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USAPhO 2016, Полуфинал — Задача 5. Question B1

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Всички материали от USAPhO 2016

USA Physics Olympiad Exam — Part A / Part B · 50 т.

Внимание: Бележка към темата

The native worked key contains printed mathematical inconsistencies at A1(b), B1 and B2(a)(iii)/(c)(ii); the source formulas and adjacent reasoning are retained as printed. The A3 key graphs omit the requested event-letter labels and do not draw the stopped-rod world-line changes; no new answer labels or lines have been invented. Exact originals and detailed source-defect evidence accompany this transcription.

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 3-6), Part B (pages 8-9), and several answer sheets for one of the questions in Part A (pages 11-12). 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, 2016.
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, 2016.

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 B

A uniform solid spherical ball starts from rest on a loop-the-loop track. It rolls without slipping along the track. However, it does not have enough speed to make it to the top of the loop. From what height $h$ would the ball need to start in order to land at point P directly underneath the top of the loop? Express your answer in terms of $R$ , the radius of the loop. Assume that the radius of the ball is very small compared to the radius of the loop, and that there are no energy losses due to friction.

Full native ball/ramp/loop track, starting ball, h dimension double arrow, R radius arrow, dotted vertical and P landing-point label.

Решение

We fix the origin at $P$ . Assume the ball leaves at an angle $\theta$ away from the vertical. At this point, the $x$ and $y$ coordinates are

x=R\sin\theta,\quad y=R(1+\cos\theta).

By energy conservation, we have

mg(h-y)=\frac12mv^2+\frac12I\omega^2=\frac12m(1+\beta)v^2

where $\beta=2/5$ , and we used the fact that the ball rolls without slipping.

Let $v$ be the speed of the ball when it leaves the loop. Then its velocity components at that moment are

v_x=-v\cos\theta,\quad v_y=v\sin\theta.

Assuming the ball impacts $P$ at time $t$ ,

y=\frac12gt^2-v_yt,\quad x=-v_xt.

The second equation yields

t=\frac{R\sin\theta}{v\cos\theta}

and plugging this into the first equation gives

R+R\cos\theta=\frac12g\left(\frac{R\sin\theta}{v\cos\theta}\right)^2-v\sin\theta\frac{R\sin\theta}{v\cos\theta}

which simplifies to

1+\cos\theta=\frac{gR\sin^2\theta}{2v^2\cos\theta}.

Now, the ball leaves the surface when the normal component of the force of the loop on the ball just drops to zero. This happens when

mg\cos\theta=m\frac{v^2}R\quad\Rightarrow\quad\frac{v^2}{gR}=\cos\theta

and plugging this into the previous equation gives

1+\cos\theta=\frac12\frac{1-\cos^2\theta}{\cos^2\theta}\quad\Rightarrow\quad 2\cos^2\theta=1-\cos\theta.

This is a quadratic equation with solutions

\cos\theta=\frac{-1\pm\sqrt{1+8}}4=-\frac14\pm\frac34

Only the positive answer of $\cos\theta=1/2$ is relevant here, though the negative answer is still physical!

Now that we know $\theta$ , getting the final answer is straightforward. We combine the energy conservation equation and the condition

mg\cos\theta=m\frac{v^2}R

to find

h=1+\left(\frac12(1+\beta)+1\right)R\cos\theta=\frac{37}{20}R.

Оригинал в Архива: USAPhO-2016.pdf · официални решения: USAPhO-2016-Solutions.pdf