Particle free to slide along a frictionless rotating curve

[Nathanael]

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.

 

[TSny]

In the E-L approach, did you take into account that $y'$ is a function of $x$?

 

[Nathanael]

In the E-L approach, did you take into account that $y'$ is a function of $x$?

Ohh, so it should have been: $\frac{ \partial L}{\partial x}=m(\omega^2x-gy'+\dot x^2y'y'')$

Then by the chain rule: $\dot x y''=\dot y'$

Thus: $\frac{ \partial L}{\partial x}=m(\omega^2x-gy'+\dot x\dot y'y')$

Which leaves the equation the same as in the Newtonian approach.

I'm used to the discrepancy being due to a mistake in the Newtonian approach... Now I know to be careful with the E-L approach too! Thanks TSny!

 

[TSny]

Looks good.

 

[TSny]

Even though they are equivalent, I like the form $\dot x^2y'y''$ rather than $\dot x\dot y'y'$ because $y'y''$ is just a function of $x$ that you can easily get from the curve $y(x)$.