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

## Hot Pocket
a. It's winter and you want to keep warm. The temperature is $T_{0}=263 \mathrm{~K}$ outside and $T_{1}=290 \mathrm{~K}$ in your room. You have started a fire, which acts as a hot reservoir at temperature $T_{2}=1800 \mathrm{~K}$.



You want to add a small amount of heat $d Q_{1}$ to your room. The simplest method would be to extract heat $-d Q_{2, \mathrm{dump}}=d Q_{1}$ from the fire and directly transfer it to your room. However, it is possible to heat your room more efficiently. Suppose that you can exchange heat between any pair of reservoirs. You cannot use any external source of work, such as the electrical grid, but the work extracted from running heat engines can be stored and used without dissipation.

Question: i. What is the minimum heat extraction $-d Q_{2, \text { min }}$ required by the laws of thermodynamics to heat up the room by $d Q_{1}$ ?

Solution:
The second law of thermodynamics implies that, no matter what you do, you must have $\mathrm{d} S_{\text {universe }} \geq 0$, and if your process is to be as efficient as possible, we can assume it is reversible, so



$$

\mathrm{d} S_{\text {universe; reversible }}=0

$$



If we do extract any work while allowing heat to transfer between reservoirs, we will later use that work to transfer more heat. So in the entire process, there are only heat transfers, and by conservation of energy,



$$

\mathrm{d} Q_{0}+\mathrm{d} Q_{1}+\mathrm{d} Q_{2}=0

$$



The entropy change associated with each reversible heat transfer is $d S=d Q / T$, so our assumption of zero entropy production becomes



$$

\frac{\mathrm{d} Q_{0}}{T_{0}}+\frac{\mathrm{d} Q_{1}}{T_{1}}+\frac{\mathrm{d} Q_{2}}{T_{2}}=0

$$



By combining these equations, we can eliminate $d Q_{0}$ and solve for $d Q_{2}$, giving



$$

-\mathrm{d} Q_{2 ; \min }=\frac{T_{2}}{T_{1}} \frac{T_{1}-T_{0}}{T_{2}-T_{0}} \mathrm{~d} Q_{1}

$$



For the provided numbers, this happens to be about $0.11 d Q_{1}$. That is, a heat pump can be much more efficient than direct heating. This problem was inspired by Jaynes, E. T, "Note on thermal heating efficiency.", American Journal of Physics 71.2 (2003): 180-182. (You can also solve the problem by considering an explicit procedure using Carnot engines. But since Carnot engines are reversible, all such procedures will just give the same answer.)

Final answer: $\frac{T_{2}}{T_{1}} \frac{T_{1}-T_{0}}{T_{2}-T_{0}} \mathrm{~d} Q_{1}$
Subfield: Thermodynamics

b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection.



Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$.



You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$.

Question: i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.

Solution:
The temperature of the air is higher than its surroundings by a fractional amount of order $\Delta T / T$. Thus, by the ideal gas law, the density is lower than its surroundings by a fraction of order $\Delta T / T$, which means the upward acceleration due to the buoyant force is of order $a=g \Delta T / T$. Since this is roughly uniformly accelerated motion, $v_{0}^{2} \propto a h$,







which implies



$$

v_{0} \sim \sqrt{g h \frac{\Delta T}{T_{0}}}

$$



Note that because $d P / d z=-\rho g$ in hydrostatic equilibrium, the pressure of the surrounding air varies between the bottom and top of the container, by a fractional amount of order $\rho_{0} g h / P_{0}$. But since we assumed $\rho_{0} g h / P_{0} \ll \Delta T / T_{0}$, we can neglect this effect.

Final answer: $\sqrt{g h \frac{\Delta T}{T_{0}}}$
Subfield: Thermodynamics

b. When the air at the bottom of a container is heated, it becomes less dense than the surrounding air and rises. Simultaneously, cooler air falls downward. This process of net upward heat transfer is known as convection.



Consider a closed, rectangular box of height $h$ filled with air initially of uniform temperature $T_{0}$. Next, suppose the bottom of the box is heated so that the air there instantly reaches temperature $T_{0}+\Delta T$. The hot parcel of air at the bottom rises upward until it hits the top of the box, where its temperature is instantly reduced to $T_{0}$.



