Modeling the dynamics of a unit mass sliding down the curve 'y = f(x)' under gravity

For a unit mass sliding down the stationary curve $y = f(x)$ at the point $(x, y)$, we can consider our mass to be sliding down a stationary inclined plane which is tangential to the curve at the point $(x, y)$. The slope of this inclined plane is thus $\frac{dy}{dx}(x)$.

For the remainder of this thread, $y'$ means $\frac{dy}{dx}$

For the purposes of this excersize, I want the coefficient of static friction to be zero.
Let $\mu_k$ be the coefficient of kinetic friction
Let $g$ be the acceleration due to gravity
Let $\vec{F_g} = mg$ be the force exerted on our unit mass due to gravity.
Let $\vec{F_{||}}$ be the force exerted on our mass parallel to the plane.
Let $\vec{F_\bot}$ be the force exerted on our mass perpendicular to and into the plane.
Let $\vec{F_N} = - \vec{F_\bot}$ be the force exerted on our mass perpendicular to and out of the plane .
Let $\vec{F_f} = \mu_k \vec{F_\bot}$ be the force of friction which resists $\vec{F_{||}}$
Then $\vec{F_Ʃ} = \vec{F_{||}} + \vec{F_f} + \vec{F_\bot} + \vec{F_N}$ is the net force exerted on our mass. This is our objective.
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From the vector analysis, I have found $\vec{F_Ʃ}$ to be given by:
$\frac{1}{g} \vec{F_Ʃ} = [cos(θ) sin(θ) - \mu_k cos^2(θ)] \ \hat{i} + [sin^2(θ) - \mu_k cos(θ) sin(θ)] \ \hat{j}$

Where:
$θ = tan^{-1} (y')$

The model behaves as expected whenever $\mu_k$ = 0.
However, consider the trivial cases 1. and 2. below:

Case 1.
$\ \ \ g = -9.81$ // arbitrarily
$\ \ \ \mu_k = 0.5$ // arbitrarily
$\ \ \ y = x$ // the standard inclined plane

Case 2.
$\ \ \ g = -9.81$ // same as Case 1.
$\ \ \ \mu_k = 0.5$ // same as Case 1.
$\ \ \ y = -x$ // this is simply observing Case 1. from the other side of the plane

These parameters result in the following dynamics:

Case 1.
$\ \ \ \frac{d^2 x}{d t^2} = -2.4525$
$\ \ \ \frac{d^2 y}{d t^2} = -2.4525$

Case 2.
$\ \ \ \frac{d^2 x}{d t^2} = 7.3575$
$\ \ \ \frac{d^2 y}{d t^2} = -7.3575$

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For the previous cases, my intuition told me to expect the following:

a) That the sign on the horizontal acceleration would be negative in Case 1. and positive in Case 2.
b) That the sign on the vertical acceleration would be negative in Case 1. and negative in Case 2.
c) That the magnitude of the horizontal acceleration in Case 1. would be equal to the magnitude of the vertical acceleration in Case 1.
d) That the magnitude of the horizontal acceleration in Case 2. would be equal to the magnitude of the vertical acceleration in Case 2.
e) That the magnitude of horizontal and vertical acceleration in Case 1. would be equal to those of Case 2.

As we can see, the results are in agreement with a), b), c), d). However, hypothesis e) has failed.

Question:
What is the flaw in the model (or hypothesis e) ) that made the results diverge from e) when $\mu_k$ was non-zero (holding all other parameters constant)?

**If it is helpful, I can post my derivation of $\vec{F_Ʃ}$ in a followup.**

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 In case 2, your friction force has the same direction as the component of gravity parallel to the plane.

 Quote by willem2 In case 2, your friction force has the same direction as the component of gravity parallel to the plane.
Hi willem2. Thanks for pointing that out.

In trying to find out why that is the case, I found an angle error in my derrivation of the force of friction. I hope this is the only bug...

Modeling the dynamics of a unit mass sliding down the curve 'y = f(x)' under gravity

Well it seemed like a good idea, but it appears as though we can't have signed angles
ie $θ ≠ tan^{-1}(y')$ but rather $θ = |tan^{-1}(y')| = tan^{-1}(|y'|)$

 That's sounds like a classic inverse trig obstacle that we encounter all the time in computer science (I'm a programmer by trade). Have you considered using the multi-parameter version of the tangent function? i.e. arctan(x, y) rather than arctan(y/x) It's used in programming libraries to correct the deficiencies of the traditional arctan. arctan(y/x) essentially loses some of the information of the system. Human beings will often subconsciously reinsert lost information into the system when they use arctan(y/x) by hand, if my memory is correct. Also, there are other ways of finding the normal vector than using trig. For example, in 2 dimensions you can convert a vector to a complex number (x = real component, y = imaginary component) and then multiply by i in order to rotate the vector by 90 degrees to get the normal vector. Quaternions can achieve a similar function for 3 dimensions I believe, from what I've heard. Hope that helps. I only glanced over the other parts of your post but these few things caught my eye quick.
 Thank you for taking the time to offer your suggestions WraithGlade. Unfortunately the problem with the model that I have presented is that displacement instead of distance was used for the purpose of determining theta. It is because of this that I am getting signed angles which are problematic for the sine function being that it is odd. It looks inevitable that fixing this bug is going to introduce some form of absolute value function. I will have to play around with it now to solve for x(t). However, it looks like Newton-Raphson will still be king at the end of the day!

