Equations of Motion for the Multi Degree of Freedom (MDOF) Problem Using LaGrange's Equations

[@Caffeinegoesinface]

Thanks for making these videos! I like the complexity of the questions because they are indicative of exam type questions. Much appreciated

[@Freeball99]

You're very welcome! Thanks for the feedback.

[@SkaifeRocks2]

From a uni student in Adelaide, South Australia. Your in depth explanation is much appreciated! I think i might just pass my Vibrational Analysis course now! Thank you @Freeball

[@Freeball99]

You're very welcome and good luck!

[@anitagofradump5195]

Absolutely terrific explanation! Who woulda known online youtube class would be so much better than online real class in such a shorter period of time

[@speedspeed121]

My mechanic professor's lectures and homework sets: OK, we will be solving 3 mass and 2 pulley systems, and incline plane and pendulum systems. My mechanics professor's exams: OK, spring to a double radius pulley, with connections to two more pulleys and a few masses mixed around.

[@AJ-et3vf]

Great video as usual. Thank you! Awesome to see that the resulting mass matrix isn't a diagonal matrix anymore.

[@Freeball99]

The diagonalizion of the matrix depends on the choice of coordinate system. If the velocities are coupled, then the mass matrix will not be diagonal. Similarly, if the coordinates are coupled, then the stiffness matrix will not be diagonal.

[@christopherhill1081]

Thank you for this explanation. This helped me to understand Lagrange more so. The only thing I found different or difficult to comprehend was that you did the partial of T and V separate but in all of my lectures it was just the partial of L from the equation of L = T - V. I know that its trivial but I just have the strange feeling that the negative sign will come back to bite me somewhere in the future.

[@b-rythescienceguy6633]

It’s because L=T-V and the Lagrangean is d/dt(dL/dx’) - dL/dx so it becomes d/dt (d(T-V)/dx’) - d(T-V)/dx dV/dx’ = 0 and dT/dx = 0 d/dt (dT/dx’) - d(0-V)/dx d/dt (dT/dx’) - d(-V)/dx d/dt (dT/dx’) + dV/dx

[@steveshaver4000]

At 25:33, it looks like you have a system of 3 differential equations and 6 unknowns. Is this true? How would you solve this system? Could you combine them all, write out three second order odes, and then solve that system of three second order odes for 3 unknowns?

[@Freeball99]

The unknowns are the three coordinates, x1(t), x2(t) & x3(t). By, using Lagrange's equation, you will always end up with an equation of motion for each coordinate. So you will always end up with the same number of equations and unknowns thus allowing you to solve the system.

[@greenfoodpower6961]

Excellent presentation, as usual!!!! Please keep it up.

[@AlanMW1996]

How can X2-X1 = the arc length if the arc length doesn't equate to a direct vertical distance? Like say theta changes to 90 degrees, then x2-x1 will be equal to 3r, not 3r*theta. Talking about at 2:35 .I guess if theta was really small it wouldn't make much of a difference? Or am I missing something?

[@Freeball99]

I think pretty much all of my videos on Introductory Mechanical Vibrations deal with small displacements which yield linear equations of motion (I think the pendulum problem might be the only nonlinear problem I've demonstrated so far). I probably should have mentioned it explicitly that we are dealing with "small vibrations", but at this point I have just kinda' taken it for granted. That said, you are correct that this equation holds only for small values of theta. In the limit (as theta becomes infinitesimal) then the arc length and the linear displacement become identical. Technically what I should have said was this: X2 - X1 = 3r · sin (theta) for theta

[@AlanMW1996]

That make sense, I was thinking about it more and i realized i was interpreting it wrong. I was picturing it as if the rope was connected to that point of the circle and as the circle rotated, the rope was anchored to that point would rotate too. When it should have been thought as the rope was being unwound from the pulley, meaning that the arc length referred to the amount of rope being unwound from the pulley and the 3m block would lower as it unwound. Thanks for the reply though! Fantastic videos that really helped me for my exam!

[@NYKYADU]

I think there are no approximations here. R*theta should hold fine. So in the question asked by the gentleman, when pulley rotates 90, X2 - X1 = 3R*pi/2, which is 4.7*R.