You may neglect any heat transfer and friction between the parcel of air and the surrounding air, and assume that the temperature difference is not too large. In addition, you may assume the height $h$ is small enough so that the pressure $P_{0}$ and density $\rho_{0}$ of the surrounding air are very nearly constant throughout the container. More precisely, assume that $\rho_{0} g h / P_{0} \ll \Delta T / T_{0} \ll 1$. Express your answers in terms of $P_{0}, g, h, \Delta T$, and $T_{0}$.
Context question:
i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.
Context answer:
\boxed{$\sqrt{g h \frac{\Delta T}{T_{0}}}$}


Question: ii. In the steady state, warm parcels of air are continuously moving upward from the bottom, and cold parcels of air are continuously moving downward from the top. Find a rough estimate for the net rate of upward energy transfer per area.

Solution:
The extra energy carried by a parcel of gas is



$$

n C_{p} \Delta T \sim n R \Delta T \sim P_{0} V \frac{\Delta T}{T_{0}}

$$



where $V$ is the volume of the parcel. The net volume of warm air transported upward per unit time is of order $A v_{0}$, where $A$ is the cross-sectional area of the box. Thus, the average power per area is roughly



$$

P_{0} v_{0} \frac{\Delta T}{T_{0}} \sim P_{0} \sqrt{g h}\left(\frac{\Delta T}{T_{0}}\right)^{3 / 2}

$$



This is a simplified version of the mixing length theory of convection, which is essential for modeling the interiors of stars.

Final answer: $P_{0} v_{0} \frac{\Delta T}{T_{0}}$
$ P_{0} \sqrt{g h}\left(\frac{\Delta T}{T_{0}}\right)^{3 / 2}$
Subfield: Mechanics


Question: i. Indicate which of the following trajectories the mass takes for a short time after $t=0$ and briefly explain your reasoning. The differences between the paths are exaggerated.



![](https://cdn.mathpix.com/cropped/2023_12_21_de76d5258eb7a0a3064dg-1.jpg?height=372&width=664&top_left_y=2023&top_left_x=771)

Solution:
The correct path is option $C$. Several explanations would work here. Here are two.



- If we go to the rotating frame of reference, there is an outward centrifugal force that the mass experiences, pushing it down the sphere.

- For the mass to go in a circle around the sphere, the spring force not only has to compensate for the gravitational force but also must provide centripetal acceleration. Therefore, the spring must get longer.

Final answer: Several explanations would work here. Here are two.


- If we go to the rotating frame of reference, there is an outward centrifugal force that the mass experiences, pushing it down the sphere.

- For the mass to go in a circle around the sphere, the spring force not only has to compensate for the gravitational force but also must provide centripetal acceleration. Therefore, the spring must get longer.
Subfield: Mechanics


Question: c. A cylinder of radius $r \ll R \theta_{0}$ is placed on top of the sphere. Suppose the mass is launched at an angle $\alpha$ away from the direction of the spring's displacement with kinetic energy $K$, as shown.



What is the maximum angle $\alpha_{\max }$ at which the mass can be launched such that it can still hit the cylinder? Express your answer in terms of $K, m, g, \theta_{0}, r$, and $R$. You may assume $K$ is large enough for the mass to reach the cylinder for $\alpha=0$.



![](https://cdn.mathpix.com/cropped/2023_12_21_ab96511c90603f8e400bg-1.jpg?height=615&width=618&top_left_y=408&top_left_x=775)



(view from above)

Solution:
The initial energy of the system is given by



$$

m g R \cos \theta_{0}+\frac{1}{2} k R^{2} \theta_{0}^{2}+K

$$



Suppose the mass is launched with speed $v$. Then, the speed in the $\theta$ direction is $v \cos \alpha$ and the speed in the $\phi$ direction is $v \sin \alpha$, and therefore, the $z$-component of the angular momentum of the mass is



$$

L=m v \sin \alpha(R \sin \theta)

$$



We now compute the distance of closest approach. If the distance of closest approach is equal to $r$ (as it does for $\alpha_{\max }$, then at $r$, the motion of the mass has no inward component, and the speed of the object at $r$ is given by conservation of angular momentum:



$$

m u r=m v \sin \alpha(R \sin \theta) \Longrightarrow u=\frac{v \sin \alpha(R \sin \theta)}{r}

$$



Because $r \ll R$, at the point of closest approach, the energy of the system is roughly



$$

m g R+\frac{1}{2} m u^{2} \approx m g R+\frac{1}{2} \frac{m v^{2} \sin ^{2} \alpha R^{2} \sin ^{2} \theta}{r^{2}} \approx m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta}{r^{2}}\right)

$$



Equating with the initial energy gives us



$$

m g R \cos \theta_{0}+\frac{1}{2} k R^{2} \theta_{0}^{2}+K=m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta}{r^{2}}\right)

$$



Before finishing the calculation, we now compute the required $k$ for the object to be at equilibrium (since our answer cannot contain $k$ ). Setting $k R \theta_{0}=m g \sin \theta_{0}$ gives us $k=$







