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
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.
- 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).
- 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.
- 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.
- 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.
- 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√(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.
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.
Frequently asked questions.
How do I find the moment of inertia about the pivot?
Why does mass appear when a pendulum's period shouldn't depend on it?
What happens if the moment of inertia is entered smaller than m × d²?
How far off is the small-angle period for bigger swings?
What is the centre of oscillation, and why does it matter?
Does the local value of gravity change the period noticeably?
References& sources.
- [1]OpenStax, University Physics Volume 1, section 15.4, Pendulums.
- [2]Young and Freedman, University Physics with Modern Physics, 15th ed. (PRINT).
- [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 3 cited below
- Method
- T = 2π sqrt(I_p / (mgd))
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
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