A bead sliding with friction down a flexible wire

Question

Problem Statement:

I would like to model the following configuration:

Here, a bead of mass $m$ slides down a wire under tension. The wire has density $\rho$. The bead experiences friction in proportion to its speed $v$ along the wire. I would like to describe the velocity of the bead through time.

Large tension limit:

If the tension in the wire is very large, the dynamics of the bead will decouple from those of the wire, and the motion of the bead would be just like a block on a plane with velocity-dependent friction (by assumption, not Coulomb friction). In coordinates along the wire, $$ m \dot{v} = - \gamma v + mg \sin\theta,$$ so the velocity approaches $mg/\gamma\sin\theta$: $$ v(t) = v_0 e^{-\gamma t/m} + \frac{m g \sin\theta}{\gamma}(1-e^{-\gamma t/m}).$$

Moderate tension:

When the tension is not too much larger than $mg$, the wire will be deformed by the mass, so it no longer slides down a straight line. Here's where the problem gets challenging and I have trouble setting up the governing equations. I believe the equation of the wire is something like $$ \rho \frac{\partial^2 \psi}{\partial t^2} = T \frac{\partial^2 \psi}{\partial x^2} +\rho g + m g \delta(x-x_p(t))$$

However the particle coordinates are now very complicated to describe. I believe the most convenient way might be to describe them in the coordinates defined by the shape of the wire, in imitation of the large tension limit, but it is not clear to me exactly how to do this.

If anyone could offer any guidance as to whether my equation for the string is correct and how to set up the equations of motion for the particle I'd be very grateful!

Edit:

In the absence of friction the Lagrangian for the particle is $$ \mathcal{L} = \frac{m}{2}\Big(\dot{x}^2\big[1+\psi_x^2\big]+2\dot{x}\psi_x \psi_t + \psi_t^2 \Big) - m g \psi(x,t).$$ This gives the following equation of motion for the particle constrained to the string (athough there may be some small errors): $$ \ddot{x}[1+\psi_x^2] + 2 \dot{x}\big(\psi_x[\psi_x \dot{x} + \psi_t] + \psi_t[\psi_{xx}\dot{x}+\psi_{xt}] + \psi_x[\psi_{xt}\dot{x}+\psi_{tt}]\big) = \psi_x \dot{x}^2 + \dot{x}[\psi_{xx}\psi_t + \psi_x \psi_{xt}] + \psi_t \psi_{xx} - g\psi_x.$$ This should be solved in conjunction with the driven wave equation above to describe the dynamics. Maybe one can neglect some terms for small displacements to derive an approximate solution?

Answer 1

I propose a scheme - quasi-static approximation, by assuming the string always maintains in a static solution as the bead is moving. Let me start with describing the string without a bead. $$ \rho \frac{\partial^2 y(x, t)}{\partial t^2} = T \frac{\partial^2 y(x,t)}{\partial x^2} -\rho g. $$ For a static solution, $\frac{\partial y(x)}{\partial t} = 0$, the solution is a quadratic function for given two boundary conditions $y(x_1) = y_1$ , and $y(x_2) = y_2$: \begin{align*} y(x) &= \frac{g}{2v^2}x^2 + A x + B.\tag{1}\\ A &= \frac{y_2 - y_1}{x_2 - x_1} - \frac{g}{2v^2} (x_2 + x_1);\tag{2} \\ B &= \frac{y_1 x_2 - y_2 x_1}{x_2 - x_1} + \frac{g}{2v^2} x_1 x_2.\tag{3} \end{align*} where $v=\sqrt{\frac{T}{\rho}}$ is the string wave speed. This provides us the basic function for the next step. An example is given in following figure with parameter $\rho=1$, $T=20$. A rather loose string for clear observation of the quadratic function. (note that this is different from the catenary problem, free hanging rope, due to the assumption of a constant tension.)

