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Rotating disc of radius R spinning at constant angular velocity

  1. Feb 12, 2016 #1
    Ok, so here's the deal. I'm working on something that I SHOULD know the equations for after 5 years of school and a degree in mechanical engineering, but then again I can't remember why I walked into a room most times. So if ya'll could give me some guidance and at least a starting point I would be most appreciative.

    So here goes:

    I have a rotating disc of radius R spinning at constant angular velocity W. I have a point moving outward on the disc in a straight line with linear velocity V. What I need is an equation for linear velocity as a function of R such that the point will spend the same amount of time on each finite area of the disc as it travels outward.
     
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  3. Feb 12, 2016 #2

    berkeman

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    Welcome to the PF. :smile:

    What do you mean by "spend the same amount of time on each finite area of the disc"?
     
  4. Feb 12, 2016 #3
    Yeah, i didn't word that very well. So its a surface finishing process. I want the head to spend an even amount of time on the surface as it moves out. The end result should have V decreasing as R increases.
     
  5. Feb 12, 2016 #4

    berkeman

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    Ah, so you want the linear velocity to be inversely proportional to the circumference of the circle at that radius. Does that help?
     
  6. Feb 12, 2016 #5
    Helps explain what I'm trying to say, yes. Helps me get the function... no. Granted, my brain is fried, but deadlines don't wait for rested minds do they?
     
  7. Feb 12, 2016 #6

    berkeman

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    How wide (in the radial direction) is the polishing head? How many passes in the radial direction does the polishing head move?

    Circumference is 2πR(t), R(t)=R(0)+Vr(t)*t, Vr(t)=Vr(0)*1/(R(t)-R(0)) or something like that. I'll have to play with it some to try to get you the equation for Vr(t) in terms of R(0)...

    EDIT -- that equation isn't quite right yet...
     
  8. Feb 12, 2016 #7

    berkeman

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    You want Vr(t) to ratio with the ratio of the two circumferences, so when you are at R(0), you are at Vr(0). When you are at 2R(0), you want to be at Vr=(1/2)Vr(0). When you are at 3R(0), you want Vr=(1/3)Vr(0), and so on.

    That looks more like Vr(R)=Vr(0)*(R(0)/R(t))

    How are you controlling the radial motion of the polishing head? You have a linear actuator that you can control the linear speed and can read back the position?
     
  9. Feb 12, 2016 #8
    Assuming that the radial motion is much slower than the angular motion and the point sweeps close enough to each point many times, W doesn't matter.
    r is the position of the point
    dA/dt is the desired cure area per time.
    ##r = \sqrt{\frac{t}{2\pi}} \sqrt{dA/dt}##
    Or if you want the velocity of the point
    ##dr/dt = \frac{1}{2\sqrt{2\pi t}} \sqrt{dA/dt}##
     
  10. Feb 12, 2016 #9

    Nidum

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    In most machining processes this requirement is met the other way around - constant radial feed rate and variable spinning speed . If set up properly this means that the cutting speed is constant and surface finish is same at different radii .
     
  11. Feb 12, 2016 #10
    I agree with Nidium that the better way of doing it would be to slow down the part for a constant finish, since as you go outward you're going to have to remove more material, and you may need to slow down the feed speed more than the surface finish requirements to prevent it from heating or loading up.

    Regardless of which you're slowing down, it would be proportional to 1/R, meaning you can't (theoretically) do the center point.
     
  12. Feb 13, 2016 #11
    Unfortunately, this is a very unique process. I do have complete control of both the part rotation and the radial sweep, however i am limited by how fast the part can spin because of other processes. In this case i HAVE to vary the radial motion to make this work.

    I think you guys have given me enough that I can start building the basic structure. Thank you. Now the fun part... theory VS application :)
     
  13. Feb 13, 2016 #12
    Good luck with it
     
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