Sticky Friction. An alternative for the damping friction model F = -by′ discussed in Section 4.1 is the “sticky friction” model. For a mass sliding on a surface as depicted in Figure 5.18, the contact friction is more complicated than simply -by′. We observe, for example, that even if the mass is displaced slightly off the equilibrium location y = 0, it may nonetheless remain stationary due to the fact that the spring force -ky is insufficient to break the static friction’s grip. If the maximum force that the friction can exert is denoted by m, then a feasible model is given by

\({{\bf{F}}_{{\bf{friction}}}}{\bf{ = }}\left\{ \begin{array}{l}{\bf{ky,if}}\left| {{\bf{ky}}} \right|{\bf{ < }}\mu {\bf{andy' = 0}}\\\mu {\bf{sign(y),if}}\left| {{\bf{ky}}} \right| \ge {\bf{0andy' = 0}}\\ - \mu {\bf{sign(y'),ify'}} \ne 0.\end{array} \right.\)

(The function sign (s) is +1 when s 7 0, -1 when s 6 0, and 0 when s = 0.) The motion is governed by the equation (16) \({\bf{m}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - ky + }}{{\bf{F}}_{{\bf{friction}}}}\)Thus, if the mass is at rest, friction balances the spring force if \(\left| {\bf{y}} \right|{\bf{ < }}\frac{\mu }{{\bf{k}}}\)but simply opposes it with intensity\(\mu \)if\(\left| {\bf{y}} \right| \ge \frac{\mu }{{\bf{k}}}\). If the mass is moving, friction opposes the velocity with the same intensity\(\mu \).

1. Taking m =\(\mu \) = k = 1, convert (16) into the firstorder system y′ = v (17)\({\bf{v' = }}\left\{ \begin{array}{l}{\bf{0,if}}\left| {\bf{y}} \right|{\bf{ < 1andv = 0}}{\bf{.}}\\{\bf{ - y + sign(y),if}}\left| {\bf{y}} \right| \ge {\bf{1andv = 0}}\\{\bf{ - y - sign(v),ifv}} \ne 0\end{array} \right.\) ,
2. Form the phase plane equation for (17) when v ≠ 0 and solve it to derive the solutions\({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\).where the plus sign prevails for v>0 and the minus sign for v<0.
3. Identify the trajectories in the phase plane as two families of concentric semicircles. What is the centre of the semicircles in the upper half-plane? The lower half-plane?
4. What are the critical points for (17)?
5. Sketch the trajectory in the phase plane of the mass released from rest at y = 7.5. At what value for y does the mass come to rest?

Let y′ = v.Therefore\({\bf{v' = y''}}\). The equation in the form of a system is:

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = v}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = - y + }}{{\bf{F}}_{{\bf{friction}}}}{\bf{ = }}\left\{ \begin{array}{l}0{\bf{,if}}\left| {\bf{y}} \right|{\bf{ < }}1{\bf{andv = 0}}\\{\bf{ - y + sign(y),if}}\left| {\bf{y}} \right| \ge 1{\bf{andv = 0}}\\{\bf{ - y - sign(v),ifv}} \ne 0.\end{array} \right.\end{array}\)

When \({\bf{v}} \ne 0\)and\(\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = - y - sign(v)}}\).

\(\begin{array}{c}\frac{{{\bf{dv}}}}{{{\bf{dy}}}}{\bf{ = }}\frac{{{\bf{ - y - sign(v)}}}}{{\bf{v}}}\\{\bf{ = }}\left\{ \begin{array}{l}\frac{{{\bf{ - y - 1}}}}{{\bf{v}}}{\bf{,v > 0}}\\\frac{{{\bf{ - y + 1}}}}{{\bf{v}}}{\bf{,v < 0}}\end{array} \right.\\{\bf{vdv = }}\left\{ \begin{array}{l}{\bf{ - y - 1,v > 0}}\\{\bf{ - y + 1,v < 0}}\end{array} \right.\\\smallint {\bf{vdv = }}\left\{ \begin{array}{l}\int {{\bf{ - y - 1dy}}} {\bf{,v > 0}}\\\int {{\bf{ - y + 1}}} {\bf{dy,v < 0}}\end{array} \right.\end{array}\)

\(\begin{array}{c}\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = }}\left\{ \begin{array}{l}{\bf{ - }}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - y + C,v > 0}}\\{\bf{ - }}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + y + C,v < 0}}\end{array} \right.\\{\bf{C = }}\left\{ \begin{array}{l}\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - }}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - y,v > 0}}\\\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - }}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + y,v < 0}}\end{array} \right.\\\left\{ \begin{array}{c}{{\bf{v}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + 2y,v > 0}}\\{{\bf{v}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + 2y,v > 0}}\end{array} \right.{\bf{ = 2C}}\\\left\{ \begin{array}{c}{{\bf{v}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + 2y + 1,v > 0}}\\{{\bf{v}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + 2y + 1,v > 0}}\end{array} \right.{\bf{ = 2C + 1}}\\\left\{ \begin{array}{c}{{\bf{v}}^{\bf{2}}}{\bf{ + (y + 1}}{{\bf{)}}^{\bf{2}}}{\bf{,v > 0}}\\{{\bf{v}}^{\bf{2}}}{\bf{ + (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{,v > 0}}\end{array} \right.{\bf{ = 2C + 1}}\end{array}\)

So, the solution is \({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\).where the plus sign prevails for v>0 and the minus sign for v<0.

In the \({\bf{y - v}}\) plane, when\({\bf{v > 0}}\), I have the semicircle. From the equation,\({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\) it is clear that the centre of these circles is (-1, 0). When \({\bf{v < 0}}\) I have the semicircles, from the equation \({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\)clear that the centre of these circles is (1, 0).

\({\bf{v' = }}\left\{ \begin{array}{l}{\bf{0,if}}\left| {\bf{y}} \right|{\bf{ < 1andv = 0}}{\bf{.}}\\{\bf{ - y + sign(y),if}}\left| {\bf{y}} \right| \ge {\bf{1andv = 0}}\\{\bf{ - y - sign(v),ifv}} \ne 0\end{array} \right.\)

V coordinate is always 0. In case\(\left| {\bf{y}} \right|{\bf{ < 1}}\)then \({\bf{0 = 0}}\) so, all points (y, 0).

In case \(\left| {\bf{y}} \right| \ge {\bf{1}}\)then

\(\begin{array}{c}{\bf{ - y + sign(y) = 0}}\\{\bf{y = sign(y)}}\end{array}\)

This is only true when\({\bf{y = 1 or y = - 1}}\). The critical points are (-1, 0), (1, 0).

The mass will come to rest if it hits the segment \({\bf{v = 0}}\),\({\bf{ - 1}} \le {\bf{y}} \le 1\). If I zoom on the graph it happens for\({\bf{y = - 0}}{\bf{.5}}\).

Therefore, the mass comes to rest for\({\bf{y = - 0}}{\bf{.5}}\)