The Atwood Machine with Lagrangian Mechanics

[@victoriawhite9441]

thank you for saving my career

[@CURSEYOUBAYLE688]

I appreciate these videos. Helped me in my classical mechanics course and now I want to go through them again for a better understanding of the methods. Thank you :)

[@DotPhysics]

Glad to help!

[@mungorn]

Minor quibble: There is a sign mistake at 5:11, but it doesn't matter when you go to the Euler-Lagrange equation since it is a term that doesn't depend on any positions or velocities.

[@DotPhysics]

oh! That's two sign errors in two straight videos! I'm going to fail this KZbin course. Thanks for catching that.

[@mariogguazzelli5929]

@@DotPhysics thank you for the great videos, i understood your video better than i did a whole engineering semester worth of classes. the best teachers always make sign mistakes!

[@mygoogleaccount4404]

@@DotPhysics sign mistakes are important! it tests if the students are paying attention!

[@christomashofski9160]

@@mariogguazzelli5929 I co-authored a few journal articles with a friend who is one of the chief scientists at NASA Ames Res. Center, and a favorite, humorous saying of his which I shamelessly borrow often is "The two banes of every engineer's existence are dropped minus signs and missed unit conversions!" LOL I can attest to the truth of your statement! ;)

[@woosix7735]

it's a classic, forget to distribute the minus sign

[@muhammadashfaq3021]

teaching method is really nice..... have you video explanation about tensors too....

[@SD-it5ko]

y1+ y2=c1? Is this a very particular type machine where the sum of the distances of the weights above ground is equal to the lenght of the cord?

[@TitanOfClash]

Yeah, that's a mistake. But when you add another constant in, like y1+y2+p=c1 it won't make a difference to the calculations

[@cathalmcdonough5247]

if m2 is large should the m2's acceleration be a negative (-g) since it's going downwards?

[@ВОЛОДИМИРК-д4о]

L=1/2(m1+m2)(dy1/dt)^2+y1g(m2+m1) - (not +) m2gc. BUT IT DOES NOT INFLUENCE ON A RESULT

[@st_s3lios860]

It's actually the same to do: Em=Ep+Ec=cst (Ep: potential energy, Ec: cinetic energy, Em: mechanic energy) You derive with respect to time and divide by the derive of the parameter with respect to time. Lagrangian is just an other presentation of this no? (sorry for my bad english)

[@hhoopplaa]

Sometimes yes, but not quite though. Lagrangian is just L=Ec-Ep , which is not necessarily constant like Em even though the sum might be (e.g. 5+2=4+3=7, but 5-2=/=4-3). What we ask of the Lagrangian is that its integral from one point in time to another is minimized (i.e. a minimum) which gives us the Euler-Lagrange equation. From then on you get the Equation of Motion. Lagrangian works the same way even when mechanical energy is not constant, however I think there is something to account for if there was friction involved. So if energy is not conserved, it may not be possible to use conservation of mechnanical energy but you can still use Lagrangian. Is using mechanical energy conservation equivalent to using Euler-Lagrange equation in the cases where they both are possible? My gut says no but I can't give a reason so yea

[@RiccardoCattaneo-uu5it]

​@@hhoopplaaLandau's phisics book would be of use to answer your question

[@pascaltran4398]

Erreur sur L: -m2gC1