Recognitions:
Homework Help
 Quote by LuculentCabal Let $\vec{F_\bot}$ be the force exerted on our mass perpendicular to and into the plane.
You mean, that component of gravity, yes?
 Let $\vec{F_N} = - \vec{F_\bot}$ be the force exerted on our mass perpendicular to and out of the plane .
This one is exerted by the wire, yes?
They're only going to be equal and opposite if the acceleration is parallel to the wire.
Where the wire is curved, there will be acceleration normal to the wire.
I note the equation you end up with has no term involving y'', which cannot be right.

 Hello haruspex. Thank you for your review. Yes, $\vec{F_\bot}$is due to gravity. I would be grateful if you (or someone) could explain why our mass would experience acceleration normal to the curve on the points where $\frac{d^2 y}{dx^2} ≠ 0$. I had imagined that at the point (x,y), our mass would experience acceleration parallel to a tangent at the point (x,y). As our mass moves instantaneously along the curve, our tangent would be instantaneously changing too as it moves along with the mass. Why is this not the case?

 Quote by LuculentCabal Hello haruspex. Thank you for your review. Yes, $\vec{F_\bot}$is due to gravity. I would be grateful if you (or someone) could explain why our mass would experience acceleration normal to the curve on the points where $\frac{d^2 y}{dx^2} ≠ 0$. I had imagined that at the point (x,y), our mass would experience acceleration parallel to a tangent at the point (x,y). As our mass moves instantaneously along the curve, our tangent would be instantaneously changing too as it moves along with the mass. Why is this not the case?
Well, as for arbitrary two dimensional curvilinear motion, the acceleration components are:
$$a_{n}=\frac{v^2}{r}$$
$$a_{t}=\frac{dv}{dt}$$
The first refers to the acceleration components normal to the tangent and thus the second refers to the acceleration components parallel to the tangent.The r is radius of curvature, which, for an arbitrary two dimensional curve, has a specific formula for it.So here you may see that you have to know the velocity of the mass so as to get the $a_{n}$

Recognitions:
Homework Help
 Quote by LuculentCabal I would be grateful if you (or someone) could explain why our mass would experience acceleration normal to the curve on the points where $\frac{d^2 y}{dx^2} ≠ 0$.
Just to add a little to what azureth wrote:
Acceleration in the direction of travel will change the speed but not the direction. To travel along a curve requires a component of acceleration normal to the velocity, a centripetal force.

 Sorry guys, but I have no idea how to find v if all that I know is y(x), g, mu_k, and the initials v0, x0. It seems v cannot be found without first either knowing x(t) or F_x(t) and neither x(t) or F_x(t) can be found without knowing v...
 Recognitions: Homework Help Science Advisor Because there is friction involved, I think you have to start with the acceleration equations. Otherwise you could have used the energy equation to relate y and v.

 Quote by LuculentCabal For a unit mass sliding down the stationary curve $y = f(x)$ at the point $(x, y)$, we can consider our mass to be sliding down a stationary inclined plane which is tangential to the curve at the point $(x, y)$. The slope of this inclined plane is thus $\frac{dy}{dx}(x)$.
If by "sliding" you really mean that the mass can move only along the curve, it would be easier to consider the parametric representation of the curve, $x = x(s), y = y(s)$, where $s$ is the length along the curve. Then what you would need to find is $s = s(t)$. The magnitude of the velocity at any point is $\dot{s} = \frac {ds}{dt}$, so $m\ddot{s} = F_{\tau}$, where the entity on the right is the tangential component of the net force. $F_{\tau} = G_{\tau} - R$, where $G_{\tau}$ is the tangential component of the gravitational force, and $R$ is the magnitude of the force of friction. $R = \mu_kN$, where $N$ is the magnitude of the reaction force. The reaction force is always normal to the trajectory, so $ma_n = N$, where $a_n$ is the normal acceleration. Now, $a_n = v^2\kappa$, where $v$ is the magnitude of the velocity and $\kappa$ is the curvature, so $R = \mu_k m \dot{s}^2 \kappa = \mu_k m \dot{s}^2 \left| x'y'' - y'x'' \right|$

 Continuing, $G_{\tau} = - mg sin \theta = -mg y'$. Putting everything together: $$m\ddot{s} = -mg y' - \mu_k m \dot{s}^2\left|x'y'' - y'x''\right|$$ Let's see how that works for a unit circle. $x = cos(s), y = sin(s)$. Then:$$\ddot{s} = -g cos(s) - \mu_k m \dot{s}^2$$ In the vicinity of $s = \frac {3 \pi } {2}$, $cos (s) \approx s$, and assuming small velocities $\dot{s}^2 \approx 2\dot{s}$, thus $$\ddot{s} + 2\mu_k \dot{s} + gs = 0$$ which is the well-known equation of damped harmonic oscillations, as is to be expected physically.

 Tags algorithms, inclined plane, root finding