George Egely - Nano-Dust Cold Fusion - Qantitative Tests in Classical Mechanics, 15p

7/27/2019 George Egely - Nano-Dust Cold Fusion - Qantitative Tests in Classical Mechanics, 15p

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Quantitative Tests in Classical Mechanics

Classical mechanics is the cornerstone of physics, yet it is usually missing one crucial point:

quantitative results. Though the conceptual models developed during introductory courses are

obviously essential, it would be equally desirable to have quantitative checks of some simple

(or more complicated) but fundamental dynamic phenomenon.Having quantitative test results of some selected one, two or three degrees of freedom-type

movement makes the process of the education more interesting.

Students can compare theoretical (numerical) tests with actual test data, asses the effect of

some usual idealization, like the neglect of friction or other non-linearity.

It is no wonder why there is a scarcity of numerical tests of actual dynamic movement. It is

quite cumbersome follow even the movement of a mass point, to say nothing of its velocity or

acceleration.

There are methods today to follow the spatial and temporal dynamics of one or two mass

points, but they are usually not enough: one might like higher derivates in order to check the

conservation principles.

Eight years ago we set about to study the quantitative dynamics of systems with two degreesof freedom.

It took about eight years and five generations of test devices and data acquisition systems to

solve this problem up to a degree of about 1% of accuracy, when the acceleration is not higher

then about 100-150 m/s2.

The whole task of quantitative data acquisition even at these modest parameters is unsolvable

without fast and specialized electronic data acquisition and processing.

We had to learn some painful lessons during these years, and to make some compromises

with costs. First of all one has to fight friction forces. Using tiny ball bearings could reduce

friction substantially, but makes the shape of moving bodies more complicated, thus more

cumbersome to calculate e.g. their moment of inertia.

The size of test machines should not usually exceed about 20 cm. Above that size and at high

accelerations, potential stress energy could be accumulated in the moving active masses or in

passive ballast parts as elastic deformations. Monitoring these deformations continuously is

not impossible but very expensive.

An ordinary strain gauge will not do the job their amplifiers are too slow for the job, and

piezoelectric ones are too expensive.

The most crucial device used in all tests is an incremental transducer which has an

optoelectronics system inside, and converts the temporal dynamics of rotation into digital

signals. The most convenient type for our tests had 2000 slits for a 360 degree rotation gauge,

manufactured by HP with optical lithography. There are also linear incremental transducers

but they are less reliable and prone to mechanical damage and distortion during dynamicoperations. Though the application of rotational incremental transducers seems to be a severe

restriction, in fact one is able to track linear movement with the help of fine flexible steel

cables or ribbons.

Special care was taken to use sturdy frames with minimal distortion, in order to reduce the

effect of external or internal vibrations and dynamic distortions.

Five test devices are shown in the first photograph, which gives some impression about the

practical size of the test devices.

Only one device will be detailed in this article, a heavy gyroscope, as this is the most

complicated and least understood device of classical mechanics.

Students could penetrate to the depth of classical mechanics if they are able to understand the

qualitative and quantitative behaviour of a heavy gyro.


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An additional advantage of this method is that one can trace the dissipation of energy due to

friction which is usually calculated by order of magnitude only.

In general these devices, and the data aquisition hardware and the software is a useful tool to

study a wide variety of movements but only within a rather narrow range of parameters.

The test device and test procedure

Fig.1

The layout of the test device is shown in Fig. 2. The layout is the following: There is solid

aluminum frame which is tied to the desk. It contains two horizontal coaxial axles, with an

upper and lower ending.

The flywheel is connected to a U-shaped arm, which rotates the lower half of the vertical axle

via a wire cable.


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Fig. 2

The rotation angle around the vertical axes () and the b tilting angle around the horizontalaxis are measured by a two incremental angle transducers. The noiseless, correct transfer of

the rotation angle is carried out by a thin metal cable wire, moving along two cable drams.

(The possible errors due to this method are discussed during the description of the test

process). The device contains a third electronic circuit, (not shown in Fig. 2) which measures

the angular velocity of the flywheel, (24 signal/revolutions.)

Due to design considerations, the 3rd

incremental transducer is able to rotate only 300,

otherwise its wires would be wound up during the rotations by the U-shaped holding arm.

Data acquisition: At the beginning of the test procedure, we start the test software, and at the

end of the process it is switched off. In between, the software collects the signals from theincremental transducers.

