L= mrv
L= m x 30 x 10
L = m x 15 x 20
The puck would accelerate to 20 cm per second in order to conserve angular momentum given to it.
If you did this on your table, you should have measured the velocity of the puck doubling. How did you measure the puck velocity?
My partial summary of pequaide's "energy producing experiments" thread
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eccentrically1
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re: My partial summary of pequaide's "energy producing
I timed with video.
If you melt your string is the puck going to drop back to 10 cm / sec? (go from 20 cm/sec to 10 cm/sec)
What if the big circle had a radius of 2 m?
How does the puck in the smaller circle know what velocity to drop back to. Does the puck know where its motion came from?
If you melt your string is the puck going to drop back to 10 cm / sec? (go from 20 cm/sec to 10 cm/sec)
What if the big circle had a radius of 2 m?
How does the puck in the smaller circle know what velocity to drop back to. Does the puck know where its motion came from?
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eccentrically1
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How do you measure velocity with video?
If you release the puck from 15 cm circular motion, it will not drop back to its original, 10 cm/sec, velocity, although it would then go towards the edge of the table and bounce off at some angle.
If the the big circle had a radius of 2 m, and you gave the puck the same push, it would have a slower velocity than it did from a string of 30 cm length because of the larger, 200 cm radius.
The puck doesn't know anything. It does what the laws of motion predict.
If you release the puck from 15 cm circular motion, it will not drop back to its original, 10 cm/sec, velocity, although it would then go towards the edge of the table and bounce off at some angle.
If the the big circle had a radius of 2 m, and you gave the puck the same push, it would have a slower velocity than it did from a string of 30 cm length because of the larger, 200 cm radius.
The puck doesn't know anything. It does what the laws of motion predict.
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Wubbly
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re: My partial summary of pequaide's "energy producing
pequaide wrote:
Here's an interesting link about a skater being pulled toward the center of the rink as a rope wraps around a center cylinder. In the example where no force is applied, linear momentum is being conserved, and angular momentum is being lost.
http://www.lhup.edu/~dsimanek/scenario/ ... htm#skater
You will probably never back down from your belief that angular momentum does not apply in the lab, even when shown an experiment that clearly demonstrates it does.
And you will probably never back down from your stance that you are creating energy in the lab.
We know you are not creating energy in the lab.
Even you know you are not creating energy in the lab.
Seriously pequaide, when are you going to stop your charade.
.
Well good. Galileo's pendulum is not an example of conservation of angular momentum. To use it to disprove Conservation of Angular Momentum is just silly.Galileo's pendulum proves that the halving of the radius does not change the linear velocity of the mass.
You don't know? It is obvious that he is applying a force. Also, in a gravitational example, a force is being applied between the two masses: the force of gravity. In the student experiment, a force must be applied to pull the masses together. Perhaps you missed the part where angular momentum was conserved in the lab.pequaide wrote:I don't know if the student is adding motion when he PULLS it in ...
No, the student knows he is applying a force. You are the one assuming that no force should be applied.pequaide wrote:The student assumes that he is adding no force
Well good. Hopefully angular momentum will not be conserved in this experiment. That is because this experiment you proposed should not (and does not) conserve angular momentum. The tangential linear velocity is the same immediately before and immediately after the string hits the pin. If your radius is suddenly half as small but with the same tangential velocity, then you just lost half your angular momemtum. To use this example to disprove conservation of angular momentum is just silly.pequaide wrote: ...Arrange for the string to be interrupted by a pin that extends from the table. ...
Plug these numbers into the angular momentum formula and it will not be conserved.
Here's an interesting link about a skater being pulled toward the center of the rink as a rope wraps around a center cylinder. In the example where no force is applied, linear momentum is being conserved, and angular momentum is being lost.
http://www.lhup.edu/~dsimanek/scenario/ ... htm#skater
Nobody is trying to impress anyone. We are trying to stop another unsuspecting soul from getting sucked into the pequaide vortex.pequaide wrote:It is just the difference between knowing the truth or impressing your friends.
You will probably never back down from your belief that angular momentum does not apply in the lab, even when shown an experiment that clearly demonstrates it does.
And you will probably never back down from your stance that you are creating energy in the lab.
