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Cylider and oil level

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A tank with oil has the shape of a straight circular cylinder. Its diameter is 1.24 m and its height is 2.44 m. The tank lies on one side so that the circular base surfaces is at a right angle to the horizontal plane. It is filled with 0.0045 m3 per second. How fast does the level of oil in the tank rise if the depth in that particular moment is 0.32 m?

    3. The attempt at a solution

    http://www.filehive.com/files/0127/container.png http://www.filehive.com/files/0127/sideways.png

    Since the volume increase is already given, along with the value for d, the only thing I need to do to compute the area of surface of the oil is to find the base of the triangle in the second image and then find the increase in the oil level.

    The base of the triangle is:

    [tex]2r~sin 0.5 \theta[/tex]

    The initial surface area of the oil is:

    [tex]2.44 \cdot 2r~sin 0.5 \theta[/tex]


    [tex]A~(r,~ \theta)~ =~ 4.88r~sin 0.5\theta~~~(1)[/tex]

    To find the height of the triangle [itex]\beta[/itex]:

    [tex]\beta~=~r~-~d~ =~ 0.62~-~ 0.32~ =~ 0.305[/tex]

    This in turn mean that we can find [itex]0.5\theta[/itex] by

    [tex]0.5\theta~ =~ cos^{-1} \frac {0.305}{0.62}[/tex]

    After substituting the value for [itex]0.5\theta[/itex] into (1), I get

    [tex]A~(r)~ =~ 4.88r~sin cos^{-1} \frac {0.305}{0.62}[/tex]

    By using the above equation, I find A(0.62) to be approx. 2.634...

    And finally, computing the increase in level of oil by

    [tex]\frac {0.0045}{A(0.62)}[/tex]

    With the result being 1.7 millimeter. This seems to be a quit reasonable answer to the problem, taking into account that the amount of oil that is filling the container every second.

    Does this sound like a reasonable answer? Did I make any logical or computational errors? Thank you for your time, have a nice day.
  2. jcsd
  3. Jan 27, 2007 #2


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    Homework Helper

    The answer is correct, but I would suggest a few alterations.

    You've defined an area function A (r,theta). But since r is a constant for this problem, you should simply define A as a function of one variable (theta).

    I would prefer a neater approach starting with [tex]\frac{dV}{dt} = \frac{dV}{dh}.\frac{dh}{dt}[/tex] where I've used h in place of your d to avoid confusion. You're trying to find [tex]\frac{dh}{dt}[/tex] given [tex]\frac{dV}{dt}[/tex] at a particular h.

    To get the symbolic expression for [tex]\frac{dV}{dh}[/tex] (which equals A, the surface area of the top of the oil layer), use Pythagoras' theorem instead of trig, [tex]A = 2L\sqrt{Dh - h^2}[/tex] where L is the length of the cylinder and D the diameter.

    Just a suggestion. :smile:
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