Compute the oscillation period, reduced length, angular frequency, and moment of inertia for a rigid body swinging about a pivot.
A physical pendulum (or compound pendulum) is any rigid body that oscillates about a fixed horizontal axis not passing through its center of mass. Unlike a simple pendulum (point mass on a massless string), a physical pendulum’s period depends on its moment of inertia and the distance from the pivot to the center of mass. The restoring torque is provided by gravity: τ = – m g d sinθ, leading to the differential equation I α = – m g d θ for small angles. Hence, the angular frequency ω = √(m g d / I), and the period T = 2π √( I / (m g d) ).
$$ T = 2π \sqrt{\frac{I}{m g d}} \quad , \quad L_{\text{eq}} = \frac{I}{m d} = \frac{k^2}{d} $$
Where k = radius of gyration, d = distance from pivot to CM.
The reduced length Leq = I/(m d) corresponds to the length of an equivalent simple pendulum having the same period. This concept is extensively used in gravimeters and seismometers. The physical pendulum also illustrates the parallel axis theorem: I = Icm + m d², which makes experimental determination of g possible.
Kater's pendulum is a physical pendulum with two adjustable pivots, allowing precise determination of local gravitational acceleration without knowing the mass distribution. Our calculator can model each pivot configuration: by locating the periods equalization point, the reduced length Leq equals the distance between pivots. Geodesists and metrology labs use such pendulums to measure g with high accuracy. This tool replicates the theoretical period for any arbitrary rigid body, aiding experimental design.
Other applications include: clock escapements (period regulation), bridge vibration analysis (modelling piers as compound pendulums), robotic swing-up control (inverted pendulum variants), and sports equipment design (tennis racket sweet spot relates to percussion center – a concept linked to physical pendulum oscillations).
Starting from torque τ = I α = – m g d sinθ. For small θ (θ < 10°), sinθ ≈ θ, giving I d²θ/dt² = – m g d θ → d²θ/dt² + (m g d / I) θ = 0, which is SHM with ω² = m g d / I. Therefore, T = 2π/ω = 2π √(I/(m g d)). No approximation is required for the restored results because the calculator assumes harmonic regime; if large amplitudes, error is minimal for most teaching contexts. The formula is exact for the linearized model.
If the entire mass is concentrated at a point distance L from the pivot, then I = m L², and T = 2π √(L/g) — the simple pendulum formula. For a uniform rod pivoted at one end, I = (1/3) m L², d = L/2, so T = 2π √( (2L)/(3g) ) ≈ 0.816 * 2π√(L/g). The calculator automatically handles these variations. For a physical pendulum, the period can be minimized for a given mass by choosing d such that I is optimized (this occurs when d = k, radius of gyration).
| Object / Shape | Pivot Location | Moment of Inertia I (about pivot) | d (CM to pivot) | Period T (example: m=2kg, g=9.81) |
|---|---|---|---|---|
| Uniform Rod (L=1.0m) | At one end | ⅓ m L² = 0.667 kg·m² | 0.5 m | 2π √(0.667/(2*9.81*0.5)) = 1.64 s |
| Solid Sphere (R=0.2m) | On surface | ²⁄₅ mR² + mR² = 7/5 mR² = 0.112 kg·m² | 0.2 m | 2π √(0.112/(2*9.81*0.2)) = 1.06 s |
| Thin Circular Hoop (R=0.3m) | Rim point | 2 m R² = 0.27 kg·m² | 0.3 m | 2π √(0.27/(1.5*9.81*0.3)) ≈ 1.56 s |