_I can teach you to bewitch the mind and ensnare the body_

You would need to define the Rayleigh's Dissipation Function, R. In this case, R = 0.5 * b * x_dot^2 and then use the extended form of Lagrange's equations. I have explained it in this video: kzfaq.info/get/bejne/hLxnlsdenLjeXX0.html

I will put this on my list, however, solving this is almost identical to what I have shown in several other videos (particularly the animation series). Solving for x and θ uses the identical method to what I have shown before.

I combined this into one step because I wanted to fit it all on the same page. If you write out the last part on some paper an go through the derivatives one step at a time, it will make sense.

F(t)

This was a remake of a previous video which was popular, but had a couple of errors. Since KZfaq no longer supports annotations to correct these errors, I decided to redo it.

Not sure exactly what you are trying to achieve, but the spherical pendulum might be a good starting point. You could model something like this attached to a rotating base perhaps. kzfaq.info/get/bejne/h9VgfLpqltrSmas.html

According to the problem statement, there is an external load, F(t) being applied to the cart in the x-direction. There is no mention of an external moment being applied at the hinge. Lagrange's equations require us to apply the load to the right hand side of the equation corresponding to the coordinate that the load is applied to (this is derived from the fact that the external load/moment will do work as a result of that displacement changing). Consequently, the external force is applied to the right hand side of the "x" equation of motion and since there is no external moment, there will be a zero on the right hand side of the θ equation. Regarding conservative/non-conservative forces....you can ALWAYS add the external load/moment to the right-hand side of the equation regardless of whether is it conservative or non-conservative. However, IF the load is conservative then it can always be written in terms of the derivative of a potential function. As a result, for a conservative load (e.g. gravity), you can either add it to the right-hand side of Lagrange's equation as a load OR you can include it in the potential function and add it to the left-hand side of Lagrange's equation.

First take the partial derivative with respect to xdot (so xdot disappears, because d(a • xdot)/dxdot=a). Then take the derivative with respect to time, which requires the product rule giving the two terms

I've never used Simulink so I'm not familiar with the process, but I'm guessing that you need to write the equations in vector/matrix form.

Yes. If the pendulum is to the right, you can just use a negative θ.

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Yes, you can use Euler (which is really just RK1) or RK4 to get numerical solutions. The method you have described is not correct... You first need to rewrite the equations of motion in state-space form. In doing so, you are converting the system of equations from two 2nd order differential equations to a system of four 1st order differential equations. I have several videos on how to do this. kzfaq.info/get/bejne/m5ehZ9l1l7HOoHk.html kzfaq.info/get/bejne/iaygfcWera-5cmQ.html kzfaq.info/get/bejne/r7OVlM5pt72vpH0.html

Freeball Oh, I remember how to do that. It’s just a bunch of substitutions, which is similar to what I ended up doing anyway. Regardless, I saw what I’d call physically accurate behavior in my simulation with my implementation of solution approximation. Thanks!

I modeled the spherical pendulum in this video: kzfaq.info/get/bejne/h9VgfLpqltrSmas.html - you should be able to combine this with the model above to get the model you are describing.

The model as described in the problem really just consists of two point masses connected by a rod. As a result of this: 1) the pendulum bob has no rotational inertia (so no rotatory kinetic energy) and 2) the cart, wheels and applied force (F) are all located at a single point located at the hinge of the pendulum - so, the wheels have no radius and the force F produces no moment about the hinge. If you wanted to model the control input to the wheels, you would need to expand the model to include assumptions about the wheels, eg what is the radius, the mass and the moment of inertia of the wheel? What is the offset between the hinged and the wheels? Etc.

If the pendulum was hanging down, you would use for its change in height l - lcos θ OR you can just define it's y-location as -lcos θ and call this the height depending on where you define you define y to be zero. These are the same thing just shifted by a constant, L. Both will produce identical results when differentiated in Lagrange's Eqns because the constant, L, disappears when you differentiate it. For the inverted pendulum, it would just be the negative of this.

The equations must be solve numerically (as far as I am aware). In order to do this, the equations must first be written in state-space form. I have shown how to solve the equations for the double pendulum in this video (kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html). Explaining control systems are outside the scope of this material, however, I would recommend Steve Brunton's channel for more on controlling the inverted pendulum. If you are going to add a control system, then you will first need to write the equations in state-space form.

