Съдържание

Условие Отговори Решение

USAPhO 2002, Полуфинал — Задача 5. B1

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Всички материали от USAPhO 2002

2002 Semi-Final Exam

Указания

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.
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	$k = 1/4\pi\varepsilon_o = 9.0 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_B = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_A = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_Ak_B = 8.31\ \mathrm{J/(mol\cdot K)}$
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 $\lvert x\rvert \ll 1$

Условие

Two diagrams of a semicircular hoop resting on a horizontal line. Left: the hoop in equilibrium, with a radius arrow labelled R from the center of curvature to the hoop and a vertical arrow labelled 2R/π from the center of curvature down to a cross marked cm. Right: the hoop tilted, touching the line at a point below the center of curvature, with the radius arrow R, a vertical line from the center of curvature to the contact point, and a tilted arrow 2R/π to the cross marked cm.

A thin semicircular hoop of radius $R$ and mass $m$ rocks back and forth without slipping on a rough surface. The center-of-mass of this hoop lies a distance $2R/\pi$ below its center of curvature when it is in its equilibrium position. See diagrams above.

a. What is the semicircular hoop’s moment of inertia about its center of mass? [10 т.]

b. Find $T_{\mathrm{no\ slip}}$ the period of oscillation for small amplitude oscillations. Ignore all terms higher than first order in the oscillation angle $\theta$ . [20 т.]

c. The hoop now undergoes oscillations on a frictionless surface with the period of oscillation $T_{\mathrm{slip}}$ . Ignore all terms higher than first order in the oscillation angle $\theta$ . Find the ratio of $T_{\mathrm{slip}}$ to $T_{\mathrm{no\ slip}}$ . [20 т.]

Отговори

Решение

(10) a. The semicircular hoop’s moment of inertia about its center is $I = mR^2$ .

Using the parallel axis theorem to find the moment of inertia about the center of mass $I_{cm}$

I = I_{cm} + mh^2

where $h$ is the perpendicular distance between an axis through the center of mass and a parallel axis through an arbitrary point.

In this case

h = \frac{2R}{\pi}

I_{cm} = I - mh^2 = mR^2 - m\left(\frac{2R}{\pi}\right)^2 = mR^2\left(1 - \frac{4}{\pi^2}\right)

(20) b. Finding the moment of inertia about the stationary point, the point where the hoop is in contact with the surface.

I = I_{cm} + mL^2 = mR^2\left(1 - \frac{4}{\pi^2}\right) + mL^2.

Using the Law of cosines to find $L$

L^2 = R^2 + \left(\frac{2R}{\pi}\right)^2 - \frac{4R^2}{\pi}\cos\theta.

Substituting into the expression for $I$

I = mR^2\left(1 - \frac{4}{\pi^2}\right) + m\left(R^2 + \left(\frac{2R}{\pi}\right)^2 - \frac{4R^2}{\pi}\cos\theta\right)

I = mR^2\left(1 - \frac{4}{\pi^2} + 1 + \frac{4}{\pi^2} - \frac{4}{\pi}\cos\theta\right) = mR^2\left(2 - \frac{4}{\pi}\cos\theta\right).

For small amplitude oscillations (keeping only terms through first order in $\theta$

\cos\theta \approx 1 \qquad \text{and} \qquad I \approx 2mR^2\left(1 - \frac{2}{\pi}\right).

The restoring torque about the stationary point is

\tau = -mgd = -mg\left(\frac{2R}{\pi}\right)\sin\theta \approx -\frac{2mgR}{\pi}\theta,

where the small angle approximation $\sin\theta \approx \theta$ has been used.

Combining the last two equations in $\tau = I\alpha$ ,

yields

-\frac{2mgR}{\pi}\theta \approx 2mR^2\left(1 - \frac{2}{\pi}\right)\alpha.

Comparing this with the defining equation of simple harmonic motion $-\omega^2x = a$ , gives an expression for $\omega^2$ .

\omega^2 = \frac{2mgR}{\pi 2mR^2\left(1 - \frac{2}{\pi}\right)} = \frac{g}{R\pi\left(1 - \frac{2}{\pi}\right)} = \frac{g}{R(\pi - 2)}.

For the period $T_{no\ slip}$

T_{no\ slip} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{g}{R(\pi - 2)}}} = 2\pi\sqrt{\frac{R(\pi - 2)}{g}}

(20) c. There is no horizontal force. The center of mass does not move from side to side. The y-coordinate of the center of mass is

y = R - \frac{2R}{\pi}\cos\theta = R - \frac{2R}{\pi}\left(1 - \frac{\theta^2}{2} + \ldots\right) \approx R - \frac{2R}{\pi}

including only terms through first order in $\theta$ . To this approximation, the vertical position of the center of mass is constant and its acceleration is zero as well. Thus

\sum F_y = N - mg = ma_y \approx 0

N \approx mg.

Taking torques about the center of mass

\tau = -Nd = -N\frac{2R}{\pi}\sin\theta \approx -\frac{2mgR}{\pi}\theta

where we have once again used the small angle approximation $\sin\theta \approx \theta$ .

\tau = I_{cm}\alpha

-\frac{2mgR}{\pi}\theta \approx mR^2\left(1 - \frac{4}{\pi^2}\right)\alpha

Comparing this with the defining equation of simple harmonic motion $-\omega^2x = a$ , gives an expression for $\omega^2$ .

\omega^2 = \frac{2mgR}{\pi mR^2\left(1 - \frac{4}{\pi^2}\right)} = \frac{2g}{R\pi\left(1 - \frac{4}{\pi^2}\right)} = \frac{2g\pi}{R\left(\pi^2 - 4\right)}.

For the period $T_{slip}$

T_{slip} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{2g\pi}{R\left(\pi^2 - 4\right)}}} = 2\pi\sqrt{\frac{R\left(\pi^2 - 4\right)}{2g\pi}}.

The ratio of periods is

\frac{T_{slip}}{T_{no\ slip}} = \frac{2\pi\sqrt{\frac{R\left(\pi^2 - 4\right)}{2g\pi}}}{2\pi\sqrt{\frac{R(\pi - 2)}{g}}} = \sqrt{\frac{R\left(\pi^2 - 4\right)g}{2g\pi R(\pi - 2)}} = \sqrt{\frac{(\pi - 2)(\pi + 2)}{2\pi(\pi - 2)}} = \sqrt{\frac{\pi + 2}{2\pi}}

Tilted semicircular hoop on a horizontal surface: a vertical line from the center of curvature to the contact point, a line of length 2R/π at angle θ from the vertical to the center of mass (cm), the weight arrow mg at the cm, the horizontal lever arm d, the radius R and the distance L from the cm to the contact point.

Tilted semicircular hoop on a frictionless surface: the line of length 2R/π at angle θ from the vertical to the center of mass (cm), the weight arrow mg, the normal force N pointing up at the contact point, the lever arm d and the radius R.

Автори

The 2002 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Warren Turner – Physics Teacher at Massachusetts Academy of Math and Science, Worcester, MA; Robert Shurtz – Physics Teacher at the Hawken School, Gates Mills, OH; Boris Zbarsky –undergraduate student at MIT, Cambridge, MA; Andrew Lin – undergraduate student at Yale University, New Haven, CT. Questions A2, A4, B1, and B2 were suggested by Leaf Turner and Questions A1 and A3 by Mary Mogge.

Оригинал в Архива: Semifinal 2002.pdf · официални решения: Semifinal 2002 Solutions.pdf