1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating Uncertainty Error

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data
    I am writing a lab report and I need to calculate the uncertainty error for Gravitational Potential Energy and Kinetic Energy. The Professor told us that the margin of uncertainty for time is +- 0.025 seconds, margin for height is +- 0.001 meters, margin for mass is +-0.001 kg, margin for velocity is +- 0.2.

    2. Relevant equations
    Eg= mgh
    Ek=1/2 mv^2

    3. The attempt at a solution
    I do not understand how to calculate the margin of error for either equations with the given variables.

    Any Help Is Appreciated.
     
  2. jcsd
  3. Nov 21, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello Rishuv, :welcome:

    I'll answer with an example: suppose you measured 1.000 +/- 0.001 m for the height and 2.000 +/- 0.001 kg for the mass. What is the error for the product?
    well, the maximum due to the uncertainty in the height is 2.002 (and the miniimum 1.998) so a contribution of 0.002 (kg m).
    Similarly the contribution due to the uncertainty in the mass is 0.001 kg m.

    Now there are two ways to add these contributions due to the factors:
    • Pessimists and some engineers say the answer for
    (1.000 +/- 0.001) m x (2.000 +/- 0.001) kg = (2.000 +/- 0.003) kg m. So they add linearly.​

    • Statisticians and physicists say that the errors in the two measurements are independent, so they don't always add up in the most unfavourable way. It turns out that their (correct :smile:) addition rule is that you have to add the squares of the contributions and then take the square of the sum, so ##\sqrt {0.001^2 + 0.001^2} = 0.001 \sqrt 5##, a value of 0.00224, which rounds off to 0.002 :
    (1.000 +/- 0.001) m x (2.000 +/- 0.001) kg = (2.000 +/- 0.003) kg m​
    The rounding off is indicated because the errors are estimates and not that precisely known.

    --

    Does that answer your question ? You can also check here or here
     
  4. Nov 21, 2015 #3
    Thanks for replying, i am still having a little trouble understanding though. For example the experiment i did was that we dropped a 0.2 kg mass from 1.5 meters. How would i calculate the uncertainty for that in terms of gravitational and kinetic energy, as well as the mechanical energy(which I think is just the sum of the uncertainty for gravitational and kinetic energy). It's distance traveled is always changing as well as the height. Sorry if I am asking too much. This is just really confusing me.
     
  5. Nov 21, 2015 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No need to apologize! It's all a bit bewildering in the beginning. You want to do it all perfectly, but it's new territory. Don't be afraid to make mistakes: you learn from them.

    Dropping a .2 kilogram mass from 1.5 m to me means the distance travelled is always 1.5 m, so I must assume there is more going on. Did you drop it from 1.5 m and record the time it took to fall on to something that was varied in height ? For a correct 'advice'on how to deal with the measurement uncertainties it is in fact important to know precisely how the experiments were conducted: what was measured for every observation, what was measured only onde, what factors were taken from elsewhere and so on.

    This is because in experiments there are statistical errors (they may be reduced by repeating the measurement: an average is determined more accurately than one single observation) and there are systematic errors: a value for g of 9.81 may in reality be 9.82 but you use the same value every time you use g, so it doesn't average out"".

    Never mind: if the mass is dropped, the initial kinetic energy is zero and all mechanical energy is potential energy from gravity. I think that the estimated accuracy of 1 mm is way too optimistic for the distance dropped - but it's up to you.

    There is a conservation law that says mechanical energy is constant (provided there is no friction eating away from the energy). With a steel ball in ai you can reasonable well ignore the friction.

    You have an intial energy mgh as the product of ##m\pm \Delta m##, ##h\pm \Delta h## and ##g \pm \Delta g##. Now: you measure m only once, h for the start only once (?) and g is the same for all measurements. Then what? Where does "distance traveled is always changing as well as the height" come from ? And where does the error in time come in ?

    --

    You can see I'm a researcher: you ask for the answer to one question and what you get back is a whole bunch of other questions :smile: !

    --
     
  6. Nov 21, 2015 #5
    haha thanks for the response. So yes we dropped the mass from 1.5 meters. We recorded the time it took to get to the top which is 1.5 meters so that is 0 seconds upper half which is 1.125m the halfway point which is .75 meters, lower half which is 0.375 and bottom which is 0 meters. As you can see the height is constant because we divided it into 5 parts. We got the velocity at each point by multiplying the time it took to get to each point by 9.8 which was the acceleration due to gravity. So for instance lets say we are at the top of 1.5 meters Eg=Em. Eg=mgh. So it would be .2kg*9.8*1.5 =Eg. With these number how do I calculate the uncertainty of Eg.

    Thanks for all your help so far ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculating Uncertainty Error
Loading...