no shit sherlock

yeah well ofcourse that is going to happen lmao

@@random22453Only if you dont do anything wrong😂

By the way, which textbook were you using for this course?

Your Playlist is very nice in which college u study?

It's just some algebra. Substitute the 2, 4 into 9, then expand it all and then simplify. If your write it out, you'll see it immediately.

This comes from squaring the velocities on the previous line (though I am noticing that I am missing a squared sign on the last term). If you write out the previous line and expand it, I think it will become clear.

Yes, you can divide through by m2 to simplify. Usually, when deriving these equations of motion, I like to leave it in this form because I find the equation more meaningful when each term has units of force (or force/length).

Yes, you can derive these equations by drawing the free body diagrams for each mass and then applying Newton's 2nd Law. This can be a little tricky as it's easy to make mistakes if you're not careful (which is why using Lagrange's Equations is preferable). Unfortunately, I don't have a video on this, but I'm sure you'll be able to find one online.

I commented without finish the video, you fixed it minutes later, my bad.

This is due to the fact that the second pendulum, due to its acceleration, exerts a force (moment) on the first pendulum.

This video address it: kzfaq.info/get/bejne/m6iXa8qgy5e2cps.html

I've never seen it, but you could derive it!

The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil.

You would incorporate this on the right-hand side of the equations as a moment. If it's a tangential force, then this would only affect the 2nd equation. If it is truly a horizontal force, then it would affect both equations of motion.

If we're going to get technical with this, I'd say the following...the EOM's and the mathematical model are not the same thing, but rather the EOM's are derived from the mathematical model. From the EOM's, we can calculate the response of the system.

dhanyawaad

There you go. kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html Additionally, this model can be incorporated into a control's problem and then the combined system can be solved...but that's for another class.

Yes, although the equations would, in effect, reduce to a single degree of freedom system.

There are two negatives due to applying the chain rule, so they cancel out. The first negative comes from differentiating cos( ) and we get a second negative when differentiating the (θ1 - θ2) part with respect to θ2.

I showed this mathematically in going from equation 8 to equation 9. If you write out on the substitutions on a piece of paper, you'll find it's just algebra. But one way that you can think about it physically, it was when Θ1 increases, it lifts, both m1 and m2, however, when Θ2 increases it lifts only m2.

Freeball Okay thank you 👍

Check out my video on Coulomb damping which explains how to incorporate friction in the equation of motion. kzfaq.info/get/bejne/h7uimKR-nLC4p58.html

Often in structural vibration theory, one must deal with structures which are NOT statically determinate. This means that one cannot use Newton's 2nd Law to get meaningful results and needs to turn to other methods - namely energy methods which deal with conservation of energy amongst other things. Lagrange's equations are used to find the equations of motion from the expressions for the energy of the system and these equations of motion can then be solved to determine exactly how the system moves in time. Lagrange's equations are derived using something called Hamilton's Principle which relies on a branch of mathematics called Variational Calculus. It's advanced stuff and typically deriving Lagrange's Equations is saved for advanced structures classes. HOWEVER, the resulting Lagrange's Equations are easy to implement and are very powerful tools so their use typically gets introduced in lower level classes - long before most folks have the required background to derive them.

@@Freeball99 Ok great thanks, just another question I have is what the next step would be because I know that in your video involving the simple pendulum, you had a differential equation that you solved and got θ(t) = Acoswt. How would you go from what you have in this video to something like θ(t) = Acoswt? Thanks again.

