What are Euler's equations of motion

Euler's equations (gyro theory)

This article explains Euler's gyroscopic equations, for other meanings see Euler's equations.

The Euler's gyroscopic equations or short Euler's equations are equations of motion for the rotation of a rigid body. There are three coupled differential equations for the components of the angular velocity of the top in the body-fixed (co-rotating) coordinate system, the axes of which are the main axes of inertia.

Euler's gyroscopic equations are not to be confused with Euler's angles, which describe the orientation of a body-fixed coordinate system in relation to a spatially fixed coordinate system. Furthermore, the equations of motion of ideal liquids in fluid mechanics are also referred to as Euler's equations.

Euler's gyroscopic equations

In the main axis system rotating with the body, these are Euler's gyroscopic equations

$ \ begin {align} M_1 = & I_1 \ dot {\ omega} _1 + (I_3-I_2) \ omega_2 \ omega_3 \ M_2 = & I_2 ​​\ dot {\ omega} _2 + (I_1-I_3) \ omega_3 \ omega_1 \ M_3 = & I_3 \ dot {\ omega} _3 + (I_2-I_1) \ omega_1 \ omega_2. \ end {align} $

With

  • ω1,2,3: Angular velocities around the main axes
  • I.1,2,3: Principal moments of inertia around the center of mass
  • M.1,2,3: torques acting in the center of mass and
  • the point on top denotes the time derivative, so that $ \ dot {\ omega} _ {1,2,3} $ represent angular accelerations around the main axes.

The quadratic terms in the angular velocities represent inertial forces in the rotating frame of reference. If the movement is known, then these equations can be used to calculate the moments that must be introduced into the body in its center of mass so that the body executes the specified movement.

In the case of a plane movement, for example in the 1-2 plane, rotations and moments about the 1 and 2 axes are omitted, and the equations are reduced to

$ M_3 = I_3 \ ddot \ varphi \ ,, $

if φ is the angle of rotation around the 3-axis.

Euler's gyroscopic equations follow from the angular momentum balance of a rigid body around its center of mass

$ \ vec {M} = \ mathbf {I} \ cdot \ dot {\ vec {\ omega}} + \ vec {\ omega} \ times \ mathbf {I} \ cdot \ vec {\ omega} $

where $ \ vec {M} $ is the torque acting in the center of mass, $ \ vec {\ omega} $ is the angular velocity and $ \ mathbf {I} $ is the inertia tensor with respect to the center of mass. Occasionally this general equation, which applies in every coordinate system, is also called Euler's gyroscopic equation. In the main axis system $ \ hat {g} _ {1,2,3} $, which is fixed to the body, as a special local reference system, the inertia tensor is constant over time and has a diagonal shape:

$ \ mathbf {I} = \ sum_ {i = 1} ^ 3 I_i \ hat {g} _i \ otimes \ hat {g} _i = \ begin {pmatrix} I_1 & 0 & 0 \ 0 & I_2 ​​& 0 \ 0 & 0 & I_3 \ end {pmatrix} _ {\ hat {g} _i \ otimes \ hat {g} _j} \ ,. $

The arithmetic symbol "$ \ otimes $" forms the dyadic product of the basis vectors. The diagonal links I.1,2,3 are the main moments of inertia, which are calculated from the solution of the eigenvalue problem of the inertia tensor. In this system, the particularly simple component equations given above result from the angular momentum balance.

Derivation in coordinate space
From the coordinate-free vector equation to the coordinate equation
Euler's gyroscopic equations follow from the angular momentum theorem, which is given by

$ \ dot {\ vec L} = \ vec {M} $,
where $ \ vec L $ is the angular momentum and $ \ vec M $ is the sum of all external torques acting on the body at the center of mass. If one puts into this equation the formula for the (intrinsic) angular momentum $ \ vec {L} = \ mathbf {I} \ cdot \ vec \ omega $ with the inertia tensor $ \ mathbf {I} $ with respect to the center of mass and the angular velocity $ \ vec \ omega $ is obtained
$ \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (\ mathbf {I} \ cdot \ vec \ omega \ right) = \ vec {M} $.
This vector equation is valid in every coordinate system, but the components of the inertia tensor are time-independent only in one that is fixed to the body. The derivation of the gyroscopic equations by means of vector / tensor calculation can be looked up under angular momentum balance on the rigid body. The proof in the coordinate space is to be specified here.
A body-fixed coordinate system with the orthonormal basis $ \ hat {g} _ {1,2,3} $ is defined, so that the components $ I ^ {ij} _ {\ rm L} $ of the inertia tensor $ \ mathbf {I} = I ^ {ij} _ {\ rm L} \ hat {g} _i \ otimes \ hat {g} _j $ are time-independent with respect to this basic system. In this section, Einstein's summation convention is to be used, according to which indices appearing twice in a product - here i and j - is to be summed up from one to three. In the inertial system the vectors $ \ hat {g} _ {1,2,3} $ rotate. By representing the inertial tensor in the inertial system with standard basis $ \ hat {e} _ {1,2,3} $, the components of the inertial tensor of the rigid body are generally time-dependent due to the rotation:
$ \ begin {align} \ mathbf {I} = & I ^ {ij} _ {\ rm L} \ hat {g} _i (t) \ otimes \ hat {g} _j (t) = I ^ {ij} ( t) \ hat {e} _i \ otimes \ hat {e} _j \ \ rightarrow I ^ {kl} = & \ hat {e} _k \ cdot \ mathbf {I} \ cdot \ hat {e} _l = \ hat {e} _k \ cdot I ^ {ij} _ {\ rm L} \ hat {g} _i \ otimes \ hat {g} _j \ cdot \ hat {e} _l = (\ hat {e} _k \ cdot \ hat {g} _i) I ^ {ij} _ {\ rm L} (\ hat {g} _j \ cdot \ hat {e} _l) \ end {align} $
Because the base $ \ hat {g} _ {1,2,3} $ emerges from the standard base by a rotation, is the transformation matrix R. with components $ R_ {ij} = \ hat {e} _i \ cdot \ has {g} _j $ a rotation matrix and $ RR ^ \ top = E $ with the identity matrix applies E. and the sign “┬” for the transposition. With the transformation matrix, the above relationship between the components of the inertia tensor can be written as a matrix equation:
$ I = RI _ {\ rm L} R ^ \ top. $
There are $ I $ and $ I _ {\ rm L} $ 3 × 3-matrices with the components $ I ^ {ij} $ and $ I ^ {ij} _ {\ rm L} $, respectively. An analogous transformation relationship also applies to other matrices, for example: $ \ Omega = R \ Omega_LR ^ \ top $, where the index L indicates the local, body-fixed coordinate system. For a vector the transformation map is:
$ \ vec v = v_i \ hat {e} _i = v ^ j _ {\ rm L} \ hat {g} _j \ quad \ rightarrow \ quad v_i = \ hat {e} _i \ cdot \ vec v = (\ hat {e} _i \ cdot \ hat {g} _j) v ^ j _ {\ rm L} \ quad \ rightarrow \ quad v = R v _ {\ rm L} \ ,. $
Here $ v $ and $ v _ {\ rm L} $ are column matrices with the components $ v_ {1,2,3} $ and $ v ^ {1,2,3} _ {\ rm L} $ of the vector im Inertial or body-solid system. Because the vectors and tensors are assigned to matrices in this way and are then literally robbed of their basis, the equations established with the matrices are no longer independent of coordinates. Column matrices are usually also referred to as vectors and noted with an arrow, which will also be done in the following.