The vertical and horizontal angles (as a function of time) are determined by the polynomial

regression method usually at one millisecond intervals and smoothed for both angles. It is

important that we should have the interpolated angle as a function of time by one millisecond,

and the smoothed original angular functions. They can be compared with each other and can

be compared with the original raw signals. This data is taken to an Excel software. The

angular velocities are calculated then the energies are calculated.

During the course of the energy calculations, the total energy of the flywheel is separated into

two parts: the internal one and the external one.

The internal energy is Ep = pap2/2, where pa is the inertia of the flywheel for the axis of

rotation, and p is the angular velocity of the flywheel. At the usual parameters of the test, theamount of internal energy is about 1-5 times higher than the external energy, made of

frame

2nd

incremental

transducer

upper axis

fl wheel

1st incremental

transducer

cable drums

steel cable

horizontal axis

U shaped arm

lower axis


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precession and nutation. The p angular velocity of the flywheel is steady according to thetest results.

The external energy of the flywheel consists of several parts. The first part of this energy is

the ballast energy of the upper and lower axles, and the rotational and tilting energy of the

U-shaped holding arms.

The second part of the external energy of the flywheel is the energy of rotation, calculatedfrom the angular velocity vector of the flywheel, and from the inertia of the flywheel,

projected to the angular velocity vector.

The third part is the gravitational potential energy of the U-shaped holder assembly of the

flywheel. (One must take into account that the angular velocity of the horizontal axis

db/dt=b is the difference between the 1 lower half of the axles angular velocity, and the2 upper angular velocity. These energies are calculated in the following manner:

1. Ballast energies:

Efb1 = fb112/2 where fb1 = lawer+0 cos2b + 90 sin2b the rotational ballast

energy depending on the tilting angle

Efb2 = upper 22/2 where upper is the ballast of the upper axis

Ebb = bb b2/2 where b = 2-1, s bb is the tilting ballast of the U-shaped arm

2. Energy of rotation:

Ef = ee2/2

Let us decompose the resultant moment of inertia and the angular velocity:

Ef = 2 + 2 = (1sinb+p)2 + (1cosb)2+ b2

1

b

p

b

Where and are the parallel and the perpendicular components of the moment of inertia,taken from the point of rotation, and consequently and are the parallel and

perpendicular components of the p angular velocity.

The term 1 sin b should be taken as zero, as this will not move the flywheel via the ball

bearings. The first term is the internal energy part of the flywheel, which tests show has a

steady value.

The second and third terms could be combined, as the moments of inertia are equal for the

mutually perpendicular directions. Considering the above, the rotational energy of theflywheel will have the following form:


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Ef=1/2(b2+1

2cos

2b)

3. The potential energy of the flywheel, due to gravity:

Egrav = (ms rs + mr rr) (sin b + 1)

In the second bracket, the presence of 1 has the sole purpose of making possible a positive

gravitational potential energy even in case of negative b angles (Fig.2.)

The actual test procedure

One has to fine tune the calculated crude values of the moments of inertia to find their

actual values. These tests are started by a gravity pendulum procedure, and then by a

horizontal physical pendulum test series, by fixing the pendulum axis, show in Fig. 1 in order

to measure the moments of inertia in the horizontal and the vertical planes. The test results are

shown in Diagrams 1 and 2.

On the first occasion when the gravity type pendulum test was carried out, the holding U-

shaped armed is fixed temporarily, and when the horizontal plane oscillation test was carried

out, the vertical axis was fixed. The actual test results were evaluated by the calculated

flywheel parameters and then compared to the test results.

These tests render possible the fine tuning the parameters of the moments of inertia. Usually a

1-2% modification of the calculated values was necessary. They are caused by small errors in

the manufacturing, and tiny uncalculated deviations from the ideal shape, due to nuts and

bolts. Several thousands of tests have shown the validity of the conservation of energy for

these cases. The energy as a function of time monotonously decreases and this tendency isdetermined by the dissipation. When the velocity of the pendulum is maximum, the tangent of

this curve is maximum. When the velocity is zero, at upper extreme positions, the dissipation

is also zero. Thus the total energy has a local maximum.

As an additional test, we checked the output of the data acquisition device against the signal

of a high speed oscilloscope. No difference was found.

