Equations of Motion for a Pendulum on a Cart (2DOF) Using Method of Lagrange's Equations
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- Опубликовано: 8 фев 2017
- Deriving the equations of motion for a pendulum attached to a cart using the method of Lagrange's Equations. Two degree of freedom system.
Everybody can be a hero without wearing a cape.You are one of them thank you so much.
truly a hero for all of us. thank you very much for making an elegant and understandable video
thank for your video. I totally understand these system by lagragian. theta is the changing value by time. It helps me a lot
Perfectly explained, thanks
best video of mechanical vibration that i've ever seen
Thanks a lot for this lecture; the explanation was very clear and I'm planning to run a numerical simulation.
Just out of curiosity, you mentioned in 0:30 that the weight of the system can be ignored, even though the displacement x is in the same axis as gravity. Does this mean that x=0 is not the resting displacement when the spring is unloaded, but when it is loaded by the system?
Here is a discussion on gravitational effects and when to ignore them. ruclips.net/video/TpZMTo-Lig4/видео.html
In this case, just assume that at equilibrium the spring is in its (slightly) stretched position. You can calculate this position by equating the entire weight to the spring force. Then assume that this is you zero point.
Alternatively, just include the gravitational effects. What you will then see is that you have a constant term that gets added to your solution for x(t). This is the gravitational offset.
Been watching your videos for sometime now to go back to the basics continuously... Do you have recommended books you use for conceptual references?
I want to dive into more examples/problems. Derivation for me is usually to understand all the influencing parameters, and quantify what I already have speculated, but i don't like getting bogged down in them for way too long.
When I fix a frame on a garage door, I don't go about deriving how to add adjustments to bolster rigidity. I just drill some wooden rods with the right fasteners in the right places if you know what I mean.
Anyways, I've been trying to combine different concepts for applications in aerospace & structural design.
Current books/material , are less textbooky (which are awesome btw & i love penguin classics) :
-The New Science of Strong Materials, J.E. Gordon
- Structures : Or Why Things Don't Fall Down, J.E. Gordon
- Tom Irvine : ruclips.net/video/vWjizHma16Y/видео.html
- Success through Failure: The Paradox of Design, Henry Petroski
Sorry, But I missed this until now. I don't really have a single book that is good for everything. I think back in the day, I used a book by Rao, which was pretty good but had some mistakes.
@Freeball Sir but X is about/from the equilibrium position so if we are taking that position as our reference point then (M+m)*g*x should be considered in the potential energy... because X will be below the reference point.because X is the additional displacement from the equilibrium position
In general, X is not from the equilibrium position, but rather from a prescribed zero position.
My suggestion to you is to include the term you have proposed above in your potential energy expression and put it into Lagrange's equation and see what happens.
What you will find is simply a constant term in your x-equation (because the x disappears after you differentiate this term). What you will find it that you end up with a forcing function which is simply a constant...which adds a constant to the total response - this constant is equal to the initial elongation of the spring. I have explained it here: ruclips.net/video/TpZMTo-Lig4/видео.html
@@Freeball99 Ok sir , i have done it and come up with -(M+m)*g ,and by your suggested video I understood that why Mg shouldn't be taken,but what about pendulum potential energy, it should be included as -mgx as initial dispace ment was 0 and after a time t pendulum makes an angle theta and M makes displacement x so i think that in h=l-(x+lcos(theta)) sould come.
I believe the argument is the same as before - i.e. you can ignore the gravitational effects by using the equilibrium position as the zero point. You can test this for yourself and determine whether or not to include.
Solve the equations of motion (using my integrator that I provided in a previous video). Run it once while including the effects of the term in question. Then re-run it after zeroing out the terms. Plot both results on the same graph and see if the results differ by a fixed, constant amount.
Thanks a lot sir
I need the solution of this equations
I don't believe that this can be solved in closed form. You will need to solve it using numerical methods. I have shown how to this in several videos -here's an example for the case of a simple pendulum.
ruclips.net/video/_eZyTNthJG4/видео.html&list=PL2ym2L69yzkZJ1fY3SQ1JCyvZIoJYXQGZ
hey I wonder if you can do a brief video on tuned mass dampers, where the spring and mass are moving horizontally, and the pendulum attached below, thanks a lot!
