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Homework Help: Absolute uncertainty of calculated density

  1. Sep 23, 2015 #1
    1. The problem statement, all variables and given/known data
    An experiment to measure density, d, of a cylindrical object uses the equation



    m=mass=0.029 +-.005kg
    r=radius=8.2 +-.1mm
    l-length=15.4+- .1

    What is the absolute uncertainty of the calculated value of the density?

    2. Relevant equations

    3. The attempt at a solution
    d=.029/((pi)(8.2)^2(15.4))=8.9 * 10^-6

    Relat. Uncertainty(m)=.005/.025=.172
    Relat. Uncertainty(r)=(.1/8.2)+(.1/8.2)=.024
    Relat. Uncertainty(l)=.1/15.4=.006

    Relat. Unc(m) +Relat. Unc.(r) +Relat. Unc.(l)=.202=20.2%
  2. jcsd
  3. Sep 23, 2015 #2


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    Gold Member

    In this case, it is way easier to just "brute force" the uncertainty in the result. You have uncertainty ranges on each of the inputs. And you have a simple formula for the result in terms of the inputs. So you know, for each +/- of each input, which of the + or - produces a larger density value output. So you put all the "tends to produce a bigger" input values together, to get the maximum possible density. And all the "tends to produce a smaller" input values together to get the minimum possibly density.

    Example: R = (a+b)/c

    Each of a, b, c, is positive, and their measured values are:

    a = a_m +/- a_u
    b = b_m +/- b_u
    c = c_m +/- c_u

    Bigger a and bigger b give bigger R. But smaller c gives bigger R. So the extremes of R are just the following.

    R-max = (a_m + a_u + b_m + b_u)/(c_m - c_u)
    R-min = (a_m - a_u + b_m - b_u)/(c_m + c_u)

    You can easily figure out your max and min from this pattern.
  4. Sep 23, 2015 #3


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    You know, I presume, that dividing by a larger number makes the quotient smaller while dividing by a smaller number makes quotient smaller.

    The largest that m can be is 0.029 +.005kg= 0.034 kg and the smallest that r can be is 8.2 - 0.1mm= 8.1 mm. So the largest d can be is [itex]\frac{m}{\pi r^2}= \frac{0.034}{(3.14)(8.1)^2}[/itex]. Calculate that.

    The smallest is m can be is 0.029- 0.005= 0.024 kg and the largest that r can be is 8.2+ 0.1= 8.3 mm. [itex]\frac{0.024}{(3.14)(8.3)^2}[/itex].

    There is an "engineer's rule of thumb" that when measured quantities are added (or subtracted), the errors add, and when quantities are multiplied (or divided) the relative errors add. Here the "relative error" in m is [itex]\frac{0.005}{0.029}= 0.1724[/itex] and the relative error in m is [itex]\frac{0.1}{8.2}= 0.012195[/itex] so the relative error in p is 0.1724+ 0.0122= 0.1846. Multiply [itex]p= 0.29/((3.14)(8.2))[/itex] by that to get an approximation to the actual error.
  5. Sep 24, 2015 #4
    Gracias my friend! What I figured out was that I needed to convert the mm to meters. When I did that I cam up with approximately 1800kg/m^3 which was the correct answer. I will definiteley remember your method though.
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