Wave Trajectory on Cycloid


September 2015
Jaroslav Kopernicky
Jan Onderco

Abstract

The purpose of this paper is to analyze acceleration of circular, cycloid motions and acceleration of a harmonic oscillator, a body on springs that is attached to a rotating wheel, the wheel axle is perpendicular to the ground and the wheel is in a uniform circular motion. The axle is firmly attached to a train car. The train moves in a straight line at a constant velocity.
The body of the harmonic oscillator will follow a wave trajectory. The acceleration analysis is done from two inertial reference frame systems. The ground and the moving train. The position ${\to}$ velocity ${\to}$ acceleration calculations using normal and tangential coordinates clearly show discrepancy between the accelerations of these two inertial reference frame systems when Coriolis acceleration is accounted for properly in the ground reference system.

Introduction

The relativity is beautiful! Many times we, Jaro and myself, were joking that if Albert Einstein lived now and we had a chance to talk to him we would be good friends, we would share the same passion, we want to discover the truth. We would have good discussions about the physics, the reality around us. We have a huge respect for what Einstein achieved, the relativity is here to stay! The very simple reason is that as of now the relativity is the best approximation and solution to many challenges in physics. Having said that the relativity leaves us with many paradoxes and puzzles. The relativity is not a complete theory.
The velocity of stars in spiral galaxies and the satellite flyby anomalies are a couple of examples. The following lines of this paper point to a solution of the relativity puzzles. We believe Einstein is smiling on all of us from up there and saying: “Come on, all of you physicists, what took you so long :-)”

The cycloid motion in the ground inertial reference frame system $K’$ is a simple constant uniform circular motion in an inertial reference frame system $K$ that moves at a constant velocity, no change in magnitude or direction of the $K$ system velocity vector.
The Galilean transformation gives us:
$x’=x+vt$      $y’=y$      $t’=t$
Let us use subsripts $\cap$ and $\circ$ for better reading.
We can write
$K’=K_{\cap}$      $x’=x_{\cap}$      $y’=y_{\cap}$      $t’=t_{\cap}$
and
$K=K_{\circ}$      $x=x_{\circ}$      $y=y_{\circ}$      $t=t_{\circ}$

Nomenclature summary

$\cap$  –  cycloid subscript referencing cycloid equations and values in the ground inertial reference frame $K_{\cap}$
$\circ$  –  circle subscript referencing circle equations and values in the moving inertial reference frame $K_{\circ}$
$\theta$  –  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
$r$  –  the radius of the rotating wheel

The Cycloid

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

In the $K_{\cap}$ frame – Eq.1$_{\cap}$
$x_\cap=vt-r\sin (\omega t)$
$y_\cap=r-r\cos (\omega t)$
The textbook cycloid equations.
In the $K_{\circ}$ frame – Eq.1$_{\circ}$
$x_\circ=-r\sin (\omega t)$
$y_\circ=r-r\cos (\omega t)$
The textbook circle equations.

 

The first derivatives are the velocities;

In the $K_{\cap}$ frame – Eq.2$_{\cap}$
$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)$
In the $K_{\circ}$ frame – Eq.2$_{\circ}$
$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.

In the $K_{\cap}$ frame – Eq.3$_{\cap}$
$v_{mag\cap}=\sqrt{(v-r\omega\cos(\omega t))^2+(r\omega\sin(\omega t))^2}\\=\sqrt{v^2+r^2\omega^2(\sin ^2(\omega t) + \cos ^2(\omega t))-2vr\omega\cos(\omega t)}\\=\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}$
In the $K_{\circ}$ frame – Eq.3$_{\circ}$
$v_{mag\circ}=\sqrt{(-r\omega\cos(\omega t))^2+(r\omega\sin(\omega t))^2}=\sqrt{r^2\omega^2(\sin ^2(\omega t) + \cos ^2(\omega t))}\\=\sqrt{r^2\omega^2}=r\omega$

Cycloid vmag plot
Figure 1: The cycloid

Cycloid vmag plot
Figure 2: The cycloid and the tangential velocity magnitudes plotted on the same graph for convenience

