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Error Transmission (Moment of Inertia)

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data

    Compute the moment of intertia Io of the two cylinders, per their axis of symmetry, and then, using Steiner's formula, the moment of inertia Ib, as per the axis of rotation.

    kMnuxKL.jpg

    2. Relevant equations

    Io = 1/2*m*R2
    Ib = 2*(Io + m*x2)

    3. The attempt at a solution

    Now, this is one of those huge exercises that you do at the Lab, which give you hours 3, plus work at home. So, this is just a part of it, but I'll type down what I've found already.

    See, the problem here is computing the error. The formula for an f(x,y,z) fraction, is:

    δf = √[(∂f/∂x*δx)2 + (∂f/∂y*δy)2+ (∂f/∂z*δz)2]

    So far so good, right? From the previous questions, I have (relevant to this exercise):

    R +- δR = (2,10000 +- 0,00025)*10-2 m
    x +- δx = (5,00000 +- 0.00025)*10-2 m
    m +- δm = (435,150 +- 0.005)*10-3 kg
    Io +- δIo = (9,595 +- 0.002)*10-5 kgm2

    Plus, from the above formula: Ib = 2,36690115 * 10-3 kgm2

    From our professor, we know Io should be in the range of 10-5 and Ib in the rang of 10-3, so that's all good. The problem I'm facing is when I try to find the error of Ib.

    First of, I'm not sure if I should break down Io into Io = 1/2*m*R2 and have x = m, y = R & z = x (going back to the general error transmission formula). At first I tried this, and then the next time I kept it as is, and went x = m, y = Io & z = x. Still, the problem persists.

    See, the SI units don't add up when I do the partial derivatives, apart from ∂Ib/∂m. And so, I can't add up the results of each derivation, and put them into the square root to get a result of [...] kgm2.

    So, what do I do now? In later questions, wehen we have to compute something else, certain of its quantities, we are instructed to use without their errors, as simple constants.

    Any help is appreciated!
     
  2. jcsd
  3. Apr 29, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    You should be able to go either way, expanding Io or not. The units will work, you just have to realize that the units are just constants that multiply the value of the variable.

    When you wrote the formula:

    Ib = 2*(Io + m*x2)

    then Io becomes a variable that has units "kg m2" When you take the partial derivative w.r.t. Io the units associated with Io are constants, and get carried along with the constant 2 in the derivative.
     
  4. Apr 29, 2017 #3
    I've actually never heard about that. Not knocking down or anything, but I've not seen it in any of my books, or had some professor tell it to us. In previous exercises, when it came to derivatives, anything that was x and nor xy we just went ahead and wrote as 1, without its SI Units. And then the units worked out.

    Okay, let's say, I break dowrn Io into Io = 1/2*m*R2. So, now I have:

    ϑIb/ϑm = R2 + 2*x2, both of which, after doing the necessary acts, have m2 SI units. When multiplied by δm, which is in kg, I get the kgm2 which I'm searching for.

    Moving on, and carrying the SI Units:

    ϑIb/ϑR = 2*m*R + 2*m*x2 || The problem here is, that, 2*m*R has kg*m SI Units, and 2*m*x2 has kg*m2 SI Units. When I multiply by δR, whose units are m, the units aren't the same, and I can't add them up. Same thing happens with ϑIb/ϑx.

    Now, if I keep Io as is, and I keep the SI units like you said, I have:

    ϑIb/ϑIo = 2*kg*m2 + 2*m*x2 || The SI Units now add up, but since I have to multiply by δIo, whose SI Units are kg*m2, then I get kg2*m4. Then I have to raise that to the second power, so I end up with kg4*m8, which makes the SI Units not add up again.

    If I keep Io, but without its SI Units, they again don't add up.

    Generally, each derivative, when multiplied by its error, and then raised to the second power, should give me kg2*m4, since the end result will be under a square root, and in the end, will give me as a result, a value measured in the desirable kg*m2 SI Units.
     
    Last edited: Apr 29, 2017
  5. Apr 29, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    When you take a partial derivative, anything that isn't the variable you're differentiating with respect to is a constant. So here you're differentiating wither respect to R and so the term 2*m*x2 becomes a constant that becomes zero upon the differentiation. Hence:

    ϑIb/ϑR = 2*m*R
     
  6. Apr 29, 2017 #5
    Oh yeah, that's it. I haven't come across anything like this: f(x,y,z) = 5*x +7y*z in quite a while, so I got carried away and treated it like a case of f(x,y,z) = x*y*z. Yup, my bad... Well, that was embarassingly simple...

    Thanks for the help though, I really appreciate it!
     
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