Solving an ODE-45 from Euler-Lagrange Diff. Eqn.

  • Thread starter Thread starter Axecutioner
  • Start date Start date
  • Tags Tags
    Euler-lagrange
AI Thread Summary
The discussion focuses on deriving the equations of motion for a double pendulum using Euler-Lagrange differential equations. The user has formulated two equations but seeks assistance in solving them, particularly without MATLAB expertise. Key points include the importance of defining generalized coordinates, applying the chain rule for differentiation, and calculating kinetic and potential energy accurately. The user emphasizes the chaotic nature of the double pendulum system and expresses a preference against using small angle approximations due to the project’s requirements. Overall, the conversation highlights the complexity of obtaining analytical solutions for such systems.
Axecutioner
Messages
31
Reaction score
0
I need to find the equation of motion of a double pendulum, as shown here:

dimg270.gif


I've gotten as far as the two euler-lagrange differential equations, simplified to this:

K1\ddot{θ}1 + K2\ddot{θ}2cos(θ1 - θ2) + K3\dot{θ}22sin(θ1 - θ2) + K4sin(θ1) = 0
K5\ddot{θ}2 + K6\ddot{θ}1cos(θ1 - θ2) + K7\dot{θ}12sin(θ1 - θ2) + K8sin(θ2) = 0

Assuming initial conditions \ddot{θ}1o, \dot{θ}1o, θ1o, \ddot{θ}2o, \dot{θ}2o, θ2o

What would these equations of motion be?
θ1(t) =
θ2(t) =

I was told it could be done in MATLAB but I don't have the software or know how to use it yet so any help would be appreciated. Step-by-step solution would be even better. Thanks in advance.
 
Physics news on Phys.org
Have you considered the rotational energy aswell? You have to have a moment of inertia dependent term.
The procedure for solving this type of problems is as follows:
1. Define your generalized coordinates for each of the masses: e.g. (x1,y1) = (l1sin(theta(t),l1cos(theta(t)) ...
2. Find the velocity and simplify it. Remember the chain rule for differentiation.
3. Find the Potential energy V and kinetic T(T = 1/2m_1v_1^2+1/2m_2v_2^2+1/2I_1 \omega_1^2+1/2I_2\omega_2^2) where omega = dtheta/dt
and the lagrangian is L = T-V
4. Write down your dynamical equations, 2 masses gives 2 equations. and there you go.
\dfrac{\partial L}{\partial \theta_1} - \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta_1}}=0
Summary: Be strict defining coordinate system. Remember the chain rule and product rule for differentiation. Concentrate it is hard algebra.
The equations of motion comes from solving this differential equation system. Sometimes it is assumed that sin(theta) = theta for theta<<1 and that simplifys a lot.
But in general this problem is chaotic in nature, and you would probably solve nummericaly in matlab.
 
Last edited:
Here's what I've done so far, before I posted the OP:

http://i.imgur.com/c1Sdb.jpg

Some of the K values are the same but just keep each one different, since this entire double pendulum case is a simplification of part of a project I'm working on.

And Rotational Kinetic energy and Kinetic energy are the same for a point mass ( 1/2mv^2 = 1/2mL^2w^2 = 1/2Iw^2) so adding both would be wrong.

And I do not want to use the small angle approximation because the actual pendulum in my project is being torqued with an angle dependent force, and the angles involved cover a wide range.
 
Iv'e tried solving the system in maple nummerically without luck. I don't think i am able to get an analytical solution, sorry. And I've never seen a analytical solution.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top