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USAPhO 2021, Финален кръг · подбор на отбора — Задача 1. The Jet Stream

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Всички материали от USAPhO 2021

2021 USAPhO+ — Traveling Team Selection Exam · 8 май 2021 г.

Information About The 2021 USAPhO+

The 2021 USAPhO+ is a 5-hour exam taking place on Saturday, May 8 from noon to 5 PM, Eastern time.
The exam is hosted by AAPT on the platform provided by Art of Problem Solving. It will be proctored by the US Physics Team coaches via Zoom.
Before you start the exam, make sure you are provided with blank paper, both for your answers and scratch work, writing utensils, graph paper and a ruler, a hand-held scientific calculator with memory and programs erased, and a computer for you to log into the USAPhO+ testing page.
At the end of the exam, you have 20 minutes to upload solutions to all of the problems for that part. For each problem, scan or photograph each page of your solution, combine them into a single PDF file, and upload them on the testing platform.
USAPhO+ graders are not responsible for missing pages or illegible handwriting. No late submissions will be accepted.

Congratulations again on your qualification for the USAPhO+. We wish you the best of luck on the challenging problems to follow.

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

JiaJia Dong, Mark Eichenlaub, Abijith Krishnan, Kye W. Shi, Brian Skinner, Mike Winer, and Kevin Zhou.

Условие

The jet stream is an eastward wind current that moves over the continental United States at an altitude of $23{,}000$ to $35{,}000$ feet (the range of typical cruising altitudes of commercial airlines). This strong current affects flight times significantly: flights traveling eastward fly significantly faster than flights traveling westward.

This problem consists of two independent parts. In the first part, you will consider a simple model for airplane flight. In the second part, you will determine the jet stream speed on a fictitious planet called Orb.

1

The power that a plane expends is used both to combat drag and to generate lift. Throughout this part of the problem, you may assume that the plane travels with horizontal velocity $\mathbf{v}_{\mathrm{rel}}$ relative to the air, the density of air is $\rho_{\mathrm{air}}$ , the mass of the plane is $m$ , and the cross-sectional area of the plane is $A_{\mathrm{cs}}$ .

(a) The drag force on an airplane is given by

\mathbf{F}_{\mathrm{drag}} = -\frac{1}{2}c_d\rho_{\mathrm{air}}A_{\mathrm{cs}}|\mathbf{v}_{\mathrm{rel}}|\mathbf{v}_{\mathrm{rel}},

where $c_d$ is the drag coefficient (which depends on the shape of the plane). Write an expression for the power expended by the airplane to combat the drag force from the air.

(b) Airplanes generate lift by deflecting air downward.

i. Estimate the air mass per unit time which is deflected by the wings of the plain.

ii. Estimate the power expended by the plane for lift.

(c) Estimate the speed at which an airplane flies relative to the air by minimizing the power expended by the plane. To get a numeric answer, you may use the following parameters:

m_{\mathrm{plane}} \sim 8\times10^4\ \mathrm{kg},\quad \rho_{\mathrm{air}} \sim 1\ \mathrm{kg/m^3},\quad c_d \sim 10^{-2},\quad A_{\mathrm{cs}} \sim 100\ \mathrm{m^2}.

2

Next, we estimate the jet stream speed using flight times. Because the jet stream speed on Earth varies greatly with location, time of year, and climate effects (such as El Niño and La Niña), you will instead consider the fictitious planet Orb, where the jet stream is eastward and uniform in the region of interest. At the end of the problem is a map of the region, whose area is much smaller than the surface area of Orb (i.e., you can neglect the curvature of Orb).

At cruising altitude, we assume all airplanes travel at a fixed speed $v_{\mathrm{rel}}$ relative to the air. (This is not necessarily the same as your answer to 1(c), which was just a rough estimate.) Additionally, we assume that flights occur in three stages – (1) taxi and takeoff, (2) flight at cruising altitude, (3) landing and taxi – and that stages (1) and (3) take a fixed total time $t_0$ for every flight.

Map of the region on Planet Orb drawn on a square grid, with a compass rose (N up, E right, S, W) and a 200 km scale bar. Cities marked with dots: Rubinstead and Boltzmannburg in the north row, Newton City, Einsteinopolis, Maxwellbury, Noethersville, Curieton, Franklinport and Planck Town further south.

(a) Suppose a plane, at cruising altitude, is traveling at an angle $\theta$ away from due east relative to the ground. What is the speed of the plane relative to the ground? Give your answer in terms of $v_{\mathrm{rel}}$ , $\theta$ , and $v_w$ , the speed of the jet stream relative to the Earth’s surface.

(b) If the plane travels a distance $D$ , what is the total travel time $t$ , including taxi, takeoff, and landing?

(c) Below, we present some data on airplane flights on Planet Orb. Each of the flight times shown below has an independent uncertainty of $\Delta t = 5\,\mathrm{min}$ . From the data and the map, determine $v_w$ and $v_{\mathrm{rel}}$ , giving your answers in km per hour with uncertainties. Indicate clearly what two quantities you are plotting against each other on each graph that you plot.