Derivation of the angular momentum balance in coordinate space
If one uses this transformation mapping from the inertial system into the local reference system in the angular momentum balance, the matrix equation results:

$ \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (I \ vec \ omega \ right) = \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left ( RI _ {\ rm L} R ^ \ top \ vec \ omega \ right) = \ dot R I _ {\ rm L} R ^ \ top \ vec \ omega + R I _ {\ rm L} \ dot R ^ \ top \ vec \ omega + I \ dot {\ vec \ omega} = \ dot RR ^ \ top I \ vec \ omega + IR \ dot R ^ \ top \ vec \ omega + I \ dot {\ vec \ omega} = \ vec M \ ,. $
As usual, dots were used as an abbreviation for time derivatives. Since $ R $ is orthogonal, $ E = RR ^ \ top $, where E. denotes the identity matrix. If one derives this equation according to the time, one obtains
$ O = \ frac {\ mathrm {d}} {\ mathrm {d} t} E = \ frac {\ mathrm {d}} {\ mathrm {d} t} \ left (RR ^ \ top \ right) = \ dot {R} R ^ \ top + R \ dot {R} ^ \ top, $
with the zero matrix O, from which it follows that the matrix $ \ Omega: = \ dot {R} R ^ \ top = -R \ dot {R} ^ \ top = - \ left (\ dot {R} R ^ \ top \ right) ^ \ top $ is skew symmetric. The meaning of this matrix can be seen if the transformation $ \ vec {a} = R \ vec {a} _ {\ rm L} $ of an arbitrary constant body-fixed vector $ \ vec a _ {\ rm L} $ into the inertial system according to the Time derives. Taking into account that $ \ vec {a} _ {\ rm L} $ is constant, we get:
$ \ dot {\ vec {a}} = \ dot R \ vec {a} _ {\ rm L} = \ dot RR ^ \ top R \ vec {a} _ {\ rm L} = \ Omega \ vec { a} \ ,. $
If one compares the last equation with the formula $ \ dot {\ vec {a}} = \ vec \ omega \ times \ vec {a} $ for the orbital velocity of a rotating vector $ \ vec {a} $, one recognizes that $ \ Omega = \ dot R \, R ^ \ top $ just those for the cross product $ \ vec {a} \ in \ mathbb {R} ^ 3 \ mapsto \ vec {\ omega} \ times \ vec {a} = [ \ vec {\ omega}] _ \ times \ cdot \ vec {a} $ is the appropriate skew-symmetric cross product matrix $ [\ vec {\ omega}] _ \ times $. Thus $ \ dot R \, R ^ \ top = \ Omega $ and $ R \, \ dot {R} ^ \ top = - \ Omega $ in the last equation for $ \ vec {M} $ can be replaced by the cross product replace with the angular velocity $ \ vec \ omega $:
$ \ vec {M} = \ vec {\ omega} \ times \ left (I \ vec {\ omega} \ right) -I \ left (\ vec \ omega \ times \ vec \ omega \ right) + I \ dot {\ vec \ omega} $
With $ \ vec {\ omega} \ times \ vec {\ omega} = \ vec {0} $ this results
$ \ vec {M} = \ vec {\ omega} \ times \ left (I \ vec {\ omega} \ right) + I \ dot {\ vec \ omega} \ ,. $
This equation was derived here for the coordinates in the inertial system, in which the coefficient matrix $ I $ of the inertia tensor can be time-variable.
When transforming into the local reference system L, the time derivative of $ \ vec \ omega $ must be taken into account. With $ \ vec {\ omega} = R \ vec {\ omega} _ {\ rm L} $ results
$ \ dot {\ vec {\ omega}} = \ dot R \ vec {\ omega} _ {\ rm L} + R \ dot {\ vec {\ omega}} _ {\ rm L} = \ dot RR ^ \ top R \ vec {\ omega} _ {\ rm L} + R \ dot {\ vec \ omega} _ {\ rm L} = \ vec {\ omega} \ times \ vec {\ omega} + R \ dot {\ vec \ omega} _ {\ rm L} = R \ dot {\ vec {\ omega}} _ {\ rm L}. $
That is, the time derivative of the angular velocity vector transformed into co-ordinates moved along is precisely the transformed angular acceleration.
This means that the last equation for $ \ vec {M} $ can be transferred directly to main axis coordinates.
$ \ vec {M} _ {\ rm L} = \ vec {\ omega} _ {\ rm L} \ times \ left (I _ {\ rm L} \ vec {\ omega} _ {\ rm L} \ right ) + I _ {\ rm L} \ dot {\ vec \ omega} _ {\ rm L} \ ,. $
In the force-free case, $ \ vec {M} _ {\ rm L} = \ vec0 \ ,. $

Alternative derivation
Let $ \ vec {a} _ {\ rm L} $ be an (arbitrary) time-dependent vector-valued quantity in local coordinates (index L) and $ R $ the rotation matrix for the transformation $ \ vec {a} = R \ vec {a} _ {\ rm L} $ from local coordinates in an inertial system. The angular velocity belonging to $ R $ in local coordinates is $ \ vec {\ omega} _ {\ rm L} $. The time derivative of the transformation rule $ \ vec {a} = R \ vec {a} _ {\ rm L} $ can be expressed in the form

$ \ dot {\ vec {a}} = R \ left (\ vec {\ omega} _ {\ rm L} \ times \ vec {a} _ {\ rm L} + \ dot {\ vec {a}} _ {\ rm L} \ right) $
represent. If you apply this rule to the law of angular momentum
$ \ vec {M} = \ frac {\ mathrm {d}} {\ mathrm {d} t} \ vec {L} $
with the representation of the angular momentum in the local reference system $ \ vec {L} = R \ vec {L} _ {\ rm L} $, one obtains
$ \ vec {M} _ {\ rm L} = \ vec {\ omega} _ {\ rm L} \ times \ vec {L} _ {\ rm L} + \ dot {\ vec {L}} _ { \ rm L}. $
If one sets $ \ vec {L} _ {\ rm L} = I _ {\ rm L} \ vec {\ omega} _ {\ rm L} $ with the constant coefficient matrix $ I _ {\ rm L} $ of the inertia tensor in principal axis representation one, then Euler's equation results
$ \ vec {M} _ {\ rm L} = \ vec {\ omega} _ {\ rm L} \ times \ left (I _ {\ rm L} \, \ vec {\ omega} _ {\ rm L} \ right) + I _ {\ rm L} \ dot {\ vec {\ omega}} _ {\ rm L}. $

Addendum
A short explanation for the equation $ \ dot {\ vec {a}} = R \ left (\ vec {\ omega} _ {\ rm L} \ times \ vec {a} _ {\ rm L} + should be given later \ dot {\ vec {a}} _ {\ rm L} \ right) $.

