Simple and physical pendulums

For a point bob you can resolve the weight into components along and across the string. For a swinging door, a metre rule or a leg you cannot, because the parts of the body sit at different radii and travel at different speeds. Take moments about the pivot instead. In a uniform gravitational field the whole weight Mg may be replaced by a single force acting at the centre of mass — that is precisely what the centre of mass is for. With the body displaced by θ from the vertical, the centre of mass lies a distance d from the pivot along a line at θ to the vertical, so the perpendicular distance from the pivot to the weight's line of action is d sin θ, and the torque is τ = −Mgd sin θ, the minus sign recording that it drives θ back towards zero. The force the pivot exerts on the body is unknown, and choosing the pivot as the axis makes that irrelevant: its moment arm about the pivot is zero, so it never appears in the equation.

The rotational second law about a fixed axis is Στ = I d²θ/dt², with I the moment of inertia about that axis. So I d²θ/dt² = −Mgd sin θ, that is d²θ/dt² = −(Mgd/I) sin θ. Set this beside d²x/dt² = −ω²x. The difference is sin θ against θ, and it is not cosmetic: the pendulum equation is nonlinear, so superposition fails and the period depends on amplitude. The Taylor series sin θ = θ − θ³/6 + θ⁵/120 − … says how much. At θ = 0.10 rad, sin θ/θ = 0.9983; at 0.349 rad (20°) it is 0.9798; at 1.047 rad (60°) it is 0.8270. Replacing sin θ by θ is therefore a controlled approximation with a quantified error, not a definition. Once it is made, ω² = Mgd/I, θ(t) = θ₀ cos(ωt + φ), and every result already proved for displacement, velocity, phase and energy in SHM carries over unchanged, with θ in place of x and I in place of m.

Write T = 2π√(I/(Mgd)). The quantity under the root is a length divided by g, and that length is L_eq = I/(Md). The parallel-axis theorem, I = I_cm + Md², unpacks it: L_eq = d + I_cm/(Md) = d + k²/d, with k the radius of gyration about the centre of mass. For a uniform rod of length ℓ, k² = ℓ²/12. Pivot it at one end, so d = ℓ/2, and L_eq = ℓ/2 + ℓ/6 = 2ℓ/3: a 1.000 m rod gives T = 2π√(0.6667/9.81) = 1.638 s, well short of the 2.006 s of a 1.000 m simple pendulum. The form d + k²/d has a consequence worth keeping. It is smallest where d = k, that is at d = ℓ/√12 = 0.2887ℓ, where L_eq = 2k = 0.5774ℓ and the metre rod's period bottoms out at 1.524 s. Slide the pivot closer to the centre of mass than that and the period grows again, diverging as d → 0 — at the centre of mass there is no restoring torque at all and nothing oscillates.

The equivalent length names a physical point, the centre of oscillation, lying at L_eq from the pivot on the line through the centre of mass. Now re-hang the body from that point. Its new pivot-to-centre-of-mass distance is d′ = L_eq − d = k²/d, so its new equivalent length is L_eq′ = d′ + k²/d′ = k²/d + d = L_eq. Unchanged. The two points are conjugate: each is the other's centre of oscillation and both give exactly the same period. A 1.000 m rod pivoted at one end has d = 0.500 m and L_eq = 0.667 m, so re-hanging it 0.667 m from that end — 0.167 m past the centre — still gives 1.638 s. Kater's reversible pendulum turns this into an instrument: adjust the masses until both knife edges time alike, measure the separation, and g = 4π²L_eq/T² follows without ever knowing I or locating the centre of mass. Struck at the same point, the body delivers no impulsive jolt to the pivot, which is why a bat has a sweet spot.

Energy conservation gives the exact period without solving the nonlinear equation. Taking the potential zero at the lowest point, ½I(dθ/dt)² = Mgd(cos θ − cos θ₀), so dθ/dt = √(2Mgd/I) √(cos θ − cos θ₀) and a quarter period is the definite integral ∫₀^θ₀ dθ / [√(2Mgd/I) √(cos θ − cos θ₀)]. The substitution sin(θ/2) = sin(θ₀/2) sin u turns it into a complete elliptic integral of the first kind of modulus sin(θ₀/2), whose expansion gives T = T₀[1 + (1/4)sin²(θ₀/2) + (9/64)sin⁴(θ₀/2) + …] = T₀(1 + θ₀²/16 + 11θ₀⁴/3072 + …). In numbers the excess is +0.048% at 5°, +0.19% at 10° and +1.74% at 30°; at 60° the elliptic integral gives +7.32% while the two-term series returns +7.28%, so the series is 0.03 percentage points light. Amplitude independence is thus an approximation of the same family as sin θ ≈ θ, not a separate law: a regulator whose swing decays from 6° to 4° shortens its period by 0.038% and gains about 33 s a day.

The word simple hides three claims: a point bob, a string of no mass, and a small amplitude. Ranking them is itself a physical-pendulum calculation. A sphere of radius r on a light wire, its centre a distance L below the support, has I = m(L² + 2r²/5) and d = L, so L_eq = L(1 + 2r²/(5L²)) and T = 2π√(L/g) √(1 + 2r²/(5L²)). With L = 1.000 m and r = 20 mm that lengthens the period by 0.0080%. A wire of 1% of the bob's mass gives L_eq/L = (1 + μ/3)/(1 + μ/2) with μ = 0.01, shortening the period by 0.083%. Swinging through 10° adds 0.19%. So amplitude beats string mass by a factor of about two, and string mass beats bob size by a factor of about ten. Finish with the standing checks: I/(Mgd) has units kg m² / (kg × m s⁻² × m) = s², so its root is a time; setting I = mL² and d = L recovers T = 2π√(L/g); and ω² = Mgd/I mirrors ω² = k/m, with Mgd the small-angle torsional stiffness and I the inertia.