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

- 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: What is the smallest number of $1 \Omega$ resistors needed such that when arranged in a certain arrangement involving only series and parallel connections, that the equivalent resistance is $\frac{7}{6} \Omega$ ?

Solution:
We can write:
$$
\frac{7}{6}=\frac{1}{2}+\frac{2}{3}
\tag{2}
$$

It takes two resistors (connected in parallel) to create a $\frac{1}{2} \Omega$ resistor. If we write $\frac{2}{3}=\frac{2 \cdot 1}{2+1}$, then it takes three resistors to create a $\frac{2}{3} \Omega$ (this is accomplished by connecting a 2 resistor in parallel with a 1 resistor.

Combining the $\frac{1}{2} \Omega$ element in series with the $\frac{2}{3} \Omega$ element gives us our desired amount. To prove this is the minimum, we can easily check all possible combinations using 4 or fewer resistors.

Final answer: 5
Subfield: Mechanics


Question: In a typical derby race, cars start at the top of a ramp, accelerate downwards, and race on a flat track, and are always set-up in the configuration shown below.
![](https://cdn.mathpix.com/cropped/2023_12_21_72dfd40d081d9f2832ecg-1.jpg?height=634&width=1461&top_left_y=1090&top_left_x=337)

A common technique is to change the location of the center of mass of the car to gain an advantage. Alice ensures the center of mass of her car is at the rear and Bob puts the center of mass of his car at the very front. Otherwise, their cars are exactly the same. Each car's time is defined as the time from when the car is placed on the top of the ramp to when the front of the car reaches the end of the flat track. At the competition, Alice's car beat Bob's. What is the ratio of Bob's car's time and Alice's car's time?

Assume that the wheels are small and light compared to the car body, neglect air resistance, and the height of the cars are small compared to the height of the ramp. In addition, neglect all energy losses during the race and the time it takes to turn onto the horizontal surface from the ramp. Express your answer as a decimal greater than 1 .

Solution:
The speed at which Alice's car hits the ground is given by $v_{A}=\sqrt{2 g(1)}=4.43 \mathrm{~m} / \mathrm{s}$ from energy conservation. The speed at which Bob's car hits the ground is given by $v_{B}=$ $\sqrt{2 g\left(1-0.1 \sin \left(45^{\circ}\right)\right.}=4.27 \mathrm{~m} / \mathrm{s}$ since its center of mass is at the very front of the car.


The time in which Alice and Bob's cars are going down the ramp (and wheels are both on the ramp) is given by:

$$
\sqrt{2}-0.1=\frac{1}{2} g \sin \left(45^{\circ}\right) t^{2} \Longrightarrow t=0.616 \mathrm{~s}
\tag{3}
$$

And the time that Alice's car takes on the flat path (and wheels are both on the floor) is given by $t_{A}=\frac{20}{v_{A}}=4.51$, and for Bob, $t_{B}=\frac{20}{v_{B}}=4.68$.

We ignore the time it takes for the car to transition. The justification behind this is that this time would be on the order of magnitude of $\frac{0.1}{v_{A}}=0.02 \mathrm{~s} \ll t_{A}$. In fact, it is smaller than the $1 \%$ margin allowed (so teams who did not read the clarification would still have gotten the correct answer). The ratio asked for was therefore:

$$
\frac{0.616+t_{B}}{0.616+t_{A}}
\tag{4}
$$

Final answer: 1.03
Subfield: Mechanics


Question: A simple crane is shown in the below diagram, consisted of light rods with length $1 \mathrm{~m}$ and $\sqrt{2} \mathrm{~m}$. The end of the crane is supporting a $5 \mathrm{kN}$ object. Point $B$ is known as a "pin." It is attached to the main body and can exert both a vertical and horizontal force. Point $A$ is known as a "roller" and can only exert vertical forces. Rods can only be in pure compression or pure tension.
![](https://cdn.mathpix.com/cropped/2023_12_21_ed61957af961179eb7adg-1.jpg?height=794&width=591&top_left_y=1186&top_left_x=756)

In $\mathrm{kN}$, what is the force experienced by the rod $C D$ ? Express a positive number if the member is in tension and a negative number if it is in compression.

