NBPhO 2026

Part i) (1.5 points)

Starting at the bottom with $|\omega| = \omega_0$, Kirill (standing, radius $a$) swings up; $|\omega|$ falls smoothly to $0$ at the turning point, after a quarter-period $T_a/4 = (\pi/2)\sqrt{a/g}$. At the turning point Kirill squats (radius $a \to b$, no effect on the plot since $\omega = 0$). Swinging back while squatting, $|\omega|$ rises smoothly from $0$ to $\eta_1$ (see part c; $\eta_1 = \omega_0\sqrt{a/b} < \omega_0$) after a quarter-period $T_b/4 = (\pi/2)\sqrt{b/g}$, which is longer than the first since $b > a$. At the bottom Kirill stands up: $|\omega|$ jumps from $\eta_1$ to $\omega_1 = \eta_1(b/a)^2 = \omega_0(b/a)^{3/2} > \omega_0$. The second half of the period repeats the pattern with new amplitudes $\omega_1$, $\eta_2 = \omega_1\sqrt{a/b} > \omega_0$, $\omega_2 = \omega_0(b/a)^3$.

The sketch should therefore show:

• smooth arches during swinging, cusps at the bottom (jumps) and smoothness at turning points;
• the “squatting” arches taking longer than the “standing” ones;
• growth of successive peaks.

Part (i): Angular speed vs time, first full period.

Part ii) (2 points)

The trajectory in the $(\alpha, \omega)$ plane pieces together arcs of pendulum-like phase curves at two radii, $a$ and $b$. Each half-swing traces an arc of the energy conservation curve for the relevant radius. Stand-ups at $\alpha = 0$ appear as vertical segments where $|\omega|$ jumps from $\eta_i$ to $\omega_i = \eta_i(b/a)^2$. Squats at the turning points ($\omega = 0$) appear as smooth transitions (the radius-$a$ arc meets the radius-$b$ arc tangentially at $\omega = 0$).

The motion is an outward spiral: each full cycle brings Kirill back to $\alpha = 0$ with a larger $|\omega|$. When $\omega$ at the bottom (standing, radius $a$) first exceeds $\omega_{\min} = 2\sqrt{g/a}$, the next upward arc does not close: it reaches $\alpha = \pm\pi$ with $\omega > 0$ and continues past from there. This is the open curve on the sketch. Part (ii): Phase diagram ($\omega$ vs $\varphi$), first three half-swings.

Key features for grading:

• lens-shaped/pendulum-like arcs (not ellipses);
• vertical line segments only at $\alpha = 0$ (stand-ups), smooth meeting at turning points (squats);
• spiral grows outward;
• final open curve goes over the top with non-zero $\omega$.

Failure to use the phase diagram of angular velocity and angle, deducts 0.5 pts from the sum of part ii).

Grading (part i and ii)

• Approximately sinusoidal arcs between jumps: 0.3 pts
• Instantaneous jumps at the maximum values of $\omega$ (at the bottom of the swing, during the stand-up): 0.6 pts
• Smaller amplitude during “squatting” arches, or equivalently a longer quarter-period while $r = b$: 0.3 pts
• Overall growth of peak $\omega$ over the period: 0.3 pts

A sketch of signed $\omega$ (rather than $|\omega|$) receives full credit if the curve correctly corresponds to the axis labels and all other qualitative features are preserved. Failure to correctly mark $|\omega|$, where necessary, deducts 0.5 pts. The same goes for not completing the full period of the first swing.

Part iii) (4.5 points)

Growth factor per stand-up. Let $\omega_i$ denote the angular velocity just after the $i$-th stand-up (at the bottom, standing, radius $a$); let $\eta_i$ denote the angular velocity just before that stand-up (at the bottom, squatting, radius $b$).

Energy conservation on the way up (standing, radius $a$) to turning point at angle $\alpha$:

$$\tfrac{1}{2}(\omega_i a)^2 = ga(1-\cos\alpha) \implies \omega_i^2 a = 2g(1-\cos\alpha).$$

At the turning point $\omega = 0$, so the squat ($a \to b$) leaves $\alpha$ and $\omega$ unchanged. Swinging back down (squatting, radius $b$):

$$\tfrac{1}{2}(\eta_{i+1}b)^2 = gb(1-\cos\alpha)$$ $$\implies \eta_{i+1}^2 b = 2g(1-\cos\alpha).$$

Equating: $\eta_{i+1} = \omega_i\sqrt{a/b}$. The stand-up is a purely radial motion, so angular momentum about the pivot is conserved: $\eta_{i+1}b^2 = \omega_{i+1}a^2$, whence

$$\frac{\omega_{i+1}}{\omega_i} = \sqrt{\frac{a}{b}}\cdot\frac{b^2}{a^2} = \left(\frac{b}{a}\right)^{3/2}.$$

After $N$ stand-ups, $\omega_N = \omega_0(b/a)^{3N/2}$.

Minimum angular velocity for a loop. 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 $2a$:

$$\tfrac{1}{2}(\omega_{\min}a)^2 = 2ga \implies \omega_{\min} = 2\sqrt{g/a}.$$

Counting. Kirill first loops over when $\omega_N \geq \omega_{\min}$, i.e.

$$N \geq \frac{2}{3}\frac{\ln(\omega_{\min}/\omega_0)}{\ln(b/a)}.$$

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

$$b/a = 1.2, \quad (b/a)^{3/2} \approx 1.3145,$$ $$\omega_{\min} = 2\sqrt{10/2.5} = 4\,\text{rad/s},$$ $$N \geq \frac{2}{3}\frac{\ln 4}{\ln 1.2} = \frac{2}{3}\cdot\frac{1.386}{0.1823} \approx 5.07$$ $$\implies \boxed{N = 6}.$$

Check: $\omega_5 = 1.2^{7.5} \approx 3.93\,\text{rad/s}$ (just short); $\omega_6 = 1.2^9 \approx 5.16\,\text{rad/s}$ (exceeds $\omega_{\min}$). So Kirill loops over on the swing-up following his sixth stand-up.

Grading (part c)

• Angular momentum conservation during the stand-up: 0.7 pts
• Energy conservation during the swing: 0.7 pts
• Growth factor $(b/a)^{3/2}$ per stand-up: 0.7 pts
• Minimum angular velocity $4\sqrt{g/a}$ for the rigid-rod swing to loop over the top: 0.8 pts
• Correct inequality for the over-the-top condition ($\omega_N \geq \omega_{\min}$ after the $N$th stand-up): 0.4 pts
• Solving the inequality for $N$ using the growth law $\omega_N = \omega_0(b/a)^{3N/2}$, yielding $N \geq [4g/(a\omega_0^2)]/[3\log(b/a)]$: 0.7 pts
• Numerical answer $N = 6$: 0.5 pts