di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Mechanics

 Swoosh!



In 1851, Léon Foucault built a pendulum 67 metres tall with a 28-kg weight. He connected it to the top of the Panthéon in Paris with a bearing that enabled it to freely change its plane of oscillation. Because of the Earth's rotation, the plane of oscillation slowly moved over time: if we imagine a large horizontal clock under the pendulum, if initially the oscillations went from " 12 " to " 6 ", later on they would move to the " $3-9$ " plane, for example, as shown in the figure below. Perhaps surprisingly, the time it took the oscillations to go back to their original plane is longer than 12 hours. In this problem we will investigate why this is the case, and what the shape the pendulum traces out.

<img_4534>



Figure 1: Left: A schematic of Foucault's pendulum. Right: The pendulum motion projected on a horizontal plane in the rotating lab frame.



First, consider the case of a Foucault pendulum installed precisely at the North Pole, with length $l$. We denote $\sqrt{g / l}=\omega$. The angular velocity of the Earth is $\Omega$.



John is an observer looking at the pendulum from a fixed point in space. At $t=0$, he sees the pendulum at position $(A, 0)$ and with velocity $(0, V)$ in the $x-y$ (horizontal) plane.
Context question:
a. For John, what are the approximate equations describing the motion of the pendulum in the $x-y$ plane? You may assume that the amplitude of the oscillations is small. We define the coordinates of the pendulum at rest as $(0,0)$.
Context answer:
\boxed{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$}


Context question:
b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ?
Context answer:
$x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$


Context question:
c. Ella, an observer resides at the North Pole, is also looking at the pendulum. What are the coordinates, $\tilde{x}(t)$ and $\tilde{y}(t)$, as observed by Ella? Assume that at time $t=0$, the coordinate systems of John's and Ella's overlap.
Context answer:
\boxed{$\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t)$,$\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t)$}

Question: d. What is the speed of the pendulum bob observed by Ella at $t=0$ ?

Solution:
In John's frame, the velocity at this time is $(0, V)$. To get the velocity in Ella's frame, we can either take the derivative of the result of part (c) directly, or transform the velocity obtained in John's frame to Ella's frame, not forgetting to add the term $-\Omega A$ to the initial velocity in the $y$ axis. This gives



$$

\tilde{v}_{x} \approx(V-\Omega A) \sin (\Omega t) ; \tilde{v}_{y}=(V-\Omega A) \cos (\Omega t)

$$



At $t=0, \tilde{v}_{x}=0$ and $\tilde{v}_{y}=V-\Omega A$.



Final answer: $\tilde{v}_{x}=0$ , $\tilde{v}_{y}=V-\Omega A$
Subfield: Mechanics


Question: e. Find the initial conditions for $A, V$, such that as measured in Ella's frame:
i. the pendulum passes precisely through its resting position.

Solution:
Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$.

Final answer: Considering the motion in John's frame, clearly the pendulum will pass through the resting position if and only if $V=0$.
Subfield: Modern Physics

## Spin Cycle



Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.

Question: a. How much time elapses on Carla's stopwatch with each revolution?

Solution:
From time dilation, her clock ticks slower by a factor $\gamma$. Therefore, each revolution takes



$$

\frac{2 \pi R}{\gamma v}=\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}

$$



when measured by Carla's stopwatch.

Final answer: $\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$
Subfield: Modern Physics

## Spin Cycle



Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.
Context question:
a. How much time elapses on Carla's stopwatch with each revolution?
Context answer:
\boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$}


Extra Supplementary Reading Materials:

Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training.



For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time.

Question: b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?

Solution:
Carla's motion is perpendicular to the displacement between Clock $A$ and Clock $B$ when they are synchronized in CIRF. Therefore, the simultaneous synchronization in CIRF is also simultaneous in the lab frame. Thus, the offset is 0 .



To understand why this offset is 0 , you can also imagine placing an lightbulb halfway between the two clocks and having it send a light pulse at some known time. In both Carla's frame and the lab frame, the light pulse reaches the two clocks simultaneously.

Final answer: 0
Subfield: Modern Physics

## Spin Cycle



Cosmonaut Carla is preparing for the Intergalactic 5000 race. She practices for her race on her handy race track of radius $R$, carrying a stopwatch with her. Her racecar maintains a constant speed $v$ during her practices. For this problem, you can assume that $v>0.1 c$, where $c$ is the speed of light.
Context question:
a. How much time elapses on Carla's stopwatch with each revolution?
Context answer:
\boxed{$\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$}


Extra Supplementary Reading Materials:

Carla decides to do a fun experiment during her training. She places two stationary clocks down: Clock A at the center of the race track, i.e. the origin; and Clock B at a point on the race track denoted as $(R, 0)$. She then begins her training.