Sounds like you're talking about the type of damper they have begun using in skyscrapers (like in the "Taipei 101" building).
justfunfacts.com/wp-content/uploads/2016/05/taipei-101-tuned-mass-damper.jpg
Did you take a look yet at my video on Dynamic Vibration Absorbers (which is the same thing as a tuned mass damper)?
ruclips.net/video/7T6pQnNBph0/видео.html
I realize this is for a sightly different system, but the mathematics and the concept are identical - namely that in order to tune the pendulum, the isolated pendulum should have the same natural frequency as that of the isolated mass-spring system (i.e. for tuning purposes treat each mass like it is its own single-degree-of-freedom system).
I remind you that the frequency of a pendulum is independent of the pendulum's mass and is equal to SQRT(g/L) where "L" is the length of the pendulum.
Here is a video that demonstrates this (non-mathematically).
ruclips.net/video/f1U4SAgy60c/видео.html
If this is unclear, let me know and I'd be happy to make an explainer video.
Thanks, your comment helped me in my experiment, sorry for the late reply though!
Hi! Thanks for the nice video! I was wondering how you would go about solving this problem assuming small angles?
The idea here is that we take the equations of motion and we change sin θ to θ and cos θ to 1 THEN we eliminate any higher-order terms. These are any terms that contain two or more coordinates or their derivatives multiplied together and can be eliminated because they are insignificantly small in relation to the first order terms (a result of the small-displacement assumption).
Therefore, the resulting equations of motions for the linear system are as follows:
(M+m)x_ddot + kx = 0 and
L θ_ddot + g θ = 0
@@Freeball99 Okay, Thanks! That helped alot!
@@Freeball99 much clean equations, i like it
also solutions for x and theta is on same format with given constants
So it's, somehow, implied that the spring has a "infinity stiffness" in the horizontal?
I guess you could. We don't consider the horizontal direction at all, so you could think of it as moving along a frictionless track in the vertical direction.
why you didnt use the gravitational potencial of the mass M for the expresion V?
It can be ignored in this case with the assumption that oscillations of the cart are about its static equilibrium position. I have explained it in this video: ruclips.net/video/TpZMTo-Lig4/видео.html My suggestion to you is to include the gravitational potential and then compare the equations of motion (and more importantly, the response) with and without it. You’ll find that your equation of motion for the x-direction just has a constant on the RHS of the equation = (M+m)g which simply has the effect of shifting the equilibrium position by (M+m)g / k.
Thanks a lot, plz what app you use for writing
The app is called "Paper" by WeTransfer. It is running on my iPad Pro 13 inch and I am using an Apple Pencil.
sir what tools you have used to make this video i mean which application yu used for writing and voice recording..
I use an iPad Pro (13 inch) with an Apple Pencil and the app is called "Paper" by 53. I record the video by connecting my iPad to my Mac with USB and then using Quicktime to record video input from the iPad. Microphone is a Blue Yeti.
How would this be written in matrix form, or is it impossible to do so as it is everything's far to coupled
also would one have to linearise the system?
You cannot do it in its current form. This is not due to the coupling, but rather due to the way that the nonlinearities enter the equations, namely sin (theta) and cos (theta).
For small vibrations (i.e. small theta) you can linearize the equations by replacing sin (theta) with theta and replacing cos (theta) with 1. Then you can write the EOM's in matrix notation.
sir, i want to ask something. this is just a dummy question, but i really don't get about the potential energy. i've always learned that we need a fulcrum to decide the position of thing. i don't get why you wrote the position of the pendulum as l(1-cos theta) and not x+lcos theta just like the position of the pendulum (xm). thank you
It was a shortcut. I have explained in previous videos when you can ignore gravitational effects on the cart. My suggestion to you is to leave in in the and work the math. You'll find that the gravity on the cart just causes an offset effect. There are other comments here that deal with this. Also, take a look at: ruclips.net/video/TpZMTo-Lig4/видео.html
how does the potential energy equation differ if there was another spring connected underneath the cart to another datum point?