@@cankaraca456 Since these equations are nonlinear in their current form, one cannot write them in a closed-form solution (ie you cannot write the response as a function). Therefore, the equations must be integrated directly using numerical methods. I have showed an example of this here: kzfaq.info/get/bejne/iaygfcWera-5cmQ.html HOWEVER, the equations can be linearized IF the displacements are assumed to be small (θ's of less than about 15°). In this case, you can make the assumption that Sin θ = θ and that Cos θ = 1 and apply this to the equations of motion. The response can then be written as a function (as you have described) using the method shown in this video: kzfaq.info/get/bejne/sNVigM9pnbfdonU.html

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This goes back to the derivation of the Euler-Lagrange Equation and Lagrange's Equation which I show in this video: kzfaq.info/get/bejne/gr9dfcamvte3dZ8.html At one point in the derivation, we take the variation of the Lagrangian (at the 2:30 mark in the attached video) and this involves that the TOTAL DERIVATIVE of the Lagrangian. When doing this, we treat the dependent variables (the coordinates) and their derivatives as independent quantities. We later resolve the fact that there are related quantities by using integration by parts. It's a little hard to explain in this text, but if you watch the video, it covers this part.

If there is a moment applied about the first hinge point, then this will appear on the right hand side of the first equations and if there is a moment applied about the second hinge point then this will appear on the right hand side of the second equation.

Equation 19 is correct as written. I multiplied out the (theta_dot1 - theta_dot2) and then the terms with the theta_dot2 cancel out. What remains is just the theta_dot1 which appears in eqn 19. If you write it out, it will become apparent.

You could model this directly in polar coordinates if you prefer. However, from my experience, in most case where you're dealing with pendulums and gravitational effects, it tends to be easier to define your kinematics in cartesian coordinates.

@@Freeball99 weird, I've only been taught the polar method. I found it difficult transforming between the two. I still prefer the polar, due to movement being predominantly circular. Thanks for your reply

I skipped a step here just to save a little time. If you expand out the equation (ie distribute the (theta dot 1 - theta dot 2) ), you will find that terms cancel out. Try expanding out the previous step and a piece of paper and you will see which terms cancel out. It is correct as written.

In equation 19 the + sign is correct. It's actually the line before this where I should have put a + sign instead of a -ve. Substituting eqn. 18 into eqn. 15 is what causes the sign to flip to +ve. Thanks for catching it.

This is due to assumption #1 that the masses are point masses so the have no rotational kinetic energy. The rotational KE effects are generally omitted for teaching purposes just in order to keep it a little more simple and easy to follow. Also, for a typical setup, the rotational KE effects tend to be less important (unless the diameter of the masses is very large). The compound pendulum then demonstrates how to include these effects. For a compound pendulum, the rotational KE effects tend to be far more important than for the case of the simple pendulum.

This is due to the fact that this system is highly nonlinear. As a result, the response of the system is extremely sensitive to the initial conditions. A slight change to the initial condition will result in a very different response. This is one of the hallmarks of a chaotic system.

In the case of a simple pendulum, it is a common assumption. What we are really saying is that the mass of the rod is so much less than the mass of the pendulum bob that its contribution to the kinetic and potential energies is insignificant. If we want to take the mass of the rod into account (as it grows larger), then this becomes a compound pendulum. I have some videos on compound pendulums with explain how to model those.

Because theta is considered a function of time. That means we have to apply the "chain rule": multiply the expression we get (in this case exactly that cos(theta sub 1) which you mentioned) by the derivative of the function inside the sine (theta sub 1)

I'm not sure I understand. Where does it show that the energy of the system is equal to zero?

Sorry, but I no longer use MATLAB since deciding to take the plunge into Python several years ago. Much cheaper.

The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil.

I have a video with a pendulum on a cart (kzfaq.info/get/bejne/fNVnntiGpq7OY6c.html). While it is not exactly what you're asking about, it is pretty close. Using this video and the one above, you should be able to derive the equations of motion for a double pendulum attached to a cart.

@@Freeball99 I will see. Thank you so much.

This comes down to where you set your zero-point. In reality, the potential energy is only zero at the center of the earth, however, we can arbitrarily set a point to be zero by shifting our axes. The truth is that we really only care about change in the potential energy rather than the absolute value. In this problem, I the axes we placed at the top hinge, so any masses below this would, in effect, have -ve potential energy. If the axes had, instead, been attached to the bottom mass, then that would be zero potential energy and all potential energies would be positive.

