Find angular velocity given velocity of attached block

1. The problem statement, all variables and given/known data

The block at C is moving downward at vc = 4.8 ft/s. Determine the angular velocity of bar AB at the instant shown.

2. Relevant equations

x = cross product
r = distance to point from frame of reference
Boldface will indicate vectors
v = ω x r
vB = vA + ω x rB/A = vA + vB/A

3. The attempt at a solution

vB = ωAB x rB/A = -2*ω[itex]\vec{j}[/itex]
vC = vB + ω x rC/B
-4.8[itex]\vec{j}[/itex] = -2*ω[itex]\vec{j}[/itex]+(-ω[itex]\vec{k}[/itex] x (3cos(30)[itex]\vec{i}[/itex]+3sin(30)[itex]\vec{j}[/itex]))

Taking the cross product and setting the y-components equal to each other, I get ωAB=1.04 rad/s. The correct answer is ωAB = 2.40 rad/s. From working backwards, I believe the correct answer can be obtained by setting vB equal to vC:

-4.8[itex]\vec{j}[/itex] = -2*ω[itex]\vec{j}[/itex]

If this is correct, my question is why can they be set equal to each other while disregarding ωBC and rC/B?

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