- #1

- 52

- 0

A particle of mass m is constrained to move under gravity with no friction on the surface xy=z. What is the trajectory of the particle if it starts from rest at (x,y,z) = (1,-1,-1) with z-axis vertical?

The lagrangian is

[tex]L=T-V=\frac{1}{2}m\left(\dot{x}+\dot{y}+\dot{z}\right)-mgz[/tex]

the constrain is

[tex]g(x,y,z) = xy-z = 0[/tex]

for the x-component,

[tex]

\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

\frac{d}{dt}\pd{L}{\dot{x}}{} - \pd{L}{x}{} + \lambda \pd{g}{x}{} = 0

[/tex]

where [tex]\lambda[/tex] is the lagrange multiplier.

[tex]m \ddot{x}+\lambda y = 0

[/tex]

Similarly,

[tex]m \ddot{y}+\lambda x = 0

[/tex]

for z-component,

[tex]m\ddot{z}+mg-\lambda = 0[/tex]

however, i find the system is very difficult to solve. I define

[tex]\phi = x + y[/tex]

then

[tex]\dot{\phi} = \dot{x} + \dot{y}[/tex]

[tex]\ddot{\phi} = \ddot{x} + \ddot{y}[/tex]

then adding the first 2 equations,

[tex]m\ddot{\phi} + \lambda \phi = 0[/tex]

which is the same as simple harmonic equation. but this is wrong because the answer of the question is [tex]x = - y = \sqrt{-z}[/tex]

so, how can I solve the system of equations? and, is [tex]\lambda[/tex] a constant, of a function of [tex]t[/tex]?

thanks for your help