I have enjoyed watching your presentation of the Lagrangian and have found your explanation and examples simple and complete. Just one comment if I may. When solving a linear second order differential equation, there should be two constants of integration. A third order requires 3 constants and a fourth order requires 4. For example, the solution for d2(theta)/dt2 + g/l *theta requires two constants of integration. The solution can be written as either theta(t) = A.Cos(w.t) + B.Sin(w.t) or theta(t) = A.Cos(w.t + phi); just as long as there are two constants of integration either A and B or A and phi. This is similar to solving for a falling body when solving the second order DE mg+m*d2x/dt2=0. The first constant of integration is the initial velocity V(t=0) and the next constant is the initial position X(t=0). Your effort in spreading education in mathematics is very much appreciated; keep up the excellent work.
I think he made that mistake deliberately because he knows that many students tend to forget the chain rule and that theta is a function of time. It helps instil it in our minds :)
شكرا جزيلا سيدي على هذا العمل الرائع بقي لدي سؤال، حينما اردنا ان نستخرج قيمه السرعه، يتم ذلك باخذ مركبه x y بينما انت قمت باخذ مركبة x. مع الارتفاع h؟ كيف تم ذلك؟
Hello!! When taking the partial derivatives of L wrt to theta why is the chain rule not applied to sin(theta)? I would have thought there would be an extra theta dot term there. I am only asking as earlier when you made the 'mistake' the chain rule was applied when determining the components of velocity. Any help would be appreciated!!!
Hi I am trying to get a context of this advanced mechanics. I remember well my high school physics, which is as far as I ever got in physics. I am wondering how we got from F=ma to the Lagrangian and Hamiltonian approach. I have been trying to learn about the principle of least action. Maybe a short video about this transition. ?? Thanks.
How you arrived at Theta = A * sin (wt) is not clear. It feels like you just jumped to it in the last steps. Could you please explain that a little bit?
The best way to check is to plug it back into the differential equation and see that it is a solution to the equation. That is a standard solution to a differential equation like that.
@@MichelvanBiezen I wish I had come across this channel in my first year. Currently doing my final year and wrote a robotics test on this last night. I really struggle knowing how to find the general coordinates of a given system.
That is a standard differential equation of an oscillator and thus the solution is the standard solution. The best way to check is to use the result to get back the differential equation. Take a look at this video: Differential Equation - 2nd Order Linear (9 of 17) Homogeneous with Constant Coeff: Free Oscillator ruclips.net/video/V9bl02Ffo_o/видео.html
No, when you take the partial derivative with respect to theta, you do not take the derivative of theta dot (which is essentially the angular velocity)
I asked this in my classical mechanics class earlier this semester. Apparently it requires some advanced mathematical tools usually taught at graduate level.
@@EagleLogic Thanks a lot. I did my research too and it appears that this Diff Equation comes down to solving an elliptic integral and has some connection to the Jacolbi amplitude function
I haven't finished all 25 videos yet so I do not have entire picture. My question at this point is -- why do we define L and use partial derivatives? How about simply d/dt(d(KE)/dx') + d(PE)/dx = 0 which is more intuitive.
@@MichelvanBiezen Thanks for the reply...my thinking was If KE only includes x' terms and PE only x terms, the original Lagrange equation dt/d(δL/δx') - δL/δx = dt/d(δ(KE)/δx' + δ(PE)/δx = 0 so we can deal with KE and PE directly, without using L = KE - PE. In one variable case, the above becomes dt/d(d(KE)/dx') + d(PE)/dx = 0 So why would we bother to use L?
Hello, is there anyone that could please help me [I really need help :( ] using the Lagrangian equation to find a function of time and theta for a cycloid? Any help is appreciated!
Still enjoying your videos. Sometimes your body blocks the view of the work. Perhaps be more conscious of body position and/or pause so we can see the board. Thanks.
Beginning at 8:33, I got a little lost. Justifying substitution of theta for sine theta is done how? And then introducing omega happens fast. Clarifying that would be helpful. THanks.
omega being equal to the square root of g/l is the standard equation for the angular frequency of a pendulum which I assumed that anyone looking at the advanced concepts of Lagrangian mechanics should know. (But then I often forget equations as well so we have to go back and look them up again).
I am a hobbyist and college dropout from the seventies. So I never had college physics, though I was a math major for 2 years before I quit school. Since RUclips came along I have been able to dabble in advanced math and physics which always interested me. Thanks for the work. Where do you live?
At 3:40, how come we also multiply by theta dot? I'm unsure where this comes from since you mention it is the derivative of the angle wrt. time. Since we have already differentiated the whole thing wrt. time.
Because of the chain rule from calculus. Since x is a function of theta and theta is a function of time, we are differentiating x wrt theta and then, differentiating theta wrt time.
