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Interesting cylinder problem

  1. Oct 12, 2014 #1
    Good day all,

    someone gave me this problem to try to figure out (for fun) and i'm having a bit of a hard time:

    A closed cylinder of radius R and length L is filled with liquid. The volume of the liquid is equal to 1/3 the total volume of the cylinder. If the cylinder is layed flat (horizontal - long ways), how high up the sides does the liquid go?
    The answer involves an inverse trig function. The answer is not in terms of elementary functions - find an approximate solution graphically.


    Well,
    I im not sure my approach is correct, but I tried integrating "slices of water" from y= -R up to X - with X being the height of water along the side in terms of R (i think the bounds on my integral are wrong).

    I ended up with :
    [tex] \frac{13\pi }{12}= \frac{x\sqrt{R^{2}+x^{2}}}{2R^{2}}+sin^{-1}(\frac{x}{R})(\frac{1}{2})[/tex]

    ...which seems horribly wrong. Any help on how to approach this problem?

    Thanks!
     
  2. jcsd
  3. Oct 12, 2014 #2

    Matterwave

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    Well, the length part is easy, you just have to multiply by L, so the only hard part of the problem is figuring out the area of a cut-off circle. You can do this by integration as you suggest. What we have is an arc of ##x^2+y^2=R^2## from -R up to X where ##X>-R##, that's correct. We can look at just the top half of the arc, and then the bottom half of the arc is symmetric so the area as a function of X is:

    $$A(X)=2\int_{-R}^X \sqrt{R^2-x^2} dx$$

    Is this the integral you had? Wolfram is giving me a function with an inverse tangent in it when I integrate this...
     
  4. Oct 12, 2014 #3
    ahh...well my integral looked a little different . I was integrating volume so I was integrating slices of water...but this looks better. so I integrate this (sin trig sub - note: invese tan comes from just integrating "1" with d(theta) so a "theta" comes out in the answer which converts back invese trig function - I chose sin inverse, but I guess wolfram likes tan inverse. Let me work with this a little bit.

    So I will compute the integral, multiply by L and set it equal to......? [tex] \pi R^{2} L (1/3)[/tex]?

    I guess im also a little confused on how to make my volume i equal to one third to total volume of the cylinder while only having X in terms of R..
     
  5. Oct 12, 2014 #4

    Matterwave

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    So, once you have ##A(X)## you can get a volume simply by multiplying by ##L##, the volume of water will be ##A(X)L## and you have to set that equal to ##\frac{1}{3}\pi R^2 L## and the ##L## will cancel on both sides, so what you really have to do is simply find when ##A(X)## is equal to ##\pi R^2/3##.

    To check if your integration is correct, make sure you get ##A(R)=\pi R^2##.
     
  6. Oct 12, 2014 #5
    Ah ha! Thanks!
     
  7. Oct 12, 2014 #6

    ehild

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    The volume of the water in the " horizontal" cylinder is that of a cylinder of length L but with base a segment of the circle of height X. No need to integral.The area of the segment is area of sector - area of triangle.

    ehild

    layingcylinder.JPG
     
  8. Oct 12, 2014 #7

    Matterwave

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    If that's what your integral looked like, then we are just off by the factor of 2, and you are doing exactly the same thing I was doing. Notice the arc ##y=\sqrt{R^2-x^2}## is only the top half of the circle, we multiply by 2 to get the bottom half of the arc. In other words, the multiplication by 2 is so we don't have to calculate the integral of ##y=-\sqrt{R^2-x^2}## which is the other valid root in the equation ##y^2=R^2-x^2##
     
  9. Oct 12, 2014 #8

    Matterwave

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    Do you know of a nice formula for the area of the base "a segment of the circle of height X"? I saw no way to calculate that except with an integral.
     
  10. Oct 12, 2014 #9

    Matterwave

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    Answered in the post right above your post here.
     
  11. Oct 12, 2014 #10

    ehild

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    The area of the sector is ##R^2\theta / 2##. The area of the triangle is ##R^2\sin(\theta/2)cos(\theta/2)=R^2sin(\theta)/2##. The area of the segment is the difference ##0.5R^2(\theta-sin(\theta))##.
    And ##X=R(1-\sin(\theta/2))##.
     
  12. Oct 12, 2014 #11
    hello ehild,

    the answer must be in terms of R and x... with your method, is there any way to get rid of theta?
     
  13. Oct 12, 2014 #12
    Hello guys:

    so I ended up with an eq:

    [tex]-\frac{\pi}{3}= \frac{x}{R^{2}}\sqrt{R^{2}-x^{2}}-arcsin(-\frac{x}{R}) [/tex]

    ...when my friend sent this to me he said:
    You should be able to find a formula involving inverse trig functions that can be solved
    for the result, but it does not have solutions in terms of elementary functions. Find
    an approximate solution graphically.

    Not even sure what to do.......
     
  14. Oct 12, 2014 #13

    OmCheeto

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    What your friend meant was, that there is no explicit solution.
    You need to solve for volume, in terms of R, graph the function, and then divide the graph into thirds.
    The answer is the point where volume is 1/3, obviously.
     
  15. Oct 12, 2014 #14

    ehild

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    You get rid of theta if you solve the equation volume of segment-based cylinder = 1/3 volume of cylinder.
    Expressed with theta, this means
    [tex]0.5R^2L(\theta-sin(\theta))=LR^2\pi/3[/tex]
    simplified: [tex]\theta-sin(\theta)=2\pi/3[/tex] The approximate solution is about 2.6. And X=R(1-cos(theta/2)). (My previous X was wrong)
     
  16. Oct 12, 2014 #15
    Ehild,

    Sorry im a little confused, I understand getting rid of theta, but 2.6 confuses me. 2.6 what?
     
  17. Oct 12, 2014 #16
    using a graphical method, I got x = 9/16R (which seems a little small..)
     
  18. Oct 12, 2014 #17

    ehild

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    theta=2.6 radian.
     
  19. Oct 12, 2014 #18

    ehild

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    Are you sure your x is what you think is?

    waterincylinder.JPG
     
  20. Oct 12, 2014 #19

    Matterwave

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    Ah, that's quite ingenious. I like this solution! :)
     
  21. Oct 12, 2014 #20

    ehild

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    Note that X = R (1-cos(theta/2)) I made a mistake in that post.
     
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