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

- 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, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A raindrop of mass $M=0.035 \mathrm{~g}$ is at height $H=2 \mathrm{~km}$ above a large lake. The raindrop then falls down (without initial velocity), mixing and coming to equilibrium with the lake. Assume that the raindrop, lake, air, and surrounding environment are at the same temperature $T=300 \mathrm{~K}$. Determine the magnitude of entropy change associated with this process (in $J / K$ ).

Solution:
The total heat gain is equal to the change in potential energy of the raindrop, which spreads through out the whole environment at thermally equilibrium temperature $T$ (the environment is very large so any change in $T$ is negligible). The entropy gain $\Delta S$ is thus generated by the dissipation of this potential energy $M g H$ to internal energy $\Delta U$ in the environment (given that the specific volume of water doesn't change much, $\Delta U \approx M g H$ ). Hence the entropy change associated with this process can be estimated by:
$$
S=\frac{\Delta U}{T} \approx \frac{M g H}{T} \approx 2.29 \times 10^{-3} \mathrm{~J} / \mathrm{K}
\tag{8}
$$

Final answer: $2.29 \times 10^{-3}$ J/K
Subfield: Electromagnetism


Question: The following information applies for the next two problems. A circuit has a power source of $\mathcal{E}=5.82 \mathrm{~V}$ connected to three elements in series: an inductor with $L=12.5 \mathrm{mH}$, a capacitor with $C=48.5 \mu \mathrm{F}$, and a diode with threshold voltage $V_{0}=0.65 \mathrm{~V}$. (Of course, the polarity of the diode is aligned with that of the power source.) You close the switch, and after some time, the voltage across the capacitor becomes constant. (Note: An ideal diode with threshold voltage $V_{0}$ is one whose IV characteristic is given by $I=0$ for $V<V_{0}$ and $V=V_{0}$ for $I>0$.)

![](https://cdn.mathpix.com/cropped/2023_12_21_1c0bc651808d926dacd3g-1.jpg?height=347&width=618&top_left_y=1450&top_left_x=751)



How much time (in seconds) has elapsed before the voltage across the capacitor becomes constant?

Solution:
When current is flowing clockwise, the circuit is equivalent to an LC circuit with a power source $\mathcal{E}-V_{0}$. Thus, the voltage $U$ is sinusoidal about its equilibrium voltage $U_{0}=\mathcal{E}-V_{0}$ with frequency $\omega=1 / \sqrt{L C}$.

When the switch is closed, $I=0$ and $U=U_{\min }=0$. Afterwards, $I$ increases to $I_{\max }$ and decreases back to 0 , completing half a period of a sine wave. However, $I$ can not go negative due to the presence of the diode. Instead, a reverse voltage builds up on the diode (so that the voltage across the inductor becomes 0 ), and $I$ stays at 0 . At this point, $U$ becomes constant as well.

The time it took until the system became static was half a period of the LC circuit oscillation, i.e.,

$$
\frac{\pi}{\omega}=\pi \sqrt{L C}=2.446 \times 10^{-3} \mathrm{~s}
$$

Final answer: $2.446 \times 10^{-3}$ s
Subfield: Electromagnetism


Question: At Hanoi-Amsterdam High School in Vietnam, every subject has its own flag (see Figure A, taken by Tung X. Tran). While the flags differ in color, they share the same central figure. Consider a planar conducting frame of that figure rotating at a constant angular velocity in a uniform magnetic field (see Figure B). The frame is made of thin rigid wires with same uniform curvature and same resistance per unit length. What fraction of the total heat released is released by the outermost wires?![](https://cdn.mathpix.com/cropped/2023_12_21_50c30c0db8a001b34ea6g-1.jpg?height=836&width=1002&top_left_y=1193&top_left_x=562)

Solution:
We call the current looping in the wires $I_{1}, I_{2}, I_{3}, I_{4}$ as shown in the attached Fig., and define the resistance of every quarter-circular section (radius $R$ ) of the wires to be $\rho$, then considering the



EMF on every loop gives:

$$
\begin{aligned}
& \left(2 I_{1}+\left(I_{1}-I_{2}\right)-I_{4}\right) \rho=\frac{d B_{\perp}}{d t} \times \frac{\pi+2}{2} R^{2} \\
& \left(2 I_{2}-I_{3}-\left(I_{1}-I_{2}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times \frac{\pi+2}{2} R^{2} \\
& \left(2 I_{4}+2\left(I_{1}-I_{2}-I_{3}+I_{4}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times 2 R^{2} \\
& \left(2 I_{3}-2\left(I_{1}-I_{2}-I_{3}+I_{4}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times 2 R^{2}
\end{aligned}
\tag{9}
$$

in which $B_{\perp}$ is the perpendicular component of the magnetic field.

