i = mrr vs i=mr experiment

[Tarsier79]

Broli, My opinion is that linear or rotaional masses still conform to the same rules. COAM, COM... They are true for like masses. When you try to accelerate something twice as fast... Ke=mvv/2, So it takes twice the energy to do so. Elastic colisions don't account for the increase in leverage required, nor do they account for the increased inertia at radius.

What is important here, is inertia is what you have to overcome to accelerate the smaller mass to the greater velocity. This experiment shows that to achieve this takes significantly more time or energy. It also shows the relationship between radius and inertia. So, if you take the conclusions for this experiment, or any of the youtube vids shown that also prove the same, Is that the relationship will hold true for other experiments, like the cylinder and spheres, The rim flinging ball, etc.


Peq, Im not sure what exactly you mean by going too fast for mrr.
Summary of experiment: If both weights had the same inertia then the wheel would provide the same resistance to the falling weight in both instances. IE, if I=mr, then the time would be equal in both cases.

Does the experiment support I=mrr? Providing the information supplied is correct (and as far as I am concerned, it is), then IMHO it completely disproves I=mr. But agreed, it doesn't absolutely confirm I=mrr either. I didn't think about it at the time, but I could prove I=mrr by simply putting on 1/4 weight at double the distance, then the time should be equal. Would that be convincing enough to prove I=mrr?

[jim_mich]

You guys are attempting to do experiments using different weights and distances, BUT you are somewhat ignoring the weight and size of the wheel/lever. All rotating structures have a radius of gyration. If you were to compress all of the mass of a rotating object into a ring having zero thickness (shaped like a paper tube/ring but still having the same mass) then this hypothetical ring would act and react exactly like the actual rotating object which you are observing. This is like the theoretical calculations using point mass pendulums. From the radius of gyration value you can determine how fast the rotating object is moving where it moving in a straight line.

So, your setup with a rod/wheel having no weights has a mass and a RoG (Radius of Gyration) and a velocity. When you add your first weight then it has its own mass, RoG, and velocity. There are formulas whereby you can merge the two objects together and determine the combined new mass, RoG, and velocity. Then if you swap out new weights at new radius you can again determine the new mass, new RoG, and new velocity.

When such RoG calculations are used then inertial momentum equals IM = M×V and KE = 1/2×M×V×V.

RoG is simply a means of averaging the more outward molecules with the more inward molecules so as to produce a radius distance where the actions of the outer molecules balance the actions of the inner molecules.

If you guys continue to massacre the numbers and ignore RoG like you've been doing then you will never agree as to what formula is correct, because neither formula will be right without knowing the radius of gyration so that you know the actual average speed of all of the mass. The only way your numbers will match one or the other formula is if you use a weightless rod/wheel setup and use very compact (point mass type) weights.

Rant over.

[path_finder]

Dear Tarsier79,

When you try to accelerate something twice as fast... Ke=mvv/2, So it takes twice the energy to do so

I guess you are not in accordance with the common law saying: for doubling the speed, we need an energy FOUR times greater than the previous one.
See here: http://en.wikipedia.org/wiki/Kinetic_energy
Excerpt from the paragraph 'Newtonian kinetic energy/Kinetic energy of rigid bodies':

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force.

[Fletcher]

I'm not sure that's really relevant Jim - all formulas for inertia are based on point mass assumptions - that's what integration analysis does IINM - it's also why the actual inertia has to be determined by experiment because only sims can calculate to point mass sizes & have rods or disks with virtual no mass.

In this case it was proposed that I = mr - therefore two objects can have the same inertia quotient made up of any combination of mass & radius multiplying to give the same inertia figure.

Clearly this is not correct & that was the purpose of Kaine's experiments, not to prove I = mr^2 [for point masses].

Here is the Newtonian proof that allows a comparison between Linear & Angular/Rotational quantities for the same system.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html

--------------------

Broli .. CoAM, like CoE & Conservation of Momentum are physics truisms, until proven to no longer be true.

[Tarsier79]

Thanks Path, my mistake is from trying to make too many points at once.

Jim, Fletcher makes the point much more eloquently (and probably more correctly) than I can. That being the inertia of the bar for the sake of this test is irrelevant. And as far as that is concerned, I believe the test was successful.

The test with the bar by itself was for a reference point, as I at one stage intended to do further calculations including the inertia of the bar and hub. It also shows by the difference in time that its inertia is relatively low compared to that of the weights.

Jim, you are correct, I don't fully understand the importance of the radius of gyration, but I guess now is probably a good time for me to do some research into what it exactly means. I am also confused about IM, because if IM= M*V, then it is linear, whereas Inertia / Ke is exponential( with reference to radius / angular velocity) Again I will have to look into it further before understanding exactly what IM means.

