Съдържание

USAPhO 2014, Полуфинал — Задача 1. Question A1

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Всички материали от USAPhO 2014

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

Внимание: Бележка към темата

The printed key is retained with its source defects: A3(a)(i) has a factor 1/2 in the expanded cubic term; A3(b)(ii) switches from question distance d to h; A4(b)(ii) prints r < b before describing the exterior field; B2(a)(i) drops the subscript 0 in one denominator. These are source discrepancies, not silent editorial corrections.

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

This examination consists of two parts.
Part A has four questions and is allowed 90 minutes.
Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3–5), Part B (pages 7–8), and several answer sheets for two of the questions in part A (pages 10–13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2014.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones, PDA's or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time.
Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets.
Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

AAPT ID #

Doe, Jamie

A1 - 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA's or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 15, 2014.

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

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Условие

Part A

Inspired by: a leaning motorcycle on a vertical wall

A unicyclist of total height $h$ goes around a circular track of radius $R$ while leaning inward at an angle $\theta$ to the vertical. The acceleration due to gravity is $g$ .

a. Suppose $h\ll R$ . What angular velocity $\omega$ must the unicyclist sustain?

b. Now model the unicyclist as a uniform rod of length $h$ , where $h$ is less than $R$ but not negligible. This refined model introduces a correction to the previous result. What is the new expression for the angular velocity $\omega$ ? Assume that the rod remains in the plane formed by the vertical and radial directions, and that $R$ is measured from the center of the circle to the point of contact at the ground.

Решение

a.

We work in a frame rotating with angular velocity $\omega$ , where the unicyclist is static. Four forces act on the unicyclist: a normal and frictional force at the point of contact, gravity downwards at the center of mass, and a fictitious centrifugal force.

If $h\ll R$ , all parts of the unicyclist are at a distance of approximately $R$ from the center of the circle, so the centripetal acceleration of every part of unicyclist is $\omega^2R$ . The centrifugal force can then be taken to act at the center of mass for purposes of computing the torque. If the center of mass is a distance $l$ from the point of contact, the torque about the point of contact is

\tau=m\omega^2Rl\cos\theta-mgl\sin\theta

Since the unicyclist is stationary in this frame, $\tau=0$ , and solving for $\omega$ gives

\omega=\sqrt{\frac gR\tan\theta}.

b.

The centripetal acceleration now varies along the length of the unicyclist. In the rotating frame, the torque about the point of contact is given by

\tau_c=\int\omega^2rz\,dm

where $r$ is the distance from the center of the circle, $z$ is the height above the ground, and $dm$ is a mass element. Because the mass of the unicyclist is uniformly distributed along a length $h$ ,

dm=\frac mh\,ds

where $s$ is the length along the unicyclist. Then

\tau_c=\int_0^h\omega^2(R-s\sin\theta)(s\cos\theta)\frac mh\,ds=m\omega^2h\cos\theta\left(\frac R2-\frac h3\sin\theta\right).

Gravity continues to act at the center of mass, a distance $h/2$ from the point of contact, and in the opposite direction,

\tau_g=-mg\frac h2\sin\theta.

Again, the total torque is zero, so $\tau_c+\tau_g=0$ . Solving for $\omega$ gives

\omega=\sqrt{\left(\frac gR\tan\theta\right)\left(1-\frac23\frac hR\sin\theta\right)^{-1}}.

Оригинал в Архива: USAPhO-2014.pdf · официални решения: USAPhO-2014-Solutions.pdf