We know you are not creating energy in the lab.
Even you know you are not creating energy in the lab.
Seriously pequaide, when are you going to stop your charade.
.
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Furcurequs
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re: My partial summary of pequaide's "energy producing
Hello pequaide,
I believe you understood my argument and I yours. I'm still trying to find more details about your actual experiments, though.
Thanks for directing me to the pictures. I still need to find whatever videos you may have up, however.
Just to make sure I understand your experimental setup, when you speak of the 19" plywood wheel and the 0.75 inch shaft, you are talking about the diameters, aren't you? ...and not the radii.
I would assume that's the case.
From looking at the other thread, I also saw that the wheel you use is made of two 0.75 inch thick pieces of plywood glued together, right?
Okay, after a little googling, I see that a 0.25 inch thick 4' by 8' sheet of plywood would normally be assumed to weigh about 25 pounds. ...but could, of course, actually be higher or lower depending upon the wood used and the glue content and such.
Using that figure, though, a 2 x 0.75" or 1.5" high stack of 4' by 8' plywood sheets should weigh about 150 pounds.
If we divide the area of your plywood disk by the top surface area of the stack of plywood sheets, then, we can multiply that number by 150 to get a ballpark figure for the mass of your wooden disk in lbm (pounds mass).
So, 3.14x(19/2)^2/(4x12x8x12)x150=9.2 lbm
Now, a solid disk (or cylinder) is supposed to have a moment of inertia (I) of 1/2 x m x r^2.
Since the moment of inertia of a hollow cylinder is supposed to be m x r^2, though, your disk could be replaced by a (most mass at the rim type of) flywheel with 4.6 lbm at the rim.
This seems to be fairly close to the amount of mass you are hanging on the rim in your experiments - and with a conventional understanding of things would be very significant. I have yet to do the calculations, but (again, using a conventional understanding) it looks like the inertial effects of the disk would be much greater than that of your suspended weights at the shaft.
You said something about assuming the Atwoods pulley to be massless? That is an oversimplification usually done in textbooks so they can concentrate on what the test is supposed to be measuring. ...but in an actual experiment, if one wanted decent results, the mass of the pulley would certainly have to be considered. In your case you have more of a huge flywheel than a lightweight pulley, it seems.
If you want to see how your experiment would behave with masses suspended from the shaft of a massless disk, I would suggest just removing the disk altogether to see.
My guess is that the suspended masses might move a bit faster than before.
To address some of the other posts...
In the experiment in the youtube video with the rotating hanger and batteries, when the fellow pulls down on the line(s), he is pulling in line with the spin axis - in line with the angular momentum vector. This shouldn't put a torque on the spinning hanger with the batteries and so angular momentum should be conserved.
By pulling down on the lines so that their redirected forces pull the batteries toward the center of rotation, though, he is pulling against the centrifugal force from the batteries, and so he is actually doing work and adding energy to the rotating hanger/battery system.
So, angular momentum would be conserved as not just the angular speed increased but also as the actual mass speed increased and thus as the energy of rotation (and kinetic energy of the masses) increased too.
This is similar to when ice skaters start a spin and then increase their spin rate by pulling in their arms and leg.
In pequaide's scenario on the air hockey table where a moving mass is circling a pin until the connecting string reaches another pin. The speed of the mass around the second pin (now constrained to a different path) would remain the same as before and there would be no energy gain.
You would calculate a different angular momentum for the new path around the new pin using the new radius, but it doesn't depend upon the previous calculated angular momentum for the different center of rotation.
...and if it sounds like I might have thought through these sorts of experiments before, I guess it would be fair to say that I have. ...if not maybe a little too much. ...lol.
(Wubbly, I typed this up last night before your latest post. I see you've now covered many of the same issues.)
Dwayne
I believe you understood my argument and I yours. I'm still trying to find more details about your actual experiments, though.
Thanks for directing me to the pictures. I still need to find whatever videos you may have up, however.
Just to make sure I understand your experimental setup, when you speak of the 19" plywood wheel and the 0.75 inch shaft, you are talking about the diameters, aren't you? ...and not the radii.
I would assume that's the case.
From looking at the other thread, I also saw that the wheel you use is made of two 0.75 inch thick pieces of plywood glued together, right?