![](https://cdn.mathpix.com/cropped/2023_12_21_6bda60f521822c3b3f7eg-1.jpg?height=781&width=786&top_left_y=778&top_left_x=664)

The set of equations in Eq. (9) can be solved to get the relation between currents:

$$
I_{1}=I_{2}, I_{3}=I_{4}, \frac{I_{1}}{I_{3}}=\frac{\pi+4}{4}
\tag{10}
$$

The fraction of heat released on the outermost wires can be calculated:

$$
\frac{Q_{\text {outermost }}}{Q_{\text {all }}}=\frac{\left(8 I_{1}^{2}+8 I_{2}^{2}\right) \rho}{\left(8 I_{1}^{2}+8 I_{2}^{2}+4 I_{3}^{2}+4 I_{4}^{2}\right) \rho}=\frac{(\pi+4)^{2}}{(\pi+4)^{2}+8} \approx 0.864 \mathrm{~J}
\tag{11}
$$

Final answer: $0.864$ J
Subfield: Electromagnetism


Question: An open electrical circuit contains a wire loop in the shape of a semi-circle, that contains a resistor of resistance $R=0.2 \Omega$. The circuit is completed by a conducting pendulum in the form of a uniform rod with length $\ell=0.1 \mathrm{~m}$ and mass $m=0.05 \mathrm{~kg}$, has no resistance, and stays in contact with the other wires at all times. All electrical components are oriented in the $y z$ plane, and gravity acts in the $z$ direction. A constant magnetic field of strength $B=2 \mathrm{~T}$ is applied in the $+x$ direction.
![](https://cdn.mathpix.com/cropped/2023_12_21_311e0a88c943ba95b98cg-1.jpg?height=463&width=1005&top_left_y=495&top_left_x=560)

Ignoring self inductance and assuming that $\alpha \ll 1$, the general equation of motion is in the form of $\theta(t)=A(t) \cos (\omega t+\varphi)$, where $A(t) \geq 0$. Find $\omega^{2}$.

Solution:
The area enclosed by the wire loop is
$$
A=\frac{1}{2} \ell^{2} \alpha+A_{0}
$$

for small angles $\alpha$, and $A_{0}$ is a constant number (which gets ignored since we really care about how this angle is changing). The flux is $\Phi=B A$ and from Lenz's Law, we have,

$$
\varepsilon=-\frac{d \Phi}{d t}=-\frac{1}{2} B \ell^{2} \dot{\alpha}
$$

One can verify that if $\alpha$ is increasing, the current will flow in the clockwise direction, so we set the counterclockwise direction as positive. The current through the wire is thus,

$$
i=\frac{\varepsilon}{R}=-\frac{B \ell^{2}}{2 R} \dot{\alpha}
$$

The magnetic force acting on it is $F_{B}=i B \ell$ and the resulting torque is

$$
\tau_{B}=F_{B} \frac{\ell}{2}=-\frac{B^{2} \ell^{4}}{4 R} \dot{\alpha}
$$

Please verify that the sign is correct. The gravitational torque is $\tau_{g}=-m g \frac{\ell}{2} \alpha$, so the torque equation gives us

$$
\begin{aligned}
0 & =\frac{1}{3} m \ell^{2} \ddot{\alpha}+\frac{B^{2} \ell^{4}}{4 R} \dot{\alpha}+m g \frac{\ell}{2} \alpha \\
0 & =\ddot{\alpha}+\frac{3}{4} \frac{B^{2} \ell^{2}}{m R} \dot{\alpha}+\frac{3}{2} \frac{g}{\ell} \alpha
\end{aligned}
$$

