How Do You Calculate the Uncertainty in Physical Measurements?

  • Thread starter Thread starter physicsss
  • Start date Start date
  • Tags Tags
    Uncertainty
AI Thread Summary
To calculate the uncertainty in the area of a circle with a radius of 2.0 cm, one must consider the impact of measurement errors in the radius. The area is computed using the formula A = πr², and the error in area can be expressed as δA = 2πrδr, where δr represents the uncertainty in the radius. For a radius error of ±0.1 cm, the calculated areas for the adjusted radii are 13.9 cm² and 11.3 cm², with the area at the nominal radius being 12.6 cm². Additionally, when calculating uncertainties in other measurements, such as air density using the ideal gas law, the relative error formula can be applied, incorporating uncertainties in pressure and temperature. Understanding these principles is crucial for accurately reporting measurement uncertainties in physical calculations.
physicsss
Messages
319
Reaction score
0
Can someone get me started with finding the approximate uncertainty of the area of a circle with a radius of 2.0 cm?


Thanks,
Yours
 
Physics news on Phys.org
Ask yourself by what amount will the area of the circle change if you introduced a small error in the radius of the circle?
 
um...

Can you give me the answer so I know if I'm doing this right? (I need to show work for full credit anyway)
 
Tell us what you did so far and I'm sure someone will provide further guidance! :-)
 
I found the area of both pi*(2.0+0.1)^2 and pi*(2.0-0.1)^2, which is 13.9 and 11.3, respectively. I also found the area of pi*(2.0)^2, which is 12.6. Then I got stuck.
 
That's a good start! Generally, the error in area will be related to the error in measuring the radius by \delta A = 2\pi r \delta r[/tex] which is essentially what you have for the specific case that \delta r = 0.1 and r = 2. (\delta means "error" or "deviation.")
 
So the final answer is 2.8 X 10^9 +/- 1.3?
 
Last edited:
I would refine it just a little. The radius was given to two significant digits meaning that it is given to the nearest 0.1 centimeters from which I would use \pm 0.5 for the error in r.
 
Thank you for your help, Tide!
 
  • #10
I'm being asked to compute the uncertantity of air density that i calculated using pv=nrt. I have a uncertainty of .1 on the barometer, and .5 degrees for temperature, but the equation in my fluid mechanics book is too complicated. I can't draw what they're trying to say. The formula has to do with like 3 terms squared to the ^1/2 power. I just can't get over all the complicated lingo, I know its really easy. My recitation teacher explained it, but I'm having a brain freeze, anybody?
 
  • #11
It appears you should be using the formula
D =\frac{P}{RT}
where:
D = density, kg/m3
P = pressure, Pascals ( multiply mb by 100 to get Pascals)
R = gas constant , J/(kg*degK) = 287.05 for dry air
T = temperature, degK = deg C + 273.15
 
  • #12
purduegrad said:
I'm being asked to compute the uncertantity of air density that i calculated using pv=nrt. I have a uncertainty of .1 on the barometer, and .5 degrees for temperature, but the equation in my fluid mechanics book is too complicated. I can't draw what they're trying to say. The formula has to do with like 3 terms squared to the ^1/2 power. I just can't get over all the complicated lingo, I know its really easy. My recitation teacher explained it, but I'm having a brain freeze, anybody?

I don't know whether I can make it any easier but essentially you just need to substitute numbers into the equation

\frac {\delta n}{n} = \sqrt{ \left( \frac{\delta v}{v} \right)^2 + \left( \frac{\delta p}{p} \right)^2 + \left( \frac{\delta T}{T} \right)^2}

The symbol \delta means "deviation" or "error" of a given quantity. Note that generally a quantity such as \frac {\delta n}{n} is called the "relative error."
 
Back
Top