So let’s change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.

After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.

I reposted two sentences; the 160 cm is two 80 cm radiuses, with one kg on each end. The dropped mass (or drive mass) is 40 kg. The 40 kg is accelerating itself and two 1 kg masses, but they are on the end of an 80 cm radius.   

Delburt

This would be 3.27 m/sec/sec acceleration at the surface of the shaft.[/size]

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That is what I am not sure abput

Mike

Tarsier: I removed my comments above...it was 3 o clock in the morning so had a misunderstanding
I agree :  a special construction of a pendulum-bearing can transfer torque but this is not what is commonly known as a pendulum.
You have to be more specific about the bearing in use.

Mike

Delburt


as I said im my last comment about your accelleration-calculation " That is what I am not sure about"


So I was looking up an old german Physics-Textbook Grimsehl-Tomaschek, dated 1940. In a chapter about the equivalence-condition
of rotating masses I found an experimental setup in which it is demonstrated that the accelleration of a falling mass which
rotates two masses ( see pic Mass-Equiv_02.jpg)  at both ends of a horizontal mounted bar stays always the same if the value of the moment of inertia ( m*r²)
stays the same for different symetrical positions of the two masses on the bar. In order for (m*r²) to be constant, the masses have to be adapted if their
radial posistion changes.


If you have 2 x 1 kg at r = 0.8 m, moment of intertia I = 2(1kg * 0.64 m²)-> I = 1.28 kg m²


Now you want to move the two masses to r = 0.4m then for I=1.28 kgm² to remain constant ( 1.28) the two masses have to be increased ( I use here the
term replacement-mass m') to


                       m'= 1.28 / 0.4² = 1.28/ 0.16 = 8kg in total, 4 kg at each side,


to fullfill the condition that the accelleration of the falling mass stays the same as with 2 * 1 kg masses at r = 0.8m.


The calculations made in this old book are based on the cgs-System ( centimeter-gramm-second )
and only with this system the formula for the resulting accelleration b of weight G is true :


                       G (gr) * 981 (dyn) = b * (m * r² + G) ( see pic Equiv-Mass_03.jpg) 
 
The formula is correct only for r = 1 cm, as the value of r² = 1 ....so that it is save to assume that the replacment-mass m'
experieneces the same accelleration as G  ( falling weigth )


         b = G (gr) * 981 (dyn) / (m (gr) + G (gr))                            ( Formula Nr. 1 )


......but what would be the new total replacement-mass m' on both sides of the bar at r = 1 cm ?


The moment of Inertia for the two 1kg masses ( in gramm ) at r=80 cm


      I = 2000 gr * (80cm)² = 2000 * 6400 = 12.800.000 gr cm²


In order to have the same moment of Inertia at r = 1cm the new mass m' = 12.800.000 / 1
that is two masses of 6.400.000 gr ( 6.400 kg) at each side.


According the Formula Nr.1 above, falling mass = 40 kg = 40.000 gr


                b = 40 000 g * 981 dyn / (12.800.00gr + 40.000gr) = 39.240.000 /12.840.000
               
                b = 3.05 cm/ s² = 0.0305 m/s²   !!!!
______________________________________________________________________________________




Now in order toget an idea the accekeratio of G with an unchanged replacement-mass m' at r=2 cm ( although this not correct )
we end up with b = 0.121 m/s²


From here you calculate the final velocity of 40kg falling down 1 meter : V = Square root of (2 * 0.121 * 1) = 0.492 m/s


Since the axis-radius is 2 cm the circumference of the axix = 0.126 m
With the velocity 0f 0.492 m/s / 0.126  you get 3.9 revolutions per second.


The cirumference at which the two 1kg masses rotate is : 2 * Pi * 0.8 = 5 m


With 3.9 Revolutions / s the rim-velocity of the two 1 kg masses is 19.5 m/s


Kinetic energy Ekin = 2kg/2 * (19.5)² m²/s² = 380.25 Joule


Input was : m * g * h = 40kg * 9.81 m/s² * 1m = 392,4 Joule


The missing 12 Joule are accounted to the falsy used r = 2cm with the unchanges m' in the Formula N.1


So there is no gain in this system. That is the reason why we dont have overunity-bicycles




Mike


Edit: Sorry for the big rotated pics. I tried to remove and replace them ..but to no avail

A lever arm can’t move up until it is first balanced.

The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side. 

And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.

But (for you) this 80 kg is still (1600 – 80) 1520 kg short of being balanced. If your rule worked, then the 1 kg at 40 cm will accelerate the 80 kg; not the other way around. For your math the 80 kg at 1 cm * 1 cm; should be overpowered by 1 kg * 40 cm* 40 cm.

----
Experiments collect data to establish or support a concept; so an experiment will collect data points. The concept is; If a longer lever arm with a proportional less mass is harder to rotate, then it will be harder to rotate. For example: if one side of a pivot point is 40 kg at 1 cm and the other side is 1 kg at 40 cm; then according to the mr² theory the long side is 40 time harder to rotate than the short side.

The Laws of Levers says that the two sides are equally hard to rotate. So, let’s construct an experiment that separates the true from the false.

If you apply the same force to two 40 kg masses at 1 cm on both sides; and to two 1 kg masses at 40 cm on both sides, then the harder to rotate side will take longer to rotate, or it will have a smaller acceleration.

