# Find angular velocity given velocity of attached block

1. Apr 8, 2014

### grekin

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$\vec{j}$
vC = vB + ω x rC/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 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$\vec{j}$ = -2*ω$\vec{j}$

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

2. Apr 9, 2014

### paisiello2

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

The length of rBC is constant and at the instant in time there is no component of v in the i direction. Therefore, ωBC = 0 and vB = vC.