Съдържание

Условие Отговори Решение

OPhO 2026, Единствен кръг — Задача 3. Kugel Fountain

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Всички материали от OPhO 2026

2026 Online Physics Olympiad · 21 август 2026 г.

Last updated

Last updated: 22nd August 03:35 UTC

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Corrections (all times are in UTC)

Q8: Cleared up wording (21/08 04:40 and 04:52)
Q3: improved wording (21/08 08:49)
Q19: Improved wording (21/08 09:42)
Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
Q1: Improved wording (22/08 03:35)

Указания

Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:

Short Answer — problems requiring numerical answers, submitted on the OPhO portal;
Long Answer — problems requiring full written solutions, submitted on the OPhO portal;
Simulation — computational problems requiring teams to run a provided executable locally, with solutions submitted on the OPhO portal.

All teams have access to all three parts for the entire duration of the contest. Submissions are made on the OPhO portal https://opho.physoly.org/ using the team login, and may be made at any time within the contest window.

Short Answer. Teams submit numerical answers on the portal, with up to three attempts per problem. Every answer submission must be accompanied by a file upload, on the portal, of the working used to obtain that answer. This working will not be graded, but submissions without it will not be accepted. The base score for each problem depends on the number of teams that successfully answer it, each incorrect attempt multiplies the base score by a factor of 0.8, and there is no time factor. The score for an individual problem follows:

w(N) = (0.8)^i\left[e^{-\frac{x(N)}{200}} + e^{-\frac{x(N)^2}{100}} + 1\right],

where $i \in \{0, 1, 2\}$ represents whether a team got it on the first, second, or third try, $N$ represents the number of teams that answer the problem correctly, and $x(N)$ is the sum of $(0.8)^i$ across all teams that have solved the problem. The base score is independent of problem number, but expect questions to be ordered roughly by difficulty. A team’s raw Short Answer total is the sum of these values, and all raw totals are then scaled linearly such that the top team receives 55 points.

Long Answer. Teams submit full solutions on the portal. Rubrics are predetermined, and solutions will be reviewed to determine the proper number of points to give. The points breakdown is given per problem, for a total of 35 points. It is recommended that participants write their solutions in LaTeX. Handwritten solutions (or a combination of both) are accepted too; if participants have more than one photo of a handwritten solution (jpg, png, etc.), it is required that they be organized in the correct order in a pdf before submitting.

Simulation. Teams will need to be able to run a provided executable locally. Solutions are submitted on the portal and will be graded against predetermined rubrics. The points breakdown is given per problem, for a total of 10 points.

Scoring. A team’s cumulative score is the sum of its scaled Short Answer score (out of 55 points), Long Answer score (out of 35 points), and Simulation score (out of 10 points), for a maximum score of 100 points.

Permitted Resources. Use of the internet is allowed, including textbooks, published papers, calculators, and computer algebra systems. However, the use of AI tools of any kind is strictly prohibited. This includes, but is not limited to, large language models (e.g. ChatGPT, Claude, Gemini), AI coding assistants, AI-generated search summaries, and any other AI-powered assistance.

Dishonest Behavior. Please refrain from cheating in any form. This includes, but is not limited to, creating alternate accounts to gain additional attempts, sharing answers with other teams, posting problems online, utilizing AI tools as described above, or employing any other method to gain an unfair advantage.

We have previously caught numerous teams engaging in such practices, and pursuing dishonest tactics provides no genuine sense of security. Any teams found to be acting maliciously will be notified via email, along with the opportunity to appeal. Failing to provide a valid appeal will result in a ban from participating in any future Physoly competitions. Our utmost priority is to ensure an enjoyable contest experience for all participants, and we remain committed to preventing any disruption to the fun for others.

Awards and Results. The top competitors will be recognized on the website. No cash prizes will be awarded this year.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant, $G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Solar Mass $M_\odot = 1.988 \cdot 10^{30}\ \mathrm{kg}$
Acceleration due to gravity, $g = 9.8\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant, $h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space, $\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant, $k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant, $\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

A Kugel fountain is a water feature that uses the phenomenon of hydroplaning to suspend large rocks above a thin film of water.

