Lagrangian of an equation, a particle with a time dependent constraint

Question

I'm currently studying for an exam and I came across the following question. Consider a bead of mass m constrained to move along a wire which is on the vertical $(x, y)$ plane. The wire has equation $y =x^2/a+ b cos(\omega t)$ where $a$, $b$ and $\omega$ are positive constants. The equation of the wire is interpreted physically as a parabola that vibrates vertically with a prescribed time dependence. The particle is under the effect of gravity with potential $V = mgy$. Find the Lagrangian of the system. I'm struggling to find generalized coordinates to represent it. I was thinking maybe cylindrical coordinates would fit the problem but I'm not sure how to apply it to the question and was wondering if somebody could point me in the right direction? Thank you!

Answer 1

Consider the lagrangian

$$ L(x,y,\lambda) = \lambda \left(y-\frac{x^2}{a}-b \cos (\omega t)\right)-g m y+\frac{1}{2} m \left(\dot x^2+\dot y^2\right) $$

and then after $L_X-\frac{d}{dt}L_{\dot X} = 0$ the movement equations

$$ m\ddot x + \frac{2\lambda x}{a} = 0\\ m\ddot y - m g -\lambda = 0\\ y-\frac{x^2}{a}-b \cos (\omega t) = 0 $$

now deriving twice the last equation we have the DE system

$$ m\ddot x + \frac{2\lambda x}{a} = 0\\ m\ddot y - m g -\lambda = 0\\ \ddot y = \frac{2}{a}(\dot x^2+\dot x\ddot x)-b\omega^2\cos(\omega t) $$

Finally solving for $\ddot x,\ddot y,\lambda$ we get

$$ \ddot x= \frac{2 \left(a b \omega ^2 x \cos (\omega t)-a g x-2 x \dot x^2\right)}{a^2+4 x \dot x}\\ \ddot y= -\frac{a^2 b\omega ^2 \cos (\omega t)-2 a \dot x^2+4 g x \dot x}{a^2+4 x \dot x}\\ \lambda = \frac{a m \left(a g+2 \dot x^2-a b \omega ^2 \cos (\omega t)\right)}{a^2+4 x \dot x} $$

which are the movement equations. Here $\lambda$ represents the reaction force between the bead and the curve.