Съдържание

OPhO 2020, Открит кръг — Задача 54. Fun with a String

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OPhO 2020

2020 Online Physics Olympiad (OPhO): Open Contest v1.5 · 25 май 2020 г.

2020 Online Physics Olympiad (OPhO): Open Contest v1.5

OPhO Lead Organizers and Writers: Adithya Balachandran, Eddie Chen, Ashmit Dutta, Ethan Hu, Viraj Jayam, Rishab Parthasarathy, Kushal Thaman, Max Wang, QiLin Xue

OPhO Website Developer: Ben Oommen

Advisors: Dr. Branislav Kisačanin, Kye Shi, Kevin Zhou

Указания

If you wish to request a clarification, please use this form. To see all clarifications, see this document.

Use $g = 9.81\ \mathrm{m/s^2}$ in this contest. See the constants sheet on the following page for other constants.
This test contains 55 short answer questions. Each problem will have three possible attempts.
The weight of each question depends on our scoring system found here. Put simply, the later questions are worth more, and the overall amount of points from a certain question decreases with the number of attempts and days that you take to solve a problem as well as the number of teams who solve it. This means that your score decreases with the number of tries and days you take to solve a given problem.
Any team member is able to submit an attempt. Choosing to split up the work or doing each problem together is up to you. Note that after you have submitted an attempt, your teammates must refresh their page before they are able to see it.
Answers should contain at least three significant figures, unless otherwise specified. All answers within the 1% range will be accepted.
When submitting a response using scientific notation, please use exponential form. In other words, if your answer to a problem is $A \times 10^B$ , please type $AeB$ into the submission portal.
A standard scientific or graphing handheld calculator may be used. Technology and computer algebra systems like Wolfram Alpha or the one in the TI nSpire will not be needed. Attempts to use these tools will be classified as cheating.
You are allowed to use Wikipedia or books in this exam. Asking for help on online forums or your teachers will be considered cheating and may result in a possible ban from future competitions.
Top scorers from this contest will qualify to compete in the Online Physics Olympiad Invitational Contest, which is an olympiad-style exam. More information will be provided to invitational qualifiers after the end of the Open Contest.
In general, answer in SI units (meter, second, kilogram, watt) unless otherwise specified. Please answer all questions in degrees unless otherwise specified.
If the question asks to give your answer as a percent and your answer comes out to be “ $x\%$ ”, please input the value $x$ into the submission form.
Do not put letters in your answer on the submission portal! If your answer is “ $x$ meters” , input only the value $x$ into the submission portal.
Do not communicate information to anyone else apart from your team-members before the exam ends on May 29, 2020 at 11:59 PM UTC.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant,
$G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Acceleration due to gravity, $g = 9.81\ \mathrm{m/s^2}$
1 unified atomic mass unit,
$1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant,
$h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space,
$\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant,
$k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant,
$\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure,
$1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant,
$\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

A child attaches a small rock of mass $M = 0.800\ \mathrm{kg}$ to one end of a uniform elastic string of mass $m = 0.100\ \mathrm{kg}$ and natural length $L = 0.650\ \mathrm{m}$ . He grabs the other end and swings the rock in uniform circular motion around his hand, with angular velocity $\omega = 6.30\ \mathrm{rad/s}$ . Assume his hand is stationary, and that the elastic string behaves like a spring with spring constant $k = 40.0\ \mathrm{N/m}$ . After that, at time $t = 0$ , a small longitudinal perturbation starts from the child’s hand, traveling towards the rock. At time $t = T_0$ , the perturbation reaches the rock. How far was the perturbation from the child’s hand at time $t = \frac{T_0}{2}$ ? Ignore gravity.

