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: Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the "ground" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless.

Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree.

Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$.

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Solution:
First, we must have $\omega^{2} R=g$ and $k=m \omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.
The translational force points "down" with a constant magnitude of $m g$, like gravity. The coriolis force



points perpendicular to the velocity with magnitude $2 m \omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\theta$. Note that we have to use numerical methods.

Specifically, the cycloid can be parametrized as $\frac{R}{4}(\alpha-\sin \alpha, 1-\cos \alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\alpha-\sin \alpha)^{2}+(1-\cos \alpha)^{2}=16$. Solving gives $\alpha=3.307$, and since $\cot \theta=\frac{1-\cos \alpha}{\alpha-\sin \alpha}$, we have $\theta=60.2^{\circ}$.

Final answer: $60.2$ $^{\circ}$