For parts (b) through (d), we define Carla's inertial reference frame (CIRF) as an inertial reference frame in which Carla is momentarily at rest, and which has the same origin of coordinates as the lab frame. Thus, CIRF is a new inertial frame each moment. The times on the clocks and stopwatch are all calibrated such that they all read 0 in CIRF when she passes by Clock $B$ for the first time.
Context question:
b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?
Context answer:
\boxed{0}


Question: c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF?

Solution:
By symmetry, the speed at which the center clock ticks according to CIRF cannot change. In one revolution, Carla's stopwatch measures $\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$, while the center clock measures $\frac{2 \pi R}{v}$. Then,





$$

t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}

$$

Final answer: $t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}$
Subfield: Mechanics

## String Cheese

Question: a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.

Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.

Solution:
We can use energy conservation to answer this question. For a bit of water with mass $m$, the total energy $E$ is the sum of the kinetic and gravitational potential energies,



$$

E=\frac{1}{2} m v^{2}+m g y \text {. } \tag{B1-1}

$$



(With this sign convention, $g \approx 10 \mathrm{~m} / \mathrm{s}^{2}$ is positive. As $y$ decreases, so does the potential energy.)



As the bit of water falls, its energy remains constant, and is equal to the initial value of



$$

E=\frac{1}{2} m v_{0}^{2}+m g y_{0} \tag{B1-2}

$$



Equating eliminating $E$ from equations B1-1 and B1-2, we have



$$

\frac{1}{2} m v^{2}+m g y=\frac{1}{2} m v_{0}^{2}+m g y_{0},

$$



and solving for $v$, we get



$$

v=\sqrt{v_{0}^{2}+2 g\left(y_{0}-y\right)}

$$



The equation for $v^{2}$ has three terms, but we were asked to choose the zero of $y$ such that there is only one. Evidently, two of the terms must cancel, and these must be the two constant terms, since the final term varies with $y$.



That means we need



$$

v_{0}^{2}+2 g y_{0}=0

$$



Solving for $y_{0}$, the vertical position of the spout is



$$

y_{0}=\frac{-v_{0}^{2}}{2 g}

$$







With this choice of the zero of $y$, the equation for $v$ simplifies to



$$

v=\sqrt{-2 g y} \tag{B1-3}

$$



We note that the result of this equation is real because $y<0$ at the spout, and decreases as the water falls, so this equation shows that $v$ is real and increases as the water falls.

Final answer: $y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$
Subfield: Mechanics

## String Cheese
Context question:
a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.

Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.
Context answer:
\boxed{$y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$}


Question: b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time.
In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$.


If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small.

Solution:
The same volume of water must fall through any horizontal cross-section of the stream each second because water doesn't disappear during its fall, and its density if constant. That volume per unit time $Q$ is the cross-sectional area of the stream multiplied by the speed of the water in the vertical direction. As an equation,



$$

Q=v \pi r^{2} \tag{B1-4}

$$



$Q$ is the same at all $y$, and is equal to its initial value of



$$

Q=v_{0} \pi r_{0}^{2} \tag{B1-5}

$$



Eliminating $Q$ from B1-4 and B1-5 and solving for $r$ gives



$$

r=r_{0} \sqrt{\frac{v_{0}}{v}}

$$



Plugging in our equation B1-3 for $v$,



$$

r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}

$$

Final answer: $r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}$
Subfield: Mechanics


Question: c. The water-air interface has some surface tension, $\sigma$. The effect of surface tension is to change the pressure in the stream according to the Young-Laplace equation,
$$

\Delta P=\sigma\left(\frac{1}{r}+\frac{1}{R}\right)

$$

where $\Delta P$ is the difference in pressure between the stream and the atmosphere and $R$ is the radius of curvature of the vertical profile of the stream, visualized below. ( $R<0$ for the stream of water; the radius of curvature would be positive only if the stream profile curved inwards.)



![](https://cdn.mathpix.com/cropped/2023_12_21_d83191acb7c4227786c9g-1.jpg?height=325&width=265&top_left_y=431&top_left_x=930)



For this part of the problem, we assume that $|R| \gg|r|$, so that the curvature of the vertical profile of the stream can be ignored. Also assume that water is incompressible.