Solution:
One naive method (though perfectly valid) is to solve for each member individually, starting from the two rods that connect to the $5 \mathrm{kN}$ weight. At each joint, we can write out force equilibrium equations in the vertical and horizontal directions, and solve a system of linaer equations to get the force in $C D$.
Instead, we can solve for this force in one line. Consider a horizontal slice right above point $D$.



![](https://cdn.mathpix.com/cropped/2023_12_21_ce7d6832dfafbf2eb586g-1.jpg?height=488&width=593&top_left_y=222&top_left_x=755)

Since the net force of this sub-element is still zero, we can do a force balance. The only external forces acting on this system is $E F, E C, C D$, and the $5 \mathrm{kN}$ weight. If we do a torque balance about $E$, we get:

$$
5(2 L)=C D(L)
\tag{5}
$$

where $L$ is the length of the rod. This immediately gives $C D=10 \mathrm{kN}$.

Final answer: 10 kN
Subfield: Electromagnetism


Question: A small block of mass $m$ and charge $Q$ is placed at rest on an inclined plane with a slope $\alpha=40^{\circ}$. The coefficient of friction between them is $\mu=0.3$. A homogenous magnetic field of magnitude $B_{0}$ is applied perpendicular to the slope. The speed of the block after a very long time is given by $v=\beta \frac{m g}{Q B_{0}}$. Determine $\beta$. Do not neglect the effects of gravity.
![](https://cdn.mathpix.com/cropped/2023_12_21_dafd88496563c6b3fc11g-1.jpg?height=580&width=813&top_left_y=477&top_left_x=661)

Solution:
Create a free body diagram. The direction of the magnetic field (into or out of the page) does not matter as we only need to know the magnitude of the terminal velocity. In dynamic equilibrium, we have three forces along the plane: the component of gravity along the plane, friction, and the magnetic force. The component of gravity along the plane is $m g \sin \alpha$. Friction has the magnitude $f=\mu m g \cos \alpha$. The magnetic force has magnitude $F_{B}=Q v B_{0}$. At the terminal velocity, these forces are balanced. To finish, note that friction and the magnetic force are perpendicular so the magnitude of their vector sum is equal to the component of gravity along the plane. By the Pythagorean Theorem,
$$
F_{B}^{2}+f^{2}=(m g \sin \alpha)^{2} \Longrightarrow v=\frac{m g}{Q B_{0}} \sqrt{\sin ^{2} \alpha-\mu^{2} \cos ^{2} \alpha}
$$

Final answer: 0.6
Subfield: Mechanics


Question: A sprinkler fountain is in the shape of a semi-sphere that spews out water from all angles at a uniform speed $v$ such that without the presence of wind, the wetted region around the fountain forms a circle in the $X Y$ plane with the fountain centered on it.
Now suppose there is a constant wind blowing in a direction parallel to the ground such that the force acting on each water molecule is proportional to their weight. The wetted region forms the shape below where the fountain is placed at $(0,0)$. Determine the exit speed of water $v$ in meters per second. Round to two significant digits. All dimensions are in meters.

![](https://cdn.mathpix.com/cropped/2023_12_21_bc101ee6c7ad2225c8a6g-1.jpg?height=1258&width=1567&top_left_y=661&top_left_x=279)

Solution:
Method One: Let us look at only the major axis $x$. The wind provides a constant acceleration along $x$, which makes this problem equivalent to throwing a ball up and down a ramp tilted at an angle of $\beta=\tan \frac{a}{g}$. The furthest point along the $x$ axis represents the maximum distance you can throw an object up and down this ramp. This distance can be derived to be:

$$
d_{\mathrm{optimal}}=\frac{v^{2} / g}{1+\sin \beta}
\tag{6}
$$



where $\beta$ is negative if it's tilting downwards. We get the systems of equations:

$$

\frac{v^{2} / g_{\mathrm{eff}}}{1+\sin \beta}=0.25+0.25 \cdot \frac{13}{20}
\tag{7}
$$
$$
\frac{v^{2} / g_{\mathrm{eff}}}{1-\sin \beta}=1

\tag{8}
$$

where:

$$
g_{\mathrm{eff}}=g / \cos \beta
\tag{9}
$$

and solving gives $v=2.51 \mathrm{~m} \mathrm{~s}^{-1}$.

