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I Relationship between the radius and area of liquid in a cylinder

  1. Apr 21, 2017 #1
    Hello,

    I did a calculation to determine whether a liquid with a fixed volume ##V##, would be spread over a larger surface area ##A## on the inside mantle of a cylinder, if the cylinder has a larger radius ##r##. So I’d like to find a relationship between the radius ##r## and the area ##A## over which the liquid is spread in the cylinder. The length of the cylinder ##L## and the volume of liequid ##V## are fixed.

    Cylinder.jpg
    If the liquid with volume ##V## is put in the cylinder with a radius ##r##, it would have the following picture when cut through;

    Circular Segment.jpg

    - The area ##O## is formulated as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) = O##.
    - The volume ##V## of the liquid would therefore be defined as: ##\frac{r^2}{2} \cdot (θ – sin(θ)) \cdot L =V##.
    - The area ##A## of the inside “mantle” of the cylinder over which the liquid is spread is defined as ##L \cdot r \cdot θ = A##.

    Combining these formulas will give the relationship of V, θ, r and A in the following form:
    $$\frac{2V}{r(1-\frac{sin(θ)}{θ})} = A$$
    The radius ##r## can be formulated in terms of θ as well since ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r##. The end formula being:
    $$\frac{2V}{\sqrt{\frac{2V}{L(θ-sin(θ)}} \cdot (1-\frac{sin(θ)}{θ})} = A$$
    Plotting the relationship ##\sqrt{\frac{2V}{L(θ-sin(θ))}}=r## with θ being the variable, ##V## and ##L## being fixed values, shows a peculiar graph:

    Graph.jpg

    There are points in the graph of which the slope is actually zero, which means that a change in θ doesn’t give a change in radius. Looking at the graph as if the radius is the variable (which is my initial question) thus says that there are values of ##r## in which the change θ is vertical, as if it means that the radius ##r## doesn’t need to change to make θ change. I’m not sure how to interpret this.
    How is this possible if the length ##L## and the volume of the liquid ##V## are fixed?
     
    Last edited: Apr 21, 2017
  2. jcsd
  3. Apr 21, 2017 #2

    mfb

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    Staff: Mentor

    ##\theta## is only meaningful up to 2 pi, the first point where the derivative goes to zero. Your cylinder is nearly filled close to this value, increasing the radius a tiny bit leads to a large change in the angle, with a diverging ratio where the cylinder is exactly full.

    There is an angle that minimized the area for a given volume. Here is a plot.
     
  4. Apr 22, 2017 #3
    Ah, can't believe I missed this. Furthermore, shouldn't ##\theta## only be meaningful up to ##\pi## instead of ##2\pi## since an angle higer than ##\pi## would give negative values? So with the mentioned formula I can only calculate the area up until the liquid is filling half of the circle?
     
  5. Apr 22, 2017 #4

    mfb

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    Staff: Mentor

    Negative values of what? The sine gets negative, but ##\theta - \sin \theta## stays positive. All your formulas work up to 2 pi. 2 pi corresponds to a fully filled cylinder.
     
  6. Apr 22, 2017 #5
    Just checked, and it does indeed work. The truth is that for the picture:

    Circular Segment.jpg

    I derived a formula myself first, for the area ##O##, which is:
    $$r^2 (\frac{\theta}{2} - sin(\frac{\theta}{2}) \cdot cos(\frac{\theta}{2}))$$
    I thought that if ##\theta ≥ \pi##, this formula would give an erroneous area of ##O## since I'm deriving the side lengths of triangles with an angle ##\theta##, and thus an angle can not exceed ##\pi##.

    However, it does give correct answers even if ##\theta ≥ \pi## and plots the exact same graph as ##\frac{r^2}{2}(\theta - sin(\theta))##.
    But I can't really deduce how:
    $$r^2 (\frac{\theta}{2} - sin(\frac{\theta}{2}) \cdot cos(\frac{\theta}{2})) = \frac{r^2}{2}(\theta - sin(\theta))$$
    Which means that:
    $$cos(\frac{\theta}{2}) \cdot sin(\frac{\theta}{2}) = \frac{1}{2} \cdot sin(\theta))$$
    Is there a way to prove this equation?
     
  7. Apr 22, 2017 #6
  8. Apr 25, 2017 #7
    Thanks for the link. So I was still a bit surprised about how the formula still works if ##\theta > \pi## and I've drawn this scenario where the liquid has fillled more than half the circle:

    Circle 2.jpg
    At first glance, to calculate the circle area ##O## of the liquid I'd say that:
    $$r^2 \cdot (cos(\frac{2\pi - \theta}{2}) \cdot sin(\frac{2\pi - \theta}{2}) + \frac{\theta}{2}) = O$$
    This formula seems to give the same plot as the initial formula for ##O## that I've posted in my opening post, so that

    $$r^2 \cdot (cos(\frac{2\pi - \theta}{2}) \cdot sin(\frac{2\pi - \theta}{2}) + \frac{\theta}{2}) = r^2 (\frac{\theta}{2} - sin(\frac{\theta}{2}) \cdot cos(\frac{\theta}{2}))$$
    Thus, the initial formula still gives valid answers for ##\theta > \pi## up until ##2\pi##

    Bet there's also a way to prove this equation?
     
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