1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    2017 Award

    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

    User Avatar
    2017 Award

    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?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...