Okay, after a little googling, I see that a 0.25 inch thick 4' by 8' sheet of plywood would normally be assumed to weigh about 25 pounds. ...but could, of course, actually be higher or lower depending upon the wood used and the glue content and such.
Using that figure, though, a 2 x 0.75" or 1.5" high stack of 4' by 8' plywood sheets should weigh about 150 pounds.
If we divide the area of your plywood disk by the top surface area of the stack of plywood sheets, then, we can multiply that number by 150 to get a ballpark figure for the mass of your wooden disk in lbm (pounds mass).
So, 3.14x(19/2)^2/(4x12x8x12)x150=9.2 lbm
Now, a solid disk (or cylinder) is supposed to have a moment of inertia (I) of 1/2 x m x r^2.
Since the moment of inertia of a hollow cylinder is supposed to be m x r^2, though, your disk could be replaced by a (most mass at the rim type of) flywheel with 4.6 lbm at the rim.
This seems to be fairly close to the amount of mass you are hanging on the rim in your experiments - and with a conventional understanding of things would be very significant. I have yet to do the calculations, but (again, using a conventional understanding) it looks like the inertial effects of the disk would be much greater than that of your suspended weights at the shaft.
You said something about assuming the Atwoods pulley to be massless? That is an oversimplification usually done in textbooks so they can concentrate on what the test is supposed to be measuring. ...but in an actual experiment, if one wanted decent results, the mass of the pulley would certainly have to be considered. In your case you have more of a huge flywheel than a lightweight pulley, it seems.
If you want to see how your experiment would behave with masses suspended from the shaft of a massless disk, I would suggest just removing the disk altogether to see.
My guess is that the suspended masses might move a bit faster than before.
To address some of the other posts...
In the experiment in the youtube video with the rotating hanger and batteries, when the fellow pulls down on the line(s), he is pulling in line with the spin axis - in line with the angular momentum vector. This shouldn't put a torque on the spinning hanger with the batteries and so angular momentum should be conserved.
By pulling down on the lines so that their redirected forces pull the batteries toward the center of rotation, though, he is pulling against the centrifugal force from the batteries, and so he is actually doing work and adding energy to the rotating hanger/battery system.
So, angular momentum would be conserved as not just the angular speed increased but also as the actual mass speed increased and thus as the energy of rotation (and kinetic energy of the masses) increased too.
This is similar to when ice skaters start a spin and then increase their spin rate by pulling in their arms and leg.
In pequaide's scenario on the air hockey table where a moving mass is circling a pin until the connecting string reaches another pin. The speed of the mass around the second pin (now constrained to a different path) would remain the same as before and there would be no energy gain.
You would calculate a different angular momentum for the new path around the new pin using the new radius, but it doesn't depend upon the previous calculated angular momentum for the different center of rotation.
Wubbly, maybe I should tie a safety line to my waist and then if you see it moving a little too fast, you might want to grab it real quick and try to extract some energy from it before it's too late. If it were long enough maybe you could even power your house for a few years. ..lolWubbly wrote:Try not to get dragged too far down pequaide's rabbit hole.
People have been trying to talk to him for years with no success.
...and if it sounds like I might have thought through these sorts of experiments before, I guess it would be fair to say that I have. ...if not maybe a little too much. ...lol.
(Wubbly, I typed this up last night before your latest post. I see you've now covered many of the same issues.)
Dwayne
I don't believe in conspiracies!
I prefer working alone.
I prefer working alone.
re: My partial summary of pequaide's "energy producing
Wow: I thought Laws were true all of the time not just some of the time.
If Galileo's pendulum is an event where angular momentum is not conserved then Angular Momentum Conservation is not a law. That is the point: that is why I used it: that is why angular momentum conservation in the lab is not a law.
You dismiss the puck experiment as being one of those experiments that does not conserve angular momentum. Well ya; that is the point; the experiment does not conserve angular momentum.
Further: since when are you allowed the add energy or motion and then still claim it is a conserved quantity. You admit that you are adding energy and motion to the batteries but you still claim it is a conserved quantity. I thought you had to end with what you started with, not add more along the way and then still claim conservation. You have to add motion to conserve your quantity: I thought that was cheating. Next you will allow Bessler to have a rope running to a hidden room and being pulled by a servant.