Circle and Cycloid – Normal and Tangential Coordinates Analysis

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

In the $K_{\cap}$ frame Eq.4$_{\cap}$

$a_{t\cap}\space\space$$=\frac{dv_{mag\cap}}{dt}=\frac{d\sqrt{(v^2+r^2\omega^2-2vr\omega\cos(\omega t))}}{dt}\\=\frac{1}{2\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}\frac{d(v^2+r^2\omega^2-2vr\omega\cos(\omega t))}{dt}\\=\frac{-2vr\omega\frac{d\cos(\omega t)}{dt}}{2\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}=\frac{-vr\omega^2\space(-\sin(\omega t))}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}\\=\frac{vr\omega^2\sin(\omega t)}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$

In the $K_{\circ}$ frame Eq.4$_{\circ}$

$a_{t\circ}\space\space$$=\frac{dv_{mag\circ}}{dt}=\frac{dr\omega}{dt}=$$\space 0$

 

Here is a plot with the tangential acceleration magnitudes.
Cycloid amag plot
Figure 3: The cycloid, the tangential velocity magnitude and the tangential acceleration magnitude

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.8O
$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}$

In the $K_{\cap}$ frame – Eq.9$_{\cap}$
$a_{n\cap}=v^2_{mag\space\cap}K_\cap$
$=\frac{|r^2\omega^3-vr\omega^2\cos(\omega t)|}{\sqrt{v^2+r^2\omega^2-2vr\omega\cos(\omega t)}}$
In the $K_{\circ}$ frame – Eq.9$_{\circ}$
$a_{n\circ}=v^2_{mag\space\circ}
K_\circ\space\space$$=\frac{v^2_{mag\space\circ}}{r}=$$\space\space r\omega^2$

 
The acceleration magnitudes are;

In the $K_{\cap}$ frame – Eq.10$_{\cap}$
$a_{mag\cap}=\sqrt{a^2_{t\cap}+a^2_{n\cap}}$
In the $K_{\circ}$ frame – Eq.10$_{\circ}$
$a_{mag\circ}=\sqrt{a^2_{t\circ}+a^2_{n\circ}}=r\omega^2$

 

The cycloid acceleration in the ground reference frame is
$a_{mag\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 direction of the accelerations is the same as well, the vector sum of the tangential and normal accelerations in the ground reference $K_{\cap}$ system points in the same direction as the normal acceleration in the moving reference system $K_{\circ}$.

Here is an example of tangential and normal accelerations.

Cycloid vmag plot
Figure 4: The Cycloid; red – cycloid; black – $a_{t\cap}$; green – $a_{n\cap}$ when $v=1m/s$; $r=0.8m$; $\omega=1rad/s$

The Wave Trajectory

The textbook wave motion equation is
$x(t) = A cos(\omega t + \phi)$ Eq.11

The motion is periodic with the amplitude $A$. Let us replace $\omega$ with $\beta$ so it’s not confused with the circle and cycloid $\omega$.

$\displaystyle \beta = \sqrt{\frac{k}{m}}$ Eq.12

The velocity of the spring periodic motion is

$\displaystyle v_s = \frac{dx(t)}{dt} = – A\beta sin(\beta t + \phi)$ Eq.13

The acceleration of the spring periodic motion is

$\displaystyle a_s = \frac{d^2x(t)}{dt^2} = – A\beta^2 cos(\beta t + \phi)$ Eq.14

A wave trajectory on a circular and cycloid motion gives us the following equations.

In the $K_{\cap}$ frame – Eq.15$_{\cap}$
$x_\cap=vt-(r+A_r cos(\beta_r t + \phi_r))\sin (\omega t +\phi+ A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$
$y_\cap=r-(r+A_r cos(\beta_r t + \phi_r))\cos (\omega t +\phi+ A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$

In the $K_{\circ}$ frame – Eq.15$_{\circ}$
$x_\circ=-(r+A_r cos(\beta_r t + \phi_r))\sin (\omega t +\phi+ A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$
$y_\circ=r-(r+A_r cos(\beta_r t + \phi_r))\cos (\omega t +\phi+ A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$

Cycloid vmag plot
Figure 5: The wave on a circle when $r=1$, $\omega=1$, $A_r=0.2$, $\beta_r=4$, $\phi_r=0$, $A_{\omega}=0.2$, $\beta_{\omega}=4$, $\phi_{\omega}=0$

