# Example 2.10: Disc with a sliding collar

For the mechanism in the figure (February 2014 exam), knowing that the disc rotates at a constant 90 rpm clockwise, use the analytical method to find the velocity of the end of the bar (point 2).

Data: all dimensions are in millimeters.

<img src="figs/problemas_cin_an/examen2013_p4.png" alt="Disc with a sliding collar" width="550px">

:::{dropdown} Solution

We use an $XY$ coordinate system with its origin at the center of the disc ($O_2$).

Since the angular velocity of the disc is known, we can compute the linear velocity of any of its points.
We therefore need a connection point between the disc and bar $O_4 2$ to find the bar's angular velocity about $O_4$, which will obviously differ from that of the disc.

The connection is point $1$, where we distinguish three different physical points:
$1_D$ (point 1 on the disc), $1_B$ (point 1 on the bar) and $1_C$ (point 1 on the collar).
Looking at how the collar slides along the bar while pivoting on the disc, it is clear that $1_D$ and $1_C$ always coincide (same coordinates, velocities and accelerations), while $1_B$ can only move relative to them along the direction of the bar.

With this in mind, we can compute the velocity of point $1_B$ in two ways:

- As a point of rigid body $O_4 2$:

$$
\vec{v}_{1B} = \vec{v}_{O_4} + \omega_{4}\,\hat{k} \times \vec{r}_{O_4 1}, \qquad \vec{v}_{O_4} = 0
$$

- As a point with a velocity relative to the disc of unknown magnitude and known direction (that of the bar, at $15^\circ$):

$$
\vec{v}_{1B} = \vec{v}_{O_2} + \vec{\omega}_{disc} \times \vec{r}_{O_2 1} + v_{rel} \left(\cos 15^\circ\,\hat{i} + \sin 15^\circ\,\hat{j}\right), \qquad \vec{v}_{O_2} = 0
$$

From the figure, $\vec{r}_{O_2 1} = 45 \left(\cos 74.6^\circ\,\hat{i} + \sin 74.6^\circ\,\hat{j}\right) = 11.95\,\hat{i} + 43.38\,\hat{j}$ and $\vec{r}_{O_4 1} = \vec{r}_{O_4 O_2} + \vec{r}_{O_2 1} = 161.95\,\hat{i} + 43.38\,\hat{j}$ (mm).
The angular velocity of the disc is $\vec{\omega}_{disc} = -90 \cdot \frac{2\pi}{60}\,\hat{k} = -9.42\,\hat{k}~(rad/s)$.

Equating both expressions, since they describe the same physical point, gives one planar vector equation, equivalent to two scalar equations with two unknowns ($\omega_4$ and $v_{rel}$):

$$
\omega_{4}\,\hat{k} \times (161.95\,\hat{i} + 43.38\,\hat{j}) =
(-9.42\,\hat{k}) \times (11.95\,\hat{i} + 43.38\,\hat{j}) + v_{rel} (0.966\,\hat{i} + 0.2588\,\hat{j})
$$

Solving:

$$
\omega_{4} = -1.28~rad/s = -73.34~{}^\circ/s~(\circlearrowright), \qquad v_{rel} = -365.8~mm/s
$$

The velocity of point $2$ then follows from the equation of a point belonging to a rigid body:

$$
\begin{aligned}
\vec{v}_2 &= \vec{v}_{O_4} + \omega_{4}\,\hat{k} \times \vec{r}_{O_4 2} = -1.28\,\hat{k} \times (250 \cos 15^\circ\,\hat{i} + 250 \sin 15^\circ\,\hat{j}) \\
&= 82.8\,\hat{i} - 309.1\,\hat{j}~(mm/s) = 320~(mm/s)~\angle -75^\circ
\end{aligned}
$$
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