Metaphysics (Alternate View)/Creation/Conical Pendulum

Book Equation

The accepted analysis of a conical pendulum results in the following equation.

• $Cos\theta =g/(L\omega ^{2})$
  • Where $\theta$ is the displacement angle
    • g is the local gravity
    • L is the length of the pendulum’s arm and
    • $\omega$ is the constant horizontal angular velocity of the bob.

Testing suggestion

Let the arm’s length equal the numerical value of the local gravity. For example, imagined a gravity value of thirty-two feet per second squared and a thirty-two foot arm. This reduces the book equation to the following.

• $Cos\theta =1/(\omega ^{2})$

Slow the angular rotation to where its value is less than one radian per second, and the cosine value becomes greater than one.

Energy Law

(An alternative view by Calgea 19:51, 1 June 2007 (UTC))

The conservation of energy law states: Energy cannot be created or destroyed. Assume this law is valid. Let’s look at a physical example where a bullet hits a stationary block of wood and their combined mass moves at a constant velocity in the direction the bullet was traveling. The kinetic energy equation is as follows.

• $KE=(1/2)mV^{2}$

Applying the law produces the following equation.

• $(1/2)m_{1}(V_{1})^{2}+(1/2)m_{2}(V_{2})^{2}=(1/2)m_{3}(V_{3})^{2}$
  • where $(V_{2})=0$ and
  • the mass $m_{3}=m_{1}+m_{2}$

Why does this conservation equation produce the wrong result?

Water Tanks

For an alternative view of this problem, consider two water tanks one above the other. The top tank has several valves to allow its water to flow into the bottom tank. Open the valves for a short time. How much water flowed from the top tank?

One usually measures the quantity, before and after, in either tank, and the difference in quantity is the answer. However, it is also possible count the number of valves, n, time the duration, t, the valves were open, and determine the rate of flow, r.

• $quantity=trn$

Applying this form to the above bullet energy problem, one has the following.

• $(trn)_{1}+(trn)_{2}=(trn)_{3}$

By writing the energy law in this form, one sees the time of transition, t, is the same on both sides of the equation. Dividing both sides by t gives the following equation.

• $(rn)_{1}+(rn)_{2}=(rn)_{3}$

Using this reasoning on the above bullet energy problem provides the following momentum equation.

• $m_{1}(V_{1})+m_{2}(V_{2})=m_{3}(V_{3})$
  • where $t=(1/2)V$

One now says the momentum equation is a special case of the conservation law with the time of transition ommited.

Transition

Conical Pendulum Energy Circle

Energy resides in a field. In the case of a simple pendulum, energy converted from the field of Spin (Sp) to the field of Linear motion (Lm) and back again. In the case of the above bullet example, energy transitioned from Linear motion in the bullet to Linear motion in the combined mass. In the case of a conical pendulum, energy transitions from Linear motion in one direction to Linear motion in a perpendicular direction. This is possible because Linear motion converts to Spin and immediately to Linear motion in a perpendicular direction. The constant displacement angel theta represents the Spin dimension. In this case, the Spin dimension acts as a catalyst. To see this, imagine two simple pendulums. One swings front to back and the other swings left to right. As one bob swings inward, the other bob swings outward from the center. To a viewer, the bob swings in a circle.

Convert is the term used when energy changes from one field to a different field. Transition is the term used when energy changes from one field to the same field. And, when energy transitions, one needs to use a momentum equation. Thus, the conical pendulum is energy in transition and requires a momentum equation.

Energy Circle

The energy circle for a conical pendulum lies in the horizontal plane of the bob’s orbit.

Conical Pendulum's Transition Vector

The hypotenuse (L) of the right triangle represents the total energy of the conical pendulum $E_{cp}$. The vertical leg represents the Spin energy $(E_{Sp})$ which is $E_{cp}$ times the cosine of the constant displacement angle theta $(\theta )$.

• $E_{Sp}=E_{cp}cos(\theta )$ ____constant vertical Spin energy
• $E_{Sp}=mgh_{o}$

• $E_{cp}=(mgh_{o})/cos(\theta )$ ____System energy of conical pendulum
  • Where m = the mass of the bob
    • g = local gravity
    • $h_{o}=L(1-cos(\theta ))$ ____the height of the horizontal plane

Transition Vector

The transition vector (TV) represents the energy in the horizontal plane, and it equals the sine component of the displacement angle.

• $TV=E_{cp}sin(\theta )$ ____Horizontal energy

• $TV=(mgh_{o})sin(\theta )/cos(\theta )$

Angular Momentum

The angular momentum of the system is the sine of the transition vector divided by the time factor.

• $AM_{o}=(TVsin(\theta ))/((\omega _{H})/2)$
• $mR_{H}^{2}(\omega _{H})=mgh_{o}sin^{2}(\theta )/(cos(\theta ))((\omega _{H})/2)$
• $R_{H}=Lsin(\theta )$ ____the radius of the horizontal circle

• $L^{2}sin^{2}(\theta )(\omega _{H})^{2}=2gh_{o}sin^{2}(\theta )/(cos(\theta ))$
• $h_{o}=L(1-cos(\theta ))$

• $L^{2}(\omega _{H})^{2}=2gL[1-cos(\theta )]/(cos(\theta ))$

• $L(\omega _{H})^{2}=2g[1-cos(\theta )]/(cos(\theta ))$

• $L(\omega _{H})^{2}cos(\theta )=2g[1-cos(\theta )]$

• $L(\omega _{H})^{2}cos(\theta )+2gcos(\theta )=2g$

Field Equation

• $cos(\theta )(L(\omega _{H})^{2}+2g)=2g$

• $cos(\theta )=2g/(L(\omega _{H})^{2}+2g)$
• $cos(\theta )=1/(1+(L(\omega _{H})^{2}/2g))$ ____Alternative view
• $cos(\theta )=g/(L(\omega _{H})^{2})$ ____ Book solution

Comments

The alternative view:

1. As omega approaches infinity, the denominator approaches zero, theta approaches ninety degrees, and the cosine approaches zero.
2. As omega approaches zero, theta approaches zero, and the cosine approaches one.

Note: The above treatment is not the accepted view.

See Also

• Simple Pendulum
• Gyroscope