Method Two: Looking at the minor axis is much easier. The wind doesn't contribute in this direction, so we know the furthest point must have a vertical displacement of $d_{\max }=\frac{v^{2}}{g}$. We measure $d_{\max }=0.50+0.25 \cdot \frac{11}{20}$ and solving for $v$ gives $d_{\max }=2.50 \mathrm{~m}$.

Note that the answers aren't exactly the same due to measurement inaccuracies, so we asked for two significant digits. An earlier version of the question had the same diagram, but the labels were accidentally shifted. This did not affect the final answer.

Final answer: 2.5 m/s
Subfield: Mechanics


Question: Poncho is a very good player of the legendary carnival game known as Pico-Pico. Its setup consists of a steel ball, represented by a point mass, of negligible radius and a frictionless vertical track. The goal of Pico-Pico is to flick the ball from the beginning of the track (point $A$ ) such that it is able to traverse through the track while never leaving the track, successfully reaching the end (point $B)$. The most famous track design is one of parabolic shape; specifically, the giant track is of the shape $h(x)=5-2 x^{2}$ in meters. The starting and ending points of the tracks are where the two points where the track intersects $y=0$. If $\left(v_{a}, v_{b}\right]$ is the range of the ball's initial velocity $v_{0}$ that satisfies the winning condition of Pico-Pico, help Poncho find $v_{b}-v_{a}$. This part is depicted below:
![](https://cdn.mathpix.com/cropped/2023_12_21_c2af4dafb1688ec8aa90g-1.jpg?height=501&width=398&top_left_y=1343&top_left_x=858)

Solution:
Using conservation of energy, the minimum initial velocity of the ball needed to pass the top of the track is $v_{a}=\sqrt{2 g h}=9.9045 \frac{\mathrm{m}}{\mathrm{s}}$. To find $v_{b}$, the centripetal force at all points on the track must be determined given the initial velocity.
$$

F_{c} =\frac{m v^{2}}{R}
\tag{11}
$$
$$
=\frac{m\left(v_{b}^{2}-2 g h\right)}{\frac{\left|1+\left(\frac{d}{d x} h(x)\right)^{2}\right|}{\frac{d^{2}}{d x^{2}} h(x)}}
\tag{12}
$$
$$
=\frac{m\left(v_{b}^{2}-2 g h\right)}{\frac{\left|1+16 x^{2}\right|^{\frac{3}{2}}}{4}}
\tag{13}
$$

For the boundary condition, the ball leaves if the normal force from the track on the ball $N=m g \cos \theta-F_{c}$ becomes 0.

$$
\begin{aligned}
m g \cos \theta-F_{c} & =0 \\
m g \cos \arctan (-4 x) & =\frac{4 m\left(v_{b}^{2}-2 g h\right)}{\left|1+16 x^{2}\right|^{\frac{3}{2}}} \\
\frac{g}{\left|1+16 x^{2}\right|^{\frac{1}{2}}} & =\frac{4\left(v_{b}^{2}-2 g h\right)}{\left|1+16 x^{2}\right|^{\frac{3}{2}}} \\
g & =\frac{4\left(v_{b}^{2}-2 g h\right)}{1+16 x^{2}} \\
v_{b \max } & =\sqrt{\frac{g+16 g x^{2}}{4}+2 g h}
\end{aligned}
$$

From the derivation, $v_{b \max }$ is the lowest at $x=0$. Thus,

$$
\begin{aligned}
v_{b \max } & =\sqrt{\frac{g}{4}+2 g h} \\
& =10.0276 \frac{\mathrm{m}}{\mathrm{s}}
\end{aligned}
$$

which is our desired $v_{b}$. The final answer, $v_{b}-v_{a}$, can be calculated.