Accounting for the pressure in the stream, find a new equation relating for $r(y)$ in terms of $\sigma, r_{0}, v_{0}$, and $\rho$, the density of water. You do not need to solve the equation for $r$.

Solution:
Our conservation of energy approach from part (b) needs to be modified to account for the work done against pressure. As we look further down in the stream, the radius is smaller. This means the pressure is higher there, and the water is slowed compared to when we assumed only gravity acted on the water.



The result of accounting for changes in pressure in a flow where no energy is dissipated is the Bernoulli equation,



$$

\frac{1}{2} \rho v^{2}+\rho g y+P=\frac{1}{2} \rho v_{0}^{2}+\rho g y_{0}+P_{0}

$$



where $P_{0}$ is the pressure in the stream at the spout.



Using the Young-Laplace equation to replace $P$ and $P_{0}$, we have



$$

\frac{1}{2} \rho v^{2}+\rho g y+\frac{\sigma}{r}=\frac{1}{2} \rho v_{0}^{2}+\rho g y_{0}+\frac{\sigma}{r_{0}}

$$



If we substitute in $y_{0}=-\frac{v_{0}^{2}}{2 g}$ and $v=v_{0} \frac{r_{0}^{2}}{r^{2}}$, this becomes



$$

\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y+\frac{\sigma}{r}=\frac{1}{2} \rho v_{0}^{2}-\rho g \frac{v_{0}^{2}}{2 g}+\frac{\sigma}{r_{0}}

$$



This may be simplified to



$$

\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y=\sigma\left(\frac{1}{r_{0}}-\frac{1}{r}\right)

$$

Final answer: $\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y=\sigma\left(\frac{1}{r_{0}}-\frac{1}{r}\right)$
Subfield: Optics

## Mirror Mirror on the Wall



Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$.



<img_4432>



*Remember that the phase of light reflected by a mirror changes by $180^{\circ}$.

Question: a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well.

Solution:
This setup is essentially a double-slit experiment with the second slit being the image of the point source on the other side of the mirror, with the additional phase shift from the mirror. The distance between the source and a spot $y$ on the wall is given by $\sqrt{(d-y)^{2}+(L / 2)^{2}}$ and the distance between the image and a spot $y$ is given by $\sqrt{(d+y)^{2}+(L / 2)^{2}}$. Subtracting the two distances and adding in the phase shift gives us approximately



$$

L / 2\left(\frac{2(d+y)^{2}}{L^{2}}-\frac{2(d-y)^{2}}{L^{2}}\right)+\lambda / 2

$$



This distance must be a multiple of $\lambda$ for interference to occur. Then,



$$

\frac{4 d y}{L}+\lambda / 2=m \lambda

$$



Substituting $m=1$ gives us $y=\frac{\lambda L}{8 d}$.

Final answer: $y=\frac{\lambda L}{8 d}$
Subfield: Optics

## Mirror Mirror on the Wall



Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$.



<img_4432>



*Remember that the phase of light reflected by a mirror changes by $180^{\circ}$.
Context question:
a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well.
Context answer:
\boxed{$y=\frac{\lambda L}{8 d}$}


Question: c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell.

hemispherical shell



<img_4523>



At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.)

Solution:
Now the optical distance between the source and a spot $y$ on the wall is increased by $(n-1) s$. Then, we need



$$

\frac{4 d y}{L}-(n-1) s+\lambda / 2=m \lambda

$$



To minimize $y$, we take $m$ to be $-\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor$. Then,



$$

y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)

$$



Because $(n-1) s$ is just an offset, the spacing between the fringes does not change, i.e., the spacing is still $\lambda L /(4 d)$.

Final answer: $y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)$
Subfield: Optics


Question: c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell.

hemispherical shell



![](https://cdn.mathpix.com/cropped/2023_12_21_051b35decfa15bf56402g-1.jpg?height=187&width=358&top_left_y=1370&top_left_x=927)



At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.)

Solution:
Now the optical distance between the source and a spot $y$ on the wall is increased by $(n-1) s$. Then, we need



$$

\frac{4 d y}{L}-(n-1) s+\lambda / 2=m \lambda

$$



To minimize $y$, we take $m$ to be $-\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor$. Then,



$$

y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)

$$



Because $(n-1) s$ is just an offset, the spacing between the fringes does not change, i.e., the spacing is still $\lambda L /(4 d)$.

