Integrate a physical pendulum and test the best pivot distance

Retain the actual mass distribution

The rod has M=3 kg,L=2 m and lambda(x)=lambda0(1+beta x/L), normalized by lambda0=M/[L(1+beta/2)]. Its center is xCM=L(1/2+beta/3)/(1+beta/2), endpoint inertia Iend=ML²(1/3+beta/4)/(1+beta/2), and central inertia Icm=Iend−M xCM². A nonuniform rod cannot automatically use the uniform central-inertia formula.

Move to the selected parallel pivot

Let d be the perpendicular distance from pivot to center of mass. The parallel-axis theorem gives Ip=Icm+Md². Gravity acts at the center, so torque about the pivot is −Mgd sin(theta). For small radians this becomes −Mgd theta=Ip theta double-dot. Thus the small-angle period is T=2pi sqrt(Ip/(Mgd)). The restoring lever arm is d, not necessarily the rod length or the pivot-to-end distance.

Differentiate the competing effects

Substitute Ip to write T²=(4pi²/g)[d+(Icm/M)/d]. Increasing d both increases the gravity lever arm and increases pivot inertia. Differentiation with respect to positive d gives a stationary point at d=sqrt(Icm/M). The derivative changes from negative to positive, so this is the continuous minimum. This optimization is a calculus extension of the core physical-pendulum relation.

Check the limiting assumption

As d approaches zero while Icm stays positive, the gravitational restoring torque vanishes and the small-angle period grows without bound. Exactly at the center, the gravity pendulum model has no restoring torque; substituting d=0 into the period expression is not allowed. The plot deliberately stays at positive offsets and does not claim a realizable mounting geometry or safe apparatus.

Worked original case

For the uniform rod beta=0, Icm=1 kg m². At d=.5 m, Ip=1.75 and T=2pi sqrt(1.75/15)≈2.146116 s. The continuous optimum is d=1/sqrt(3)≈.577350 m, not necessarily one of the selectable offsets.

Investigation

Integrate the beta=0 and beta=1 mass distributions independently. For each, compare at least three pivot offsets, derive the continuous optimum and test the sign of the period-squared derivative on either side. Explain why moving the pivot closer to the center does not always shorten the period.

Public numerical transfer practice

Try a new practice case

A physical pendulum has Ip=2 kg m²,M=2 kg,d=.5 m and g=10 m/s². What is its squared angular frequency in s⁻²?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax harmonic motion and pendulums. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.