How this instrument works
A physical pendulum is any rigid body that swings about a pivot under gravity — a swinging door, a metronome arm, a connecting rod, a human leg mid-stride. Unlike the idealized simple pendulum, which treats all the mass as a single point on a massless string, a physical pendulum keeps the mass exactly where it actually sits, and that distribution is captured by I, the moment of inertia about the pivot axis itself.
The formula comes straight from rotational dynamics: gravity acting through the center of mass creates a restoring torque of magnitude mgd sinθ, and for small swings sinθ is close enough to θ that the motion obeys the same second-order equation as a mass on a spring. Solving it gives T = 2π√(I ⁄ mgd) — swap I for mL² and d for L and it collapses exactly to the familiar simple-pendulum formula, which confirms a simple pendulum is just the special case where all the mass sits at one radius.
Two edge cases are worth knowing before you trust a reading. First, I has to be measured about the actual pivot, not the center of mass — if a datasheet only lists the center-of-mass moment of inertia, the parallel-axis theorem, I_pivot = I_cm + md², converts it. Second, the small-angle approximation behind the formula starts drifting once the amplitude passes roughly 20°, and if the pivot happens to sit exactly at the center of mass, d becomes zero, gravity produces no restoring torque at all, and the body simply balances instead of swinging.
- Enter the body's Moment of inertia about the pivot, in kg·m² — the rotational inertia measured about the actual hinge or pivot axis, not the center of mass.
- Enter the Mass of the rigid body in kilograms.
- Enter the Distance from pivot to center of mass in metres, or switch the unit menu to centimetres if that's easier to measure.
- Read the Oscillation period in seconds — the time for one full swing out and back, accurate while the amplitude stays small.
Worked example — timing a 2 kg physical pendulum
Take a rigid arm swinging on a hinge: its moment of inertia about that pivot is 0.5 kg·m² (found with a torsion-swing test), its total mass is 2 kg, and its center of mass sits 0.3 m from the pivot, located by balancing the arm across a knife edge.
Multiply mass, standard gravity, and pivot distance: mgd = 2 × 9.80665 × 0.3 = 5.88399 kg·m²/s². Divide the moment of inertia by that, 0.5 ⁄ 5.88399 = 0.084976, take the square root, 0.291506, and multiply by 2π to get T = 1.831593 s, matching the instrument's readout of about 1.83 seconds per full swing.
That period equals what a simple pendulum with length L = I ⁄ (md) = 0.833 m would give — nearly triple the true 0.3 m pivot-to-CoM distance, because mass spread away from the center of mass, which is exactly what I measures, still has to be carried through the swing, and that extra inertia stretches the period well past what a naive point-mass guess at 0.3 m would predict.
Questions
How is a physical pendulum different from a simple pendulum?
A simple pendulum idealizes all the mass as a point at radius L on a massless string, so I = mL² by definition. A physical pendulum is any rigid body swinging on a real pivot, and its I comes from the actual mass distribution — a baseball bat, a swinging door, or a robot arm rarely matches the point-mass assumption, so this formula is the general case and the simple pendulum is only the special case where I happens to equal mL².
Why does the formula need I about the pivot, not the center of mass?
Because torque and rotational inertia are only meaningful about the actual axis of rotation — the pivot, in this case. If you only have I measured about the center of mass, convert it first with the parallel-axis theorem, I_pivot = I_cm + md², before entering it here; using I_cm directly by mistake understates the true period.
Does this formula still work for large swings?
Only approximately. The 2π√(I⁄mgd) result comes from replacing sinθ with θ, which holds well under about 20° of amplitude, with error staying under roughly 1%. Beyond that the true period is longer and needs an elliptic-integral correction; a grandfather clock or lab pendulum kept to small arcs stays safely inside this formula's accuracy.
What happens if the pivot sits at the center of mass?
Then d = 0, gravity produces zero restoring torque about that axis, and the body doesn't swing at all — it just balances, like a wheel spinning on a perfectly centered axle. The formula's denominator goes to zero, which is the mathematics agreeing with the physics: no torque, no oscillation, no defined period.
What is this formula used for in practice?
It's the physics behind Kater's reversible pendulum, a 19th-century instrument that located two pivot points giving identical periods to measure g to five-figure precision without ever needing I or d individually, and it's also how clockmakers and instrument designers predict the true period of a pendulum bob with real size and shape rather than a point mass.
Can I use this for a swinging door or gate?
Yes, provided you know its moment of inertia about the hinge line, its mass, and how far its center of mass sits from the hinges. A plain slab door is roughly I = mw² ⁄ 3 about a vertical edge, where w is the width, but a door with glass panes, a closer, or hardware mounted off-center needs its real I, not that shortcut, or the predicted period will be off.