Rod AB sliding over wheel C

Setting up. The rod slides over the wheel at $C$, so the point of the rod that is at $C$ moves along the rod. Let $A$ be the origin and $\theta$ the angle of the rod with the horizontal, assuming $AB = 508$ mm measured from $A$ along the rod.

$$\tan\theta = \frac{178}{254} \;\Rightarrow\; \theta = 35.02^\circ, \qquad AC = \sqrt{254^2 + 178^2} = 310.2\ \text{mm}$$

            B
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       C o  <- wheel, 178 mm high
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    A *--+----> vA = 635 mm/s
      254 mm

(a) Angular velocity of the rod

Velocity of the rod point at $C$: $\vec v_C = \vec v_A + \vec\omega\times\vec r_{C/A}$. It must be along the rod, so its component perpendicular to the rod is zero. The component of $\vec v_A$ perpendicular to the rod is $v_A\sin\theta$, and the rotation contributes $\omega\,(AC)$ in the opposite sense:

$$v_A\sin\theta = \omega\,(AC)$$ $$\omega = \frac{v_A\sin\theta}{AC} = \frac{635\times\sin 35.02^\circ}{310.2} = 1.1749\ \text{rad/s (counter-clockwise)}$$

(b) Velocity of end B

$$\vec v_B = \vec v_A + \omega\hat k\times\vec r_{B/A}, \qquad \vec r_{B/A} = 508(\cos\theta\,\hat i + \sin\theta\,\hat j)$$ $$\vec v_B = 635\hat i + 1.1749\,(508)(-\sin\theta\,\hat i + \cos\theta\,\hat j) = (292.5\,\hat i + 488.8\,\hat j)\ \text{mm/s}$$ $$v_B = 569.6\ \text{mm/s}$$

at 59.1° above the horizontal, to the right.

Answer: $\omega = 1.1749$ rad/s counter-clockwise; $v_B = 569.6$ mm/s at 59.1° above the horizontal. If the 508 mm is measured from $C$ instead of from $A$, replace $r_{B/A}$ by $AC + 508 = 818.2$ mm.