Dimensional analysis of a physical pendulum

[BruceSpringste]

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?

 

[Dr.D]

What variables did you include in your dimensional analysis?

 

[BruceW]

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?

 

[Dr.D]

No, BruceW, I think you have missed the point.

The OP asked about physical pendula, but his result is for an ideal pendulum.

 

[BruceSpringste]

Exactly, I am looking for the general physical pendulum. The analysis i have done is for the pendulum with a point concentrated mass.

 

[Dr.D]

So, I repeat, what variables did you include in your analysis?

What else do you think you might need to include?

 

[BruceSpringste]

So, I repeat, what variables did you include in your analysis?

What else do you think you might need to include?

The inertia of the pendulum. But how do I include this? As a factor?

 

[Chestermiller]

The inertia of the pendulum. But how do I include this? As a factor?

What is your definition of inertia, and how does that relate to the mass m?

 

[BruceSpringste]

What is your definition of inertia, and how does that relate to the mass m?

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.

 

[Chestermiller]

The inertia is the resistance of any physical object to any change in its state of motion.

OK. You want to include "inertia." What are the units of "inertia?"

Chet

 

[BruceSpringste]

OK. You want to include "inertia." What are the units of "inertia?"

Chet

\begin{equation}
m\times{\overrightarrow{a}}
\end{equation}

 

[Chestermiller]

\begin{equation}
m\times{\overrightarrow{a}}
\end{equation}

These are not units. This is mass times acceleration. Is that what you are calling inertia?

chet

 

[BruceSpringste]

These are not units. This is mass times acceleration. Is that what you are calling inertia?

chet

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.

 

[Chestermiller]

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.

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

 

[BruceSpringste]

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

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.

 

[Chestermiller]

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.

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

 

[BruceSpringste]

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

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?

 

[BruceSpringste]

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...

 

[Chestermiller]

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

 

[BruceSpringste]

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

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.

 

[Dr.D]

I don't think I fully understand the point of your experiment, but the factor you are missing in your analysis is the mass moment of inertia (often denoted I or J) which accounts for both mass and the fact that the mass is distributed. See if this gets you going.

 

[Chestermiller]

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.

Thanks. Please tell more about the piece of wood, such as shape etc.

Also, how did you come up with that strange and unexpected functional form for the period. Please try plotting the period versus the length on a log-log plot and seeing how close the data is to a straight line with a slope of 1/2. Also note that the ideal pendulum equation is not the B/L term, but rather the CL term in your equation. (Where did the B/L term come from?)

Chet

 

[Chestermiller]

I can see now where that other term is coming from. For a rigid body pendulum, the moment of inertia about the center of rotation is determined by the parallel axis theorem:

I = I_C + m L²

where I_C is the moment of inertia about the center of mass and L is the distance of the center of mass from the center of rotation. The first term ultimately gives rise to the B/L term.

I didn't notice Dr.D's post #21. He said the same thing there that I am saying here. You should be able to determine the constants B and C in your equation from the analysis of a pendulum with distributed mass.

Chet

 

[BruceSpringste]

Sorry for not replying. I finished the experiment with a correct analysis and totally forgot this post. To anyone viewing this in the future, use Buckinghams PI - Theorem. From that you will reduce the amount of parameters you need to study in order to come to find the formula. The point of the experiment is to use Dimensional analysis (where Buckinghams PI - Theorem is the most powerful tool) and then determine the relations of the remaining parameters through experiment.