Yes; the grey disk and grey puck are the same item, all the motion is given to one 456g grey steel disk (or puck).
The white nylon cord is to the left of the center (of the side) of the gray puck (in the picture) and is in the center of the side of the bottom half of the white disk. Yes it is hard to see.
The white disk is made of two layers of 1/2 inch sheets of HDPE (high density polyethylene). The top half of the white disk is a solid disk and is used to wrap the ribbon, the bottom half of the white disk is in pieces forming a seat for the gray puck and a ring circumference which is used to wrap the nylon string.
I have since replaced the nylon string with gray fishing line. I am trying to figure out a means of releasing the fishing line when the white disk and drive wheel are stopped, and then the final motion would be linear. Then the puck can be caught as a pendulum bob and it will rise higher than what would be necessary to restart the motion.
energy producing experiments
Moderator: scott
re: energy producing experiments
I switched sides, now the fishing line (the white nylon cord was on the left) is on the right side of the gray puck, and of course the white disk spins the opposite direction.
I changed to a narrower ribbon.
I drilled a hole in the white disk and placed the fishing line through it, this holds the tether to a length of 4 3/32 inches to the center of mass of the gray puck.
I also added 700g at 7.75 inches from the center of the 9 inch radius drive wheel. This is roughly equal to 600 additional grams moving at the same speed as the ribbon. The white disk still appeared to stop; this I think puts us at 7.8 times the mass of the gray puck. This means that the gray puck is capable of finishing with 7.8 times the original energy. Final kinetic energy 1/2 * 456g * 7.8m/sec * 7.8m/sec; original ½ * 3556g *1m/sec * 1 m/sec
I changed to a narrower ribbon.
I drilled a hole in the white disk and placed the fishing line through it, this holds the tether to a length of 4 3/32 inches to the center of mass of the gray puck.
I also added 700g at 7.75 inches from the center of the 9 inch radius drive wheel. This is roughly equal to 600 additional grams moving at the same speed as the ribbon. The white disk still appeared to stop; this I think puts us at 7.8 times the mass of the gray puck. This means that the gray puck is capable of finishing with 7.8 times the original energy. Final kinetic energy 1/2 * 456g * 7.8m/sec * 7.8m/sec; original ½ * 3556g *1m/sec * 1 m/sec
- Attachments
-
-
re: energy producing experiments
I replaced the drive wheel with a sled of approximately equal inertia. The sled and the white disk were stopped as the gray puck swung out on the end of the tether (fishing line). You just quickly see the stop; it was not quite as distinct as the wheel. I found out why when I measured the mass of the sled. The sled had a mass of around 3.4 kg, much greater than the estimated inertial of the wheel (2.8kg). Video tapes will reveal the true nature of the observed stops. The wheel stop may actually be reversing, or the sled stop may not be quite coming to a complete stop (but it is close).
This should be a fairly good proof that linear motion and circular motion is one and the same thing. How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?
This should be a fairly good proof that linear motion and circular motion is one and the same thing. How are the angular momentum conservationists going to calculate angular momentum conservation when most of the original momentum is linear?
- Attachments
-
- linear motion stopped by the cylinder and spheres machine concept (sled, disk, and puck machine?)
re: energy producing experiments
I think if you take the reaction of the earth into account, while ignoring its daily rotation, the initial and final momenta will be zero, both linearly and angularly.
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.
re: energy producing experiments
Newton walked on the same Earth we do and he said F = ma. Newton did not place Earth motion in his formula. Newton’s momentum conservation was linear. You absolutely can not conserve both linear and angular and Newton did enough experiments to know this.
Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time. This will be linear Newtonian momentum; it is not angular momentum being produced. The angular momentum of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.
