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USAPhO 2022, Полуфинал — Задача 1. Question A1

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Всички материали от USAPhO 2022

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

Both original question and worked-key packets have been read in full. They refer to a separate answer-sheet packet, including a blank plotting grid, which is not present in the retained question/key sources or the matching catalogue entry. All native question data and the actual key graph are retained. Printed variable/sign/wording defects and grading allowances remain source faithful and are documented.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination has two parts. Part A has three questions and lasts for 90 minutes. Part B also has three questions and lasts for 90 minutes.
For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–7), and Part B questions (pages 8–10).
Begin by giving students the instructions and all of the answer sheets. Let the students read the instructions and fill out their information on the answer sheets. They can keep the instructions for both parts of the exam. Also give students blank sheets of paper to use as scratch paper throughout the exam.
Students may bring calculators, but they may not use symbolic math, programming, or graphing features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones or other electronics may not be used during the exam or while the exam papers are present. Students may not use books or other references.
To start the exam, give students the Part A questions, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the Part A questions and answer sheets.
Then give students a 5 to 10 minute break. Then give them the Part B questions, and allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, collect everything, including the questions, the instructions, the answer sheets, and the scratch paper. Students may not keep the exam questions, but they may be returned to the students after April 19th, 2022.
After the exam, sort each student’s answer sheets by page number. Scan every answer sheet, including blank ones.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvilli, Abijith Krishnan, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the table on the next page, answer sheets, and blank paper for scratch work. Read this page carefully before the exam begins.
You may use a calculator, but its memory must be cleared of data and programs, and you may not use symbolic math, programming, or graphing features. Calculators may not be shared. Cell phones or other electronics may not be used during the exam or while the exam papers are present. You may not use books or other outside references.
When the exam begins, your proctor will give you the questions for Part A. You will have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. If you finish all of the questions, you may check your work, but you may not look at Part B during this time.
After 90 minutes, your proctor will collect the questions and answer sheets for Part A. You may then take a short break.
Then you will work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. When the exam ends, you must return all papers to the proctor, including the exam questions.
Do not discuss the questions of this exam, or their solutions, until after April 19th, 2022. Violations of this rule may result in disqualification. Below are instructions for writing your solutions.
This year, all of your solutions will be written on official answer sheets. Nothing outside these answer sheets will be graded. Before the exam begins, write your name, student AAPT number, and proctor AAPT number as directed on the answer sheets.
There are several answer sheets per problem. If you run out of space for a problem, you may use the extra answer sheets, which are at the end of the answer sheet packet. To ensure this work is graded, you must indicate, at the bottom of your last answer sheet for that problem, that you are using these extra answer sheets.
Only write within the frame of each answer sheet. To simplify grading, we recommend drawing a box around your final answer for each subpart. You should organize your work linearly and briefly explain your reasoning, which will help you earn partial credit. You may use either pencil or pen, but sure to write clearly so your work will be legible after scanning.
You may have to fit a line to data, which you can plot on the answer sheet with grid lines. You may use a ruler, pencil, pen, or piece of paper as a straightedge.

Reference table of possibly useful information

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm C$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$	$(1+x)^n\approx1+nx$ for $\lvert x\rvert\ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\lvert\theta\rvert\ll1$	$\cos\theta\approx1-\theta^2/2$ for $\lvert\theta\rvert\ll1$

You may use this sheet for both parts of the exam.

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Important Instructions for the Exam Supervisor — Key Packet Pagination

For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–13), and Part B questions (pages 14–22).

Условие

Part A

Moment of Clarity

Hold your pencil horizontally by its tip. To keep it still, you will have to exert a combination of forces on its bottom and top. These forces can be viewed as a superposition of a net upward force, and a pair of opposite forces. The former ensures the forces on the pencil are balanced, while the latter provides a torque, called the bending moment, which ensures the torques on the pencil are balanced. Since the bending moment arises from a pair of opposite forces, it doesn’t depend on the choice of origin.

a. Consider a rod of length $\ell$ and mass per length $\lambda$ . What is the bending moment you must exert to hold the rod horizontally at its end?

Just as each piece of a string exerts a tension on neighboring pieces in equilibrium, each piece of a solid rod also exerts a bending moment on its neighbors. For thin rods under heavy loads, this bending moment can be the limiting factor that causes them to break.

Both complete native bridge support configurations, full rods, triangular supports, both ell dimensions and d dimension/arrow endpoints.

Suppose a rod is supported at both ends, so that it forms a bridge, as shown at left above. Assume the supports are simple, so that they only provide an upward force, and no bending moment. In equilibrium, a bending moment will appear throughout the rod. The magnitude of the maximum bending moment the rod can exert at any point without breaking is $M_0$ , and the length of the rod is $\ell$ . The bridge is loaded uniformly, with a mass per length of $\lambda$ (including its own mass).

b. Find the maximum possible value of $\lambda$ before the bridge collapses.

c. Now suppose that one support remains at the left end, while the other is a distance $d>\ell/2$ away from the left end, as shown at right above. In static equilibrium, find the bending moment $M(x_0)$ at a distance $x_0<d$ from the left end.

d. Find the value of $d$ that maximizes the load $\lambda$ that the bridge can take before collapsing.