In order to linearize the final equations, use the small angle approximation. So, sin θ --> θ and cos θ --> 1.

When taking time derivatives to find velocities, these derivatives must be taken with respect to an INERTIAL REFERENCE FRAME. This cross-product term in the kinetic energy is, indeed a coupling effect, and arises from considering the pendulum's motion from the perspective of a non-inertial reference frame - the moving cart. As the cart accelerates, so does the hinge point of the pendulum, which means that if you just consider the first two terms of the kinetic energy, then you're observing the pendulum's motion from a platform that itself is moving. This non-inertial reference frame thus introduces complexities in how the pendulum's kinetic energy is calculated. Intuitively, the cross-product term accounts for the additional velocity and, therefore, kinetic energy that the pendulum mass has due to its rotation around the hinge point, which isn't stationary but moving with the cart. This is why it contains both x_dot and θ_dot. This is probably a good topic for a video!

Both are right. This is fundamentally just a difference of the coordinate system. In this case, I have the origin at the hinge. In the other problem, I placed the origin at the equilibrium position of the mass. As a result, these expressions for potential energy differ by a constant, HOWEVER, the constant get differentiated out when the potential energy is substituted into Lagrange's equations - so, as one might expect, you can arbitrarily select the point of zero potential energy depending on where you place the origin of you coordinate system and it should make no difference to the final result.

You can linearize the equations of motion by making the small angle assumption. Under this assumption, sin θ = θ and cos θ = 1. Then to write them in suitable form for use in a controls problem, you must rewrite the equations in state-space form. I have several videos demonstrating how to do this. Here is one: kzfaq.info/get/bejne/iaygfcWera-5cmQ.html

I first differentiate with respect to θ and THEN I differentiate with respect to t. This is why the order of the derivative on X increases.

@@Freeball99 Thank you for the clarification.

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I didnt mention that my consideration was made because x=constant and theta= pi/2 inst a equilibrium of the system. It's very strange, the stick of the pendulum just don't want to rest?

It's a little hard to tell without looking at your work, but my sense is that you have not squared your velocity properly. Make sure you: 1. locate the position of the mass(es) using the coordinates and geometry of the problem, 2. Differentiate with respect to time to get the velocity of the mass(es), 3. Use this value for the velocity for the kinetic energy expression AND SQUARE IT! When dealing with rotations, do try to do this intuitively as it is easy to miss a term (usually a cross-product type of term) the result is that there is a coupling in the motion of the two masses that you are missing.

@@Freeball99 thank you very much for your answer. Well, I basically just naively summed up all the kinetic energies that I thought are involved which come from both translation of both masses and the rotation of the pendulum, which equaled to T =1/2* (m+M)*v^2 + 1/2*m*L^2 *w^2 (assuming point mass). So if I would here simply substitute v for dx/dt and w for dFi/dt, I would got something similar, but with discussed term missing (-2*L*w*v*cos(fi)). I guess one can't just sum up what he assumes to be energies, but rather has to start bottom up and define radius vectors and calculate their derivative to obtain velocities and kinetic energies, but I am wondering if there is some interpretation of this missing term here. I thought it has to do something with coordinate definitions perhaps, but what you mention as "coupling in the motion of the two masses" sounds interesting. Especially since this term does resemble cross product of linear and angular velocity (depending on the cos(fi)). I am not 100% sure what are you refering to by this, could you perhaps point me somewhere where I could read more about it? Thank you once again for you clarifications, they have been most helpful.

I differentiated wrt x in the previous term. Since both x_dot and Cos θ are functions of t (as you have pointed out) I need to apply the product rule when differentiating this. So first I differentiate the x_dot and multiply it by Cos θ and then I differentiate the Cos θ and multiply it by the x_dot. This is what I did in terms 2 and 3.

@@Freeball99 Hi, I think what he/she said is that , why d (-ml theta_dot multiply x_dot cos theta) not = -ml x_double_dot theta_dot cos_theta + ml multiply theta_dot_square x_dot sin_theta - ml theta_double_dot x_dot multiply cos_theta. I am stuck on there too. :(

The are 2 degrees of freedom in this system ( corresponding to each of the generalized coordinates x and θ). That is to say that if we know x and θ, then we can completely describe the location of each of the masses in the system.

