Lagrangian -> Equation of motion derivation

AI Thread Summary
The discussion focuses on verifying that a specific Lagrangian, L=\frac{1}{12}m^{2}\dot{x}^{4}+m\dot{x}^{2}V-V^{2}, produces the same equations of motion as the simpler L=\frac{1}{2}m\dot{x}^{2}-V. The user attempts to derive the equations using Lagrange's equations but encounters difficulties in simplifying the resulting expression. They suspect a sign error in their algebra might be the cause of the issue. The community suggests rechecking the calculations to identify any potential mistakes. The conversation emphasizes the importance of careful algebraic manipulation in deriving equations of motion from the Lagrangian.
Simonelis
Messages
1
Reaction score
0

Homework Statement


I teach myself classical mechanics from David Tong
http://www.damtp.cam.ac.uk/user/tong/dynamics.html
From the homework set
I should verify that the Lagrangian

L=\frac{1}{12}m^{2}\dot{x}^{4}+m\dot{x}^{2}V-V^{2}

Yields the same equations as the mere L=\frac{1}{2}m\dot{x}^{2}-V



Homework Equations



Lagranges equation of motion

The Attempt at a Solution



This seems kinda trivial exercise, straightforward derivative computation yields

something like
m\dot{x}^{2}\left(\frac{\partial V}{\partial x}-m\ddot{x}\right)-2V\left(\frac{\partial V}{\partial x}+m\ddot{x}\right)=0

which should somehow factor out and give simple equation.

Still, it does not seem to...

Could anyone help with this one?
 
Physics news on Phys.org
I think you just made a sign error somewhere. Recheck your algebra.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top