Parabolic Motion: Analyzing Forces & Investigating Results

[danielakkerma]

Greetings!
I've recently embarked upon an attempt to compute the various aspects of (rather) simple Newtonian motion on the surface of a parabola, encompassing alternating conditions(such as friction, intial conditions, etc).
I start with a trivial parabolic representation of:
y(t)=ax(t)^2
Therefore, as illustrated in the first diagram appended(though roughly), the forces at play are detailed there.
Thence, I had acquired the following equations of motion:
<br /> Ncos(\alpha)+fsin(\alpha)-mg=m\ddot{y}<br />

<br /> Nsin(\alpha)-fcos(\alpha)=-m\ddot{x}<br />

\alpha = tg^{-1}(2ax)<br />
<br /> f = \mu N <br />

<br /> \frac{dy}{dx} = 2ax = tg(\alpha)<br />
(where mu is taken as the friction constant).
Of course, one must first differentiate y, which yields:
<br /> \frac{d^2y}{dt^2} = 2a(\dot{x}^2+x\ddot{x})<br />
Finally, assembling all these together grants us:
<br /> (2a(\dot{x}^2+x\ddot{x})+g)\frac{(sin(\alpha)-\mu cos(\alpha))}{(\mu sin(\alpha)+cos(\alpha))}=-\ddot{x}<br />
Analytically, this differential equation is of course insoluble; But, numerically, it can very well be evaluated. So, intuitively, I fed it to an excellent numerical processor("Mathematica"), expecting to see some physically sound results. But "Lo' and behold", to my dismay and horror, when the product of all these efforts had been but venial.
You'll find two charts demonstrating my perplexity; Both depict a manipulated and interchanged \mu.
But as you can see, for different(or any!, excepting 0) values of that constant(mu) the amplitude of the oscillations(which should be present) does not change! one would've thought, that under the constraints of energy loss, there would be no gains to accommodate such a behavior, and I am left, as a result really puzzled.
In any case, I leave it to you, with gratitude for your time and attention,
What have I done wrong?
Thankful,
Beholden,
Daniel Akkerman

 

[dgOnPhys]

Hi Daniel,

I can spot a couple of issues:
- your equations do not account for the dependence of the sign of friction on velocity (your equation is valid only if the body is coming down the parabolic track not going up)
- I am not following you derivation formula where you end up with
<br /> \frac{d^2y}{dt^2} = 2a(\dot{x}^2+x\ddot{x})<br />
It looks a bit off

 

[danielakkerma]

Hey!
First, let me thank you for a very prompt reply!
Now, let's start from the bottom of your post, with respect to your query:
<br /> y(t)=ax(t)^2 \\<br />
But y however is dependant on x, which is in itself squared, leading to:
<br /> \frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}=2ax\dot{x}<br />
Here, I have a product of two functions(x and x-dot), therefore the second derivative has to envelop that, ultimately arriving at:
<br /> \frac{d^2y}{dt^2} = 2a(\dot{x}^2+x\ddot{x})=\frac{d}{dt}(\frac{dy}{dx} \frac{dx}{dt})<br />
And regrading your first suggestion, have you recommended my adding a "sign" element?, as in:
<br /> f = sgn(v(t))\mu N<br />
I must say that I had tried this appendage, but I am afraid, to no avail! The graphs remain EXACTLY(though, a slight shift is visible, and is post below;But I am, still, most alarmed by it) the same!
What else is there?
Very grateful,
Daniel Akkerman
P.S
Note, in the adjoined file, the inexplicable augmentation of its amplitude; What do you think is behind that?

 

[arildno]

I think I'll derive this for myself, and we may compare the results afterwards:

1. Parametric representation of parabolic path:
\vec{X}(t)=(x(t),ax(t)^{2})
Unit tangent vector, in the direction of velocity:
\vec{t}=\frac{\dot{x}}{|\dot{x}|}\frac{(1,2ax)}{\sqrt{1+4a^{2}x^{2}}}
Upwards unit normal vector:
\vec{n}=\frac{(-2ax,1)}{\sqrt{1+4a^{2}x^{2}}}

2. Components of gravity:
-mg\vec{j}=-(\frac{mg}{\sqrt{1+4a^{2}x^{2}}}\vec{n}+\frac{\dot{x}}{|\dot{x}}\frac{2axmg}{\sqrt{1+4a^{2}x^{2}}}\vec{t}

3. Scalar normal force:
N\vec{n}=\frac{mg}{\sqrt{1+4a^{2}x^{2}}}\vec{n}

4. Frictional force:
\vec{f}=-\mu{N}\vec{t}=-mu\frac{mg}{\sqrt{1+4a^{2}x^{2}}}\vec{t}

5. x-component of Newton's second law of motion:
m\frac{d^{2}x}{dt^{2}}=-\frac{2axmg}{1+4a^{2}x^{2}}-\frac{\dot{x}}{|\dot{x}|}\frac{\mu{mg}}{1+4a^{2}x^{2}}
Note that it is the component of the normal force that provides oscillatory behaviour, whereas the component of the frictional force is strictly dampening.

