# 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 v_{c} = 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**

**v _{B}** =

**v**+

_{A}**ω**x

**r**=

_{B/A}**v**+

_{A}**v**

_{B/A}**3. The attempt at a solution**

**v _{B}** =

**ω**x

_{AB}**r**= -2*ω[itex]\vec{j}[/itex]

_{B/A}**v**=

_{C}**v**+

_{B}**ω**x

**r**

_{C/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 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[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 ω_{BC} and **r _{C/B}**?

http://ift.tt/OzHc8J

## Leave a comment