A brief preliminary consideration of the transformation of the cross product will help. Let $ \ vec {a}, \ vec {b}, \ vec {c} \ in \ mathbb {R} ^ 3 $ be any three vectors. The spar spanned by $ \ vec {a}, \ vec {b}, \ vec {c} $ does not change its volume under a rotation $ R \ in \ operatorname {SO} (3) $. So $ \ vec {a} ^ \ top \ left (\ vec {b} \ times \ vec {c} \ right) = (R \ vec {a}) ^ \ top \ left ((R \ vec { b}) \ times (R \ vec {c}) \ right) = \ vec {a} ^ \ top \ left (R ^ \ top \ left ((R \ vec {b}) \ times (R \ vec { c}) \ right) \ right). $ Since $ \ vec {a} $ is arbitrary in this equation, $ \ vec {b} \ times \ vec {c} = R ^ \ top \ left ((R \ vec {b}) \ times (R \ vec {c}) \ right) $, i.e. $ R \ left (\ vec {b} \ times \ vec {c} \ right) = (R \ vec {b}) \ times (R \ vec {c} ) $.
The time derivative of $ \ vec {a} = R \ vec {a} _ {\ rm L} $ results in
$ \ dot {\ vec {a}} = \ dot {R} \ vec {a} _ {\ rm L} + R \ dot {\ vec {a}} _ {\ rm L} = \ dot {R} R ^ \ top R \ vec {a} _ {\ rm L} + R \ dot {\ vec {a}} _ {\ rm L} = \ vec {\ omega} \ times \ vec {a} + R \ dot {\ vec {a}} _ {\ rm L} $
and with the preliminary consideration of the transformation of the cross product one obtains
$ \ dot {\ vec {a}} = R \ left (\ vec {\ omega} _ {\ rm L} \ times \ vec {a} _ {\ rm L} + \ dot {\ vec {a}} _ {\ rm L} \ right). $

Applications

Rotating spherical top

A spherical top is a top with three identical main moments of inertia I.. The gyro equations are reduced to a spherical top

$ \ vec {M} = I \ dot {\ vec {\ omega}} \ ,. $

The angular acceleration is therefore parallel to the moment.

Kick off a billiard ball

Fig. 1: Bump of a billiard ball parallel to the table top

A billiard ball with a radius should be parallel to the table top r, Dimensions m and mass moment of inertia I. be bumped in such a way that it does not slip over the table, see Fig. 1. The question arises as to what height H above the plate the force F. must be initiated so that no friction force on the table is necessary for the slip-free rolling.

The horizontal force acting eccentrically on the ball develops a moment M.=-(Mr)F.that the sphere according to the gyroscopic equation

$ M = - (h-r) F = I \ ddot \ varphi $

set in rotation. The moment is negative because it is opposite to the counting direction of the angle of rotation φ works. In addition, the force accelerates the ball according to the law "force equals mass times acceleration":

$ F = m \ ddot x \ ,. $

The acceleration $ \ ddot x $ is parallel to the table in the direction of the force. The condition for slip-free rolling

$ \ ddot x = -r \ ddot \ varphi $

closes the system of equations for the three unknowns H, φ and x from. This is calculated

$ - (hr) F = I \ ddot \ varphi = - \ frac {I} {r} \ ddot x = - \ frac {I} {mr} F \ quad \ rightarrow \ quad h = r + \ frac {I} {mr} \ ,. $

For a massive homogeneous sphere the mass moment of inertia is $ I = \ tfrac {2} {5} mr ^ 2 $ and thus

$ h = \ frac {7} {5} r = \ frac {7} {10} d \ ,, $

if d is the diameter of the sphere.

A wheel rolling down an incline

Fig. 2: A wheel rolling down an inclined plane.

A wheel with a radius roll in a plane motion r, Dimensions m and mass moment of inertia I. one with the angle α inclined plane under the influence of an acceleration due to gravity G down, see Fig. 2. Because the wheel also moves in a translatory manner, its mass is also included in the acceleration. However, the acceleration increases when the mass moment of inertia decreases.

Due to the slip-free rolling, a frictional force arises at the contact point of the wheel R.that sets the wheel in rotation because it corresponds to a moment M.=-r R. It's negative because it's counter to the counting direction of the angle of rotation φ is working.Thus the gyro equation reads in the plane case:

$ -r R = I \ ddot \ varphi \ ,. $

The component acting on the wheel downhill F. the weight mg has the size F.=mgsin (α). Opposite it is the frictional force, so that according to Newton's second law:

$ F-R = m \ ddot x \ quad \ rightarrow \ quad R = mg \ sin (\ alpha) -m \ ddot x \ ,, $

where $ \ ddot x $ is the acceleration of the wheel counting down the slope and sin is the sine function. The condition for slipless rolling $ \ ddot x = -r \ ddot \ varphi $ closes the system of equations for the three unknowns R., φ and x which ultimately result in:

$ \ begin {align} \ ddot \ varphi = & - \ frac {mr ^ 2} {I + mr ^ 2} \ frac {g} {r} \ sin (\ alpha) = - \ frac {r} {I + mr ^ 2} F \ R = & \ frac {I} {I + mr ^ 2} mg \ sin (\ alpha) = \ frac {I} {I + mr ^ 2} F \ \ ddot x = & \ frac {mr ^ 2} {I + mr ^ 2} g \ sin (\ alpha) = \ frac {r ^ 2} {I + mr ^ 2} F \,. \ end {align} $

The maximum acceleration for a given gradient occurs when the mass of the wheel is concentrated as close as possible to its center point, thus minimizing the mass moment of inertia.

A block sliding down a slope without friction experiences the acceleration $ \ ddot x = g \ sin (\ alpha) $, which is greater than that of the wheel, because the wheel converts part of the potential energy into rotational energy, which is then missing for translation .

General characteristics of the movement of force-free rotating tops

Except in weightlessness, a force-free top can be realized in a gravitational field by being rotatable in its center of gravity, for example gimbaled.

Conservation quantities of movement

In addition to the gyroscopic equations, the rotational movement of a force-free top is subject to two conditions: on the one hand, the conservation of angular momentum $ \ vec L = \ text {const.} $ And, on the other hand, the energy conservation law $ E_ \ mathrm {rot} = \ text {const.} $ Im Expressed in the local main axis system fixed to the body, these read with $ L: = | \ vec L | $:

$ \ begin {align} L ^ 2 = & (\ mathbf {I} \ cdot \ vec \ omega) \ cdot (\ mathbf {I} \ cdot \ vec \ omega) = I ^ 2_1 \ omega ^ 2_1 + I ^ 2_2 \ omega ^ 2_2 + I ^ 2_3 \ omega ^ 2_3 = \ text {const.} \ 2E_ \ mathrm {red} = & \ vec \ omega \ cdot \ mathbf {I} \ cdot \ vec \ omega = I_1 \ omega ^ 2_1 + I_2 \ omega ^ 2_2 + I_3 \ omega ^ 2_3 = \ text {const.} \ end {align} $

These two equations represent mathematically ellipsoids, the first the twist ellipsoid, the second the energy ellipsoid. So that both ellipsoids can have points in common, must at any point in time

$ 2 I_1 E _ {\ rm red} \ le L ^ 2 \ le 2 I_3 E _ {\ rm red} $ or $ \ frac {L ^ 2} {2I_3} \ le E _ {\ rm red} \ le \ frac { L ^ 2} {2I_1} $

if, as usual, the main moments of inertia according to I.1 < I.2 < I.3 are arranged.