Since this spring would still have a displacement of x, you would add the contribution of this spring to the potential energy of the system as V = 0.5 * k1 * x^2 (assuming the spring stiffness is k1). Hopefully, what you are able to tell just by looking at the problem and what I have written, that as a result of your described setup, the two spring stiffnesses would be additive (i.e. both resist the motion of the cart). As as result, instead of re-deriving everything, you could simply adjust your equations of motion by replacing all occurrences of k by (k+k1) - i.e. the net effect is that the spring you currently have would effectively be made stiffer.
BTW - It would make no difference whether you attached the spring to the bottom of the cart, as you have described, or whether you attach it to the top of the cart in parallel with the existing spring - in both cases it will resist the displacement of the mass with the same force.
@@Freeball99 You are a Legend mate. Much appreciated
excuse me, sir. Doesn't M also have gravitational potential energy?
Yes it does, however, I explained briefly at 4:05 in the video why this can be ignored. In the case of translational motion of the mass, the effect of gravity is that it simply shifts the equilibrium position of M (i.e. the position about which M oscillates). Using the assumption that M is oscillating about this equilibrium position, the effect of gravity on M can be ignored. It is explained in detail in this video: ruclips.net/video/TpZMTo-Lig4/видео.html
+Freeball
Thanks a lot for these videos. :D
For some reason it's surprisingly hard to find videos about this specific Lagrangian method (most videos are obsessed with Lagrange Multipliers for some reason), but you have lots of videos on the formula _L = T - V,_ which is what I am always looking for. ^_^
@@Freeball99 Sir but X is about/from the equilibrium position so if we are taking that position as our reference point then (M+m)*g*x should be considered in the potential energy... because X will be below the reference point.because X is the additional displacement from the equilibrium position
Sir given these two equation can we now use small angle approximations to do modal analysis and find natural frequencies?
Yes, sure. You can linearize the problem as you described and then solve for the mode shapes and frequencies for the linearized problem.
Sir, hi. I have a question. If we take the datum the very top of the sistem h becomes as h = -x-l*cos(theta) which changes the equation. What am I doing wrong?
Not sure if I understand completely...what you have done, in effect, is to shift the origin of your coordinate system from the mass (at equilibrium) to the attachment point of the spring. It should have no net effect on your equations of motion.
Hello. How can you define the function of angle in such problems? I mean, what's the function of angle, derivative of which we take in order to find general 'dot'-equations of the coordinates?
A closed-form solution for this does not exist which means there is no such function that can be written out. As a result, the response must be determined by numerically integrating the equations of motion.
@@Freeball99 Oh, thank you. Thus, we have to calculate (approximately) the value of the final function at each point of time, for each value of the angle, haven't we?
what's software used in presentation ?
App is "Paper" by WeTransfer. Running on a iPad Pro 13-inch and using an Apple Pencil. Video is captured using Quicktime.
is is possible to get the oscillation period from this?
Since this is a nonlinear problem, the frequency of oscillation (and therefore the period) is a function of θ (and thus the initial angle). Also, if the vibrations are small, then we can find the frequency/period for the linear case - which is constant. You can calculate it if given a specific initial condition. In practice what happens is we end up with limit-cycle oscillations in the nonlinear case.
How we can Linearized both equation of motion ?
Apply the small angle formula. So, replace sin θ with θ and replace cos θ with 1 in your equations of motion.
sir
how can i get a pdf of all question
I'm going to look into making the pdf's available for download. However, I recommend writing it down yourself. This is the best way to understand and retain the information.
Why is xdot = xdot - l theat dot sintheta? Instead of xdot = xdot - l sintheta?
Which equation are you referring to?
@@Freeball99 the equation of the bob in motion in the x direction. Thank you
hi can you solve my problem, do you have time
What is your problem?
@@Freeball99 thanks but time is over