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Surface of a rotating liquid

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data A cylindrical container of water with a radius of 6.0 cm is placed on a phonograph turntable so that its outer edge just touches the outer edge of the turntable. The radius of the turntable is 14.5 cm, and the turntable rotates at 33 and a 1/3 revolutions per minute. How much is the water depressed in the centre compared to the outer edge of its container?


    2. Relevant equations gz − 1/2 ω2 (x2 + y2 ) = constant


    3. The attempt at a solution I know that x2+y2 is equal to the radius squared. I think from my notes it's equal to the radius squared of the water in the container. So plugging this in, and omega I calculated to be 3.49 rad/s, I get a height of 0.0022 m. I'm not sure where to go from here though.
     
  2. jcsd
  3. Dec 3, 2015 #2
    The height is an answer to the problem, so what more do you want to do?
     
  4. Dec 3, 2015 #3
    So you don't need to calculate another answer using the 14.5 cm?
     
  5. Dec 3, 2015 #4
    It is not clear to me what center the problem is referring to, as the glass with water is not over the center of the turntable.
    So probably the center of the glass with water.
     
  6. Dec 3, 2015 #5
    How do you get an omega of 3.49 rev/s, when the problem says omega is 33.33 rev/min?

    Omega in units of seconds is 0.56 rev/s.
     
  7. Dec 3, 2015 #6
  8. Dec 3, 2015 #7
    my bad, i read "rad" as "rev".
     
  9. Dec 3, 2015 #8
    So I get a depression can be characterized as w^2 R^2 /4g, which results in 0.65 cm.
     
  10. Dec 3, 2015 #9
    @DuckAmuck Might I ask how did you get that, if that's ok?
     
  11. Dec 3, 2015 #10
    The depression is defined as original height of the liquid when it is not spinning, minus the central height when it is spinning?
     
  12. Dec 3, 2015 #11
    Why is it 4g though? Should it not be 2g?
     
  13. Dec 3, 2015 #12
    So there are two forces acting on the surface, gravity and centripetal force. The acceleration from gravity is g, the acceleration from centripetal force is a.
    a is defined as a = rw^2.
    The infinitesimal elements of the surface are dr (radial distance), dz (vertical distance).
    The ratio between the forces and the elements are equal such that:
    dz/dr = rw^2/g

    Now integrate to get z(r). Then keep in mind the volume is still the same as when it was standing still to find the depression.
     
  14. Dec 3, 2015 #13
    It's possible I missed a factor of 2.
     
    Last edited: Dec 3, 2015
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