I had the hardest time understanding this when doing related rates. I went back and drilled it so hard till I understood the chain rule fully. Since x = l*sin(θ) when you do dx/dt you gotta understand that θ is a function of t. You can even write the equation as x = l*sin(θ(t)) so then dx/dt is l*cos(θ)*θ' You gotta do the derivative of the outside keeping the inside the same. Which is why he made it cos(θ) but then you gotta multiply by the derivative of the stuff on the inside which is θ'.
I have enjoyed watching your presentation of the Lagrangian and have found your explanation and examples simple and complete. Just one comment if I may. When solving a linear second order differential equation, there should be two constants of integration. A third order requires 3 constants and a fourth order requires 4. For example, the solution for d2(theta)/dt2 + g/l *theta requires two constants of integration. The solution can be written as either theta(t) = A.Cos(w.t) + B.Sin(w.t) or theta(t) = A.Cos(w.t + phi); just as long as there are two constants of integration either A and B or A and phi. This is similar to solving for a falling body when solving the second order DE mg+m*d2x/dt2=0. The first constant of integration is the initial velocity V(t=0) and the next constant is the initial position X(t=0). Your effort in spreading education in mathematics is very much appreciated; keep up the excellent work.
SUPPORT MICHEL VAN BIEZEN ON PATREON!: www.patreon.com/user?u=3236071
The way you have derived each and every step is Crystal clear to Understand.Thank You!!!
You are welcome!
great! even as a foreign language, it's very easy to follow your words!
I think he made that mistake deliberately because he knows that many students tend to forget the chain rule and that theta is a function of time. It helps instil it in our minds :)
Lol perhaps. I tend to make that mistake very often.
This class is amazing professor! Thank you very much! A big hug from Brazil!🇧🇷
Thank you and welcome to the channel!
che ricordi, che bellezza, (Meccanica Razionale, Prof. Nappo, Roma 1988)
شكرا جزيلا سيدي على هذا العمل الرائع
بقي لدي سؤال، حينما اردنا ان نستخرج قيمه السرعه، يتم ذلك باخذ مركبه x y بينما انت قمت باخذ مركبة x. مع الارتفاع h؟ كيف تم ذلك؟
Hello!! When taking the partial derivatives of L wrt to theta why is the chain rule not applied to sin(theta)? I would have thought there would be an extra theta dot term there. I am only asking as earlier when you made the 'mistake' the chain rule was applied when determining the components of velocity. Any help would be appreciated!!!
Because it's partial derivative respect to theta instead of time.
Maybe it's too late to tell you. But i tell you if you're still wondering.
@@zahranf.a.9864 dw bout it ur only 2 years late
Awesome, lucid, comprehensive lecture! Love this.
Hi! Would just like to ask when is it ok to use sine or cosine? Love your videos!
The use of the sin of cos depends on the initial conditions. Where is the object at when t = 0 and which way is it moving?
Hi
I am trying to get a context of this advanced mechanics. I remember well my high school physics, which is as far as I ever got in physics. I am wondering how we got from F=ma to the Lagrangian and Hamiltonian approach. I have been trying to learn about the principle of least action. Maybe a short video about this transition. ?? Thanks.
It is in the plan for when we have time.
How you arrived at Theta = A * sin (wt) is not clear. It feels like you just jumped to it in the last steps. Could you please explain that a little bit?
The best way to check is to plug it back into the differential equation and see that it is a solution to the equation. That is a standard solution to a differential equation like that.
Very helpful for our learning that the mistake wasn't edited out.
Just wanted to show that everyone makes mistakes.
@@MichelvanBiezen I wish I had come across this channel in my first year. Currently doing my final year and wrote a robotics test on this last night. I really struggle knowing how to find the general coordinates of a given system.
You are a king Michel 👑👑
Far from it, but glad we could help. 🙂
Please can you tell me how you solved the last equation to get theta = Axsin(wt)? I tried and I tried but couldn't have understanding it.
That is a standard differential equation of an oscillator and thus the solution is the standard solution. The best way to check is to use the result to get back the differential equation. Take a look at this video: Differential Equation - 2nd Order Linear (9 of 17) Homogeneous with Constant Coeff: Free Oscillator ruclips.net/video/V9bl02Ffo_o/видео.html
thank u sure for helping us ....... just a note i think that u forgot the (theta dot ) in the derival of largrangien at the time 7:06.....and thank u
No, when you take the partial derivative with respect to theta, you do not take the derivative of theta dot (which is essentially the angular velocity)
Time period is equal to 2pai square root of l divided by g
If we don't substitute θ=sin, how does one solve this diff equation? Any paper to read about it?
I asked this in my classical mechanics class earlier this semester. Apparently it requires some advanced mathematical tools usually taught at graduate level.
