Finding Lagrangian of a system (Oscillations in a parabolic trough)

Question

A particle of mass m slides down on the curve $z=1+\frac{x^2}{2}$ in the $xz$-plane without friction under the action of constant gravity.Suppose $z$-axis points vertically upward.Find the Lagrangian of given system.

The position vector of the particle is given by $\overline r=(x,0,1+\frac{x^2}{2})$. Then the velocity vector is given by $ \dot {\overline r}=(\dot x,0,x\dot x)$. So the kinetic energy is given by $T=\frac {m}{2}((\dot x)^2+x^2(\dot x)^2)$. Now I am confused in potential energy. The potential energy is given by $V=mgz$ but what will be the sign of it? What does it mean that $z$-axis points vertically upward?

Answer 1

The movement is restricted to the manifold $f(x,z) = 1+\frac 12 x^2-z=0$. The potential energy is taken having a referential plane as for instance, the plane $XY$ so the potential energy is given by $m g z$. The Lagrangian is given by

$$ L = T - V +\lambda f = \frac 12 m(\dot x^2+\dot z^2) - m g z + \lambda\left(1+\frac 12x^2-z\right) $$

giving the movement equations

$$ \cases{ \ddot x = -\frac{x(g+\dot x^2)}{1+x^2}\\ \ddot z = \frac{\dot x^2-g x^2}{1+x^2}\\ \lambda = -\frac{m(g+\dot x^2)}{1+x^2} }\ \ \ \ \ \ \ \ (*) $$

Here $\lambda$ represents the reaction force along $f(x,z)=0$

NOTE

According to Euler-Lagrange formalism, the movement equations are

$$ \cases{ m\ddot x -\lambda x=0\\ m\ddot z+\lambda+mg=0\\ 1+\frac 12 x^2=z} $$

now deriving twice the $f(x,z)=0$ manifold we obtain the third equation substitution as

$$ \ddot z= x \ddot x +\dot x^2 $$

now solving

$$ \cases{ m\ddot x -\lambda x=0\\ m\ddot z+\lambda+mg=0\\ \ddot z= x \ddot x +\dot x^2} $$

for $\ddot x,\ddot z,\lambda$ we obtain $(*)$

Answer 2

The Lagrangian is set up with Mechanics energy formulation and DE set up using Euler Lagrange Equation.

Angles $ \phi$ are tangent or normal rotations positive counterclockwise.

Parabola trough coordinates are taken wlog $ R=1, z_{min}=0 $in parametric form:

$$ (x,z)=( R \tan \phi, R/2 \tan^2 \phi)\tag1$$ $$ ( \dot x, \dot z)= (R \sec^2 \phi\, \dot \phi,R \sec^2 \phi \tan \phi\;\dot\phi )\tag2$$

And the absolute value of velocity is

$$R \sec^3 \phi \;\dot\phi \tag3$$

$$ KE= \frac{m}{2}(R \sec^3 \phi \;\dot \phi)^2 \tag4 $$

PE is taken positive as shown and as usual $z>0$. When PE increases KE decreases and vice-versa.

$$ PE=mgz=mg \frac{R}{2}\dot\tan^2 \phi \tag5 $$

$$\text{Lagrangian } = L =KE -PE $$

$$ L= \dfrac{m}{2} \cdot (R^2 \sec^6 \phi\; \dot \phi ^2)- mg \dfrac{R}{2} \tan^2 \phi \tag6$$

Euler -Lagrange Equation ( after removing constants)

$$ 6 \sec^5 \sec\phi \tan \phi\;{\dot{\phi}}^2-\frac{g}{R } 2 \tan {\phi} \sec^2\phi -\frac{d(2 \dot \phi)}{dt} \sec^6 \phi =0\tag7$$

Simplifying,

$$ \dfrac{\ddot{\phi}}{\sin \phi \cos^2 \phi}+ \frac{g}{R}\cos \phi -3 \dot\phi^2=0 \tag8 $$

I have no closed form solution of the DE or any reference. So a numerical solution with the following boundary conditions is undertaken.

$$( \phi_{max}= \frac{\pi}{3}, \dot\phi_{max}= 0, g=9.8, R=6) \tag9 $$

The parabolic trough profile and time oscillation of slope is plotted below:

The formula for shallow $\phi_{max}$ trough as a simple pendulum of constant $L$ for verification :

$$ T= 2 \pi\sqrt{\dfrac{L}{g}} \tag {10} $$ is tallying alright.