Final answer: $y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)$
Subfield: Optics

## Mirror Mirror on the Wall



Consider a square room with side length $L$. The bottom wall of the room is a perfect mirror.* A perfect monochromatic point source with wavelength $\lambda$ is placed a distance $d$ above the center of the mirror, where $\lambda \ll d \ll L$.



<img_4432>



*Remember that the phase of light reflected by a mirror changes by $180^{\circ}$.
Context question:
a. On the right wall, an interference pattern emerges. What is the distance $y$ between the bottom corner and the closest bright fringe above it? Hint: you may assume $\lambda \ll y \ll L$ as well.
Context answer:
\boxed{$y=\frac{\lambda L}{8 d}$}


Context question:
c. Now suppose we place a transparent hemispherical shell of thickness $s$ and index of refraction $n$ over the source such that all light from the source that directly strikes the right wall passes through the shell, and all light from the source that strikes the mirror first does not pass through the shell.

hemispherical shell



<img_4523>



At what $y$ is the fringe closest to the bottom-most corner now? (You may find it convenient to use $\lfloor x\rfloor$, the largest integer below $x$.)
Context answer:
\boxed{$y=\frac{L}{4 d}\left((n-1) s-\lambda\left\lfloor\frac{(n-1) s}{\lambda}-\frac{1}{2}\right\rfloor-\frac{\lambda}{2}\right)$}


Question: d. Now, suppose the hemispherical shell is removed, and we instead observe the interference pattern on the top wall. To the nearest integer, what is the total number of fringes that appear on the top wall? You may assume that $d \ll L$.

Solution:
Now, the distance between the source and a spot $x$ on the wall is given by $\sqrt{(L-d)^{2}+x^{2}}$ and the distance between the image and a spot on the wall is $\sqrt{(L+d)^{2}+x^{2}}+\lambda / 2$. We do not assume $x \ll L$ this time. Subtracting the two distances gives us roughly



$$

\sqrt{L^{2}+x^{2}} \sqrt{1+\frac{2 d L}{L^{2}+x^{2}}}-\sqrt{L^{2}+x^{2}} \sqrt{1-\frac{2 d L}{L^{2}+x^{2}}}+\lambda / 2=m \lambda

$$



Taylor expanding gives us



$$

\frac{2 d L}{\sqrt{L^{2}+x^{2}}}=(m-1 / 2) \lambda

$$



Then,



$$

x= \pm L \sqrt{\frac{4 d^{2}}{(m-1 / 2)^{2} \lambda^{2}}-1}

$$



For $x$ to be physical, we require that $m-1 / 2 \leq 2 d / \lambda$.



The maximum allowed $x$ is $L / 2$. Then,



$$

\sqrt{\frac{4 d^{2}}{(m-1 / 2)^{2} \lambda^{2}}-1} \leq \frac{1}{2}

$$



so



$$

\frac{4 d^{2}}{(m-1 / 2)^{2} \lambda^{2}} \leq \frac{5}{4}

$$



Thus, we have that



$$

m-1 / 2 \geq \frac{4 d}{\sqrt{5} \lambda}

$$



Then, the number of fringes is



$$

2 \cdot \frac{2 d}{\lambda}\left(1-\frac{2}{\sqrt{5}}\right)

$$



where the extra factor of 2 comes from there being two sides to the interference pattern.

Final answer: $2 \cdot \frac{2 d}{\lambda}\left(1-\frac{2}{\sqrt{5}}\right)$
Subfield: Thermodynamics

## Real Expansion



Consider a "real" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this "real" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.

Question: a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.

Solution:
The density of the gas is given by $N / V$. In a spherical shell of radius $r$ and thickness $\Delta r$, there are $\left(4 \pi r^{2} \Delta r\right) N / V$ atoms. The potential energy is given by



$$

\Delta U=-\left(4 \pi r^{2} \Delta r\right) N / V \epsilon d^{6} / r^{6}

$$



Then, the total potential energy is given by



$$

U=\int_{d}^{\infty}-\left(4 \pi r^{2} \mathrm{~d} r\right) N / V \epsilon d^{6} / r^{6}=-2 a^{\prime} N / V

$$

Final answer: $-2 a^{\prime} N / V$
$-2\cdot \frac{2 \pi d^{3} \epsilon N}{3V}$