Furcurequs (aka Dwayne) questions Jim_Mich

[pequaide]

http://video.mit.edu/watch/mit-physics- ... hine-3138/

Yes: the 10 grams is causing all the motion; it is the only unbalanced force. This is 10 grams accelerating 1110 grams.

I did not say their experiment made energy. But add to this the experimental concept proven by the double and triple Atwood's and you have a means of making energy. This concept is that the Laws of Levers applies to Atwood's pulleys connected over the same axis but with different radii. As described below.

Let’s say that the radius of the pulley is 8cm.

550 grams on the left side at 8 cm places a certain torque on the point of rotation.  This torque is perfectly countered with an equal amount of torque from 550 grams on the right side at 8 cm.

The extra 10 grams on the right side is the accelerating force of .01 kg * 9.81 N/kg = .0981N which is expressed as torque upon the center point of rotation. These 10 grams of measured force remains constant throughout the acceleration. There is not one force for the static position and then another force for the acceleration. Acceleration can be determined from F = ma; F/m = a; .0981 N / 1.110 kg = .08838 m/sec²

The individual torque of each 550 grams was caused by the gravitational pull on the mass. The one side’s torque canceled the others side’s torque but the inertia of the 550 (1100 grams) grams is not canceled.  This inertia is proportional to the previously existing torque.

In Newtonian Physics inertia is measured by the change in momentum (linear Newtonian momentum) caused by an applied force. The change in linear Newtonian momentum of both 550 gram sides is equal. The bobs are viewed as going in the same positive direction.

55 grams at 80 cm has an equal amount of torque as 550 grams at 8 cm. If you try to move the 55 grams at 80 cm you will find that it acts like it has an equal amount of inertia, from the point of rotation, as 550 grams at 8 cm.  When moved; it also has an equal amount of linear Newtonian momentum, and an equal amount of centrifugal force. The system is balanced and it acts like it.

The 10 gram will accelerate the 55 g at 80 cm just as easily as 550 g at 8 cm; in fact it will accelerate two 55 grams on each side at 80 cm just as easily as it accelerates two 550 grams at 8 cm on both sides. The final velocity for the 10 grams (left at the 8 cm position) with two 55 grams at 80 cm will be exactly the same as when you had two 550 grams at 8 cm; with the additional 10 grams at the 8 cm position. And the velocity of the 10 grams will be .42 m/sec. Velocity can be determined by a rearrangement of the distance formula: v = the square root of (2 * d * a) = the square root of (2 * 1 m * .08837m/sec²) = .42 m/sec

After a drop of one meter for the 10 grams; the 110 (55 + 55) grams on a 80 cm wheel will be moving 4.2 m/sec; this (1/2mv²) is .971 joules of energy. The input energy was (10 grams dropped one meter.01 kg * 9.81 N/kg *1 m =) .0981 joules: N * m. Roughly 10 times the energy

For a logical proof place 550 grams on one side at 8 cm and 54 grams on the other side at 80 cm. The 550 grams will rotate the 54 grams.  Place 550 grams on one side at 8cm and 56 grams on the other side at 80 cm. The 550 grams will be rotated by the 56 grams. With 55 gram on the other side at 80 cm there will be no rotation.

Now place 550 grams on one side at 8cm and 549 grams on the other side at 8 cm. The 550 grams will rotate the 549 grams.  Place 550 grams on one side at 8 cm and 551 grams on the other side at 8 cm. The 550 grams will be rotated by the 551 grams.  With 550 grams on both sides there will be no rotation.

Conclusion: the Difficulty of rotating 55 grams at 80 cm is somewhere between the difficulty of rotating 549 grams at 8 cm and 551 grams at 8 cm.

When an outside force is asked to accelerate 55 grams at 80 cm it will find it as difficult (somewhere between 549 grams at 8 cm and 551 grams at 8 cm) as rotating 550 grams at 8 cm.  Don’t buy into the myth that a different force takes over as soon as motion begins. Their experiment proves that the 10 grams of force starts the motion and it stays with the system throughout the one meter drop. The 4.80 seconds is predicted for the 10 grams of force (.0981 N) by F = ma. d = 1/2at² = 4.757 seconds.

