August 2015

Jaroslav Kopernicky

Jan Onderco

## Abstract

The purpose of this paper is to analyze acceleration of a cycloid motion with a focus on the inflection point of a curtate cycloid. The interesting observation is that the well know equation F=ma becomes a=F/m at the inflection point.

The acceleration is a response that depends on force and mass and it is not a factor. The relativity ignores the mass present at the inflection point and therefore it cannot predict how much energy is required to change the potential energy of the present mass only by following \mbox{position $\rightarrow$ velocity $\rightarrow$ acceleration} analysis.

## Introduction

The cycloid motion in the ground inertial reference frame is a simple constant circular motion in an inertial reference frame that moves at a constant velocity. The moving inertial reference frame velocity vector does not change magnitude or direction. Let us assume a body on an elastic string in horizontally oriented circular motion, its axle is perpendicular to the ground.

## Nomenclature summary

$\cap$ – cycloid subscript referencing cycloid equations and values in the ground inertial reference frame

$\circ$ – circle subscript referencing circle equations and values in the moving inertial reference frame

$\theta$ – the angle of the rotation of the circle in radians

$\omega$$\space=\frac{d\theta}{dt}$ ; the angular velocity of the circle

$d\theta=dt$ ; $t$ corresponds to the angle $\theta$ therefore $\omega=1$ in our plots

$v$ – the velocity of the moving frame; $v=1$ in our plots

$r$ – the radius of the wheel; $r=1$ in our plots

$c$ – the speed of light

## The Cycloid

A circle of radius r, consisting of the points (x, y);

_{∩}

$x_\cap=vt-r\sin (\omega t)$

$y_\cap=r-r\cos (\omega t)$

The textbook cycloid equations.

_{O}

$x_\circ=-r\sin (\omega t)$

$y_\circ=r-r\cos (\omega t)$

The textbook circle equations.

The first derivatives are the velocities;

_{∩}

$v_{x\cap}$$\space=\frac{dx_\cap}{dt}=$$\space v-r\omega\cos(\omega t)$

$v_{y\cap}$$\space=\frac{dy_\cap}{dt}=$$\space r\omega\sin(\omega t)$

_{O}

$v_{x\circ}$$\space=\frac{dx_\circ}{dt}=$$\space -r\omega\cos(\omega t)$

$v_{y\circ}$$\space=\frac{dy_\circ}{dt}=$$\space r\omega\sin(\omega t)$

The $v_x$ and $v_y$ are components of the tangential velocity vector and the calculation of the magnitude is a straightforward vector sum.

_{∩}

$v_{mag\cap}=\sqrt{(v-r\omega\cos(\omega t))^2+(r\omega\sin(\omega t))^2}\\=\sqrt{v^2+r^2\omega^2(\sin ^2(\omega t) + \cos ^2(\omega t))-2vr\omega\cos(\omega t)}\\=\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}$

_{O}

$v_{mag\circ}=\sqrt{(-r\omega\cos(\omega t))^2+(r\omega\sin(\omega t))^2}=\sqrt{r^2\omega^2(\sin ^2(\omega t) + \cos ^2(\omega t))}\\=\sqrt{r^2\omega^2}=r\omega$

Figure 1: The cycloid and the tangential velocity magnitudes plotted on the same graph for convenience

## Normal and Tangential Coordinates Analysis

The equations Eq.3$_{\cap}$ shows us that the cycloid tangential velocity magnitude in the ground frame changes, it is a function of $t$. The circle velocity magnitude Eq.3$_O$ in the moving frame is constant. The tangential velocity magnitude derivatives are the tangential accelerations;

In the ground reference frame Eq.4_{∩}

In the moving reference frame Eq.4_{O}

Here is a plot with the tangential acceleration magnitudes.

Figure 2: The cycloid, the tangential velocity magnitude and the tangential acceleration magnitude

Let us rename $a_{mag\cap}$ to $a_{t\cap}$ and $a_{mag\circ}$ to $a_{t\circ}$.

_{∩}

$a_{t\cap}\space$$=\frac{vr\omega^2\sin(\omega t)}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$

_{O}

$a_{t\circ}=0$

The normal/radial acceleration is responsible for the velocity vector rotation. The normal acceleration keeps the velocity vector always tangential to the curved trajectory, it is a function of the velocity and the radius

$\displaystyle a_n\space\space=\frac{v^2_{mag}}{R}$ Eq.5

Radius is defined by a curvature of the trajectory.