One such fountain has a flat base. A circular hole is drilled through the center of the fountain and fitted with a vertical pipe of radius $a = 4\,\mathrm{cm}$ . The opening of the pipe is flush with the base. Water flows out of the pipe at a volume flow rate $Q = 100\,\mathrm{cm^3/s}$ . A puck-shaped rock with radius $R = 120\,\mathrm{cm}$ is placed on the fountain, with its flat side resting on the base and its center right above the pipe’s center. The water film suspends it, and thickness of the water film between the rock and the base is $h = 0.8\,\mathrm{mm}$ . Find the weight of the rock $W$ in Newtons to 5% accuracy.

Assume that the water beneath the rock is in steady-state laminar flow (although this is not very realistic due to the high Reynolds number of the flow). The viscosity of water at room temperature is $\mu = 8.9 \times 10^{-4}\,\mathrm{Pa\,s}$ .

Отговори

Решение

Because the water film is so thin, viscous effects become very significant. In fact, as we will show later, the excess pressure required to overcome viscous effects is much larger than any contribution due to Let us warm up by solving a simpler problem. Consider a water pipe with a rectangular cross-section of width $w$ and height $h \ll w$ . Water flows through the pipe at a volumetric flow rate $Q$ . Assume that the water is undergoing steady-state laminar flow. Our goal is to find the pressure gradient in the fluid.
First note that pressure is uniform across a cross-section of the pipe, as the fluid has to be flowing in parallel layers. We can ignore the contribution of gravity since that just gives an additional constant contribution proportional to depth. Also, since $w \gg h$ , we ignore the edge effects close to the vertical walls.
Performing force balance on a layer of fluid, we get (defining $z$ to be positive upwards and $x$ to be positive in the direction of fluid flow.)

\frac{d\tau}{dz} = \frac{dP}{dx}

\mu\frac{d^2v}{dz^2} = \frac{dP}{dx}

Since the LHS must be independent of x and the RHS is independent of z, the values of both must be constant.

\therefore \mu\frac{d^2v}{dz^2} = \frac{dP}{dx} = k

Defining $z = 0$ at the middle of the pipe and $v_0$ to be the velocity of the fluid at that point, we can write the speed of fluid $v$ as a function of $z$

v = v_0 + \frac{k}{2\mu}z^2

The fluid velocity must be 0 at $z = \pm h/2$ , so

k = -8v_0\frac{\mu}{h^2}

\therefore v = v_0\left(1 - 4\left(\frac{z}{h}\right)^2\right)

Pressure difference

\frac{dP}{dx} = -8v_0\frac{\mu}{h^2}

We express $v_0$ in terms of $Q$ :

Q = \int v\,\mathrm{d}A = 2wv_0\int_0^{h/2}\left(1 - 4\left(\frac{z}{h}\right)^2\right)\mathrm{d}z = \frac{2}{3}whv_0

\frac{dP}{dx} = -12\frac{Q\mu}{wh^3}

Applying this to our problem, at every radius $r$ the width is $2\pi r$ , so

\frac{dP}{dr} = -\frac{6\mu Q}{\pi h^3 r}

Integrating from $r = R$ to an arbitrary $r$ , we find that the excess pressure at a radial distance $r$ is

\Delta P = \frac{6\mu Q}{\pi h^3}\ln\frac{R}{r}

Integrating this over area in polar coordinates from $r = a$ to $r = R$ , the force on the rock, which is equal to its weight, is

W = 3\frac{\mu Q R^2}{h^3} = \boxed{751\,\mathrm{N}}

Note that $a^2 \ll R^2$ , so the contribution to the force due to the water in the $r < a$ region is insignificant.
Now, let us estimate the effect of dynamic pressure. For this part, we are just obtaining an estimate, so we assume that the velocity profile is uniform in the vertical axis.

P_{\mathrm{dyn}} = -\frac{1}{2}\rho v^2 = -\frac{1}{2}\rho\left(\frac{Q}{2\pi r h}\right)^2

Integrating over the area $a < r < R$ ,

F_{\mathrm{dyn}} = \frac{1}{4\pi}\frac{\rho Q^2}{h^2}\ln\frac{R}{a} = 4.2\,\mathrm{N}

This is around 0.5% of the answer, so it is definitely insignificant.

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf