Along with animation 🙏🙏🙏

Khud bhi kuch karlo .....

I just found it more convenient. From my experience, it makes the equations look a little cleaner that way - this is not uncommon for pendulum problems. But could have just as easily pointed it up.

By adding the constraint, it will constrain θ1 to be equal to θ2 at all times, which effectively turns this into a single DOF model. It seems to me that you should be using two separate models. The first is the original 2 DOF model and the second is a 1 DOF model that starts at the moment that θ1 = θ2. The first model is used to generate the initial conditions for the second model - ie run the first model, then at the moment that θ1 = θ2, use that θ1 and θ1_dot as the initial conditions for the second model. For what it's worth, you would add the constrains by adding to the Lagrangian + λ(θ1 - θ2). Then, since you've added the coordinate, λ you will produce a third equation from Lagrange's Equations you would differentiate with respect to λ.

Including the spring was just for fun. No reason for it except to show how it could be included. Might be useful for modeling something like a prosthetic knee joint, but can also easily be ignored by setting k_t=0.

In the case of the compound pendulum the rotatory kinetic energy effects become significant (when compared with the simple pendulum). In this case each bar is NOT rotating about its center of mass, but rather about each hinge point.

It's a little hard to explain in the comments - I'll probably need to make a video on it, but basically, you would do the following...instead of using the kinematic constraints to relate the potential and kinetic energies to θ1 and θ2, you would write the energies in terms of x, y and their derivatives and then you would add two constraint equations to relate the coordinates to one another. It makes it slightly easier to derive the energy expressions by using the Lagrange multiplier methods, but it increases the number of equations and unknowns.

@@Freeball99 thank you for your answers, i’ll ask again if i still confuse🥺

This is because I used J about the C.G. - so I had to include the translation of the C.G. If I had just used the correct J's (about the hinge points) then I could have ignored the translational velocity, but as I showed, it can be tricky to get this correct, so it helps to be able to get there step-by-step from basic principles.

Oops! Found a bug. My kinetic energy was off because an angle was used instead of a velocity. My expression is much simpler. The animation reminds me of a circus trapeze act. My animation still gains a little energy, but is much more stable. I haven't added the forcing moment as yet. What do you suppose I should use? Mo*cos(wt)? The moment of inertia from the end of a rod is m*L**2/3 instead of M*L**2/12 as it is at the center of mass. However, I dunno if a forcing moment needs to include that or not. I will probably add a control to change omega on the fly, just for the fun of it.

I'm about to release a video (part 2) that shows the rest of the derivations of the equations of motion. Should be ready by this weekend.

hello @greg Hoke, would you mind sending me the file of your work (its fine if you're not comfortable with share it). Iam currently struggling with an assignment, and I just need an example for reference. sorry for the trouble

You need to add (superimpose) these effects. Take the kinetic energies from both problems and add them. Similarly, take the potential energies from both problems and add them (but eliminate the torsional spring). Look at the 2nd problem and see how I added the effects of a discrete spring. In a similar manner, you will be adding the effects of the discrete masses to the compound pendulum solution.

@@Freeball99 i am building an Acrobot..and where i need the concept of compound pendulum with bob

This is the moment of inertia for a rod about it's center of mass.

At this stage of the problem, we have not stated anything about the constraints of the problem, so we simply use the moment of inertia about its center of mass. In the next step, we apply the kinematic constraints. As a result of this, and if you complete the math (which I did not do), you will find that the ML^2/12 becomes ML^2/3 when you simplify the algebra - ie as a result of the way that the motion of the bars are constrained (hinged at the tip), the equations of motions will show that the effective moment of inertia is as you have stated.