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Homework Help: A question on Dimensional analysis

  1. Jun 12, 2014 #1
    This is actually an example from Physics for Scientists and Engineers by Serway. I am confused about the way they solved it.

    1. The problem statement, all variables and given/known data

    Suppose we are told that acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r^m, and some power of v, say v^m. Determine the values of n and m and write the simplest form of an equation for the acceleration.

    2. Relevant equations

    The problem is solved in the book. But I dont understand the solution properly.This problem was solved in the example as below:

    [tex]a = k(r^n)(v^m)[/tex][tex]L/T^2 = (L^n)(L/T)^m = L^{n+m}/T^m[/tex][tex]n+m = 1 [/tex][tex] m = 2[/tex][tex]n = -1[/tex][tex]a = kr^{-1}v^2[/tex]

    3. The attempt at a solution

    From the question we don't know if k has any dimension or not. So how did they (in the second line) write [tex]L/T^2=(L^n)(L/T)^m ?[/tex] Where did the k go??
  2. jcsd
  3. Jun 12, 2014 #2


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    k is by definition dimensionless. It's the constant of proportionality (meaning it's a constant that changes it from being "proportional" to being "equal")

    The definition of "a" being proportional to "b" is a=kb where k is some constant (hence it's a pure, dimensionless, number)
    Last edited: Jun 12, 2014
  4. Jun 12, 2014 #3
    OOPPss... Suddenly I understand. k is not anything like length mass or time, right? Maybe thats the reason why they left out k?
  5. Jun 12, 2014 #4
    Wait, I have another question. In law of gravitation we know F = GmM/r^2 Here G is a proportionality constant, isn't it? Then how come it is not a pure number (ie it has units like) 6.67×10^−11 N·(m/kg)2 ?
    Last edited by a moderator: Jun 12, 2014
  6. Jun 12, 2014 #5


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    Correct it's a unitless number. (I think such a number is ofted called a "pure number")

    Good question. I don't know a satisfactory answer. It seems I was wrong when I said "k is by definition dimensionless," I suppose it doesn't necessarily have to be dimensionless.

    (Which of course brings us back to your original question, how do we know k is dimensionless in your original post?)

    Sorry about the misinformation, I'm not sure why it must be dimensionless in your OP.
    Last edited: Jun 12, 2014
  7. Jun 12, 2014 #6


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    Perhaps it's because they asked for "the simplest form of an equation for the acceleration."

    (Or perhaps it's just to avoid the infinite possibility that comes with k being of choose-able units)
  8. Jun 12, 2014 #7
    It never seems to be the case. Serway's book describes it as a general procedure. Here is the excerpt from Serway's book:(as this page can be previewed in amazon; I think it never breaks any copyright)


    Now this certainly works for x ∞ (a^n)(t^m); but what happens when we apply this for, say - the law of gravitation? Here, F ∞ Mm/r^2 certainly doesn't pass the test of dimensional analysis (?). I am confused!
  9. Jun 12, 2014 #8


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    Well, why don't you apply dimensional analysis to the law of gravitation? Hint: not every constant of proportionality is unitless, like the π in A = π r[itex]^{2}[/itex] for the area of a circle. And besides, 'constant' means the value doesn't change; it doesn't mean there are no units.
  10. Jun 12, 2014 #9
    So I conclude - dimensional analysis works for equations; but it will work for proportionality only if the constant is unit less. (The book was a bit confusing.)
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