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Circular motion of water in a glass

  1. Sep 18, 2008 #1
    Hey all,

    I'm attempting a lab where I want to have water moving in circular motion in a glass. I realized that the water will climb the side of the glass creating a sort of conical shape, if you will. I know that there is a way to calculate the change in height that occurs as the speed of the water increases. I was thinking centrifugal force, but I don't remember my motion well enough for this sort of thing.

    Any and all help would be freakin' awesome:smile:

  2. jcsd
  3. Sep 18, 2008 #2


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    The centripetal force necessary to hold something in a circle of radius R with constant speed v (and so angular speed [itex]\omega[/itex] is mv2/R or [itex]m\omega^2R[/itex]. That vector force, <-mv2/R,0>, added to the gravitational force <0, -mg> gives total force m<-v2/R, -g>, in the xz-plane. More generally, it is [itex]m<-(v^2/R)cos(\theta), -(v^2/R)sin(\theta), -g>. It is the "equilibrium" condition, that that vector be perpendicular to the surface of the water that determines its form.
  4. Mar 11, 2011 #3
    As I recall, It forms a paraboloid
  5. Mar 11, 2011 #4

    Andy Resnick

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    Are you rotating the glass, or are you moving the glass in orbital motion- moving the glass in a circle without rotating the glass? There's a big difference.
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