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: 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 $\left.\mathrm{rad} / \mathrm{s}\right)$.
<img_4273>

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 $L / \sqrt{3}$.
<img_4248>

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: $4.1411$ $\mathrm{rad} / \mathrm{s}$