The issue is actually the line before, the sign should be a plus since we are subtracting eq. 18 both of the minus signs become a plus.

When taking the derivative with respect to θ2, we get two negatives in the first term which combine to form a positive. Applying the chain rule to cos(θ1 - θ2) gives us -sin(θ1 - θ2) · (-1) = +sin(θ1 - θ2) . So, this term should have a positive sign.

App is "Paper" by WeTransfer. Running on iPad Pro 13-inch. Using an Apple Pencil.

Definitely not a stupid question...both are correct. The difference is just due to a constant (L1) which is a result of where one assumes the potential energy is zero. In this problem, I have assumed that zero potential is where the system attaches to the ceiling. In your description of the problem, zero potential (of m1) is assumed to be at the rest position of m1. The point is that we don't really care about the absolute value of the potential energy, but rather WE CARE ABOUT ITS DERIVATIVE (ie the CHANGE in potential energy with each coordinate). Once we substitute the potential energy expression into Lagrange's equations and differentiate it, the constant disappears and both our descriptions of potential energy yield the same results.

The Q_i represent the generalized force. It is zero if there is no externally applied force/moment in the direction of a particular generalized coordinate.

The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple Pencil.

This is due to the assumption that these are point masses so they have no rotatory inertia. For the inverted pendulum problem, take a look at this video which explains it: kzfaq.info/get/bejne/fNVnntiGpq7OY6c.html - You need to be careful with just flipping your coordinate system. Fundamentally, it is the potential energy expression that will flip signs, but you need to be careful to be consistent with your definition of the positive direction of θ.

​@@Freeball99 Thanks for the explanaiton. The thing is that I need to do like a human spine and I was looking for a mathematical model or at least an example, because it feels so difficult to do a 3D model, don't know if you have any idea... And sorry for bothering you.

Clearly your best option is to watch all these videos immediately so that you've got the next test covered!!

@@Freeball99 ahhh mate I will but unfortunately we're doing electrodynamics now so these won't show up anytime soon. Love the content tho

also, another weird thing is, it specifically asks that i derive the equations of motion in terms of theta1 and theta2, these being the angles of rotation of the two joints, instead of in terms of the x,y cords of the two points, which is yet more confusing to me

Do you have a figure that you can post using Dropbox or similar?

The difference here is just a matter of the coordinate system used and where you place the origin. This comes down to the question of where does the system have zero potential energy. Some might define that the point of zero potential energy is the center of the earth, others might define it as the height of a mass in its equilibrium position. In my case, I placed the origin at the hinge (whereas yours places it at the mass). Shifting the location of the origin, will simply change the potential energy expression by a constant. So the observation that you made is the correct one, namely, that for the purpose of Lagrange's Equations, the actual value of the potential energy is irrelevant. What is relevant is the DERIVATIVE of the potential energy - as long as these match, then you're good. As a result, you should pick whichever coordinate system you find most comfortable.

Generally I find it simpler to derive pendulum problems in cartesian coordinates and it is less prone to errors. You can certainly derive it in polar coordinates. Try it and see if you can get the same results.

Yes, the motion does depend on m1 as shown in equation 19.

According to Lagrange's Equations, the generalized forces, Q_i appear on the right hand side of the equation. So in the case of a moment applied to mass 1, this would appear on the right hand side of equation 1. For a moment on mass 2, this would appear on the right hand side of equation 2 - this includes both conservative and non-conservative forces/moments.

@@Freeball99 yeah, and i if i am not wrong on the right hand side we only consider the non conservative forces because the delta W (work done) should be of non conservative forces. plss correct me if i am wrong as at this part i am confused. i get confused on what to put when we come to RHS of Lagrange's Equation

@@nihargandhi3740 Any non-conservative forces should appear on the right-hand side. Additionally, if you have conservative forces, you can EITHER add these on the right-hand side as work OR can be included on the left-hand side as part of the system potential which is just the negative of the external work produced by the conservative force (V = -We).

