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Homework Help: Mass momentum of inertia in lab, error analysis

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi, I've a lab assingment, and the labwork must be planned beforehand, but I have some trouble figuring out some parts of my error analysis.

    So, I'm supposed to measure (as in not use integrals to find out) the mass momentum of a ball. The plan is to place the ball on an inclined plane, and let it roll to the ground. I'm going to measure the mass, radius, the velocity just before the ball hits the ground, and the distance to ground from the beginning position.

    This is done with 10 or so different starting distances (and consecutively with 10 different end velocities).

    2. Relevant equations

    First I derive the formula for my calculations from the principle of concervation of mechanical energy:
    [tex]mgh &= \frac{1}{2}I_{cm} \omega _{cm}^2+\frac{1}{2}mv_{1}^2 \newline \ldots \Leftrightarrow mgh = \frac{I_{cm}+mr^2}{2r^2}v_{1}^2\\[/tex], where m is the mass, r radius, g is the acceleration due gravity, h is the initial position when the ball is at rest, v is the velocity just before the ground, and I_cm is the mass momentum of inertia.

    Now, I have 10 different results, I use linear fit on a set of points [ m*g*h, v^2], so I get an equation like [tex]U(v^2) = kv^2+b[/tex]. Now, computer gives me k with it's error limit (and b, which is just the systematic error).

    [tex]k=\frac{I_{cm}+mr^2}{2r^2} \Leftrightarrow 2kr^2 = I_{cm}+mr^2 \Leftrightarrow I_{cm} = 2kr^2-mr^2[/tex]

    Now I have the equation for my mass moment of inertia.

    3. The attempt at a solution
    The problem is, the when using the propagation of uncertainty (http://en.wikipedia.org/wiki/Error_propagation): [Broken]
    [tex]\displaystyle\delta I_{cm} &= \sqrt{\sum_{i}^n (\frac{\partial I_{cm}}{\partial x_i} \delta x_i)^2} =
    \sqrt{(\frac{\partial I_{cm}}{\partial k} \delta k)^2+(\frac{\partial I_{cm}}{\partial r} \delta r)^2+
    (\frac{\partial I_{cm}}{\partial m} \delta m)^2}
    \\ &= \sqrt{4r^4\delta k^2+4r^2(2k-m)^2\delta r+r^4\delta m^2} = r\sqrt{4r^2\delta k^2+4(2k-m)^2\delta r+r^2\delta m^2 }.[/tex]

    k is not an independent variable, as it depends on mass m and radius r. So the big question is, how am I supposed to change the last equation, when taking into account that k depends on m and r?

    As for my bad english, I apologize, please ask, if I didn't make myself clear. Thank you.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 1, 2008 #2


    User Avatar
    Homework Helper

    Welcome top PF.

    Actually I might consider trying to use a different k,

    k = k'm

    such that you could write your equation as

    I = mr2*(2k' - 1)

    Then you might be able to use the RSS of the relative errors of m,2r,k'

    (Note I think your r2 term should be carried as ∂r/r twice because it should be treated as r*r for taking the RSS.)
  4. Dec 1, 2008 #3
    Still thinking about this, but the equation you have for propigation of uncertenty assumes you have an explicit function for moment of Inertia I as a function of all your variables. You don't have this. If you wanted to to this, you could solve for I and you would have both of your data values (h and v) on the left side of the equation. This, however would eliminate using the graphical method to find a slope. Each run would yield values for height and velocity to plug into an equation, and you would have ten different values for moment of inertia. You could simply average these.

    For error analysis, don't forget to put some uncertenty in height into your equation too.

    To more acurately measure height in the lab, I would consider measuring the distance up the ramp and using trig to find it's actual height. More accurate
  5. Dec 1, 2008 #4
    Ok. I think I've got it. Basically two ways to do this error analysis.

    1. Solve explicitly for moment of inertia and use your propigation of uncertenty you found from wikipedia. This eliminates the graphical method and leaves you with 10 values for moment of inertia to average in the end. With this model, you are GUESSING theoretical uncertenties for all variables in you equation for moment of inertia.

    2. still thinking
  6. Dec 2, 2008 #5
    If I solve explicitly for moment of inertia:
    [tex]I=mr^2 \frac{2gh-v^2}{v^2}[/tex]
    (did I understand you correctly?)

    Then mass moment of inertia depends on m,r,h and v. But v depends on h :O How to take this into consideration?

    And lowlyPion's solusion was somewhat confusing: I =mr2*(2k' - 1), but k still depends on m and r, doesnt it?
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