Because a point that lies on both ellipsoids must meet the conditions

$ \ begin {align} 1 = & \ frac {1} {L ^ 2} (I_1 ^ 2 \ omega_1 ^ 2 + I_2 ^ 2 \ omega_2 ^ 2 + I_3 ^ 2 \ omega_3 ^ 2) = \ frac {1} {2E _ {\ rm red}} (I_1 \ omega_1 ^ 2 + I_2 \ omega_2 ^ 2 + I_3 \ omega_3 ^ 2) \ \ rightarrow 0 = & (2E _ {\ rm red} I_1-L ^ 2) I_1 \ omega_1 ^ 2 + (2E _ {\ rm red} I_2-L ^ 2) I_2 \ omega_2 ^ 2 + (2E _ {\ rm red} I_3-L ^ 2) I_3 \ omega_3 ^ 2 \ = & 2 \ left (E_ { \ rm red} - \ frac {L ^ 2} {2I_1} \ right) I_1 ^ 2 \ omega_1 ^ 2 +2 \ left (E _ {\ rm red} - \ frac {L ^ 2} {2I_2} \ right) I_2 ^ 2 \ omega_2 ^ 2 +2 \ left (E _ {\ rm rot} - \ frac {L ^ 2} {2I_3} \ right) I_3 ^ 2 \ omega_3 ^ 2 \ end {align} $

fulfill. In the last two equations, all factors except the parentheses are zero or positive. A nontrivial solution exists if the smallest expression in brackets in the equations is not positive and the largest is not negative. With the assumed proportions of the main moments of inertia, this is ensured by the above limits for the angular momentum square and the rotational energy. Then the rotational energy and the amount of angular momentum are compatible with a rotational movement of the body under consideration.

Geometric interpretation of the conservation quantities

A clear representation of the areas defined by the conservation of energy and angular momentum results from the transition to the local components of the angular momentum according to L.i=I.iωi are the product of the principal moment of inertia with the corresponding component of the angular velocity. The equation system arises from the conservation of energy and the amount of angular momentum:

Fig. 4: Carrier curves of the solutions of the gyro equations with I.1= 2 kgm², I.2= 4 kg m² and I.3= 10 kg m². The curves of the same rotational energy have values ​​of 1.5… 6 N m in 0.5 N m steps with an angular momentum $ | \ vec L | = 5 \, \ mathrm {N \, m \, s} $ drawn.
$ \ begin {align} \ dot {L} _1 = & - \ left (\ frac {1} {I_2} - \ frac {1} {I_3} \ right) L_2L_3 \ \ dot {L} _2 = & \ left (\ frac {1} {I_1} - \ frac {1} {I_3} \ right) L_3L_1 \ \ dot {L} _3 = & - \ left (\ frac {1} {I_1} - \ frac {1 } {I_2} \ right) L_1L_2. \ L ^ 2 = & L_1 ^ 2 + L_2 ^ 2 + L_3 ^ 2 = \ text {const.} \ 2E_ \ mathrm {red} = & \ frac {L_1 ^ 2} {I_1} + \ frac {L_2 ^ 2} {I_2} + \ frac {L_3 ^ 2} {I_3} = \ text {const.} \ End {align} $

The last two conservation equations are automatically fulfilled by the first three gyro equations, which can be proven by their time derivative. The angular momentum, which remains constant in the space-fixed inertial system, lies in the local, body-fixed, rotating reference system on a sphere, which is shown in Fig. 4 viewed from three directions. Because here the components of the angular momentum can occupy any point on the sphere, even if the angular momentum in the inertial system always points in the same direction. The additional condition of energy conservation forces the components of the angular momentum onto an ellipsoid. The intersection of sphere and ellipsoid are closed curves (red and blue in the picture) that can be circular, elliptical or taco-shaped.

Stability considerations

The curves of the same energy and the same angular momentum degenerate into points when the rotation takes place around one of the main axes of inertia. If the angular momentum is in the vicinity of the main axis of inertia with the largest or smallest moment of inertia (blue or red points in the picture), then it also remains in its vicinity, because these points are surrounded by the curves. Therefore these axes of rotation are stable axes of rotation of free rotation. Their points of intersection with the energy ellipsoid are elliptical fixed points of an autonomous differential equation.

Insofar as the components of the angular momentum are Lipschitz continuous, the trajectories of the angular momentum cannot intersect (Picard-Lindelöf theorem). This condition is on the Separatrix with 2 I.2E.red = L.² injured (dashed black in the picture) if like here I.2 according to I.1 < I.2 < I.3 is the mean major moment of inertia. If the angular momentum lies exactly on the 2-axis (black point), then it remains there. If it is in another point in the vicinity of the intersection, the angular momentum can only move away from the intersection and does not stay near the intersection either. The 2 axis is one unstable Axis of rotation, it meets the energy ellipsoid in a hyperbolic fixed point or saddle point of the associated autonomous differential equation (see also # Stability of the movement of asymmetrical tops below).

If two main moments of inertia coincide, what about the symmetrical Gyro is the case, then the energy ellipsoid is an ellipsoid of revolution, the intersection curves with the sphere are then small circles and the separatrix becomes a great circle. An unstable axis of rotation then does not exist.

The movements of the angular momentum in the local frame of reference

The angular momentum wanders through these curves - with the exception of the separatrix - completely without ever standing still or even changing the direction of rotation. Because on the curves away from the main axes of inertia, at most one component of the angular momentum disappears. Therefore the local velocities $ \ dot {L} _ {1,2,3} $ cannot all three be zero at once. In the special case of the separatrix, an aperiodic movement develops because the angular momentum cannot exceed the intersection points on the 2-axis. It turns out that the main axis of inertia with the mean main moment of inertia on a loxodrome asymptotically approaches the axis given by the angular momentum, see #Movement on the separatrix below.

If the rotational energy decreases, for example through dissipation, the axis of rotation will move in the direction of the axis with the greatest moment of inertia, which is the 3-axis in the picture, because there the energy ellipsoid with the lowest energy touches the angular momentum sphere.

Force-free symmetrical top

At the symmetrical By definition, gyroscopes are two of the main moments of inertia. Without loss of generality, here is from I.1=I.2=:I.0 and rotation around the 3-axis - the figure axis - assumed.

Consideration in the rotating frame of reference

In the case of a symmetrical top, the third top equation is simplified in the force-free case to $ I_3 \, \ dot {\ omega} _3 = 0 $, so that the angular velocity $ \ omega_3 $ is constant. The other two gyroscopic equations form the linear ordinary differential equation system

$ \ begin {align} 0 = & \ dot {\ omega} _1 + \ frac {I_3-I_0} {I_0} \ omega_2 \ omega_3 = \ dot {\ omega} _1 + \ Omega \ omega_2 \ 0 = & \ dot { \ omega} _2 + \ frac {I_0-I_3} {I_0} \ omega_3 \ omega_1 = \ dot {\ omega} _2- \ Omega \ omega_1 \ end {align} $

with constant coefficient $ \ Omega: = \ tfrac {I_3-I_0} {I_0} \ omega_3 $. The general solution of these equations can be represented as follows:

$ \ begin {pmatrix} \ omega_1 (t) \ omega_2 (t) \ end {pmatrix} = \ begin {pmatrix} \ cos (\ Omega \, t) & - \ sin (\ Omega \, t) \ \ sin (\ Omega \, t) & \ cos (\ Omega \, t) \ end {pmatrix} \ begin {pmatrix} \ omega_1 (0) \ omega_2 (0) \ end {pmatrix} $

The values ω1,2(0) are initial conditions at the time t= 0. If ω3(0) = 0 and / or ω1(0)=ω2(0) = 0 holds, so stay ω1 and ω2 constant and the gyro performs a constant rotary movement or remains in special cases ω1,2,3(0) = 0 at rest.

Fig. 4: Components of movement in a force-free top

For the sketching of the general movement, the center of mass of the top is at the point in time t= 0 a Cartesian coordinate system with x, y and z axes is placed so that the figure axis and the angular velocity lie in the x-z plane, see Fig. 4. Let the angle between the angular velocity and the figure axis be λ. Then ω1(0)=ω sin (λ), ω2(0) = 0 (shown differently in Fig. 4) and ω3(0)=ω cos (λ) with the amount $ \ omega: = | \ vec \ omega | $ the angular velocity. The main axes of inertia belonging to the angular velocities are denoted by $ \ hat {g} _ {1,2,3} $.