Recall that for a damped harmonic oscillator in the form of $\ddot{\alpha}+\gamma \dot{\alpha}+\omega_{0}^{2} \alpha=0$, the frequency of oscillations is $\omega^{2}=\omega_{0}^{2}-\gamma^{2} / 4$, so in our case, we have

$$
\omega^{2}=\frac{3}{2} \frac{g}{\ell}-\frac{9}{64}\left(\frac{B^{2} \ell^{2}}{m R} \dot{\alpha}\right)^{2}=145 \mathrm{~s}^{-1}
$$

Final answer: $145$ $\mathrm{~s}^{-1}$
Subfield: Thermodynamics


Question: For black body radiation, Wien's Displacement Law states that its spectral radiance will peak at
$$
\lambda_{\text {peak }}=\frac{b}{T}
$$

where $b=2.89777 \times 10^{-3} \mathrm{mK}$, and $T$ is the temperature of the object. When QiLin tried to reproduce this in a lab, by working with a tungsten-filament lightbulb at $2800 \mathrm{~K}$, he computed a different value for $b$ by measuring the peak wavelength using a spectrometer and multiplying it with the temperature. He hypothesizes that this discrepancy is because tungsten is not an ideal black body. The graph below, courtesy of the CRC Handbook of Chemistry and Physics, shows the emissivity of tungsten at various conditions (the units for wavelength is $\mathrm{nm}$ ).



![](https://cdn.mathpix.com/cropped/2023_12_21_d5c8b11775292675c006g-1.jpg?height=881&width=1762&top_left_y=205&top_left_x=173)

Assuming QiLin's hypothesis is correct, and assuming there were no other errors in the experiment, how off was his value for $b$ ? Submit $\frac{\left|b_{\text {theory }}-b_{\text {experiment }}\right|}{b_{\text {theory }}}$ as a decimal number, to one significant digit (giving you room to estimate where the points are).

Solution:
Recall Planck's Law, which says the spectral radiance of a black body is given by
$$
B_{0}(\lambda, T)=\frac{2 h c^{3}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{B} T}\right)-1}
$$

The regular Wien's Displacement Law can be derived by finding the peak by computing $\frac{\partial B_{0}}{\partial \lambda}$, to find the wavelength associated with the maximal radiance. For a nonideal body with emissivity $\epsilon(\lambda, T)$, we can write the radiance as

$$
B(\lambda, T)=B_{0}(\lambda, T) \epsilon(\lambda, T)
$$

We can estimate $\epsilon(\lambda, T)$ by looking at the given graph. The tungsten is at $2800 \mathrm{~K}$, so we will use the red line, and assuming it is near a black body, the peak wavelength should be around $1000 \mathrm{~nm}$. Performing a linear approximation around $1000 \mathrm{~nm}$, we get

$$
\epsilon(\lambda, T)=-173333\left(\lambda-1000 \cdot 10^{-9}\right)+0.366
$$

where $\lambda$ is in meters. Numerically finding the maximum of $B(\lambda, T)$ with respect to $\lambda$ (i.e. with a graphing calculator), we get the new peak wavelength to be $\lambda_{\text {new }}=949 \mathrm{~nm}$, while the old peak wavelength (assuming a perfect blackbody) is $\lambda_{\text {old }}=1035 \mathrm{~nm}$, and their percent difference (rounded to 1 significant digit) is 0.08

Final answer: 0.08
Subfield: Mechanics


Question: A coin of uniform mass density with a radius of $r=1 \mathrm{~cm}$ is initially at rest and is released from a slight tilt of $\theta=8^{\circ}$ onto a horizontal surface with an infinite coefficient of static friction. The coin has a thicker rim, allowing it to drop and rotate on one point. With every collision, the coin switches pivot points on the rim, and energy is dissipated through heat so that $k=0.9$ of the coin's prior total energy is conserved. How long will it take for the coin to come to a complete stop?
![](https://cdn.mathpix.com/cropped/2023_12_21_21db419ae935527412b4g-1.jpg?height=135&width=469&top_left_y=2014&top_left_x=752)

A cross-sectional view of the coin before release. The rim can be seen on the edges of the coin.