[Fletcher]

Kaine .. jim_mich has a lot of knowledge of inertia & momentum.

IM = mv ?

IINM, Inertial Mass [IM] is the mass found by applying a known force to an unknown mass & measuring the resultant acceleration.

Gravitational Mass [M] is basically the same thing determined by comparing a known mass against an unknown mass on a balance beam in a gravitational field.

p = mv => Momentum [p] = inertial mass [im] x velocity [v]

F = ma => Force [F] = gravitational mass [m] x acceleration [a]

EDIT: sorry, penny just dropped - IM means Inertia Momentum not Inertial Mass

Inertia & Momentum are different.

[Tarsier79]

I looked up inertial momentum on google, and it doesn't seem to match anything. The closest I could find is moment of inertia. Perhaps it is a typo , as momentum = m * v.

[jim_mich]

Inertial

Definition:
[adj] of or relating to inertia

Inertia

Webster's 1913 Dictionary

Definition:

\In*er"ti*a\, n. [L., idleness, fr. iners idle. See
{Inert}.]
1. (Physics) That property of matter by which it tends when
at rest to remain so, and when in motion to continue in
motion, and in the same straight line or direction, unless
acted on by some external force; -- sometimes called {vis
inerti[ae]}.

Momentum

Webster's 1913 Dictionary

Definition:

\Mo*men"tum\, n.; pl. L. {Momenta}, F. {Momentums}. [L.
See {Moment}.]
1. (Mech.) The quantity of motion in a moving body, being
always proportioned to the quantity of matter multiplied
into the velocity; impetus.

The symbol 'P' is usually used for momentum because the symbol 'm' is already taken to mean mass. We use the double symbol 'KE' to indicate Kinetic Energy. So awhile back I suggested we use the symbol 'IM' to indicate momentum. The word 'inertial' is simply an adjective to the noun momentum. If I were to write P = M×V then few people here on the forum would understand that the 'P' means momentum.
From Wikipedia

The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference. In physics, the usual symbol for momentum is a bold p (bold because it is a vector); so this can be written

p = m × v

where p is the momentum, m is the mass and v is the velocity.

From From Wikipedia

Physics and mathematics appear to be less inclined to use the original concept of inertia as "a tendency to maintain momentum" and instead favor the mathematically useful definition of inertia as the measure of a body's resistance to changes in momentum or simply a body's inertial mass.

This was clear in the beginning of the 20th century, when the theory of relativity was not yet created. Mass, m, denoted something like amount of substance or quantity of matter. And at the same time mass was the quantitative measure of inertia of a body.

The mass of a body determines the momentum p of the body at given velocity v; it is a proportionality factor in the formula:

p = mv

The factor m is referred to as inertial mass.

But mass as related to 'inertia' of a body can be defined also by the formula:

F = ma

Here, F is force, m is mass, and a is acceleration.

So... what symbols should we use when discussing such things? I try very hard to preface all formula with a simple phrase describing my meaning and the 1st instance of abbreviations with their meaning.

When such RoG calculations are used then inertial momentum equals IM = M×V and KE = 1/2×M×V×V.

So, your setup with a rod/wheel having no weights has a mass and a RoG (Radius of Gyration) and a velocity.

The reason that Google does not find anything for 'inertial momentum' is because the two word phase is seldom use. It is like saying a male man. The two words are redundant. The phrase is not proper English. I used it in an attempt to make it very clear as to my meaning, but it seems I failed. Maybe from now on I'll use the more scientifically proper p = m × v. In my mind IM equates to inertial momentum in the same way that KE equates to kinetic energy. The two are similar but different.

To understand Bessler's principle of perpetual motion you need to understand how inertial momentum relates to kinetic energy. You need to understand the difference between transferring KE and transferring IM. The vast majority of energy transfers involve KE. But in certain minor cases IM can be transferred from object to object.

Or, the above paragraph re-written using the symbol 'p' for momentum...

To understand Bessler's principle of perpetual motion you need to understand how momentum relates to kinetic energy. You need to understand the difference between transferring KE and transferring p. The vast majority of energy transfers involve KE. But in certain minor cases p can be transferred from object to object.

So, should I use 'p' or 'IM' ?

[john]

Tarsier,
I was wondering, did you notice if momentum was constant ? Basically how much rotation there is after there is nothing causing your set up to rotate ?

John

[Fletcher]

Jim .. from my position you can use either, IM if you prefer - I should have picked up earlier what you meant when you had it equaling mv - only one thing does that so it has to be p.

You obviously believe you have found a way to transfer completely momentum in certain circumstances, from a fast moving object to a slower moving object & visa versa - since the energy quotient is a v^2 relationship then using your integration analysis from the 'jim_mich formula' then there shows mathematically an increase in useable KE i.e. a Net KE increase.

