Physics Quest

[Matos de Matos]

Hi everybody:

The bob will stop at which position?
1, 2, or 3 for A and B.
The explanation and some math will elucidate us, but the poll is open to all.
Thank you
David

[Sephiroth]

Guessed 1 for both, but I would have guessed 2 if you mentioned that the system was without losses

3 would be nice though

[Matos de Matos]

Guessed 1 for both, but I would have guessed 2 if you mentioned that the system was without losses

3 would be nice though

Sorry, it is a frictionless (magnetic bearing) fulcrum and perfect elastic collision.
The velocity in will be same as velocity out.
I think that is physically possible with conventional materials.

1 is for before fulcrum, 2 in the fulcrum and 3 over the fulcrum
David


"An initially stationary object which is allowed to fall freely under gravity drops a distance which is proportional to the square of the elapsed time"
Gravitation - Wikipedia, the free encyclopedia


"In physics, and more specifically kinematics, acceleration is the change in velocity over time."
Acceleration - Wikipedia, the free encyclopedia

[Cloxxki]

A3
B2

The rope was not stated to be massless? They in A, it had potential energy not totally used up in A2. A3 is an exaggeration of the situation, but correct.

B2 seems simplest.

[sebosfato]

A2
B2

because its potential energy will return the same.

[Matos de Matos]

A-3
B-3

Gravity is time dependent and not height dependent.
The velocity attained at the bottom due to gravity will be higher than the velocity lost going up.
The trajectory down is longer, takes more time to travel (longer gravity effect) than the direct trajectory up, and both directions are influenced only by gravity.
Children do that all the time pumping swings.
David
Pumping of a Swing - Physics | Grinnell College

[Cloxxki]

A-3
B-3

Gravity is time dependent and not height dependent.
The velocity attained at the bottom due to gravity will be higher than the velocity lost going up.
The trajectory down is longer, takes more time to travel (longer gravity effect) than the direct trajectory up, and both directions are influenced only by gravity.
Children do that all the time pumping swings.
David
Pumping of a Swing - Physics | Grinnell College

You forget that when you are not falling straight down, you also don't receive the full 9.81m/s². The extra time you take vs. the straight down path, is what makes you end up at the same KE at the lowest point.

In the end, it's all about height and mass. Time and angle cancel each other out.

[Armagdn03]

A-3
B-3

Gravity is time dependent and not height dependent.
The velocity attained at the bottom due to gravity will be higher than the velocity lost going up.
The trajectory down is longer, takes more time to travel (longer gravity effect) than the direct trajectory up, and both directions are influenced only by gravity.
Children do that all the time pumping swings.
David
Pumping of a Swing - Physics | Grinnell College

I really hate to leave negative remarks, but this is wrong.

1) Gravity is time and Height dependent, it is an inverse square field.
2) Think in terms of potential energy, a loss less system with such redirection will end up at the same potential it started at.
3) Children parametrically pump their oscillatory system on swings.

If you disagree, perhaps you could show a mathematical proof rather than posting a link to a website, this way people who are not familiar with what you are trying to show will be able to learn and grow, or perhaps you will learn something yourself!

Take care, and thanks for the brain teaser.

[gasman]

1 on both

One of the tricks in this setup is air resistance. Since it is not in a hypothetical vacuum (which, of course, would be just as impossible as the assumptions in this physics puzzle), there will be air resistance. Eventually there will be a terminal velocity reached, and there will be no more acceleration after that point (until something changes... like the ground). Up until terminal velocity is reached, acceleration is velocity/terminal velocity*gravity coefficient. So 2 in both situations is impossible.