Final answer: 0.1231 m/s
Subfield: Mechanics


Question: Anyone who's had an apple may know that pieces of an apple stick together, when picking up one piece a second piece may also come with the first piece. The same idea is tried on a golden apple. Consider two uniform hemispheres with radius $r=4 \mathrm{~cm}$ made of gold of density $\rho_{g}=19300 \mathrm{~kg} \mathrm{~m}^{-3}$. The top half is nailed to a support and the space between is filled with water.
![](https://cdn.mathpix.com/cropped/2023_12_21_8e722241e5b95bed0c25g-1.jpg?height=298&width=317&top_left_y=458&top_left_x=904)

Given that the surface tension of water is $\gamma=0.072 \mathrm{~N} \mathrm{~m}^{-1}$ and that the contact angle between gold and water is $\theta=10^{\circ}$, what is the maximum distance between the two hemispheres so that the bottom half doesn't fall? Answer in millimeters.

Solution:
Let $h$ be the difference in height. There are 3 forces on the bottom hemisphere. The force from gravity, which has magnitude $\frac{2}{3} \pi \rho_{g} g r^{3}$, the force from the surface tension, and the force from the pressure difference at the top and bottom. The pressure difference is given by the young-laplace equation,
$$
\Delta P=\gamma\left(-\frac{1}{r}+\frac{2 \cos \theta}{h}\right) \approx \frac{2 \gamma \cos \theta}{h}
$$

The radii are found by some simple geometry. It is likely that $r$ will be much larger than the height, so we can neglect the $1 / r$ term. Now the force from surface tension is $2 \pi r \gamma \sin \theta$, since we take the vertical component. So we can now set the net force to 0 ,

$$
\frac{2}{3} \pi \rho_{g} g r^{3}=\pi r^{2} \Delta P+2 \pi r \gamma \sin \theta \Longrightarrow \frac{2}{3} \rho_{g} g r^{2}=(2 r \gamma \cos \theta) \frac{1}{h}+2 \gamma \sin \theta
$$

This is simple to solve,

$$
h=\frac{2 r \gamma \cos \theta}{2 \rho_{g} g r^{2} / 3-2 \gamma \sin \theta}=2.81 \times 10^{-5} \mathrm{~m}=0.0281 \mathrm{~mm}
$$

This is very small so our approximation from earlier is justified.

Final answer: 0.0281 mm
Subfield: Mechanics


Question: In the following two problems we will look at shooting a basketball. Model the basketball as an elastic hollow sphere with radius 0.1 meters. Model the net and basket as shown below, dimensions marked. Neglect friction between the backboard and basketball, and assume all collisions are perfectly elastic.

![](https://cdn.mathpix.com/cropped/2023_12_21_a1e75b3588890e36fda2g-1.jpg?height=1225&width=1634&top_left_y=426&top_left_x=191)


For this problem, you launch the basketball from the point that is 2 meters above the ground and 4 meters from the backboard as shown. You attempt to make a shot by hitting the basketball off the backboard as depicted above. What is the minimum initial speed required for the ball to make this shot?
Note: For this problem, you may assume that the size of the ball is negligible.

Solution:
For this part, we regard the basketball as a point mass. Now to account for the backboard bounce, we reflect the hoop over the backboard. The problem is now equivalent to making it into the hoop that is 4.15 meters away now. So now we need to find the smallest velocity to reach this point. One way to do it is to use the safety parabola, which makes it so minimal calculations are needed, so we present that method here. However, this problem is also readily solvable with only basic kinematics.
The point that the ball must hit (4.15 meters in front and 1 meter above) lies on the safety parabola when it is thrown at minimum speed. Since the focus of the safety parabola is the point of launch,



the directrix is

$$
\sqrt{1^{2}+4.15^{2}}=4.27 \mathrm{~m}
$$

away from the top of the hoop. The distance from the focus to the vertex of the parabola is then $(4.27+1) / 2=2.63 \mathrm{~m}$. So the minimum velocity to go this height is given by

$$
\frac{v^{2}}{2 g}=2.63 \mathrm{~m}
$$

and that gives $v=7.19 \mathrm{~m} / \mathrm{s}$. There is one last bit we need to check to complete this problem. The basketball's trajectory has to actually be possible (i.e. it doesn't go underneath the rim). It is clearly less optimal intuitively. If it passes through the underside, it means that its apex of the trajectory is very near its target.