$\frac{m g \sin \theta_{0}}{R \theta_{0}}$. Then,



$$

m g R \cos \theta_{0}+\frac{1}{2} m g R \theta_{0} \sin \theta_{0}+K=m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta_{0}}{r^{2}}\right)

$$



Solving for $\alpha$ gives us



$$

\alpha=\frac{r}{R \sin \theta_{0}} \sqrt{1-\frac{m g R\left(1-\cos \theta_{0}\right)-(1 / 2) m g R \theta_{0} \sin \theta_{0}}{K}}

$$



Solving for $\alpha$ without using the small angle approximation for $\alpha$ also earned full credit. (The answer $\alpha=\pi$, which is technically also correct, earned partial credit.)

Final answer: $\alpha=\frac{r}{R \sin \theta_{0}} \sqrt{1-\frac{m g R\left(1-\cos \theta_{0}\right)-(1 / 2) m g R \theta_{0} \sin \theta_{0}}{K}}$
Subfield: Mechanics

## Braking up



An infinitely long wire with linear charge density $-\lambda$ lies along the $z$-axis. An infinitely long insulating cylindrical shell of radius $a$ is concentric with the wire and can rotate freely about the $z$-axis. The shell has moment of inertia per unit length $I$. Charge is uniformly distributed on the shell, with surface charge density $\frac{\lambda}{2 \pi a}$.



The system is immersed in an external magnetic field $B_{0} \hat{\mathbf{z}}$, and is initially at rest. Starting at $t=0$, the external magnetic field is slowly reduced to zero over a time $T \gg a / c$, where $c$ is the speed of light.



<img_4365>

Question: a. Find an expression of the final angular velocity $\omega$ of the cylinder in terms of the symbols given and other constants.

Solution:
From Faraday's law, you can find the induced electric field inside the cylinder at a distance $r$ from the wire:



$$

\begin{aligned}

\oint \vec{E}_{\text {ind }} \cdot d \vec{\ell} & =-\frac{d \Phi_{B}}{d t} \tag{A1-1}
\end{aligned}
$$
$$
\begin{aligned}

E_{\text {ind }}(r) & =-\frac{r}{2} \frac{d B}{d t} \tag{A-2}

\end{aligned}

$$



This induced field exerts a torque on the cylinder, causing it to rotate:



$$

\begin{aligned}

\tau & =2 \pi a \cdot \frac{\lambda}{2 \pi a} \cdot E_{\text {ind }}(a) \cdot a=I \cdot \frac{d \omega}{d t}
\end{aligned} \tag{A1-3}
$$
$$
\begin{aligned}

& \Rightarrow \frac{d \omega}{d t}=-\frac{\lambda a^{2}}{2 I} \cdot \frac{d B}{d t}
\tag{A1-4}
\end{aligned}

$$



Integrate on both sides, and noting that $\omega(t=0)=0$, we have:



$$

\omega(T)=-\frac{\lambda a^{2}}{2 I}\left(B(T)-B_{0}\right) \tag{A1-5}

$$



It is important to note that $B(T) \neq 0$. Even though the external field decreases to zero, the now-rotating charged cylinder generates a magnetic field. Using Ampere's Law, you can find at $t=T$, the magnetic field is:



$$

\begin{gathered}

\oint \vec{B}_{\mathrm{ind}} \cdot d \vec{\ell}=\mu_{0} I_{\mathrm{enc}},
\end{gathered} \tag{A1-6}
$$
$$
\begin{gathered}

\text { where } I_{\mathrm{enc}}=\frac{\lambda}{2 \pi a} \cdot \omega(T) \cdot a 
\end{gathered} \tag{A1-7}
$$
$$
\begin{gathered}

\Rightarrow B(T)=\mu_{0} \frac{\lambda}{4 \pi^{2} a} \omega(T)

\end{gathered} \tag{A1-8}

$$



Combining equations A1-5 and A1-8, we have:



$$

\omega(T)=\frac{\frac{\lambda a^{2}}{2 I} B_{0}}{1+\mu_{0} \frac{\lambda^{2} a}{8 \pi I}}

$$

Final answer: $\omega(T)=\frac{\frac{\lambda a^{2}}{2 I} B_{0}}{1+\mu_{0} \frac{\lambda^{2} a}{8 \pi I}}$
Subfield: Mechanics

## Braking up



An infinitely long wire with linear charge density $-\lambda$ lies along the $z$-axis. An infinitely long insulating cylindrical shell of radius $a$ is concentric with the wire and can rotate freely about the $z$-axis. The shell has moment of inertia per unit length $I$. Charge is uniformly distributed on the shell, with surface charge density $\frac{\lambda}{2 \pi a}$.