Construct two equal mass vertically mounted rim mass wheels with different radii. If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time. This will be linear Newtonian momentum; it is not angular momentum being produced. The angular momentum of the two rims are not equal. F = ma does not make angular momentum, and in fact no such angular momentum is ever conserved in the laboratory.
re: energy producing experiments
An example of the conditions under which angular momentum is conserved is where gravitation from the Sun increases the linear velocity of Halley’s Comet from .908 km/sec to 54.6 km/sec. Since your object is in the laboratory there is no analogous gravitational acceleration and therefore your angular momentum will not be conserved.
Only if you keep the radius of the circle the same will angular momentum remain the same, and under these conditions you have conducted no experiment.
Halley's Comet has a radius change but it also has a linear velocity change, this is he only way that angular momentum conservation works.
You can call it angular momentum if you wish but it will not be conserved in the laboratory, so what is the point of the terminology? In the lab the terminology is a hoax.
Only if you keep the radius of the circle the same will angular momentum remain the same, and under these conditions you have conducted no experiment.
Halley's Comet has a radius change but it also has a linear velocity change, this is he only way that angular momentum conservation works.
You can call it angular momentum if you wish but it will not be conserved in the laboratory, so what is the point of the terminology? In the lab the terminology is a hoax.
re: energy producing experiments
Kepler devised a formula to predict or describe the motion of satellites (planets, comets, etc). The formula takes the mass of the satellite m times the radians per second and then multiplies by the radius twice. m * radians/sec * radius * radius; this is the angular momentum formula. (a radians is equal to the number of radii of displacement per sec around the circumference of the circle, for instance; if an object is moving one meter per second around the circumference of a one meter (radius) circle it is moving at one radian per sec.; if an object is moving one meter per second around the circumference of a ten meter (radius) circle it is moving at 1/10 radian per sec. if an object is moving ten meter per second around the circumference of a one meter (radius) circle it is moving at ten radian per sec)
For objects moving at the same speed a large radius means the object will have a small radians per second quantity. For objects moving at the same speed a small radius means the object will have large radians per second quantity.
So by multiplying radians by radius you have the equivalent of linear motion. A motion of 5 m/sec in a 1 meter radius circle is 5 radians per second times a 1 m radius = 5 A motion of 5 m/sec in a 5 meter radius circle is 1 radians per second times a 5 m radius = 5
Looks good so far; but the formula requires that you multiply by one more radius. So the 1 meter circle above still has an angular momentum of 5, but the five meter circle has an angular momentum 25. Even though the object is moving at five meter per sec in both cases the angular momentums are different. So why were they equal for Kepler?
For Kepler’s satellites when the radius changed the velocity changed, at 5 time farther away the comet was moving 1/5 as fast. This change in velocity caused by the gravitational acceleration of the Sun is what makes the formula work. But the puck on the end of a 5 meter string that has the string hit a stationary pin one meter from the puck has no velocity change caused by the Sun. The puck was in a five meter (radius) circle but is now in a 1 meter circle and the velocity of the puck has not changed. What works in space will not work in the lab.
For objects moving at the same speed a large radius means the object will have a small radians per second quantity. For objects moving at the same speed a small radius means the object will have large radians per second quantity.
So by multiplying radians by radius you have the equivalent of linear motion. A motion of 5 m/sec in a 1 meter radius circle is 5 radians per second times a 1 m radius = 5 A motion of 5 m/sec in a 5 meter radius circle is 1 radians per second times a 5 m radius = 5
Looks good so far; but the formula requires that you multiply by one more radius. So the 1 meter circle above still has an angular momentum of 5, but the five meter circle has an angular momentum 25. Even though the object is moving at five meter per sec in both cases the angular momentums are different. So why were they equal for Kepler?
For Kepler’s satellites when the radius changed the velocity changed, at 5 time farther away the comet was moving 1/5 as fast. This change in velocity caused by the gravitational acceleration of the Sun is what makes the formula work. But the puck on the end of a 5 meter string that has the string hit a stationary pin one meter from the puck has no velocity change caused by the Sun. The puck was in a five meter (radius) circle but is now in a 1 meter circle and the velocity of the puck has not changed. What works in space will not work in the lab.
re: energy producing experiments
I'm probably wasting my breath, but it occurred to me that a mechanism I invented was designed, and only works, if angular momentum is conserved.
