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Find the angular velocity of the Absolute Motion System

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Point A is given a constant acceleration a to the right starting from rest with x essentially zero. Determine the angular velocity w of link AB in terms of x and a.

    I have attached an image of the question

    2. Relevant equations

    w = θ'

    α = θ''

    Pythagoras and trig


    3. The attempt at a solution

    x = 2bcos(θ)

    x' = -2bθ'sin(θ)

    x'' = -2bθ''sin(θ)-2bθ'2cos(θ)

    a = -2bαsin(θ) - 2bw2cos(θ)

    I know that bsin(θ) = √b2-x2/4

    and I know that bcos(θ) = x/2

    Substituting these is

    a = -2α√b2-x2/4 - 2w2x/2

    After here is where I get stuck. I'm not sure how to deal with the angular acceleration α. I can see that I have part of the final answer (shown in the image) but I'm not sure about the last few steps.

    Any help would be appreciated. Thank you
     

    Attached Files:

  2. jcsd
  3. Mar 30, 2013 #2

    TSny

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    Gold Member

    Maybe you can get θ' from this. But you'll need to find x' in terms of x and a.
     
  4. Mar 30, 2013 #3
    Looking at the problem again, I had thought that maybe I could use

    x' = V0 + at

    where V0 is zero, but this introduces another unknown. Am I over thinking this?
     
  5. Mar 30, 2013 #4
    Ahh, wait. I got it. I was overthinking it. I only needed to take the first derivative. From there I can solve for θ'

    θ' = -x'/2bsin(θ)

    Knowing that a is constant I can use:

    x' = v

    v2 = v02 + 2ax

    where v0 = 0

    plug in √2ax for x'

    Then i can also see that 2bcos(θ) = √4b2-x2

    Cheers for your help TSny

    But I still have a negative in my final answer which shouldn't be there...
     
    Last edited: Mar 30, 2013
  6. Mar 30, 2013 #5

    TSny

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    The signs are going to depend on your choice of positive directions. If you choose counterclockwise as positive ##\omega##, then note that ##\omega = -\dot{\theta}##.
     
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