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: 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?
<img_4298>

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