NBPhO 2026

Difficulty medium

Prerequisites

• Pendulum motion and the small-angle approximation
• Angular momentum $L = m r^{2}\dot\theta$ and conditions for its conservation
• Energy conservation in rotational motion
• Phase-space portraits of 1D conservative systems
• Distinction between rigid-rod and string-suspended pendulums

Learning objectives

• Recognise that an instantaneous radial motion at $\dot\theta = 0$ preserves $\dot\theta$, while one at $\theta = 0$ preserves $L = m r^{2}\dot\theta$ and so multiplies $\omega$ by $(b/a)^{2}$
• Derive the per-half-cycle gain factor $\omega_{n+1}^{2} = (b/a)^{3}\omega_{n}^{2}$ and identify it with parametric pumping
• Apply the rigid-rod over-the-top criterion $\omega^{2} \geq 4g/a$ rather than the string criterion $\omega^{2} \geq 5g/a$
• Sketch a phase trajectory whose successive loops are joined by vertical jumps at $\theta = 0$ and identify when it crosses the separatrix
• Account for energy added by the rider's legs at the bottom and absorbed by them at low-amplitude turning points

Watch out for

• Confusing the roles of squatting and standing. The squat at the turning point does not pump the swing; only the stand-up at the bottom does (because that is where angular momentum conservation forces $\omega$ to grow). Reversing the protocol — squatting at the bottom and standing at the top — damps the swing.
• Using the string-pendulum criterion $\omega^{2} \geq 5g/a$ instead of the rigid-rod criterion $\omega^{2} \geq 4g/a$. With a rigid rod the constraint can both pull and push, so any non-negative $\omega$ at the top is enough.
• Off-by-one in counting. The answer is the smallest $n$ for which $\omega_{n}^{2} \geq 4g/a$, where $\omega_{n}$ is the bottom-of-swing angular speed in standing config after the $n$-th stand-up. Counting half-cycles, full cycles, or squats instead of stand-ups gives a different number.
• Overlooking that during the squat at low-amplitude turning points ($\theta_n < \pi/2$), Kirill's muscles absorb energy rather than add it. The clean recursion $(b/a)^{3}$ already nets these losses against the gains from the stand-ups; you do not need to add a separate dissipation term.
• Treating the squat as conserving angular momentum even when $\dot\theta \ne 0$. It does so here only because the squat happens at a turning point ($\dot\theta = 0$); in general gravity exerts a finite torque and a finite-time radial motion would change $L$.