Conserved means nothing added.
Kepler's angular momentum conservation works in space, but it is a rabbit trail in the lab. A trail that many go down.
I have made free energy in the lab in two different ways. Both methods us Newtonian Physics.
Yes the wheel has mass. But the wheel contains the same quantity of motion with the two 20.54 kg at the shaft as it does for the two .900 kg at the circumference. The wheel is moving at the same speed in both arrangements so the momentum or energy is the same. The bearing resistance is the same as well.
Yes the 19 inch and .75 inches are diameters.
Moment of inertia goes hand in hand with angular momentum conservation.
I calculate inertial mass by actually spinning the wheels (F = ma). I can't remember measuring the inertia of the 19 inch wheel. But it is the same in both arrangements. I sometimes measure bearing resistance, but again it is the same with two 20.54 kg at the shaft or two .900 kg at the circumference.
Wubbly: Thanks for posting Galileo's pendulum.
If Galileo's pendulum is an event where angular momentum is not conserved then Angular Momentum Conservation is not a law. That is the point: that is why I used it: that is why angular momentum conservation in the lab is not a law.
You dismiss the puck experiment as being one of those experiments that does not conserve angular momentum. Well ya; that is the point; the experiment does not conserve angular momentum.
Further: since when are you allowed the add energy or motion and then still claim it is a conserved quantity. You admit that you are adding energy and motion to the batteries but you still claim it is a conserved quantity. I thought you had to end with what you started with, not add more along the way and then still claim conservation. You have to add motion to conserve your quantity: I thought that was cheating. Next you will allow Bessler to have a rope running to a hidden room and being pulled by a servant.
Conserved means nothing added.
Kepler's angular momentum conservation works in space, but it is a rabbit trail in the lab. A trail that many go down.
I have made free energy in the lab in two different ways. Both methods us Newtonian Physics.
Yes the wheel has mass. But the wheel contains the same quantity of motion with the two 20.54 kg at the shaft as it does for the two .900 kg at the circumference. The wheel is moving at the same speed in both arrangements so the momentum or energy is the same. The bearing resistance is the same as well.
Yes the 19 inch and .75 inches are diameters.
Moment of inertia goes hand in hand with angular momentum conservation.
I calculate inertial mass by actually spinning the wheels (F = ma). I can't remember measuring the inertia of the 19 inch wheel. But it is the same in both arrangements. I sometimes measure bearing resistance, but again it is the same with two 20.54 kg at the shaft or two .900 kg at the circumference.
Wubbly: Thanks for posting Galileo's pendulum.
re: My partial summary of pequaide's "energy producing
Thanks Wubbly for posting the link to Donald Simanek's insight page.
It is easy to see that in the case of a rope winding around a capstan the skaters tangential velocity is unchanged, but in the case of the rope being wound in over a central rotating pivot that work is being done on the skater & their tangential velocity increases accordingly.
My personal experience of this was when once paragliding behind a speed boat & being winched in to land on the deck whilst in a gentle turn.
Most reading this thread I believe understand that the simple rotating hanger & shifting batteries video was an example of CoAM & that the deflection to the puck example was not.
It is easy to see that in the case of a rope winding around a capstan the skaters tangential velocity is unchanged, but in the case of the rope being wound in over a central rotating pivot that work is being done on the skater & their tangential velocity increases accordingly.
My personal experience of this was when once paragliding behind a speed boat & being winched in to land on the deck whilst in a gentle turn.
Most reading this thread I believe understand that the simple rotating hanger & shifting batteries video was an example of CoAM & that the deflection to the puck example was not.
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Wubbly
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re: My partial summary of pequaide's "energy producing
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Dwayne, his videos can be accessed from here: http://www.besslerwheel.com/forum/viewt ... 9797#99797
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Dwayne, his videos can be accessed from here: http://www.besslerwheel.com/forum/viewt ... 9797#99797
I'm sorry. It's already too late.Dwayne wrote:maybe I should tie a safety line to my waist and then if you see it moving a little too fast, you might want to grab it real quick .. before it's too late.