Departure City	Arrival City	$t$ (min)
Noethersville	Rubinstead	185
Rubinstead	Noethersville	286
Curieton	Franklinport	107
Franklinport	Curieton	244
Planck Town	Maxwellbury	143
Maxwellbury	Planck Town	256
Rubinstead	Boltzmannburg	92
Boltzmannburg	Rubinstead	190
Einsteinopolis	Maxwellbury	160
Maxwellbury	Einsteinopolis	384
Planck Town	Franklinport	128
Franklinport	Planck Town	266
Einsteinopolis	Franklinport	188
Franklinport	Einsteinopolis	431
Boltzmannburg	Maxwellbury	135
Maxwellbury	Boltzmannburg	150

Departure City	Arrival City	$t$ (min)
Noethersville	Einsteinopolis	68
Einsteinopolis	Noethersville	74
Franklinport	Newton City	144
Newton City	Franklinport	129
Curieton	Rubinstead	186
Rubinstead	Curieton	175
Planck Town	Curieton	95
Curieton	Planck Town	102
Planck Town	Rubinstead	249
Rubinstead	Planck Town	250

Отговори

Решение

1

(a) We have that $P = \mathbf{F}\cdot\mathbf{v}$ , so

P_{\mathrm{drag}} = \frac{1}{2}c_d\rho_{\mathrm{air}}A_{\mathrm{cs}}v_{\mathrm{rel}}^3.

(b) i. The mass flux is given by $\rho v_{\mathrm{rel}}$ , so the rate is

\rho_{\mathrm{air}}v_{\mathrm{rel}}A_{\mathrm{cs}}.

ii. Suppose the deflected air has velocity $u$ downward. Then the lift force is given by

mg = \rho v_{\mathrm{rel}}Au.

The power is given by

P = Fu = \frac{m^2g^2}{\rho_{\mathrm{air}}v_{\mathrm{rel}}A_{\mathrm{cs}}}.

This is the power that goes into accelerating the air downward, so by energy conservation it must have come from the plane’s engine.

(c) The total power is given by

P \sim \frac{m^2g^2}{\rho_{\mathrm{air}}v_{\mathrm{rel}}A_{\mathrm{cs}}} + \frac{1}{2}c_d\rho A_{\mathrm{cs}}v_{\mathrm{rel}}^3.

The minimum power occurs when the derivative vanishes, which roughly gives

v_{\mathrm{rel}} \sim \left(\frac{m^2g^2}{c_d\rho^2A^2}\right)^{1/4} \sim 300\ \mathrm{m/s}.

2

(a) The velocity of the plane relative to the ground, $\mathbf{v}$ , the velocity of the plane relative to the air, $\mathbf{v}_{\mathrm{rel}}$ , and the jet stream velocity, $\mathbf{v}_w$ , all form a triangle under tip-tail addition. Then, from law of cosines,

v_{\mathrm{rel}}^2 = v^2 + v_w^2 - 2vv_w\cos\theta.

Solving gives us

v = v_w\cos\theta + \sqrt{v_{\mathrm{rel}}^2 - v_w^2\sin^2\theta}.

Of course, we assumed $v_w < v_{\mathrm{rel}}$ , as otherwise flying westward would not even be possible.

(b) The answer is given by

t = \frac{D}{v} + t_0,

where $v$ is our answer to the previous part.

(c) Note that the pairs of cities given in the second table are directly north/south of each other. For such cities, we expect the times in either direction to be equal, up to the timing uncertainty, and the expression for the time simplifies to

t_{NS} = \frac{D}{\sqrt{v_{\mathrm{rel}}^2 - v_w^2}} + t_0.

We can read $t_{NS}$ off the second table, and calculate $D$ from the figure. The unknown quantity $t_0$ goes into the intercept, while the slope is

\frac{1}{\sqrt{v_{\mathrm{rel}}^2 - v_w^2}} = 0.0768 \pm 0.0048\,\mathrm{min/km}.

Next, consider the pairs of cities given in the second table. We can cancel out the effect of $t_0$ by considering the difference in flight times in the two directions,

t_{12} - t_{21} = -\frac{D}{v_w\cos\theta + \sqrt{v_{\mathrm{rel}}^2 - v_w^2\sin^2\theta}} + \frac{D}{-v_w\cos\theta + \sqrt{v_{\mathrm{rel}}^2 - v_w^2\sin^2\theta}}.

After some simplification, we find

t_{12} - t_{21} = \frac{2Dv_w\cos\theta}{v_{\mathrm{rel}}^2 - v_w^2}.

Therefore, plotting the time difference versus $D\cos\theta$ , where $\theta$ can also be measured from the graph, gives

\frac{2v_w}{v_{\mathrm{rel}}^2 - v_w^2} = 0.0722 \pm 0.0003\,\mathrm{min/km}.

Combining these results gives the answers,

v_{\mathrm{rel}} = 860 \pm 50\,\mathrm{km/h},\quad v_w = 370 \pm 30\,\mathrm{km/h}.

Uncertainty on the times should be propagated through the calculation using the usual rules, while the uncertainties on the slopes are found by drawing the steepest and shallowest fit lines.

Оригинал в Архива: 2021-USAPhO_plus.pdf · официални решения: 2021-USAPhO_plus_Solutions_v2.pdf