Отговори

Решение

Solution: Let $x$ be the distance from a point on the unstretched elastic string to the center of rotation (child’s hand). Note that $x$ varies from 0 to $L$ . However, the string stretches, so let $r$ be the distance from a point on the stretched string (in steady state) to the center of rotation. Let $T$ be the tension in the string as a function of position. Let $\lambda = \frac{m}{L}$ . Consider a portion of the string $dx$ . We know that the portion as spring constant $k\frac{L}{dx}$ and it is stretched by $dr - dx$ , so by Hooke’s Law, we have $T = k\frac{L}{dx}(dr - dx) = kL(\frac{dr}{dx} - 1)$ . Also, by applying Newton’s Second Law on the portion, we get $dT = -\lambda dx\cdot\omega^2 r$ , which implies $\frac{dT}{dx} = -\lambda\omega^2 r$ . Combining the two equations, we obtain

T = -kL\left(\frac{1}{\lambda\omega^2}T'' + 1\right).

We know that

T'(x = 0) = 0,

since $r = 0$ when $x = 0$ . The general solution is

T = A\cos(\frac{\omega}{L}\sqrt{\frac{m}{k}}x) - kL,

for some constant $A$ . Thus, we have

r = -\frac{1}{\lambda\omega^2}T' = \frac{A}{\omega\sqrt{km}}\sin(\frac{\omega}{L}\sqrt{\frac{m}{k}}x).

Also, we have that

T(x = L) = M\omega^2\int_0^L rxdx = M\omega^2\int_0^L\left(\frac{T}{kL} + 1\right)x.

Plugging in our general solution, we can get $A\cos\left(\omega\sqrt{\frac{m}{k}}\right) - kL = M\omega^2\cdot\frac{A}{kL}\frac{L}{\omega}\sqrt{\frac{k}{m}}\sin\left(\omega\sqrt{\frac{m}{k}}\right)$ . Solving for $A$ , we obtain

A = \frac{kL}{\cos\left(\omega\sqrt{\frac{m}{k}}\right) - \frac{M\omega}{\sqrt{km}}\sin\left(\omega\sqrt{\frac{m}{k}}\right)}.

We now introduce a claim:

Claim. The speed of a longitudinal wave on a spring with spring constant $k$ , length $L$ , and mass $m$ is given by $v = L\sqrt{\frac{k}{m}}$

Proof. Let a spring with spring constant $k$ and mass $m$ be stretched to length $L$ . The spring constant of a small portion $dx$ of the spring is $k\frac{L}{dx}$ , and the excess tension is $\delta T = k\frac{L}{dx}ds = kL\frac{ds}{dx}$ , where $s$ is the displacement from equilibrium. By Newton’s second law on the portion, we get $dT = \frac{m}{L}dx\cdot\frac{d^2s}{dt^2}$ , or $\frac{dT}{dx} = \frac{m}{L}\frac{d^2s}{dt^2}$ . Thus, $L^2\frac{k}{m}\frac{d^2s}{dx^2} = \frac{d^2s}{dt^2}$ , which we recognize as the wave equation with speed $v = L\sqrt{\frac{k}{m}}$ and the time it takes to traverse the spring is $\sqrt{\frac{m}{k}}$ . $\square$

Thus, we have

t = \int dt = \int_0^x\sqrt{\frac{\lambda dx}{k\cdot\frac{L}{dx}}} = \int_0^x\sqrt{\frac{\lambda}{kL}}dx = \sqrt{\frac{m}{k}}\frac{x}{L}.

Since we know $x = L$ when $t = T_0$ , we have $x = \frac{L}{2}$ when $t = \frac{T_0}{2}$ . Therefore, our answer is

r = \frac{A}{\omega\sqrt{km}}\sin\left(\frac{\omega}{L}\sqrt{\frac{m}{k}}x\right) = \frac{1}{\omega}\sqrt{\frac{k}{m}}\frac{L\sin\left(\frac{1}{2}\omega\sqrt{\frac{m}{k}}\right)}{\cos\left(\omega\sqrt{\frac{m}{k}}\right) - \frac{M\omega}{\sqrt{km}}\sin\left(\omega\sqrt{\frac{m}{k}}\right)} = \boxed{1.903\ \mathrm{m}}

Оригинал в Архива: open.pdf · официални решения: open_sol.pdf