Subfield: Thermodynamics

- 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: The following information applies for the next two problems. Pictured is a wheel from a 4wheeled car of weight $1200 \mathrm{~kg}$. The absolute pressure inside the tire is $3.0 \times 10^{5} \mathrm{~Pa}$. Atmospheric pressure is $1.0 \times 10^{5} \mathrm{~Pa}$. Assume the rubber has negligible "stiffness" (i.e. a negligibly low Sheer modulus compared to its Young's modulus).

<img_4309>


The rubber on the bottom of the wheel is completely unstretched. The rubber has a thickness of $7 \mathrm{~mm}$. Based on this information, find the Young's Modulus of the rubber. Is this answer reasonable?

Solution:
Let $P=2.0 \times 10^{5}$ Pa be the gauge pressure of the tire, $w=20 \mathrm{~cm}$ be the width of the tire, $r=20 \mathrm{~cm}$ be the radius of the tire, $t=7 \mathrm{~mm}$ be the thickness of the rubber, $M=1200 \mathrm{~kg}$ be the mass of the car, $Y$ be the Young's modulus, and $\theta$ be the angle subtended by the portion of the tire in contact with the ground. We have
$$
\begin{gathered}
\frac{M g}{4}=2 r w P \sin \left(\frac{\theta}{2}\right) \\
\theta=2 \arcsin \left(\frac{M g}{8 R w P}\right)=0.3696
\end{gathered}
$$
Let $T$ be the tension in the stretched portion of the rubber. Balance forces on a small section of the tire subtending an angle $d \theta$.
$$
\operatorname{Pr} w d \theta=2 T \sin \left(\frac{d \theta}{2}\right)
$$
Using the approximation $\sin \theta=\theta$,
$$
T=\operatorname{Pr} w
$$
The stress on the rubber is
$$
\sigma=\frac{T}{w t}=\frac{P r}{t}
$$
The strain is
$$
\epsilon=\frac{\theta}{2 \sin \left(\frac{\theta}{2}\right)}-1
$$
The Young's modulus is
$$
Y=\frac{\sigma}{\epsilon}=\frac{\frac{P r}{t}}{\frac{b}{2 \sin \left(\frac{b}{2}\right)}-1}=1.00 \times 10^{9} \mathrm{~Pa}
$$

Final answer: $1.00 \times 10^{9}$ Pa