N.B. IM/p is a vector & CoE, CoM & CoAM well accepted in science - KE is a scalar quantity having magnitude only - that does not mean that these Conservation Laws can not be broken but as you know to be accepted as fact a robust experiment must be demonstrated to prove the theory is new fact - the 'Eureka' thread was 5 years ago IINM - at some level you support some parts of pequiade's theoretical approach but with a different mechanism to extract Net Energy - there is no question to ask other than will you ever demonstrate that experiment so that the rest of us can stop looking & get on with our lives ? - we know your formula shows a mathematical gain in Energy but it is the vector quantities of momentum [linear & angular] that causes the real world problem in seeing that possibility manifest, IMO.

--------------------

For those who might be wondering about the seeming dichotomy between the rotational inertia formula for point mass & say the centripetal force, I will attempt to explain, as I understand it.

I = mr^2

Cp = mv^2/r

Both are rotational environments yet the Inertia formula doesn't appear to 'gel' with the force derivation formula.

As radius of rotation increases the Inertia gets greater by the square of the radius - it's natural to try to extrapolate this formula & say what happens when the circumference curve is straight line i.e. infinite r - the Inertia 'quality' would be impossibly huge, therefore the formula must be wrong as this doesn't make sense - it does if you remember that it is just for the rotational, angular momentum situation, so as r increases so does the inertia as long as its forced to rotate - IOW's it takes more force to rotate that point mass in the same time as a point mass at a lesser r.

Cp is a force which is velocity dependent & r dependent - as r increases the Cpf decreases, all else being equal - this is opposite of Rotational Inertia.

So I = mr^2 looks to be at odds with Cp = mv^2/r by saying that as r increases Inertia increases & Cp decreases - the symmetry of yin & yang & the symmetry of the Conservation Laws ?!

---------------

Kaine .. I will be interested in your interpretation of the importance of RoG ?

[Tarsier79]

Ho

Basically how much rotation there is after there is nothing causing your set up to rotate ?

The bearing has very little friction, so it will continue to rotate at nearly the same speed, and would do so for a long time, except that there is a string attached that stops the rotation.

I will be interested in your interpretation of the importance of RoG ?

Since this thread is the first time I have taken any notice of ROG, here is where my understanding comes from almost exclusively:
http://www.youtube.com/watch?v=UXAwEneKEFM (first 2 minutes only)

It fits perfectly with this experiment, and if you considerer the masses on their own, this is what is being used. I understand ROG refers to the entire system though, and in that case, Jim is correct, I am sure we could calculate the ROG of the entire system then use it to confirm I=Mrr.... But due to the relationship of these two:

ROG = root( i / m ) : ROG = root( mrr / m)

We would basically be using mrr to prove itself, so I think this current method is perfectly acceptable.




How does inertia mathematically affect acceleration of the bar?

f=ma (so we can calculate the force of the falling mass)

Then for the bar assebly with weights, we calculate ROG, and using f above, and sub m as I:
(F = i x a) to give us acceleration of the bar

(I hope)




On another note, when comparing inertia and Cp, to make things easier to compare, I think you should use them both in reference to angular velocity... So

I = mr^2

Cp = mrW^2/r

When I said above CF is a product of inertia, I should have said it is a product of IL (inertial angular momentum) ;)

[Fletcher]

RoG

http://van.physics.illinois.edu/qa/listing.php?id=330


Radius of Gyration [Rg] in dynamics is a distance from the axis of rotation say, where an objects Moment of Inertia is the same for both the true voluminous object & a point mass equivalent - if this is a disk or ring then the point mass circle equivalent etc.

This means the two objects would behave the same after torque is applied.

It's not the same as Center of Mass for objects, when the Center of Rotation is within the object instead of at one end, e.g a pivoted disk v's an end pivoted rod.

Every object has its own Moment of Inertia, this has to be found or calculated.

To find the Radius of Gyration you apply the equation Rg^2 = I/M where I = Moment of Inertia, M = Mass

therefore Rg = (I/M)^1/2 or the sqrt of Moment of Inertia divided by the Mass.


Lets follow the disk example in the link ...

lets say we have radius 1 m , mass 2 kg

I = 1/2 m r^2 => I = 1 kg m^2

Rg = sqrt (I/M) => sqrt (1/2) = 0.707 meters

N.B. any different radius length will be already accounted for in the Moment of Inertia calculation therefore doesn't need to be applied again in the Rg calculation & that's why r is missing.


------------------


Theory ?: two masses can rotate & move, one closer to the axis & the other further from the axis - normally this has a penalty of changing the Moment of Inertia i.e. two objects moving apart relative to a radius create greater inertia which slows the angular velocity & momentum - when they come together again this is reversed e.g. the ice skater example.