Final answer: 7.19 m/s
Subfield: Mechanics


Question: Consider a toilet paper roll with some length of it hanging off as shown. The toilet paper roll rests on a cylindrical pole of radius $r=1 \mathrm{~cm}$ and the coefficient of static friction between the role and the pole is $\mu=0.3$.

![](https://cdn.mathpix.com/cropped/2023_12_21_20d264993198809dd70dg-1.jpg?height=612&width=480&top_left_y=1464&top_left_x=817)

The length of the paper hanging off has length $\ell=30 \mathrm{~cm}$ and the inner radius of the roll is $R_{i}=2 \mathrm{~cm}$. The toilet paper has thickness $s=0.1 \mathrm{~mm}$ and mass per unit length $\lambda=5 \mathrm{~g} / \mathrm{m}$. What is the minimum outer radius $R_{o}$ such that the toilet paper roll remains static? Answer in centimeters.

Solution:
Due to the length of toilet paper hanging off, the toilet paper will be slightly tilted, in order for torques to balance. The tilt isn't shown in the diagram, since it is meant to be found. So the tilted normal force has to be compensated by the frictional force, and just before slippage,


it will look something like:

![](https://cdn.mathpix.com/cropped/2023_12_21_9bf394e5a782e8f11601g-1.jpg?height=618&width=553&top_left_y=304&top_left_x=778)

Let $\theta=\arctan \mu$. So now balancing torques about the contact point, if $m$ is the mass of the toilet paper roll,

$$
m g R_{i} \sin \theta=(\ell \lambda) g\left(R_{o}-R_{i} \sin \theta\right) .
$$

And now we need $m$. The mass per unit area is $\sigma=\lambda / s$, so $m=\pi\left(R_{o}^{2}-R_{i}^{2}\right) \lambda / s$. So substituting this in,

$$
\frac{\pi\left(R_{o}^{2}-R_{i}^{2}\right) \lambda}{s} g R_{i} \sin \theta=\ell \lambda g\left(R_{o}-R_{i} \sin \theta\right)
$$

Simplifying a bit,

$$
\pi\left(R_{o}^{2}-R_{i}^{2}\right) R_{i} \sin \theta=s \ell\left(R_{o}-R_{i} \sin \theta\right) .
$$

This is a quadratic and we can just plug in the numbers and use the quadratic formula or use a graphing calculator to finish, yielding $R_{o}=2.061 \mathrm{~cm}$.

Final answer: 2.061 cm
Subfield: Electromagnetism


Question: In the circuit shown below, a capacitor $C=4 \mathrm{~F}$, inductor $L=5 \mathrm{H}$, and resistors $R_{1}=3 \Omega$ and $R_{2}=2 \Omega$ are placed in a diamond shape and are then fed an alternating current with peak voltage $V_{0}=1 \mathrm{~V}$ of unknown frequency. Determine the magnitude of the maximum instantaneous output voltage shown in the diagram.
![](https://cdn.mathpix.com/cropped/2023_12_21_29b1038a6a3b0fabe55eg-1.jpg?height=469&width=616&top_left_y=443&top_left_x=749)

Solution:
We use the method of phasors. Consider the following phasor diagram:
![](https://cdn.mathpix.com/cropped/2023_12_21_29b1038a6a3b0fabe55eg-1.jpg?height=387&width=707&top_left_y=1064&top_left_x=709)

