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Single piston engine problem - position of piston

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Consider the simplified single-piston engine in the figure. The wheel rotates at a constant angular speed of 1 rad/s and the radius of the wheel is 2 m. If the piston is fully extended at time t=0, find the position of the piston at 2s. The figure depicts a wheel with a peg that draws a piston in and out of a bore as the wheel rotates (counter clockwise, with the piston horizontal to the right).


    2. Relevant equations
    x=A cos (omega*t)
    where A is the radius of the wheel



    3. The attempt at a solution

    omega*t = 2rad
    cos 2 * 2m = x.
    since 2 > pi/2, I add 2meters to the resulting x.

    The answer I get is reportedly wrong. what's wrong with my approach?
     
  2. jcsd
  3. Nov 19, 2008 #2

    berkeman

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    Staff: Mentor

    It would help to see a diagram, but are you taking into account the angle of the connecting rod? When the pin is not at 0 or PI, you lose some of the stroke distance since the pin is above or below the axis of the cylinder.
     
  4. Nov 19, 2008 #3

    berkeman

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    Actually it looks like you are trying to include that factor, but the distance x is not cos(2rad*2m). I don't think that's the angle that you should be taking the cos of...
     
  5. Nov 19, 2008 #4
    after doing some searching, I was able to find another problem with an identical diagram.

    http://faculty.ksu.edu.sa/alkurtass/Tut102/Tut102-9.pdf

    problem number 7 shows the piston and the wheel. my thinking of the angle being 2 rad stems from the piece that catches on the piston assembly following the path of the wheel. Since the piston is fully extended at time t=0, I believe that puts the initial angle at 0. am I wrong in thinking so?
     
  6. Nov 20, 2008 #5

    berkeman

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    At t=0, the pin is at the far right. At 2s into rotation, the wheel is 2 radians CCW, which is most of the way to PI radians which is the farthest left. The x displacement to the left of the piston is most of the way to max displacement possible, but a little less. To calc the displacement x, draw a triangle. The theta that you take the cos of is most definitely not 2 radians. Draw the triangle on that figure, and show us your calculation...
     
  7. Nov 20, 2008 #6

    berkeman

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    And actually, I don't think it's a simple cosine. It's a simple angle calculation, though...
     
  8. Nov 20, 2008 #7
    Sorry to intrude, but what is it? I have the same problem, with the same omega = 1 rad/s, but my r = 1.3m.
    I am trying x = Acos(pi - 2rad) but this is not working. I have drawn a little triangle and concluded that the angle is pi minus omega(t), but no bueno! Any suggestions would be greatly appreciated
     
  9. Nov 20, 2008 #8

    berkeman

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    Can you post a sketch showing your work? That would make it easier to help you.
     
  10. Nov 21, 2008 #9
    For the sake of posterity and future students facing this kind of problem, I thought I'd post an update. It turns out that because the piston is fully extended at time t=0, the answer is simply x=A cos (omega*t). I was making the problem more difficult than it needed to be. In this case, the answer is -.83 meters
     
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