On the 15th you said,
"If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time."
But they will have different speeds, and it will take different times; I'm sure you can check. They must be different because the eccentric weight falls through different heights, since the radii are different.
On the 15th you said,
"If you hang the same mass from each rim the force of the hanging mass will accelerate each rim to the same circumference velocity in the same period of time."
But they will have different speeds, and it will take different times; I'm sure you can check. They must be different because the eccentric weight falls through different heights, since the radii are different.
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.
re: energy producing experiments
Take a 9 kg rim (or ring) that has a 2 meter diameter and place it vertically on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the ring will be roughly equal; you could place the string in a grove in the rim to make the speeds even closer. After one second all parts will be moving .981 m/sec. After one second everything will have moved .4905 m, the dangling mass will have moved down .4905 m and the mass in the ring will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum of the rim would be (mass * radians/sec * radius * radius) 9kg * .981 radians/sec *1m * 1m = 8.829
Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414
I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.
I am sure you could find similar experiments on the internet if you knew what the experimenter called it.
It is similar to an Atwood’s machine.
In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.
I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).
Take a 9 kg thin wall pipe that has a .1 meter diameter and place its length horizontally so that it rotates in a vertically plane on dry ice. Wrap a string around it and suspend a 1 kg mass from the string. This will cause 9.81 newtons of force to accelerate 10 kg for an acceleration rate of .981 m/sec². The speed of the string and the speed of the particles in the pipe will be roughly equal; you could place the string in a grove in the pipe to make the speeds even closer. After one second all particles will be moving .981 m/sec. After one second everything will have moved .4905 m, the hanging mass will have moved down .4905 m and the mass in the pipe will have moved around the circle .4905 m. This is from F = ma. After one second the angular momentum would be (mass * radians/sec * radius * radius) 9kg * 19.62 radians/sec *.05m * .05m = .4414
I did this experiment with about a 40 kg wheel 1.05 m in diameter. The calculations were a little more complex because the wheel had bearings with a small steel hub and steel spokes. The majority of the mass was in the 2 inch by 1-1/4 inch steel rim. The resulting data proved that F = ma; I probably did not even keep the data because the results where what I expected.
I am sure you could find similar experiments on the internet if you knew what the experimenter called it.
It is similar to an Atwood’s machine.
In the picture: I cut away the lower circle (forward of the entry hole) that allowed the fishing line to enter inside the circumference which determined tether length. This slows the reacceleration because now the line swings free and is attached closer to the point of rotation. This slowed reacceleration allows the experimenter to more easily observe the motion of the system as it comes to a stop.
I also lowered the center of mass of the sled by using rusty bars instead of the lead anchor ball. The sled has a mass of 2758g; the 456g gray puck easily stops the sled and white disk (300g rotational mass).
re: energy producing experiments
Pictured: Four masses (brass bushings) were used to accelerate the wheel and a flag (hexagonal wrench) past two photo gates held at a uniform distance. This should be enough information to determine if F = ma. Do you agree?
I took four consecutive readings that were within 2/10,000th of a second from each other, amazing isn’t it.
I took four consecutive readings that were within 2/10,000th of a second from each other, amazing isn’t it.
- Attachments
-
re: energy producing experiments
F = ma is a mathematical relationship between force, mass, and acceleration. To prove that the statement is true you must show that a proportional change in force causes a proportional change acceleration. For example if force increases by 20% then acceleration increases by 20%, etc.
