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Homework Statement
A particle (of mass m) is free to move along a frictionless curve y(x) which is rotating about the y-axis at a constant angular speed ω. A uniform gravitational field (of strength g) acts along the negative y direction. Find the equation of motion of the particle.
(That's not the entire problem, but it's enough to explain the discrepancy I'm coming across.)
Homework Equations
##y'\equiv \frac{dy}{dx}##
3. The attempt at a Newtonian Solution:
This picture should help explain the force equation:
The net force along the curve is due to gravity, namely ##mg\sin(\tan^{-1}(y'))=\frac{mgy'}{\sqrt{1+y'^2}}##
This has to account for the net acceleration in this direction, which comes from two things; the component of xω2 along the curve, and dv/dt the time derivative of the speed of the particle.
Since ##v=\frac{\dot x}{\cos(\tan^{-1}(y'))}=\dot x\sqrt{1+y'^2}## we have ##\dot v = \ddot x\sqrt{1+y'^2}+\frac{\dot x\dot y' y'}{\sqrt{1+y'^2}}##
Therefore the net-force equation in the direction along the curve is:
##\frac{mgy'}{\sqrt{1+y'^2}}=\frac{x\omega ^2}{\sqrt{1+y'^2}}-\big ( \ddot x\sqrt{1+y'^2}+\frac{\dot x\dot y' y'}{\sqrt{1+y'^2}} \big ) ##
Which can be written a little more nicely:
##x\omega^2-gy'=\ddot x(1+y'^2)+\dot x \dot y'y'##
4. The attempt at a Lagrangian Solution:
For the Lagrangian, I get ##L=\frac{m}{2}(x^2\omega^2+v^2)-mgy##
Where v is the speed as before: ##v=\frac{\dot x}{\cos(\tan^{-1}(y'))}=\dot x\sqrt{1+y'^2}##
So ##L=\frac{m}{2}(x^2\omega^2+\dot x^2(1+y'^2))-mgy##
Then the E-L equation gives:
##\frac{ \partial L}{\partial x}=m(\omega^2x-gy')=\frac{d}{dt}\big (\frac{ \partial L}{\partial \dot x} \big ) = \frac{d}{dt}\big ( m\dot x(1+y'^2)\big ) =m\big (\ddot x(1+y'^2)+2\dot x \dot y' y' \big )##
##x\omega^2-gy'=x(1+y'^2)+2\dot x \dot y' y'##
The right-most term differs by a factor of 2 from the Newtonian method. I can't seem to find the reason for this discrepancy; the force equation seems right to me.