Съдържание

Условие Решение

USAPhO 2006, Полуфинал — A1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2006

2006 Semi-Final Exam · 25 т.

2006 Semi-Final Exam — Instructions

AAPT / AIP — United States Physics Team 2006.

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 March 20th.
Good luck!

Possibly Useful Information – (Use for both part A and for part B)

Quantity	Value
Gravitational field at the Earth’s surface	$g=9.8$ N/kg
Newton’s gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb’s constant	$k=1/4\pi\epsilon_0=8.99\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.00\times10^8$ m/s
Boltzmann’s constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro’s number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/(mol·K)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.602\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.602\times10^{-19}$ J
Planck’s constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.109\times10^{-31}\text{ kg}=0.511\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\approx1+nx$ for $\lvert x\rvert\ll1$
Small angle approximations	$\sin\theta\approx\theta$ , $\cos\theta\approx1-\tfrac12\theta^2$

Условие

An otherwise uniform disk of radius $R$ has a circular hole of radius $R/2$ cut out so that the hole touches both the center and the edge of the disk. The disk has a mass $M$ after the hole is cut out. The disk is placed above a wheel so that when the lower wheel rotates, the disk will rotate with a constant angular velocity. The disk is constrained so that it can only move in the up and down direction or rotate freely. If the angular velocity of the disk exceeds a certain value, $\omega_{\max}$ , the disk will bounce up and down on the lower wheel. Find the total kinetic energy of the disk when the angular velocity is $\omega_{\max}$ . Express your answer in terms of $M$ , $R$ , and $g$ (but not in terms of $\omega_{\max}$ ).

Disk with an off-center circular hole resting on a lower wheel.

Решение

Source note: the key says “to the right” for the center of mass, while its algebra uses $-r$ and the question figure places the hole on the right. The printed wording is retained.

Sooner or later we need to find both the location of the center of mass $r$ and the rotational inertia $I$ about the axis of rotation. Let’s do these two things first.

We can treat the top disk as a solid disk of mass $M_s$ with a hole of mass $M_h$ cut out. Since the radius of the hole is half the radius of the disk, we can conclude that

M_h=\frac{\pi(R/2)^2}{\pi R^2}M_s=\frac14M_s,\qquad (A1-1)

M=M_s-M_h=\frac34M_s=3M_h.\qquad (A1-2)

By symmetry the center of mass will be on a line connecting the centers of both the hole and the disk, and must be located a distance $r$ to the right of the center of the disk, where

M_s(0)=M(-r)+M_h\frac R2.\qquad (A1-3)

Then

r=\frac R6.\qquad (A1-4)

We can apply a similar approach to find the rotational inertia about the axis of rotation:

I_s=I+I_h,\qquad (A1-5)

where $I_s$ is the rotational inertia of a solid disk and $I_h$ is the rotational inertia of the part that used to be in the hole, both measured about the axis of rotation. $I_s$ is easy,

I_s=\frac12M_sR^2=\frac23MR^2,\qquad (A1-6)

while $I_h$ requires an application of the parallel axis theorem,

I_h=\frac12M_h\left(\frac R2\right)^2+M_h\left(\frac R2\right)^2=\frac18MR^2.\qquad (A1-7)

Finally,

I=I_s-I_h=\frac{13}{24}MR^2.\qquad (A1-8)

Now consider the motion of the center of mass. It is rotating about the axis of rotation with an angular speed $\omega$ , so that the magnitude of the acceleration of the center of mass is given by

a=r\omega^2.\qquad (A1-9)

The net force on the disk when the center of mass is at the top is then given by

F_{\rm net}=Mg-F_N=Mr\omega^2,\qquad (A1-10)

where $F_N$ is the normal force. The disk begins to hop when the normal force is zero, so

\omega^2=\frac gr=\frac{6g}{R}.\qquad (A1-11)

The total kinetic energy is then

K=\frac12I\omega^2=\frac12\left(\frac{13}{24}MR^2\right)\left(\frac{6g}{R}\right)=\frac{13}{8}MgR.\qquad (A1-12)

Оригинал в Архива: Semifinal 2006.pdf · официални решения: Semifinal 2006 Solutions.pdf