# Sine curve trajectory

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1. Jul 20, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Every small part of the trajectory can be taken as a circular arc.
Then,
$$\frac { m v^2} R ≤ kmg$$

But how to find out R from the sine-curve?

2. Jul 20, 2017

### Hiero

In general (for not simply circular motion) the radius of curvature (the R in your centripetal acceleration formula) is given by (one over) the magnitude of acceleration of a particle moving along the curve at the point with (always) unit speed.

For example a particle moving unit speed along a circle has acceleration 1/R so the radius of curvature is (surprise!) R.

Thus, to find the radius of curvature at any point on the sinusoid, we could parameterize x and y in time such that (dx/dt)^2+(dy/dt)^2=1 at all times, then the radius of curvature will be (one over) the acceleration in that parameterization.

But... we only care about the minimum radius of curvature, (i.e. maximum acceleration/curvature,) we don't care about it's value everywhere. So maybe we can save some trouble by realizing; where the curvature is maximum?

3. Jul 25, 2017

### Pushoam

Please give me a reference for this information .
$\{{\frac {dx}{dt}}\}^2 +\{ {\frac {dy} {dt}}\}^2 = 1$
$\{{\frac {dx}{dt}}\}^2 + \{\left ({ {\frac {dy} {dx}} {\frac {dx} {dt}} }\right )^2$ $=\left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }} {\frac {dx} {dt}}\right ) ^2\} = 1$

$\{{\frac {dx}{dt}}\}^2 =$ $\frac 1{1+ \left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }}\right)^2}$

$\{{\frac {dy}{dt}}\}^2 =$ $\frac { \{ {\frac a {\alpha} } \cos {\frac x { \alpha} } \}^2 } {1+ \left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }}\right)^2}$

Differentiating both sides wrt t,
$2 v_x \ddot x = \frac{ -1} {\{ {1+\left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }}\right)^2}\}^2 } {\frac a {{\alpha}^2}} \sin {\frac{2 x} \alpha}$

$2 v_y \ddot y = \{1 - \{ {\frac a {\alpha} } \cos {\frac x { \alpha} } \}^2\} \frac{ 1} {\{ {1+\left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }}\right)^2}\}^2 } {\frac a {{\alpha}^2}} \sin {\frac{2 x} \alpha}$

$2 v_y \ddot y = \{1 - \{ {\frac a {\alpha} } \cos {\frac x { \alpha} } \}^2\} 2 v_x \ddot x$

$\frac {dx} {dt}= \sqrt {\frac 1{1+ \left ( {\frac a {\alpha}} \cos {\frac x{ \alpha }}\right)^2}}$

Is this correct so far?

Last edited: Jul 25, 2017
4. Jul 25, 2017

### haruspex

5. Jul 25, 2017

### Hiero

Sorry I think that wasn't the simplest way to put it, but @haruspex's link is worth looking at; it is what I had in mind in my first reply.

The question is asking, if we let a car travel in this certain path at a constant speed, what is the maximum constant speed at which it won't slip? So then at some point(s) on the curve we must want static friction to be acting at it's maximum (or else the car could've gone a bit faster). If we can confidently determine which point(s) along the curve it should be maximum, then we can save ourselves trouble by considering the motion at a single point rather than considering the entire curve.

6. Jul 26, 2017

### Pushoam

$\frac{ m v^2} R ≤ kmg$
$v≤ \sqrt {k g R}$
So, the maximum allowed speed is $v_{mx} = \sqrt { k g R_{mn} }$
where Rmn is the minimum radius of curvature
If the curve is given in Cartesian coordinates as y(x), then the radius of curvature is (assuming the curve is differentiable up to order 2):
$R = | \frac { \left ( 1 + y' ^2\right )^ \frac 3 2 } { y" }|$

Rmn, turns out to be radius of curvature at that point where Sine function has maximum value. So, the curve is a straight line. And radius of curvature of straight line is infinity, right? So, where am I wrong?

