# Example 2.13: Carousel horse (3D relative motion)
The following mechanism is a fairground carousel horse (February 2010 exam), whose rocking motion is driven by link $AB$ rotating clockwise at angular velocity $\omega_0$ ($\vec{\omega}_0 = -\omega_0\,\hat{i}$, with the $x$ axis pointing out of the page), while point $C$ slides along a vertical slot.
Knowing that at the initial instant the angle at $A$ is a right angle ($\phi=\frac{\pi}{2}$), that $\theta=\frac{\pi}{3}$, and that point $D$ lies on bar $AC$ (of length $|AC|=R$) at a distance $\frac{1}{3}R$ from $A$, compute:
1. The angular velocity $\omega_{AC}$ of bar $AC$.
2. The velocity of the horse (point $D$).
3. The acceleration of the same point, knowing that $\vec{a}_A = \omega_0^2 \frac{R}{8} (-\sqrt{3}\,\hat{j} + \hat{k})$ and that the motion of bar $AC$ can be considered uniform.
In addition to the rocking motion of the horse, now consider the uniform counterclockwise rotation $\Omega$ of the carousel platform about a vertical axis through its center.
Assuming the horse is 2 meters away from the rotation axis (with the axis to its left), compute:
4. The absolute velocity of the horse.
5. The absolute acceleration of the horse.
:::{dropdown} Solution
**1. Angular velocity $\omega_{AC}$ of bar $AC$**
Because of the constraint at point $C$, for each position of bar $AB$, bar $AC$ is forced into a single compatible position.
Therefore, an angular velocity $\omega_0$ of $AB$ induces an angular velocity of $AC$ in the opposite sense:
To find $\omega_{AC}$ we relate it to the velocity of point $C$, $\vec{v}_C$, and impose that its horizontal component must be zero (because of the slot).
Point $C$ belongs to a body (bar $AC$) rotating with angular velocity $\omega_{AC}$, for which we know the velocity of another point ($A$), so:
$$
\vec{v}_C = \vec{v}_A + \vec{\omega}_{AC} \times \vec{r}_{AC}
$$
The velocity of $A$ is the time derivative of its position vector $\vec{r}_{BA}$ (of magnitude $\sqrt{3}R$), a constant-magnitude vector rotating with angular velocity $\vec{\omega}_0$:
$$
\begin{aligned}
\vec{v}_A &= \frac{d\vec{r}_{BA}}{dt} = \vec{\omega}_0 \times \vec{r}_{BA}
= (-\omega_0\,\hat{i}) \times \left(\sqrt{3}R \cos 60^\circ\,\hat{j} - \sqrt{3}R \sin 60^\circ\,\hat{k}\right) \\
&= (-\omega_0\,\hat{i}) \times R \left(\tfrac{\sqrt{3}}{2}\,\hat{j} - \tfrac{3}{2}\,\hat{k}\right)
= \omega_0 \tfrac{\sqrt{3}R}{2} \left(-\sqrt{3}\,\hat{j} - \hat{k}\right)
\end{aligned}
$$
In addition, from the geometry of the problem:
we know that, at the instant of interest:
$$
\vec{r}_{AC} = R \left(-\sin 60^\circ\,\hat{j} - \cos 60^\circ\,\hat{k}\right) = -\tfrac{R}{2} \left(\sqrt{3}\,\hat{j} + \hat{k}\right)
$$
Since $\vec{\omega}_{AC} = \omega_{AC}\,\hat{i}$, substituting:
$$
\begin{aligned}
\vec{v}_C &= -\omega_0 \tfrac{\sqrt{3}R}{2} (\sqrt{3}\,\hat{j} + \hat{k}) + (\omega_{AC}\,\hat{i}) \times \left[-\tfrac{R}{2} (\sqrt{3}\,\hat{j} + \hat{k})\right] \\
&= -\omega_0 \tfrac{\sqrt{3}R}{2} (\sqrt{3}\,\hat{j} + \hat{k}) - \omega_{AC} \tfrac{R}{2} (\sqrt{3}\,\hat{k} - \hat{j}) \\
&= \left(-\tfrac{3R}{2}\omega_0 + \tfrac{R}{2}\omega_{AC}\right)\hat{j} + \left(-\tfrac{\sqrt{3}R}{2}\omega_0 - \tfrac{\sqrt{3}R}{2}\omega_{AC}\right)\hat{k}
\end{aligned}
$$
We now apply the constraint on point $C$, which can only move vertically: $\vec{v}_C = v_C\,\hat{k}$.
Equating component by component gives a system of two equations with two unknowns ($v_C$ and $\omega_{AC}$):
$$
\left\{
\begin{aligned}
0 &= -\tfrac{3R}{2}\omega_0 + \tfrac{R}{2}\omega_{AC} \\
v_C &= -\tfrac{\sqrt{3}R}{2}\omega_0 - \tfrac{\sqrt{3}R}{2}\omega_{AC}
\end{aligned}
\right.