The solution of the gyroscopic equations given above results with the given initial conditions:

$ \ begin {pmatrix} \ omega_1 (t) \ omega_2 (t) \ end {pmatrix} = \ begin {pmatrix} \ cos (\ Omega \, t) & - \ sin (\ Omega \, t) \ \ sin (\ Omega \, t) & \ cos (\ Omega \, t) \ end {pmatrix} \ begin {pmatrix} \ omega \ sin (\ lambda) \ 0 \ end {pmatrix} = \ begin { pmatrix} \ omega \ sin (\ lambda) \ cos (\ Omega \, t) \ omega \ sin (\ lambda) \ sin (\ Omega \, t) \ end {pmatrix} $

The amplitude $ \ omega_ \ bot: = \ sqrt {\ omega_1 ^ 2 + \ omega_2 ^ 2} = \ omega | \ sin (\ lambda) | $ is constant, so that the figure axis and the angular velocity always have the same angle, namely λ, lock in. The difference vector $ \ vec {\ omega} _ \ bot: = \ vec \ omega- \ omega_3 \ hat {g} _3 = \ omega_1 \ hat {g} _1 + \ omega_2 \ hat {g} _2 $ has the amount $ \ omega_ \ bot $ and rotates around the figure axis with the speed $ \ tfrac {\ Omega} {2 \ pi} $. The angular velocity therefore executes a rotary movement around the figure axis in the main axis system that is fixed to the body and thereby forms the fixed axis Gear pole cone (red in Fig. 4 and 5). This movement of the axis of rotation is known as precession.

The angular momentum is in the spatially fixed system

$ \ begin {align} \ vec {L} _s = & \ mathbf {I} \ cdot \ vec {\ omega} = I_0 \ omega_1 \ hat {g} _1 + I_0 \ omega_2 \ hat {g} _2 + I_3 \ omega_3 \ hat {g} _3 = I_0 (\ vec {\ omega} - \ omega_3 \ hat {g} _3) + I_3 \ omega_3 \ hat {g} _3 \ = & I_0 \ vec {\ omega} + (I_3 -I_0) \ omega_3 \ hat {g} _3 \ end {align} $

constant around the center of mass (green in Fig. 4). The latter decomposition shows that the angular momentum lies in the plane spanned by the figure axis and the angular velocity $ \ vec {\ omega} $. The movement of the force-free symmetrical top can only consist in the fact that the figure axis and the angular velocity rotate together around the axis fixed in space, which is defined by the angular momentum. The movement of the figure axis is called nutation (black ellipse in Figs. 4 and 5).

The coordinate system can now - as in Fig. 4 - be aligned so that the angular momentum points in the z-direction and thus $ \ vec {L} _s =: L \ hat {e} _z $ applies. Because the rotational energy

$ E _ {\ rm red} = \ frac {1} {2} \ vec {\ omega} \ cdot \ mathbf {I} \ cdot \ vec {\ omega} = \ frac {1} {2} \ vec {\ omega} \ cdot L \ hat {e} _z = \ frac {1} {2} \ omega_z L $

also does not change, so is the z-component ωz the angular velocity constant. The angular velocity thus also moves around the z-direction that is fixed in space on a cone, the one that is fixed in space Locking pole cone (blue in Fig. 4 and 5, there called “spatially fixed gang pole cone”.) The angle β between the figure axis and the angular momentum as well as the z-component of the angular velocity can be determined with the mechanical analysis in the following section.

Fig. 5: Form of movement of an oblate, force-free top

The three vectors angular momentum, figure axis and angular velocity do not change their relative position: the gear pole cone rolls on the detent pole cone. At the prolates The (slim) top is $ I_0> I_3 $ and the gear pole cone rolls as in Fig. 4 Outside on the locking pole cone. At the oblaten The (sturdy) top is $ I_3> I_0 $ and the gear pole cone rolls as in Fig. 5 Inside on the locking pole cone.

The rolling is even slip-free, because the joint surface line of the detent pole and gear pole cone is the momentary axis of rotation set by the angular velocity, which goes through the stationary center of mass (shown differently in Fig. 5). Euler's equation of velocity $ \ dot {\ vec {x}} = \ dot {\ vec {s}} + \ vec {\ omega} \ times (\ vec {x} - \ vec {s}) $ is reduced to $ \ dot {\ vec {x}} = \ vec {\ omega} \ times \ vec {x} $, if the center of mass in the origin of the coordinate system is chosen as the reference point $ \ vec s $. Thus, the particles of the top stand still on the axis of rotation (for $ \ vec x \ parallel \ vec \ omega $), the detent pole cone is at rest anyway, and slip between the gear pole and detent pole cone is therefore excluded.

Movement function of the symmetrical top

Fig. 6: Euler's basic system (green) indicates the axes around which the Euler angles α, β and γ rotate.

The computation of the gyroscopic motion in the fixed-space reference system succeeds with the Euler angles, for example in the standard x-convention (z, x ', z "), see Fig. 6. The unit vectors $ \ hat {e} _ {x, y , z} $ the standard basis fixed in space (blue in Fig. 6) and $ \ hat {e} _ {X, Y, Z} = \ hat {g} _ {1,2,3} $ the one rotating with the body, moving base (red in Fig. 6), then the moving base vectors with respect to the spatially fixed base are:

$ \ begin {align} \ hat {g} _1 = & \ begin {pmatrix} \ cos (\ alpha) \ cos (\ gamma) - \ sin (\ alpha) \ cos (\ beta) \ sin (\ gamma) \ \ sin (\ alpha) \ cos (\ gamma) + \ cos (\ alpha) \ cos (\ beta) \ sin (\ gamma) \ \ sin (\ beta) \ sin (\ gamma) \ end { pmatrix} \ \ hat {g} _2 = & \ begin {pmatrix} - \ cos (\ alpha) \ sin (\ gamma) - \ sin (\ alpha) \ cos (\ beta) \ cos (\ gamma) \ \ - \ sin (\ alpha) \ sin (\ gamma) + \ cos (\ alpha) \ cos (\ beta) \ cos (\ gamma) \ \ sin (\ beta) \ cos (\ gamma) \ end { pmatrix} \ \ hat {g} _3 = & \ begin {pmatrix} \ sin (\ alpha) \ sin (\ beta) \ - \ cos (\ alpha) \ sin (\ beta) \ \ cos (\ beta) \ end {pmatrix} \,. \ end {align} $

They can be found in the rows of the rotation matrix. If, as in the previous section, the angular momentum points in the direction of the z-axis and the angular velocity ω as well as the angle λ are given, then the angular momentum is calculated

$ L = \ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)} \, \ omega \ ,, $

the angular velocities

$ \ begin {align} \ Omega = & \ frac {I_3-I_0} {I_0} \ omega \ cos (\ lambda) \ \ omega_1 = & \ dot \ alpha \ sin (\ beta) \ sin (\ gamma) = \ omega \ sin (\ lambda) \ cos (\ Omega t) \ \ omega_2 = & \ dot \ alpha \ sin (\ beta) \ cos (\ gamma) = \ omega \ sin (\ lambda) \ sin ( \ Omega t) \ \ omega_3 = & \ dot \ alpha \ cos (\ beta) + \ dot \ gamma = \ omega \ cos (\ lambda) \ end {align} $

and the angles

$ \ alpha = \ frac {\ pi} {2} + \ frac {1} {I_0} \ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)} \ , \ omega t \ ,, \ quad \ beta = \ arctan \ left (\ frac {I_0} {I_3} \ tan (\ lambda) \ right) \ ,, \ quad \ gamma = \ frac {\ pi} {2 } - \ Omega t \ ,. $

