NBPhO 2026

Overview

The swing is a planar pendulum whose effective length jumps between two values, $a$ and $b > a$, at carefully chosen moments. Two configuration-change events drive the dynamics:

• Squat at a turning point ($\omega = 0$). The angular impulse $\int\tau\,dt = -\int mgr\sin\theta\,dt$ vanishes for an instantaneous radial motion, so angular momentum $L = mr^2\dot\theta$ is conserved. Since $L=0$ both before and after, the swing remains at $\omega = 0$. Energy is exchanged (gravity does work as $r$ changes at fixed $\theta$), but it goes to/from Kirill’s muscles, not to the swing’s kinetic energy.
• Stand-up at the vertical ($\theta = 0$). The gravitational torque vanishes identically (gravity is along the rod), so angular momentum is again conserved. But now $L \ne 0$, and conserving it while shrinking $r$ forces $\omega$ to increase by the factor $(b/a)^2$. This is where Kirill pumps energy into the swing — the work is done by his legs against the centrifugal force.

Between these events, $r$ is fixed and the swing is an ordinary pendulum: energy is conserved.

The geometric idea is that of a Foucault-style parametric pump: shortening the pendulum at the bottom (where $\omega^2$ is largest) and lengthening it at the turning point (where $\omega = 0$) extracts net work each cycle. We will see that the kinetic energy at the bottom multiplies by

$$\left(\frac{b}{a}\right)^{3}$$

per half-period, regardless of amplitude — a clean exponential growth.

The condition for going over the top depends on the constraint type. The rod is rigid, so it can both pull and push; the swing only has to reach $\theta = \pi$ with $\omega \geq 0$. Compared to a string-supported swing (which needs $\omega^2 \geq g/a$ at the top, i.e. $\omega_{\mathrm{bottom}}^2 \geq 5g/a$), the rigid-rod threshold is

$$\omega_{\mathrm{bottom}}^{2} \;\geq\; \frac{4g}{a},$$

a factor $4/5$ lower in $\omega^2$.

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Part (i) — Angular speed during the first full period

Let $\omega_0$ be the initial angular speed (at the bottom, standing). One full period contains two up-and-back excursions — one to the right, one to the left — connected by a stand-up at the bottom.

Going right and coming back

1. Up-swing, standing ($r=a$). Pendulum motion from $(\theta,\omega)=(0,\omega_0)$ to $(\theta_1, 0)$, where $\cos\theta_1 = 1 - a\omega_0^2/(2g)$. Speed decreases from $\omega_0$ to $0$.
2. Squat at right top. Instantaneous; angular speed stays at $0$.
3. Down-swing, squatting ($r=b$). Pendulum motion from $(\theta_1, 0)$ to $(0, -\omega_b^{(0)})$, with $$\omega_b^{(0)} \;=\; \sqrt{\tfrac{a}{b}}\,\omega_0 \;<\; \omega_0,$$ from energy conservation: $\tfrac12 b^2 (\omega_b^{(0)})^2 = gb(1-\cos\theta_1) = \tfrac12 a^2\omega_0^2 \cdot (b/a) \cdot (b/a)\cdot\dots$ — see Part (iii) for the algebra. Squatting is slower (period $\propto \sqrt{r}$), so this leg takes longer than the standing up-swing.
4. Stand-up at the bottom. Angular momentum conservation $b^2\omega_b^{(0)} = a^2\omega_1$ gives a discontinuous jump $$\omega_1 \;=\; \left(\frac{b}{a}\right)^{2}\omega_b^{(0)} \;=\; \left(\frac{b}{a}\right)^{3/2}\omega_0 \;>\; \omega_0.$$ The stand-up finishes the half-cycle with the swing moving to the left, so $\dot\theta = -\omega_1$ at this moment.

Going left and coming back

The next half-cycle mirrors the first but with a larger amplitude $\theta_2$ (since $\omega_1 > \omega_0$). It ends with $\dot\theta = +\omega_2 = (b/a)^{3}\omega_0$, the angular speed after the second stand-up.

