NBPhO 2026

The official scheme labels its first table “Grading (part i and ii)” with four explicit criteria summing to 1.5 pts; the remaining 2.0 pts of part (ii) is graded against the four “key features for grading” listed without per-line weights. Below, that 2.0 pt budget is distributed evenly (0.5 pts per feature) as a derived split, and the combined section is scored against 3.5 pts total — matching the problem’s (i) + (ii) weight.

Part (i) + (ii) — 3.5 / 3.5 pts

Criterion	Points	Result
Approximately sinusoidal arcs between jumps	0.3	✓ Pendulum arcs at fixed $r$ identified as energy-conserving level curves; the standing and squatting legs are described as smooth excursions between turning points and stand-ups
Instantaneous jumps at the maximum values of $\omega$ (at the bottom of the swing, during the stand-up)	0.6	✓ “A vertical jump upward in $\lvert\omega\rvert$ at the bottom (stand-up): from $\omega_b^{(0)}$ to $\omega_1 > \omega_0$”, with the same jump repeated at every crossing of $\theta=0$
Smaller amplitude during squatting arches, or equivalently a longer quarter-period while $r=b$	0.3	✓ “Squatting is slower (period $\propto \sqrt{r}$), so this leg takes longer than the standing up-swing”; the squatting peak $\omega_b^{(0)} = \sqrt{a/b}\,\omega_0 < \omega_0$ is given explicitly
Overall growth of peak $\omega$ over the period	0.3	✓ “The peaks at the bottom (just after each stand-up) form a geometric sequence with ratio $(b/a)^{3/2}$ per half-cycle”; second-period peak quoted as $\omega_2 = (b/a)^3\omega_0$
Phase-plane arcs lens-shaped/pendulum-like (not ellipses)	0.5	✓ Each arc identified as a level curve $\omega^2 = (2g/r)(\cos\theta - \cos\theta_*)$ — the pendulum oval pinching to $\omega=0$ at $\theta = \pm\theta_*$ and bulging at $\theta=0$
Vertical line segments only at $\alpha=0$ (stand-ups), smooth meeting at turning points (squats)	0.5	✓ The stand-up at $\theta=0$ is ”$\omega$ jumps by $(b/a)^2$ vertically in the diagram”; the squat at the turning point is called out as a “change of curve at the point $(\theta_*, 0)$, not a jump”
Spiral grows outward	0.5	✓ “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”
Final open curve goes over the top with non-zero $\omega$	0.5	✓ Last standing arc described as crossing the separatrix $\omega^2 = (4g/a)\cos^2(\theta/2)$ on the going-over half-cycle, with $\omega_{\mathrm{top}} = \sqrt{\omega_n^2 - 4g/a} \approx 3.26\,\text{rad/s}$ at $\theta=\pi$ for $n=6$

Part (iii) — 4.5 / 4.5 pts

Criterion	Points	Result
Angular momentum conservation during the stand-up	0.7	✓ "$mb^2\omega_b = ma^2\omega_{n+1}$", with explicit justification that the gravitational torque vanishes at $\theta=0$ since gravity is along the rod
Energy conservation during the swing	0.7	✓ Two energy-conservation steps: $\tfrac12 ma^2\omega_n^2 = mga(1-\cos\theta_n)$ on the standing up-swing and $\tfrac12 mb^2\omega_b^2 = mgb(1-\cos\theta_n)$ on the squatting down-swing
Growth factor $(b/a)^{3/2}$ per stand-up	0.7	✓ Recursion $\omega_{n+1}^2 = (b/a)^3\omega_n^2$ derived and boxed; equivalent to $\omega_{n+1}/\omega_n = (b/a)^{3/2}$, also quoted as $\omega_1 = (b/a)^{3/2}\omega_0$ in the part (i) framing
Minimum angular velocity $\omega_{\min}^2 = 4g/a$ for the rigid-rod swing to loop over the top	0.8	✓ "$\omega_n^2 \geq 4g/a$" derived from the rigid-rod over-the-top criterion $1 - \cos\theta_n \geq 2$, with the rod-vs-string distinction (push and pull) called out up front and the factor-$4/5$-versus-$5g/a$ comparison made explicit
Correct inequality for the over-the-top condition ($\omega_N \geq \omega_{\min}$ after the $N$th stand-up)	0.4	✓ "$(b/a)^{3n}\omega_0^2 \geq 4g/a$" stated as the going-over inequality, with the explicit observation that the crossing must happen during a standing up-swing
Solving the inequality for $N$ using the growth law $\omega_N = \omega_0 (b/a)^{3N/2}$, yielding $N \geq \log\!\bigl[4g/(a\omega_0^2)\bigr]/[3\log(b/a)]$	0.7	✓ $n = \left\lceil \ln(4g/(a\omega_0^2))/[3\ln(b/a)] \right\rceil$ in closed form, boxed
Numerical answer $N = 6$	0.5	✓ $n = 6$, with $\omega_5^2 = 1.2^{15} \approx 15.41 < 16$ and $\omega_6^2 = 1.2^{18} \approx 26.62 > 16$ as the bracketing check

