Physics Quest

[Harvey]

Let's compare the two for a height of 100 meters:

A straight line fall can be calculated using this:
Description of MotionPendulumLarge Amplitude Pendulum (notice this one is pendl)
So that is a little more accurate at 23.603624209842796 seconds for a full period or about 5.9 seconds to reach BDC from the starting point.

And what of it's velocity? Borrowing an equation from here we get:

v = √{2gL[1-cos(a)]}

where:

• v is the velocity of the weight at the bottom of the swing
• g is the acceleration due to gravity
• L is the length of the wire
• a is the angle from the vertical
• cos(a) is the cosine of angle a

v = sqrt(2 * 9.8 * 100 * [1 - Cos(90)])

So the velocity is 44.271887242357310647984509622058 m/s

same velocity, longer period.



Cheers!

[Matos de Matos]

same velocity, longer period.



Cheers!

[Matos de Matos]

My intention is to model a hypothetical mechanism like the picture below.

[Harvey]

My intention is to model a hypothetical mechanism like the picture below.
[ATTACH]6134[/ATTACH]

You have an interesting model there to evaluate.

We know that the acceleration due to gravity will be consistent regardless of the starting velocity. However, we also know that the starting velocity reduces the overall time required to travel between the two reflectors.

Here is a fun little app: Pendulum Physics Simulation

And here is full treatise of the amplitude dependent formulations:
Large Amplitude Period of a Physical PedulumBDC

But I am not so sure this is the right way to do that since that equation is really an approximation. But it might be good enough for your evaluation

For an accurate result, we would need to apply calculus and integrate force vectors along the arc path to arrive at the VBDC.

To determine the final velocity at TDC (VTDC ) we simply need to apply the negative acceleration that results from gravity pulling against the vertical bounce. In this case it is consistent and straight forward just like dropping an object, only in reverse. You can think of it as deceleration.

The expectation is that VTDC will = 20 m/s supposing that there are no losses in the system and the process would continue indefinitely. But using the straight motion calculator here and plugging in the 64.27 starting velocity for VBDCVTDC.

[Matos de Matos]

You have an interesting model there to evaluate.

We know that the acceleration due to gravity will be consistent regardless of the starting velocity. However, we also know that the starting velocity reduces the overall time required to travel between the two reflectors.

Here is a fun little app: Pendulum Physics Simulation

And here is full treatise of the amplitude dependent formulations:
Large Amplitude Period of a Physical PedulumBDC

But I am not so sure this is the right way to do that since that equation is really an approximation. But it might be good enough for your evaluation

For an accurate result, we would need to apply calculus and integrate force vectors along the arc path to arrive at the VBDC.

To determine the final velocity at TDC (VTDC ) we simply need to apply the negative acceleration that results from gravity pulling against the vertical bounce. In this case it is consistent and straight forward just like dropping an object, only in reverse. You can think of it as deceleration.

The expectation is that VTDC will = 20 m/s supposing that there are no losses in the system and the process would continue indefinitely. But using the straight motion calculator here and plugging in the 64.27 starting velocity for VBDCVTDC.

Hi Harvey

Did you see ?

Can you help math model it?

Thankyou
David

[Harvey]

The error exists in the equation itself which supposes a definitive time embedded in the 'g'.

In order to allow for a starting velocity, we must bring time back into the equation because the overall swing is driven and overrides the resonant period of the pendulum causing a shorter period and thus reducing the overall acceleration of gravity.

Therefore, the equation:

v = √{2gL[1-cos(a)]}

needs to be rewritten to include time, or we need to find a formula for a driven oscillator rather than a harmonic oscillator that will give us the velocity of a rigid arm arc in a gravitational field with a starting velocity.

[Matos de Matos]

The error exists in the equation itself which supposes a definitive time embedded in the 'g'.

In order to allow for a starting velocity, we must bring time back into the equation because the overall swing is driven and overrides the resonant period of the pendulum causing a shorter period and thus reducing the overall acceleration of gravity.

Therefore, the equation:

v = √{2gL[1-cos(a)]}

needs to be rewritten to include time, or we need to find a formula for a driven oscillator rather than a harmonic oscillator that will give us the velocity of a rigid arm arc in a gravitational field with a starting velocity.

http://www.energeticforum.com/renewa...tml#post105923

Thank you
David

[boguslaw]

Most obvious example and proof of OU :
ball rolling from the top of the hill

all is fine under unity unless you take snow ball into consideration.


Read Tesla autobiography.

[Matos de Matos]

Most obvious example and proof of OU :
ball rolling from the top of the hill

all is fine under unity unless you take snow ball into consideration.


Read Tesla autobiography.