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Find angular velocity given velocity of attached block

  1. Apr 8, 2014 #1
    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?
  2. jcsd
  3. Apr 9, 2014 #2
    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.
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