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Shell containing liquid

  1. Apr 30, 2016 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=9560d9156199d743d9a5fd6c3bf9723d.png

    2. Relevant equations


    3. The attempt at a solution

    Considering the upper hemispherical liquid part ,the forces acting on it are , force due to the shell and that due to the bottom liquid .

    Doing a force balance , force due to upper hemispherical shell = ##\pi R^3ρg - \frac{2}{3}\pi R^3ρg = \frac{1}{3}\pi R^3ρg## (downwards)

    Force due to the liquid on upper hemisphere would be ## \frac{1}{3}\pi R^3ρg## (upwards) gives option A) for Q 13 .

    For Q 14 , I think rotating the shell doesn't rotate the liquid which means the pressure at point P should remain unaffected i.e option A )

    Is that correct ?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Apr 30, 2016 #2

    TSny

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    I guess we assume that the pressure is zero at the point of the liquid at the top of the shell.

    I agree with your answer for 13.

    I agree with your answer for 14 if the liquid doesn't rotate. But that's not very interesting. I suspect that the problem assumes that the liquid rotates with the shell.
     
  4. Apr 30, 2016 #3
    In that case do you get any of the four options ?

    Would the assumption of liquid rotating suitable for an intro physics problem :smile: ?

    Assuming the liquid rotates , how should I proceed ?
     
  5. Apr 30, 2016 #4

    TSny

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    I used simple integration. Maybe there's another way. I'll have to think about it. I'm off to bed for now.
     
  6. Apr 30, 2016 #5

    haruspex

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    A walk sometimes helps. Forces one to look for ways so easy they can be done in the head. I arrived at one of the options. I did use my knowledge that in an open tank of uniformly rotating liquid the surface is parabolic.
     
  7. Apr 30, 2016 #6
    Is it possible for the liquid to have parabolic surface in this problem ( considering that the shell is completely filled ) ?

    Could you please share your reasoning .
     
  8. Apr 30, 2016 #7

    haruspex

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    In this case, the sphere is full, so no exposed surface. But does it tell us anything about the surfaces of constant pressure within the liquid?
     
  9. Apr 30, 2016 #8
    o_O
     
  10. Apr 30, 2016 #9

    haruspex

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    Umm... are you hinting that you don't know what I mean by a surface of constant pressure, or you cannot figure out how it relates to the parabolic surface of a partly filled vessel?
     
  11. Apr 30, 2016 #10
    Both .
     
  12. Apr 30, 2016 #11

    haruspex

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    A surface of constant pressure just means the locus of points within the liquid where the pressure is constant. Like a line on a contour map, or like an equipotential surface. If the vessel is not full, the exposed surface is the surface at (gauge) pressure zero, i.e. ambient pressure.
    Given that in a rotating vessel that is not full the exposed surface would be parabolic, can you see how to show that all constant pressure surfaces within it are parabolic? Think about the force balance on a small parcel of liquid, and which way the pressure gradient points.
     
    Last edited: Apr 30, 2016
  13. Apr 30, 2016 #12

    TSny

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    Yes, indeed!
    Very nice. I guess that pushes the integration back to getting the parabolic equipotentials. But if you already know the formula for the parabolic shape, then this is a very clever way to get the answer to the problem.
     
  14. Apr 30, 2016 #13
    Are you referring to the equation ##y = \frac{ω^2r^2}{2g} + constant ## ? Is this the "simple integration" you were referring to in post#4 ?

    :eek: I don't see any parabolic shapes in the liquid. The shell is completely full of liquid .

    Sorry , I am clueless . What is the clever way you are alluding to ?
     
  15. Apr 30, 2016 #14

    TSny

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    I was thinking of starting from scratch. Take a small element of the rotating fluid as shown below. It has a width dx and the areas of the vertical faces perpendicular to the x-axis are A. Apply F = ma in the x direction to find the change in pressure as you move a distance dx along the x-axis.

    haruspex's method is clever. But it presupposes that you are already familiar with a well-known result about rotating fluids.
     

    Attached Files:

  16. Apr 30, 2016 #15
    ## dP = ρω^2x dx ##

    ##P(x) = \frac{1}{2}ρ ω^2 R^2 + P_{center} ##

    ##P_{center} = ρgR##

    ##P(x) = \frac{1}{2}ρ ω^2 R^2 + ρgR ##

    ##Pressure(P) = \frac{3}{2}ρgR ## i.e option C)

    Is this what you are suggesting ?
     
  17. Apr 30, 2016 #16

    haruspex

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    Very good.
     
  18. Apr 30, 2016 #17
    Thanks .

    Just want to clear few things . The sphere is rotating about an axis coming out of the plane of the page . Right ?
     
  19. Apr 30, 2016 #18

    haruspex

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    No, it says about the axis z1-z2, which makes it vertical in the diagram. With a horizontal axis, your Pcenter=ρgR would be doubtful.
     
  20. Apr 30, 2016 #19
    In that case the pressure at the topmost point of the shell on z1-z2 axis would be 0 and that at the bottommost point would be ##2ρgR## ??

    Don't you think the dotted lines at the top and bottom of the shell in the figure indicate that Z1-Z2 axis is coming out of the plane of the page ?
     
    Last edited: Apr 30, 2016
  21. Apr 30, 2016 #20
    Why ? Is it because pressure "ρgh" holds under hydrostatic conditions which would no longer be the case ?
     
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