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 concentric isolated rings of radius $a=1 \mathrm{~m}$ and $b=2 \mathrm{~m}$ of mass $m_{a}=1 \mathrm{~kg}$ and $m_{b}=2 \mathrm{~kg}$ are kept in a gravity free region. A soap film of surface tension $\sigma=0.05 \mathrm{Nm}^{-1}$ with negligible mass is spread over the rings such that it occupies the region between the rings. The smaller ring is pulled slightly along the axis of the rings. Find the time period of small oscillation in seconds.

Solution:
The force on the two rings when they are a distance $L$ apart follows as
$$
F=4 \pi r \sigma \sin \theta
$$

In small displacements, the change in $\theta$ is small. Therefore,

$$
\begin{aligned}
F & =4 \pi r \sigma \theta \\
\frac{F}{4 \pi r \sigma} & =\frac{\mathrm{d} y}{\mathrm{~d} r} \\
\int_{a}^{b} \frac{F}{4 \pi \sigma r} \mathrm{~d} r & =\int_{0}^{L} \mathrm{~d} y \\
\frac{F}{4 \pi \sigma} \ln (b / a) & =L
\end{aligned}
$$

Now let $a_{1}$ and $a_{2}$ be acceleration of a and $\mathrm{b}$ respectively. We have that

$$

a_{\text {net }}=a_{1}+a_{2}
\tag{12}
$$
$$
a_{\text {net }}=F\left(\frac{m_{1}+m_{2}}{m_{1} m_{2}}\right)
\tag{13}
$$
$$
L \omega^{2}=\frac{4 \pi \sigma}{\ln (b / a)} L\left(\frac{m_{1}+m_{2}}{m_{1} m_{2}}\right)
\tag{14}
$$
$$
T=2 \pi \sqrt{\frac{\ln (b / a) m_{1} m_{2}}{4 \pi \sigma\left(m_{1}+m_{2}\right)}}
\tag{15}
$$
$$
T=2 \pi \sqrt{\frac{10 \ln (2)}{3 \pi}}=5.388 \mathrm{~s}
\tag{16}
$$

Final answer: $2 \pi \sqrt{\frac{10 \ln (2)}{3 \pi}}$ s