Exponential Wire Dynamics - A Strange Anomaly

Consider the diagram shared by @DARK ANGEL in his solution. One can see that:

$\tan{\theta} = -\frac{dy}{dx}=\mathrm{e}^{-x}$

Now, the way I approach these problems is slightly different. I do not exploit the notion of an instantaneous radius of curvature. Instead, I apply Newton's laws along the cartesian directions as such. At any instant of time $t$ let the normal force be $N$ and the friction force be $f$ . Applying Newton's second law along the X and Y directions gives:

$m\ddot{x} = N \sin{\theta} - f \cos{\theta}$ $m\ddot{y} = N \cos{\theta} + f \sin{\theta}-mg$ $f = \mu \ N$

The constraint equation is:

$y = \mathrm{e}^{-x}$

Double differentiating the above gives:

$\ddot{y} = -\mathrm{e}^{-x} \ddot{x} + \mathrm{e}^{-x} \dot{x}^2$

Solving for $\ddot{x}$ , $\ddot{y}$ and $N$ using all the above equation leads to the answer:

$\ddot{x} = \frac{\left({\mathrm{e}}^{-x}\,\dot{x}^2+10\right)\,\left(10\,{\mathrm{e}}^{-x}-3\right)}{10\,\left({\mathrm{e}}^{-2\,x}+1\right)}$

Simplifications have been left out. Having found this equation of motion, the next step is to solve it. The first set of results is when $\mu=0$ , as follows:

One can see that the result without friction aligns perfectly with expectations. The final speed of the particle can be verified using the energy conservation principle. The first set of results is when $\mu=0.3$ , as follows:

Here, things get strange. The particle appears to be oscillating. @Steven Chase provided me with the insight that the motion must only be considered until the particle comes to rest for the first time when $t>0$ .

We start out by examining general expression for acceleration of a particle: $\frac{d\vec{v}}{dt} = \frac{d(v\hat{T})}{dt} = \dot{v}\hat{T} + v^{2}\frac{d\hat{T}}{ds} = \dot{v}\hat{T} + v^{2}\kappa \hat{N}.$ In our case, denoting $s$ to be the distance traveled along the wire, we have: $\frac{d\vec{v}}{dt} = \ddot{s}\hat{T} + \dot{s}^{2}\kappa \hat{N}.$ Now, by Newton's second law there must be forces which generate both tangential and normal acceleration and hence we obtain two equations: $\begin{aligned} m\ddot{s} &= mg\sin{\alpha} - f \\ m\dot{s}^{2}\kappa &= N - mg\cos{\alpha}\end{aligned}$ Here $\alpha$ is the angle which tangent subtends with a horizontal line and $\kappa$ is curvature of the wire. Since we have $f = \mu N$ , we are left with: $\begin{aligned} \ddot{s} &= g\sin{\alpha} - \mu\dot{s}^{2}\kappa - \mu g\cos{\alpha}\end{aligned}.$ We are now ready to switch to $x$ perspective: $\begin{aligned} \frac{d}{dt}\left(s'\dot{x} \right ) = s'\ddot{x} + \dot{x}^{2}s'' &= g\left | \frac{y'}{s'} \right | - \mu\dot{x}^{2}s'^{2}\frac{y''}{s'^{3}} - \mu g\left |\frac{1}{s'} \right | \\ s'^{2}\ddot{x} + \dot{x}^{2}s''s' &= g|y'| - \mu\dot{x}^{2}y'' - \mu g \\ s'^{2}\ddot{x} &=g|y'| - \frac{\dot{x}^{2}(s'^{2})'}{2} - \mu\dot{x}^{2}y'' - \mu g \\ (1+e^{-2x})\ddot{x} &= ge^{-x} + \dot{x}^{2}e^{-2x} - \mu\dot{x}^{2}e^{-x} - \mu g \\ \ddot{x} &= \frac{(\dot{x}^{2}e^{-x} + g)(ge^{-x} - \mu g)}{g(1+e^{-2x})}.\end{aligned}$ We read $a = 2,\,b = 10,\, c = 3$ making the result $15$ .

Although you could find a potential for friction force if you treat $x$ and $\dot{x}$ as independent variables, I don't think there is a way to add it to Lagrangian so that you retain the way of deriving of equations of motion. This is because the friction force depends on $\dot{x}^{2}$ and when you apply $\frac{d}{dt}\frac{\partial }{\partial \dot{x}}$ you will obtain a term involving $\ddot{x}$ which contradicts physics. Consequently, in the case of inclined plane, the reason why you can import friction term to the Lagrangian is because the object travels in a straight line so there's no normal acceleration to be provided by normal force.