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eccentrically1
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re: My partial summary of pequaide's "energy producing
I went out to the lab and removed both 20.54 kg masses. That left the ropes hanging with only the 7.5 pounds on the one side of the shaft and nothing at the circumference. The wheel rotated through one half rotation, just as in all the runs, in an average of 1.46 second. Now this includes bearing resistance so you still can't calculate the exact inertia of the wheel, but it gives you a rough idea. Yes it has inertia. With two 20.54 kg at the shaft it slows to about 2.00 second. and with two .900 kg at the circumference it slows to about 2.00 second.
Free fall for .0334 m is only .08 second so much of the force is used upon the wheel inertia and bearing resistance. Is this a dilemma or a challenge? Well that depends on who you are. They are thick grease industrial bearings, much like that ball with honey. And yes the wheel has mass. But the concept remains sound. Use better bearing (or none) and reduce the wheel mass (or use the mass).
I did this with a huge wheel and the results were the same. I did use the photo gates on that experiment.
I also did this experiment with a tube, with a cord for a bearing. the photo gate were used here too; same results.
Free fall for .0334 m is only .08 second so much of the force is used upon the wheel inertia and bearing resistance. Is this a dilemma or a challenge? Well that depends on who you are. They are thick grease industrial bearings, much like that ball with honey. And yes the wheel has mass. But the concept remains sound. Use better bearing (or none) and reduce the wheel mass (or use the mass).
I did this with a huge wheel and the results were the same. I did use the photo gates on that experiment.
I also did this experiment with a tube, with a cord for a bearing. the photo gate were used here too; same results.
re: My partial summary of pequaide's "energy producing
I went to the 19 inch plywood wheel and suspended 2.5 pounds from the shaft. The 2.5 pounds would not rotate the wheel. If you started it moving it would slowly stop. So I am going to guess that 2.5 pound is the force necessary to overcome the bearing resistance.
I then placed 5 pounds on the wheel at the shaft and it rotated one half rotation in 1.9 seconds. This means that half of the 5 pounds is used to overcome bearing resistance and the other 2.5 pounds is used to accelerate the inertia of the wheel (I know very rough). If I did the math correctly that places the inertia of the wheel at about 1.2 kilograms at the circumference. Or it acts like a 1.2 kilogram ring. Very roughly equal to two .900 kilogram masses at the circumference.
I also noted a statement at the top of page 89.
I think the inertia of the wheel is less than 1.8 kilograms.
This is consistent with the fact that we can place another 2.5 pounds (total of 7.5 pounds) on the shaft and another 1.8 kilograms (2 * .900 kg) on the circumference and we return to the 1.9 second territory at 2.00 second.
Do you think a wheel could be made of Styrofoam?
I then placed 5 pounds on the wheel at the shaft and it rotated one half rotation in 1.9 seconds. This means that half of the 5 pounds is used to overcome bearing resistance and the other 2.5 pounds is used to accelerate the inertia of the wheel (I know very rough). If I did the math correctly that places the inertia of the wheel at about 1.2 kilograms at the circumference. Or it acts like a 1.2 kilogram ring. Very roughly equal to two .900 kilogram masses at the circumference.
I also noted a statement at the top of page 89.
I think the inertia of the wheel is less than 1.8 kilograms.
This is consistent with the fact that we can place another 2.5 pounds (total of 7.5 pounds) on the shaft and another 1.8 kilograms (2 * .900 kg) on the circumference and we return to the 1.9 second territory at 2.00 second.
Do you think a wheel could be made of Styrofoam?
re: My partial summary of pequaide's "energy producing
Hi pequaide
Styrofoam......hmmm maybe, but as well as making lighter wheels why not spend some time reducing the friction?
This would surely increase the accuracy of your experiments and negate a stumbling block to acceptance of your results.
The early pioneers went to a lot of trouble in order to achieve accuracy.
http://physics.kenyon.edu/EarlyApparatu ... chine.html
Just washing out the grease and running dry (assuming you're using ball races) should be worthwhile.
Regards
Mick
Styrofoam......hmmm maybe, but as well as making lighter wheels why not spend some time reducing the friction?
This would surely increase the accuracy of your experiments and negate a stumbling block to acceptance of your results.
