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Derive the period of an extended pendulum from its torque

An extended pendulum rotates about a pivot, so its inertia is not generally Md² even when its center lies distance d below that pivot. For a uniform rod of length 2 m and mass 3 kg, the center inertia is 1 kg m² and the parallel-axis result is Ipivot = 1 + 3d².

Learning outcome: Derive the small-angle equation, compare the point-mass approximation, and differentiate the period-squared expression to locate its minimum.

Starting recap

Derive equations with their initial or boundary conditions, integrate distributed profiles, and check model domains independently of numerical outputs.

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Try

At d = 1 m, inertia is 4 kg m² and angular frequency is sqrt(7.5) rad/s, giving period about 2.29429 s. The longer pivot distance does not monotonically reduce the period because it also increases inertia.

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Predict

Which inertia belongs in the physical-pendulum period?

Your prediction and reason

Experiment

Derive the small-angle equation, compare the point-mass approximation, and differentiate the period-squared expression to locate its minimum.

Derive a physical pendulum from its distributed inertia

Worked teaching model · synthetic values · Uniform rigid rod L=2 m,M=3 kg with fixed pivot distance d from its center. Stable orientation has center below pivot. Small-angle linearization, g=10, initially at rest; no physical swinging or torque rig.

Pivot distance from mass center: 0.5 m Initial small angular amplitude: 0.1 rad

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Small-angle physical-pendulum period

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: 2pi sqrt((1+3d²)/(30d)) (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Small-angle physical-pendulum period

2pi sqrt((1+3d²)/(30d))

pivot-to-center distance (m)	period (s)
0.1	3.681611
0.11125	3.502561
0.1225	3.350528
0.13375	3.219764
0.145	3.106113
0.15625	3.006478
0.1675	2.918504
0.17875	2.840356
0.19	2.770583
0.20125	2.70802
0.2125	2.651718
0.22375	2.600896
0.235	2.554901
0.24625	2.513186
0.2575	2.475287
0.26875	2.440805
0.28	2.4094
0.29125	2.380774
0.3025	2.354668
0.31375	2.330856
0.325	2.309138
0.33625	2.289336
0.3475	2.271293
0.35875	2.254869
0.37	2.239937
0.38125	2.226383
0.3925	2.214105
0.40375	2.20301
0.415	2.193014
0.42625	2.184041
0.4375	2.176019
0.44875	2.168886
0.46	2.162582
0.47125	2.157054
0.4825	2.152251
0.49375	2.148129
0.505	2.144644
0.51625	2.141757
0.5275	2.139433
0.53875	2.137637
0.55	2.136339
0.56125	2.135509
0.5725	2.13512
0.58375	2.135147
0.595	2.135566
0.60625	2.136355
0.6175	2.137494
0.62875	2.138963
0.64	2.140744
0.65125	2.142821
0.6625	2.145176
0.67375	2.147796
0.685	2.150666
0.69625	2.153772
0.7075	2.157103
0.71875	2.160647
0.73	2.164392
0.74125	2.168328
0.7525	2.172445
0.76375	2.176734
0.775	2.181186
0.78625	2.185793
0.7975	2.190547
0.80875	2.195439
0.82	2.200464
0.83125	2.205615
0.8425	2.210885
0.85375	2.216268
0.865	2.221758
0.87625	2.227351
0.8875	2.23304
0.89875	2.238821
0.91	2.24469
0.92125	2.250641
0.9325	2.256671
0.94375	2.262776
0.955	2.268952
0.96625	2.275195
0.9775	2.281502
0.98875	2.28787
1	2.294295

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Inertia about pivot	1.75	kg m²
Linearized angular frequency	2.9277	rad/s
Linearized period	2.146116	s
Linearized initial oscillation energy	0.075	J
Point-mass-at-center approximate period	1.404963	s
Period-minimizing pivot distance	0.5773503	m

The full pivot inertia, not only center distance, sets a physical pendulum period.

Torque−Mgd sinθ linearizes to−Mgd θ, giving I θddot+Mgd θ=0. Parallel-axis inertia is ML²/12+Md². Differentiating T² proportional to L²/(12d)+d gives its minimum at d=L/sqrt12. This equilibrium is stable with center below pivot; the formula assumes small angles and excludes pivot at the center.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The full pivot inertia, not only center distance, sets a physical pendulum period.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Derive the period of an extended pendulum from its torque

Gravity provides torque −Mgd sinθ. At small angle this gives Ipivot θddot + Mgd θ = 0, so angular frequency is sqrt(Mgd/Ipivot). The stable orientation has the mass center below the pivot; an inverted orientation would reverse the restoring sign.

At d = 0.5 m and g = 10, inertia is 1.75 kg m², frequency is sqrt(15/1.75) ≈ 2.92770 rad/s and period is about 2.14612 s. Treating all mass as a point at the center would give 2πsqrt(0.5/10) ≈ 1.40496 s and omit the rod’s distributed inertia.

For a general uniform rod, T² is proportional to L²/(12d) + d. Differentiating this expression places the shortest small-angle period at d = L/sqrt(12), not at the center or necessarily at the end. The d = 0 limit is excluded from this pendulum-frequency formula: gravity then supplies no orientation-restoring torque. Amplitude-dependent corrections beyond the small-angle model are not silently included.

Changed-case check

Predict before reading the comparison. At d = 1 m, inertia is 4 kg m² and angular frequency is sqrt(7.5) rad/s, giving period about 2.29429 s. The longer pivot distance does not monotonically reduce the period because it also increases inertia.

Retained investigation

Derive the small-angle equation, compare the point-mass approximation, and differentiate the period-squared expression to locate its minimum. Why can increasing the gravitational lever arm eventually increase the period?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All prose, functions and evidence packets are original. Bounded mathematical models do not establish measured apparatus, supervised practical skill, independent learner mastery or institutional credit. No physical loading, liquids, heat or sound exposure is requested.

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Why can increasing the gravitational lever arm eventually increase the period?

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Previous: Recover oscillator phase from both initial conditions Next: Keep the same initial conditions across three damping regimes

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