Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Physical Pendulum Calculator

Find the small-angle period of a rigid swinging body from its moment of inertia about the pivot, its mass, gravity and the pivot-to-centre-of-mass distance.

Physical Pendulum Calculator

kg·m²
kg
m/s²
m
Small-angle period
2.0064
The time in seconds for one complete back-and-forth oscillation, valid for small swings of a rigid, frictionless pendulum.
Model scope
Rigid-body, frictionless, small-angle period; large amplitude, damping, flexible bodies, moving pivots, distributed buoyancy, and uncertainty in I or centre of mass are outside it.

Background.

A clock pendulum with a heavy bob on a rod, a ruler hung from a nail, a leg swinging through a stride, a gate rocking on its hinges — most real pendulums are not a point mass on a weightless string. Their mass is spread out, and the spread matters: some of it sits near the pivot and barely moves, some swings wide. The physical, or compound, pendulum formula handles this by replacing the string length with two properties of the body: its moment of inertia about the pivot, I, and the distance d from the pivot to its centre of mass.

For small swings the period is T = 2π√(I/(mgd)). Gravity pulls on the centre of mass with a lever arm d, which supplies the restoring torque; the moment of inertia resists being turned. More inertia slows the swing, while more mass, stronger gravity or a longer lever arm speeds it up.

A handy way to read any physical pendulum is through its equivalent simple pendulum, of length I/(md). The default case — 2 kg, I = 0.8 kg·m², centre of mass 0.4 m below the pivot — behaves exactly like a simple pendulum 1 m long, with a period of about 2.006 s at standard gravity, close to the seconds pendulum of a longcase clock. The point that distance below the pivot, the centre of oscillation, is special: hang the body from there instead and the period is unchanged, which is how Kater's reversible pendulum measured g.

The formula assumes a rigid body, no friction and small swings. At an amplitude of 20° the true period is already about 0.8% longer than the figure given here.

What is physical pendulum calculator?

A physical pendulum is any rigid body that swings freely about a fixed horizontal axis that does not pass through its centre of mass. Its small-angle period is the time for one complete back-and-forth oscillation when the swing is only a few degrees. Unlike a simple pendulum, whose period depends only on string length and gravity, a physical pendulum's period depends on how the body's mass is distributed around the pivot, which is captured by its moment of inertia.

How to use this calculator.

  1. Enter the moment of inertia about the pivot axis in kg·m², not about the centre of mass. If you only know the centre-of-mass value, add m × d² to it (the parallel-axis theorem).
  2. Enter the total mass in kilograms and the pivot-to-centre-of-mass distance in metres. For a flat or slender object, the balance point on a knife edge locates the centre of mass.
  3. Leave gravity at the standard 9.80665 m/s² unless you need a local value or another world; on the Moon, use about 1.62.
  4. Check that the moment of inertia is at least m × d². No real body can have less, and a smaller entry gives a period that is impossibly short.
  5. Read the period in seconds for one full back-and-forth swing; halve it for the time to swing from one side to the other.

The formula.

T = 2π sqrt(I_p / (mgd))

T = 2π√(I / (m·g·d)), where I is the moment of inertia about the pivot in kg·m², m the mass in kilograms, g the gravitational acceleration in m/s² and d the pivot-to-centre-of-mass distance in metres. It comes from the rotational form of Newton's second law. Turned through a small angle θ, the body feels a restoring torque m·g·d·sin θ, which is close to m·g·d·θ when θ is small; I·θ'' = −m·g·d·θ is simple harmonic motion with angular frequency √(m·g·d/I), and the period is 2π divided by that. Everything sits under a square root, so quadrupling I doubles the period and quadrupling g halves it. Setting I = m·L² and d = L, a point mass on a string, recovers the simple pendulum, T = 2π√(L/g). The step sin θ ≈ θ is the main approximation.

A worked example.

Example

The default pendulum has a mass of 2 kg, a moment of inertia of 0.8 kg·m² about its pivot and its centre of mass 0.4 m below the pivot, under standard gravity of 9.80665 m/s². The restoring term is m·g·d = 2 × 9.80665 × 0.4 = 7.84532 N·m per radian. Dividing, I ÷ (m·g·d) = 0.8 ÷ 7.84532 = 0.101972 s², whose square root is 0.31933 s. Multiplying by 2π gives a period of 2.0064 s for one full oscillation. Its equivalent simple pendulum is I ÷ (m·d) = 0.8 ÷ 0.8 = 1 m long, and a 1 m simple pendulum gives the same 2.0064 s. Taken to the Moon, at 1.62 m/s², the same body would take about 4.94 s per oscillation.

pivot Moment Of Inertia Kg M20.8
mass Kg2
gravity Mps29.807
pivot To Center Of Mass M0.4

Frequently asked questions.

How do I find the moment of inertia about the pivot?
Start from the centre-of-mass value for the shape and add m × d², the parallel-axis theorem. A uniform rod of length L pivoted at one end has I = mL²/12 + m(L/2)² = mL²/3. For a 1 m rod that gives a period of 1.638 s, whatever its mass.
Why does mass appear when a pendulum's period shouldn't depend on it?
For a body of fixed shape, the moment of inertia is proportional to its mass, so the two cancel and the period depends only on shape and gravity. The fields here are independent, so if you change the mass you must scale the moment of inertia with it to describe the same object.
What happens if the moment of inertia is entered smaller than m × d²?
The result is physically impossible. With 2 kg and d = 0.4 m, no body can have less than 0.32 kg·m², which is all the mass at the centre of mass, giving 1.269 s. Entering 0.2 kg·m² returns 1.003 s, faster than any real pendulum of that size; the page does not catch this.
How far off is the small-angle period for bigger swings?
The true period grows with amplitude. Compared with the figure here it is about 0.19% longer at 10°, 0.77% longer at 20° and 4.0% longer at 45°. For timing work, keep swings under about 10°, or apply a large-amplitude correction yourself.
What is the centre of oscillation, and why does it matter?
It is the point a distance I ÷ (m·d) below the pivot, the length of the equivalent simple pendulum. Swing the body from that point instead and the period is the same, the property Kater's reversible pendulum used to measure g. It is also the centre of percussion, where a blow causes no jolt at the pivot.
Does the local value of gravity change the period noticeably?
Slightly, but it adds up. Gravity at sea level ranges from about 9.78 m/s² at the equator to 9.83 at the poles, which moves the default period from 2.0091 s to 2.0040 s. A clock regulated at the pole would lose roughly 3.7 minutes a day at the equator.

How this page was produced

Published by
Quanta Calculator
Primary sources
3 cited below
Method
T = 2π sqrt(I_p / (mgd))
Published
Last verified

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