Subfield: Mechanics


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.
![](https://cdn.mathpix.com/cropped/2023_12_21_aed9cb2b56be78867ca6g-1.jpg?height=407&width=610&top_left_y=1561&top_left_x=752)

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,


![](https://cdn.mathpix.com/cropped/2023_12_21_16605de808d6d47cbc7bg-1.jpg?height=580&width=716&top_left_y=222&top_left_x=691)

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