(m·g·d)) is the small-angle approximation, accurate for amplitudes below roughly 10°. At larger amplitudes the true period is longer (the exact solution needs an elliptic integral), so a stopwatch measurement on wide swings will read slightly higher than this calculator. Keep test swings small when calibrating with the Inverse tab.
Which gravity value should I use?
The calculator offers presets: Earth 9.80665 m/s², Moon 1.62 m/s², Mars 3.71 m/s², Jupiter 24.79 m/s², plus a custom field. Period scales as 1/√g: the 1 m / 1 kg end-pivoted rod with T = 1.6382 s on Earth swings with T = 4.0307 s on the Moon. If you need another body or an exact local g, choose Custom.
When should I not use this calculator?
Not for a point mass on a string (simple pendulum, T = 2π√(L/g)) — use our natural-frequency calculator at nrtechstudio.com/tools/natural-frequency/ instead. Not for torsional pendulums (twisting wire or balance wheel), which follow T = 2π√(I/κ) with a torsion constant κ. And not for driven or damped systems where you need resonance or decay behavior rather than the free small-swing period.