Sketch

|ω|
 ω_2|·                                           ·-- after 2nd stand-up
    |\                                          /|
    | \                                        / |
 ω_1|  \              ·-- jump (1st stand-up) /  |
    |   \           ·/|                      /   |
 ω_0|·   \         / |                      /    |
    |\    \       /  |                     /     |
    | \    \     /   \                    /      |
 ω_b|  \    \---/-- ·  \-- 1st squat-end / ω_b'  |
    |   \         (squat at top: |ω|=0)\_____   /
    |    \___________________________________\_/
    +---------------+---------+---------+----------- t
        up-stand   down-squat up-stand  down-squat
       (turn pt)              (turn pt)

The features that must appear in the qualitative sketch:

• A first hump dropping from $\omega_0$ to $0$ at the right turning point.
• A flat segment at $|\omega|=0$ representing the squat (idealised as instantaneous; in practice a brief plateau).
• A second segment rising back to $\omega_b^{(0)} < \omega_0$ at the bottom — lower than the original peak, because the swing is squatting.
• A vertical jump upward in $|\omega|$ at the bottom (stand-up): from $\omega_b^{(0)}$ to $\omega_1 > \omega_0$.
• Repeat for the second half-period, with peaks at $\omega_1$ and $\omega_2 = (b/a)^3\omega_0$.

The peaks at the bottom (just after each stand-up) form a geometric sequence with ratio $(b/a)^{3/2}$ per half-cycle.

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Part (ii) — Phase diagram

In the $(\theta, \omega)$ plane, a constant-$r$ pendulum trajectory is a level curve of $\tfrac12 r^2\omega^2 - gr\cos\theta$. For amplitude $\theta_*$ this is

$$\omega^{2} \;=\; \frac{2g}{r}\bigl(\cos\theta - \cos\theta_{*}\bigr),$$

a “pendulum oval” pinching to $\omega=0$ at $\theta=\pm\theta_*$ and bulging to $\omega^2 = (2g/r)(1-\cos\theta_*)$ at $\theta=0$.

Two key observations:

• For the same amplitude $\theta_*$, the squatting oval ($r=b$) is smaller in $\omega$ than the standing oval ($r=a$) by a factor $\sqrt{a/b}$. So the trajectory transitions from a larger oval to a smaller one when squatting at the turning point — in phase space this is a change of curve at the point $(\theta_*, 0)$, not a jump.
• At the bottom, the stand-up jumps angular momentum’s geometric meaning: $r^2\omega$ is conserved, so $\omega$ jumps by $(b/a)^2$ vertically in the diagram.

Trajectory of the swing

Starting at $(0, +\omega_0)$ in the upper half-plane, the trajectory traces (counterclockwise):

1. The standing arc to $(+\theta_1, 0)$ — an outer-shape oval segment.
2. Continues onto the squatting arc from $(+\theta_1, 0)$ to $(0, -\omega_b^{(0)})$ — inside the standing oval, since the squatting oval is smaller for the same amplitude.
3. Vertical downward jump at $\theta = 0$ from $(0, -\omega_b^{(0)})$ to $(0, -\omega_1)$ — outside the previous oval, since $|\omega_1|>|\omega_b^{(0)}|>$ even possibly $|\omega_0|$.
4. The standing arc to $(-\theta_2, 0)$ with $\theta_2 > \theta_1$.
5. The squatting arc to $(0, +\omega_b^{(1)})$.
6. Vertical upward jump to $(0, +\omega_2)$.
7. Repeat, each loop reaching higher $|\theta|$ and higher $|\omega|$.

The trajectory is therefore an outward-spiralling pair of half-loops, each composed of two arcs (one standing, one squatting) joined at a turning point and broken at the bottom by a vertical jump.