The function tan is the tangent and arctan is its arc function. The plane spanned by the z-axis and the figure axis, which also contains the angular velocity vector, closes the angle with the x-z plane

$ \ varphi = \ frac {I_3 \ cos ^ 2 (\ lambda) + I_0 \ sin ^ 2 (\ lambda)} {\ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)}} \, \ omega t $

a, a movement known as nutation.

proof
The proof succeeds with the Lagrange formalism as follows. The unit vectors $ \ hat {e} _ {x, y, z} $ denote the space-fixed standard basis (blue in Fig. 6) and $ \ hat {e} _ {X, Y, Z} = \ hat {g} _ {1,2,3} $ the moving base rotating with the body (red in Fig. 6), then the vectors are $ \ hat {u} _ {\ alpha, \ beta, \ gamma} $ des Eulerian frame of referencethat indicates the axes around which the angles α, β or. γ turn (green in Fig. 6):

$ \ begin {align} \ hat {u} _ \ alpha = & \ hat {e} _z \ = & \ sin (\ beta) \ sin (\ gamma) \ hat {g} _1 + \ sin (\ beta) \ cos (\ gamma) \ hat {g} _2 + \ cos (\ beta) \ hat {g} _3 \ \ hat {u} _ \ beta = & \ cos (\ alpha) \ hat {e} _x + \ sin (\ alpha) \ hat {e} _y \ = & \ cos (\ gamma) \ hat {g} _1- \ sin (\ gamma) \ hat {g} _2 \ \ hat {u} _ \ gamma = & \ sin (\ alpha) \ sin (\ beta) \ hat {e} _x- \ cos (\ alpha) \ sin (\ beta) \ hat {e} _y + \ cos (\ beta) \ hat {e} _z \ = & \ hat {g} _3 \ end {align} $
The angular velocity in the moving system is thus:
$ \ begin {align} \ vec \ omega = & \ dot {\ alpha} \ hat {u} _ \ alpha + \ dot {\ beta} \ hat {u} _ \ beta + \ dot {\ gamma} \ hat {u } _ \ gamma \ = & [\ underbrace {\ dot \ alpha \ sin (\ beta) \ sin (\ gamma) + \ dot \ beta \ cos (\ gamma)} _ {\ omega_1}] \ hat {g } _1 + [\ underbrace {\ dot \ alpha \ sin (\ beta) \ cos (\ gamma) - \ dot \ beta \ sin (\ gamma)} _ {\ omega_2}] \ hat {g} _2 + [\ underbrace {\ dot \ alpha \ cos (\ beta) + \ dot \ gamma} _ {\ omega_3}] \ hat {g} _3 \,. \ end {align} $
The angular momentum is constant in the absence of external moments and wise in the z-direction:
$ \ begin {align} \ vec {L} = & L \ hat {e} _z = \ mathbf {I} \ cdot \ vec \ omega = I_0 \ omega_1 \ hat {g} _1 + I_0 \ omega_2 \ hat {g} _2 + I_3 \ omega_3 \ hat {g} _3 \ L = & \ vec {L} \ cdot \ hat {e} _z = I_0 \ omega_1 \ sin (\ beta) \ sin (\ gamma) + I_0 \ omega_2 \ sin (\ beta) \ cos (\ gamma) + I_3 \ omega_3 \ cos (\ beta) \ = & I_0 \ dot \ alpha \ sin ^ 2 (\ beta) + I_3 [\ dot \ alpha \ cos (\ beta ) + \ dot \ gamma] \ cos (\ beta) \ end {align} $
The Lagrange function here is the kinetic energy of the top:
$ \ mathcal {L}: = \ frac12 \ vec {\ omega} \ cdot \ mathbf {I} \ cdot \ vec {\ omega} = \ frac {I_0} {2} (\ omega_1 ^ 2 + \ omega_2 ^ 2 ) + \ frac {I_3} {2} \ omega_3 ^ 2 = \ frac {I_0} {2} [{\ dot \ alpha} ^ 2 \ sin ^ 2 (\ beta) + {\ dot {\ beta}} ^ 2] + \ frac {I_3} {2} [\ dot \ alpha \ cos (\ beta) + \ dot \ gamma] ^ 2 \ ,. $
According to the Lagrange formalism, the angles are α and γcyclical and your conjugate impulses constant:
$ \ begin {align} p_ \ alpha: = & \ frac {\ partial \ mathcal {L}} {\ partial \ dot {\ alpha}} = I_0 \ dot \ alpha \ sin ^ 2 (\ beta) + I_3 [ \ dot \ alpha \ cos (\ beta) + \ dot \ gamma] \ cos (\ beta) = L = \ text {const.} \ [1ex] p_ \ gamma: = & \ frac {\ partial \ mathcal { L}} {\ partial \ dot {\ gamma}} = I_3 [\ dot \ alpha \ cos (\ beta) + \ dot \ gamma] = I_3 \ omega_3 = \ vec {L} \ cdot \ hat {g} _3 = L \ cos (\ beta) = \ text {const.} \ End {align} $
The first equation confirms the conservation of angular momentum and the second shows that the angular velocity ω3 around the figure axis and the angle β are constant between the figure axis and the z-direction.
The third equation of motion results from the Lagrange function for β:
$ \ begin {align} \ frac {\ mathrm {d}} {\ mathrm {d} t} \ frac {\ partial \ mathcal {L}} {\ partial \ dot \ beta} = & \ frac {\ partial \ mathcal {L}} {\ partial \ beta} \ \ rightarrow I_0 \ ddot \ beta = 0 = & I_0 {\ dot \ alpha} ^ 2 \ sin (\ beta) \ cos (\ beta) -I_3 [\ dot \ alpha \ cos (\ beta) + \ dot \ gamma] \ dot \ alpha \ sin (\ beta) = [I_0 \ dot \ alpha \ cos (\ beta) -I_3 \ omega_3] \ dot \ alpha \ sin (\ beta) \ \ rightarrow \ dot \ alpha = & \ frac {I_3 \ omega_3} {I_0 \ cos (\ beta)} = \ text {const.} \ end {align} $
For the angular momentum this means:
$ L = I_0 \ dot \ alpha \ sin ^ 2 (\ beta) + I_3 \ omega_3 \ cos (\ beta) = I_0 \ dot \ alpha \ sin ^ 2 (\ beta) + I_0 \ dot \ alpha \ cos ^ 2 (\ beta) = I_0 \ dot \ alpha = \ frac {I_3 \ omega_3} {\ cos (\ beta)} \ ,. $
From $ \ omega_3 = \ dot \ alpha \ cos (\ beta) + \ dot \ gamma $ it follows
$ \ dot \ gamma = \ omega_3- \ dot \ alpha \ cos (\ beta) = \ omega_3- \ frac {I_3 \ omega_3} {I_0 \ cos (\ beta)} \ cos (\ beta) = \ frac {I_0 -I_3} {I_0} \ omega_3 = - \ Omega = \ text {const.} $
In Euler's reference system, $ \ vec \ omega = \ dot \ alpha \ hat {e} _z + \ dot \ gamma \ hat {g} _3 $ because of $ \ dot \ beta = 0 $. The angular velocity around the z-axis is calculated as follows
$ \ omega_z: = \ vec {\ omega} \ cdot \ hat {e} _z = \ dot \ alpha + \ dot \ gamma (\ hat {g} _3 \ cdot \ hat {e} _z) = \ dot \ alpha + \ dot \ gamma \ cos (\ beta) = \ frac {L} {I_0} + \ frac {I_0-I_3} {I_0} \ frac {I_3} {L} \ omega_3 ^ 2 \ ,. $

