A question on Dimensional analysis

In summary, the conversation discusses dimensional analysis and the use of constants of proportionality in equations. It is explained that in order for dimensional analysis to work for proportionality, the constant must be dimensionless. The example from Serway's book is used to illustrate this concept, and the law of gravitation is also discussed as an example.
  • #1
Amio
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This is actually an example from Physics for Scientists and Engineers by Serway. I am confused about the way they solved it.

Homework Statement



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.

Homework Equations



The problem is solved in the book. But I don't 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]

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??
 
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  • #2
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)
 
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  • #3
OOPPss... Suddenly I understand. k is not anything like length mass or time, right? Maybe that's the reason why they left out k?
 
  • #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 ?
 
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  • #5
Amio said:
OOPPss... Suddenly I understand. k is not anything like length mass or time, right? Maybe that's the reason why they left out k?

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

EDIT:
Amio said:
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 ?

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.
 
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  • #6
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)
 
  • #7
Nathanael said:
Perhaps it's because they asked for "the simplest form of an equation for the acceleration."

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)

Y3YP2Rd.png


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!
 
  • #8
Amio said:
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)

Y3YP2Rd.png


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!

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.
 
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  • #9
Thanks.
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.)
 

1. What is dimensional analysis?

Dimensional analysis is a mathematical approach used in science to understand and analyze physical quantities and their units. It involves breaking down complex physical equations into simpler components to better understand their relationship and make predictions.

2. How is dimensional analysis used in science?

Dimensional analysis is used in various fields of science, including physics, chemistry, and engineering. It is particularly useful in solving problems related to unit conversions, understanding relationships between physical quantities, and verifying the correctness of equations.

3. What are the benefits of using dimensional analysis?

Dimensional analysis helps scientists and engineers to avoid errors in calculations and understand the physical meaning behind mathematical equations. It also allows for easier unit conversions and provides a more intuitive understanding of complex physical phenomena.

4. Can dimensional analysis be used for any type of physical quantity?

Yes, dimensional analysis can be applied to any physical quantity as long as it has units assigned to it. This includes both fundamental quantities such as length, mass, and time, as well as derived quantities like velocity, acceleration, and energy.

5. What are some real-world applications of dimensional analysis?

Dimensional analysis has many practical applications, including designing experiments, developing new theories, and solving engineering problems. It is also commonly used in industries such as aerospace, medicine, and construction to ensure accuracy and safety in calculations.

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