The separatrix and going over the top

The standing-config separatrix (the boundary between oscillation and full rotation, $r=a$) is

$$\omega^{2} \;=\; \frac{2g}{a}\bigl(1 + \cos\theta\bigr) \;=\; \frac{4g}{a}\cos^{2}(\theta/2),$$

passing through $(\pm\pi, 0)$ and reaching $|\omega|_{\max} = 2\sqrt{g/a}$ at $\theta=0$. The swing first goes over the top when one of its standing arcs crosses this separatrix, i.e. when $\omega_n^{2} \geq 4g/a$.

Sketch

           ω
           ↑
       2√(g/a)·····  ← standing separatrix (crossed when going over)
        ω_2 ·-·   _.- this arc crosses past θ=π
            | \_/                         _
        ω_1 |        _,-- larger spiral _-
            |      ,'                   `\
        ω_0 |    ,'  __                  `,
            |   /  .'   `.                |
         ω_b|  ;  ;       ;               ;
       _____|__|_._________|_____θ=π____  → θ
   -π  '  ' \  \         /              /
            \  \       /              /
            '  '\   _/`              /
            -ω_b'   ¯¯              /
       -ω_1 '·_           jump at  /
            '   `··-..__   θ=0    /
       -ω_2 ·                    /
            ↓

Each half-loop is broken at $\theta=0$ by a vertical jump (the stand-up); the squatting and standing arcs join at the turning points $(\pm\theta_n, 0)$. The last standing arc reaches the separatrix at $(\pm\pi,0)$ and continues into rotational motion.

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Part (iii) — How many stand-ups before going over

Setting up the recursion

Let $\omega_n$ denote the angular speed at the bottom in the standing configuration just after the $n$-th stand-up (so $\omega_0$ is the initial push; $\omega_1$ is reached after one stand-up; etc.). We want the smallest $n$ such that the next up-swing reaches $\theta=\pi$.

One half-cycle consists of: up-swing standing → squat → down-swing squatting → stand-up. We track how $\omega_n$ transforms into $\omega_{n+1}$.

(a) Up-swing in the standing configuration

Constant $r=a$, energy conservation between $(0,\omega_n)$ and $(\theta_n, 0)$:

$$\tfrac{1}{2}m a^{2}\omega_{n}^{2} \;=\; mga\bigl(1-\cos\theta_{n}\bigr) \quad\Longrightarrow\quad 1-\cos\theta_{n} \;=\; \frac{a\omega_{n}^{2}}{2g}.$$

A real $\theta_n$ exists only if $\omega_n^2 \leq 4g/a$; otherwise the swing never turns around (and that is the going-over condition we are after).

(b) Squat at the turning point

$r$ jumps $a \to b$ at angle $\theta_n$, with $\omega = 0$ before and after (angular momentum conservation, $L = mr^2\omega = 0$). The angular position and velocity are unchanged.

Why the energy bookkeeping does not enter the recursion. The squat changes the gravitational PE by $-mg(b-a)\cos\theta_n$ but leaves $\omega$ alone. The corresponding work is done on (or by) Kirill’s body via his legs — his muscles dissipate the gravitational work for $\theta_n < \pi/2$, and add work for $\theta_n > \pi/2$. Either way, the swing’s kinetic state does not change, which is all we need.

(c) Down-swing in the squatting configuration

Constant $r=b$, energy conservation between $(\theta_n, 0)$ and $(0, \omega_b)$:

$$\tfrac{1}{2}m b^{2}\omega_{b}^{2} \;=\; mgb\bigl(1-\cos\theta_{n}\bigr) \;=\; mgb \cdot \frac{a\omega_{n}^{2}}{2g} \;=\; \tfrac{1}{2}m a b\,\omega_{n}^{2}.$$

Thus

$$\omega_{b}^{2} \;=\; \frac{a}{b}\,\omega_{n}^{2}.$$

Note $\omega_b < \omega_n$: at the bottom in the squatting configuration, the swing is slower than it was in the standing configuration. Energetically, $\tfrac12 m b^2\omega_b^2 = \tfrac12 m a b\,\omega_n^2$, which is larger than $\tfrac12 m a^2\omega_n^2$ by the factor $b/a$ — gravity has done extra work on Kirill during the squat-and-down phase, working over the larger height $b(1-\cos\theta_n)$ rather than $a(1-\cos\theta_n)$.

