Movement of Lagrange of the spinning top

The movement of Lagrange of the spinning top is the movement of a heavy Toupie around a point O of its axis, the reaction of axis not having moment compared to O (perfect kneecap).

It is about the movement of an ordinary spinning top, with this close in a spinning top, the nail of the spinning top is round and slips while rubbing on the plan where it " repose" : it follows by application of the Théorème of the gyroscopic moment which it rectifies and comes in position from dormant spinning top.

The case of Lagrange is visualized better by the movement of a gyroscope with two axes of Cardan joint, overloaded: if rotation is very fast, the precession is directly proportional to the weight. There is thus a gyroscopic Balance! It is advisable to initially study this easier case, before approaching the general case.

gyroscopic Balance: Precession ~ mga/Cr (0)

geometrical Description: one adopts following conventions for the Angles of Euler: the rotation of the first Cardan joint is defined by the angle $\ psi (T)$ ; the rotation of the second Cardan joint, interior with the first is nutation $\ theta (T) = angle (\ vec {K}, \ vec {K})$ ; the spinning top is finally mobile around its axis of revolution O, K , according to the clean swing angle $\ phi (T)$ .
kinematic Description: that is to say O, U the axis of the nodes. the vector rotation is thus:

$\ dowry {\ psi} \ vec {K} + \ dowry {\ theta} \ vec {U} + \ dowry {\ phi} \ vec {K}$

kinetic Description: own inertias with rotation have, have, C;

energy cinétique.2 is thus $A \ dowry {\ theta} ^2 + has \ dowry {\ psi} ^2sin^2 \ theta + C (\ dowry {\ psi} cos \ theta + \ dowry {\ phi}) ^2$

dynamic Description:

the theorem of the kinetic moment written according to U, K \ U, K gives the three equations:

$\ frac {D} {dt} (\ dowry {\ theta} has, has \ dowry {\ psi} sin \ theta, C (\ dowry {\ psi} cos \ theta + \ dowry {\ phi})) ^t + (\ dowry {\ theta}, \ dowry {\ psi} sin \ theta, \ dowry {\ psi} cos \ theta + \ dowry {\ phi}) ^t \ wedge (has \ dowry {\ theta}, has \ dowry {\ psi} sin \ theta, C \ dowry {\ psi} cos \ theta+ C \ dowry {\ phi}) ^t = (mga sin \ theta, 0,0) ^t$

Integration in the gyroscopic approximation (number of very large clean revolutions):

the third equation gives Lz = =cste = Cr (0) =~ C $\ dowry {\ phi}$ .

The second equation gives $\ theta = \ theta_0$ = cste: nutation is constant.

The first equation gives $\ dowry {\ psi}$ = mga/Cr (0).

The precession “thus weighs” linearly on mga, independently of nutation.

One can as consider as that is only artificial, because the moment is in sin $\ theta$ : then, it is more natural to consider the couple M and to write $\ dowry {\ psi} sin \ theta_0$ = M/Cr (0): = Precession: = Pr in rad/s

To find that by the Theorem of the gyroscopic moment:

Cr (0) of K' /dt = M = mga K \ K ; that is to say Cr (0). Pr. K \ K +mga K \ K = 0

Precession of the equinoxes of Hipparque

This time, the moment M is that of the sun on the equatorial bulge. That is to say Q the terrestrial Quadrupole = 2 (AC) <0: M = 3/4

\ omega_o^2

.cos

\ theta

.Q. K \ K (cf Gravimetry):

Precession = - $\ frac {3} {2} \ cdot \ frac {\ omega_o^2} {r_o}$ .cos $\ theta \ cdot \ frac {CA} {C}$ .

Value of the precession (26 000 years), one draws the value from (CA) /C of the Earth.

Actually, it is also necessary to take account of the influence of the Moon: this one intervenes approximately by the double (but, conversely, the Moon undergoes the moment in return, whereas the Sun which also undergoes it is influenced in a negligible way following its mass): with the 16" /an is thus added 34" /an had with the moon, is 50" /an, is a turn in 26.000 years:

Hipparque could not measure the Gamma point with the precision of 1" of arc, but it had 2 centuries old former measurements, that is to say 10.000" of arc, which was enough.

The celestial pole is not eternally close to pole star. In 26.000 years, it describes in the sky a circle around the point Q, of direction perpendicular to the ecliptic , not Q located approximately in the middle of the Dragon (to look on a celestial sphere the ecliptic circle (the Zodiaque is enough) and to take the perpendicular with this ring): the circle of 23°26' is very broad and will take along in 13.000 years the pole towards the Constellation of the Swan (to be checked!) : then from the climate point of view, the northern Hemisphere will receive this time a little more heat than the Southern hemisphere (~ 7%), which is important in the theory of Milancovitch of the climate (it is necessary to add there the periodic variation of eccentricity E (T) and the periodic variation of slope I (T) (stabilized by the existence of the Moon!) One thus obtains a remarkable agreement with the glacial cycles of the Holocène.

It should be noted that the orbital plan of the Moon is shifted compared to the terrestrial equator and the ecliptic; it results a light shift from it from 9" of arc and a period of 1 saros (18.6 years); this effect is negligible on the climate, certainly!