[@steveshaver4000]

Hi! At 9:56, you define Lagrange’s equations for the system. It seems like you have two definitions of L, the Lagrangian, in the equation. If you look in a textbook or Wikipedia, there can only be one L. Do you have an explanation for this?

[@Freeball99]

I have used the definition of the Lagrangian, L = T - V which is the difference between the kinetic and potential energies and substituted it into Lagrange's equations to re-write it in this form (in terms of the kinetic and potential energies). This form is valid if the kinetic energy, T, depends only on the derivatives of the coordinate variables (q_dot_i) and NOT directly on the variables themselves (ie. dT/dq_i = 0) which is the case for most of the problems you'll encounter.

[@NYKYADU]

Hey freeball, @4:37 Why is it that for potential energy, 3k spring has a deflection X3 and for 2K we consider the effect of X1 and theta along with X3?

[@Freeball99]

You need to look at the displacement of each end of the spring. Based on the diagram, in the case of spring, 3k, the only displacement is due to x3. However, in the case of spring 2k, the one end displaces by x3 (another way of say this is that both springs 3k and 2k add to the stiffness of the x3 coordinate) while the other end displaces by an amount due to a combination of effects: the first is due to the displacement x1 and the other due to the rotation θ. I have explained the reason for the negative sign in the video. Consider the effects of x1 and θ separately and you should be able to convince yourself this is correct based on the kinematics of the problem.

[@andrealiu8650]

This video just fill my blank in pulley. Very helpful. I have a question here. If we know the relationship between x1, x2 and theta, like x1= 2*x2= 0.5*r*theta, the generalized coordinates will be reduced from three to one. Will the mdof system reduced to sdof?

[@Freeball99]

Not sure if I understand the question exactly. Hopefully this covers it. The number of DOF's is equal to the MINIMUM REQUIRED number of coordinates to uniquely describe the system for any configuration. Usually this involves looking at each mass and assigning a coordinate(s) the account for its translation and/or rotation (up to 6 DOF's for each mass). Often, the system can be written in terms of other coordinates, but this doesn't change the DOF's for the system. As long as masses can move independently of one another, then this requires a coordinate to describe it.

[@ahmedibrahim4412]

Hello Sir, I was told that if we use the Newtonian method (drawing the FBD for each mass) to write down the equations of motion, we will get 4 equations: one for mass "3m", another for the mass "m", another for the vertical motion of the pulley and another for the rotational motion of the pulley. But when we use LaGrange's equations, we only get 3 equations of motion. How is that?

[@Freeball99]

This problem has only 3 degrees of freedom - since theta is related to x1 and x2 as I showed at 3:00. That said, you can certainly add coordinates, but then those coordinates must be coupled with a constraints equation. This is what you are describing here. So, if you have 4 unknowns, then you need 4 equations. The 4th equations will, in fact, be a constraint equation that relates the dependent coordinate to the independent coordinates. Take as an example the simple pendulum. Clearly, this is only a single degree of freedom problem (with theta as the coordinate). However, if you use x and y coordinates instead to describe the pendulum, you will end up with two equations, BUT x & y are not independent for the pendulum. It requires a constraint equation that relate the two.

[@eduardodossantos3099]

When you do the partial derivative in x1, (x2-x1)^2. It would be 2(x2dot-x1dot)*x1double dot ?

[@Freeball99]

Yes, that is correct (except you left out the -ve sign). This is what I have done, but then in the same step, I simplified the expression by grouping together the coordinate variables.

[@martinzupet3359]

Partial derivative is done in respect to x1dot (Lagrange equations). When you take a derivative of (x2dot-x1dot)^2 in respect to x1dot (following the chain rule) you get 2(x2dot-x1dot)*(-1). After that you take a derivative of 2(x2dot-x1dot)*(-1) in respect to time: -2(x2double_dot-x1double_dot).

[@steveshaver4000]

I think there are two coordinate systems here. Theta is a polar coordinate, and the other variables are in Cartesian. How would you implement this in programming or when you go to actually solve the equations of motion?

[@Freeball99]

In this case, we just used the rotational coordinates as an intermediate step in solving the problem. The final equations of motion contain only translations and no rotations. However, in general, one can have mixed coordinates in the equations of motion and this makes no difference when solving them. We use the idea of "generalized coordinates" to include both translations and rotations. As long as the coordinates are independent then we are fine, if they are not independent, then we need an additional constraint equation.

[@marcogaruffo]

I will try to make an animation of this system using Lagrange equation...

[@martinandre]

did you do that?

[@Ninja-ok7gv]

why did you not consider the mass of the pulley M in the potential energy equation?

[@mrsush1994]

Ignore gravity is mentioned in question

[@samgdotson]

The only thing that is unfamiliar to me here is the matrix form you put it in at the end. I have already taken linear algebra so that part is all well and good. I do have a question about solving the differential equation at the end, however (I have never seen a matrix differential equation before). Usually when solving a diff eq of the form MX" + KX = C we will get something like X(t) = Acos(wt) + Bsin(wt) where w (omega) squared is equal to K/M. Omega (w) is simply the oscillation frequency (for small angles of theta). In this case, you have a matrix differential equation at the end. Could it be solved in the same way? I'm thinking that I could find a matrix of oscillation frequencies that would be: [M]^-1 * [K] = W^2 It is probably more nuanced than that, but those are my initial thoughts. Thanks for making this video!

[@Freeball99]

Take a look at this video: kzbin.info/www/bejne/sKCVfp9ubbasqKM I think this pretty much covers it. Using matrix math will make your life much easier also matrices are very well-suited for computer implementation and solution. As a structural engineer, you MUST get comfortable with matrix math. There is no way around it. That said, there is nothing to be afraid of here. This is very similar to the single degree of freedom problem and you seems to have a good grasp of how to solve that. This video will show you how to extend the solution to the multi-degree-of-freedom case. You just need to remember that: 1) the order of matrix multiplication is important (i.e. A.B ≠ B.A) and 2) you cannot divide by a matrix, instead multiply by its inverse and 3) review what the determinant of a matrix is. If you work a couple of examples you will likely get the hang of it very quickly. My suggestion is to watch the video (the one that I linked to in this comment) and actually write it all down while following the video. THEN, do it again, only instead of watching the video, try to derive the solution again just using your previous notes. This whole process with take you maybe 30 mins and by then you should be comfortable enough for any of the materials in my videos. Happy to answer questions as you go down this path.

[@ThoTochRMm28]

Hello I have a question : why here your definition of LG eq. is partial T and not partial L ?

[@Freeball99]

I skipped a step here. Since L = T - V, if I substitute this into Lagrange's Equation, then it reduces to the form I have in the video. It reduces to this because in this problem, the kinetic energy, T, is a function of q_dot only and does not explicitly depend on q.

[@opstinabackatopola4848]

Hi! I'm very happy to find your solution! Can I get one more help, how to calculate it in MATLAB? Can you help me?

[@Freeball99]

It has been a very long time since I last used MATLAB (I'm a Mathematica user) so, unfortunately, I cannot advise on this.

[@steveshaver4000]

Does each point in the system that we are modelling have a unique energy value? Can you make a video about this?

[@Freeball99]

I don't believe there is anything unique about the energy of an individual point but there is something unique about the configuration of the system the minimizes the Lagrangian at any given time.

[@sayanjitb]

Thank you sir for providing this ingenious contents. I have a question in which cases we need to include generalized forces and some cases where we equate it to 0?

[@Freeball99]

This is based on the external loads that are given in the problem. If there is a load (force/moment) being applied directly to a particular coordinate, then it appears on the right-hand side of the equation. If not, then we equate it to 0.

[@sayanjitb]

@@Freeball99 Thank you sir, I understood

[@soubhadramaiti1996]

Very good video 👍

[@alvaromendoza3337]

Wow, thank you sir!

[@bobjo3549]

Thank you really helped

[@mrsush1994]

Hello sir Sir i did not use the energy method to write the equation of motion but wrote it through newton's second law. I got same equations as you wrote but sir that's only when i don't consider F1 torque about pulley. Will be consider F1 torque about pulley or not ?? Tension = tension in string connecting pully and 3m Jo* (theta double dot) = (3*F1*r) + (3*Tension* r) - (2*k*( x3 - x1 + r*theta )*r ) I am getting correct result when i am not taking F1 torque Please help.. or any way i can share my solution with you

[@Freeball99]

sushant sharma the external force acts directly on the coordinate itself (ie the center of the pulley). As a result it does NOT produce any external torque on the pulley. The diagram is a little confusing. I probably should have made this point during the video.