$\displaystyle R\space\space=\frac{1}{K}$ Eq.6

The curvature$^{[1]}$ is

$\displaystyle K\space\space=\frac{|x”y’-x’y”|}{((x’)^2+(y’)^2)^{3/2}}=\frac{|a_x v_y-v_x a_y|}{(v^2_{mag})^{3/2}}=\frac{|a_x v_y-v_x a_y|}{v^3_{mag}}$ Eq.7

The cycloid curvature is Eq.8_{∩}

$K_\cap\space\space$$=\frac{|r\omega^2\sin(\omega t)\space r\omega\sin(\omega t)-(v-r\omega\cos(\omega t))\space r\omega^2\cos(\omega t)|}{((v-r\omega\cos(\omega t))^2+(r\omega\sin(\omega t))^2)^{3/2}}\\=\frac{|r^2\omega^3(\sin^2(\omega t)+\cos^2(\omega t))-vr\omega^2\cos(\omega t)|}{(v^2-2vr\omega\cos(\omega t)+r^2\omega^2\cos^2(\omega t)+r^2\omega^2\sin^2(\omega t))^{3/2}}\\=\frac{|r^2\omega^3-vr\omega^2\cos(\omega t)|}{(v^2+r^2\omega^2-2vr\omega\cos(\omega t))^{3/2}}$

The circle curvature is Eq.8_{O}

$K_\circ\space\space$$=\frac{|r\omega^2\sin(\omega t)\space r\omega\sin(\omega t)-(-r\omega\cos(\omega t))\space r\omega^2\cos(\omega t)|}{((-r\omega\sin(\omega t))^2+(r\omega\sin(\omega t))^2)^{3/2}}\\=\frac{|r^2\omega^3(\sin^2(\omega t)+\cos^2(\omega t))|}{(r^2\omega^2\cos^2(\omega t)+r^2\omega^2\sin^2(\omega t))^{3/2}}=\frac{|r^2\omega^3|}{(r^2\omega^2)^{3/2}}\\=\frac{|\omega|}{\sqrt{(r^2\omega^2)}}=\frac{1}{r}$

_{∩}

$a_{r\cap}=v^2_{mag\space\cap}K_\cap$

$=\frac{|r^2\omega^3-vr\omega^2\cos(\omega t)|}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$

_{O}

$a_{r\circ}=v^2_{mag\space\circ}

K_\circ\space\space$$=\frac{v^2_{mag\space\circ}}{r}=$$\space\space r\omega^2$

The total net accelerations are;

_{∩}

$a_\cap=\sqrt{a^2_{t\cap}+a^2_{r\cap}}$

_{O}

$a_\circ=\sqrt{a^2_{t\circ}+a^2_{r\circ}}=r\omega^2$

The cycloid acceleration in the ground reference frame is

$a_\cap\space\space$$=\sqrt{(\frac{vr\omega^2\sin(\omega t)}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}})^2+(\frac{|r^2\omega^3-vr\omega^2\cos(\omega t)|}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}})^2}\\=\sqrt{\frac{(vr\omega^2\sin(\omega t))^2+(r^2\omega^3-vr\omega^2\cos(\omega t))^2}{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}\\=\sqrt{\frac{v^2r^2\omega^4\sin^2(\omega t)+r^4\omega^6-2r^2\omega^3vr\omega^2\cos(\omega t)+v^2r^2\omega^4\cos^2(\omega t)}{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}\\=\sqrt{\frac{r^2\omega^4(v^2(\sin^2(\omega t)+\cos^2(\omega t))+r^2\omega^2-2vr\omega\cos(\omega t))}{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$

$\space\space\space\space=\sqrt{\frac{r^2\omega^4(v^2+r^2\omega^2-2vr\omega\cos(\omega t))}{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$$\space=\sqrt{r^2\omega^4}=r\omega^2$Eq.10_{∩}

We come to a conclusion that both accelerations, cycloid and circular have the same magnitude. The next step is to investigate the direction of the accelerations, especially at the inflection point of a curtate cycloid.

