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Centripetal force + uncertainty

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    The centripetal force is given by F= (mv2) / r. If the mass is measured to be 2.8 +/- 0.1 kg, the velocity is 14 +/- 2 ms^-1, and the radius 8.0 +/- 0.2 m, find the force on the mass, including the uncertainty.


    2. Relevant equations
    F= (mv2) / r


    3. The attempt at a solution
    0.1/2.8 = .04
    2/14 = .14
    .2/8.0 = .025

    add all up and i got .205

    I did the equation and got 69 (2 sig figs) and finally got F = 69 +/- .21

    Is that the proper uncertainty answer? If not please help.
     
  2. jcsd
  3. Sep 6, 2011 #2
  4. Sep 6, 2011 #3
    edit: nvm

    final answer then: F = 270 +/- .21 ? (2 sig figs both)
     
    Last edited: Sep 6, 2011
  5. Sep 6, 2011 #4
    I did not say the force was wrong but you changed it; moreover, you still don't have the uncertainty right as I understand it.

    I guess I'm not helping.
     
  6. Sep 6, 2011 #5
    I'm sorry I kind of don't get it cause I'm really new to this stuff but my final answer without changing the force is:

    F = 69 +/- 14

    is that better? thanks
     
  7. Sep 7, 2011 #6
    You were closer in your original post.
     
  8. Sep 8, 2011 #7
    Why is no one helping with this? I'm about to give him the answer as I think it is--it's really not something I do day-to-day but IAW to site I posted previously.

    We are surpassing Winston Chruchill adage here "The Americans and the British are a single people divided only by a common language."
     
  9. Sep 8, 2011 #8
    Well he has to get the percentage uncertainties for each of the values.

    So for example, for velocity it would 2/14*100 = 14.28% (quite a large uncertainty there)

    Since it's squared, the uncertainty would be double: 28.57%

    Do this for each of the variables, and add up the percentages. Then your final percentage of the calculated value for force should be your uncertainty in Newtons.
     
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