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 bicycle wheel of mass $M=2.8 \mathrm{~kg}$ and radius $R=0.3 \mathrm{~m}$ is spinning with angular velocity $\omega=5 \mathrm{rad} / \mathrm{s}$ around its axis in outer space, and its center is motionless. Assume that it has all of its mass uniformly concentrated on the rim. A long, massless axle is attached to its center, extending out along its axis. A ball of mass $m=1.0 \mathrm{~kg}$ moves at velocity $v=2 \mathrm{~m} / \mathrm{s}$ parallel to the plane of the wheel and hits the axle at a distance $h=0.5 \mathrm{~m}$ from the center of the wheel. Assume that the collision is elastic and instantaneous, and that the ball's trajectory (before and after the collision) lies on a straight line.

<img_4251>

Find the time it takes for the axle to return to its original orientation. Answer in seconds and round to three significant figures.

Solution:
After the collision, let the wheel have speed $v_{1}$ and the ball have speed $v_{2}$. Conserving momentum, energy, and angular momentum gives:
$$
\begin{gathered}
m v=M v_{1}+m v_{2} \\
\frac{1}{2} m v^{2}+\frac{1}{2} M R^{2} \omega^{2}=\frac{1}{2} M v_{1}^{2}+\frac{1}{2} m v_{2}^{2}+\frac{1}{2} M R^{2} \omega^{2}+\frac{1}{2} \cdot \frac{1}{2} M R^{2} \omega_{1}^{2} \\
m\left(v-v_{2}\right) h=\frac{1}{2} M R^{2} \omega_{1}
\end{gathered}
$$
where $\omega_{1}$ is the angular velocity (after collision) of the wheel in the direction perp. to the axis and the velocity of the ball.



Solving for $\omega_{1}$, we get
$$
\omega_{1}=\frac{4 h m v}{m\left(R^{2}+2 h^{2}\right)+M R^{2}}
$$
Now, we realize that the angular momentum of the wheel is given by $I_{x} \omega \hat{x}+I_{y} \omega_{1} \hat{y}$ where the wheel's axis is the $\mathrm{x}$-axis and the $\mathrm{y}$-axis is in the direction of $\omega_{1}$. Since angular momentum is conserved, the wheel must precess about its angular momentum vector. Let $\hat{L}$ represent the direction of the angular momentum vector. To find the rate of precession, we can decompose the angular velocity vector $\omega \hat{x}+\omega_{1} \hat{y}$ into a $\hat{L}$ component and a $\hat{x}$ component. Since $I_{x}=2 I_{y}$, the $\hat{L}$ component is $\sqrt{(2 \omega)^{2}+\omega_{1}^{2}}$, resulting in a precession period of
$$
T=\frac{\pi}{\sqrt{\omega^{2}+\frac{\omega_{1}^{2}}{4}}}=0.568 s
$$

Final answer: 0.568 s