I copied these latex codes from mathematica. Now how to bring it in the standard form?
\documentclass{article}
\usepackage{amsmath, amssymb, graphics, setspace}

\newcommand{\mathsym}[1]{{}}
\newcommand{\unicode}[1]{{}}

\newcounter{mathematicapage}
\begin{document}

\begin{doublespace}
\noindent$$\pmb{\text{}}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{\text{}}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{\text{}}\\ \pmb{}\\ \pmb{y= a \text{Sin}[x/\alpha ]}$$
\end{doublespace}

\begin{doublespace}
\noindent$$a \text{Sin}\left[\frac{x}{\alpha }\right]$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{y'= D[y,x]}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\frac{a \text{Cos}\left[\frac{x}{\alpha }\right]}{\alpha }$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{y\text{''} = D[y',x ]}$$
\end{doublespace}

\begin{doublespace}
\noindent$$-\frac{a \text{Sin}\left[\frac{x}{\alpha }\right]}{\alpha ^2}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{R = \{(1+ y' {}^{\wedge}2){}^{\wedge}(3/2 ) \}/y\text{''}}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\left\{-\frac{\alpha ^2 \left(1+\frac{a^2 \text{Cos}\left[\frac{x}{\alpha }\right]^2}{\alpha ^2}\right)^{3/2} \text{Csc}\left[\frac{x}{\alpha }\right]}{a}\right\}$$
\end{doublespace}

\begin{doublespace}
\noindent$$\pmb{\text{DSolve}[R'[x] =0, x]}$$
\end{doublespace}

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Last edited: Jul 26, 2017
7. Jul 26, 2017

### TSny

Yes.
OK

Yes. It's not hard to prove it without a computer. If you inspect the expression for R as a function of x for this specific y(x), you can see that the denominator attains its maximum value at the same point that the numerator attains its minimum value.

I'm not sure what you are saying here. What curve are you referring to?

Last edited: Jul 26, 2017
8. Jul 26, 2017

### Pushoam

I am referring to that curve for which I have calculated Rmn.
I have got Rmn at that point for which Sine function has its local maximum. At the point of local maximum the slope is zero. Slope being zero implies that the curve is a straight line.

Last edited: Jul 26, 2017
9. Jul 26, 2017

### TSny

Why wouldn't a slope of 1 also imply that the curve is a straight line?
Knowing the value of the slope of a curve at some point does not tell you anything about the curvature at that point. Rotating the whole curve rigidly would change the slope values but would not change the curvature values.

10. Jul 26, 2017

### Pushoam

What I want to say that even the upper part of a circle has maximum at some point say(x0 , y0 ). And if I use the above formula for calculating the Radius of curvature , then around (x0 , y0 ), the curve is a straight line with slope zero and a straight line has an infinte radius of curvature. I am having some confusion here.

11. Jul 26, 2017

### TSny

Approximating a small region of a curve by a straight line is going too far when you are interested in the curvature. Curvature is related to approximating a small region of a curve by a "best fitting" circular arc. The radius of curvature is then the radius of this arc. The slope involves only the first derivative of the function. But curvature requires knowledge of the second derivative (that is, how the slope is changing as you move along the curve).

Note that at a point where $y'(x) = 0$, $R = \frac 1 {|y''(x)|}$.

12. Jul 26, 2017

### Nidum

Do you actually need to find the radius of curvature ?

Can this problem be solved more elegantly by considering the lateral acceleration of the car directly ?

13. Jul 26, 2017

### haruspex

Yes, the slope is zero, but that is just y'. Your straight line there is just the tangent. Putting y'=0 in your formula does not give infinity. You need to calculate y" and plug that in too. (If that were zero then it would be a straight line.)

14. Jul 27, 2017

### Pushoam

What is a lateral acceleration and how to relate it with the problem?

What about the following in post #6?

Last edited: Jul 27, 2017
15. Jul 27, 2017

### haruspex

Do you mean $\ddot y$? The constant speed condition makes that a bit tricky.