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No. This is a common mistake - DON'T DO IT!! The velocity of m is x_dot - L·θ_dot·cos θ and you need to square this. This gives you THREE components of velocity for the mass, m - the two that you have included plus a rotational effect that you are not including. If you look at Eqn. 6, you will find that you are missing the last term. When you are dealing with rotating systems, this is generally nonlinear stuff. Solve it using basic principles and math rather than using intuition which is often wrong when dealing with rotations.

@@Freeball99 Hey! We don't see a term that involves moments of inertia in these equations. So has the effect of inertia been neglected for simplicity, or is it embedded in the cross term?

@@reubenthomas1033 When modeling a simple pendulum, it is common to treat the mass as a point mass and ignore and rotational kinetic energy of the mass. Really what we are saying here is that if the diameter of the mass is much less than the length of the rod, L, then the contribution of the rotational kinetic energy is negligible. This also keeps the math a little more simple - which is helpful for teaching purposes. I probably should have mentioned this at the start of the problem (I think I mentioned it for some of the other simple pendulum examples). I have separately showing an example of a compound pendulum which does require one to include the rotational kinetic energy (kzfaq.info/get/bejne/m6iXa8qgy5e2cps.html)

@@Freeball99 Awesome! Got it. Thanks a lot :D

This is really just a question of where you are drawing your axes. Truth of the matter is that zero potential energy is only at the center of the earth, so if any problem we arbitrarily chose a zero potential point. In my case, the x-axis is on the cart and this is defined as zero potential. To be clear, all these different descriptions of the potential energy vary by a constant. When you substitute the potential energy into Lagrange's Equation, the constant gets integrated away. The result is that it make no difference what the absolute value of the potential energy is. Rather what is important, is the DERIVATIVE of the P.E. with respect to the particular coordinate. Also, this means that negative values of potential energy are fine. Hope this clears it up.

In this case we are differentiating with respect to t and NOT with respect to θ_dot.

The problem shows that f(t) is the force applied to the system in the x-direction. Therefore, it appears as a generalized force on the right-side of the equation of motion for the x-coordinate.

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@@Freeball99 thank you!

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The app is called "Paper" by WeTransfer. It is running on an iPad Pro 13 inch and I am using an Apple Pencil.

Yes, provided the angle is, in fact, small (typically less than 10 degrees). Linearization can be used when examining the control problem of keeping the pendulum balanced and upright.

Yes, exactly right. Typically we model a simple pendulum as a point mass. This assumption is fine as long as the dimensions of the mass are relatively small compared to the length of the rod.

@@Freeball99 Thanks. I'm so happy to have (re)discovered the joys of Lagrange Motion equations. Together with Runge Kutta solvers, I get excellent non-linear simulation behaviour. I am modelling the balancing of a a non-slipping ball on a glass plate. After linearizing and Kalman filtering, I can measure the position of the ball by the torque needed to keep the glass plate level. Next step: position control. Without position sensors ;-)

@@abumohandes4487 Sounds interesting. Would love to see it when you're done.

Thank you for this thread. I was wondering about the rotational kinetic energy as well.

I have several videos on this channel that demonstrate how to solve problems using Lagrange's Equations. Also, there are undoubtedly numerous other KZfaq channels which contain videos on using Lagrange's Equations to solve problems.

This is a 2nd order differential equation in time, so it requires 2 initial conditions. Typically it is assumed that the initial angle and the initial angular velocity are known. This is be based on the setup.

We are considering only external forces in the case of generalized forces, so it's just F in this case (we can ignore forces from the rod) No, the force does not apply a moment to the small mass - it is purely applied to the cart and in the x-direction. If the force were applied directly to the small mass, m, then you would be correct that it would also produce an external moment. For the generalized force, we only consider those forces that are acting directly on the particular coordinate. In the case, the external force is assumed to act act through the location of the hinge (the cart is assumed to be infinitely thin) and so cannot produce a moment about this point.

This is due to the displacement of the cart. If the pendulum angle remained zero at all. times, then the mass, m, will still be displaced by the same amount as the cart, so we need to add X.