With this basic knowledge, we add the bead at the fixed position, $x_p$. The static equation: $$ 0 = T \frac{\partial^2 y(x,t)}{\partial x^2} -\rho g - mg \delta(x-x_p). $$ We divide the solution into two region: $$ y(x) = \Big\{ { \begin{array}{rr} \frac{g}{2v^2} x^2 + A_1 x + B_1 & \text{ for } x_1 < x < x_p \\ \frac{g}{2v^2} x^2 + A_2 x + B_2 & \text{ for } x_p < x < x_2 \end{array} } \tag{4} $$

Similar as Eq.2 and Eq.3, find $A$s and $B$s parameters in the interval $[x_1, x_p]$, and $[x_p, x_2]$ assume $y(x_p) = y_p$.

\begin{align*} A_1 &= \frac{y_p - y_1}{x_p - x_1} - \frac{g}{2v^2} (x_p + x_1); \\ B_1 &= \frac{y_1 x_p - y_p x_1}{x_p - x_1} + \frac{g}{2v^2} x_1 x_p.\\ A_2 &= \frac{y_2 - y_p}{x_2 - x_p} - \frac{g}{2v^2} (x_p + x_2); \\ B_2 &= \frac{y_p x_2 - y_2 x_p}{x_2 - x_p} + \frac{g}{2v^2} x_2 x_p. \end{align*}

Then, the connection formula at $x= x_p$ \begin{align*} \left[\frac{\partial y}{\partial x}\right]_{x_p^+} -\left[\frac{\partial y}{\partial x}\right]_{x_p^-} & = \frac{mg}{T}\\ A_2 - A_1 & = \frac{mg}{T} \tag{5}\\ \end{align*}

From Eq.5, we find the function $y_p(x_p)$: $$ y_p(x_p) = y_2\frac{x_p-x_1}{x_2-x_1} +y_1\frac{x_2-x_p}{x_2-x_1} -\frac{mg}{T} \frac{(x_2-x_p)(x_p-x_1)}{x_2-x_1} - \frac{g}{2v^2} (x_2-x_p)(x_p-x_1)\tag{6} $$

The following figure shows the Eq.4, and Eq.6 for $\rho=1$, $T=200$, $m=10$ with $x_p = 0.7$ For large tension ($200$), the qudratic functions in both regions are very closed to linear lines.

In this figure, the force directions are marked: the tension $T_1$ , and $T_2$ in the tangential direction of Eq.4 with discontinue directive between them, the gravitational $mg$ in the $-\hat y$ direction, and dragging force $-bv$ in the tangential of $y_p(x_p)$. All these forces may be decomposed in $x$ and $y$ components for computing the force along the tangential of $y_p(x_p)$, which is the moving direction of the bead $m$. $$ \begin{array}{llc} \text{Moving direction} & \hat t = \frac{1}{\sqrt{1+\left(\frac{dy_p}{dx_p}\right)^2}}\left(\hat x + \hat y\frac{dy_p}{dx_p}\right)& \text{ Eq.6} \\ \text{Tension up } &\vec T_1= \frac{T}{\sqrt{1+\left(\frac{dy}{dx}\right)_{x_p^-}^2}}\left(\hat x + \hat y\frac{dy}{dx}\right)_{x_p^-}& \text{ Eq.4 (a)} \\ \text{Tension down } &\vec T_2=\frac{T}{\sqrt{1+\left(\frac{dy}{dx}\right)_{x_p^+}^2}} \left(\hat x + \hat y\frac{dy}{dx}\right)_{x_p^+} &\text{ Eq.4 (b)}\\ \text{Bead mass } & \vec F_g = mg \hat y & \\ \text{Dragging } & \vec f_d =-\gamma v \hat t & \\ \end{array} $$ where \begin{align*} \frac{dy_p}{dx_p} &= \frac{y_2-y_1}{x_2-x_1}+\frac{mg}{T}\frac{2x_p-x_2-x_1}{x_2-x_1} +\frac{g}{2v^2}\left(2x_p - x_2- x_1\right);\\ \left(\frac{dy}{dx}\right)_{x_p^-} &= \frac{g}{v^2} x_p + A_1;\\ \left(\frac{dy}{dx}\right)_{x_p^+} &= \frac{g}{v^2} x_p + A_2; \end{align*}

Finally, the equation of motion for the bead along the $y_p(x_p)$ curve: $$ m\frac{dv}{dt} = -\gamma v + \vec F_g \cdot \hat t - \vec T_1 \cdot \hat t + \vec T_2 \cdot \hat t. $$ The bead is constrained to move along the cureve $y_p(x_p)$, therrfore, there are constrain forces in the vertical direction of $\hat t$, which are not shown in the figure.