End B of the 2 m long rigid rod $AB$ is constrained to move horizontally in the slot as shown in the figure and has a velocity of $1.0\,\mathbf{i}$ m/s. The angular velocity of the rod at the instant shown is $2$ rad/s. The unit vectors along $x$ and $y$ directions are denoted by $\mathbf{i}$ and $\mathbf{j}$. The velocity of point A in m/s is then given by:
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For rigid-body rotation, use $\vec{v} = \vec{v}_B + \vec{\omega} \times \vec{r}_{A/B}$. The perpendicular direction of $\omega \times r$ is key to getting the correct signs.
The rod has length $AB = 2$ m and makes an angle of $60^\circ$ with the horizontal.
Point $B$ moves horizontally with velocity:
\[
\vec{v}_B = 1\,\mathbf{i}
\]
The relative velocity of point $A$ with respect to point $B$ is given by the rigid-body motion relation:
\[
\vec{v}_{A/B} = \vec{\omega} \times \vec{r}_{A/B}
\]
Here, $\omega = 2$ rad/s (counterclockwise).
The position vector of $A$ from $B$ is:
\[
\vec{r}_{A/B} = 2(\cos60^\circ\,\mathbf{i} + \sin60^\circ\,\mathbf{j})
= 2\left(\tfrac{1}{2}\mathbf{i} + \tfrac{\sqrt{3}}{2}\mathbf{j}\right)
= \mathbf{i} + \sqrt{3}\mathbf{j}
\]
Now compute the cross product:
\[
\vec{v}_{A/B} = \omega \times r_{A/B}
= 2(-\sqrt{3}\mathbf{i} + 1\mathbf{j})
= -2\sqrt{3}\mathbf{i} + 2\mathbf{j}
\]
Finally, the velocity of point A is:
\[
\vec{v}_A = \vec{v}_B + \vec{v}_{A/B}
= 1\mathbf{i} + (-2\sqrt{3}\mathbf{i} + 2\mathbf{j})
\]
So,
\[
\vec{v}_A = (1 - 2\sqrt{3})\mathbf{i} + 2\mathbf{j}
\]
