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Dimensional analysis of a physical pendulum

  1. Nov 19, 2014 #1
    Hi! Sorry if this is the wrong section to post this:

    I am doing a laboration on physical pendulums and I have a bit of trouble making sense of it all and I am in need of some guidance. When I do the analysis I get the standard mathematical pendulum.

    [T]=[m]^a*[l]^b*[g]^c, where a = 0, b = -c, c=-1/2

    Any help?
     
  2. jcsd
  3. Nov 19, 2014 #2
    What variables did you include in your dimensional analysis?
     
  4. Nov 19, 2014 #3

    BruceW

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    Homework Helper

    yeah... what is your question? It looks like you have got the answer there, so I'm guessing you'd like some help in understanding the method?
     
  5. Nov 19, 2014 #4
    No, BruceW, I think you have missed the point.

    The OP asked about physical pendula, but his result is for an ideal pendulum.
     
  6. Nov 19, 2014 #5
    Exactly, I am looking for the general physical pendulum. The analysis i have done is for the pendulum with a point concentrated mass.
     
  7. Nov 19, 2014 #6
    So, I repeat, what variables did you include in your analysis?

    What else do you think you might need to include?
     
  8. Nov 20, 2014 #7
    The inertia of the pendulum. But how do I include this? As a factor?
     
  9. Nov 20, 2014 #8
    What is your definition of inertia, and how does that relate to the mass m?
     
  10. Nov 20, 2014 #9
    The inertia is the resistance of any physical object to any change in its state of motion.

    With the data from the laboration we get an equation with matlab by linearisation so we made the change to the dimensional analysis to get:

    \begin{equation}
    T = \sqrt{\frac{B}{L} + CL}
    \end{equation}

    Where the dimensions:
    \begin{equation}
    B=T^2L \\
    C=\frac{T^2}{L}
    \end{equation}

    However the line of thought seems very wrong and unscientific.
     
  11. Nov 20, 2014 #10
    OK. You want to include "inertia." What are the units of "inertia?"

    Chet
     
  12. Nov 20, 2014 #11
    \begin{equation}
    m\times{\overrightarrow{a}}
    \end{equation}
     
  13. Nov 20, 2014 #12
    These are not units. This is mass times acceleration. Is that what you are calling inertia?

    chet
     
  14. Nov 20, 2014 #13
    Yes that is what I am calling inertia. And it seems that this was our problem. If we take into account the dimensions of inertia our intial analysis will then make sense. Or am I mistaken?

    The dimensions of inertia are [M][T^(-2)] which we haven't taken into account.
     
  15. Nov 20, 2014 #14
    Well, in the original analysis, you have mass m and g (which has units of acceleration). So both sets of units of what you call inertia are accounted for.

    I'm having trouble understanding what you are trying to accomplish. Are you trying to identify the reason that the form of the equation for the period from your experiments is not consistent with the form of the equation for the period for an ideal pendulum, and finding out what physical effects hove been omitted which could explain the difference? If so, what are your thoughts?

    Chet
     
  16. Nov 20, 2014 #15
    Exactly. The form of the equation is consistent with the ideal pendulum. But the pendulum we are studying has a spread mass and not a point mass. When we linearised our data we get an extra constant which we haven't taken into account. With the help of a proper dimensional analysis we can find the dimensions of this constant and determin what it means. How do we make an analysis on a physical pendulum?

    Sorry if I am being unclear, english is not my native language.
     
  17. Nov 20, 2014 #16
    I want to first check your understanding. Which term under the square root radical corresponds to the ideal pendulum analysis, and which term does not?

    Chet
     
  18. Nov 20, 2014 #17
    B/L is the term which equals that of an ideal pendulum whilst the added term C*L should be that of the moment of the inertia?
     
  19. Nov 20, 2014 #18
    If \begin{equation}
    T=A\times{L^a}\times{M^b}\times{(\frac{L}{T^2})^c}\times{(\frac{ML}{T^2})^d}
    \end{equation}

    Then
    1=-2d-2c
    0=a+c+d
    0=b+d
    which has infinite solutions.

    Edit:
    However if we could assume that the value of a and c are unchanged from the analysis of an ideal pendulum it is solvable. However we would get that d must be 0 since the dimensions of time wouldn't be correct...
     
    Last edited: Nov 20, 2014
  20. Nov 20, 2014 #19
    Before we get into this any further, I'd like to clear up a couple of questions.

    1. Regarding the actual functional form for the period that you are using, is this a form that your professor is telling you to use, or is it something that you deduced based on the experimental data? How well does this functional form fit the data?

    2. What are the possible physical factors that could affect the period and that are not included in the ideal analysis? Please make a list.

    3. Is the only parameter you are varying the length of the pendulum?

    4. Is it a ball on a string, or is there significant mass to the shaft that joins the ball to the center of rotation?

    5. Would you know how to analyze the problem using force (and possibly moment) balances on the system to derive the differential equations?

    Chet
     
  21. Nov 20, 2014 #20
    First and foremost thank you for taking your time and helping me.

    1. No function is given. We simply made a dimensional analysis of the physical parameters that are involved, mass, time and length. And tried to make sense of it dimensionally. The form of this function does not fit the data. This is our problem.

    2. Friction, air resistance and the fact taht our pendulum is not ideally rigid. However these parameters are considered small and therefor ignored.

    3. The only varying parameter is the length, yes.

    4. It is not a ball nor is there any significant mass that joins the ball to the center of rotatition. The study was made on a piece of wood with holes in it. We found the center of mass and studied the period of time with a photocell when the piece of wood was connected to a rotational axis length l from the center of mass with an angle of rotation small enough where :
    \begin{equation}
    sin(\theta)\approx{\theta}.
    \end{equation}

    5. Yes but the point of the experimen is to study the harmonic motion of this piece of wood without involving any underlying theory. Instead we had to rely on data.
     
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