## The Inflection Point

The cycloid at the inflection point $T_{1}$ and after a short interval at point $T_{2}$

Figure 3: The cycloid at the inflection point $T_{1}$ and $T_{2}$

Zoomed in cycloid at the inflection point $T_{1}$ and at after a short interval at point $T_{2}$.

Figure 4: Zoomed in cycloid at the inflection point $T_{1}$ and $T_{2}$

These are boundaries of our system. Let us use arbitrary units of measure.

There is a body of mass $1$ at the point $1$ at the time $T_{1}$ and external force $1\, 000\, 000$ pulling the body in the direction to the center of the circle; to the axis. The force line is parallel to the equipotential line at the point $1$ at the time $T_{1}$.

Let us assume potential energy quanta change of $1$ between two equipotential lines at the point $1$ at time $T_{1}$ and the point $2$ at the time $T_{2}$. There is no other potential energy in between. There is $0$ potential energy at the point $1$ and potential energy $1$ at the point $2$. The force available to pull in the normal direction is $1$ at point $2$ at the time $T_{2}$ and the force available to pull in the tangential direction is $999\, 999.9999995$ at the point $2$ at the time $T_{2}$. The external force does not change, it is $1\, 000\, 000$.

The external force normal component $1$ represents the **exact amount** of energy required to move mass $1$ body from $0$ to $1$ potential energy. The force normal component generates the normal acceleration $1$ for the mass $1$ body.

The normal acceleration **is a function of the angle** between the field force line and the equipotential line.

Zoomed in cycloid at the point $T_{2}$.

Figure 5: The cycloid at the point $T_{2}$

Let us assume a body of mass $100$ at the point $1$ at the time $T_{1}$. The second body has $100$ mass $1$ bodies in them. It is a rigid body and the internal forces are orders of magnitude greater than the external force. The force available to pull in the normal direction is still $1$ at the time $T_{2}$ because it depends on the angle between the field force line and the equipotential line and it would be able to move **only one** of the bodies to the potential $1$ but **NOT** the other $99$ of them. There is not enough force in the normal direction to move them all.

What is going to happen? Assuming elasticity in the external force (body on an elastic string) the second heavier body is going to stay at the potential $0$ at the time $T_{2}$. The mass/inertia/energy of the $99$ other bodies is not going to allow quanta jump to the potential energy $1$. The second body is going to end up at the point $3$ at the time $T_{2}$ with $0$ potential energy. **The velocity and acceleration vectors are not going to change direction.** There is not enough energy to do so. The heavier body is not going to move in the normal direction therefore there is no normal acceleration at the time $T_{2}$.

Once the angle between the equipotential line and the force field line is big enough when there is enough force in the normal direction; that is the point when the second body can jump to the potential energy 1. This will happen at the time $T_{3}$ and $T_{3}$ > $T_{2}$. The force that was supposed to be pulling in the normal direction is still being applied in the tangential direction during the $T_{2}-T_{3}$ interval. The total net acceleration does not point to the center of the wheel during this interval.

It appears that the normal acceleration is not only function of the angle between the equipotential line and the force field line but **the normal acceleration is also a function of the mass/inertia/energy present at the point $1$**. This mass changes the curvature of the space at the inflection point.

## The Conclusion

Itâ€™s logical to acknowledge that in our case at the inflection point there is no normal acceleration until there is big enough force, greater angle between the equipotential and force field lines.

Heavier mass $m$ requires greater force $F$ (greater angle) to generate the same acceleration.

The force and mass are factors; the acceleration is a response: $\textbf{a=F/m}$.

The acceleration is a response to energy/mass/inertia.

The relativity cannot be right because it cannot predict the same angle between the equipotential lines and the field lines in different inertial reference frames for all scenarios of motion. **The relativity ignores the mass present at the inflection point and therefore it cannot predict how much energy is required to change the potential energy of the present mass only by following position $\rightarrow$ velocity $\rightarrow$ acceleration analysis.**

## References

[1] 18.02A Topic 22: Vector derivatives: velocity, curvature; from http://web.mit.edu/dvp/18.01A/topic22.pdf

Thank you, to all the people behind the GNU Octave software that was used for the plotting.