[@mrsush1994]

@@Freeball99 thank you sir

[@Freeball99]

sushant sharma I am realizing that I misread your question. Just want to clear up any confusion I may have caused. F2 produces no external torque on the pulley. However, F1 and F3 do produce torques on the pulley.

[@mrsush1994]

@@Freeball99 but sir if we draw F.B.D of pulley then F3 won't act directly on pulley. So F3 will not cause any external torque. Rather the spring force in spring with stiffness 2K will cause the torque on pulley. Right ? Also sir will i take F1 external torque for right equation of motion ?

[@Freeball99]

@@mrsush1994 I'm sorry, I transposed F1 & F2 - it's what happens when I try to respond from a phone instead of my computer! I meant to say that F1, which acts through the center of the pulley (i.e. it acts on the x1 coordinate), produces no external torque. The other two forces, F2, & F3, which are both offset from the center of the pulley by 3r and r respectively, do produce torques on the pulley. Whew! Sorry about that.

[@sudaratchairattanamanokorn356]

What if there are dampers, what's the formula of lagrange equation?

[@Freeball99]

Assume we place a damper with damping constant c between the pulley and the 3m mass. So, the ends of the damper move with x1 and x2 respectively, then we use the Rayleigh's Dissipation Function, which is defined as: R = 0.5 * c * (x2_dot - x1_dot)^2 where x2_dot - x1_dot is the relative velocity across the damper (ie the difference in velocity between the two end of the damper). Then simply add to Lagrange's equations (to the side of the equation which contains the Lagrangian): dR / dq_dot where q_dot is the generalized velocity associated with that particular equation of motion. It's a little difficult to write equations in this comments section. Hope this makes sense. I should probably make a video on this.

[@sudaratchairattanamanokorn356]

@@Freeball99 I tried but I get c*x_double dot so I was really confused

[@Freeball99]

@@sudaratchairattanamanokorn356 There should be nothing with a double-dot in it. Take a look at this: www.dropbox.com/s/zlq2mdsekwb5c7u/RayleighDissipationExample.png?dl=0

[@amrnaser5284]

really helpful ,thanks

[@SimranSingh-yg1xm]

Thank you Sir!

[@pratikrai3404]

thanks a lot

[@heymiyanar1123]

watch this video if you want to fail your class.

[@Freeball99]

I'm sorry to learn that you're struggling with this class. Perhaps you would be good enough to clarify what it is that you don't seem to understand and I can explain it.

[@heymiyanar1123]

Is the equation you used same with this: nerdwisdom.files.wordpress.com/2007/10/lagrange001.jpg if it is same then sorry for miss understanding.

[@Freeball99]

These are the same equations. Consider two things: 1) In the link you posted, the equation is written using the Lagrangian (L) which is defined as L = T - V (i.e. the difference between the kinetic energy and the potential of the system). If you make this substitution into the equation you posted, then the left-hand-side ("LHS") of your equation reduces to the left-hand side of my equation BECAUSE the potential (V) is a function of the generalized coordinates (q_i) only AND BECAUSE the kinetic energy (T) is a function of the generalized velocities (q_dot_i) only (as can be seen in equations 1 & 2 in the video). This is true for many problems, but not all - like in the pendulum problem (kzbin.info/www/bejne/jZPNYYaXqZ2jmLs). So what I have used on the LHS is slightly less general than the LHS of the equation you have posted, but applies fine to this problem. 2) I have included the generalized forces (Q_i) on the right-hand-side ("RHS") of my equation. This is a sightly more general form than the RHS of the equation than what you posted because it allows for external forces (both conservative and non-conservative) to be included. These Q_i's come from deriving Lagrange's Equations from d'Alembert's Principle (which is fundamentally the application of the Principle of Virtual Work but includes both static and dynamic forces). Here are a couple of articles that explain it: kestrel.nmt.edu/~raymond/classes/ph321/notes/lagrange/lagrange.pdf www.quora.com/Classical-Mechanics-What-is-the-Lagrangian-of-a-non-conservative-force

[@Ninja-ok7gv]

why did you not consider the mass of the pulley M in the potential energy equation?

[@Freeball99]

We're told explicitly that gravity is to be ignored. As a result, none of the masses contribute to the potential of the system (just the springs).