The experiment; four different descending masses were hung from the ribbon, The force they produce is proportional to their mass, this mass is only a small portion of the accelerated mass. The four descending masses were a brass bushing or a combination of several brass bushings. Bushing 1 was 37.4g, bushing two was 26.7g, three was 26.5g, and the fourth was 73.0g
Descending mass 1 was 37.4g, Descending mass 2 was 64.1g (37.4 + 26.7), descending mass 3 was 90.6g (37.4 + 26.7 + 26.5g), descending mass 4 was163.6g (37.4 + 26.7 + 26.5g + 73.0g),
The time interval for each force to make the flag cross the same distance between the photo gates (from the same starting point) was; F1 .6957 sec, F2 .5336 sec, F3 .4556 sec, and F4 .3420 sec.
All forces worked over the same distance but they did not take the same amount of time to do it. To find acceleration from time we can us the formula d = ½ a * t * t because all the distance are the same the only variable are time and acceleration. Rearranging the formula to solve for a we get 2 * d / t * t = a, again distances are all the same so the proportional acceleration of F1 is 1 / .6957 *.6957 = 2.078, F2 is 1 / .5336 * .5336 = 3.512, F3 is 1 / .4556 *.4556 = 4.818, F4 is 1 / .3420 * .3420 = 8.55.
The proportion of force to the proportion of acceleration is close to as follows.
F1 37.4g 2.078
F2 64.1g 3.512
F3 90.6 g 4.818
F4 163.6g 8.55
But there is one more correction to make. The fourth force F4 is accelerating a larger mass than the first force. This is in a proportion to its total mass and the total mass accelerated by the smaller force F1, that proportion is the rotational mass (inertia of the wheel, 2600g) of the wheel plus each mass that caused the force. F4 accelerates the larger mass slower in a proportion of 2637.4g / 2763.6g = .954; this means that F4 would have accelerated the smaller mass 2637.4 faster giving it a higher acceleration by 1 / .954 = 1.048 * 8.55 = 8.96.
The proportion of force to the proportion of acceleration is as follows.
F1 37.4g 2.078
F2 64.1g 3.512 corrected for the greater mass being accelerated 2664.1g / 2637.4g to 3.55
F3 90.6 g 4.818 corrected for the greater mass being accelerated 2690.6g / 2637.4 to 4.92
F4 163.6g 8.55 corrected for the greater mass being accelerated 2763.6 / 2637.4 to 8.96
So does a proportional change in force (which equals a proportional change in descending mass) equal a proportional change in acceleration? Or does: 37.4g / 163.6g = 2.078 / 8.96
37.4g / 163.6g = .2286, 2.078 / 8.96 = .2319 .2319 / .2286 = 1.0145
37.4g / 64.1g = .5834, 2.078/3.55 = .5853 .5853/.5834 = 1.0033
37.4g / 90.6g = .4128, 2.078/4.92 = .4223 .4223/.4128 = 1.023
These are within two percent of a perfect F = ma relationship. So yes; F = ma is true for tangent forces working on the circumference of a circle. This is true even though the wheel is not a rim mass wheel, and this is Linear Newtonian Momentum.
The experiment; four different descending masses were hung from the ribbon, The force they produce is proportional to their mass, this mass is only a small portion of the accelerated mass. The four descending masses were a brass bushing or a combination of several brass bushings. Bushing 1 was 37.4g, bushing two was 26.7g, three was 26.5g, and the fourth was 73.0g
Descending mass 1 was 37.4g, Descending mass 2 was 64.1g (37.4 + 26.7), descending mass 3 was 90.6g (37.4 + 26.7 + 26.5g), descending mass 4 was163.6g (37.4 + 26.7 + 26.5g + 73.0g),
The time interval for each force to make the flag cross the same distance between the photo gates (from the same starting point) was; F1 .6957 sec, F2 .5336 sec, F3 .4556 sec, and F4 .3420 sec.
