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OPhO 2020, Теория — Задача 6. Flat Earth

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Всички материали от OPhO 2020

2020 Online Physics Olympiad: Invitational Contest v1.4 · 22 т.

Внимание: Бележка към темата

The v1.4 question paper, native Partial Solutions and separately attributed organizer clarifications are retained. The key is incomplete: Problem 1 is only marked TBD, Problem 8 has no key for parts f–h or an explicit critical power in e, and Problem 10 has no key for Part B f–h. Problems 2 and 9 also have source gaps noted below. Printed mathematical defects and differences between repeated questions in the key and the question original are not silently repaired. Problem2(e) gives a time-dependent spin result but no explicit answer in orbital angle; key2(c) refers to an absent diagram. Problem9(a) gives its ODE and initial conditions without an explicit speed, and 9(b) does not state whether the pressure increases or decreases. Native source typography is retained, including the Planck-constant value, missing electron-charge unit, missing kappa in key3(a), mismatched labels in key4, inconsistent theta indices and ring factors in key6, the negative probability/energy signs and r_B/K_B/k variants in key7, radiation expressions and fusion-rate/proton-rate difference in key8, and image/flux signs and field factors in key10. All these are printed-source issues, not repaired answers. Question2 omits the word define while the repeated key question includes it; the repeated key question7 adds For simplicity, ignore spin. The 4-page organizer clarification contradicts question1(d)’s frictionless table and also contains both only one final submission and multiple submissions instructions. The native original and the separately attributed clarifications remain visible.

Original credits

2020 Online Physics Olympiad (OPhO): Invitational Contest v1.4

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, Qilin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

July 3 - 6, 2020

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

This test contains 10 long answer questions.
The total base score for the exam is 300 points; a factor incorporating the number of teams who solved the question will be added in the marking scheme. The final scores of all the teams will be available a few days after the contest ends.
The team leader should submit their final solution document in this here. We don’t anticipate a tie, but in the rare circumstance that there is one, the time you submit will be used to break it.
You can write on paper, type it up online, or a mix of both. If you wish to use a pre-made LaTeX template, we made one which you can choose to use here.
Since this is a long answer response, you will be judged on the quality of your work. To receive full points, you need to show your work, including deriving equations. As a general rule of thumb, any common equations (such as the ones in this formula sheet) can be cited without proof. Remember to state any approximations made, which system of equations were solved after every step. Explicitly showing every step of algebra is not necessary.
You may leave all final answers in symbolic form (in terms of variables) unless otherwise specified.2
Be sure to state all assumptions.

Sponsors

WOLFRAM · AwesomeMath (making x,y,z as easy as a,b,c) · Physics Olympiad · AoPS.com Art of Problem Solving

List of Constants

Proton mass, $m_p=1.67\cdot10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n=1.67\cdot10^{-27}\ \mathrm{kg}$
Electron mass, $m_e=9.11\cdot10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0=6.02\cdot10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R=8.31\ \mathrm{J/(mol\cdot K)}$
Boltzmann’s constant, $k_B=1.38\cdot10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e=1.60\cdot10^{-19}$
1 electron volt, $1\ \mathrm{eV}=1.60\cdot10^{-19}\ \mathrm J$
Speed of light, $c=3.00\cdot10^8\ \mathrm{m/s}$
Universal Gravitational constant, $G=6.67\cdot10^{-11}\ \mathrm{(N\cdot m^2)/kg^2}$
Acceleration due to gravity, $g=9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm u=1.66\cdot10^{-27}\ \mathrm{kg}=931\ \mathrm{MeV}/c^2$
Planck’s constant, $h=6.63\cdot10^{-34}\ \mathrm{J\cdot s}=4.41\cdot10^{-15}\ \mathrm{eV\cdot s}$
Permittivity of free space, $\epsilon_0=8.85\cdot10^{-12}\ \mathrm{C^2/(N\cdot m^2)}$
Coulomb’s law constant, $k=\frac1{4\pi\epsilon_0}=8.99\cdot10^9\ \mathrm{(N\cdot m^2)/C^2}$
Permeability of free space, $\mu_0=4\pi\cdot10^{-7}\ \mathrm{T\cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi}=1\cdot10^{-7}\ \mathrm{(T\cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm}=1.01\cdot10^5\ \mathrm{N/m^2}=1.01\cdot10^5\ \mathrm{Pa}$
Wien’s displacement constant, $b=2.9\cdot10^{-3}\ \mathrm{m\cdot K}$
Stefan-Boltzmann constant, $\sigma=5.67\cdot10^{-8}\ \mathrm{W/m^2/K^4}$

