gracie
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Please help, I've put this is true on my personal statement, but I'm now fairly sure that it is not, but a more expert opinion would be extremely helpful.
You are absolutely right, and it is embarrassing to know that I forgot to add the dots! As I punishment, I've written out the full procedures for Gracie.nasu said:The first term in that Lagrangian should have the derivatives of x and y in respect to time.
What about the opportunity for gacie to discover this for herself ? "Give it a try and PF will help you further..." Then it woudn't have been a waste either !Rescy said:Of course! Lagrangian is equivalent to Newton's second law of motion in Cartesian coordinates.
$$L= {1\over 2}m(x^2+y^2)-mgy$$ apply Euler-Lagrange equation for x and y and you derive a differential equation whose solutions are SUVAT.
It is a waste for Lagrangian Mechanics to be applied on SUVAT!
May I ask if you are applying to UK or US?gracie said:Thank you so much everybody, I thought I knew the answer but wanted proof from experts, I really appreciate it :)
BvU said:Hello Gracie, welcome to PF!
Impressed you should be concerned with Lagrangians at 17 !
But, to answer your question: yes you can. Simply write down the Lagrangian and then the Lagrange equations of motion follow.
Give it a try and PF will help you further...
Rescy said:May I ask if you are applying to UK or US?
BvU said:Hello Gracie, welcome to PF!
Impressed you should be concerned with Lagrangians at 17 !
But, to answer your question: yes you can. Simply write down the Lagrangian and then the Lagrange equations of motion follow.
Give it a try and PF will help you further...
And how can one derive distance, initial velocity and time, knowing only the lagrangian?BvU said:Google to the rescue:
s = distance (metres, m)
u = initial velocity (metres per second, ms^-1)
v = final velocity (metres per second, ms^-1)
a = acceleration (metres per second squared, ms^-2)
t = time (seconds, s)