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Wize University Physics Textbook (Master) > Foundations / Introduction / Measurement

Dimensional Analysis

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By looking at just the physical quantities, ignoring the numbers and the units, we are doing dimensional analysis. There are symbols that we can represent each physical quantity with, and we use square brackets to show that we are talking about the dimension. Here is a list of dimensions, notice how similar they are with units:


In dimensional analysis, we expect our equations and statements in physics to be dimensionally consistent.
  • If I asked you: "What's the temperature like outside?" and you answered, "42 metres," I would say, "Wow! That's dimensionally inconsistent!"
  • In an equation, we would expect both sides to have the same dimensions in order for the equation to be dimensionally consistent.








Example

Is this equation dimensionally consistent? (cc is the speed of light)
E=mc2E = mc^2
Answer is yes!








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Let's use dimensional analysis to answer the following problems:

1. You are in the last few minutes of an exam and you are rushed. You forget the exact formula for centripetal acceleration. You remember a few options of what it might look like. Which formula is correct?

a=vr         a=vr2        a=v2r      a=v2r2a=\frac{v}{r}\ \ \ \ \ \ \ \ \ a=\frac{v}{r^2}\ \ \ \ \ \ \ \ a=\frac{v^2}{r}\ \ \ \ \ \ a=\frac{v^2}{r^2}
Write out the units of velocity, m/s, and r, s, for each option. Square them in the options where the quantities are squared. We need to have a term that gives us units of acceleration, ms2.\frac{m}{s^2}.

Only the third option gives us this: a=v2r  →(m/s)2(m)=m2ms2=ms2a=\frac{v^2}{r}\ \ \rightarrow \frac{(m/s)^2}{(m)}=\frac{m^2}{m s^2}=\frac{m}{s^2}


Based on units, the correct formula is
the third choice
.


2. You re-arrange a kinematic formula into the following form. Use dimensional analysis to determine if you successfully re-arranged the formula.

vf−vo2a=Δx\frac{v_f-v_o}{2a}=\Delta x

Considering the units (velocity on the numerator, acceleration on the denominator):

m/sm/s2=s≠m\frac{m/s}{m/s^2}=s\neq m
vf−vo2a=Δx\frac{v_f-v_o}{2a}=\Delta x
We wanted units of distance (m). So this is not the correct formula. (The correct formula is vf2−vo22a=Δx\frac{v_f^2-v_o^2}{2a}=\Delta x).



Based on units, this
IS NOT
(is/is not) a correct formula.






3. Show that the following two terms use the same units: (PS, g has units of acceleration, m/s2)

12mv2          mgh\frac{1}{2}mv^2\ \ \ \ \ \ \ \ \ \ mgh
Units of the left-hand term: (careful: "m" in a formula is a mass. "m" as a unit is a distance)

(kg)(m/s)2=kg m2s2(kg)(m/s)^2=\frac{kg~ m^2}{s^2}
Units of the right-hand term:

(kg)(m/s2)(m)=kg m2s2(kg)(m/s^2)(m)=\frac{kg~ m^2}{s^2}
These are the same, so they have the same units. So we could add these terms together! By the way, this unit is called the Joule (J), and these terms describe energy.

X=YZ32WnX = \frac{Y Z}{32 W^n}

If X=[Mθ]X = [M\theta] , Y=[MT2]Y = [MT^2] , Z=[θ/L2]Z = [\theta / L^2] , and W=[T/L]W = [T/L] , what is the exponent n?