Rod End Velocity

I can think of two ways to address this problem:

1) Analyze the problem in terms of length invariance
2) Represent the motion as a superposition of translation and rotation

Length invariance method:

$L^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \\ \frac{d L^2}{dt} = 0 \implies 2(x_2 - x_1)(v_{x_2} - v_{x_1}) + 2(y_2 - y_1)(v_{y_2} - v_{y_1}) = 0 \\ \implies v_{y_2} = v_{y_1} + \frac{-2(x_2 - x_1)(v_{x_2} - v_{x_1})}{2(y_2 - y_1)} = 6 + \frac{-2(4 - 1)(7 + 2)}{2(5 - 3)} = \boxed{-7.5}$

Velocity superposition method:

Represent the velocity as a superposition of translation and rotation. Here, $\vec{v_t}$ is the translational velocity of the rod, and $\vec{n}$ is a vector normal to the rod's length (associated with rotation).

$\vec{v_1} = \vec{v_t} + \alpha \, \vec{n} \\ \vec{v_2} = \vec{v_t} - \alpha \, \vec{n}$

Breaking into $x$ and $y$ components.

$v_{1_x} = v_{t_x} + \alpha \, n_x \\ v_{1_y} = v_{t_y} + \alpha \, n_y \\ v_{2_x} = v_{t_x} - \alpha \, n_x \\ v_{2_y} = v_{t_y} - \alpha \, n_y$

The unknowns are $v_{t_x}$ , $v_{t_y}$ , $\alpha$ , and $v_{2_y}$ . There are four equations and four unknowns, allowing for unique solution of $v_{2_y}$ . Solving these equations (left as an exercise for the reader) results in $v_{2_y} = \boxed{-7.5}$ .