# Metaphysics (Alternate View)/Creation/Simple Pendulum
## System Circle

* (An alternative view by Calgea 19:10, 19 May 2007 (UTC))
* A system circle has parameters equal to those of the physical system it represents. In this case, the physical system is a simple pendulum.
* The radius, R, of the circle equals the length, L, of the pendulum’s arm, i.e., R = L.
* The period, T, of the circle equals the period, T, of the pendulum, i.e., T = T.

### For the circle:

* $\omega =2\pi /T$
** where $\omega$ is the angular velocity in radians per second.
* The period is constant, and the angular velocity is constant.
* The radial acceleration, Ar, is a constant.
** $Ar=R\omega ^{2}$

### For the pendulum:

* The local gravity, g, is a constant.
Ergo, $g=Ar=R\omega ^{2}=L(2\pi /T)^{2}$
* and $T=2\pi {\sqrt {}L/g)$
lknn; gns;K KNF KJF FDNFK;e vd'ond iE FOOIEH ELKKB IUFBSD ;AJBF IUB'

### Testing Suggestion

Construct two identical pendulums. Do not use elastic material for the arms. Use a displacement angle of five degrees for one pendulum and say forty degrees for the other pendulum. Mechanically release them at the same time. The arms will move through the vertical at the same time.
* Note: This is not an accepted view. However, this alternative view says the period is independent of the displacement angle, i.e., not restricted to small angles. Therefore, it is testable.

## Energy Content

(An alternative view by Calgea 17:50, 25 May 2007 (UTC))
* Start with the pendulum’s arm in the vertical position. Move the arm to the right through a small angle $(\theta _{i})$. The energy (E) equation for the system is as follows.
* E = mad
** Where m = the mass of the bob
*** d = the distance the bob moved along the arc
**** $d=L\theta _{i}$
**** ($s=r\theta$)
**** s = arc length, r = L
**** L = length of pendulum’s arm
*** a = acceleration
**** $g(sin(n(\theta _{i})))=$ tangential acceleration
**** g = the local gravity
**** n = number of small angles
The first average acceleration over the first arc is given by the folowing equation.
* $a_{1}=(1/2)g(sin(\theta _{1})-sin(0))$
The set of energy equations for the curve are as follows.
* $E_{1}=m(1/2)g(sin(\theta _{i})-sin(0))L(\theta _{i})$
* $E_{2}=m(1/2)g(sin(2(\theta _{i}))-sin(\theta _{i}))L(\theta _{i})$
* $E_{n}=m(1/2)g(sin(n(\theta _{i}))-sin((n-1)(\theta _{i})))L(\theta _{i})$
When summing these incremental energies, the inner sine terms cancel.
* $E_{s}=m(1/2)g(sin(\theta _{0})-sin(0))L(\theta _{0})$)
* $E_{s}=m(1/2)g(sin(\theta _{0}))L(\theta _{0})$
** where $E_{s}=$ System Energy
*** $(\theta _{0})$ = Initial displacement angle in radians
Note: This alternative view of the equation for the System Energy $(E_{s})$ is not the accepted view. The Potential Energy, PE, is the accepted view.
* $PE=mgL(1-cos(\theta _{0}))$
* $E_{s}=mgL(k_{1})$
** Where $(k_{1})=(1/2)(\theta _{0})sin(\theta _{0})$
One equation measures the energy needed to move the mass in a vertical direction. The other measures the energy needed to move the mass along a curve. Which is correct?

## Energy Circle

* An alternative view by Calgea 20:53, 27 May 2007 (UTC))
Each energy form resides in a field. The two fields of interest for a simple pendulum are linear motion (Lm) and angular motion or Spin (Sp).
An energy circle represents these two fields. The vertical diameter is the Linear motion dimension and the horizontal diameter is the Spin dimension.
The radius (R) equals the length (L) of the pendulum. This radius turns at a constant angular velocity ($\omega$) as in the System Circle.

