Physics Quest

[Harvey]

Interesting, because I would say that the free fall would reach the lower point first and that the 90º would reach it latter but at higher velocity.
Thanks Cloxxki
David

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

A straight line fall can be calculated using this:
Description of Motion

by inserting 100m for the distance and 9.8 in the acceleration box for 9.8m/s² (gravitational acceleration at sea level) and an initial velocity of 0, we get the result of 44.271887242357316 m/s final velocity and a time lapse of 4.5175395145262565 seconds. So that's easy, straight algebra.

Now lets compare that to a simple pendulum bob on a massless string:
Pendulum

Now is that correct, 20 seconds for a complete period (from 90°, to BDC to 180° and back to 90°) so from 90° to BTC is approx 5 seconds? - ah...the formula is only good for small angles . . . we need this one:
Large 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!

Hi Harvey

Thank you for the math and the Hyperphysics site. Very informative.
If you have an initial velocity of 20m/s, 100 m vertically, it will reach the Bottom at a velocity of 48.579831205964474 m/s in 2.9163093067310686 s.
With same initial velocity of 20m/s the pendulum will have the same velocity of vertically throw, at bottom?
I can’t find a formula for the pendulum with initial velocity.
David

[Matos de Matos]

My intention is to model a hypothetical mechanism like the picture below.
Quest2.JPG

[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 Pedulum

The problem you have posed is a bit more complicated than the norms because it presupposes a switched massless rod that becomes flexible (or none existent) at BDC and become rigid at TDC. Or to put it another way, we turn off centripetal force during the vertical reflection period and turn it on during the swing period.

It is very tempting to just add our initial velocity to the result velocity using the equation I posted earlier:

so that BDC Velocity = (initial velocity) + sqrt(2 * gravitational acceleration * length * [ 1-cos(angle)]

(20 m/s) + (√ 2 * 9.8 m/s² * 50 * [ 1 - cos(180)]) = 64.27 m/s VBDC

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 VBDC and -9.8m/s² we don't get 20 m/s, but instead get 46.59 m/s as our VTDC.

So as tempting as it is to just add the velocities together . . . I think there is a problem with that. Otherwise your model would produce an OU of about 26.59 m/s on the first cycle and that would grow exponentially in time.

Since that is not possible, I have obviously made an error somewhere. (probably in trying to add velocities, or applying angles greater than 90° to the equation or something like that)

EDITED: For a 50m rod length and 100m overall height

[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 Pedulum

The problem you have posed is a bit more complicated than the norms because it presupposes a switched massless rod that becomes flexible (or none existent) at BDC and become rigid at TDC. Or to put it another way, we turn off centripetal force during the vertical reflection period and turn it on during the swing period.

It is very tempting to just add our initial velocity to the result velocity using the equation I posted earlier:

so that BDC Velocity = (initial velocity) + sqrt(2 * gravitational acceleration * length * [ 1-cos(angle)]

(20 m/s) + (√ 2 * 9.8 m/s² * 50 * [ 1 - cos(180)]) = 64.27 m/s VBDC

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 VBDC and -9.8m/s² we don't get 20 m/s, but instead get 46.59 m/s as our VTDC.

So as tempting as it is to just add the velocities together . . . I think there is a problem with that. Otherwise your model would produce an OU of about 26.59 m/s on the first cycle and that would grow exponentially in time.

Since that is not possible, I have obviously made an error somewhere. (probably in trying to add velocities, or applying angles greater than 90° to the equation or something like that)

EDITED: For a 50m rod length and 100m overall height

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.

Hi Harvey:
Yes, you are right.
I think that will balance out, because we are shortening the radius with a force against centripetal, and “no gain”.

I found a way to divide the radius without applying any force, but an action. I act on the centrifugal force vector when I cut the radius.
Check the thread

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.

That is yours or Tesla´s? The snow ball comprehended, it is fantastic.
Can you recommend any sites with Tesla´s autobiography?
Thank you
David