\quad\rightarrow\quad
\omega_{AC} = 3\,\omega_0, \qquad v_C = -2\sqrt{3}\,R\,\omega_0
$$
The velocity of $C$ comes out negative because it was taken as positive along the $Z$ axis: for $\omega_0 > 0$, point $C$ moves down at the instant of interest.
**2. Velocity of point $D$**
This is the same situation as before: knowing the velocity of one point ($A$) of a rigid body (bar $AC$) rotating with angular velocity $\vec{\omega}_{AC}$, find the velocity of another point of the same body ($D$), with $\vec{r}_{AD} = \frac{1}{3}\vec{r}_{AC}$:
$$
\begin{aligned}
\vec{v}_D &= \vec{v}_A + \vec{\omega}_{AC} \times \vec{r}_{AD}
= -\omega_0 \tfrac{\sqrt{3}R}{2} (\sqrt{3}\,\hat{j} + \hat{k}) + 3\omega_0\,\hat{i} \times \left(-\tfrac{R}{3}\sin 60^\circ\,\hat{j} - \tfrac{R}{3}\cos 60^\circ\,\hat{k}\right) \\
&= -\tfrac{3}{2}\omega_0 R\,\hat{j} - \tfrac{\sqrt{3}}{2}\omega_0 R\,\hat{k} - \tfrac{\sqrt{3}}{2}\omega_0 R\,\hat{k} + \tfrac{1}{2}\omega_0 R\,\hat{j} \\
&= \omega_0 R \left(-\hat{j} - \sqrt{3}\,\hat{k}\right)
\end{aligned}
$$
**3. Acceleration of point $D$**
Differentiating the previous velocity equation, and noting that the motion of $AC$ is uniform ($\dot{\vec{\omega}}_{AC} = 0$) and that $\vec{\omega}_{AC} \perp \vec{r}_{AD}$:
$$
\begin{aligned}
\vec{a}_D &= \vec{a}_A + \dot{\vec{\omega}}_{AC} \times \vec{r}_{AD} + \vec{\omega}_{AC} \times (\vec{\omega}_{AC} \times \vec{r}_{AD})
= \vec{a}_A - \omega_{AC}^2\,\vec{r}_{AD} \\
&= \omega_0^2 \tfrac{R}{8} (-\sqrt{3}\,\hat{j} + \hat{k}) + 9\omega_0^2 \tfrac{R}{6} (\sqrt{3}\,\hat{j} + \hat{k}) \\
&= \omega_0^2 R \left(\tfrac{11\sqrt{3}}{8}\,\hat{j} + \tfrac{13}{8}\,\hat{k}\right)
\end{aligned}
$$
**4. Absolute velocity of $D$, including the rotation $\Omega$**
$\vec{v}_D$ is now a velocity relative to the carousel platform, which rotates with $\vec{\Omega} = \Omega\,\hat{k}$.
With $\vec{r} = 2\,\hat{j}$ the vector from the rotation axis to $D$ (in meters), the transport term is $\vec{\Omega} \times \vec{r}$, and the origin of the moving frame (on the rotation axis) is fixed:
$$
(\vec{v}_D)_{abs} = \vec{v}_D + \vec{\Omega} \times \vec{r} = \vec{v}_D + \Omega\,\hat{k} \times 2\,\hat{j}
= -2\Omega\,\hat{i} - \omega_0 R\,\hat{j} - \sqrt{3}\,\omega_0 R\,\hat{k}
$$
Only the distance to the rotation axis matters, not the exact position of the reference point chosen on that axis: the vertical component of the relative position vector is parallel to $\vec{\Omega}$ and does not contribute to the cross product.
**5. Absolute acceleration of $D$**
We just substitute, into the acceleration formula for a moving point in a moving reference frame, the velocity and acceleration of $D$ relative to the carousel obtained in parts 2 and 3 (with $\dot{\vec{\Omega}} = 0$):
$$
\begin{aligned}
(\vec{a}_D)_{abs} &= \vec{a}_D + \dot{\vec{\Omega}} \times \vec{r} + \vec{\Omega} \times (\vec{\Omega} \times \vec{r}) + \underbrace{2\,\vec{\Omega} \times \vec{v}_D}_{\text{Coriolis}} \\
&= \vec{a}_D + \Omega\,\hat{k} \times (\Omega\,\hat{k} \times 2\,\hat{j}) + 2\,\Omega\,\hat{k} \times \omega_0 R (-\hat{j} - \sqrt{3}\,\hat{k}) \\
&= \omega_0^2 R \left(\tfrac{11\sqrt{3}}{8}\,\hat{j} + \tfrac{13}{8}\,\hat{k}\right) - 2\Omega^2\,\hat{j} + 2\,\Omega\,\omega_0 R\,\hat{i} \\
&= 2\,\Omega\,\omega_0 R\,\hat{i} + \left(\tfrac{11\sqrt{3}}{8}\omega_0^2 R - 2\Omega^2\right)\hat{j} + \tfrac{13}{8}\omega_0^2 R\,\hat{k}
\end{aligned}
$$
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