energy producing experiments
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re: energy producing experiments
I often use photo gates. And in one of my versions of the aluminum tube experiment I halved the mass when I moved it to double the distance. And the photo gates recorded the same rate of acceleration to a few ten thousandths of a second. It is mr.
re: energy producing experiments
I then reduced the moved mass to one fourth, and the photo gate read out was way off. It is not mrr.
re: energy producing experiments
So what Victorian back street child’s game is Bessler alluding to?
What goes around, comes around.
re: energy producing experiments
Possibly it was a hinged board.
The hinged board has 33% more energy than the same mass dropped straight down. Changing the distribution of mass can give you even more energy.
The force is evenly distributed down the board and so of course is the mass; but the motion is forced into the end portion of the board because the near end is the point of rotation. The board falls slower at first because it is a Cos * Force causing the acceleration. This is made up for by the fact that the board falls a longer distance through the arch. The Force is applied for a longer time. But I agree with the professor; by the time the board reaches the floor the center of mass of the board is moving at the same velocity as an object that had dropped straight down the same distance.
Or you could of course drop the board straight down and then grab it on its end; and the center of mass rule would come into effect. The center of mass is the point of the board that continues at the same velocity when an object is directed into a spin. And that spin motion is provided by the linear momentum of the board.
Anyway this is yet another why to make energy. The board makes energy.
The hinged board has 33% more energy than the same mass dropped straight down. Changing the distribution of mass can give you even more energy.
The force is evenly distributed down the board and so of course is the mass; but the motion is forced into the end portion of the board because the near end is the point of rotation. The board falls slower at first because it is a Cos * Force causing the acceleration. This is made up for by the fact that the board falls a longer distance through the arch. The Force is applied for a longer time. But I agree with the professor; by the time the board reaches the floor the center of mass of the board is moving at the same velocity as an object that had dropped straight down the same distance.
Or you could of course drop the board straight down and then grab it on its end; and the center of mass rule would come into effect. The center of mass is the point of the board that continues at the same velocity when an object is directed into a spin. And that spin motion is provided by the linear momentum of the board.
Anyway this is yet another why to make energy. The board makes energy.
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daanopperman
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re: energy producing experiments
Hi pequaide ,
In the following drawing I have set up a mass less board with a weight in the center of the board and one at the end with the board pivoted at lh bottom . For the board to make energy , it would need to not only lift A , but olso lift B to the same height or higher as it started out . which it cannot do . It will only happen if B was left behind after the fall and not raised to the top . Any hinged board/lever , no matter the weight or COM would then make us some energy which is not the case .
In the following drawing I have set up a mass less board with a weight in the center of the board and one at the end with the board pivoted at lh bottom . For the board to make energy , it would need to not only lift A , but olso lift B to the same height or higher as it started out . which it cannot do . It will only happen if B was left behind after the fall and not raised to the top . Any hinged board/lever , no matter the weight or COM would then make us some energy which is not the case .
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re: energy producing experiments
I would not be so sure it could not rise higher than dropped.
There are several forms of energy; the energy that you have chosen is gravitational potential energy; or simply, height. Which is perfectly fine; GPE is a good choice.
You can transfer one form of energy into another; and the form I was thinking of was KE, kinetic energy. KE is equal to GPE; but let us look at KE first.
There is a good deal of evidence to support the concept that the final velocity of the center of mass for the board is the same as the ball dropped the same distance. Oh; by the way the ball and the board are not dropped the same distance in the video; the board is only dropped half as far. But what was stated, by the professor, was that the acceleration of the ball and the center of mass of the board are the same; and that is what we are focusing on. And lets put a number on that velocity: lets say the center of mass of the board has a final velocity of one meter per second as it strikes the floor.
With the center of the board moving 1m/sec the end of the board is moving 2 meters per second. The mass on the end of the board has four times as much energy (1/2mv²) as a mass in the center of the board. Of course there is more mass toward the center of the board. The mass toward the pivot point have almost no energy; but it does not quite balance out with the other end, you are left with a 33% increase in energy. This would be 33% more than the same mass (ball) dropped the same distance. The way to amplify this quantity of energy is to move the mass to both ends.
