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 wooden bus of mass $M=20,000 \mathrm{~kg}$ ( $M$ represents the mass excluding the wheels) is on a ramp with angle $30^{\circ}$. Each of the four wheels is composed of a ring of mass $\frac{M}{2}$ and radius $R=1 \mathrm{~m}$ and 6 evenly spaced spokes of mass $\frac{M}{6}$ and length $R$. All components of the truck have a uniform density. Find the acceleration of the bus down the ramp assuming that it rolls without slipping.

Answer in $\mathrm{m} / \mathrm{s}^{2}$.

<img_4289>

Solution:
The moment of inertia of each wheel can be thought of as a superposition of the six spokes and a ring. Therefore, we get:
$$
I_{\text {wheel }}=\frac{M}{2} R^{2}+6\left(\frac{1}{3} \frac{M}{6} R^{2}\right)=\frac{5}{6} M R^{2}
$$

The moment of inertia of four wheels is:

$$
I_{\mathrm{bus}}=4\left(\frac{5}{6} M R^{2}\right)=\frac{10}{3} M R^{2}
$$

The total mass of the bus is

$$
m=M+4\left(\frac{M}{2}+6 \cdot \frac{M}{6}\right)=7 M
$$

Using Newton's second law down the ramp,

$$
7 M a=7 M g \sin \theta-4 f
$$

if $f$ is the friction at each wheel, and the torque balance on each wheel is:

$$
\frac{5}{6} M R^{2} \alpha=f R
$$

Letting $a=\alpha r$ for the no slip condition, we can solve for $f$ to be:

$$
f=\frac{5}{6} M a
$$

so our force balance equation becomes:

$$
7 M a=7 M g \sin \theta-\frac{10}{3} M a \Longrightarrow a=\frac{21}{31} g \sin \theta=3.32 \mathrm{~m} / \mathrm{s}^{2}
$$

Final answer: 3.32 m