The 110 grams moving 4.2 m/sec (.42 m/sec * 80 cm / 8 cm) could rise .889 meters for .110 kg * 9.81 N/kg  * .889 m= N * m = .9702 joules

Detractors will try to angular momentum you; but Atwood’s are linear and Angular Momentum is for space.

[rlortie]

pequaide,

I am not insulting or insinuating, but you leave me pondering.

Is it possible that you suffer from "autism"? A disorder of neural development characterized by impaired social interaction and verbal and non-verbal communication, and by restricted, repetitive or stereotyped behavior.

You keep shelling out the same data time after time, but fail to respond to direct questions put before you. Your response is always another ploy on the same experiments.

Repetitively stacking or lining up objects is a behavior sometimes associated with individuals with autism, also effecting one's social behavior.

If you are autistic, it is not something to be ashamed of, you are not alone by any standards. There are other autistic members here One exploits his autistic skills, and readily admits that he has autism, he has stated that his social skills are limited, so do not expect to get a hug from him.

Ralph

[Grimer]

I think we are all nuts in one way or another, Ralph - or we wouldn't be here. ;-)

[rlortie]

Grimer,

A search for "nuts" definition other than reference to seeds is hard to come by, but I did find the following:

Signifies rejection of a proposal or idea, as in forget it, no way, or nothing doing.

I do not believe we are nuts, we are judged as such by those who reject our pursuit and beliefs. It is a personal trait of closed minded teachings.

Ralph

[Fletcher]

'nuts' means crazy, insane, 'nutty as a fruit cake', a fruit loop, bottle short of a six pack, moron.

[pequaide]

Unbelievable childish behavior; F an R

Actually R and F I was responding to mark, He seems to have a hard time thinking that only 10 grams could cause so much motion. But 10 g / 1110 g * 9.81 m/sec for an acceleration; gives you 4.757 seconds. Which is confirmed by the experiment.

You don't have to read anything rlortie; and I will save my time by not reading your posts.

[Tarsier79]

http://en.wikipedia.org/wiki/Na%C3%AFve_physics :

55 grams at 80 cm has an equal amount of torque as 550 grams at 8 cm. If you try to move the 55 grams at 80 cm you will find that it acts like it has an equal amount of inertia, from the point of rotation, as 550 grams at 8 cm. When moved; it also has an equal amount of linear Newtonian momentum, and an equal amount of centrifugal force. The system is balanced and it acts like it.

55g at 80cm will have the same force as 550g at 8cm, but different inertia. This has been proven by experiment. Your own biased views are making you blind.

[ovyyus]

'nuts' means crazy, insane, 'nutty as a fruit cake', a fruit loop, bottle short of a six pack, moron.

Lucky you didn't use the.......Retard word Fletcher... touchy subject in the asylum :D

[rlortie]

Actually I check in with Crank Dot Net http://www.crank.net/
daily to see if "R" has made the grade yet.

It is pleasing to see that the whole forum made it!

[rlortie]

Anticrank:

'Perpetual Motion? What a foolish notion! Only cranks and crackpots think such a thing is possible; only fools are interested in such things.' That's the reaction of many people, and most scientists. But let's consider the matter by looking at the long history of this subject and its relation to science. The notion of perpetual motion is interesting from the perspective of history of science and technology, for it illustrates both the search for a perfect machine and the development of our modern understanding of the energy principle in physics. Besides, perpetual motion machines and their inventors provide a case study of human psychology: ingenuity, persistence, optimism and fanatism, even in the face of repeated failures."

[Mark]

So..... I guess us self-proclaimed Eccentrics are a subset of cranks and crackpots. Good to know.

:P

[Mark]

Pequaide

http://video.mit.edu/watch/mit-physics- ... hine-3138/

This is only 10 gram accelerating 1110g for 5 seconds.

Sorry pequaide, but I disagree. You infer that the 10 Gram differential is doing the work all by itself.

It IS NOT: 10 grams accelerating 1110 grams for 5 seconds.

