- #1
LFCFAN
- 22
- 0
Hello fellow PF members
I was wondering how one would go about finding the lagrangian of a problem like the following:
A particle is constrained to move along the a path defined by y = sin(x).
Would you simply do this:
x = x
y = sin(x)
x'^2 = x'^2
y'^2 = x'^2 (cos(x))^2
Kinetic energy = T = (1/2)m(x'^2 (1+ (cos(x))^2))
Potential = V = mgsin(x)
Therefore, the lagrangian is given by
L = T - V = (1/2)m(x'^2 (1+ (cos(x))^2)) - mgsin(x)
Or have a been totally wrong throughout?
Thanks a lot guys. Any input would be greatly appreciated.
I was wondering how one would go about finding the lagrangian of a problem like the following:
A particle is constrained to move along the a path defined by y = sin(x).
Would you simply do this:
x = x
y = sin(x)
x'^2 = x'^2
y'^2 = x'^2 (cos(x))^2
Kinetic energy = T = (1/2)m(x'^2 (1+ (cos(x))^2))
Potential = V = mgsin(x)
Therefore, the lagrangian is given by
L = T - V = (1/2)m(x'^2 (1+ (cos(x))^2)) - mgsin(x)
Or have a been totally wrong throughout?
Thanks a lot guys. Any input would be greatly appreciated.