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 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 $\alpha$.
<img_4310>

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