All forces worked over the same distance but they did not take the same amount of time to do it. To find acceleration from time we can us the formula d = ½ a * t * t because all the distance are the same the only variable are time and acceleration. Rearranging the formula to solve for a we get 2 * d / t * t = a, again distances are all the same so the proportional acceleration of F1 is 1 / .6957 *.6957 = 2.078, F2 is 1 / .5336 * .5336 = 3.512, F3 is 1 / .4556 *.4556 = 4.818, F4 is 1 / .3420 * .3420 = 8.55.
The proportion of force to the proportion of acceleration is close to as follows.
F1 37.4g 2.078
F2 64.1g 3.512
F3 90.6 g 4.818
F4 163.6g 8.55
But there is one more correction to make. The fourth force F4 is accelerating a larger mass than the first force. This is in a proportion to its total mass and the total mass accelerated by the smaller force F1, that proportion is the rotational mass (inertia of the wheel, 2600g) of the wheel plus each mass that caused the force. F4 accelerates the larger mass slower in a proportion of 2637.4g / 2763.6g = .954; this means that F4 would have accelerated the smaller mass 2637.4 faster giving it a higher acceleration by 1 / .954 = 1.048 * 8.55 = 8.96.
The proportion of force to the proportion of acceleration is as follows.
F1 37.4g 2.078
F2 64.1g 3.512 corrected for the greater mass being accelerated 2664.1g / 2637.4g to 3.55
F3 90.6 g 4.818 corrected for the greater mass being accelerated 2690.6g / 2637.4 to 4.92
F4 163.6g 8.55 corrected for the greater mass being accelerated 2763.6 / 2637.4 to 8.96
So does a proportional change in force (which equals a proportional change in descending mass) equal a proportional change in acceleration? Or does: 37.4g / 163.6g = 2.078 / 8.96
37.4g / 163.6g = .2286, 2.078 / 8.96 = .2319 .2319 / .2286 = 1.0145
37.4g / 64.1g = .5834, 2.078/3.55 = .5853 .5853/.5834 = 1.0033
37.4g / 90.6g = .4128, 2.078/4.92 = .4223 .4223/.4128 = 1.023
These are within two percent of a perfect F = ma relationship. So yes; F = ma is true for tangent forces working on the circumference of a circle. This is true even though the wheel is not a rim mass wheel, and this is Linear Newtonian Momentum.
re: energy producing experiments
I have been using the ribbon to drop different masses from the edge of the wheel. By using the photo gates to record their acceleration rate I can determine the rotational inertia of the wheel. The rotational inertia of the wheel is about 2,500g. That means the dropping mass accelerates as if it were accelerating a sled (using a pulley) with a mass of 2,500g.
For example; I dropped a 137g mass .5715m. This mass and the average mass of the wheel had a velocity of about .7692 m/sec. This is an acceleration rate of .5176m/sec². 9.81 /.5176 tells us the relative mass of the wheel to dropped mass.
So after a 137g mass has dropped only .5715m we have 2.02 (2,630g * .7692 m/sec) units of momentum.
By using the cylinder and spheres principal we can give all the motion to the 137g. For .137 kilograms to have 2.03 unit of momentum it will have to be moving 14.8 m/sec
The 137g object can rise 11.1 meters with a velocity of 14.8m. d = 1/2v²/a.
The 137g mass was only dropped .5715m. This is an energy increase of 19.4 times the original energy. 11.1m / .5715m.
For example; I dropped a 137g mass .5715m. This mass and the average mass of the wheel had a velocity of about .7692 m/sec. This is an acceleration rate of .5176m/sec². 9.81 /.5176 tells us the relative mass of the wheel to dropped mass.
So after a 137g mass has dropped only .5715m we have 2.02 (2,630g * .7692 m/sec) units of momentum.
By using the cylinder and spheres principal we can give all the motion to the 137g. For .137 kilograms to have 2.03 unit of momentum it will have to be moving 14.8 m/sec
The 137g object can rise 11.1 meters with a velocity of 14.8m. d = 1/2v²/a.
The 137g mass was only dropped .5715m. This is an energy increase of 19.4 times the original energy. 11.1m / .5715m.