Cycloid vmag plot
Figure 6: The wave on a cycloid $r=1$, $\omega=1$, $A_r=0.2$, $\beta_r=4$, $\phi_r=0$, $A_{\omega}=0.2$, $\beta_{\omega}=4$, $\phi_{\omega}=0$

The first derivatives are the velocities;

In the $K_{\cap}$ frame – Eq.16$_{\cap}$
$v_{x\cap}\space=\frac{dx_\cap}{dt}=\space A_r\beta_r\sin(\beta_r t+\phi_r)\sin(\omega t +\phi +A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))\\ \hspace{60pt}-(r+A_r\cos(\beta_r t+\phi_r))(\omega-A_{\omega}\beta_{\omega} sin(\beta_{\omega} t+\phi_{\omega}))\cos(\omega t + \phi+A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega})) + v$
$v_{y\cap}\space=\frac{dy_\cap}{dt}=\space A_r\beta_r\sin(\beta_r t+\phi_r)\cos(\omega t + \phi + A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))\\ \hspace{60pt}+(r+A_r\cos(\beta_r t+\phi_r))(\omega-A_{\omega}\beta_{\omega} sin(\beta_{\omega} t+\phi_{\omega}))\sin(\omega t + \phi+A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))$

In the $K_{\circ}$ frame – Eq.16$_{\circ}$
$v_{x\circ}\space=\frac{dx_\circ}{dt}=\space A_r\beta_r\sin(\beta_r t+\phi_r)\sin(\omega t +\phi +A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))\\ \hspace{60pt}-(r+A_r\cos(\beta_r t+\phi_r))(\omega-A_{\omega}\beta_{\omega} sin(\beta_{\omega} t+\phi_{\omega}))\cos(\omega t + \phi+A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))$
$v_{y\circ}\space=\frac{dy_\circ}{dt}=\space A_r\beta_r\sin(\beta_r t+\phi_r)\cos(\omega t + \phi + A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))\\ \hspace{60pt}+(r+A_r\cos(\beta_r t+\phi_r))(\omega-A_{\omega}\beta_{\omega} sin(\beta_{\omega} t+\phi_{\omega}))\sin(\omega t + \phi+A_{\omega}\cos(\beta_{\omega} t+\phi_{\omega}))$

The velocity vector has its magnitude;

In the $K_{\cap}$ frame – Eq.17$_{\cap}$
$\displaystyle v_{mag\cap}\space=\sqrt{v_{x\cap}^2+v_{y\cap}^2}$
In the $K_{\circ}$ frame – Eq.17$_{\circ}$
$\displaystyle v_{mag\circ}\space=\sqrt{v_{x\circ}^2+v_{y\circ}^2}$

 

Cycloid vmag plot
Figure 7: The wave on a circle with the $v_{mag}$

Cycloid vmag plot
Figure 8: The wave on a cycloid with the $v_{mag}$

Wave on Circle and Cycloid – Normal and Tangential Coordinates Analysis

The tangential velocity magnitude derivatives are the tangential accelerations;

In the $K_{\cap}$ frame – Eq.18$_{\cap}$
$\displaystyle a_{t\cap}\space=\frac{dv_{mag\cap}}{dt}=\frac{d\sqrt{v_{x\cap}^2+v_{y\cap}^2}}{dt}
=\frac{1}{2\sqrt{{v_{x\cap}^2+v_{y\cap}^2}}}\frac{d[v_{x\cap}^2+v_{y\cap}^2]}{dt}\\$
$=\frac{2v_{x\cap}\frac{dv_{x\cap}}{dt}+2v_{y\cap}\frac{dv_{y\cap}}{dt}}{2\sqrt{{v_{x\cap}^2+v_{y\cap}^2}}}=\frac{v_{x\cap}a_{x\cap}+v_{y\cap}a_{y\cap}}{\sqrt{{v_{x\cap}^2+v_{y\cap}^2}}}$

In the $K_{\circ}$ frame – Eq.18$_{\circ}$
$\displaystyle a_{t\circ}\space=\frac{dv_{mag\circ}}{dt}=\frac{d\sqrt{v_{x\circ}^2+v_{y\circ}^2}}{dt}
=\frac{1}{2\sqrt{{v_{x\circ}^2+v_{y\circ}^2}}}\frac{d[v_{x\circ}^2+v_{y\circ}^2]}{dt}\\$
$=\frac{2v_{x\circ}\frac{dv_{x\circ}}{dt}+2v_{y\circ}\frac{dv_{y\circ}}{dt}}{2\sqrt{{v_{x\circ}^2+v_{y\circ}^2}}}=\frac{v_{x\circ}a_{x\circ}+v_{y\circ}a_{y\circ}}{\sqrt{{v_{x\circ}^2+v_{y\circ}^2}}}$