The major trick to this thought puzzle, though, is that many who do not know about physics could easily assume that they calculate the length of the dotted line (effectively straightening it), and base bounce height off of that. What someone who chooses 2 or 3 (on either) fails to see is that the dotted line stays the same length the whole time. This rigid line means is that there is only one (theoretical) point at which nearly full gravitational acceleration (approx. 9.8 m/s/s) is achieved (which, in the graphs, is the right-most point). The rest of the time, the gravity affecting the "ball" at any time is a function of the distance from vertical center. Vertical being exactly 0%, and right most being nearly 100%. This also means that, in graph 2, there must be some force added into the system to even begin. The rest of the gravitational force (100%-value from above) is exerted down/through the rigid line, into the fulcrum. So, when compared to a ball simply dropped vertically, this added time will cause much more atmospheric resistance, making the ball's velocity be significantly lower. There are other things, but this is enough to prove that the max bounce height is lower than the original ball height. Since there is only one option less, there is no reason to go further, or do accurate calculations.

[Matos de Matos]

1) Gravity is time and Height dependent, it is an inverse square field.
2) Think in terms of potential energy, a loss less system with such redirection will end up at the same potential it started at.
3) Children parametrically pump their oscillatory system on swings.

Take care, and thanks for the brain teaser.

I am sorry, but I am a Portuguese spoken and sometimes linear translations do not work well.
I meant Gravity is not mass (Weight- height) dependent.
I thought that perfect elastic collisions could change the moment of inertia of a bob, without any kinetic energy losses, and that mechanical parametric pumping was not a force but the variation of its resonance frequency in time.
Changing the amplitude from 90º to 180º with elastic collisions, in my opinion will stop over the fulcrum line in both situations, unless the only way to change the path is to apply a force without changing its kinetic energy.
Perfect elastic collisions transform all the energy into potential and then all back to kinetic that will reach the top faster, with a left over that will allow it to go higher.
It is easier for me to copy and paste exerts from Wikipedia, than the mental exercise of translating to English my thoughts and write them. (Automatic grammar is always correcting).
Thank you
David
“Parametric resonance occurs in a mechanical system when a system is parametrically excited and oscillates at one of its resonant frequencies. Parametric excitation differs from forcing since the action appears as a time varying modification on a system parameter. This effect is different from regular resonance because it exhibits the instability phenomenon

Remarkably, if the parameters vary at roughly twice the natural frequency of the oscillator (defined below), the oscillator phase-locks to the parametric variation and absorbs energy at a rate proportional to the energy it already has. Without a compensating energy-loss mechanism provided by β, the oscillation amplitude grows exponentially. (This phenomenon is called parametric excitation, parametric resonance or parametric pumping.) However, if the initial amplitude is zero, it will remain so; this distinguishes it from the non-parametric resonance of driven simple harmonic oscillators, in which the amplitude grows linearly in time regardless of the initial state.
A familiar experience of parametric oscillation is playing on a swing. By alternately raising and lowering their center of mass (and thereby changing their moment of inertia, and thus the resonance frequency) at key points in the swing, children can quickly reach large amplitudes provided that they have some amplitude to start with (e.g., get a push). Doing so at rest, however, goes nowhere.
Parametric oscillator - Wikipedia, the free encyclopedia

[Matos de Matos]

Hi everybody:

I did a simple test with a bucket, a valve and tubing.
This was what I found.
Even against all the friction the water rose over the initial water level, for an instant and then flow back in the tubing to the water level
There is other explanation for this?
On your experiences, did somebody tried to do a collision similar to the quest?

Thank you
David

water level.JPG

[Harvey]

Hi David,

There are several factors involved in the various posts above, but they all have to do with energy conversions.

First of all, regarding the Bob - it is important to recognize that the energy being converted between gravitational potential (starting point of the Bob) and kinetic (motion of the Bob) is not consistent. At the start, the conversion is 100% because the gravitational force is tangent to the rotation - but 90° later it is 0% because the force is orthogonal to the rotation. Therefore, throughout the path, the force applied is diminishing along the path integral.

Secondly, there is a finite quantity of energy stored in the Bob at its starting point. If no other positive forces exist, then the Bob cannot contain more energy than it started with and therefore cannot exceed a height of position 2 for either A or B.

As a third caveat, gravity is applied as a negative force to the upward travel and this is 100% for the entire vertical path. So any kinetic energy stored as momentum, becomes canceled by this action prior to exceeding position 2.

