Balancelle

That is to say a simple Pendulum, whose mass, m, via a mechanism intern with itself, can rise or go down to its liking along the bar (without mass) (it is to some extent a all small Guernazelle). The length of the pendulum is variable (see simple Pendule variable length).

While working skilfully, small the ciron increases the amplitude regularly movement: it is the principle of the balancelle, very simple case of a pseudo-parametric resonance (see simple Pendule with parametric resonance).

Amplification

Let us take the simplest case of amplification:

The ciron is during the first quarter of the pendular motion, on the circle of radius L = a+h. Parti M1 (of maximum elongation, of altiude z1), it low reaches the point B (altitude z=0) with the quarter of period T/4 .f ( $\ theta_1$ ).

Then, abruptly, it develops an infinite power during an infinitely short time, producing a FINISHED work, W, which enables him to climb B out of H vertically (of altitude Z = H).

It is let balance on the second quarter of the pendular mouvemnt, on the circle of radius OH = has, traversing the arc of circle HN2, in a time T'/4 .f ( $\ theta_2$ ).

At the point N2, it is motionless. It is then let instantaneously slip along the bar at the point m2 (of altitude z2).

It is ready then to carry out the return according to the same principle way: arc M2B, BH, arc HN3, N3M3): 'is a kind of figure of real Scot, in form of eight (" carré") lengthened not closed.

The profit of energy is obviously primarily due to the effect sling of the shortening of the ray of a+h to has, in the passage of B with H (for more, to see discussion). To apply the theorem of the energy-power to the equation of the theorem of the kinetic moment: dL/dt = - mgl sin $\ theta$ , while multiplying by L and while integrating: $(1-cos \ theta_ {n+1}) a^3 = (1-cos \ theta_n) (a+h) ^3$ , is:

$z_1 \ cdot {(a+h) ^3} = z_2 \ cdot {a^3}$

Its altitude Z = (a+h) (1-cos $\ theta$ ) does not cease increasing (with each return ticket, of a factor (1+h/a)^6): when it exceeds 2 (a+h); it passes in mode of whirling, which we will treat later on.

Analyzes energy

It is clear that the ciron provided the energy which raises it: W = Mg (z2-z1).

Demonstration:

The ciron has a constant mechanical energy of M1 with B (not low): mgz1. In this point, it develops an infinite power to move on the vertical of has in A': at this time, the moment of gravity is null, nor of course that of the reaction out of O, therefore the kinetic moment, L, SE PRESERVES, therefore the angular velocity increases brutally: it is the effect sling

$\ omega_2 a^2 = \ omega_1 (a+h) ^2$

The conservation of energy then gives: $\ omega_2^2 has = G (1-cos \ theta_2)$ , and thus:

$z_1 \ cdot {(a+h) ^3} = z_2 \ cdot {a^3}$

It is well the preceding reasoning.

Is to be evaluated the total work of the ciron W : to consider cycle BHN2M2B.

The work of gravity is null, since the force Mg is conservative.

The only moment when the ciron works is on segment BH, and segment N2M2.

With share the energy of recovered gravity (W1 = - mgh cos (theta2)), the ciron does not provide any N2 work in m2.

On the other hand of B out of H, the kinetic moment is preserved L1 = L2 = L and the kinetic energy prodigiously quickly increases finished value (at infinite power) of Ec1 (B) = L ² /2m (a+h) ² with Ec' 2 (H) = L ² /2m (a) ²; that is to say an increase in Ec1 (B) = Mg. (a+h) . (1-cos $\ theta_1$ ) with Ec' 2 (H) = Mg. has . (1 - cos $\ theta_2$ ); that is to say W2 = Ec' 2 (H) - Ec1 (B); and moreover the ciron W3 provides = mgh.

Thus on the whole, W = W1 + W2 + W3 = mgh $(1-cos \ theta_2)$ +Ec' 2 (H) - Ec1 (B) = Mg ( $z_2-z_1$ ),

fine of demonstration .

To say in short, the sling effect is carried out at kinetic time L= cste; thus the angular velocity $\ omega$ increases like ~1/l ², therefore $\ omega^2 l$ like 1/l ³; however $\ omega^2 l^2$ like Z: from where the result.

Reasoning in the reference frame related to the bar (of which the angular velocity undergoes a discontinuity that we will suppose " régularisée"): the work of the interior forces not depending on the reference frame, one must find same work W for the ciron:

Ec2relatif (B) - Ec1relatif (B)=0 = W +W (gravity) no one + W (coriolis) no one + Wer (Radial sweeping force, axifuge) + Weo (sweeping force orthoradial).

While always proceeding on the same course as previously, the tavail Wer on arc HN2 and arc M2B is null (orthogonality of the force and displacement), and also on the arc N2M2, mgh cos (theta2) because the radial force is Mg cos there (theta2). On arc BH, work Wer is negative and enormous: with L =cste, we already saw that it was worth Ec1 (B) - E' c2 (H) (and it is advisable to add there - mgh because the centrifugal force must take into account the weight.

Weo work is null since the force moves only orthogonally.

The preceding result well is found.

More general case

We took this course in eight " carré" , but it is quite obvious that, by taking again the preceding results, energy will be increased each time the integral

\ int L (\ theta) ^3 sin \ theta D \ theta

taken on an outward journey return is positive (either a kind of eight (figure of reality in Scottish dance) described in the same direction as the " eight carré" precedent).

It is clear that if the " huit" is described in contrary direction, the ciron recovers the energy of the swing and deadens the movement.

Synchronization to be adapted

It should well be noticed that this amplification takes place in manner not isochrone : often, when the amplitude becomes close to 180°, one sees wax them which operates badly; once again it is L (theta) which counts and not L (T) in this type of amplification. In this direction, there is difference with parametric amplification of Mathieu-Hill.

Lastly, nothing prevents from continuing this mechanism when one is in mode of whirling: it is necessary simply that the ciron, with the vertical is projected immediately at the distance (a+h); it will then describe more and more quickly the trajectory (made of two half-circles of little rat R and R+h, joined by vertical segments length H).

The opposite mechanism makes it possible more not to whirl and stabilize little by little its attitude.

Case of loss of energy by friction

It is clear that if the energy lost by friction (either fluid, or solid) equalizes W, the young lady will balance herself with constant amplitude: a child adopts this control when his/her parents judge the amplitude of his movement sufficient.

See too

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