The early pioneers went to a lot of trouble in order to achieve accuracy.
http://physics.kenyon.edu/EarlyApparatu ... chine.html
Just washing out the grease and running dry (assuming you're using ball races) should be worthwhile.
Regards
Mick
re: My partial summary of pequaide's "energy producing
I see it totally differently. Having no bearing resistance would be impressive, but bearing resistance does not place a question mark on the experimental results. If the bearing resistance was erratic then you would have a problem. But the bearing resistance is consistent; and of course the wheel inertia is consistent. The 7.5 pounds is used to accelerate the wheel to .38 m/sec/sec in all runs. What portion of the 7.5 pound force that has been used to overcome bearing resistance and the inertia of the wheel is consistent and can be factored out of the equation.
With 41.08 kilograms rotating at the shaft the wheel does one half rotation in about 2.00 second. If you run the test 20 time each data point will be a little over or under of two second. Maybe one or two data points will actually hit 2.00, but it is consistent not erratic.
Now you remove the 41.08 kilograms and replace them with 1.800 kilograms at the 19 inch circumference. Then you make the runs again and exactly the same thing happens. If you run the test 20 time each data point will be a little over or under of two second. Maybe one or two data points will actually hit 2.00, but it is consistent not erratic.
If the 1.8 kilogram at 19 inches gave you more resistance than the 41.08 kilograms at the shaft then the wheel would have to rotate slower, because nothing else has changed. The bearing resistance is the same and the wheel inertia is the same. The same amount of force has to be used to overcome bearing resistance and wheel inertia in order to remain at the 2.00 second rotation. The only way to get the same results is for the 41.08 kg to 1.800 kg change to also be the same.
It is sometimes comforting to remember what the opponents are proposing; which is that the 1.8 kg at 19 inches is 22.8 times harder to rotate. Ya right; and you are still at 2.00 seconds!
Reducing the bearing resistance is a worthy goal, but it is not necessary to prove that the concept is correct. The concept has already been proven to be correct.
It is just as easy to accelerate 10 kilograms to one meter per second per second as it is to accelerate 1 kilogram to ten meters per second per second. Or F = ma
With 41.08 kilograms rotating at the shaft the wheel does one half rotation in about 2.00 second. If you run the test 20 time each data point will be a little over or under of two second. Maybe one or two data points will actually hit 2.00, but it is consistent not erratic.
Now you remove the 41.08 kilograms and replace them with 1.800 kilograms at the 19 inch circumference. Then you make the runs again and exactly the same thing happens. If you run the test 20 time each data point will be a little over or under of two second. Maybe one or two data points will actually hit 2.00, but it is consistent not erratic.
If the 1.8 kilogram at 19 inches gave you more resistance than the 41.08 kilograms at the shaft then the wheel would have to rotate slower, because nothing else has changed. The bearing resistance is the same and the wheel inertia is the same. The same amount of force has to be used to overcome bearing resistance and wheel inertia in order to remain at the 2.00 second rotation. The only way to get the same results is for the 41.08 kg to 1.800 kg change to also be the same.
It is sometimes comforting to remember what the opponents are proposing; which is that the 1.8 kg at 19 inches is 22.8 times harder to rotate. Ya right; and you are still at 2.00 seconds!
Reducing the bearing resistance is a worthy goal, but it is not necessary to prove that the concept is correct. The concept has already been proven to be correct.
It is just as easy to accelerate 10 kilograms to one meter per second per second as it is to accelerate 1 kilogram to ten meters per second per second. Or F = ma
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Wubbly
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re: My partial summary of pequaide's "energy producing
pequaide, do you really understand your system?
What force is being applied to the system? It is the 7.5 pound (3.4 kg) mass times the acceleration due to gravity (9.81 m/s²) which gives you 33.4 Newtons of force.
Force = 33.4 N
The 7.5 pound mass (3.4 kg mass) is hung at the 3/4" diameter (0.009535 meter radius). If the wheel turns through 180 degrees, this means the 3.4 kg (7.5 pound) mass dropped 2.99 centimeters (0.0299 meters).
We know the time is 2.00 seconds, so plugging the distance and time into the linear acceleration equation gives us an linear acceleration of a=0.01495 m/sec² at the 0.009535 meter (3/8") radius, which equates to pequaide's linear acceleration of 0.38 m/sec² at the 9.5" (or 0.2413 meter) radius.
