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 boy is riding a tricycle across along a sidewalk that is parallel to the $x$-axis. This tricycle contains three identical wheels with radius $0.5 \mathrm{~m}$. The front wheel is free to rotate while the last two wheels are parallel to each other and to the main body of the tricycle. See the diagram.
<img_4304>

The front wheel is rotating at a constant angular speed of $\omega=3 \mathrm{rad} / \mathrm{s}$. The child is controlling the tricycle such that the front wheel is making an angle of $\theta(t)=0.15 \sin ((0.1 \mathrm{rad} / \mathrm{s}) t)$ with the main body of the tricycle. Determine the maximum lateral acceleration in $\mathrm{m} / \mathrm{s}^{2}$. Assume a massless frame. The marked plus sign implies CoM. The degree is in radians.

Solution:
Assuming the wheels roll without slipping, the cart will instantaneously rotate about a fixed point. This fixed point can be constructed by drawing perpendicular lines from all the wheels and seeing where they intersect. Consider the below diagram,


<img_4299>

We have:
$$
\sin \theta_{s}=\frac{L}{R}, \quad \quad \sin \theta_{s}^{\prime}=\frac{L}{2 R^{\prime}}
\tag{6}
$$
The angular frequency of every point on the robot is the same $\omega_{n}=\frac{V_{0}}{R}$. Therefore,
$$
\begin{aligned}
V_{C M} & =\omega_{n} R^{\prime} \\
& =\frac{V_{0}}{R} \frac{L}{2 \sin \theta_{s}^{\prime}} \\
& =\frac{V \sin \theta_{s}}{2 \sin \theta_{s}^{\prime}} .
\end{aligned}
$$
The naive idea here is to say the lateral velocity is $V_{C M} \sin \theta_{s}^{\prime}$, but note that the robot could already be rotated. Let the angle of rotation with respect to the horizontal line be $\alpha$, so we have:
$$
V_{C M, l a t}=V_{C M} \sin \left(\theta_{s}^{\prime}+\alpha\right)
$$
Using the sine addition formula, we can simplify this to
$$
\begin{aligned}
V_{C M, l a t} & =V_{C M} \sin \theta_{s}^{\prime} \cos \alpha+V_{C M} \cos \theta_{s}^{\prime} \sin \alpha \\
& =\frac{V_{0}}{2} \sin \theta_{s} \cos \alpha+\frac{V_{0} \sin \theta_{s}}{2 \tan \theta_{s}^{\prime}} \sin \alpha
\end{aligned}
$$
To simplify this, note that we can write:
$$
\frac{\sin \theta_{s}}{\tan \theta_{s}^{\prime}}=2
$$
where we used the fact that
$$
\sin \theta_{s} \approx \tan \theta_{s}=2 \tan \theta_{s}^{\prime} \approx 2 \sin \theta_{s}^{\prime}
$$
Then,
$$
\dot{y}_{C M}=\frac{V_{0}}{2} \theta_{s}+V_{0} \alpha
$$
At the center of the two back wheels (marked with an X), the speed is
$$
V_{b a c k}=R \cos \theta_{s} \omega_{n}=V_{0} \cos \theta_{s} \approx V_{0}
$$

so its lateral velocity is
$$
\dot{y}_{b a c k}=V_{0} \sin \alpha \approx V_{0} \alpha .
$$
Note that the angle $\alpha$ can be written as
$$
\alpha \approx \sin \alpha=\frac{y_{C M}-y_{b a c k}}{(L / 2)} \Longrightarrow \dot{\alpha}=\frac{2}{L}\left(\dot{y}_{C M}-\dot{y}_{b a c k}\right)=\frac{V_{0}}{L} \theta_{s}
$$
Taking higher derivatives of $\dot{y}_{C M}$, we have
$$
\begin{aligned}
\ddot{y}_{C M} & =\frac{V_{0}}{2} \dot{\theta}_{s}+V_{0} \dot{\alpha} \\
& =\frac{V_{0}}{2} \dot{\theta}_{s}+\frac{V_{0}^{2}}{L} \theta_{s} .
\end{aligned}
$$
Plugging in $\theta(t)=0.15 \sin (0.1 t)$ and $V_{0}=1.5 \mathrm{~m} / \mathrm{s}$, we can optimize $\ddot{y}_{\mathrm{CM}}$ to get $0.16912 \mathrm{~m} / \mathrm{s}$.

Final answer: $0.16912$ $\mathrm{~m} / \mathrm{s^2}$