The Force is held to be the same by using the same suspended mass at the same location. In the 40 to 1 we used an extra 40 kg on the short side.

You find data points to determine if the acceleration is the same. A 26 mm gate trip at the same times after the two Atwood’s (two 40 kg at 1 cm, and two 1 kg at 40 cm) has moved the same distance (real numbers .0628 and .0631 sec for one of many set ups).  Having the same final velocity after moving the same distance assures us that the different Atwood’s have the same acceleration.

Having the same F divided by the same ‘a’ for both Atwood’s (40 kg at 1 cm both side; and 1 kg at 40 cm both sides) assures us that the inertia ‘m’ has to be the same also.

Real numbers for a certain Atwood’s are:  two 20.7 kg at a 1.03 cm r; compared to two .8628 kg at 24.13 cm. Both took 3.3 sec; in another set up. They had the same F and the same 'a' and the same inertia 'm'.

Delburt

The 80 to 1 lever arm, according to your math, can’t be moved up until 6,400 kg are placed on the opposite short (1 cm) side. [/size]And: according to your math, then the long side of a 40 to 1 lever arm can’t be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.

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[/size]
No, this is not what is meant please reread my description.[/size]

[/size]
We have to distinguish between a static situation ( Nm)  and a dynamic one-> accelleration..--> mr² ..resistance to rotary accelleration.[/size]

[/size]
"By the way: this is not "my math",  I just try to understand the basics and s[/size]ince you changed to a lever-system ( away form the Cyl & sphers) in order to find an easier way ( compared to the Cyl. & Sph) I was forced to dive into the depth of what was known to this subject back then.[/size]

[/size]
I cannot and will not argue whether this is true of not but I understood the principle of this experiment. [/size]

[/size]
One thing is sure that these guys back then did extensive and very elaborated experiments. I spent a few days searching the internet ...didnt find anything and then looked the subject up in my old book. I have no time to translate all this to english but the experiment was basis in university
teachings and sure enough they demonstrated this in front of students.
If I had the time and the means I would build this experiment.


I am a pragmatic person, i.e. I build, experiment and analyse...like you to. This I did with the Cylinder & Spheres experiment. Something went wrong
with a brass-bearing and I have to re-engineer the setup. However I could see and film that momentum is conserved not energy. I will continue with this small wheel setup and build it according to sound mechanical means and do not let myself distracted by other setups because I have not reached the core-function of the Cyl & Sphere system.


I agree: They do everything to control attention to protect their narrative.

Levine on moment of inertia

https://www.youtube.com/watch?v=lvfzdibrUFA

Mike

Demonstrating Rotational Inertia (or Moment of Inertia) - YouTube     flipping physics

Question for you:   Why is the torque formula one r and (your) inertia formula is two r s? Because changing the location of torque is exactly the same thing as changing the location of inertia. It is a relationship between the two radii not an actual distance.

I don’t think they (guts back then) did do extensive and elaborate research; and that is why you cannot find the experiments. Just like the flipping physic guys; they don’t do data: just; it is faster it is slower, etc.

You have a 80 cm wheel with a 2 cm shaft and you have 40 kg suspended from the left side of the shaft and 1 kg suspended from the right side of the wheel. It is balanced: but you then add 5 grams to the 1 kg on the right. We will let those 5 grams accelerate the wheel to .1 m/sec. At one tenth meter per sec we add 200 grams to the 40 kg on the shaft left side.    What happens to the acceleration?

Energy?? Tarsier; since when does F and m make energy?  F = ma makes linear momentum. “Rotation gained or lost” is acceleration. So you have the three components of Newtonian motion; and Newton never made energy.

Torque is an applied force upon a resistance; if there is no resistance there is no torque. From Newton Third Law we know that the inertia is applying a force upon the torque: equal and opposite but entirely in the same manner. It is a reverse torque upon the applied torque. This inertial resistance is force being applied at a radial distance.

I will give two arguments:

Make a horizontal lever arm with one kilogram attached (at rest) at one meter. You then (perpendicularly) collide 2 kg moving 5 m/sec at the .5 meter location. What is the resultant motion?

Second argument:

In the flipping physics ‘Demonstrating Rotational Inertia (or Moment of Inertia) - YouTube’ lets assume that the torque can be placed at two locations, on the pulley, where one location is twice the radius of the other. And we know that we can place the inertia at twice the distance; one location toward the middle, and then one location on the end of the rods.

Now we have four possible arrangements of the spoke wheel: small pulley torque to inertia position on the end of the rod, small pulley torque to inertia position in the middle of the rod; large pulley torque position to inertia position on the end of the rod; large pulley torque to inertia position in the middle of the rod.

Now ask yourself; is the small to small equal to the large to large? And I hope you say yes.

So if you halve the torque radius you double the inertial resistance.

But (for your law): if you double the inertia radius you quad the inertial resistance.

So you can move from ‘large radius torque and middle radius inertia’: to ‘large r torque to large r inertia’. And you say it will be one fourth the acceleration. 

Or you can move from large radius torque and middle radius inertia: to small r torque to middle r inertia. And it will be one half the acceleration.

But the radius torque / radius inertia is the same in these two changes.

You can’t have two rules.