The derivative, d/dθ (sin θ) = cos θ HOWEVER, we are taking the derivative with respect to t and NOT with respect to θ. Therefore the chain rule must be applied. d/dt (sin θ) = cos θ · d/dt (θ)

​@@Freeball99 dear sir, thanks a lot for the reply, i have figured that out just a few minutes ago, and you have confirmed it. although initially i thought it has something to do with derivative of a complex function, like d sin(f(t))/dt which is equal to cos(f(t))f(t) + f(t)'sin(f(t))

As an example, a human being can be modeled as an inverted pendulum (since we're all top-heavy). Our ability to move our feet allows us to stay balanced and upright. This could be modeled as an inverted pendulum on a cart.

Yes, but it will take me a little while to get there. I have several other videos to make first.

We have not been given any external moment in the problem. It is assumed that F(t) is acting directly through center of the hinge so it produces no moment about it.

You can linearize the equations by assuming small displacements and then eliminating all higher-order terms - i.e. fully expand out the equations and eliminate all terms that contain x^2, θ^2 or x.θ (or higher powers). Also you will make the approximation that sinθ=θ and cosθ=1.

@@Freeball99 yes i've learnt some tricks in 4 months. Thanks anyway. Appreciate it

There are actually 3 negatives here and they combine to produce a -ve sign. You are correct that the negative sign from the derivative of cos will cancel out the negative sign that already exists for this term in the Lagrangian. HOWEVER, if you look at eqn 7 (we are dealing with the 2nd term here) dL/dq_i must be SUBTRACTED. This is why we have negative signs before the last 2 terms of this equation.

@@Freeball99 Ouh I see sir, Thank You Very Much 🙏.

Typically, we treat a simple pendulum as a point mass (though I should have made this more clear at the start), in which case its rotational KE is zero. In general, you are correct that the rotational kinetic energy should be included. In reality, what you'll find is that in most cases, the rotational KE is negligible for a simple pendulum UNLESS the radius of the mass is very large. In the case of a compound pendulum (or in cases where the mass of the rod/string is not negligible) then the rotational kinetic energy must be included - I have another video showing this.

Yes, you could say that. A better explanation would be that the force, F, acts only in the x-direction and produces no moment in the θ direction.

@@Freeball99 thanks for getting back to me. I realize my confusion was that considering the whole system, pulling the cart would cause the arm to rotate, but we are not considering the whole system but rather just the virtual work done by F, which is zero in the theta coordinate.

Adam Johnston correct. That’s an even better explanation.

1. The mass, M, has zero potential energy since it is located at y=0 and it's y location doesn't change. 2. This is correct as written. The 2nd term in eqn 10 comes from the THIRD term in eqn 9 - because the 2nd term in equation 9 does not contain an 'x' so it does not appear in eqn 10. 3. If m is to the right, then according to my current setup, θ would be negative. However, if you choose to make θ positive to the right instead, then you will need to adjust the kinematics and flip the sign of the 2nd term to positive in eqns 1 & 3.

This problem exists in many references, however, this came from my own notes. The diagram I re-drew comes from Wikipedia though it too exists in many places online and in many textbooks. The inverted pendulum is a fairly common problem in vibrations and controls classes.

@@Freeball99 Yes, thank you. What is one source of the book, sir, because I really love working on problems like this

@@mhd.arifin2904 You can try Mechanical Vibrations by S. S. Rao, 5th Edition. It does not contain the inverted pendulum problem, however, there are many other examples in there.

@@Freeball99 thanks very much sir

@@Freeball99 I was reading a survey paper the other day that said the inverted pendulum problem has been used as an educational tool in controls for 62 years !!! web.mit.edu/klund/www/keeling/Pendulum2009.pdf

We place the generalized force/moment on the right-hand side of the equation. In the case of the x-coordinate, there is a force, F, in this direction, so this appears on the right side of the x-equation. In the case of the θ-coordinate, there is no external moment applied, so the right side of the θ equation is 0.

No, this is not correct. If I take the derivative of m*L*Q_dot* x_dot*cos(Q) according to x_dot, then I get m*L*Q_dot*cos(Q) according to x_dot. I then need to take the derivative of this with respect to t. This is where the product rule come in because both Q_dot AND cos(Q) depend on time. This results in the TWO terms in the parentheses.