Native key title

2020 Online Physics Olympiad (OPhO): Invitational Contest Partial Solutions

The same organizer, writer, developer and advisor credits appear on this key cover.

Organizer clarification document — major notes

Separately quoted from the organizer clarification document, titled 2020 OPhO Invitational Clarifications:

MAJOR NOTES

Hi everyone! All team members are welcome to fill out this (optional) feedback form. Everyone who submits (and consents) will be entered in a draw to receive a $25 Amazon gift card.
Use this link to submit clarification requests. The problem writers may or may not deem it necessary to answer all clarification requests.
Link to the overleaf template (the link has been fixed) for answer submission.
Reminder: There is only one final submission at the end of the competition via the submission form only. Do NOT email us a copy of your submission. In case of internet connectivity issues, any team member may email us the file.
Make sure to read over the instructions. You are expected to give symbolic answers unless explicitly otherwise stated. If the numerical values for a symbol are stated, you do not need to solve the problem for a general case.
Uploading a .jpeg on the overleaf for a graph is allowed.
Please check the newest version v1.4 for revisions.
Please provide clear reasoning or derivations of any obscure results. Directly citing them would not necessarily fetch full points.
We will be awarding a special prize for whichever team produces the most creative and/or elegant solution to certain problems. We will not be disclosing which problems until after the competition has concluded.

Note on submission of solutions:

It is highly recommended that the solutions to the invitational round be submitted in LaTeXed form. We recommend using Overleaf. It is preferred that the solutions are quite detailed; this means that while you do not need to explicitly write all your algebra, all the relevant equations must be clearly written. A brief wording on how you solve a given system of equations is also optimal. Clearly state any approximations you make. Solutions typed in .doc/.docx, and ones neatly written by hand will also be accepted. Some students have a handwriting that’s hard to read; if that’s you, please read this. You can submit solutions multiple times. Only the latest submission will count.

Условие

In this problem, we will explore the true gravitational model of the earth, not the one that is claimed in most textbooks. Contrary to popular belief, the Earth is a flat circle of radius $R$ and has a uniform mass per unit area $\sigma$ . The Earth rotates with angular velocity $\omega$ .

a) (5 pts) A pendulum of length $\ell$ that is constrained to only move in one plane is placed on the ground at the center of the Earth. The pendulum has more than one angular frequency of small oscillations. Find the value of each angular frequency of small oscillations $\Omega(0),\Omega_1(0),\ldots$ in terms of $\sigma$ , $\omega$ , $\ell$ , and physical constants and the equilibrium angle $\theta,\theta_1,\ldots$ that the frequency occurs at. Assume for all parts that $\ell\ll R$ .

An equilibrium angle corresponds to the angle with respect to the vertical where there is an equilibrium point.

b) (2 pt) Investigate the stability of each equilibrium position with varying angular velocity of the Earth.

c) (12 pts) The entire pendulum is moved a horizontal distance $r\ll R$ away from the center of the Earth. It is oriented so that it is constrained to only move in the radial direction. Now, find the new angular frequency $\Omega(r)$ of small oscillations about the lowest equilibrium point in terms of the given parameters, assuming that $\omega^2r$ is much less than the local gravitational acceleration.

d) (3 pts) The angular frequencies $\Omega(0)$ and $\Omega(r)$ are both measured and the difference is found to be $\Delta\Omega$ . Assuming that $\Delta\Omega\ll\Omega(0)$ and $\omega^2\ll\frac g\ell$ , determine $\sigma$ in terms of $\omega,r,\Omega(0),\Delta\Omega$ , and physical constants.

