- #1
jameshaley
- 5
- 0
Homework Statement
Consider the following Lagrangian in Cartesian coordinates:
L(x, y, x', y') = 12 (x^ 2 + y^2) -sqrt(x^2 + y^2)
(a) Write the Lagrange equations of motion, and show that x = cos(t);
y =sin(t) is a solution.
(b) Changing from Cartesian to polar coordinates, x = r cos ; y = r sin ,
show that the Lagrangian becomes
L(r;θ ; r';') = 1/2 (r'^2 + r^2θ^2)- r;
and hence fi nd the Lagrange equations in polar coordinates.
(c) Show that the solution given in Cartesian coordinates in (a) is still a
solution when expressed in polar coordinates.
Homework Equations
d/dt(∂L/∂x')-∂L/∂x
L=T-V
The Attempt at a Solution
d/dt(∂L/∂x')-∂L/∂x=x''-x/sqrt(x^2+y^2)=0
d/dt(∂L/∂y')-∂L/∂y=y''-y/sqrt(x^2+y^2)=0
got both as i think there is 2 degrees of freedom
put in x=cost, y=sint
x'=-sint, x''=-cost
y'=cost y''=-sint
put these it above i get
-cost-cost/sqrt(sint^2+cost^2)=-2cost
-sint-sint/sqrt(sint^2+cost^2)=-2sint
for a this is where i think i get stuck.
b) x=rcosθ,x'=cosθr'-rsinθθ'
y=rsinθ,y'=sinθr'+rcosθθ'
x'^2+y'^2=r'^2+r^2θ'^2
sqrt(x^2+y^2)=r
there for
L(r,θ,r',')=(1/2)(r'^2+r^2θ'^2)-r
and so
d/dt(∂L/∂r')-∂L/∂r=r''-rθ'^2+1
d/dt(∂L/∂θ')-∂L/∂θ=r'^2θ''
and that's where i get stuck. This is a practise exam question, and i want to make sure I am going the right way about it. Any help would be much appropriated