Why does the period of the rod change when I move the pivot point?

The period T = 2π √( I / (mgd) ) depends on both d (distance from pivot to center of mass) and I (moment of inertia about the pivot). As you move the pivot, both d and I change according to the parallel axis theorem. The interplay between these two quantities creates a non-linear relationship, resulting in a period that is long when the pivot is near an end, decreases to a minimum, and then increases again as the pivot approaches the center.

What is the 'equivalent length' (L_eq) and what does it tell us?

The equivalent length L_eq = I/(md) is the length of a simple pendulum (a point mass on a massless string) that would have the same period as the physical pendulum. It is a useful conceptual tool because it reduces the complex motion of an extended object to a simpler, familiar system. For a physical pendulum, L_eq is always longer than the distance d from the pivot to the center of mass.

Does this model work for large swing angles?

No, the derived formula for the period T = 2π √( I / (mgd) ) relies on the small-angle approximation (sinδ ≈ δ, where δ is in radians). For larger amplitudes, the restoring torque is no longer proportional to the angular displacement, and the motion is not simple harmonic. The actual period increases with amplitude, a correction not included in this simulator's core model.

Where is the pivot location for the minimum period of oscillation?

For a uniform rod, the minimum period occurs when the pivot is placed at a distance of L/√12 ≈ 0.289L from the center of the rod (or about 0.211L from one end). This point minimizes the ratio I/(md) in the period formula. It is a specific result of applying calculus to the period equation using the parallel axis theorem.

Physical Pendulum (Rod)

Thin uniform rod: pivot along L, I and T(δ), equivalent length L_eq.

Live graphs

About this model

A physical pendulum, unlike a simple pendulum with a point mass, is an extended rigid body free to rotate about a fixed horizontal pivot. This simulator models a classic example: a thin, uniform rod of length L and mass m, pivoted at a point a distance d from one end. The system's motion is governed by the rotational analog of Newton's second law, τ = Iα, where τ is the restoring torque due to gravity, I is the moment of inertia about the pivot, and α is the angular acceleration. For small angular displacements δ from the vertical equilibrium, the torque is approximately τ ≈ −mgd sinδ, leading to the small-angle approximation sinδ ≈ δ. This yields the equation for simple harmonic motion: α = d²δ/dt² = −(mgd / I) δ. The angular frequency is therefore ω = √(mgd / I), and the period is T = 2π/ω = 2π √( I / (mgd) ). For a uniform rod, the moment of inertia about its center of mass is I_cm = (1/12)mL². Using the parallel axis theorem, I = I_cm + md², where d is the distance from the pivot to the center of mass. A key learning point is the concept of the 'equivalent length' or 'length of a simple pendulum with the same period,' L_eq = I / (md). Students can explore how the period T changes with the pivot position d, discovering that it reaches a minimum and that there are two different pivot distances for the same period, symmetrically placed about the center of mass. The simulator simplifies reality by assuming a frictionless pivot, no air resistance, a perfectly uniform rod, and strictly small-angle oscillations where the motion is truly harmonic. By interacting with it, students solidify their understanding of rotational dynamics, moment of inertia, the parallel axis theorem, and the conditions for simple harmonic motion in a rotational context.

Who it's for: Undergraduate physics students studying rotational dynamics and oscillatory motion, and advanced high school students in AP Physics C: Mechanics.

Key terms

Physical Pendulum
Moment of Inertia
Simple Harmonic Motion
Parallel Axis Theorem
Restoring Torque
Small-Angle Approximation
Equivalent Length
Period of Oscillation

How it works

A rigid uniform rod pivots about a point along its length. The parallel-axis theorem gives I = mL²/12 + mδ² about the pivot, where δ is the distance from pivot to the center of mass. For small oscillations about the hanging vertical, the motion matches a simple pendulum with equivalent length L_eq = I/(mδ), so T ≈ 2π√(L_eq/g). Moving the pivot closer to the CM (smaller δ) reduces the restoring torque and lengthens the period; placing it at the end (δ = L/2) recovers the classic rod end formula T = 2π√(2L/(3g)).

Key equations

I = mL²/12 + mδ² · ω₀² = mgδ / I · L_eq = I/(mδ)

End pivot: δ = L/2 → T = 2π√(2L/(3g))

Frequently asked questions

Why does the period of the rod change when I move the pivot point?: The period T = 2π √( I / (mgd) ) depends on both d (distance from pivot to center of mass) and I (moment of inertia about the pivot). As you move the pivot, both d and I change according to the parallel axis theorem. The interplay between these two quantities creates a non-linear relationship, resulting in a period that is long when the pivot is near an end, decreases to a minimum, and then increases again as the pivot approaches the center.
What is the 'equivalent length' (L_eq) and what does it tell us?: The equivalent length L_eq = I/(md) is the length of a simple pendulum (a point mass on a massless string) that would have the same period as the physical pendulum. It is a useful conceptual tool because it reduces the complex motion of an extended object to a simpler, familiar system. For a physical pendulum, L_eq is always longer than the distance d from the pivot to the center of mass.
Does this model work for large swing angles?: No, the derived formula for the period T = 2π √( I / (mgd) ) relies on the small-angle approximation (sinδ ≈ δ, where δ is in radians). For larger amplitudes, the restoring torque is no longer proportional to the angular displacement, and the motion is not simple harmonic. The actual period increases with amplitude, a correction not included in this simulator's core model.
Where is the pivot location for the minimum period of oscillation?: For a uniform rod, the minimum period occurs when the pivot is placed at a distance of L/√12 ≈ 0.289L from the center of the rod (or about 0.211L from one end). This point minimizes the ratio I/(md) in the period formula. It is a specific result of applying calculus to the period equation using the parallel axis theorem.

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