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Условие Отговори Решение

USAPhO 2004, Полуфинал — Задача 2. A2

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Всички материали от USAPhO 2004

AAPT UNITED STATES PHYSICS TEAM AIP 2004 — 2004 Semi-Final Exam · 25 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

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.
Do not discuss the contents of this exam with anyone until after the submission deadline.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

A hot air balloon with average density 1.20 kg/m $^3$ is in equilibrium at a height of 1.00 km above sea level. A gust of wind pushes the balloon to a height of 1.10 km above sea level and releases it at rest.

a. How long does it take for the balloon to return to its equilibrium position? [20 т.]

b. How fast is it going when it passes through equilibrium? [5 т.]

Assume that: The atmospheric density is a linear function of height.

The balloon volume and average density are constant.

Variations in the gravitational field with height are negligible.

The density of air at sea level is 1.29 kg/m $^3$ .

Drag is negligible.

Отговори

Решение

A2. First find the forces on the balloon. The weight force is

W = mg = \rho_b Vg

where $\rho_b = 1.20\ \mathrm{kg/m^3}$ is the density of the balloon, $V$ is its volume, and $g$ is the gravitational field strength. The buoyant force is

B = \rho_a Vg

where $\rho_a$ is the density of air. Since it is assumed to be a linear function of height, $\rho_a = \rho_0 - \alpha h$ where $\rho_0 = 1.29\ \mathrm{kg/m^3}$ is the density of air at sea level and $h$ is the height above sea level. The forces are in equilibrium at $h_0 = 1.00\ \mathrm{km} = 1.00 \times 10^3\ \mathrm{m}$ .

\rho_b Vg = \rho_a Vg = (\rho_0 - \alpha h_0)Vg. \qquad (\mathrm{A2-1})

Solving this for $\alpha$ ,

\alpha = \frac{\rho_0 - \rho_b}{h_0} = \frac{1.29\ \mathrm{kg/m^3} - 1.20\ \mathrm{kg/m^3}}{1000\ \mathrm{m}} = 9 \times 10^{-5}\ \mathrm{kg/m^4}

a. After being blown to a height of $h = 1.10$ km, the forces are no longer balanced.

ma = B - W

\rho_b Va = (\rho_0 - \alpha h)Vg - \rho_b Vg.

Substituting (A2 – 1) into the above equation

\rho_b Va = (\rho_0 - \alpha h)Vg - (\rho_0 - \alpha h_0)Vg = -\alpha(h - h_0)Vg = -\alpha\Delta h Vg.

Solving for the acceleration $a$

a = -\left(\frac{\alpha g}{\rho_b}\right)\Delta h.

The acceleration is proportional to the displacement. The motion is simple harmonic motion with

\omega = \sqrt{\left(\frac{\alpha g}{\rho_b}\right)} = \sqrt{\frac{(9 \times 10^{-5}\ \mathrm{kg/m^4})(9.8\ \mathrm{m/s^2})}{1.20\ \mathrm{kg/m^3}}} = 0.0271\ \mathrm{rad/s}.

The balloon is released at rest at amplitude $A = h - h_0 = 100$ m and first passes through its equilibrium position at a time equal to one fourth its period.

t = \frac{T}{4} = \frac{2\pi}{4\omega} = \frac{\pi}{2(0.0271\ \mathrm{rad/s})} = 57.9\ \mathrm{s}.

b. The balloon passes through its equilibrium position with maximum velocity

v = \omega A = (0.0271\ \mathrm{rad/s})(100\ \mathrm{m}) = 2.71\ \mathrm{m/s}.

Free-body diagram of the balloon: a circle with an upward arrow labelled B and a downward arrow labelled W.

Оригинал в Архива: Semifinal 2004.pdf · официални решения: Semifinal 2004 Solutions.pdf