Find angular velocity given velocity of attached block

[grekin]

Homework Statement

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

Homework Equations

x = cross product
r = distance to point from frame of reference
Boldface will indicate vectors
v = ω x r
v_B = v_A + ω x r_B/A = v_A + v_B/A

The Attempt at a Solution

v_B = ω_AB x r_B/A = -2*ω$\vec{j}$
v_C = v_B + ω x r_C/B
-4.8$\vec{j}$ = -2*ω$\vec{j}$+(-ω$\vec{k}$ x (3cos(30)$\vec{i}$+3sin(30)$\vec{j}$))

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

-4.8$\vec{j}$ = -2*ω$\vec{j}$

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

 

[paisiello2]

In your attempt at a solution you assumed that ω_AB = ω_BC which is not the case.

The length of r_BC is constant and at the instant in time there is no component of v in the i direction. Therefore, ω_BC = 0 and v_B = v_C.