Решение

a.

Balance torques about the end of the rod. The weight is $\lambda g\ell$ , applied uniformly along the rod, so the torque due to gravity is $\lambda g\ell^2/2$ . This can only be balanced by the bending moment, so it is $\lambda g\ell^2/2$ .

b.

Placing the origin at the left support, consider the system consisting of the points on the rod at $0<x<x_0$ . The external forces are:

An upward force $\lambda g\ell/2$ at the left end, due to the left support.
A weight of $\lambda gx_0$ .
An upward force $\lambda g(x_0-\ell/2)$ at the right end, due to the rest of the rod.

Now consider balancing torques on this system about $x=x_0$ . The external torques are:

A clockwise torque $\lambda g\ell x_0/2$ due to the left support.
A counterclockwise torque $\lambda gx_0^2/2$ due to the weight.
A bending moment $M(x_0)$ due to the rest of the rod.

For torque balance, the bending moment must be

M(x_0)=\frac{\lambda gx_0^2}2-\frac{\lambda g\ell x_0}2=\frac{\lambda gx_0(x_0-\ell)}2.

The rod is most likely to break where $M$ is the largest, which occurs at the center, so

M_0=\frac{\lambda g(\ell/2)^2}2.

Solving for $\lambda$ gives

\lambda=\frac{8M_0}{g\ell^2}.

c.

To keep the expressions simple, let $W=\lambda g\ell$ be the weight of the rod. By balancing torques about the left end of the rod, the external forces on the entire rod are

A weight force $W$ .
An upward force $W\ell/2d$ from the right support.
An upward force $W-W\ell/2d$ from the left support.

Note that the reason we specified $d>\ell/2$ is because otherwise, the force from the left support would become negative, in which case the bridge would tip over to the right.

Now consider the system consisting of the points on the rod with $0<x<x_0$ , where $x_0<d$ . To balance the forces, the upward force from the rest of the rod must be

F_r=W\left(\frac{x_0}\ell+\frac\ell{2d}-1\right).

The external torques on this system, about the left end, are

A counterclockwise torque $Wx_0^2/2\ell$ due to the weight.
A clockwise torque $F_rx_0$ due to the rest of the rod.
A bending moment $M(x_0)$ due to the rest of the rod.

Balancing the torques thus gives

M(x_0)=-W\left(\frac{x_0^2}{2\ell}+\frac{\ell x_0}{2d}-x_0\right)=-\lambda g\left(\frac{x_0^2}2+\frac{\ell^2x_0}{2d}-x_0\ell\right).

The sign of $M$ is convention-dependent, so either sign is acceptable.

d.

If the bridge doesn’t collapse, then the magnitude of the bending moment must be less than $M_0$ everywhere throughout the rod. Consider how the bending moment varies as one moves from the left end of the rod to the right. It begins at zero at $x_0=0$ , then reaches a positive maximum somewhere between the two supports, when $M'(x_0)$ vanishes. This occurs when

x_0=\ell-\frac{\ell^2}{2d}

and which point there is a bending moment of

M_1=\frac{\lambda g}2\left(\ell-\frac{\ell^2}{2d}\right)^2.

As a check, we recover the answer to part (b) when $d=\ell$ .

Continuing rightward, the bending moment decreases, passes through zero at some point $x_0<d$ , and then becomes negative. (This change in sign is necessary to support the part of the rod hanging to the right of the right support.) At $x=d$ , the bending moment does not change discontinuously, because the supports provide no bending moment. It then smoothly returns to zero as $x_0$ approaches $\ell$ . Therefore, the most negative value of the bending moment occurs at $x_0=d$ , giving a bending moment of magnitude

M_2=-M(d)=\frac{\lambda g}2(\ell-d)^2.

As another check, this matches the answer to part (a) for a rod of length $\ell-d$ .

As we move the right support closer to the center, the middle of the bridge becomes more stable, since $M_1$ decreases. At the same time, $M_2$ increases, because more of the bridge is hanging off the edge of the right support. The bridge will not collapse as long as both of these quantities are less than $M_0$ . Thus, when the right support is placed most efficiently, both of them will become equal to $M_0$ just before the bridge collapses, which means we should have $M_1=M_2$ . Equating them yields

1-\frac\ell{2d}=1-\frac d\ell

which gives the answer,

d=\frac\ell{\sqrt2}.

Оригинал в Архива: 2022-USAPhO-Exam.pdf · официални решения: 2022-USAPhO-solutions.pdf