The normal accelerations using Eq.7 for curvature $\kappa$ are;

In the $K_{\cap}$ frame – Eq.19$_{\cap}$
$\displaystyle a_{n\cap}=v_{mag\cap}^2\kappa_{\cap}\,\,=v_{mag\cap}^2\frac{|a_{x\cap} v_{y\cap}-v_{x\cap} a_{y\cap}|}{v^3_{mag\cap}}$

In the $K_{\circ}$ frame – Eq.19$_{\circ}$
$\displaystyle a_{n\circ}=v_{mag\circ}^2\kappa_{\circ}\,\,=v_{mag\circ}^2\frac{|a_{x\circ} v_{y\circ}-v_{x\circ} a_{y\circ}|}{v^3_{mag\circ}}$

The $a_{x\cap}$, $a_{y\cap}$, $a_{x\circ}$ and $a_{y\circ}$ acceleration components are;

In the $K_{\cap}$ frame – Eq.20$_{\cap}$
$\displaystyle a_{x\cap}\space=\frac{dv_{x\cap}}{dt}=(r+A_r cos(\beta_r t+\phi_r))(\omega – A_{\omega}\beta_{\omega} sin(\beta_{\omega} t+\phi_{\omega}))^2 sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}+ A_r \beta_r^2 cos(\beta_r t + \phi_r) sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}+ 2(\omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}+(r+A_r cos(\beta_r t + \phi_r))A_{\omega} \beta_{\omega}^2 cos(\beta_{\omega} t + \phi_{\omega}) cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$

$\displaystyle a_{y\cap}\space=\frac{dv_{y\cap}}{dt}=(r+A_r cos(\beta_r t + \phi_r))(\omega – A_{\omega}\beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega}))^2 cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}+ A_r \beta_r^2 cos(\beta_r t + \phi_r) cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}- 2(\omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))\\ \hspace{60pt}-(r+A_r cos(\beta_r t + \phi_r)) A_{\omega} \beta_{\omega}^2 cos(\beta_{\omega} t + \phi_{\omega}) sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$

Equations Eq.20$_{\circ}$ for the $x$ and $y$ acceleration components in the $K_{\circ}$ frame are equal to Eq.20$_{\cap}$ equations in the $K_{\cap}$ frame.

In the $K_{\circ}$ frame – Eq.20$_{\circ}$
$\displaystyle a_{x\circ}\space=a_{x\cap} $
$\,$
$\displaystyle a_{y\circ}\space=a_{y\cap} $

 
The acceleration magnitudes are;

In the $K_{\cap}$ frame – Eq.21$_{\cap}$
$\displaystyle a_{mag\cap}\space=\sqrt{a_{t\cap}^2 +a_{n\cap}^2} $
In the $K_{\circ}$ frame – Eq.21$_{\circ}$
$\displaystyle a_{mag\circ}\space=\sqrt{a_{t\circ}^2 +a_{n\circ}^2} $

 
When equations 18, 19, 20, and 21 are entered into a plotting software, for example GNU Octave, then these are the acceleration plots.

Cycloid vmag plot
Figure 9: The wave trajectory, velocity and acceleration on a circle $r=1$, $\omega=1$, $A_r=0.2$, $\beta_r=4$, $\phi_r=0$, $A_{\omega}=0.2$, $\beta_{\omega}=4$, $\phi_{\omega}=0$

Cycloid vmag plot
Figure 10: The wave trajectory, velocity and acceleration on a cycloid $r=1$, $\omega=1$, $A_r=0.2$, $\beta_r=4$, $\phi_r=0$, $A_{\omega}=0.2$, $\beta_{\omega}=4$, $\phi_{\omega}=0$

The frames $K_{\cap}$ and $K_{\circ}$ appear to be equal for the acceleration analysis.
Having said that this might not be the proper analysis of the reality. It takes just one experiment to undermine the above analysis. That experiment could be the satellite flyby anomalies. The General Theory of Relativity in combination with Kepler’s laws and Newtonian mechanics come short in the explanation of the anomalies. It appears satellites are sometimes moving on cycloids around the Earth when the Earth orbits the Sun. We need to go deeper in an attempt to understand what is happening in the real world.

