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Water Drops Shaken off an Umbrella

  1. Jul 18, 2008 #1
    Hi there,

    I received this question and was rather confused as to how to apporach solving it:

    Water drops shaken off the rim of a rotating umbrella meet the ground in a circle. If you rotated an umbrella making one revoution per second, what would you expect the radius of the circle formed by the drops to be. Justify all values used and define all variables.

    I tried to draw a diagram labelling the radius of the umbrella as x and the radius of the drops as x+y and use Keplers's third law of planetary motion, but was unsuccessful. Any help would be greatly appreciated.

  2. jcsd
  3. Jul 18, 2008 #2


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    Kepler's laws of planetary motion have nothing to do with this. Equations of motion under a uniform gravitational field and rotational kinematics likely do. Can you make a better try at this?
  4. Jul 19, 2008 #3
    i will look it up thanks
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