@@Freeball99 I get X-dot derivation, what about theta dot. cause the derivation of theta dot for mine is -mL*Xdot*cos theta (only derivation of theta dot), with respect to time, isnt the formula mLX_dott*sin(theta)?

I agree that something seems off here. The cart should move whenever the pendulum swings, regardless of its initial condition. This is not related to the EOM's having no x_dot terms (these would come from adding damping).

@@Freeball99 Hey there, I worked on it and figured out this was a non linear problem so the mass matrix( being part of 2n x 2n matrix A from your video on Direct time integration) and B matrix (comprising of damping and stiffness matrix) were also varying with time and I had considered them constant. This has solved the above problem of theta and x coupling for theta = pi/2. The reason I asked about x_dot was that the absence of x_dot in the EOMs made it feel like the system behaved independent of cart velocity. So assume I give an initial velocity to the cart, the pendulum hence theta should change but it's not.

@@NYKYADU This problem is a semi-definite problem because the system is not restrained and can move as a rigid body without oscillating. For example, if initially the cart had some velocity, x_dot and the pendulum was pointing perfectly straight up (theta = 0) then the cart would just continue to move along at its initial velocity with the pendulum pointing straight up. The initial velocity of the cart doesn’t affect the pendulum (unless there is damping applied to the cart), but rather it is the accelerations of the cart that impart a force/moment to the pendulum and hence the coupling. Pro Tip: Any time you are dealing with geometric nonlinear problems, you will have to update the stiffness matrix with each time step. You will also find that updating the mass matrix with each time step becomes very expensive because you need to invert M. Inverting a matrix requires in the order of n^2 iterations. As a result, as n grows, inverting the mass matrix becomes prohibitively expensive (time consuming). The good news is that in many instances you actually don’t need to update the mass matrix (unless mass is being lost by the system) and this results in very little loss of accuracy, but a huge improvement in speed. You should alway test whether not updating the mass matrix produces similar/accurate results.

@@Freeball99 That clears things up. Never crossed my mind that it was a semi definite system. Trying to understand balancing of inverted pendulum now. Thanks for the tip. You have been phenomenal. Thanks a lot.

@@NYKYADU the balancing of the pendulum, which is a controls problem, requires you to first linearize the EOM, by assuming that the displacements are small and then throwing away the higher-order terms and making appropriate substitutions i.e. Sin θ = θ and Cos θ = 1.

Technically you are correct! However, for the case of the simple pendulum, it is typically assumed that we are dealing with point masses which is why we can ignore the rotational kinetic energy. I mentioned this in the video on the double pendulum, but should have been more clear here. The reality is, that when dealing with a simple pendulum, the rotational kinetic energy is typically negligible when compared to the translational kinetic energy (except for cases when the diameter of the mass is very large and the connecting rod is very short). Also in reality, the connecting rod is not massless and it will also contribute to the rotational kinetic energy (and is usually more significant than the rotational KE of the mass). I have shown this for the case of the so-called "compound pendulum" in this video: kzfaq.info/get/bejne/m6iXa8qgy5e2cps.html So, for the simple pendulum, you can ignore the rotational KE with little loss of accuracy.

@@Freeball99 Thanks for the informative video. But continuing with Enise's question, if we include rotational kinetic energy as well, what I (moment of inertia) should be taken. Is it about the centre of mass? or is it about the hinge?

@@ashishnegi9663 You need to find the rotational KE about the center of mass. The effects of the mass rotating about the hinge are already taken into account by considering its translational velocity relative to the hinge.

@@Freeball99 ok, thanks. I’ll think about it.

Yes, that is correct. But according to Lagrange's Equation, I need to subtract this term. This flips the sign again. So we have three negative signs: 1) the original sign of the term, 2) the sign we get from taking the derivative 3) the sign we get from the 2nd term of Lagrange''s Equation after substitution.

@@Freeball99 I want to connect with you via whatsapp, sir I am Control Engineer

@@perceptronsaber4479 I don't use WhatsApp. You can email me at: apf999@gmail.com

@@Freeball99 I was also trying to figure out why we got that negative. Thank you so much for the video and all the explanations