Potential Energy Density of Hanging String (Lagrangian)

[yngstr]

Homework Statement

Find the potential energy density of a hanging string of mass density m/L that has been displaced from equilibrium at a point a distance d up from the bottom of the string. This point is displaced a distance X in the x direction, and a distance Y in the y direction. The string cannot be stretched.

The Attempt at a Solution

I'm a bit confused as to the actual shape the string makes after displacement. Is it a simple shape such that the top part of the string (L-d) forms an angle to equilibrium, and the bottom part (d) hangs vertically? If this is the case, what is the need for a potential density? This could just be solved discretely...

 

[kuruman]

This is how I would interpret what the question is asking. Consider a string hanging straight down in a vertical line. Put your origin at the free end of the string. Now grab the free end and move it horizontally and "up" to a new point (x,y) such that sqrt(x²+y²) = d. I would assume that the string bends in a catenary type curve. You can try this at home with a real string and see for yourself what shape you get.

 

[yngstr]

This is how I would interpret what the question is asking. Consider a string hanging straight down in a vertical line. Put your origin at the free end of the string. Now grab the free end and move it horizontally and "up" to a new point (x,y) such that sqrt(x²+y²) = d. I would assume that the string bends in a catenary type curve. You can try this at home with a real string and see for yourself what shape you get.

Hmm, I'm fairly certain this is not the case. My ultimate goal is to construct a Lagrangian Density, and the displacement from equilibrium occurs at a point on the rope that is not the at the end points.

X and Y are functions of time and displacement along the rope too, it's field theory...

 

[kuruman]

Hmm, I'm fairly certain this is not the case. My ultimate goal is to construct a Lagrangian Density, and the displacement from equilibrium occurs at a point on the rope that is not the at the end points.

X and Y are functions of time and displacement along the rope too, it's field theory...

Sorry, I guess I misinterpreted the question. Nevertheless, if it involves Lagrangian Density and field theory, do you really think this post belongs in "Introductory Physics"?

 

[yngstr]

This is how I would interpret what the question is asking. Consider a string hanging straight down in a vertical line. Put your origin at the free end of the string. Now grab the free end and move it horizontally and "up" to a new point (x,y) such that sqrt(x²+y²) = d. I would assume that the string bends in a catenary type curve. You can try this at home with a real string and see for yourself what shape you get.

Sorry, I guess I misinterpreted the question. Nevertheless, if it involves Lagrangian Density and field theory, do you really think this post belongs in "Introductory Physics"?

I had posted this in Advanced, but it got moved. It seems that I have to find some sort of relationship between X and Y, but I'm not sure how to do this...it isn't a simple geometric relationship is it?

 

[kuruman]

I will see if someone else can help you with this.

 

[yngstr]

I will see if someone else can help you with this.

Thanks. I've already written down my kinetic energy density and my potential energy density, but I need the relationships between X' and Y' to get the Lagrange Density down to one coordinate.

KE =Integral of : (1/2)*mu*(Xdot^2) dy + (1/2)*mu*(Ydot^2) dy

PE = Integral of: mu *g* Y' dy (tex messing up)

 

[diazona]

Is the string moving? It doesn't seem to say anything about that in the problem.

My interpretation of it when I read it was that you grab on to a point on the string a distance d above the free end, and move that point by (x, y). The portion of the string below that point (the length d at the bottom) will still hang straight down; the portion of the string above the point you grab will probably form some sort of catenary. You can use variational calculus to identify the shape of the top part of the string.