It IS: Gravity accelerating a 560 gram mass while simultaneously decelerating a 550 gram mass through the distance of 1 meter.

There is NO creation of energy!

Yes: the 10 grams is causing all the motion; it is the only unbalanced force. This is 10 grams accelerating 1110 grams.
I did not say their experiment made energy. But add to this the experimental concept proven by the double and triple Atwood's and you have a means of making energy.

I didn't say that you said that there was creation of energy, but experience suggests that that's where you were most likely headed with it. It's certainly the theme of the majority of your posts.

You then go into an 800 word tutelage [which you needn't have done, I already "got it"] and end with:

Detractors will try to angular momentum you; but Atwood’s are linear and Angular Momentum is for space.

I would very much like to see your explanation as to why you believe the pulley in an Atwood's does not have angular momentum.


Now, to get back to the root of this issue. A couple of hours later, after some comments by other members, you also say:

... I was responding to mark, He seems to have a hard time thinking that only 10 grams could cause so much motion.

Not at all, Pequaide. But what I do have a hard time with, is that sometimes the things that you write don't match the facts that you say are correct.

For example, you still insist "This is only 10 gram accelerating 1110g for 5 seconds."
But I say that it's really "Gravity accelerating a 560 gram mass while simultaneously decelerating a 550 gram mass through the distance of 1 meter."

Look at the attached image while thinking of each of our two statements.
Then look at the video again while thinking of each of our two statements.

Now, tell me Pequaide, which quote most accurately describes the attached image?
And, which quote most accurately describes the demonstration in the video?

[Mark]

Note to everyone that goes to the trouble of posting simulator results --

My previous comments about wm2d were not in regards to any ill-will towards simware, as I have none, but were just to convey the thought that it does help those of us that don't have simware when you also post frame captures so that we can see what you're talking about. The additional work involved does not go unappreciated.
My apologies for not stating that more clearly before, and for any misunderstanding that may have arisen from it.

[pequaide]

Figure A is most like the Atwood's.

In figure A the center of mass of the 1,100 grams is on a frictionless plane and is not rising or lowering. The 1100 g is being accelerated by 10 grams.

In the MIT Atwood's the center of mass (550 g on one side and 550 g on the other) is not rising or lowering, and the acceleration is identical to that of the mass on the frictionless plane.

Both accelerations will be 10 g /1110 g * 9.81 m/sec². This would be an acceleration of .08838 m/sec². The drop time (of the ten grams) for one meter would be 4.757 seconds.

On the net: I sometimes see the mass of the force (10g) being left out of the total mass (1110 g) being accelerated. This is not correct. The easiest way to remember this is to think of the mass in free fall; it is 10 grams of force accelerating 10 grams of mass; 10 / 10 * 9.81 m/sec². It is more obvious when the mass force is half of the total mass, 10 grams hanging from the pulley and ten grams on the frictionless plane. 10 g / 20 g * 9.81 for half standard gravitational acceleration.

Figure A has no angular momentum and neither does an Atwood's. All mass (1110g) in the Atwood's is going in a straight line. Pulley radius means nothing; if the force and the accelerated mass are at the same radius. If the Force or portions of the mass are in different radii; then Laws of Levers applies.

It is assumed that the pulley is almost mass-less and almost frictionless; the end data proves that both of these assumptions are correct.

Force times distance (Leibniz) and Force times time (Newton) do not concur with each other. Both men knew that one or the other was wrong. If you assume that Leibniz’s mv² is correct then you might just as well give up Bessler's or any other means of making energy. But if you think Newton's (mv) may have been correct then you need to use the momentum produced by the application of large quantities of time.

[Wubbly]

From: http://www.besslerwheel.com/forum/viewt ... 402#104402
jim_mich wrote:
Quote:
Weights on rod = 1kg each
Rod weights are 1m from the center.
Pulley has a radius of 10cm
Falling mass = 500g

Is there a formula that will tell me how much time and how far the weight would need to fall to reach 50RPM?