To sum up the test procedure, the following steps were taken:

a/ The device was fixed to a table, and b was fixed at zero.

b/ The gyro was raised to the top position, and was released in order to swing. The energy

balance is described by the following equation: E=1/2 2 + mgR(1+sin).

(The test results of the isolated, closed system are shown in the first diagram.)


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1st

diagram shows the energy components of the flywheel as a function of time as a pendulum,

in a reference test. The flywheel is not rotating, but swinging in a vertical plane.

The test error is less than 1 %. Even a 0,1 degree error of the angle severely distorts this

curve. When the pendulum stops at the upper dead point maximum, the dissipation due to

friction is zero, and the tangent of the energy-time curve is zero, as the potential energy has its

local maximum.

2nd

test diagram shows the flywheel is not rotating, but swinging along a horizontal plane.

This reference test improves the accuracy of the moments of intertia for the active

and ballast parts of the flywheel.


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When the pendulum is at its lowest point, its velocity has its local maximum and the

dissipation losses are the highest. Thus the energy-time curve monotonously decreases.

The test results of the spring-driven pendulum for the horizontal plane are shown in the 3rd

diagram. This test was carried out on the horizontal plane; the blob of the pendulum is the

wheel of the gyro.

The sum of energies:

E = 1/2 2 + 1/2 c y2 =1/2 2 + 1/2 c r22.

Here c is the spring constant of the driving spring, and r is the radius of the wire, driven

by the spring. The (t) function is the angle of rotation as a function of time, and its first timederivative is .It is worth noting that the degree of dissipation (the decrease of the total energy as a function

of time) is much steeper due to the internal friction of the wire.

The start of the tests

Once the modified, improved gyro parameters were found, we could start the tests. At first the

axis of the gyro was locked into a starter device, (shown in the photograph), in order to have

repeatable results. Then the gyro was wound by a flexible string, and by pulling it suddenly

we could rotate it up to a medium rpm. Then the rotating gyro was thrown by the starter

device toward a direction, with a given energy. In practice it is possible to test the angular

velocity of the gyro wheel for the first 300 C of rotation.

The test results shown in Fig. 3 were measured from the flywheel at 1700 rpm.. It is apparent

that the energytime diagram monotonously decreases, as it obeys the conservation of

energy.

One may start the flywheel tests with the modified, corrected inertia parameters. The spring

loaded starter device will ensure the repeatability of the test results. Thus the tip of the

flywheel with the holder arms was locked into it, then the flywheel was revved up, then the

starter spring was unloaded. This pushes the rotating flywheel to a certain direction.

This test is reliably repeatable, and one may measure the angular velocity of the wheel during

a test run.

The path of the center of axis is shown in Fig. 3 at 1700 rpm. The energies of the processes

are shown in the third diagram, (except the energy of the flywheel.) The rotations are the

following: Eb=vertical oscillation-nutation due to tilting, horizontal oscillation: Eforg or

precession. These were kinetic energies. The potential energy of the system is marked by Egrav

and their sum is Etot, the total external energy. The rounded off total energy (by 20 test points)is marked by thick line. (There is not much difference between the rounded off and the raw

data of the total energy.)

It is quite apparent that the curve monotonously decreases, as expected from the conservation

of energy.

It is quite apparent at the first stage if diagram 3, that the total value of the kinetic energies are

higher than that of the potential energy of gravity. In the middle they are equal. Then the

energy is higher again in the last phase.

One may surmise that the center of mass of the flywheel is nutating and precessing around the

equator of the sphere. On the first part it has a looped path; on the second it has a path with

peaks; at the last stage it has an elongated cycloid path.

When this path is looped or elongated, at the highest point there is still a kinetic (external)energy. At the peak of the path, the flywheel stops at the maximum points, so all the external


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energy is potential. Therefore here at that point the potential energy is the total external

energy.

The tests were carried out at different rpm. In the fourth diagram, the external energy-time

diagram is shown, for a heavy flywheel, at 700 rpm. The path of the center of axis is shown in

Fig. 4. The angular velocity of the flywheel was recorded simultaneously. Due to friction, it

decreased exponentially by one percent during the full test time period. It might be consideredconstant. Thus there is full test time period. Thus there is no flow of energy from the flywheel

to the axis.