Initial conditions
For now t= 0 is $ \ omega_3 = \ vec {\ omega} \ cdot \ hat {g} _3 = \ omega \ cos (\ lambda) $ and keeps this value ω3. The angle β can now as a function of the angle λ can be expressed:

$ \ begin {align} \ omega \ cos (\ lambda) = & \ omega_3 = \ dot \ alpha \ cos (\ beta) + \ dot \ gamma = \ dot \ alpha \ cos (\ beta) - \ Omega \ \ omega \ sin (\ lambda) = & \ omega_ \ bot = \ sqrt {\ omega_1 ^ 2 + \ omega_2 ^ 2} = \ dot \ alpha \ sin (\ beta) \ \ rightarrow \ cot (\ lambda) = & \ cot (\ beta) - \ frac {\ Omega} {\ dot \ alpha \ sin (\ beta)} = \ cot (\ beta) + \ frac {I_0-I_3} {I_0} \ omega_3 \ frac {I_0 \ cos (\ beta)} {I_3 \ omega_3 \ sin (\ beta)} = \ cot (\ beta) + \ frac {I_0-I_3} {I_3} \ cot (\ beta) \ \ rightarrow \ tan (\ beta) = & \ frac {I_0} {I_3} \ tan (\ lambda) \,. \ end {align} $
The cotangent cot is the reciprocal of the tangent. Because of $ \ cos x = (1+ \ tan ^ 2 x) ^ {- 1/2} $ and $ \ tan (\ beta) = \ tfrac {I_0} {I_3} \ tan (\ lambda) $ follows for the Angular momentum:
$ L = \ frac {I_3 \ omega_3} {\ cos (\ beta)} = \ frac {I_3 \ omega \ cos (\ lambda)} {\ cos (\ beta)} = \ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)} \, \ omega \ ,. $
The specifications
$ \ begin {align} \ omega_1 (t = 0) = & \ dot \ alpha \ sin (\ beta) \ sin (\ gamma) = \ omega \ sin (\ lambda) \ sin (\ gamma) \, \ stackrel {\ displaystyle!} {=} \, \ omega \ sin (\ lambda) \ \ omega_2 (t = 0) = & \ dot \ alpha \ sin (\ beta) \ cos (\ gamma) = \ omega \ sin (\ lambda) \ cos (\ gamma) \, \ stackrel {\ displaystyle!} {=} \, 0 \ end {align} $
can with the initial value of the angle γ be fulfilled by $ \ tfrac {\ pi} {2} $, so that $ \ gamma = \ tfrac {\ pi} {2} - \ Omega t $. The angular velocity is currently t=0:
$ \ vec {\ omega} = \ omega \ sin (\ lambda) \ hat {g} _1 + \ omega \ cos (\ lambda) \ hat {g} _3 = \ omega \ begin {pmatrix} \ sin (\ alpha) \ sin (\ beta- \ lambda) \ - \ cos (\ alpha) \ sin (\ beta- \ lambda) \ \ cos (\ beta- \ lambda) \ end {pmatrix} \ ,. $
So that this lies in the x-z plane, the initial value of α set to $ \ tfrac {\ pi} {2} $ so that
$ \ alpha = \ frac {\ pi} {2} + \ dot \ alpha t = \ frac {\ pi} {2} + \ frac {L} {I_0} t = \ frac {\ pi} {2} + \ frac {1} {I_0} \ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)} \, \ omega t $
results. If the angle of rotation of the figure axis around the z-axis with φ and has the value zero at the beginning, then follows:
$ \ varphi: = \ omega_z t = \ left [\ frac {L} {I_0} + \ frac {I_0-I_3} {I_0} \ frac {I_3} {L} \ cos ^ 2 (\ lambda) \ omega ^ 2 \ right] t = \ frac {I_3 \ cos ^ 2 (\ lambda) + I_0 \ sin ^ 2 (\ lambda)} {\ sqrt {I_3 ^ 2 \ cos ^ 2 (\ lambda) + I_0 ^ 2 \ sin ^ 2 (\ lambda)}} \, \ omega t \ ,. $

Force-free asymmetrical top

By definition, asymmetrical gyroscopes have three different main moments of inertia. If such a top rotates around the 3-axis, then this movement can be unstable or stable. In the former case, small perturbations increase exponentially and the top begins to stagger, which is explained in the next section. In the stable case, periodic forms of movement develop in the second section. At the end you will be informed about the special case of the movement on the Separatrix, which was defined in the section #General properties of the movement of force-free rotating gyroscopes.

Stability of the movement of asymmetrical tops

File: Dzhanibekov effect.ogv

The main axes with the largest or the smallest main moment of inertia are stable axes of rotation. This has been known since 1851 at the latest[1] and easy to demonstrate with a rotating table tennis bat. In English, the statement is accordingly as "sentence from the tennis racket" (tennis racket theorem)[2] common. After the Soviet cosmonaut Vladimir Dschanibekow observed the movement of a component around its unstable main axis of inertia during a space flight in 1985, the situation was examined more closely[3] and has been called the "Dschanibekow Effect" since then.

To check the stability of the rotary axes, the gyro should first rotate around the 3-axis: $ \ omega_3 \ ne 0 $ and $ | \ omega_ {1,2} | \ ll \ sqrt {| \ dot \ omega_3 | } $. The gyro equations are now

$ \ begin {align} 0 = & \ dot {\ omega} _1- \ frac {I_2-I_3} {I_1} \ omega_2 \ omega_3 \ 0 = & \ dot {\ omega} _2- \ frac {I_3-I_1 } {I_2} \ omega_3 \ omega_1 \ 0 = & \ dot {\ omega} _3- \ frac {I_1-I_2} {I_3} \ omega_1 \ omega_2 \ approx \ dot {\ omega} _3 \ end {align} $

Analogous to the consideration in the rotating reference system, the angular velocity is derived from derivatives according to time and with the approximate constancy of the angular velocity ω3:

$ \ begin {align} 0 = & \ ddot {\ omega} _1- \ frac {I_2-I_3} {I_1} \ omega_3 \ dot {\ omega} _2 = \ ddot {\ omega} _1- \ frac {I_2- I_3} {I_1} \ frac {I_3-I_1} {I_2} \ omega_3 ^ 2 \ omega_1 = \ ddot {\ omega} _1 + k \ omega_1 \ 0 = & \ ddot {\ omega} _2- \ frac {I_3 -I_1} {I_2} \ omega_3 \ dot {\ omega} _1 = \ ddot {\ omega} _2- \ frac {I_3-I_1} {I_2} \ frac {I_2-I_3} {I_1} \ omega_3 ^ 2 \ omega_2 = \ ddot {\ omega} _2 + k \ omega_2 \ & \ text {with} \ quad k: = \ frac {I_1-I_3} {I_2} \ frac {I_2-I_3} {I_1} \ omega_3 ^ 2 \ end {align} $

If k is negative, there is positive feedback of the angular velocities and thus the rotation around the 3-axis is abandoned to stagger. If k is positive, there are periodic forms of motion. For this, the main moments of inertia I.1,2 either both larger or both smaller than the third principal moment of inertia I.3 from which the above statement about the stability of the axes follows.