N.B. two masses changing radius this way will have different velocities - the summed KE of both after the shift is greater than the KE of both when close together therefore potential exits to release surplus KE to do Work [the jim_mich formula, root mean square].

This is dependent on the Radius of Gyration [Rg] being maintained so that angular velocity is not affected adversely - this means a certain ratio of distance moved & speed of movement to keep the same Rg & how KE & momentum can be manipulated.

[pequaide]

Tarsier 79 Your speed falls in between mr and mrr, but it is too fast for mrr. Too slow may be caused by the ½ masses rolling on the round shaft, or flexing of the shaft. But too fast is very hard to explain.

The spheres stop the cylinder at 90° just as the tether enters the opening in the cylinder. But it only does this if the exact quantity of momentum is present in the 'add on' cylindrical mass underneath the top cylinder. For example: a 1200 gram one inch pipe is equal to a 300 gram 4 inch pipe when both are spun at the same rate. The mrr of these two objects is no where near similar, so the experiments to prove that the mrr concept is false have already been done.

Using the same wheel and changing the radial location of the drive force is another go proof of the laws of levers and F = ma relationship.

[Tarsier79]

Gday Peq

The speed of the bar is too fast? You are correct, there will be some temporary variation in acceleration due to movement of the masses. The speed difference depends on the dropping weight and the difference in inertia of the bar/hub assembly compared to the inertial weights.

Today is my last day on shift at work, So I intend to perform the test using 1/4 weight at 2x distance, which according to mrr, should produce the same resistance to acceleration, and therefor an equal drop time. For me at least this will be successful proof of I=mrr.


Your cylinder and spheres test does provide some interesting results. In the spirit of possible reproduction of your test,...

Did you use equal weight spheres?
Did you use equal length string, and did the spheres expand to an equal radius, or an equal distance from the cylinder?
What holds the spheres in place before release?
What in your opinion is the difference between the cylinder and spheres, and single flinging weight, like Nics setup?

[pequaide]

When moment of inertia is applied within its original purpose the r is a fundamental quantity or a real distance if you will.  When moment of inertia is applied inappropriately in the lab it is a relative distance not an independent quantity. 

In space the applied force acts at no radius; it is a gravitational point source. In the laboratory application; moment of inertia is the differences in radii; it is the relationship between the radial distance to the inertia and the radial distance to the applied force. 

If the radius of the applied force is one and the radius to the inertial mass is 5 then your mrr is m * 5 * 5.

If the inertial mass is moved to 10 then your mrr is m * 10 * 10.

If the applied force is then moved to 2 then your mrr is back to 10/2  m * 5 * 5.

So to examine the results in different arrangements you can move the applied force or the inertial mass.

You can also compare different positions of different inertial masses along the shaft, as you did. But the r is always a proportion or a comparison with another force or mass at another radius.

My 18 inch (3 kilogram) wheel can be accelerated from three different radii, I used 6.028 inches and 8.91 inches. I placed a flag on the wheel and accelerated the wheel for about one third rotation. After about a third rotation the flag crossed the photo gates which were 26 mm apart. I am not concerned with actual velocity but only relative velocity.

I suspended, from a string, about 718.7 grams at 6.028 inches and the flag on the accelerating wheel tripped the photo gates in .0271 seconds.

I suspended, from a string, about 486.2 grams at 8.91 inches and it tripped the photo gates in .0279 seconds.

Within experimental error these numbers (.0271 and .0279) are right on top of each other.

Do you see the relationship? 718.7g * 6.028 inches = mr = 486.2g * 8.91 inches and .0271 sec = .0279 sec.


79; Did you use equal weight spheres? 

Yes

79; Did you use equal length string, and did the spheres expand to an equal radius, or an equal distance from the cylinder? 

Yes, yes and yes.


79; What holds the spheres in place before release? 

My fingers; and in two other models “mechanical arms'.


79; What in your opinion is the difference between the cylinder and spheres, and single flinging weight, like Nics setup?

In vertical throws you must throw fast, or the sphere is merely grabbed by gravity and plunked on the ground or at best it causes a very weak throw. Nic made some throws that were fast and he was bombarding his neighborhood with bouncy balls.

The 18 inch wheel can throw single flinging weights (bags of BBs) about 40 meters, they were accelerated by hand.

The cylinder and spheres throws horizontally and it is unaffected by gravity. The cylinder is throwing the spheres while it is dropping, I call it floating because it can also be thrown up. But of course the dropping motion adds nothing to the spinning motion and is totally independent of the spinning. I have also tossed the cylinder so that the spheres were in a vertical plane, same results, but again it is floating.

Because gravity has no affect of the cylinder and spheres the throw can be very slow or as fast as you can throw. The 90° stop occurs no mater what the rate of rotation. The faster the rotation the quicker the stop, but it always occurs just the same.