Overall score: 8.0 / 8.0 pts — full marks

Numerical results all match the official key: the boxed final answer is $n = 6$, agreeing with the official $N = 6$. Intermediate sanity checks coincide too — Claude’s $\omega_5^2 \approx 15.41 < 16$, $\omega_6^2 \approx 26.62 > 16$ (with $\omega_{\min}^2 = 4g/a = 16$) are the squared analogues of the official’s $\omega_5 \approx 3.93\,\text{rad/s}$, $\omega_6 \approx 5.16\,\text{rad/s}$ against $\omega_{\min} = 4\,\text{rad/s}$.

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads the four structural claims that would otherwise be scattered through the algebra: (1) angular impulse vanishes for an instantaneous radial motion at a turning point, so $L = 0$ stays $L = 0$ and the squat does nothing kinematically to the swing; (2) at the bottom $L \neq 0$ but the gravitational torque vanishes identically since gravity is along the rod, so conserving $L$ while $r$ shrinks forces $\omega$ to grow by $(b/a)^2$; (3) the resulting per-half-cycle factor in $\omega^2$ is $(b/a)^3$, advertised before the algebra and recovered cleanly later as $(b/a)^4 \cdot (a/b)$; (4) the rigid-rod-vs-string distinction is called out at the top — the rod can both pull and push, so the criterion is $\omega^2 \geq 4g/a$ rather than $5g/a$, a factor $4/5$ lower in the threshold. Part (iii) then adds an “energy bookkeeping” subsection that splits the per-cycle gain into a positive stand-up contribution $W_{\mathrm{up}} = \tfrac12 mb^2\omega_b^2[(b/a)^2 - 1] + mg(b-a)$ and a turn-point squat contribution $W_{\mathrm{down}} = -mg(b-a)\cos\theta_n$ that is negative for $\theta_n < \pi/2$ — a useful warning that the muscle work absorbs energy at low amplitudes, and that the clean $(b/a)^3$ recursion is the residual after that absorption is netted against the stand-up gain. Four independent consistency checks follow: the $b \to a$ limit where pumping stops and $n \to \infty$, an explicit dimensional check on $4g/a \sim 1/\text{time}^2$, the rigid-rod-vs-string comparison showing that for the keyed parameters $5g/a = 20$ also yields $n = 6$ (the answers happen to coincide here even though the thresholds differ by $5/4$), and an “energy ratio” remark — KE multiplied by $1.2^{18} \approx 26.6$ over six pumps, i.e. roughly 73% per pump, “strong parametric amplification, consistent with the practical observation that a few well-timed pumps suffice.” The closing “Educational remark” identifies the asymmetry that makes pumping work — squat at the turning point where $\omega^2 r$ is zero, stand-up at the bottom where $\omega^2 r$ is largest — and explicitly states what happens if the protocol is reversed (the swing damps because angular momentum conservation at the bottom now makes $\omega$ shrink), tying the result to the textbook parametric-resonance picture (forcing period = half the natural period, two pumps per cycle).

Where the official solution is sharper. Two places. (1) The official’s derivation of $\omega_{\min} = 2\sqrt{g/a}$ is more physically immediate: “the rods are rigid, so the condition for reaching the top is simply that the kinetic energy at the bottom suffice to raise Kirill by $2a$”, giving $\tfrac12(\omega_{\min} a)^2 = 2ga$ in one line. Claude reaches the same threshold via the up-swing energy equation $1 - \cos\theta_n \geq 2$ and reads $\theta_n = \pi$ off, which is algebraically equivalent and correctly invokes the rigid-rod insight up front but a step further from the “raise by $2a$” picture that makes the result obvious in zero algebra. (2) The official notation distinguishes $\eta_i$ (just before the $i$-th stand-up, squatting at radius $b$) from $\omega_i$ (just after, standing at radius $a$), which keeps the two-radius bookkeeping unambiguous through the recursion; Claude uses $\omega_b$ (without an index) for the pre-stand-up speed and lets the index live on $\omega_n$, which is fine in a long write-up but slightly less crisp than the official’s two-letter convention. The official also supplies the explicit quarter-period formulas $T_a/4 = (\pi/2)\sqrt{a/g}$ and $T_b/4 = (\pi/2)\sqrt{b/g}$ that quantify the “squatting arch is longer” feature; Claude states the proportionality $\propto\sqrt{r}$ but does not give the prefactor.