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Velocity of a Piston

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    In the engine, a 7-inch rod is fastened to a crank of radius 5 inches. The crankshaft rotates counterclockwise at a constant rate of 200rpm. Find the velocity of the piston at t=1/1200min.

    2. Relevant equations
    θ = 400πt
    [itex]\frac{dx}{dθ}[/itex] = -5sinθ - [itex]\frac{25sinθcosθ}{\sqrt{24 + 25cos^{2}θ}}[/itex]

    3. The attempt at a solution
    [itex]\frac{dθ}{dt}[/itex] = 400π
    [itex]\frac{dx}{dt}|_{t=\frac{1}{1200}}[/itex] = -5sin[itex]\frac{π}{3}[/itex]θ' - [itex]\frac{25sin\frac{π}{3}cos\frac{π}{3}θ'}{\sqrt{24 + 25cos^{2}\frac{π}{3}}}[/itex]
    [itex]\frac{dx}{dt}|_{t=\frac{1}{1200}}[/itex] = (-6.30)*(400π)
    [itex]\frac{dx}{dt}|_{t=\frac{1}{1200}}[/itex] = -7917in/min

    EDIT: Okay, so I think I fixed the main problem (I wasn't accounting for θ' in the equation with respect to time). I would really appreciate a check on my answer, though (this is an extra credit question that could boost my calc 1 grade up from a B to an A). Also, the maximum velocity should logically be at π/2 (correct?) but at π/2 (with respect to angle), I only come up with -5 for x'. Can anyone explain this?
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2


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    I did not go through your calculations as they are a little bit tiny to see.

    But consider that the piston velocity = 0 at the top of its stroke and also at the bottom.
    The piston moves fastest when the crank has moved an angle pi/2 measured from the top ( and on both sides by the way ).
    So at t=1/1200min, the crank could be somewhere inbetween.

    One fact that you did not put in your problem is : At what angle is the crank when t=0, and measured to what. ie where is crank angle = 0 .

    EDIT: Crank velocity is fastest in the y direction when theta = pi/2, piston velocity is fastest at another angle.
    Last edited: Oct 23, 2011
  4. Oct 23, 2011 #3
    The problem doesn't specify what the angle is at 0, so I assume that it's at 0 (I was wondering that myself).

    Anyway, at π/2, the velocity only comes out to be 5, which is smaller than 6.3. Even if that was the correct maximum velocity, though, it still seems very slow compared to the speed of the crankshaft. I just can't seem to find what went wrong and where, though. :/

    EDIT: Actually, since I am ultimately taking the derivative with respect to time (and not θ), I have to include θ' in my equation, correct? That would explain why the velocity is so low, but not why the speed at π/3 is greater than at π/3. Multiplied by 400π, the velocity comes out to -7917in/min.
    Last edited: Oct 23, 2011
  5. Oct 23, 2011 #4


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    Well, let me see.

    the old fashioned way:
    θ' = 200 rpm = 200 . 2 . pi rad/min = 1256 rad / min

    Radial velocity
    V = r . θ' = 5 inch . 1256 rad/min = 6280 inch/min

    At t= 1/1200min --> θ = θ' . t = 200rpm . 2,pi rad/rev . 1/1200 min = pi/3 ;
    Vy = V . cos (θ) = 6280 inch/min . (0.5) = 3140 inch/ min ( descending )

    Vx = V . sin(θ) = V . (0.866) = 5438 inch/min ( rod end at crank moving to the left)

    Does that make sense to you?
  6. Oct 23, 2011 #5
    No, it doesn't. Why do you have two velocities? The piston speed is just the change in x with respect to time.
  7. Oct 23, 2011 #6


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    It's not true that "The piston moves fastest when the crank has moved an angle pi/2 measured from the top".

    In this case, the piston travels fastest when the crank has rotated about 63.8° from the top. That's about π/2.82 .
  8. Oct 23, 2011 #7


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    Thanks SammyS.
    I should have known better.
    Post 4 is only crank velocity at theta=pi/2.
    Edit to post 2 also for clarification.
  9. Oct 23, 2011 #8
    Ah, okay, thanks. Back to the question, though, how does my answer look?
  10. Oct 23, 2011 #9


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    The 2 velocities are the velocities of the crank end connected to the rod in the x and y direction, and not the velocity of the piston. It was to give a ballpark idea of what the piston velocity would be.

    The crank end has a velocity component in the same direction as the piston - translation.
    The rod can be thought of as having this translation, and also a rotation as it staightens out( which is why I calculated the 2nd velocity of the rod) .
    I should have mentioned that in the same post.
  11. Oct 23, 2011 #10


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    I would say your answer looks OK, as pi/3 is near the angle of fastest rotation according to SammyS of pi/(2.82).
  12. Oct 24, 2011 #11
    Well, let's hope it's right, then. Thanks for the help.
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