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: Three unit circles, each with radius 1 meter, lie in the same plane such that the center of each circle is one intersection point between the two other circles, as shown below. Mass is uniformly distributed among all area enclosed by at least one circle. The mass of the region enclosed by the triangle shown above is $1 \mathrm{~kg}$. Let $x$ be the moment of inertia of the area enclosed by all three circles (intersection, not union) about the axis perpendicular to the page and through the center of mass of the triangle. Then, $x$ can be expressed as $\frac{a \pi-b \sqrt{c}}{d \sqrt{e}}$ $\mathrm{kg} \mathrm{m}^{2}$, where $a, b, c, d, e$ are integers such that $\operatorname{gcd}(a, b, d)=1$ and both $c$ and $e$ are squarefree. Compute $a+b+c+d+e$.

<img_4246>

Solution:
Define point $O$ as the point in the plane that the axis of rotation passes through. Since moments of inertia simply add about a given axis, we can calculate the moments of inertia of the three "sectors" whose union forms the given area and subtract twice the moment of inertia of the triangle, so our answer will be $3 I_{s, O}-2 I_{t, O}$.
Claim: The center of mass of a sector is $\frac{2}{\pi}$ away from the vertex of the sector along its axis of symmetry. Proof: We can divide the sector into arbitrarily small sectors that can be approximated as isosceles triangles. It's well known that the center of mass of one such isosceles triangle is $\frac{2}{3}$ of the way from the central vertex to the base. Therefore, the center of mass of the sector is the center of mass of the arc with central angle $\frac{\pi}{3}$ and same center with radius $\frac{2}{3}$ contained within the sector. Since the center of mass has to lie on the axis of symmetry, we set that as the $\mathrm{x}$ axis with the vertex of the sector being $x=0$. Then, the $x$-coordinate of a point on the arc whose corresponding radius makes an angle of $\theta$ with the axis of symmetry is $\frac{2}{3} \cos (\theta)$. We can integrate this over all possible angles $\left(-\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}\right)$ and then divide by the range $\left(\frac{\pi}{3}\right)$ to get the average $x$-coordinate, or the center of mass.

$$
\begin{gathered}
\frac{\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{2}{3} \cos (\theta) \mathrm{d} \theta}{\frac{\pi}{3}} \\
\frac{2}{\pi} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos (\theta) \mathrm{d} \theta \\
\frac{2}{\pi}\left(\sin \left(\frac{\pi}{6}\right)-\sin \left(-\frac{\pi}{6}\right)\right)
\end{gathered}
$$



This concludes the proof.

Now define point $X$ as the vertex of a sector and point $M$ as the center of mass of that sector. According to the parallel axis theorem,

$$
I_{s, X}=I_{s, M}+m_{s}\left(\frac{2}{\pi}\right)^{2}
$$

. It's well known that $I_{s, X}=\frac{1}{2} m_{s} r^{2}=\frac{m_{s}}{2}$, and so

$$
I_{s, M}=\frac{m_{s}}{2}-\frac{4 m_{s}}{\pi^{2}}=m_{s}\left(\frac{\pi^{2}-8}{2 \pi^{2}}\right)
$$

It's also well known that $O$ is on the line of symmetry and a distance of $\frac{1}{\sqrt{3}}$ away from $X$, and so $M X=$ $\frac{2}{\pi}-\frac{1}{\sqrt{3}}$. Therefore,

$$
I_{s, O}=I_{s, M}+m_{s}\left(\frac{2}{\pi}-\frac{1}{\sqrt{3}}\right)^{2}=m_{s}\left(\frac{5 \pi-8 \sqrt{3}}{6 \pi}\right)
$$

It's well known that, since $O$ is the center of mass of the triangle,

$$
I_{t, O}=\frac{1}{12}
$$

Now we just need to calculate $m_{s}$. Since the mass of the triangle is $1 \mathrm{~kg}$, this is equivalent to finding the ratio of the area of a sector to the area of a triangle. Through geometry, this is found to be $\frac{2 \pi}{3 \sqrt{3}}$. Finally, we get our answer to be

$$
\left(\frac{2 \pi}{\sqrt{3}}\right)\left(\frac{5 \pi-8 \sqrt{3}}{6 \pi}\right)-\frac{1}{6}=\left(\frac{10 \pi-17 \sqrt{3}}{6 \sqrt{3}}\right)
$$

and $a+b+c+d+e=10+17+3+6+3=039$

Final answer: 39