Solution:
By the parallel axis theorem, the moment of inertia of the coin around the pivot can be expressed as
$$
I=I_{x}+m \ell^{2}=\frac{1}{4} m r^{2}+m r^{2}=\frac{5}{4} m r^{2}
$$



As $\theta$ is small, that means the perpendicular force of gravity is $m g \cos \theta \approx m g$. Hence, Newton's second law to find the angular acceleration as

$$
I \alpha=\tau \Longrightarrow \alpha=\frac{4 g}{5 r} \cdot(1 \mathrm{rad})
$$

Using rotational kinematics, the time for the first collision will be $\theta=\frac{1}{2} \alpha t^{2} \Longrightarrow t_{0}=\sqrt{\frac{5 r \theta_{0}}{2 g}}$. By conservation of energy, the angular velocity at the time of collision is $\frac{1}{2} I \omega_{0}^{2}=m g r \sin \theta \approx m g r$, meaning that $\omega_{0}=\sqrt{\frac{8 g \theta_{0}}{5 r}}$. Now consider $k$. This is the ratio of the initial energy $E_{n-1}$ and next energy $E_{n}$. Then, we can say that

$$
k=\frac{\frac{1}{2} I \omega_{n}^{2}}{\frac{1}{2} I \omega_{n-1}^{2}} \Longrightarrow \omega_{n}=\sqrt{k} \omega_{n-1}
$$

By recurrence, $\omega_{n}=k^{n / 2} \omega_{0}$. The time of flight for each cycle will be $t_{n}=\frac{2 \omega_{n}}{\alpha}$. Therefore,

$$
\begin{aligned}
T & =t_{0}+\sum_{n=1}^{\infty} t_{n} \\
& =\sqrt{\frac{5 r \theta_{0}}{2 g}}+\sum_{n=1}^{\infty}\left(2 k^{n / 2} \sqrt{\frac{8 g \theta_{0}}{5 r}} \cdot \frac{5 r}{4 g}\right) \\
& =\sqrt{\frac{5 r \theta_{0}}{2 g}}+\sqrt{\frac{10 r \theta_{0}}{g}} \frac{k^{1 / 2}}{1-k^{1 / 2}}
\end{aligned}
$$

Converting $8^{\circ}$ to $\frac{2 \pi}{45}$ radians, $k=0.9, g=9.8 \mathrm{~m} / \mathrm{s}^{2}, r=0.01 \mathrm{~m}$, we find that $T=0.716 \mathrm{~s}$ which is a reasonable estimate.

Final answer: 0.716 s
Subfield: Mechanics

- 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, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Suppose all cars on a (single-lane) highway are identical. Their length is $l=4 \mathrm{~m}$, their wheels have coefficients of friction $\mu=0.7$, and they all travel at speed $v_{0}$. Find the $v_{0}$ which maximizes the flow rate of cars (i.e. how many cars travel across an imaginary line per minute). Assume that they need to be able to stop in time if the car in front instantaneously stops. Disregard reaction time.

Solution:
Suppose the maximum speed is $v^{\prime}$. Notice that $a=\mu g$ so it takes a time $t^{\prime}=\frac{v^{\prime}}{\mu g}$ amount of time to stop. This means the car will travel a distance of
$$
d=\frac{1}{2} \mu g\left(\frac{v^{\prime}}{\mu g}\right)^{2}=\frac{v^{\prime 2}}{2 \mu g}
$$

Thus for every distance $d+l=l+\frac{v^{\prime 2}}{2 \mu g}$ there is a car. This means in unit time $t$, there will be $N=\frac{d+l}{v}$ cars that passes through the line. To maximize the flow rate, we need to maximize

$$
\begin{gathered}
f\left(v^{\prime}\right)=\frac{l}{v^{\prime}}+\frac{v^{\prime}}{2 \mu g} \\
f^{\prime}=\frac{-l}{v^{\prime 2}}+\frac{1}{2 \mu g}=0 \Rightarrow v^{\prime}=\sqrt{2 \mu g l}=7.41 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$

Final answer: $7.41$ m/s
Subfield: Mechanics


Question: Here is a Physoly round button badge, in which the logo is printed on the flat and rigid surface of this badge. Toss it in the air and track the motions of three points (indicated by cyan circles in the figure) separated a straight-line distance of $L=5 \mathrm{~mm}$ apart. At a particular moment, we find that these all have the same speed $V=4 \mathrm{~cm} / \mathrm{s}$ but are heading to different directions which form an angle of $\theta=30^{\circ}$ between each pair. Determine the then angular velocity of the badge (in $\left.\mathrm{rad} / \mathrm{s}\right)$.
![](https://cdn.mathpix.com/cropped/2023_12_21_86ad3cbd2a33fe5168c3g-1.jpg?height=518&width=529&top_left_y=473&top_left_x=798)