Straight Line Acceleration and Rotational Acceleration

Let us imagine a pendulum in a straight line accelerated frame of reference.
There is a heavy wheel oriented horizontally with the vertical axle in point $C$. The wheel is on a train or truck and it is undergoing a constant acceleration $a$ in a straight line. There is no rotation around the axle $C$ in this first scenario. This is a straight line accelerated frame of reference. The magnitude of acceleration of point $B$ is $a$, the same as the axle $C$ acceleration.

There is a pendulum with point $A$ of mass $m$ that can rotate around the axle $B$. Let us assume the pendulum arm is massless, in other words the mass of the arm is much smaller compared to the mass of point $A$. The arm is locked at the beginning and when released the point $A$ will be accelerated as shown in Figure 11.

The pendulum in the straight line accelerated frame of reference
Figure 11: The pendulum in the straight line accelerated frame of reference

The pendulum accelerated in a straight line would undergo a simple pendulum acceleration as it would be in gravity of magnitude $a$.

Now, the wheel is stationary but it rotates with constant angular velocity $\Omega$ and the magnitude of $\Omega$ is such that we can write $a=R\Omega^2$ in the second scenario. The point $B$ acceleration magnitude is $a$ again but in this case it is centripetal acceleration. The pendulum arm is locked at the beginning and when released the point $A$ will be accelerated as shown in Figure 12.
Cycloid vmag plot
Figure 12: The pendulum in the rotationally accelerated frame of reference

The pendulum in the rotational frame of reference is being accelerated by additional Coriolis’ acceleration that comes from $dR$ therefore we can conclude that even though point $B$ is being accelerated by the acceleration of the same magnitude $a$ in both scenarios the magnitude of point $A$ acceleration is different for the straight line acceleration and the rotational acceleration.

Pendulum on Cycloid

Cycloid vmag plot
Figure 13: The circle and cycloid trajectories (red), the cycloid velocity (black), the cycloid radius $R$ (blue), the cycloid $\Omega$ (cyan).

The pendulum on the cycloid is accelerated in a straight line at the inflection points of the cycloid when $\Omega=0$ or in other words the radius of the cycloid is $R{\to}\infty$ and the additional Coriolis acceleration, adding $dR$ has no effect on the total acceleration. The inflection points are shown in Figure 13 when $\Omega$ – cyan line is equal $0$. When the pendulum undergoes rotational acceleration at the top of the cycloid then adding Coriolis acceleration $dR$ influences the total acceleration.

The third line of the Eq.20$_{\cap}$ for $a_{x\cap}$
$- 2(\omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$
and for $a_{y\cap}$
$- 2(\omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$
represent the Coriolis acceleration.

It appears that in order to better reflect the reality small $\omega$ should be replaced with big $\Omega$.
For $a_{x\cap}$
$- 2(\Omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) cos(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$
and for $a_{y\cap}$
$- 2(\Omega – A_{\omega} \beta_{\omega} sin(\beta_{\omega} t + \phi_{\omega})) A_r \beta_r sin(\beta_r t + \phi_r) sin(\omega t + \phi + A_{\omega} cos(\beta_{\omega} t + \phi_{\omega}))$

The following plot in Figure 14 shows how new equations with big $\Omega$ predict a resonant magnitude change of frequency $1/\pi\, Hz$. This is a reasonable prediction considering no Coriolis effect at the inflection points that are approximately at $\pi/2$ and $3\pi/2$.
Circle and Cycloid amag plot
Figure 14: The circle acceleration (green), the cycloid acceleration (magenta) when $v=1m/s$, $r=0.3m$, $\omega=1rad/s$, $A_r=0.0001m$, $\beta_r=100$, $\phi_r=0$, $A_{\omega}=0.0001m$, $\beta_{\omega}=100$, $\phi_{\omega}=0$

Discussion

The most important question that needs to be addressed in the discussion – is Coriolis acceleration different at every point on the cycloid compared to the circle?
If the answer is yes then this breaks the relativity.

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.