Just for a thought question and I know it is not quite reality; lets say you move half of all the mass to both ends; that would be half the mass at the pivot point and half the mass on the end. Now half the mass is moving twice as fast as the center of mass. An object moving 2 m/sec will rise; d = 1/2v²/a, .2038 m. An object moving 1 m/sec will rise .0509 m.
So if the center of all the mass is dropped 5 cm then half the mass on the end will rise to 20 cm. You now have to throw the mass on the end upward.
We can not forget the mass at the pivot point, it has to be lifted back to the 5 cm height as well. Okay, grab a pulley, and use the half mass at 20 cm to lift the other half mass to 10 cm and now we have both half masses at 10 cm.
We only need to drop the whole mass 5 cm and now we are at 10 cm GPE.
There are several forms of energy; the energy that you have chosen is gravitational potential energy; or simply, height. Which is perfectly fine; GPE is a good choice.
You can transfer one form of energy into another; and the form I was thinking of was KE, kinetic energy. KE is equal to GPE; but let us look at KE first.
There is a good deal of evidence to support the concept that the final velocity of the center of mass for the board is the same as the ball dropped the same distance. Oh; by the way the ball and the board are not dropped the same distance in the video; the board is only dropped half as far. But what was stated, by the professor, was that the acceleration of the ball and the center of mass of the board are the same; and that is what we are focusing on. And lets put a number on that velocity: lets say the center of mass of the board has a final velocity of one meter per second as it strikes the floor.
With the center of the board moving 1m/sec the end of the board is moving 2 meters per second. The mass on the end of the board has four times as much energy (1/2mv²) as a mass in the center of the board. Of course there is more mass toward the center of the board. The mass toward the pivot point have almost no energy; but it does not quite balance out with the other end, you are left with a 33% increase in energy. This would be 33% more than the same mass (ball) dropped the same distance. The way to amplify this quantity of energy is to move the mass to both ends.
Just for a thought question and I know it is not quite reality; lets say you move half of all the mass to both ends; that would be half the mass at the pivot point and half the mass on the end. Now half the mass is moving twice as fast as the center of mass. An object moving 2 m/sec will rise; d = 1/2v²/a, .2038 m. An object moving 1 m/sec will rise .0509 m.
So if the center of all the mass is dropped 5 cm then half the mass on the end will rise to 20 cm. You now have to throw the mass on the end upward.
We can not forget the mass at the pivot point, it has to be lifted back to the 5 cm height as well. Okay, grab a pulley, and use the half mass at 20 cm to lift the other half mass to 10 cm and now we have both half masses at 10 cm.
We only need to drop the whole mass 5 cm and now we are at 10 cm GPE.
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daanopperman
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re: energy producing experiments
Hi pequaide ,
If the mass in the center of the bar is accelerating the mass on the end of the bar , does the center mass not fall slower than 1m/s . Is this not just a case of leverage of the 3rd kind . I think if the center mass is located halfway on the bar , it will move the same amount slower as would the faster mass moves faster . If the center weight gives energy to the outside mass to accelerate beyond G , it must be robbed of some of it's own energy received from G .
If the mass in the center of the bar is accelerating the mass on the end of the bar , does the center mass not fall slower than 1m/s . Is this not just a case of leverage of the 3rd kind . I think if the center mass is located halfway on the bar , it will move the same amount slower as would the faster mass moves faster . If the center weight gives energy to the outside mass to accelerate beyond G , it must be robbed of some of it's own energy received from G .
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FunWithGravity2
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re: energy producing experiments
Dan
But what if it doesn't? "connectedness" refence frame sharing for a moment of potential GPE increase?
Crazy Dave
But what if it doesn't? "connectedness" refence frame sharing for a moment of potential GPE increase?
Crazy Dave
Si mobile in circumferentia circuli feratur ea celeritate, quam acquirit cadendo ex
altitudine, quae sit quartae parti diameter aequalis ; habebit vim centrifugam suae
gravitati aequalem.
altitudine, quae sit quartae parti diameter aequalis ; habebit vim centrifugam suae
gravitati aequalem.
Re: re: energy producing experiments
Quite right. :-)daanopperman wrote:Hi pequaide ,
If the mass in the center of the bar is accelerating the mass on the end of the bar , does the center mass not fall slower than 1m/s . Is this not just a case of leverage of the 3rd kind . I think if the center mass is located halfway on the bar , it will move the same amount slower as would the faster mass moves faster . If the center weight gives energy to the outside mass to accelerate beyond G , it must be robbed of some of it's own energy received from G .