Published organizer clarifications (separate source)

Separately quoted from the organizer clarification document:

Problem 6:

Rotation is about the center.
An equilibrium angle corresponds to the angle with respect to the vertical where there is an equilibrium point.
Equilibrium points are defined with respect to the rotating reference frame of the Earth.
For part a, you are to find the value of angular frequency of small oscillations at each one of these equilibrium angles.
For part b, just to be clear: Investigate the stability of each equilibrium position with varying angular velocity of the Earth.
For part c, make no assumptions of $r\gg l$ .
You can express your answers in terms of l if required.
The pendulum rotates with the Earth.
The pendulum is placed such that it does not touch the ground and oscillates (it is placed like a regular pendulum attached to a ceiling or you could imagine a person holding the pendulum over the center of the flat Earth).
The distance r the pendulum moves is in the horizontal direction. In other words, it moves a distance r on the surface of the flat Earth.
The flat Earth rotates about the z-axis.
For all parts you may assume the radius of the Earth R is large enough to be ignored.

Решение

Solution 6:

(a) The first thing that we note is that the acceleration due to the gravitational force is different as compared to the one on Earth. Consider a cylindrical Gaussian surface that has it’s circular faces parallel to the Earth. From Gauss’s Law of gravitation, we can calculate the acceleration due to gravity on a flat plane to be

4\pi Gm=2\times\pi r^2g\implies g=2\pi G\sigma

by substituting $m=\sigma\pi r^2$ . Let us now consider a small displacement of the mass $m$ from the origin. We see that there are two forces involved: the centrifugal force, and the gravitational force. We now draw a free-body diagram as shown below:

Pendulum over flat disk with theta/h/R/sigma labels and both force arrows, complete plane outline.

Note that the gravitational force will be restoring and is given by $mg\sin\theta\approx mg\theta$ . The centrifugal force is directed outwards and is given by $m\omega^2\ell\sin\theta\cos\theta\approx m\omega^2\ell\theta$ . Writing Newton’s Laws on the pendulum now gives us

m\ell\ddot\theta=m\omega^2\ell\theta-mg\theta\implies\ddot\theta=-\left(\frac g\ell-\omega^2\right)\theta.

An equilibrium position occurs when $\ddot\theta=0$ . This tells us that the two equilibrium positions are defined by

\theta_1=0,\qquad\theta_2=\arccos\left(\frac g{\omega^2\ell}\right).

For oscillations near $\theta_1=0$ , we can use small-angle approximations gives us

\ddot\theta=-\left(\frac g\ell-\omega^2\right)\theta.

Therefore, we see $\Omega(0)=\sqrt{\frac g\ell-\omega^2}=\boxed{\sqrt{\frac{2\pi G\sigma}{\ell}-\omega^2}}$ .

For oscillations near $\theta_2=\arccos(\frac g{\omega^2\ell})$ , we let $\theta=\theta_1+\varphi$ where $\varphi\ll\theta_1$ . Using a first order approximation tells us that

\ddot\varphi+\left(\frac g\ell+\omega^2\varphi\sin\theta_1-\omega^2\cos\theta_1\right)(\sin\theta_1+\varphi\cos\theta_1)=0.

Simplifying gives us

\ddot\varphi=-(\omega^2\sin^2\theta_1)\varphi.

This tells us that $\Omega_1(0)=\omega\sin\theta=\omega\sqrt{1-\frac{g^2}{\omega^4\ell^2}}=\boxed{\frac1{\omega\ell}\sqrt{\omega^4\ell^2-4\pi^2G^2\sigma^2}}$ .

