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Show the volume and surface area of a linear object scale using dimensional analysis

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose we consider different flying objects, and that each object is characterized by a linear dimension l.

    Part A: Use dimensional arguments to show that the volume V scales with size as
    [itex]V \sim l^{3}[/itex] and that the surface area scales as [itex]S \sim l^{2}[/itex].

    Part B: Show that this implies that [itex]S \sim V^{\frac{2}{3}}[/itex]


    2. Relevant equations

    [itex]x = [L], y = [L], z = [L][/itex]


    where x, y, and z are the length, width and height of the object respectivley.


    3. The attempt at a solution
    The volume of the object will be equal to the product of its length, width and height which all have the dimensions of length. Finding the product of these three will provide us with the scaling relationship.
    [itex]V = (x)(y)(z) \sim [L][L][L] = [L]^{3}[/itex]

    The same can be done for the surface area instead we will just use the length and width of the object:
    [itex]S = (x)(y) \sim [L][L] = [L]^{2}[/itex]

    I am not sure if this is how the problem is asking me to do the problem using dimensional analysis and scaling relations. Can anybody please verify for me if this is correct? It seems really basic (almost too basic) and I want to use the correct method to solve the problem.

    For part b, I use the Buckingham Pi theorem as shown below:

    [itex]SV^{\alpha}\sim [1][/itex]
    [itex]l^{2}[l^{3\alpha}]\sim [1][/itex]
    [itex]\Rightarrow 2 + 3\alpha = 0 \Rightarrow \alpha = -\frac{2}{3}[/itex]
    Plugging α back into the equation above gives us:
    [itex]SV^{-\frac{2}{3}}\sim [1][/itex]
    This then implies that:
    [itex]S=V^{\frac{2}{3}}[/itex]

    Can someone pleae verify this for me? I feel like I should inlclude a dimensionless constant [tex]C[/itex] in my answer but I am not sure.

    Thanks in advance.
     
  2. jcsd
  3. Jan 17, 2013 #2

    HallsofIvy

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    Science Advisor

    Re: Show the volume and surface area of a linear object scale using dimensional analy

    This is, strictly speaking, only true for a rectangular solid- and you don't know the shape of these objects (or even if they have the same shape). Of course, for any shape, the volume is a dimensionless constant times the product of three lengths, not necessarily "length, width, and height".

    Again, "area equals length times width" is true only for a rectangle. What is generally true is that the area of a figure is some dimensionless constant times the product of two lengths, not necessarily "length and width".

    Well, that is what the problem said, isn't it? "Use dimensional arguments".

    Not "almost". Assuming a rectangular solid is too basic.

     
  4. Jan 17, 2013 #3
    Re: Show the volume and surface area of a linear object scale using dimensional analy

    HallsofIvy, thank you for your reply.

    So to better describe the volume of any object could be better described using the following equation:

    [itex]V=C(l_{1})(l_{2})(l_{3})[/itex] , where C is a dimensionless constant and [itex]l_{1} l_{2}, l_{3}[/itex] are the lengths of the object.

    In the same manner we can write the formula for the surface area as shown below.

    [itex]S=C(l_{1})(l_{2})[/itex] where C once again is a dimensionless constant and [itex]l_{1} l_{2}[/itex] are the lengths of the object.

    Knowing that the dimensions of [itex]l_{1}, l_{2}, l_{3}[/itex] are all length [itex][L][/itex] I can use the same process as I did originally.

    Then we can say that independent of what the dimensionless constant C is the volume will scale with [itex][L^{3}][/itex] and the surface area will scale with [itex][L^{2}][/itex].

    I now realize that I should change my final result to:

    [itex]S=C(V^{\frac{2}{3}}) \Rightarrow S \sim V^{\frac{2}{3}}[/itex]

    Is this now correct? Did I use the Buckingham Pi theorem correctly in order to solve this problem?

    Thanks again.
     
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