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Math Help - Relative Motion / Piston Motion

  1. #1
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    Oct 2007
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    21

    Relative Motion / Piston Motion

    Hello

    I'm trying to find the velocity of the piston in the following position, but I cant seem to get the right answer? Can you please help me? I really want to learn how to do this!

    Thanks Heaps

    Edward
    Attached Thumbnails Attached Thumbnails Relative Motion / Piston Motion-pq_01.jpg  
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Using the triangle in the figure, by Law of Cosines,
    L^2 = r^2 +x^2 -2(L)(r)cos(theta)
    (68)^2 = (17)^2 +x^2 -2(17)(x)cos(theta)
    x^2 -34x*cos(theta) -4335 = 0 -----------------(i)

    We are to find dx/dt when theta = 30degrees and d(theta)/dt = 206 rpm.

    Differentiate both sides of (i) with respect to time t,
    2x(dx/dt) -34[x(-sin(theta))(d(theta)/dt) +cos(theta)(dx/dt)] = 0
    Divide both sides by 2, then expand,
    x(dx/dt) +17x*sin(theta)(d(theta)/dt) -17cos(theta)(dx/dt) = 0
    [x -17cos(theta)](dx/dt) +[17x*sin(theta)](d(theta)/dt) = 0 ----------(ii)

    When theta = 30degrees,
    x^2 -34x*cos(theta) -4335 = 0 -----------------(i)
    x^2 -29.445x -4335 = 0
    x = {29.445 +,-sqrt[(29.445)^2 -4(1)(-4335)]} / 2(1)
    x = 82.189 or -52.744 mm
    Or,
    x = 82.189 mm

    d(theta)/dt = 206 rev/min = (206rev/1min)(2pi radians /1rev) = 412pi radians/min.

    Substitute those into (ii),
    [82.189 -14.722](dx/dt) +[17(82.189)(0.5)(412pi)] = 0
    dx/dt = -904,231.664 / 67.467
    dx/dt = -13,402.577 mm/min

    Therefore, at that instant, the piston is moving towards O at the rate of 13,403 mm/min. ----------------answer.
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