Buckinghams PI-Theorem

  • Thread starter BruceSpringste
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In summary, It would be interesting to try it both ways. I think conservation of energy or a forces/acceleration/velocity approach would be a good starting point.
  • #1
Hello! I am doing an experiment on rolling cylinders with both an inner diameter and outer diameter. I.E. they are not solid. I have to determine the time it takes for a cylinder to roll down an inclined slope. I need to do a dimensional analysis with Buckinghams PI-Theorem but I am stuck and need a little help.

Parameters: Width (w) of cylinder, mass (m) of cylinder, inner diameter of cylinder (d), outer diameter (D), length of slope (l), angle of slope (v) and gravity (g).

t=(d,D,w,m,l,v,g)
We define a lengthscale L0 = d, massscale M0=m and time scale T0=(L0/g)1/2

Any thoughts on the next step?
 
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  • #2
Are you thinking of using conservation of energy or a forces/acceleration/velocity approach? It would be interesting to try it both ways.
 
  • #3
BruceSpringste said:
Any thoughts on the next step?

You should do better than making time t a function of all the fundamental parameters. Instead of

[itex] t=f(d,D,w,m,l,v,g) [/itex]

you should show some functions "inside" of [itex] f [/itex]

One way is to think of sets of parameters that can be varied while keeping crucial quantities the same. Another way is visualize solving the problem "the long way". Think of situations where you'd use some parameters to compute some intermediate quantities and not need to use parameters again.

For example, changing the parameters [itex] d [/itex] and [itex] W [/itex] doesn't seem to matter as long as we keep the cyclinders moment of inertia constant.

Use [itex] g(d,D,w,m) [/itex] as the moment of intertia within the function and eliminate [itex] d,w [/itex] from the argument list.

[itex] t = f( g(d,D,w,m) ,D,m,l,v,g). [/itex]

The forces on the cylinder are determined by [itex] v,g,m [/itex] (There is also a frictional force. I'm assuming it's sufficient to make the cylinder roll without slipping.) In terms of dynamics, the only role I see for [itex] v,g [/itex] in the problem is to compute the forces. So forces are functions of the form [itex] h(v,g,m) [/itex].

[itex] t = f( g(d,D,w,m), h(v,g,m), D,m,l ) [/itex]
 
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  • #4
Mass m is not a parameter, because it is the only one with units of mass on your list. w is not a parameter because it is a 2D problem. So,

[tex]t\sqrt{\frac{g}{l}}=f(\frac{d}{D}, v, \frac{D}{l})[/tex]

Chet
 
  • #5
Chestermiller said:
Mass m is not a parameter, because it is the only one with units of mass on your list. w is not a parameter because it is a 2D problem. So,

[tex]t\sqrt{\frac{g}{l}}=f(\frac{d}{D}, v, \frac{D}{l})[/tex]

Chet

I kind of got to the same conclusion. However I had [tex](\frac{d}{l}, v, \frac{D}{l})[/tex]How did you come to that conclusion?
Also how do you scale with l? Shouldnt time be scaled with [tex]t\sqrt{\frac{g}{sin(v)*l}}[/tex].
Since g acts vertically I mean.
 
  • #6
Stephen Tashi said:
You should do better than making time t a function of all the fundamental parameters. Instead of

[itex] t=f(d,D,w,m,l,v,g) [/itex]

you should show some functions "inside" of [itex] f [/itex]

One way is to think of sets of parameters that can be varied while keeping crucial quantities the same. Another way is visualize solving the problem "the long way". Think of situations where you'd use some parameters to compute some intermediate quantities and not need to use parameters again.

For example, changing the parameters [itex] d [/itex] and [itex] W [/itex] doesn't seem to matter as long as we keep the cyclinders moment of inertia constant.

Use [itex] g(d,D,w,m) [/itex] as the moment of intertia within the function and eliminate [itex] d,w [/itex] from the argument list.