We know the force is 33.4N.
We know the linear acceleration at the 3/8" radius is 0.01495 m/sec².
We can calculate the mass of the system.
Using F=ma
mass=F/a = 33.4[N]/0.01495[m/sec²] = 2234 kilograms.
Pequaide's system is acting like it has a mass of 2234 kilograms.
Where is all the missing mass?
You have a 7.5 pound (3.4 kg) mass hanging at a radius of 0.009535 meters (3/8").
You also have two 20.54 kg masses hanging off each side at a radius of 0.009535 meter (3/8").
That would account for (3.4 kg + 20.54 kg + 20.54 kg) = 44.48 kg.
Where is the missing 2189.52 kilograms of system mass!
The only thing left is the plywood pully and the stiff bearings.
Pequaide's 19" diameter pulley is constructed of two sheets of 3/4" plywood.
Based on the weight of 3/4" plywood, his pulley should weigh about 4.6 pounds, or about 2.1 kg.
So the total mass we can account for is the hanging driver mass, the two additional balanced hanging masses, and the pulley mass: (3.4 kg + 20.54 kg + 20.54 kg + 2.1 kg) = 46.6 kg.
pequaide, your system is accelerating like it is 2234 kg, but you can only account for 46.6 kg of the mass. Have you figured out where 97.9% of your mass is hiding?
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What force is being applied to the system? It is the 7.5 pound (3.4 kg) mass times the acceleration due to gravity (9.81 m/s²) which gives you 33.4 Newtons of force.
Force = 33.4 N
The 7.5 pound mass (3.4 kg mass) is hung at the 3/4" diameter (0.009535 meter radius). If the wheel turns through 180 degrees, this means the 3.4 kg (7.5 pound) mass dropped 2.99 centimeters (0.0299 meters).
We know the time is 2.00 seconds, so plugging the distance and time into the linear acceleration equation gives us an linear acceleration of a=0.01495 m/sec² at the 0.009535 meter (3/8") radius, which equates to pequaide's linear acceleration of 0.38 m/sec² at the 9.5" (or 0.2413 meter) radius.
We know the force is 33.4N.
We know the linear acceleration at the 3/8" radius is 0.01495 m/sec².
We can calculate the mass of the system.
Using F=ma
mass=F/a = 33.4[N]/0.01495[m/sec²] = 2234 kilograms.
Pequaide's system is acting like it has a mass of 2234 kilograms.
Where is all the missing mass?
You have a 7.5 pound (3.4 kg) mass hanging at a radius of 0.009535 meters (3/8").
You also have two 20.54 kg masses hanging off each side at a radius of 0.009535 meter (3/8").
That would account for (3.4 kg + 20.54 kg + 20.54 kg) = 44.48 kg.
Where is the missing 2189.52 kilograms of system mass!
The only thing left is the plywood pully and the stiff bearings.
Pequaide's 19" diameter pulley is constructed of two sheets of 3/4" plywood.
Based on the weight of 3/4" plywood, his pulley should weigh about 4.6 pounds, or about 2.1 kg.
So the total mass we can account for is the hanging driver mass, the two additional balanced hanging masses, and the pulley mass: (3.4 kg + 20.54 kg + 20.54 kg + 2.1 kg) = 46.6 kg.
pequaide, your system is accelerating like it is 2234 kg, but you can only account for 46.6 kg of the mass. Have you figured out where 97.9% of your mass is hiding?
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re: My partial summary of pequaide's "energy producing
bearing resistance
What is important is that 41.08 at the shaft acts like 1.8 at the circumference. I don't care what is going on in the rest of the system.
There are ways to nearly eliminate friction; I am not concerned in the least.
It acts like 2234 in both arrangements doesn't it. If it bothers you then don't experiment.
It just tells you what a thick grease bearing does to motion.
What is important is that 41.08 at the shaft acts like 1.8 at the circumference. I don't care what is going on in the rest of the system.
There are ways to nearly eliminate friction; I am not concerned in the least.
It acts like 2234 in both arrangements doesn't it. If it bothers you then don't experiment.
It just tells you what a thick grease bearing does to motion.