Finishing the method of undetermined multipliers

Рет қаралды 8,926

11 жыл бұрын

Continuing from the last video, we finish using the method of undetermined multipliers to find the equations of motion and the force of constraint.
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Пікірлер: 14

@Mario95133 5 жыл бұрын

I just gotta say...if I do well on my mechanics exam tomorrow, its all thanks to you

@PhysicsHelps 5 жыл бұрын

Glad to help! Let me know how it goes. :)

@ReidarWasenius 4 жыл бұрын

Beautiful! Thank you for creating and posting the series leading up to this video.Very clear and highly educational. Greetings from Finland.

@terrancelangston6429 2 жыл бұрын

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@casonjimmy4776 2 жыл бұрын

@Terrance Langston Instablaster ;)

@terrancelangston6429 2 жыл бұрын

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@terrancelangston6429 2 жыл бұрын

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@casonjimmy4776 2 жыл бұрын

@Terrance Langston Glad I could help :D

@kamalgurnani924 5 жыл бұрын

You make good videos, thanks for the efforts!

@Zolarmew 9 жыл бұрын

you're my hero

@orientaldagger6920 3 жыл бұрын

nice !

@thrunsalmighty 10 жыл бұрын

My hope was that using the Lagrangian, at least the first integration of the equations of motion would be automatically completed.But it appears that this is not one of the benefits of the Lagrangian method.

@AL-jg8pv 5 жыл бұрын

Ah, ive got a weird little question we put f=X-R*Theta=0 but i was just wondering i decided to take the derivative right here and said df/dX= d(X-R*Theta)/dX=0 wouldnt i get df/dX=1=0,which seems silly. Ive kinda confused myself i know i'm probably thinking of it in the wrong way maybe my assumption that df/dx=0 is wrong but could ya clarify where exactly i went wrong in my argument

@shlokdave6360 3 жыл бұрын

So, if you look at the equation of f = X - R*theta = 0 in isolation, please remember that f is a constraint condition hence theta is not independent of X. ie values of theta are constrained to have only certain values corresponding to value of X. so, df/dx is not just 1, because the constraint equation is forcing theta to take values depending on x. hence df/dx = 1 - R dtheta/dx. which again is 1/R, hence the 0. The point is that its a constraint equation hence x and theta cease to be independent.