Fig.3 Path of the center of mass for a heavy gyro. Initially

it rotated at 1700 rpm

It is worth noting that the grip of the steel wire cable could be a possible source of error. One

end was fixed by a screw, but the other was connected to a spanning spring. When the

flywheel starts from a dead point, the spring might overstretch. But in this case, the curve of

the gravity might deviate at the upper dead point from the curve of total external energies,

which would be visible in the energy-time diagrams. No such errors were observed during

these tests.


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The 3rd

diagram shows the energy-as a function of time for a heavy, nutating gyro

rotating at 1700 rpm at the start

From the spatial and temporal data the path of the center of mass has been drawn, moving on

a sphere (Fig. 3.) The largest horizontal meridian of this sphere is quite apparent.

The 4th

diagram shows the energy components of a heavy, nutating gyro

starting at 700 rpm. Here the path of the center of mass already has

peaks at the start. Its path is shown in Fig 4.


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Fig 4. Path of the center of mass, and of two points on the perimeter of

a heavy, nutating gyro. Its initial speed was 700 rpm

Based on the above practical tests method one can draw hands on conclusions from the path

data of the precessing, nutating flywheel; (See Appendix.)

The movement of the heavy gyro is truly remarkable in case of general initial and boundaryconditions. In Fig. 3, one can see a cycloid on the surface of a sphere. This is the solution of

the dynamics of a heavy gyro for the following condition: the flywheel axis angular velocity

axis p nutates around the p angular velocity (in the vertical plane), and this p acceleration

vector precesses with angular velocity around the vertical axis. The and angularvelocities can be determined approximately by just looking at Fig. 3. (Obviously the data

collected by the computer is much more accurate).

The Mnut momentum of nutation (Mnut = 2 xp) keeps the gyro around the circular path

of the axis of the angular velocity of nutation.The angular velocity of the center of mass has two components, one from the nutation and the

other from the precession. The angular velocity due to precession is nearly steady, and the

nutation takes places around it in a nearly circular manner. The change in the external

kinetic energy equals the change in the gravitational potential energy. The nearly steady

angular velocity of nutation and precession is apparent in this test case.

It is appropriate to use the term approximately, as there is a change in , due to the

momentum of centripetal force around the vertical , and partly due to the last term of

acceleration.

One can integrate from the last term Mpn = 2 x the precession-nutation momentum. Ithas a parallel component with the precession momentum of Mp = 2 x, and it has a

parallel component with the p axis, which yields the p momentum.The angular


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velocity of the heavy gyro may change on its bearings due to momentum perpendicular to its

axis.

The movement of the gyroscope is perhaps the most complicated type of rigid body

movement in mechanics. Therefore it was a complicated task to carry out the test and to

develop the the hardware and software of data acquisition equipment. The energy-time curve

of the external (and internal) energies monotonously decreases, proving that correct resultswere measured. One could study the heavy gyro movement with this device. Any test error

would have distorted the monotonous decrease of the energy-time curves. The test error is

estimated to be around 2 % during these tests. Any deviations were under this value.

Appendix: The Dynamics of a Gyroscope

There are a number of papers and books about the dynamics of the gyroscope. At first sight

their movement is unusual if not weird. We should like to present a simplified yet general

view of its dynamic relations, by analyzing a rotating, flat disk moving around two axes.

Fig.5

In the first figure of the appendix, a flat disk of radius r is shown, having angular velocity,

perpendicular to its plane of rotation. This angular velocity is rotated around an angular

velocity, having a fixed direction. We are about to examine the accelerations of a point mass

on the perimeter of the disk.

In Fig. 1 a Cartesian system of coordinates is taken, where for reasons of convenience the

vertical direction ofz is the a steady angular velocity, while direction x is fix. In thissystem the direction of is arbitrary; its angle z is noted by .

At a given time the angular velocity has an angle of to the x axis as shown in Fig. 1.


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For a given point on the perimeter of the disk v and v velocity components are shown, as

well as the unit vectors, necessary for the differentiation. The unit vector v will be e, and v

will have e as unit vector. The unit vector n it is parallel with r, and n is a unit vector is

perpendicular to axis, that is of axis z taken from the given point on the perimeter of the

disk to be examined. And at last s is the distance of the point on the perimeter from axis.

The first derivative by time of the vector r is the velocity:.

r = v

In a similar manner, the first derivative by time of is:

=

That is only its absolute value may change while its direction is permanent.

The vector after derivation consists of two components, one is by the change in direction

and the other is from the change of absolute value.