### System Energy Components

$E_{\left(Sy\right)}$ components: $E_{\left(Sp\right)}$ and $E_{\left(Lm\right)}$
** $E_{\left(Sp\right)}=E_{\left(Sy\right)}cos(\theta )$ ____Spin
** $E_{\left(Lm\right)}=E_{\left(Sy\right)}sin(\theta )$ ___Linear motion
These components are perpendicular quantities and add by the sum of squares.
* $(E_{\left(Sy\right)})^{2}=(E_{\left(Lm\right)})^{2}+(E_{\left(Sp\right)})^{2}$
* $(E_{\left(Sy\right)})^{2}=(E_{\left(Sy\right)}sin(\theta ))^{2}+(E_{\left(Sy\right)}cos(\theta ))^{2}$
* $(E_{\left(Sy\right)})^{2}=(E_{\left(Sy\right)})^{2}(sin^{2}(\theta )+cos^{2}(\theta ))$
** $1=sin^{2}(\theta )+cos^{2}(\theta )$
Take the square root of both sides.
* $E_{\left(Sy\right)}=E_{\left(Sy\right)}(sin^{2}(\theta )+cos^{2}(\theta ))$
* $E_{\left(Sy\right)}=mg(1/2)(sin(\theta _{0}))R(\theta _{0})(sin^{2}(\theta )+cos^{2}(\theta ))$
The identity says something is changing, and it could only relate to the acceleration component, radial acceleration, RA.
* $RA_{\left(Sy\right)}=g(sin^{2}(\theta )+cos^{2}(\theta ))$
The constant System Circle radial acceleration $(RA_{\left(Sy\right)})$ equals the constant value of gravity (g). This equation says the System Circle’s radial acceleration only appears constant. In fact, it comprises two accelerations.
** One acceleration is the rate of conversion of Spin energy into linear motion energy.
** The other acceleration is the rate of conversion of linear motion energy into Spin.
At the bottom of the swing, all the energy is in the Linear motion dimension and is known as Kinetic Energy (KE).
* $E_{\left(Sy\right)}=(1/2)(mg)sin(\theta _{0})R(\theta _{0})$ ____ System Energy
* $(1/2)(m)(V_{T})^{2}=(1/2)(mg)sin(\theta _{0})R(\theta _{0})$ __________Kinetic Energy
* $(V_{T})^{2}=(g)sin(\theta _{0})R(\theta _{0})$ ____________at the bottom of the swing
$V_{T}=L\omega _{p}$
* where $V_{T}$ is the tangential velocity
** $\omega _{p}$ is the angular velocity of the pendulum
** L is the Length of the arm and equal to R.
* $(V_{T})^{2}=g[sin(\theta _{0})R(\theta _{0})]$
* $(L\omega _{p})^{2}=g[sin(\theta _{0})(\theta _{0})]L$
* $L(\omega _{p})^{2}=g[sin(\theta _{0})(\theta _{0})]$
* $(\omega _{p})^{2}=(g/L)[sin(\theta _{0})(\theta _{0})]$
* $(\omega _{p})^{2}=(\omega )^{2}k_{o}$
** where $(\omega )^{2}=(g/L)$
*** $\omega =$ constant system circle angular velocity and
*** $k_{o}=sin(\theta _{0})(\theta _{0})$
Assign the sine squared term to the radial acceleration of the pendulum.
* $R(\omega _{p})^{2}=koR(g/L)sin^{2}(\theta )$
* $(\omega _{p})^{2}=k_{o}(g/L)sin^{2}(\theta )$
* $(\omega _{p})^{2}=k_{o}(\omega )^{2}sin^{2}(\theta )$
Angular velocity of Pendulum
* $\omega _{p}={\sqrt {}k_{o})(\omega )sin(\theta )$
This equation is one of a set that includes the angle $(\theta _{p})$ and angular acceleration $(\alpha _{p})$ for the pendulum.
* $\theta _{p}={\sqrt {}k_{o})cos(\theta )$ ___________Angle
* $\omega _{p}={\sqrt {}k_{o})(\omega )sin(\theta )$ ______ Angular Velocity
* $\alpha _{p}=-{\sqrt {}k_{o})(\omega )^{2}cos(\theta )$____Angular Acceleration
* $(\theta _{p})(\alpha _{p})=-k_{o}(\omega )^{2}cos^{2}(\theta )$
* $(\omega _{p})^{2}=k_{o}(\omega )^{2}sin^{2}(\theta )$
* $(\omega _{p})^{2}-(\theta _{p})(\alpha _{p})=k_{o}(\omega )^{2}(sin^{2}(\theta )+cos^{2}(\theta ))$
* $(\omega _{p})^{2}-(\theta _{p})(\alpha _{p})=k_{o}g/L$
* $L^{2}((\omega _{p})^{2}-(\theta _{p})(\alpha _{p}))=k_{o}gL$
* $(1/2)mL^{2}((\omega _{p})^{2}-(\theta _{p})(\alpha _{p}))=(1/2)k_{o}mgL$
** $(1/2)k_{o}mgL=E_{\left(Sy\right)}$ _______System Energy
It appears that the energy at the bottom of the swing depends on $(\omega _{p})^{2}$ and at the end of the swing upon $(\theta _{p})(\alpha _{p}).$
Note: The above treatment is not the accepted view.
Conical Pendulum (AV)