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Условие Отговори Решение

USAPhO 2003, Полуфинал — Задача 6. B2

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Всички материали от USAPhO 2003

AAPT / AIP United States Physics Team 2003 — 2003 Semi-Final Exam

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. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
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	$k = 1/4\pi\varepsilon_0 = 9.0 \times 10^{9}\ \mathrm{N{\cdot}m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T{\cdot}m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
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	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J{\cdot}s} = 4.14 \times 10^{-15}\ \mathrm{eV{\cdot}s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$

Условие

A spherical container of radius $R$ is filled with a light, incompressible fluid carrying uniform charge density $\rho$ . Let a point inside the container be described by $\vec{r} = r\hat{r} = x\hat{i} + y\hat{j} + z\hat{k}$ where the origin is at the center of the container and $\hat{r}$ , $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ are unit vectors in the radial, $x$ , $y$ , and $z$ directions, respectively.

a. Find the electrostatic field inside the container. [10 т.]

b. A small amount of fluid is replaced with a small, uncharged ball with mass density $\delta$ which is much, much greater than the fluid density. There is no gravitational field present. Ignore energy loss due to drag, and ignore any magnetic fields created by the moving fluid. Suppose that at time $t = 0$ the ball is released from rest at $\vec{r}_0 = x_0\hat{i}$ , $x_0 < R$ . What is $\vec{r}(t)$ ? [15 т.]

c. Repeat Part b, if there is a uniform gravitational field $\vec{g} = g\hat{i}$ . Assume that $g$ is small enough so that the ball does not hit the wall of the container. [5 т.]

d. The external gravitational field is removed, and the ball is again placed at $\vec{r}_0 = x_0\hat{i}$ . Now, however, we start the ball with some nonzero velocity $\vec{v}_0 = v_0\hat{j}$ .

d. i. What is the ball’s trajectory (before it hits the wall, if it does)? That is, what is the time-independent relationship between the ball’s $x$ and $y$ coordinates? Give the name of this trajectory, e.g. straight line, parabola, spiral… [10 т.]

d. ii. Is it possible for the ball to avoid the wall of the container? If so, what condition(s), if any, are required of $x_0$ and $v_0$ ? [5 т.]

d. iii. If you answered “yes”, is the ball’s orbit periodic when it avoids the wall, and if so, what is the period? If you answered “no”, what is the maximum time that can elapse before it hits the wall? [5 т.]

Отговори

Решение

a. The field is spherically symmetric and radially oriented. Applying Gauss’s Law to a sphere of radius $r$ ,

\oint \vec{\mathrm{E}}\cdot d\vec{\mathrm{S}} = \frac{q_{encl}}{\varepsilon_0}

4\pi r^2E(r) = \frac{\frac{4}{3}\pi r^3\rho}{\varepsilon_0}

Therefore

\vec{E}(r) = \frac{\rho}{3\varepsilon_0}\vec{r}

b. Consider the small amount of fluid before it is replaced by the ball. It is in equilibrium. The electrostatic force $\vec{F}_E$ on the small amount of fluid due to the field is balanced by a force exerted by the rest of the fluid which we shall call an electrostatic buoyant force $\vec{F}_B$ .

\vec{F}_B + \vec{F}_E = 0

\vec{F}_B = -\vec{F}_E = -q\vec{E} = -(\rho V)\left(\frac{\rho}{3\varepsilon_0}\vec{r}\right) = -\frac{\rho^2V}{3\varepsilon_0}\vec{r}

where $V$ is the volume. When the fluid is replaced by the ball, there is no longer an electrostatic force on it and the buoyant force $\vec{F}_B$ is unbalanced and the net force acting on the ball. Using Newton’s second law with $m = \delta V$ ,

\delta V\vec{a} = -\frac{\rho^2V}{3\varepsilon_0}\vec{r}

\vec{a} = -\frac{\rho^2}{3\varepsilon_0\delta}\vec{r}. \qquad \mathrm{(B2\text{-}1)}

The motion is simple harmonic motion. Examining the initial conditions, the motion is one-dimensional in the $x$ direction. $\vec{r}(t) = x(t)\hat{i}$ .

a_x = -\left(\frac{\rho^2}{3\varepsilon_0\delta}\right)x.

x(t) = x_0\cos\omega t \qquad \text{where} \qquad \omega = \frac{\rho}{\sqrt{3\varepsilon_0\delta}}.

and

\vec{r}(t) = x_0\hat{i}\cos\omega t,

c. Now the net force is

\vec{F} = -\frac{\rho^2}{3\varepsilon_0}V\vec{r} + mg\hat{i}

Since the ball is once again released from rest at $\vec{r}_0 = x_0\hat{i}$ , the motion is once again one-dimensional in the $x$ direction. $\vec{r}(t) = x(t)\hat{i}$ and

ma_x = -\frac{\rho^2}{3\varepsilon_0}\left(\frac{m}{\delta}\right)x + mg = -m\omega^2\left(x - \frac{g}{\omega^2}\right)

Letting $x' = x - \frac{g}{\omega^2}$ , and noting that $a_{x'} = a_x$ ,

a_{x'} = -m\omega^2x'

i.e., $x'$ undergoes simple harmonic motion with the same frequency. Again, examining the initial conditions, we must have

x'(t) = x'(0)\cos\omega t = \left(x_0 - \frac{g}{\omega^2}\right)\cos\omega t

i.e.

x(t) = \left(x_0 - \frac{g}{\omega^2}\right)\cos\omega t + \frac{g}{\omega^2}

\vec{r}(t) = \left[\left(x_0 - \frac{g}{\omega^2}\right)\cos\omega t + \frac{g}{\omega^2}\right]\hat{i}

d. (i) Returning to equation (B2-1)

\vec{a} = -\frac{\rho^2}{3\varepsilon_0\delta}\vec{r} = -\omega^2\vec{r} \qquad \text{where} \quad \omega = \frac{\rho}{\sqrt{3\varepsilon_0\delta}}

Writing

\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}

yields a separate simple harmonic motion equation for each component

a_x = -\omega^2x \qquad a_y = -\omega^2y \qquad a_z = -\omega^2z

The direction of the ball’s initial displacement is $\vec{r}_0 = x_0\hat{i}$ , and the direction of its initial velocity is $\vec{v}_0 = v_0\hat{j}$ . Examining these initial conditions, we must have

x(t) = x_0\cos\omega t \qquad y(t) = \frac{v_0}{\omega}\sin\omega t \qquad z(t) = 0.

Eliminating $t$ , we have

\frac{x(t)^2}{x_0^2} + \frac{\omega^2y(t)^2}{v_0^2} = \cos^2\omega t + \sin^2\omega t = 1.

Therefore the path is an ellipse.

(ii) The maximum radius reached by the ball is the semimajor axis of the ellipse, which is the greater of $x_0$ and $\frac{v_0}{\omega}$ . Since we know that $x_0 < R$ , we only need $\frac{v_0}{\omega} < R$ for the ball to avoid the wall; that is, the ball can avoid the wall so long as $v_0 < \omega R$ .

(iii) Since $x$ , $y$ , and $z$ all oscillate at the same frequency, the orbit must be periodic with that frequency, i.e. with period $\frac{2\pi}{\omega} = 2\pi\frac{\sqrt{3\varepsilon_0\delta}}{\rho}$ .

Оригинал в Архива: Semifinal 2003.pdf · официални решения: Semifinal 2003 Solutions.pdf