Radius of Gyration formula:
RoG = (a²×W+b²×w)÷(a×W+b×w)

W = 2000 g (two 1_kg weights at ends of rod)
a = 100 cm (radial distance of weight 'W')
w = 500 g (dropping weight)
b = 10 cm (radius of pulley)

Filling in the values:
RoG = (100²×2000+10²×500)÷(100×2000+10×500)

Radius of Gyration result:
RoG = 97.804878 cm

In other words, the arrangement will accelerate and rotate as if it were a single 2500g weight located at 97.804878 cm radius, sans gravity.

The 500 kg dropping weight is at 10 cm while the RoG is at 97.8 cm, so the torque force is leveraged.
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.

The desired speed is 50 RPM. The RoG is 97.804878.
Note that Radius of Gyration value must be used as the radius of the moving weight, else the calculations WILL NOT BE CORRECT.
I've seen some members just grab any old radius that they come up with from who know where.
If you want accurate calculations concerning rotating objects then you MUST reduced everything that is rotating down to equivalant Radius of Gyration values.

So the circumference at RoG is 2×pi×97.804878 = 614.526172727 cm.
The desired velocity is 614.526172727 × 50 RPM = 30726.3086364 cm/min.
Or 30726.3086364 ÷ 60_seconds = 512.10514394 cm/s velocity desired.
Or 5.1210514394 m/s velocity desired.

We want to know the time needed to accelerate 2500 grams up to a Radius of Gyration velocity of 5.1210514394 m/s using a force of 9.7804878 grams where the gravitational acceleration is 9.80665_m/s².

Time = (M × V) ÷ (Ga × F)
Time = (2500_g × 5.121_m/s) ÷ (9.80665_m/s² × 9.7804878_g)
Time = 133.48 seconds to accelerate the mechanism up to a speed of 50 RPM.

Final velocity of dropping weight = 50_RPM × 10_cm × 2 × pi = 3141_cm/min = 52.36_cm/s
Distance dropped = (V×T)÷2
Distance dropped = 52.36_cm/s × 133.48_s ÷ 2 = 3494.5_cm or about 114.65 feet.


Hopefully I've made no mistakes. I make no guarantee. I've checked and double checked and believe this is the correct answer.



I made one small error...
The line that reads:
It will be 97.804878÷10 = 9.7804878_g of force acting on 2500 g of mass.
Should be:
It will be 500_g × 97.804878÷10 = 51.1222_g of force[/color] acting on 2500 g of mass.
By mistake I failed to multiply by the 500_g.

And of course this error trickles down through the rest of the calculations.
The bottom line results should be a time of 25.5367 seconds to accelerate the mechanism up to a speed of 50 RPM.

As a double check I just now went back and simulated it using WM2D, which came up with an answer of 21.358 seconds. The discrepancy is that with my calculations the weight masses out on the ends of the rod are fixed to the rods and thus are rotated, whereas the weights of WM2D were allowed to free-wheel and thus not rotate. Fixing the weights to the ends of the rods rather than letting them free-wheel results in it taking a slightly longer time to accelerate up to speed. In this case about 4.17 seconds longer.

I suppose I could redo the WM2D simulation and fix the weights so they don't free-wheel, but I've already wasted enough time on this.

This method, which uses radius of gyration works, contrary to what Furcurequs claims.

From now on I'm never going to try to help anyone, for if I make a mistake (we are all human) then the trolls here crucify me.

You could also look at the problem as an Atwoods.
The Atwoods acceleration equation was derived in this post here: http://www.besslerwheel.com/forum/viewt ... 6980#96980

The parameters are:
m1 = 0.5 kg
m2 = 1.0 kg
m3 = 1.0 kg
Flywheel moment of Inertia = 0 (massless pulley - in this case massless rods)
r1 = 0.1 meter
scale factor = 10 (to make r2 = 1 meter)

The final output desired is an RPM of 50. At a radius of 1 meter, this would equate to m2 and m3 having a linear speed of 5.236 meters/second.

By entering various numbers into the "distance dropped" field, you can find a distance that will achieve a final velocity of 5.236 meters for m2 and m3.

The spreadsheet got a number of 21.4 seconds for the fall time. This equates with Jim's WM2D simulation that got 21.358 seconds.