+=

The velocity of the given point to be examined is a sum of two velocities as a consequence of

the rotation around two axes:

v = r+ r

Consequently the acceleration is:

( ) ( ) ( ) ( ) ( )rrrrrrrv ++++++

=

The second term could be rearranged in the following way:

( x ) x r = x ( x r) - x ( x r)

After substitution and rearrangement:

( ) ( ) ( )rrrrrv ++++

=

2

After further arrangement, using unit vectors, the acceleration could be written in the

following form:

( )rnsnReseRv ++=

222

(The vector multiplications should be repeated by the reader!) After looking at the results, one

can see that the first two terms are the well known tangential acceleration around the two axes

of rotation, while the third and fourth terms are the centripetal terms around the two axes of


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rotation. The fifth term is the coriolis acceleration. This term is responsible for the precession

momentum of the gyroscope.

The precession momentum for the whole disk is calculated by an integration on the whole

mass of the disk:

Mp = 2m {rx [ x( x r)]}dm =2m{rx [ x ( x r) - rx( x )]}dm

The vector product of the first term yields zero. This is easy to prove, when is separated to

a parallel and a perpendicular component.

Shifting forward and resolving r to components with parallel and perpendicular directions

with a axis, produces the result:

Mp = -2( x ) m(R sin )2dm, after the integration: Mp = 2 x

where =m(R sin )2dm is the moment of inertia for the axis perpendicular to the twoaxis of rotation.(Fig.1.)

The meaning of the precession momentum is the following: one should create an externalprecession momentum which yields - precession movement.When the external precession momentum is terminated, the precession is also terminated

having angular velocity. This is an important statement to grasp the character of the

movement of a gyro. (Obviously the rest of the kinetic energy will dissipate, due to friction).

If we rotate the disc along axis x with a torsion spring, then is increased, the angle willincrease. It will increase up to the point when its precession momentum will reach the sum of

the momentum of the centripetal forces around and the momentum of the torsion spring.These two momentums render possible the rotation of.One should mention the nature of the Coriolis force which has two components.

Fig.6

Their physical meaning is the following:


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1. the rotation of axis causes the gyro to rotate with vvelocity, that is it rotates v, as aconsequence of the rotation of angular velocity.

2. due to the rotation of the axis makes change the distance s and thus it forces vvelocity to change its value. Their signs and their values are identical.

Therefore the Coriolis acceleration is the result of the mutual effects of the two axes of

rotation, and it is generated as a response to the actual external momentum.

General case, rotation around three axis

The generalization is possible along similar manner for three axes of rotation. For the general,

three dimensional case, the velocity is:

rrrvr ++==

The temporal derivatives of angular velocities are:

=; +

=

;

++=

;

The relations of angular velocities are expressed in the above expressions.

One can understand further that the rotation of the axis is general in the differential form. (In

the previous test results corresponds to precession, to nutation, but the order of nutationwas reversed.)

The full acceleration will have more complicated formulas:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )rrrrrrr

rrrrrrrrv

+++++++

+++

++++=

Adding the vectorial multiplications having identical terms, and rearranged to contain unit

vector, the acceleration will follow:

( ) ( ) ( )rrr

nrnrnrerererv

+++

++=

222

222

This long formula contains three components for tangential and centripetal acceleration

around the three axes of rotation, and three Coriolis accelerations due to the mutual

interactions.

The last term is the most interesting, as there might appear a momentum parallel to the

direction of angular velocity.

Consequently the disk could be accelerated by external angular velocities, or

momentums.


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At the parameters of the test shown above, the values of and are too small compared to .Thus this effect is negligible and is too small to notice.

The above equations are applicable for the heavy flywheel, described in the tests. When the

ideal flat disk is replaced by an actual heavy disk of thickness t, then the end plates of the

gyroscope act as a heavy top, though their weight does not act upon them due to reasons ofsymmetry.

The spatial vector of a mass point on asegment of what at the distance of v/2 could be written

as follows: R=R0+r as shown in Fig 7.

Fig. 7.

The derivatives of the two vectors yield systems of equations. The first system describes the

tangential and centripetal acceleration of the center of mass of the flywheel around and .The second equation is similar to the previous one of two axis rotation. However, in general it

will yield 13 separate terms for the acceleration.

r

R

R0

v

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George Egely - Nano-Dust Cold Fusion - Qantitative Tests in Classical Mechanics, 15p