Also, it should be stated that gravity itself is mass dependent. Gravitational potential energy is distance dependent. Since acceleration is defined as Force divided by mass (a = F/m) we must add time to both sides of the equation in order for it to be relevant. Acceleration is a change in velocity, but a change in velocity can be time independent, that is it can have a constant speed, but a changing direction. So a = F/m can apply to distance and direction completely independent of time and still be relevant. This is why a rotating object having the same angular velocity is said to be in a state of constant acceleration.

=======================

As regards your water bucket test, somehow it reminds me of Boyles Self Filling Flask.

But I think the answer really is bound to the forces at work. In addition to the gravity being applied to the water in the bucket, you also have air pressure being applied to the entire surface of the water in the bucket, so there is a specific pressure at the valve before it is opened and after it is opened. Now the air total pressure being applied to the outlet opening of the hose is much less than the total pressure being applied to the water in the bucket, because they both have different areas. This is why pressure is measured in terms like Pounds per Square Inch (PSI) or Pascals which are area dependent.

So the process is like this: A pressure is applied to a valve which is opened and a flow of water begins. Since water has mass and a force is pressing on this mass, acceleration occurs. As the water accelerates, it develops kinetic energy in the form of momentum. The path for the moving water is a curved path such that Gravity on the curved path nets to zero - i.e. gains on one side are equally lost on the other so that the net momentum of the moving water is unaffected by gravity from the valve to the hose outlet. Therefore, the only negative restriction to the momentum of the water is the friction in the hose and the air pressure at the outlet.

Once the water leaves the hose, it is subject to the negative effect of the force of gravity and therefore has a finite distance it can travel based on the energy stored in its momentum.

Q1. Could you get an arrangement that would allow the exit stream to re-enter the bucket?

Q2. How different would the experiment be if the hose was filled with water rather than air prior to the experiment.

Q3. How does the answer to Q2 impact continuous operation and how does it demonstrate that extra energy is added to extract the water from the hose?

[Matos de Matos]

Q1. Could you get an arrangement that would allow the exit stream to re-enter the bucket?

Q2. How different would the experiment be if the hose was filled with water rather than air prior to the experiment.

Q3. How does the answer to Q2 impact continuous operation and how does it demonstrate that extra energy is added to extract the water from the hose?

Hi Harvey:

Thank you for your opinion.
I would love to have a HomeDepo in Luanda, but here we lack everything, experimentation is very difficult.

I understand the bob elastic collision explained by you.
I would like to do an experiment where a weight is drop vertically down and then same weight dropped at same height as a pendulum from 3:00 o’clock.
Which one will pass the lowest point first and which one will have higher velocity?

Regarding the bucket;

1- No, initially the outlet water goes over the bucket level, and then goes down to the bucket level, and stops the flow.
2- Only works with the hose empty. I had to empty the hose for each experiment.
3- I found that if we have a perfect circular path with constant radius on the hose, the water starts slowing down before bucket level and stops at level without going higher.
Your explanation for the different areas of air pressure my explain it, but I did not found it on the setup explained on 3, but my trials are very rudimental.

I would like to have your opinion on my opinion of Bessler´wheel

Thank you
David
http://www.energeticforum.com/renewa...how-i-see.html

[Cloxxki]

I would like to do an experiment where a weight is drop vertically down and then same weight dropped at same height as a pendulum from 3:00 o’clock.
Which one will pass the lowest point first and which one will have higher velocity?

While experimentation is great, I think these relationship are very reliably available described in formulae. I once looked them up for a gravity wheel design, but I'm impossible with formulae, I forget them all.

I don't know the time relationship anymore, but can tell you the end velocity is the same between free fall and 90degree pendulum fall. Just the terminal direction is 90degrees offset. Any difference in speed your experimentation would prove, would allow the simplest gravity design to run indefinitely, and produce excess energy easily harvested.
Inverse direction by the way, offer the same duration and change in velocity, just negative.

[Matos de Matos]

can tell you the end velocity is the same between free fall and 90degree pendulum fall. Just the terminal direction is 90degrees offset.

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