5. represents the differential equation you need to solve; y(t) is found by the relation:
y(t)=ax(t)^{2}

 

[danielakkerma]

Hi!
Thank you for a very detailed response.
First, an argument on the merits of my attempts: Was it, as you reckon, a dreadful error on my part to divide the forces on the x, y axes, rather, then, as you have done, select tangential and normal directions?
I had selected my method simply because it enables an immidiate derivation(i.e taking straightforwardly y-double-dot, and x-double-dot, and then applying the angles as necessary).
Do you think this was the cause of all my troubles?
Secondly, I am afraid your formulae lead to a breakdown in calculation. You see, my boundary conditions assume that \dot{x}(0) = 0 which leads to a nonexistent derivative at that moment.(there's a resultant division by zero in equation 5).
What should I therefore do?
Thanks again!
Daniel Akkerman

 

[arildno]

1. As for choice of coordinate systems, it shouldn't matter as long as you do it properly.
2. As for the "silly" behaviour of my sign function (represented in 5. as a fraction), it is easily remedied (I implicitly thought in this manner):
sgn(a)=1 if a>0, sgn(0)=0, sgn(a)=-1 if a<0

This is the proper physical sign function for the frictional force, since it vanishes for zero velocity.

 

[danielakkerma]

But, perhaps, skimming through my labors, have you found them by any chance physically plausible? I mean, I had abided by every possible precaution there should be with respect to Newton's 2nd law.
The problem is not with the Sign, apparently. I am afraid it vested directly here:
m\frac{d^{2}x}{dt^{2}}=-\frac{2axmg}{1+4a^{2}x^{2}}-\frac{\dot{x}}{|\dot{x}|}\frac{\mu{mg}}{1+4a^{2}x^ {2}}
In the portion of (\frac{\dot{x}}{|\dot{x}|}) which leads to zero at t=0.
Chaging the initial conditions(so that, say v(0)=1), does provide an accurate view, clearly demonstrating that your approach is correct.(Minus a few tumults towards the end of the plotting range, but that's merely due to the precision of the numerical algorithm).
But I was wondering, if there was anyway to circumvent the need to nudge the object abit before it starts off(i.e setting v(0)=0), and I am very much disheartened my trials proved futile. I would very much appreciate, if you could just shed light on where I had erred, if merely so that I could avoid these solecisms in the future.
In any case, your help has been most thorough and diligent,
For that, I am ever bound by gratitude,
Thanks again!
Daniel

 

[arildno]

No, it won't.
Replace that fraction with the sign function I gave you.

 

[arildno]

Your problem is probably in your choice of the arctan function.
This is commonly defined between -90 degrees and 90 degrees, as measured with respect to the positive x-axis, but in YOUR problem, the proper angle lie between 0 and 180.

Thus, weirdness is guaranteed to occur when you approach 90 degrees from either side.

 

[danielakkerma]

Well, first of all, naturally, employing the "sgn" function, solved all remaining quandries with respect to v being 0 at any rate.
I have thought about the various vascilliations of the Arctan as a possible source of "agony".
Do you think there is any way to resolve that, perhaps adding a constant Pi(to bring it about to 180 degress)?
In any case, I am very much again behold for your aid, it has been most beneficial.
Daniel Akkerman

 

[arildno]

I'll take a closer look later on.

 

[arildno]

Let us start out, Daniel, with your two equations of motion, as given by post 1&3:
m\ddot{x}=N(-\sin\alpha+sgn(v)\mu\cos\alpha)
m\ddot{y})=N\cos\alpha+sgn(v)\mu{N}\sin\alpha-mg
I'll get back to this...

 

[danielakkerma]

Great!
I am very keen on identifying the tidings of this problem.
Again, I would very much like to express my benediction for your steadfast tackling of my "whimsy" here :).
Thankful as always,
Daniel

 

[arildno]

Okay, Daniel!
I've looked a bit on your system, and it just gets too messy to work with an angle "alpha".
It doesn't help you at all, you make the very simple rule "friction works in the opposite direction of velocity" into a complicated monster.

You'll just end up getting bitten by that monster (i.e, getting the sign changes wrong), as your previous simulations have proven.

 

[danielakkerma]

Well,
I can definitely accept your point; But am I also to infer that there was nothing wrong with my direct handling of the design?
In any case, I concede with great humility that your method is by far superior to mine. I guess that addressing any question head-on is not always most preferable, although, I do have a greater proclivity for Mathematics, and that seemed to me at least, to be the most reasonable resort.
At any rate, I cannot thank you enough for all your grand advice and help,
It has been prodigious and infallible,
So again, I reiterate,
Simply grateful!
Daniel Akkerman

 

[arildno]

Well, it is nothing intrinsically wrong with seeking to describe this in terms of the tangent angle, but one of your problems there is to keep control of WHICH tangent you ought to use (at each point, you have two tangents, anti-parallell to each other).

Furthermore, it is the bother about the proper domain of the arctan function.

You do not need to delve into these intricacies if you stick to a frictional force of the form:
\vec{f}=-\mu{N}\frac{\vec{v}}{||\vec{v}||}, N\geq{0},\vec{v}\neq\vec{0},\vec{f}=\vec{0},\vec{v}=\vec{0}