@@Andy-hy8px ohhh yeah I get it. Thanks for the explanation 👍

This would, in effect, increase g. So it would increase the restoring force on the pendulum. As a result, the pendulum will oscillate at a higher frequency.

@@Freeball99 thanks 😊

errrrrrr differential equ u solve after that

basically you have a system of 2 second order differential equations and you just solve them which is massively painful because....well, just try it :D

It is shown here: kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html

Thanks for your question, I will have to save this for a different video since the text comments are too limiting.

In this problem, the Qi's are 0 because there are no externally applied forces or moments.

This is because there is no external moments being applied.

Yes, good idea. There you go. Here is part 1. The other parts will be out within a week or so. kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html

@@Freeball99 thank you😊

@@Freeball99 THANK YOUU

Vincent Tarantino software is “Paper” by WeTransfer. Using an iPad Pro 13-inch and Apple Pencil.

@@Freeball99 Awesome! It looks so good! I'm a Physics teacher and I'll have to be teaching online classes and I love the way this looks so that's what I'm going to use

Yes, there is a careless typo here. The - sign should be + here, I have it correct in the lines that follow. Which is the extra term?

Never mind I see thanks anyway

I multiplied it out (distributed it). If you look at the 3rd and 4th terms of the following equation, you will see that they are contained there.

@@Freeball99 thanks , how would I then solve these equations?

This was an error, but I corrected it a few minutes later in the video.

perchè il papa non è re ;)

Good question. I purposely did it differently to prove a point...These are, in effect, the same thing. They just differs by a constant, so really it amounts to where one decided to place the the point of zero potential energy (the origin of your axes). I have assumed that the zero potential is at the hinge while your expression puts the zero potential energy location at the lowest position of the mass. The truth is that BOTH are equally valid. It turns out that we don't really care about the absolute value of the potential energy (and the location that we call zero potential energy is somewhat arbitrary - technically the point of zero potential energy is really at the center of the earth). Rather, what we care about is the DERIVATIVE of the potential energy and the derivatives are identical (because you just added a constant, l1, which disappears when you differentiate it.

@@Freeball99 I see your point, I care about the derivative of potential energy and for object m1 it's m1gl1sin(theta1)

@@Freeball99 I have been watching different videos searching for an explanation on why people use different methods for calculating potential energy. Thanks so much for the thorough explanation!:-)

This comes from the geometry of the problem and based on the fact that θ1 is measured relative to the vertical direction. Based on the definition of the coordinate system and the location of the angle, θ1, the x1-coordinate is equal to the length of the side opposite θ1, and we already know the length of the hypotenuse (which is L1), so from trigonometry: sin(θ1) = x1 / L1 which gives us x1 = L1 * sin(θ1)

Why y using -cos not cos?

@@hikmahmaulidina333 Because the y-axis is defined as positive upwards, so this is in the negative y-direction.

kindly solve this problem

I need more information for this. Do you have a problem statement you can post?

@@Freeball99 yes

There you go, I solved it in this video: kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html

@@Freeball99 comm si fort, ti amo

Not sure exactly what you are referring to - I'm guessing the substitution into Lagrange's equations...in which case, we are taking the derivative with respect to θ and then with respect to time.

Absolutely!!

these 'solutions' are the differential equations describing the motion of the double pendulum. Indeed, this is not a conclusive solution just yet. In order to obtain this, you should solve this differential equation given certain boundary conditions.

I solve the equations in this video: kzfaq.info/get/bejne/pKhyhK6Lx5yuip8.html

This is because θ2 is a function of t. If we differentiate w.r.t. θ1, then we would treat θ2 as a constant, HOWEVER, when we differentiate w.r.t. t, then we are take the so-called "total derivative" - i.e. it is not a partial derivative.

This is just the general form of Lagrange's Equations. In this particular case, yes, it is equal to zero.

@@Freeball99 What does the Q_i mean in the general equation?

@@michaelwang1730 The Qi's are the generalize forces (ie forces or moments) associated with the ith coordinate. These are simply the forces/moments that are applied directly to the particular coordinate in question.