Remark 1: We can find the gravitational acceleration on the planet by in fact another way of analyzing the force per each infinitesimal ring. We can split the flat Earth into many tiny rings as shown below

Infinitesimal ring dR/R/sigma, conical theta/h geometry and dF vector arrow above complete disk outline.

Each ring provides a force of $d\vec F$ . If each ring has a thickness $dR$ , our force will be given by

dF=\frac{Gm\sigma}{h^2+R^2}\cos\theta\cdot(\pi(R+dR)^2-\pi R^2).

Using the fact that $\cos\theta=\frac h{\sqrt{h^2+R^2}}$ , we then find that

dF=\frac{Gm\sigma}{h^2+R^2}\frac h{\sqrt{h^2+R^2}}\cdot(\pi(R+dR)^2-\pi R^2).

Using a first order approximation gives us

\left(\pi(R+dR)^2-\pi R^2\right)\approx2RdR.

We then find by substituting that,

dF=\frac{2GmR\sigma h}{(h^2+R^2)^{3/2}}dR.

Now, integrating this force $dF$ from 0 to $\infty$ gives us

F=\int_0^\infty dF=2Gm\sigma h\int_0^\infty\frac R{(h^2+R^2)^{3/2}}dR=2\pi G\sigma m.

The gravitational acceleration is then given by

g\equiv\frac{2\pi G\sigma m}{m}=2\pi G\sigma.

$\blacksquare$

Remark 2: We can find the equation of motion with the Euler-Lagrange equations. Let us consider a rotating frame at the center of the Earth. The pendulum’s coordinates are then given by

(x,y,z)=(\ell\sin\theta,0,-\ell\cos\theta)

which implies that the velocity of the pendulum is given by

(\dot x,\dot y,\dot z)=(\ell\dot\theta\cos\theta,0,\ell\dot\theta\sin\theta).

In the fixed frame, the pendulum has an additional velocity from the centrifugal force which is given by

\vec\omega\times\vec r=(0,0,\omega)\times(\ell\sin\theta,0,-\ell\cos\theta)=(0,\omega\ell\sin\theta,0).

We now write the lagrangian of the system as

\begin{aligned}\mathcal L&\equiv T-V\\&=\frac12m\sqrt{(\ell\dot\theta\cos\theta)^2+0^2+(\ell\dot\theta\sin\theta)^2}+\frac12m\ell^2\omega^2\sin^2\theta+mg\ell\cos\theta\\&=\frac12m\ell^2\dot\theta^2+\frac12m\ell^2\omega^2\sin^2\theta+mg\ell\cos\theta.\end{aligned}

Now, using the Euler-Lagrange equations gives us

\frac d{dt}\left(\frac{\partial\mathcal L}{\partial\dot\theta}\right)=\frac{\partial\mathcal L}{\partial\theta}\implies\ddot\theta=-\left(\frac g\ell-\omega^2\cos\theta\right)\sin\theta.

$\blacksquare$

(b) We have two cases:

Case 1: (oscillations for $\Omega(0)$ ) If $\omega<\sqrt{\frac g\ell}=\sqrt{\frac{2\pi G\sigma}{\ell}}$ , we see that the oscillations are stable. The oscillations would follow the equation

\theta(t)=A_1\cos(\Omega(0)+\phi_1)

where $A_1$ and $\phi_1$ are constants to be determined from initial conditions.

Case 2: (oscillations for $\Omega_1(0)$ ) If $\omega>\sqrt{\frac{2\pi G\sigma}{\ell}}$ we see that the oscillations are stable. The oscillations would follow the equation

\varphi(t)=A_2\cos(\Omega_1(0)+\phi_2)

where $A_2$ and $\phi_2$ are constants to be determined from initial conditions.

If $\omega=\sqrt{\frac{2\pi G\sigma}{\ell}}$ the oscillations are neutrally stable but do not display stable small oscillations.

