Physical Pendulum Calculator
Find the period and frequency of a physical (compound) pendulum — a rod, disc, ring or sphere swinging about a pivot — plus its equivalent simple-pendulum length and the pivot point for the shortest period.
How it works
T = 2π √( I ÷ (m g d) ) · I = I_cm + m d² (parallel-axis theorem) · I_cm = m k²: rod k² = L²/12, disc k² = R²/2, ring k² = R², sphere k² = 2R²/5 · so T = 2π √( (k² + d²) ÷ (g d) ) · equivalent length L_eq = (k² + d²) ÷ d · shortest period at d = k: T_min = 2π √(2k ÷ g)
A physical (compound) pendulum is any rigid body swinging about a fixed pivot, not a point mass on a string. Its period depends on how the mass is spread out, which the moment of inertia about the pivot captures; the parallel-axis theorem moves the textbook moment of inertia from the centre of mass to the pivot. The mass cancels out, so only the shape, the pivot distance d and g matter. The small-angle formula is accurate to better than 0.5% for swings under about 15°.
Worked example
A uniform rod 1 m long hangs from one end, so d = 0.5 m and k² = L²/12 = 0.0833 m². With g = 9.80665 m/s²: T = 2π √((0.0833 + 0.25) ÷ (9.80665 × 0.5)) = 1.638 s, a frequency of 0.6104 Hz. It swings like a simple pendulum 0.6667 m long (two-thirds of the rod). Moving the pivot to d = k = 0.2887 m from the centre would give the shortest possible period, 1.525 s.
Sources: HyperPhysics — Physical pendulum · OpenStax University Physics — 15.4 Pendulums
Frequently asked questions
What is the difference between a simple and a physical pendulum?
A simple pendulum is an idealized point mass on a massless string, with T = 2π√(L/g). A physical pendulum is a real rigid body; its period uses the moment of inertia about the pivot, T = 2π√(I/(m g d)), and is always the same as a simple pendulum of length (k² + d²)/d.
What is the period of a rod pivoted at one end?
T = 2π√(2L/(3g)). A 1 m rod swings with a period of about 1.64 s — the same as a 0.667 m simple pendulum.
Why does the mass not matter?
Both the restoring torque (m g d sin θ) and the moment of inertia (m(k² + d²)) are proportional to the mass, so it cancels. Only the distribution of mass, through the radius of gyration k, and the pivot position change the period.
Where should the pivot be for the fastest swing?
The period is shortest when the pivot is at a distance d = k (the radius of gyration) from the centre of mass. For a 1 m rod that is 0.289 m from the middle, giving about 1.52 s. Pivoting at the centre of mass (d = 0) gives no swing at all.
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This calculator is for educational purposes. Double-check important results before acting on them.