Pendulum Physics

[spingoogl]

http://www.facebook.com/photo.php?pid=2 ... =610572676

A minimum of 16.97 m/s^2 is required to maintain the pendulum at a 60° angle. However given the potential energy between 30° and 60°,
converted to kinetic energy we have v=2.678m/s, spinning around axis A with a .366m radius, our radial acceleration is 19.6 m/s^2.

Inversely, adding 3.659 joules of kinetic energy to the rotation of the pendulum. The potential energy is more than the kinetic energy.
Can a rotating pendulum gain free potential energy from kinetic energy? or do we assume this device requires the sum of potential and kinetic
energy to rotate and raise?

[spingoogl]

Nearly a hundred people have seen this post, and no comments.
Hmmm.

Is this a little too complicated?

I guess I am asking if this is overunity.

I'll try re-explaining it.
1. Get a horizontal beam that can rotate on a vertical axis.
2. Put a pendulum on the end of the beam so that it can swing out radially, and be at the mercy of gravity and centrifugal forces.
3. Add a quantity of energy to make the pendulum and the beam rotate.
4. Measure the potential energy the risen pendulum has and the kinetic energy in its velocity.

Compare the energy levels in steps 3 and 4.

Can anyone please direct me to the information that can properly predict these results.

Will the energy added to make the system rotate go directly into the rotation as kinetic and give us some free potential energy as the pendulum raises, or will the quantity of energy added to the rotation be split between the kinetic and potential?

[rlortie]

Edit: I have said enough for one day!

[spingoogl]

Thanks for the attention.

First let us assume no friction in our merry go round. If we add energy to the rotation. The merry go round spins and keeps spinning, this is our kinetic energy. The potential energy is the distance the pendulum on the end of the beam raises up.

Is this potential energy free?

[rlortie]

No it is the creation of a pseudo force due to inertia supplied by your kinetic, true the pendulum has raised in height showing a gain in potential energy, Unless you keep a constant applied force on the I beam it will drop.

No gain or nothing free

[spingoogl]

The constant applied force on the beam is not needed in a theoretical frictionless environment. We should assume the beam will continue rotating without adding energy.

The illustration above (however confusing it may be), has the axis of the pendulum on the other side of the axis of rotation. The pendulum itslef will rest on the axis rotation when at rest. This allows a shorter radius and thus more force to drive the pendulum up. F=mv²/r is a free force created with an object traveling in a circular path.

[spingoogl]

[rlortie]

The constant applied force on the beam is not needed in a theoretical frictionless environment. We should assume the beam will continue rotating without adding energy.

For theoretical purposes you may assume that the beam will continue rotating without adding energy. But to assume it will maintain any degree of lift on the pendulum at the same time within a gravity field is pushing the envelope. A bit to far for my limited brain to understand. Maybe in space with no gravity or wind resistance?

When it comes to this type discussion, I prefer keeping my feet on the ground.

Ralph

[scott]

Spingoogl please use the file attachment feature.
Thanks,
Scott

[broli]

spingoogl, have you considered conservation of angular momentum. Although I'm not a fan of it but in this case it will make its appearance. When you start rotation and allow the pendulum mass to go up you are essentially increasing its radius. CoAM then says that increasing radius leads to a decrease in speed. Decrease in speed is reduction in energy and like any physicists will tell you that is where the potential energy came from. Pequaide has changed my views on these matters I suggest you read his thread.

[spingoogl]

CoAM then says that increasing radius leads to a decrease in speed

Increasing the radius does not decrease speed. It decreases rotations per second (or minute). The speed and energy are constant.

Ever seen a figure skater spin? They start with arms extended out their sides, twist their body as they skate, as they stop they turn in to a spin, gradually they pull their arms into their chest and the spin is faster. The energy hasn't changed, they just shortened their radius.


I also had a quick look at pequaide's thread. I am a little intimidated by 38 pages of thread, but I get the gist of the mv=momentum thing. I am not sure how we are applying this though.

I did see the bola clock. The model is similar but instead we keep the string solid and only let it swing out radially. Remove the wrapping posts, and just let the thing spin.

And I just found the file attachment feature. Thx.

But to assume it will maintain any degree of lift on the pendulum at the same time within a gravity field is pushing the envelope.

Because it is spinning, a constant force will act like it is repelling away from the axis.

[rlortie]

spingoog,

Suggest you take another look at what you just posted. It appears to me you are contradicting your self.

Increasing the radius does not decrease speed.

they pull their arms into their chest and the spin is faster.

Sounds to me that if this is so then you have found the constant gradient we have been looking for.

try turning your math logic toward Leonhard Euler and freshen up on Angular momentum and velocity. That is if you cannot find it in pequaide's thread.

http://en.wikipedia.org/wiki/Angular_momentum
http://en.wikipedia.org/wiki/Pseudovector

Ralph

[WaltzCee]

Can a rotating pendulum gain free potential energy from kinetic energy? or do we assume this device requires the sum of potential and kinetic
energy to rotate and raise?

No. You're wrong. Guess again.

You need to figure out how to get some more bubble gum.

Hello Spingoogl,

I spent a long time on the phone with a friend who's keeping me in beer money. This is the essence of what he said. The pendulum is a well understood phenomena and has been for a very long time. If there were a gain in potential by swinging a pendulum then there would be a gain by keeping a flywheel spinning. A flywheel might be seen as a multitude of pendulums.

Theory is useful. However at some point the rubber of theory has to meet the road of practical application. Unless you can practically prove your idea I'm afraid it's not going to be accepted by people who have an amazing understanding of physics.

I'm not one of them. I'm studying economics. I hope when I graduate the government needs one more economist. I'll be needing a job to pay for my education.




Walter

[broli]

CoAM then says that increasing radius leads to a decrease in speed

Increasing the radius does not decrease speed. It decreases rotations per second (or minute). The speed and energy are constant.

Angular momentum is linear momentum times radius, L=m*v*r. If you increase the radius and want to keep L constant the only thing you can do is decrease linear speed.

The skater example actually loses a lot of energy when he moves his arms out since he's not storing it in the form of gravity potential or spring potential, which the former you are doing with this pendulum.

[FunWithGravity]

'accepted by people who have an amazing understanding of physics. "

Priceless