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: During quarantine, the FBI has been monitoring a young physicists suspicious activities. After compiling weeks worth of evidence, the FBI finally has had enough and searches his room.

The room's door is opened with a high angular velocity about its hinge. Over a very short period of time, its angular velocity increases to $\omega=8.56 \mathrm{rad} / \mathrm{s}$ due to the force applied at the end opposite from the hinge. For simplicity, treat the door as a uniform thin rod of length $L=1.00 \mathrm{~m}$ and mass $M=9.50 \mathrm{~kg}$. The hinge (pivot) is located at one end of the rod. Ignore gravity. At what distance from the hinge of the door is the door most likely to break? Assume that the door will break at where the bending moment is largest. (Answer in metres.)

<img_4296>

Solution:
Let $N$ be the force from the pivot and $F$ be the applied force at the end. Let $\alpha$ be the angular acceleration. Writing the torque equation and Newton's 2nd law for the whole door, we get:
$$
\begin{aligned}
& F \cdot L=\frac{1}{3} M L^{2} \alpha \\
& N+F=\frac{1}{2} M L \alpha
\end{aligned}
$$

Solving, we get $F=\frac{1}{3} M L \alpha$ and $N=\frac{1}{6} M L \alpha$. Now, we consider the part of the door with length $x$ attached to the pivot. The rest of the door applies a torque $\tau$ and shear force $f$ on our system. (There is also tension force). Let $\lambda=\frac{M}{L}$. We can write the torque equation and Newton's 2 nd law for our system:

$$
\begin{aligned}
& \tau+f x=\frac{1}{3} \lambda x^{3} \alpha \\
& N+f=\lambda x \cdot \frac{x}{2} \alpha
\end{aligned}
$$

Solving, we get

$$
\tau=\frac{1}{6} \lambda x \alpha\left(L^{2}-x^{2}\right)
$$

and

$$
f=\frac{1}{6} \lambda \alpha\left(3 x^{2}-L^{2}\right)
$$

We maximize $\tau$ (which is equivalent to maximizing bending moment) to get $x=\frac{L}{\sqrt{3}}=0.577 \mathrm{~m}$

Final answer: $\frac{\sqrt{3}}{3}$ m