We define angles as $\angle A P B=\angle A Q B=\gamma, \angle A O B=2 \gamma, \angle B P Q=\alpha, \angle A Q P=\beta$. Note that $\overrightarrow{P A}=I_{2} \omega L, \overrightarrow{A Q}=I_{2} R, \overrightarrow{P B}=I_{1} R_{1}, \overrightarrow{B Q}=1 / I_{1} \omega C$. We seek to maximize the length of $\mathrm{AB}$. We can write via law of cosines that

$$
\mathrm{AB}=\sqrt{r^{2}+r^{2}-2 r \cos (2 \gamma)}=2 r \sin (\gamma)
$$

where $r=V_{0}$ is the radius of the circle of which phasors are inscribed in. Since $\gamma=\alpha+\beta$, we can then rewrite the length of $\mathrm{AB}$ to be

$$
\mathrm{AB}=r(\cos \alpha \cos \beta-\sin \alpha \sin \beta)=V_{0}\left(\frac{R_{1} R_{2}}{V_{0}^{2}}-\frac{L}{C V_{0}^{2}}\right) I_{1} I_{2}=V_{0}\left(R_{1} R_{2}-\frac{L}{C}\right)\left(\frac{1}{\left|Z_{1}\right|\left|Z_{2}\right|}\right)
$$

where the product of both complex exponentials is simply

$$
\left|Z_{1}\right|\left|Z_{2}\right|=\sqrt{\omega^{2} L^{2}+R_{2}^{2}} \sqrt{\frac{1}{\omega^{2} C^{2}}+R_{1}^{2}}
$$

which implies the best frequency of the circuit is $\omega=\sqrt{\frac{R_{2}^{2}}{L^{2} R_{1} C}}$. This means with algebra, the lowest value of $\left|Z_{1}\right|\left|Z_{2}\right|=\frac{L}{C}-R_{1} R_{2}$. Hence, the maximum value of $\mathrm{AB}$ is simply

$$
\mathrm{AB}=V_{0}\left(\frac{R_{1} R_{2}-\frac{L}{C}}{R_{1} R_{2}+\frac{L}{C}}\right)
$$

Final answer: 0.65 V
Subfield: Mechanics


Question: A uniform bar of length $l$ and mass $m$ is connected to a very long thread of negligible mass suspended from a ceiling. It is then rotated such that it is vertically upside down and then released. Initially, the rod is in unstable equilibrium. As it falls down, the minimum tension acting on the thread over the rod's entire motion is given by $\alpha m g$. Determine $\alpha$.
![](https://cdn.mathpix.com/cropped/2023_12_21_8778e2463d0600bda16eg-1.jpg?height=512&width=160&top_left_y=457&top_left_x=972)

Solution:
First note that the thread is given to be very long. Therefore, only vertical tension and gravitational forces act on the rod allowing for its center of mass to move in a straight vertical line. Using this fact, we can now write the acceleration of the rod in terms of angular velocity $\omega$ and angular acceleration $\varepsilon$. Let the angle of the rod at any moment to the vertical be $\varphi$. For simplicity, we define the length of the rod to be $2 l$. Then, by defining the coordinate $y$ to be the change in vertical length of the rod where $y=l \cos \varphi$, one can write for varying $\varphi \in[\varphi, \varphi+\mathrm{d} \varphi)$ that $l \cos (\varphi+\mathrm{d} \varphi)=y+\mathrm{d} y \Longrightarrow \mathrm{d} y=l \sin \varphi \mathrm{d} \varphi$. Thus, simple differentiation proves that
$$
v=l \sin \varphi \omega \Longrightarrow a=l \varepsilon \sin \varphi+l \omega^{2} \cos \varphi
$$

We can now also write conservation of energy to get another relationship between velocity and angular velocity. At any given moment, it can be written that $m g l=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}-m g l \cos \varphi$. Since $I=\frac{1}{2} m(2 l)^{2}=\frac{1}{3} m l^{2}$, then

$$
m g l(1+\cos \varphi)=m l^{2} \omega^{2}\left(\frac{1}{2} \sin ^{2} \varphi+\frac{1}{6}\right)
$$

Our third equation comes from the fact that the tension on the rod at any moment can be written as $T=m g-m a \Longrightarrow m a=m g-T$. We can finally get a fourth equation by equating torques such that $\frac{1}{3} m l^{2} \varepsilon=T l \sin \varphi$. With these four equations, simple algebra yields the tension at any point is

$$
T(\varphi)=\frac{1+(3 \cos \varphi-1)^{2}}{\left(1+3 \sin ^{2} \varphi\right)^{2}} m g \Longrightarrow T_{\min } \approx 0.165 m g
$$

Final answer: 0.165 mg