Solving Flexible Rope Eqn w/ Matlab & FlexPde - Alessandro

[alenglaro]

Hello!
I'm trying to solve a differential equations system describing the shape of a flexible rope of constant density rotating with constant rotational speed w, as shown in the attached picture, and subjected to gravity. The ends of the rope are fixed to the rotating frame. I set the equations in a cylindrical coordinate system, which seems more appropriate:$$\frac{d}{ds}\left(T\frac{dr}{ds}\right)+\rho\omega^2 r=0$$
$$\frac{d}{ds}\left(T\frac{d\theta}{ds}\right)=0$$
$$\frac{d}{ds}\left(T\frac{dz}{ds}\right)=\rho g$$
$$ds= \sqrt{1+\left(\frac{dr}{dz}\right)^2+r^2\left(\frac{d\theta}{dz}\right)^2}dz$$
(Where T is rope tension and s is the curvilinear coordinate lying on the rope)
boundary conditions:
- where z=0, r=r0 theta=theta0
-where z=H, r=r1 theta=theta1
-where dr/dz=0 r=Rmax
(without setting BCs for T if it's possible)

How can I solve that for r, theta and T in MATLAB or FlexPde? Any help would be appreciated. Thank you in advance and sorry for my English.
Alessandro

 

[jvicens]

Hello Alessandro,

Thank you for sharing your problem and equations with us. I understand the importance of accurately solving differential equations in order to understand and predict the behavior of physical systems. In this case, you are trying to model the shape of a flexible rope under the influence of gravity and rotation. This type of problem falls under the category of mechanics and can be solved using numerical methods in MATLAB or FlexPde.

First, let's break down your equations and understand what they represent. The first equation represents the balance of forces acting on the rope in the radial direction. The term on the left-hand side represents the tension force in the rope, which is equal to the product of the tension T and the rate of change of radial displacement with respect to the curvilinear coordinate s. The term on the right-hand side represents the centrifugal force due to the rotation, which is proportional to the square of the rotational speed w and the distance from the center of rotation r. This equation essentially states that the tension force and centrifugal force must be in equilibrium in order for the rope to maintain its shape.

The second equation represents the balance of forces in the tangential direction. Since there are no external forces acting in this direction, the term on the right-hand side is equal to zero. This means that the tangential component of tension, which is represented by the term on the left-hand side, must be constant along the length of the rope. This makes sense intuitively, as the tension in a rope should be the same at any point along its length.

The third equation represents the balance of forces in the vertical direction. The term on the right-hand side represents the weight of the rope, which is equal to its density rho multiplied by the acceleration due to gravity g. The term on the left-hand side represents the vertical component of tension, which must balance the weight in order for the rope to remain in equilibrium.

The fourth equation represents the relationship between the curvilinear coordinate s and the vertical coordinate z. This is necessary in order to convert the differential equations from the cylindrical coordinate system to the Cartesian coordinate system, which is more commonly used in numerical simulations.

Now, let's discuss how to solve these equations in MATLAB or FlexPde. Both of these software packages have built-in functions for solving differential equations numerically. In MATLAB, you can use the "ode45" function, which uses a Runge-Kutta method to solve the equations. In Flex