# Example 2.11: Rocker with a pin in a rotating slot

Pin $A$ of rocker $AO_3$ is constrained to move inside the rotating slot of arm $O_2D$ (February 2016 exam).
The arm's angular velocity is $\omega_2 = 2~rad/s$ clockwise, and it remains constant at the instant of interest.
For the position shown in the figure (with $AO_3$ horizontal), find:

1. Using the analytical method, the absolute velocity of pin $A$ and its velocity relative to the rotating slot of $O_2D$.
2. The angular acceleration of $AO_3$ and the acceleration of $A$ relative to the slot of the rotating arm $O_2D$.

<img src="figs/problemas_cin_an/ex2016feb_mec_enunciado.png" alt="Rocker with a pin in a rotating slot" width="450px">

:::{dropdown} Solution

From the figure, with $\hat{i}$ pointing right and $\hat{j}$ pointing up (in mm):

$$
\vec{r}_{O_3A} = 225\,\hat{i}, \qquad \vec{r}_{O_2A} = -225\,\hat{i} - 225\,\hat{j}, \qquad \vec{\omega}_2 = -2\,\hat{k}~(rad/s)
$$

**1. Absolute velocity of $A$ and velocity relative to $O_2D$**

For point $A_3$, which belongs to Body 3, we write the rigid-body equation:

$$
\vec{v}_{A_3} = (\omega_3\,\hat{k}) \times \vec{r}_{O_3A} = 225\,\omega_3\,\hat{j}~(mm/s)
$$

On the other hand, the same point $A_3$ can be seen as a point moving relative to Body 2, with relative velocity $(\vec{v}_{A_3})_2$:

$$
\begin{aligned}
\vec{v}_{A_3} &= \vec{v}_{O_2} + \vec{\omega}_2 \times \vec{r}_{O_2A} + (\vec{v}_{A_3})_2 \\
&= (-2\,\hat{k}) \times (-225\,\hat{i} - 225\,\hat{j}) + (v_{A_3})_2 \left(\tfrac{1}{\sqrt{2}}\,\hat{i} + \tfrac{1}{\sqrt{2}}\,\hat{j}\right)
\end{aligned}
$$

where $\vec{v}_{O_2}=0$, and the relative velocity has been split into an unknown magnitude, $(v_{A_3})_2$, and a known direction: that of the slot at the instant studied.
The unit vector at $45^\circ$ (pointing from $A$ towards $O_2$) has been chosen arbitrarily, so the sign of $(v_{A_3})_2$ will give the actual sense.

Equating both expressions:

$$
225\,\omega_3\,\hat{j} = \left(-450 + \tfrac{(v_{A_3})_2}{\sqrt{2}}\right)\hat{i} + \left(450 + \tfrac{(v_{A_3})_2}{\sqrt{2}}\right)\hat{j}
$$

The $\hat{i}$ component gives $(v_{A_3})_2$, and the $\hat{j}$ component gives $\omega_3$:

$$
(v_{A_3})_2 = 450\sqrt{2} = 636.4~mm/s, \qquad \omega_3 = 4~rad/s~(\circlearrowleft)
$$

The positive relative velocity means that pin $A$ moves towards $O_2$ along the slot, while rocker 3 rotates counterclockwise.
The absolute velocity of $A$ is:

$$
\vec{v}_{A_3} = (\omega_3\,\hat{k}) \times \vec{r}_{O_3A} = 900\,\hat{j}~(mm/s)~(\uparrow)
$$

**2. Angular acceleration of $AO_3$ and acceleration of $A$ relative to the slot**

We write the acceleration equations corresponding to the two velocity expressions above. The second one includes the **Coriolis** term, since $A$ moves relative to a rotating body (Body 2):

$$
\begin{aligned}
\vec{a}_{A_3} &= \vec{a}_{O_3} + (\alpha_3\,\hat{k}) \times \vec{r}_{O_3A} + \vec{\omega}_3 \times (\vec{\omega}_3 \times \vec{r}_{O_3A}) \\
\vec{a}_{A_3} &= \vec{a}_{O_2} + (\vec{a}_{A_3})_2 + \vec{\alpha}_2 \times \vec{r}_{O_2A} + \vec{\omega}_2 \times (\vec{\omega}_2 \times \vec{r}_{O_2A}) + 2\,\vec{\omega}_2 \times (\vec{v}_{A_3})_2
\end{aligned}
$$

Dropping the zero terms ($\vec{a}_{O_3}=\vec{a}_{O_2}=0$, and $\vec{\alpha}_2=0$ since $\omega_2$ is constant) and using $\vec{\omega} \times (\vec{\omega} \times \vec{r}) = -\omega^2\,\vec{r}$ when $\vec{\omega} \perp \vec{r}$, we equate both:

$$
(\alpha_3\,\hat{k}) \times \vec{r}_{O_3A} - \omega_3^2\,\vec{r}_{O_3A}
=
(\vec{a}_{A_3})_2 - \omega_2^2\,\vec{r}_{O_2A} + 2\,\vec{\omega}_2 \times (\vec{v}_{A_3})_2
$$

which is a system of two equations with two unknowns: $\alpha_3$ and the magnitude $(a_{A_3})_2$ of the relative acceleration, whose direction is again that of the slot.
Evaluating each term:

$$
\begin{aligned}
(\alpha_3\,\hat{k}) \times \vec{r}_{O_3A} - \omega_3^2\,\vec{r}_{O_3A} &= 225\,\alpha_3\,\hat{j} - 3600\,\hat{i} \\
-\omega_2^2\,\vec{r}_{O_2A} &= 900\,\hat{i} + 900\,\hat{j} \\
2\,\vec{\omega}_2 \times (\vec{v}_{A_3})_2 &= 2\,(-2\,\hat{k}) \times (450\,\hat{i} + 450\,\hat{j}) = 1800\,\hat{i} - 1800\,\hat{j} \\
(\vec{a}_{A_3})_2 &= (a_{A_3})_2 \left(\tfrac{1}{\sqrt{2}}\,\hat{i} + \tfrac{1}{\sqrt{2}}\,\hat{j}\right)
\end{aligned}
$$

Equating components:

$$
\left\{
\begin{aligned}
\hat{i}:&\quad -3600 = 2700 + \tfrac{(a_{A_3})_2}{\sqrt{2}} \\
\hat{j}:&\quad 225\,\alpha_3 = -900 + \tfrac{(a_{A_3})_2}{\sqrt{2}}
\end{aligned}
\right.
\quad\rightarrow\quad
(a_{A_3})_2 = -6300\sqrt{2} = -8910~mm/s^2, \qquad \alpha_3 = -32~rad/s^2~(\circlearrowright)
$$

The negative sign of $(a_{A_3})_2$ means the relative acceleration points from $O_2$ towards $D$, i.e., opposite to the relative velocity: the pin is slowing down in its motion towards $O_2$ along the slot.
As a check, both paths give the same absolute acceleration of $A$: $\vec{a}_{A_3} = -3600\,\hat{i} - 7200\,\hat{j}~(mm/s^2)$.
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