(d) Stand-up at the bottom

At $\theta = 0$ the gravitational torque vanishes (gravity is along the rod), so angular momentum is conserved during the radial motion:

$$m b^{2}\,\omega_{b} \;=\; m a^{2}\,\omega_{n+1} \quad\Longrightarrow\quad \omega_{n+1} \;=\; \left(\frac{b}{a}\right)^{2}\!\omega_{b}.$$

Squaring and substituting $\omega_b^2 = (a/b)\omega_n^2$ gives the central recursion:

$$\boxed{\;\omega_{n+1}^{2} \;=\; \left(\frac{b}{a}\right)^{4}\omega_{b}^{2} \;=\; \left(\frac{b}{a}\right)^{4} \cdot \frac{a}{b}\,\omega_{n}^{2} \;=\; \left(\frac{b}{a}\right)^{3}\omega_{n}^{2}.\;}$$

Iterating from $\omega_0$:

$$\omega_{n}^{2} \;=\; \left(\frac{b}{a}\right)^{3n}\omega_{0}^{2}.$$

Where the energy comes from

It is instructive to track total energy. The kinetic energy at the bottom in standing configuration is $K_n = \tfrac12 m a^2\omega_n^2$, growing by the same factor $(b/a)^3$ per half-cycle. The energy gain per half-cycle splits cleanly:

• Stand-up at the bottom does work $W_{\text{up}} = \tfrac12 m b^2\omega_b^2\bigl((b/a)^2 - 1\bigr) + mg(b-a) > 0$ on the swing — Kirill’s legs push the seat (and himself) inward against centrifugal force, then continue against gravity.
• Squat at the turning point does work $W_{\text{down}} = -mg(b-a)\cos\theta_n$ on the swing — for $\theta_n < \pi/2$ this is negative (Kirill’s legs absorb energy as gravity pulls him outward), for $\theta_n > \pi/2$ it is positive (Kirill must push outward against gravity).

The net of $W_{\text{up}} + W_{\text{down}}$ over one half-cycle is exactly the energy gain $K_{n+1} - K_n + mg(a-a) = K_{n+1} - K_n$. So the squat removes some of what the stand-up adds, and the recursion factor $(b/a)^3$ is the residual after both effects.

Going-over criterion

With a rigid rod the only condition for crossing $\theta = \pi$ on the standing leg is that $\theta_n$ would have to equal or exceed $\pi$ — equivalently, $1-\cos\theta_n \geq 2$:

$$\frac{a\omega_{n}^{2}}{2g} \;\geq\; 2 \quad\Longleftrightarrow\quad \omega_{n}^{2} \;\geq\; \frac{4g}{a}.$$

(Strict inequality for the swing actually to cross in finite time; equality gives an asymptotic approach.)

The going-over event happens during a standing up-swing, never during a squatting down-swing: a squatting leg starts with $\omega = 0$ at the turning point and is therefore decelerating-then-accelerating toward $\theta = 0$, away from $\theta=\pi$.

So the smallest number $n$ of stand-ups before Kirill goes over the top satisfies

$$\left(\frac{b}{a}\right)^{3n}\omega_{0}^{2} \;\geq\; \frac{4g}{a},$$

giving

$$\boxed{\;n \;=\; \left\lceil\,\frac{1}{3}\,\log_{b/a}\!\!\left(\frac{4g}{a\,\omega_{0}^{2}}\right)\right\rceil\;} \;=\; \left\lceil\,\frac{\ln\!\left(4g/(a\omega_{0}^{2})\right)}{3\,\ln(b/a)}\,\right\rceil.$$

Numerical evaluation

For $a = 2.5\,\text{m}$, $b = 3.0\,\text{m}$, $\omega_0 = 1.0\,\text{rad/s}$, $g = 10\,\text{m/s}^{2}$:

$$\frac{4g}{a\,\omega_{0}^{2}} \;=\; \frac{40}{2.5} \;=\; 16, \qquad \frac{b}{a} \;=\; 1.2.$$ $$\frac{\ln 16}{3\ln 1.2} \;=\; \frac{2.7726}{3\times 0.18232} \;=\; \frac{2.7726}{0.5470} \;\approx\; 5.069.$$