(c) We use the fact that, for small oscillations of an object subject to a potential $U(x)$ , the frequency of small oscillations will be defined as

\omega=\sqrt{\frac{U''(x)}{m_{\mathrm{eff}}}}.

The potential energy is due to the gravitational potential and the centrifugal potential:

U=-mg\ell\cos\theta-\frac12m\omega^2(r+\ell\sin\theta)^2

We see that

\frac{d^2U}{d\theta^2}=mg\ell\cos\theta+m\omega^2r\ell\sin\theta+m\omega^2\ell^2(\sin^2\theta-\cos^2\theta)

When evaluated at $\theta=\theta_0$ , the equilibrium position,

U''(\theta_0)=mg\ell\cos\theta_0+m\omega^2r\ell\sin\theta_0+m\omega^2\ell^2(\sin^2\theta_0-\cos^2\theta_0)=k_{\mathrm{eff}}.

We now need to find the effective mass. Note that the kinetic energy is given by

\frac12m\ell^2\dot\theta^2=\frac12m_{\mathrm{eff}}\dot\theta^2\implies m_{\mathrm{eff}}=m\ell^2.

Thus, the frequency is given by

\Omega(r)=\sqrt{\frac{k_{\mathrm{eff}}}{m_{\mathrm{eff}}}}=\sqrt{\frac g\ell\cos\theta_0+\frac{\omega^2r}{\ell}\sin\theta_0+\omega^2(\sin^2\theta_0-\cos^2\theta_0)}

Now, we have $\cos\theta_0=\frac g{\sqrt{\omega^4r^2+g^2}}\approx1-\frac{\omega^4r^2}{2g^2}$ and $\sin\theta_0\approx\frac{\omega^2r}{g}$ . Substituting,

\Omega(r)=\sqrt{\frac g\ell-\frac{\omega^4r^2}{2g\ell}+\frac{\omega^4r^2}{g\ell}+\frac{\omega^6r^2}{g^2}-\omega^2+\frac{\omega^6r^2}{2g^2}}

\Omega(r)=\sqrt{\Omega(0)^2+\frac{\omega^4r^2}{2g\ell}+\frac{3\omega^6r^2}{2g^2}}=\boxed{\sqrt{\Omega(0)^2+\frac{\omega^4r^2}{4\pi G\sigma\ell}+\frac{3\omega^6r^2}{8\pi^2G^2\sigma^2}}}

(d) Note that we can write

\Omega(r)=\sqrt{\Omega(0)^2+\frac{\omega^4r^2}{2g^2}\left(\frac g\ell+3\omega^2\right)}=\Omega(0)\sqrt{1+\frac{\omega^4r^2}{2g^2}\left(\frac{\frac g\ell+3\omega^2}{\frac g\ell-\omega^2}\right)}

Therefore, we have

\Omega(r)\approx\Omega(0)\sqrt{1+\frac{\omega^4r^2}{2g^2}}\approx\Omega(0)\left(1+\frac{\omega^4r^2}{4g^2}\right)

Therefore, $\Delta\Omega=\frac{\omega^4r^2}{4g^2}\Omega(0)$ . This implies that

g^2=4\pi^2G^2\sigma^2=\frac{\omega^4r^2\Omega(0)}{4\Delta\Omega}

We can solve for $\sigma$ to get $\sigma=\boxed{\frac{\omega^2r}{4\pi G}\sqrt{\frac{\Omega(0)}{\Delta\Omega}}}$ .

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

Uploading images: View this link to find how to upload images on overleaf.
You are allowed to mix LaTeX and handwritten PDFs for your answer sheet. Note that you need to write your solution to the problem alongside your final answers.
You can use basic formulae related to topics on the IPhO syllabus without derivation but all other formulae which go beyond that must be derived.
If you find that space is running out on your LaTeX file, try to use $\verb|\newpage|$ and see if that fixes it.

Оригинал в Архива: invitational.pdf · официални решения: invitational_sol.pdf · допълнителни данни: Clarifications.pdf