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Relative Motion Analysis: Acceleration

  1. Apr 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A slotted link AC is is driven by the peg P connected to a rotating disk. Point A is fixed. Determine the link's angular velocity, [itex] \omega_{ac}[/itex] and acceleration, [itex] \alpha_{ac}[/itex] when the angular velocity and acceleration are [itex] \omega [/itex] and [itex] \alpha [/itex], respectively.
    [itex] \omega = 6 \frac{rad}{s} [/itex] CCW
    [itex] \alpha = 10 \frac{rad}{s^2} [/itex] CCW
    [itex] l_{ap} = 0.75 \hspace{1 mm} m [/itex]
    [itex] r_{op} = 0.30 \hspace{1 mm} m [/itex]
    [itex] \theta = \frac{\pi}{6} [/itex]
    kBBmaLU.png

    2. Relevant equations


    3. The attempt at a solution
    First, I should state that I am genuinely lost on this problem. I am not sure at all how to go about finding the angular velocity for this one. For the acceleration I was thinking that I could perhaps draw an acceleration diagram and attempt to find the solution that way. Maybe I could do something similar for the velocity? I was hoping someone could sort of push me in the right direction/inform me whether there is a better way to solve this rather than a graphical approach. Any help at all is greatly appreciated. Thanks.
     
  2. jcsd
  3. Apr 11, 2015 #2

    Simon Bridge

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    The diagram uses theta for two different things... probably because they have the same size at t=0.
    Play with the setup for different angles to P ... OP has a constand angular velocity so you can redo the sketch for equal times ans get a feel for what is happening.
    There are several approaches but try finding the equation of the angle of the link as a function of time aband differentiating.
     
  4. Apr 12, 2015 #3
    Awesome, thanks for the reply. One thought I did have was to consider this as a crank and slotted lever mechanism, where the disk is the crank. Could I do an analysis this way? Or would it be too inaccurate?
     
  5. Apr 13, 2015 #4

    Simon Bridge

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    Maybe - I'd do it directly by geometry myself.
     
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