@@EagleLogic Thanks a lot. I did my research too and it appears that this Diff Equation comes down to solving an elliptic integral and has some connection to the Jacolbi amplitude function
Ahh yes! You are right. No idea how to do that haha. That's neat though
@@EagleLogic I hoped a proffesor would have answered our question, but never mind. Cheers mate. Stay curious
Likewise! I have my classical mechanics final tomorrow and I'm very nervous. Best of luck with your studies!
Cheers!
In time 2:30 we should take the dervitive of x with respect to time using products rule.if I’m not mistaken
You're taking the derivative pf y with respect to time so you treat l as a constant. Thus, no need for product rule.
At 7:04 why do we not multiply by the derivative of theta, theta dot?
https: //ruclips.net/video/paVOEi7cYrA/видео.html.( Mec.English and french)👍💐
Exactly
I like the lecture so much
We are glad it was helpful
🤗🤗, but what about when sin theta not same as theta, large angles?
Then you no longer have simple harmonic motion. The angle must be small for simple harmonic motion for a pendulum.
Can u solve this problem by using conical pendulum
I haven't finished all 25 videos yet so I do not have entire picture. My question at this point is -- why do we define L and use partial derivatives? How about simply
d/dt(d(KE)/dx') + d(PE)/dx = 0
which is more intuitive.
The partial derivatives are needed in order to isolate the terms needed for KE only and for PE only.
@@MichelvanBiezen Thanks for the reply...my thinking was
If KE only includes x' terms and PE only x terms, the original Lagrange equation
dt/d(δL/δx') - δL/δx = dt/d(δ(KE)/δx' + δ(PE)/δx = 0
so we can deal with KE and PE directly, without using L = KE - PE.
In one variable case, the above becomes
dt/d(d(KE)/dx') + d(PE)/dx = 0
So why would we bother to use L?
When you plugged in PE, you messed up: L = KE - (mgl - cos(theta)) --> L = KE - mgl + cos(theta)
https: //ruclips.net/video/UHocGHguWJI/видео.html👍👍
Good job
Thank you
how abaout double pendulum (2DOF) with m1=m2 and l1=l2 sir? have you made video about it? thanks a lot before.
No, not yet. That is something for the future.
Hello, is there anyone that could please help me [I really need help :( ] using the Lagrangian equation to find a function of time and theta for a cycloid? Any help is appreciated!
https: //ruclips.net/video/paVOEi7cYrA/видео.html.( Mec.English and french)👍💐
Still enjoying your videos. Sometimes your body blocks the view of the work. Perhaps be more conscious of body position and/or pause so we can see the board. Thanks.
Thanks for the suggestion. I have been trying to work on that.
Beginning at 8:33, I got a little lost. Justifying substitution of theta for sine theta is done how? And then introducing omega happens fast. Clarifying that would be helpful. THanks.
For small angles theta = sin (theta) (try it on your calculator, using radians).
omega being equal to the square root of g/l is the standard equation for the angular frequency of a pendulum which I assumed that anyone looking at the advanced concepts of Lagrangian mechanics should know. (But then I often forget equations as well so we have to go back and look them up again).
I am a hobbyist and college dropout from the seventies. So I never had college physics, though I was a math major for 2 years before I quit school. Since RUclips came along I have been able to dabble in advanced math and physics which always interested me. Thanks for the work. Where do you live?
We live in the Los Angeles area.
At 3:40, how come we also multiply by theta dot? I'm unsure where this comes from since you mention it is the derivative of the angle wrt. time. Since we have already differentiated the whole thing wrt. time.
Because of the chain rule from calculus. Since x is a function of theta and theta is a function of time, we are differentiating x wrt theta and then, differentiating theta wrt time.
Prof neglected to do that and had to back up around 2:45 and erase.
I had the hardest time understanding this when doing related rates. I went back and drilled it so hard till I understood the chain rule fully. Since x = l*sin(θ) when you do dx/dt you gotta understand that θ is a function of t. You can even write the equation as x = l*sin(θ(t)) so then dx/dt is l*cos(θ)*θ'
You gotta do the derivative of the outside keeping the inside the same. Which is why he made it cos(θ) but then you gotta multiply by the derivative of the stuff on the inside which is θ'.
Good to see that you are human and can make mistakes sir.
Just kidding!
Thank you for all of your 'easy to understand' lectures. They help a lot! :)
Thanks sir❤🙏
Most welcome
why did we take parial derivative wrt theta
Because position in the x and y direction as well as the velocity can be expressed as a function of theta
you may define y-direction first
isn't omega= l/g
not g/l
w^2 = g/l (it makes sense when you consider that the restoring force will be larger if g is larger.
if sinx=x then cosx=1
HOW FAR I KNOW THAT T = 2π√(l÷g)
https: //ruclips.net/video/paVOEi7cYrA/видео.html.( Mec.English and french)👍💐
👏👏👏👏
Thank you. 🙂
From where did you take coordinates.....if you are going to take coordinates as from mass than the numerical is pointless
He make a mistake. but, that no problem..