Solution:
Call the three tracking points on the Physoly badge A, B, C, and their geometrical center $\mathrm{O}$. The distance from $\mathrm{O}$ to these three points are the same and equal to $L / \sqrt{3}$.
![](https://cdn.mathpix.com/cropped/2023_12_21_86ad3cbd2a33fe5168c3g-1.jpg?height=767&width=1033&top_left_y=1178&top_left_x=554)

Due to symmetry, the velocity vector of $\mathrm{O}$ has to be perpendicular to the $\mathrm{ABC}$ plane. In the reference frame of $\mathrm{O}$, the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ both have the same speed $2 V \sin (\theta / 2) / \sqrt{3}$ but are heading to different directions which form an angle of $120^{\circ}$ between each pair. Also due to symmetry, the axis of rotation has to be perpendicular to the $\mathrm{ABC}$ plane, thus the velocity vectors of points $\mathrm{A}$, $\mathrm{B}, \mathrm{C}$ in $\mathrm{O}$ reference frame looks like described in the attached figure. For $L=5 \mathrm{~mm}, V=4 \mathrm{~cm} / \mathrm{s}$, $\theta=30^{\circ}=\pi / 12$, the angular velocity of the badge can be calculated as:

$$
\Omega=\frac{2 V \sin (\theta / 2) / \sqrt{3}}{L / \sqrt{3}}=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right) \frac{V}{L} \approx 4.1411 \mathrm{rad} / \mathrm{s}
$$

Final answer: $4.1411$ $\mathrm{rad} / \mathrm{s}$
Subfield: Thermodynamics


Question: The following information applies for the next two problems. Pictured is a wheel from a 4wheeled car of weight $1200 \mathrm{~kg}$. The absolute pressure inside the tire is $3.0 \times 10^{5} \mathrm{~Pa}$. Atmospheric pressure is $1.0 \times 10^{5} \mathrm{~Pa}$. Assume the rubber has negligible "stiffness" (i.e. a negligibly low Sheer modulus compared to its Young's modulus).

![](https://cdn.mathpix.com/cropped/2023_12_21_aa674442fff889231bc7g-1.jpg?height=436&width=653&top_left_y=1820&top_left_x=736)


The rubber on the bottom of the wheel is completely unstretched. The rubber has a thickness of $7 \mathrm{~mm}$. Based on this information, find the Young's Modulus of the rubber. Is this answer reasonable?

Solution:
Let $P=2.0 \times 10^{5}$ Pa be the gauge pressure of the tire, $w=20 \mathrm{~cm}$ be the width of the tire, $r=20 \mathrm{~cm}$ be the radius of the tire, $t=7 \mathrm{~mm}$ be the thickness of the rubber, $M=1200 \mathrm{~kg}$ be the mass of the car, $Y$ be the Young's modulus, and $\theta$ be the angle subtended by the portion of the tire in contact with the ground. We have
$$
\begin{gathered}
\frac{M g}{4}=2 r w P \sin \left(\frac{\theta}{2}\right) \\
\theta=2 \arcsin \left(\frac{M g}{8 R w P}\right)=0.3696
\end{gathered}
$$
Let $T$ be the tension in the stretched portion of the rubber. Balance forces on a small section of the tire subtending an angle $d \theta$.
$$
\operatorname{Pr} w d \theta=2 T \sin \left(\frac{d \theta}{2}\right)
$$
Using the approximation $\sin \theta=\theta$,
$$
T=\operatorname{Pr} w
$$
The stress on the rubber is
$$
\sigma=\frac{T}{w t}=\frac{P r}{t}
$$
The strain is
$$
\epsilon=\frac{\theta}{2 \sin \left(\frac{\theta}{2}\right)}-1
$$
The Young's modulus is
$$
Y=\frac{\sigma}{\epsilon}=\frac{\frac{P r}{t}}{\frac{b}{2 \sin \left(\frac{b}{2}\right)}-1}=1.00 \times 10^{9} \mathrm{~Pa}
$$

Final answer: $1.00 \times 10^{9}$ Pa