The smallest integer not less than $5.069$ is $6$.

$$\boxed{\;n \;=\; 6.\;}$$

Verification. With $n=5$: $\omega_5^2 = 1.2^{15} \approx 15.41 < 16$, so the swing reaches amplitude $\cos\theta_5 = 1 - 2.5\cdot 15.41/20 \approx -0.926$, i.e. $\theta_5 \approx 157.8^\circ$ — close to but short of upright. With $n=6$: $\omega_6^2 = 1.2^{18} \approx 26.62 > 16$, so the swing has more than enough energy to clear the top, with $\omega_{\mathrm{top}} = \sqrt{26.62-16} \approx 3.26\,\text{rad/s}$ as it passes $\theta=\pi$.

Consistency checks

• Limits. As $b \to a$, the ratio $(b/a)^3 \to 1$ and the recursion stops pumping. Indeed, formally $n \to \infty$: with no leg motion, the swing oscillates forever at fixed energy. ✓
• Dimensions. $4g/a$ has units of $1/\text{time}^2$, matching $\omega^2$. ✓
• Rigid-rod vs string. For a string-supported swing, the criterion would be $\omega_n^2 \geq 5g/a$ (so that $g$ provides centripetal at the top). With $5g/a = 20$, we would need $1.2^{3n} \geq 20$, giving $n \geq \ln 20/(3\ln 1.2) \approx 5.48$, i.e. $n = 6$ also — although for our specific parameters the answer happens to coincide, the thresholds differ by a factor $5/4$ in $\omega_n^2$.
• Energy ratio. The KE at the bottom after 6 stand-ups is $(1.2)^{18} \approx 26.6$ times the initial — a 27-fold increase, all delivered by Kirill’s legs over six pumps. Average gain per pump: $(b/a)^3 - 1 = 0.728$, i.e. each pump increases KE by $\sim 73\%$ — that is strong parametric amplification, consistent with the practical observation that a few well-timed pumps suffice.

Educational remark — what makes pumping work

The key asymmetry is where the configuration changes happen: stand-ups at the bottom (where $\omega^2 r$ is largest), squats at the turning points (where it is zero). Reversing the protocol — squat at the bottom, stand-up at the top — would damp the swing, because angular momentum conservation at the bottom would now make $\omega$ shrink (going from $r=a$ to $r=b$), and the squat at the top would do nothing useful. The protocol Kirill uses is the unique phasing that converts metabolic energy into amplitude growth.

Equivalently, one can think of the radial motion as a forcing whose period is half the swing period (two pumps per cycle). The forcing is in phase with the natural pendulum motion, hitting the parametric resonance — exactly the mechanism a child uses to pump a swing standing up.

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Summary of results

Part	Result
(i)	Two humps per period; each drops to zero at a turning point and is followed by a flat (squat) segment, with a discontinuous upward jump in $\|\omega\|$ at each crossing of $\theta=0$. The peaks (just after each stand-up) form a geometric sequence with ratio $(b/a)^{3/2}$.
(ii)	A counterclockwise outward spiral in the $(\theta,\omega)$ plane: each loop is a standing arc joined to a smaller squatting arc at the turning point, broken by a vertical jump at $\theta=0$. The trajectory crosses the standing-config separatrix $\omega^{2}=(4g/a)\cos^{2}(\theta/2)$ on the final up-swing.
(iii)	$\displaystyle n \;=\; \left\lceil \frac{\ln\bigl(4g/(a\omega_{0}^{2})\bigr)}{3\ln(b/a)} \right\rceil$. For $a=2.5\,\text{m}$, $b=3.0\,\text{m}$, $\omega_0=1.0\,\text{rad/s}$, $g=10\,\text{m/s}^2$: $\boxed{\,n=6\,}$.