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Velocity at Piston P and angular velocity of link HP help

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The instantaneous configuration of a slider crank mechanism has a crank GH 10cm long, the connecting rod HP is 50cm. The crank makes an angle of 60 degree with the inner dead centre position and is rotating at 110 rev/min. Determine the velocity of the piston P and the angular velocity of the link HP.

    cannot find any simliar examples in text books or online which will help me with this question.



    2. Relevant equations




    3. The attempt at a solution
    Really have no idea where to start but i've made a start whether its right i dont know.

    Right, first off i've converted the 110rev/min=w into 11.5Rads/s
    so w=11.5Rads/s
    i then assumed you had to find the velocity at H in order to find the velocity at P
    so Velocity at H = (Wgh) X gh
    Vh = 11.5rads/s X 0.1m
    = 1.15ms^-1

    (i dont know if that is correct for starters & how i use that to find the velocity at P then the angular velocity at HP, please help?)
     

    Attached Files:

  2. jcsd
  3. Apr 7, 2012 #2

    LCKurtz

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    Before putting all the numbers in, I would label the length of the crankshaft ##c##, and the radius of the wheel ##r##, the distance ##PG = x##. Then write an equation giving ##x## in terms of the angle of rotation ##\theta##. Solve the equations you need before plugging in numbers.
     
  4. Apr 8, 2012 #3
    i have an idea of what you mean but i'm still quite lost
     
  5. Apr 8, 2012 #4

    LCKurtz

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    You are going to have to show some effort. Drop a perpendicular down from H to the horizontal axis. That breaks ##x## into two legs of right triangles. You should be able to use trigonometry and the Pythagorean theorem to express ##x## in terms of the angle of rotation ##\theta## and the other constants ##r## and ##c##. Ultimately, the velocity of the piston will be ##\frac{dx}{dt}##.
     
  6. Apr 11, 2012 #5
    would this formula work, [ -rw(cos(60)+((sin2(60)/(2xsqrt(n²-sin²(60)))) ]
    where n = L/R
    i get a negative value when i use it, the piston is moving in the negative direction.
     
  7. Apr 11, 2012 #6

    LCKurtz

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    I'm sorry, but since you don't seem to care about my suggestions, I am resigning from this problem.
     
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