Also, we must not forget that some of the vertical acceleration energy (2nd derivative) is going into rotation energy (3rd derivative).
Who is she that cometh forth as the morning rising, fair as the moon, bright as the sun, terribilis ut castrorum acies ordinata?
re: energy producing experiments
I suspect that the mass near the bearing is being robbed of its linear momentum. I suspect that the total linear momentum of the two masses is the same as if all the mass was at the center of mass.
As one half of the mass moves a distance out the other half mass moves equally in. One has a decrease in linear momentum while the other has an equal increase. All the while the center of mass stays put; and the total linear momentum remains the same.
As one half of the mass moves a distance out the other half mass moves equally in. One has a decrease in linear momentum while the other has an equal increase. All the while the center of mass stays put; and the total linear momentum remains the same.
re: energy producing experiments
Suppose you have a pendulum that is an upside down T; so that as the T swings down all points in the arms of the T are moving at the same velocity.
As the arms of the T are at the low point of the swing the base of the T (which is now the top) moves freely along a linear bearing. As the base of the T is sliding upside down along the linear bearing the arms of the T rotate into a linear position with the trunk of the T. This rotation should cost almost no energy because all points in the T retain the same velocity.
After the T arms move into a straight line position with the trunk of the T the linear bearing catches and the linear stick pendulum swings up.
How high will the stick pendulum swing up?
As the arms of the T are at the low point of the swing the base of the T (which is now the top) moves freely along a linear bearing. As the base of the T is sliding upside down along the linear bearing the arms of the T rotate into a linear position with the trunk of the T. This rotation should cost almost no energy because all points in the T retain the same velocity.
After the T arms move into a straight line position with the trunk of the T the linear bearing catches and the linear stick pendulum swings up.
How high will the stick pendulum swing up?
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daanopperman
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re: energy producing experiments
Hi peqaide ,
If my understanding of your explanation is correct , here is my input .
If my understanding of your explanation is correct , here is my input .
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daanopperman
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re: energy producing experiments
Hi Dunebery ,
T arm rotate by gravity . Movement I don't know where come from . It must be induced .
T arm rotate by gravity . Movement I don't know where come from . It must be induced .
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re: energy producing experiments
Place the cart on a rail track so that the T can be suspended upside down from the cart. Let the big black circle be a set a barbells; you can see the front mass of the barbell but you can not see the second mass on the other side of the barbells. While locking the T in place accelerate the system to 2 m/sec and then unlock the T and jam the cart, we can assume that the barbells will rise d = 1/2v²/a or .2039 m.
For a second run lets rotate the spheres of the barbell so that one of the masses is just under the cart and the other mass is an equal distance down from the center of mass. This will leave the center of mass in the same location. While locking the barbell accelerate the system to 2 m/sec and then unlock and jam the cart, can we assume that the center of mass of the barbells will rise .2039 m?
The two position of the barbell are horizontal and vertical; but sense the center of mass remains in the same location it will take almost no energy to go from one position to the other. Therefore can not we expect them to rise the same amount.
Some point on the vertical barbell must continue moving at the same rate as the barbells swing up. Experiment like the spinning wrench or spinning hammer seem to indicate that the center of mass is the point on the barbell that continues at the same velocity.
A similar question is: will the center of mass of a barbell that is rotated from one end have a velocity of 2 m/sec after the center of mass has fallen .2039 m?
How do you make the drawings?
For a second run lets rotate the spheres of the barbell so that one of the masses is just under the cart and the other mass is an equal distance down from the center of mass. This will leave the center of mass in the same location. While locking the barbell accelerate the system to 2 m/sec and then unlock and jam the cart, can we assume that the center of mass of the barbells will rise .2039 m?
The two position of the barbell are horizontal and vertical; but sense the center of mass remains in the same location it will take almost no energy to go from one position to the other. Therefore can not we expect them to rise the same amount.
Some point on the vertical barbell must continue moving at the same rate as the barbells swing up. Experiment like the spinning wrench or spinning hammer seem to indicate that the center of mass is the point on the barbell that continues at the same velocity.
A similar question is: will the center of mass of a barbell that is rotated from one end have a velocity of 2 m/sec after the center of mass has fallen .2039 m?
How do you make the drawings?