[itex] t = f( g(d,D,w,m) ,D,m,l,v,g). [/itex]

The forces on the cylinder are determined by [itex] v,g,m [/itex] (There is also a frictional force. I'm assuming it's sufficient to make the cylinder roll without slipping.) In terms of dynamics, the only role I see for [itex] v,g [/itex] in the problem is to compute the forces. So forces are functions of the form [itex] h(v,g,m) [/itex].

[itex] t = f( g(d,D,w,m), h(v,g,m), D,m,l ) [/itex]

You're not using Buckinghams I am guessing? Also the inner diameter (d) does change the moment of inertia?
 
  • #7
BruceSpringste said:
I kind of got to the same conclusion. However I had [tex](\frac{d}{l}, v, \frac{D}{l})[/tex]How did you come to that conclusion?

The two representations are equivalent. But using d/D appealed to me more aestheticly.

Also how do you scale with l? Shouldnt time be scaled with [tex]t\sqrt{\frac{g}{sin(v)*l}}[/tex].
Since g acts vertically I mean.
That makes use of what you know about the physics of the problem. According to my understanding, you were only allowed to use the Buckingham Pi approach exclusively.

Chet
 
  • #8
Chestermiller said:
The two representations are equivalent. But using d/D appealed to me more aestheticly.That makes use of what you know about the physics of the problem. According to my understanding, you were only allowed to use the Buckingham Pi approach exclusively.

Chet

Thank you, I think I understand your line of thought now!

Edit: Altough the fact that you wrote d/D bothers me a bit. It appealed to you aestheticly. Imagine I could have used physics when I were done with the analysis. Could I have come to the conclusion d/D?
 
  • #9
BruceSpringste said:
Thank you, I think I understand your line of thought now!

Edit: Altough the fact that you wrote d/D bothers me a bit. It appealed to you aestheticly. Imagine I could have used physics when I were done with the analysis. Could I have come to the conclusion d/D?
No. It's mathematical. If d/l and D/l are dimensionless groups in your problem, then their quotient must also be an acceptable dimensionless group. But, if you include their quotient as one of your dimensionless groups, then you need to drop either d/l or D/l.

Chet
 
  • #10
Chestermiller said:
No. It's mathematical. If d/l and D/l are dimensionless groups in your problem, then their quotient must also be an acceptable dimensionless group. But, if you include their quotient as one of your dimensionless groups, then you need to drop either d/l or D/l.

Chet
Yes I understand that. But how come you didn't write D/d instead? I thought maybe you had some physical reasoning behind the choice of d/D instead of D/d.
 
  • #11
BruceSpringste said:
Yes I understand that. But how come you didn't write D/d instead? I thought maybe you had some physical reasoning behind the choice of d/D instead of D/d.
No. Either one is fine.

Chet
 

What is Buckingham's PI-Theorem?

Buckingham's PI-Theorem is a mathematical theorem that states that if a physical problem involves n variables and these variables can be grouped into k dimensionless groups, then the problem can be represented by a relationship between these groups and n-k independent dimensionless variables.

Why is Buckingham's PI-Theorem important in science?

Buckingham's PI-Theorem is important because it allows us to simplify complex physical problems and reduce the number of variables needed to analyze them. This makes it easier to understand and solve the problem.

How is Buckingham's PI-Theorem applied in practical situations?

Buckingham's PI-Theorem is commonly used in engineering and physics to reduce the number of variables in a problem and determine relationships between them. It can also be applied in other scientific fields such as chemistry and biology.

What are the assumptions of Buckingham's PI-Theorem?

The main assumptions of Buckingham's PI-Theorem are that the physical problem is described by a set of variables that can be grouped into dimensionless groups, and that these groups are independent of each other. Additionally, the variables must be measurable and have a linear relationship.

Can Buckingham's PI-Theorem be applied to all physical problems?

No, Buckingham's PI-Theorem cannot be applied to all physical problems. It is only applicable to problems that can be described by measurable variables and have a linear relationship between them. Additionally, the problem must have enough variables to be reduced into dimensionless groups.

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