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Physics Lab Centripetal Force - deriving the error in the calculated centripetal forc

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the error in the calculated centripetal force, F (dynamic), for each value of r from the estimated error in the quantities m, r, and from the standard deviation from the mean of your three measurements of f. Derive the equation for the error using Rule # 2 for the combination of errors.

    Rule #2 = When you multiply or divide two or more measurements, you add the absolute value of the fractional or percentage errors to find the fractional or percentage error in the result.

    2. Relevant equations

    FD = 4pi(squared) m f(squared)r


    3. The attempt at a solution

    Δ FD = Δm/ m + Δ f/ f + Δr/ r

    Am I on the right track???
     
  2. jcsd
  3. Mar 12, 2012 #2
    Re: Physics Lab Centripetal Force - deriving the error in the calculated centripetal

    if a quantity y depends as

    y=[(k^a)* (x^b)*(z^c)]/(t^d) where k is a fixed constant like ∏ or e or lets say 4/

    Δy/y = b*(Δx/x) + c*(Δz/z) + d*(Δt/t) (Why?)


    (its simply taking sum of all Δ(measured variable)/variable and mu;tiplying it by the power it was raised to in the formula)
    here Δy means maximum possible error in y.

    Now calculate your answer again
     
  4. Mar 12, 2012 #3
    Re: Physics Lab Centripetal Force - deriving the error in the calculated centripetal

    So would it be

    ΔFd = Δm/m + 2 Δf/f + Δr/r ???

    Thank you for replying!!
     
  5. Mar 12, 2012 #4
    Re: Physics Lab Centripetal Force - deriving the error in the calculated centripetal

    Yes now the equation is correct.

    Also part of the problem suggests you calculate Δf using the mean values.

    What we actually assume in these cases is that the mean value you calculate after repeating the experiment n times is equal to the real vale of f.

    Suppose you repeat the experiment 3 times and get value of f as 3.10,3.20, 3.20.
    The mean of all this will be (3.10+3.20+3.36)/3=3.22

    Now we assume the the real value of f is 3.22

    so Δf in case one is |3.10-3.22|(where || is for absolute value) and similarly for other cases.
    We take || to calculate maximum possible error.


    I hope you understood
     
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