Linearized movement

To take again the 3 exact equations; and to inject the following approximations:

$\ theta (T) = \ theta_0 + Y (T)$
$\ psi (T) sin \ theta_0 = t.Pr +X (T)$ , with Pr = M/Cr (0)

and to linearize in (X, Y):

X" + Pr ². X + Y'. Coupling = 0
- X'. Coupling + Y" = 0

with Coupling = constant = (Cr (0) - 2A.Pr.cos $\ theta_0$ ) /A, practically equal to C/A R (0) in rad/s, characterizing the gyroscopic coupling, which preserves energy: this one can only tranférer of X towards Y and reciprocally.

The second equation is integrated immediately: - X. Coupling + Y' = constant = has

That is to say X" + X (Pr ² + Coupling ²) + a.Couplage = 0

one thus finds X (T) = cste + oscillation of pulsation = sqrt (Pr ² + Couplage ²) ~ Couplage = C/A R (0)

and Y (T) = B .t + of the same oscillation pulsation.

One recognizes the drawing of a Trochoïde more or less lengthened drawn on the sphere unit by the vector K : the points of cycloid are always to the top (to pay attention in flat projection according to whether $\ theta_0 > or < \ pi/2$ ).

Application: m = 1kg; OG = has = 2cm; With = m R ² with R = 4cm and C=mr' ² with r'= 3.6 cm (spinning top-prolate); theta (0) = 30°; R (0) = 4500 turns/minute gives:

theta-max - theta (0) = 3 '; nutation of period = 10 ms; precession = 14°/s.

Movement of general Lagrange

It is obviously more complicated if all OdG (Order of magnitude) are recut!

One can show that it is about a Mouvement in Poinsot in a reference frame moving in Poinsot: it any more but does not remain to represent the movement mentally.

More concretely, the movement is integrable, because there exist three integrals first: since M = K \ K, L.k = constant = L1; L.K = constant = L2 and energy E = 1/2L. $\ omega$ + mga.cos $\ theta$ .

Some calculations give an equation of Newton (see time Diagramme); thus the movement is integrable near with a squaring: E = 1/2 has $\ dowry {\ theta} ^2 + U_ {EFF} (\ theta)$ , with Ueff = mga.cos $\ theta$ + L2 ² /2C + (L1-L2.cos $\ theta$ ) ² /2 (A.sin ² $\ theta$ ).

One must distinguish in possible mathematics 8 cases. But not in physics (see simple Pendulum).

The discussion is carried out primarily according to 2 great possible cases: for theta = 0, Ueff (0) infinite or are finished (what requires the condition " impossible" in physics: L1 = L2); then, in the case Ueff (0) finished, it remains to distinguish if expressed according to cos (theta), there is minimum or not. " bifurcations" and the separating ones in the space of the phases are really sensitive only to the exerted eye of which wants to look at well the detail of any movement of this spinning top.

Egg of Christophe Colomb: L ² /4A > mga

In this last case, one finds that if L1=L2 is sufficiently large, the driving position of the spinning top is stable: L ² /4A > mga is the condition known as of Colomb: the legend wants that Colomb took an egg (cooked) and made it hold vertical by making it turn quickly on a table.

If L1 different of L2, it is easy to show that theta remains blocked between two values: the values of théta lower than 90° are due, one suspects it, with the gyroscopic effect. When R (0) is not sufficient any more, theta remains always blocked between two values (thus theta = 180° is impossible: it is the object of multiple plays of fair; déquiller the skittle of the center is impossible with a gyroscopic Pendule). Conversely a vertically launched gyroscopic pendulum will have evil to change its driving position, and will answer less than the simple spherical pendulum the external disturbances.

Uniform precession

To include/understand that this condition imposes null nutation must now be rather intuitive. And the speed of precession Pr is given by the formula:

L2.Pr - A. Pr ². cos $\ theta$ = mga

What can be found geometrically:

The vector K precess without nuter, (thus L precess also without nuter), around the vertical (one says that the humming-top; it is said that she sleeps if theta is null). The dynamic moment is Pr K \ L = U - has Pr ² .cos (theta) .sin (theta) and is equal to M = U mga.sin (theta) cqfd.

The forts in mechanics will be able to also apply the course: knowing that R = $\ dowry {\ phi} + \ dowry {\ psi} cos \ theta$ , one can write directly:

0 = dynamic couple gyro +couple of déséquilibrage + M, is:

0 = - C $\ dowry {\ phi}$ .Pr.sin $\ theta$ - (1/2). (CA) Pr ² .sin2 $\ theta$ +mga.sin $\ theta$

It is well the same result This result exact specifies that found for the gyroscopic Balance.

See too

Movement in gyroscopic Poinsot
Pendulum
gyroscopic Balance
gyroscopic Compass
Gyroscope

External bonds

There exists many other case of the integrability of the movement of the solid around a point: a very thorough work is on the fabric (to seek Kovaleskaia and Goritchev). One will quote only one article of popularization:

For more, to refer to the work integrable Systems of Michele Audin, in which it has various results on the theory of the integrability, which made notable progress during the two last decades.

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