The system is immersed in an external magnetic field $B_{0} \hat{\mathbf{z}}$, and is initially at rest. Starting at $t=0$, the external magnetic field is slowly reduced to zero over a time $T \gg a / c$, where $c$ is the speed of light.



<img_4365>
Context question:
a. Find an expression of the final angular velocity $\omega$ of the cylinder in terms of the symbols given and other constants.
Context answer:
\boxed{$\omega(T)=\frac{\frac{\lambda a^{2}}{2 I} B_{0}}{1+\mu_{0} \frac{\lambda^{2} a}{8 \pi I}}$}


Question: b. You may be surprised that the expression you find above is not zero! However, the electric and magnetic fields can have angular momentum. Analogous to the "regular" angular momentum definition, the EM field angular momentum per unit volume at a displacement $\mathbf{r}$ from the axis of rotation is:


$$

\mathcal{L}(\mathbf{r})=\mathbf{r} \times \mathcal{P}(\mathbf{r})

$$



$\mathcal{P}(\mathbf{r})$ is a vector analogous to momentum, given by



$$

\mathcal{P}(\mathbf{r})=\alpha \cdot(\mathbf{E}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})) .

$$



where $\alpha$ is some proportionality constant. Find an expression for $\alpha$ in terms of given variables and fundamental constants.

Solution:
The electric field inside the cylindrical shell is given by $\mathbf{E}(r)=-\frac{\lambda}{2 \pi \epsilon_{0} r} \hat{\mathbf{r}}$ inward. The magnetic field is given by $B(t) \hat{\mathbf{z}}$. Then:



$$

\mathcal{P}(\mathbf{r})=\alpha \frac{\lambda B(t)}{2 \pi \epsilon_{0} r} \hat{\theta}

$$



The angular momentum per unit volume is then:



$$

\mathcal{L}(\mathbf{r})=-\alpha \frac{\lambda}{2 \pi \epsilon_{0}} \hat{\mathbf{z}}

$$



The angular momentum per unit length is then:



$$

\mathbf{L}=-\frac{\alpha \lambda B(t) a^{2}}{2} \hat{\mathbf{z}}

$$



Comparing this to Equation shows that $\alpha=\epsilon_{0}$.

Final answer: $\alpha=\epsilon_{0}$
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.

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 $(0,0)$.

Solution:
John's reference frame is inertial and the point of attachment stationary, so this is a freemoving pendulum obeying simple harmonic motion in each axis:



$$

a_{x}+\omega^{2} x=0 ; a_{y}+\omega^{2} y=0 \tag{A2-1}

$$

Final answer: $a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$
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{$a_{x}+\omega^{2} x=0 , a_{y}+\omega^{2} y=0$}

Question: b. What will the coordinates $x, y$ in Jonh's frame be at a later time $t$ ?

Solution:
The solution to Eq. A2-1) is the familiar simple harmonic motion. In general, if the displacement is $r=A \cos \omega t$, then the velocity is $v=-A \omega \sin \omega t$. Using the inital conditions







provided, we have:



$$

x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t) \tag{A2-2}

$$



Note that this corresponds to an ellipse.



Final answer: $x(t)=A \cos (\omega t), y(t)=\frac{V}{\omega} \sin (\omega t)$
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:
$x(t)=A \cos (\omega t) ; y(t)=\frac{V}{\omega} \sin (\omega t)$


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.

Solution:
In the rotating frame, we have $\tilde{x}=x \cos (\Omega t)+y \sin (\Omega t), \tilde{y}=-x \sin (\Omega t)+y \cos (\Omega t)$ (with $2 \pi / \Omega=24 \mathrm{hrs})$. Plugging in the form of $x(t)$ and $y(t)$ we find:



$$

\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t) \tag{A2-3}

$$



and:



$$

\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t) \tag{A2-4}

$$

Final answer: $\tilde{x}=A \cos (\omega t) \cos (\Omega t)+\frac{V}{\omega} \sin (\omega t) \sin (\Omega t)$,$\tilde{y}=-A \cos (\omega t) \sin (\Omega t)+\frac{V}{\omega} \sin (\omega t) \cos (\Omega t)$