Movement function of the asymmetrical top

Fig. 7: Time courses of the Jacobi elliptic functions sn, cn and dn at k = 0.95

The main moments of inertia are numbered in such a way that I.1< I.2< I.3 applies. Then the gyroscopic equations can be fulfilled in the force-free case with the Jacobi elliptic functions sn, cn and dn[4]. From the rotational energy and the square of the magnitude of the angular momentum

$ E_ \ text {red}: = \ frac12 \ vec \ omega \ cdot \ mathbf {I} \ cdot \ vec \ omega = \ frac12 (I_1 \ omega_1 ^ 2 + I_2 \ omega_2 ^ 2 + I_3 \ omega_3 ^ 2) \ ,, \ quad L ^ 2: = \ vec {L} \ cdot \ vec {L} = I_1 ^ 2 \ omega_1 ^ 2 + I_2 ^ 2 \ omega_2 ^ 2 + I_3 ^ 2 \ omega_3 ^ 2 \ ,. $

the angular velocities result

$ \ begin {align} \ omega_1 = & A_1 \ operatorname {cn} (at; k) \ \ omega_2 = & A_2 \ operatorname {sn} (at; k) \ \ omega_3 = & A_3 \ operatorname {dn} (at; k) \ end {align} $

with the amplitudes

$ A_1 = \ sqrt {\ frac {2I_3E_ \ text {red} -L ^ 2} {I_1 (I_3-I_1)}} \ ,, \ quad A_2 = \ sqrt {\ frac {2I_3E_ \ text {red} -L ^ 2} {I_2 (I_3-I_2)}} \ ,, \ quad A_3 = \ sqrt {\ frac {L ^ 2-2I_1E_ \ text {red}} {I_3 (I_3-I_1)}} \ ,, $

the frequency and the elliptical module

$ a = \ sqrt {\ frac {I_3-I_2} {I_1I_2I_3} (L ^ 2-2I_1E_ \ text {red})} \ ,, \ quad k = \ sqrt {\ frac {I_2-I_1} {I_3-I_2 } \, \ frac {2I_3E_ \ text {red} -L ^ 2} {L ^ 2-2I_1E_ \ text {red}}} \ ,. $

It appears

$ \ begin {align} 2I_3E_ \ text {rot} -L ^ 2 = & I_1 (I_3-I_1) \ omega_1 ^ 2 + I_2 (I_3-I_2) \ omega_2 ^ 2> 0 \ L ^ 2-2I_1E_ \ text {red} = & I_2 ​​(I_2-I_1) \ omega_2 ^ 2 + I_3 (I_3-I_1) \ omega_3 ^ 2> 0 \ ,, \ end {align} $

so that the aforementioned constants are real. The functions sn and cn are periodic with period 4K and dn with period 2K, see Fig. 7. Where is K the complete elliptic integral of the first kind:

$ K: = \ int_0 ^ {\ frac {\ pi} {2}} \ frac {\ mathrm {d} \ theta} {\ sqrt {1-k ^ 2 \ sin ^ 2 \ theta}} \ ,. $

Thus the angular velocity is periodic with the period length $ T = \ tfrac {4K} {a} $. After this time, the angular velocity has returned to its initial state: $ \ vec \ omega (T) = \ vec \ omega (0) \ ,. $

However, this does not apply to the top as a whole: it generally does not return to an initial position.[4] The Euler angles - see #Motion function of the symmetrical top - result in

$ \ begin {align} \ dot \ alpha = & L \ frac {I_3-I_2 + (I_2-I_1) \ operatorname {sn} ^ 2 (at; k)} {I_1 (I_3-I_2) + I_3 (I_2-I_1 ) \ operatorname {sn} ^ 2 (at; k)} \ \ cos (\ beta) = & \ sqrt {\ frac {I_3} {I_3-I_1} \ frac {L ^ 2-2I_1E_ \ text {rot} } {L ^ 2}} \ operatorname {dn} (at; k) \ \ tan (\ gamma) = & \ sqrt {\ frac {I_1 (I_3-I_2)} {I_2 (I_3-I_1)}} \ frac {\ operatorname {cn} (at; k)} {\ operatorname {sn} (at; k)} \ end {align} $

The formulas remain valid if the main moments of inertia are reversed I.1 > I.2 > I.3 exhibit. However, the differences return

$ \ begin {align} 2I_3E_ \ text {rot} -L ^ 2 = & I_1 (I_3-I_1) \ omega_1 ^ 2 + I_2 (I_3-I_2) \ omega_2 ^ 2 <0 \ L ^ 2-2I_1E_ \ text {red} = & I_2 ​​(I_2-I_1) \ omega_2 ^ 2 + I_3 (I_3-I_1) \ omega_3 ^ 2 <0 \ end {align} $

then change their sign. In contrast to the force-free symmetrical top, the angular velocities are $ \ omega_3, \, \ dot \ alpha, \, \ dot \ gamma $ and the angle β between the angular momentum and the 3-axis Not constant. In a special case I.1=I.2 is A.1=A.2 and k= 0, so that the elliptic functions sn and cn merge into the harmonic functions sin and cos and dn≡1. Then the local solution changes to that of the symmetrical top.

proof
First it is shown that the angular velocities satisfy Euler's gyroscopic equations. The Jacobi elliptic functions satisfy the differential equations

$ \ begin {align} \ frac {\ mathrm {d}} {\ mathrm {d} z} \ operatorname {sn} (z; k) = & \ operatorname {cn} (z; k) \ operatorname {dn} (z; k) \ \ frac {\ mathrm {d}} {\ mathrm {d} z} \ operatorname {cn} (z; k) = & - \ operatorname {sn} (z; k) \ operatorname { dn} (z; k) \ \ frac {\ mathrm {d}} {\ mathrm {d} z} \ operatorname {dn} (z; k) = & - k ^ 2 \ operatorname {sn} (z; k) \ operatorname {cn} (z; k) \ end {align} $
With $ z = at $ and $ \ dot z = a $ this results in accordance with the gyro equations:
$ \ begin {align} \ dot {\ omega} _1 = & A_1 \ frac {\ mathrm {d}} {\ mathrm {d} t} \ operatorname {cn} (at; k) = A_1 \ frac {\ mathrm { d}} {\ mathrm {d} z} \ operatorname {cn} (z; k) \ dot z = -A_1a \ operatorname {sn} (at; k) \ operatorname {dn} (at; k) = - \ frac {A_1a} {A_2A_3} \ omega_2 \ omega_3 = \ frac {I_2-I_3} {I_1} \ omega_2 \ omega_3 \ \ dot {\ omega} _2 = & A_2 \ frac {\ mathrm {d}} {\ mathrm {d} t} \ operatorname {sn} (at; k) = A_2 \ frac {\ mathrm {d}} {\ mathrm {d} z} \ operatorname {sn} (z; k) \ dot {z} = A_2a \ operatorname {cn} (at; k) \ operatorname {dn} (at; k) = \ frac {A_2a} {A_1A_3} \ omega_1 \ omega_3 = \ frac {I_3-I_1} {I_2} \ omega_1 \ omega_3 \ \ \ dot {\ omega} _3 = & A_3 \ frac {\ mathrm {d}} {\ mathrm {d} t} \ operatorname {dn} (at; k) = A_3 \ frac {\ mathrm {d}} { \ mathrm {d} z} \ operatorname {dn} (z; k) \ dot {z} = -A_3ak ^ 2 \ operatorname {sn} (at; k) \ operatorname {cn} (at; k) = - \ frac {A_3ak ^ 2} {A_1A_2} \ omega_1 \ omega_2 = \ frac {I_1-I_2} {I_3} \ omega_1 \ omega_2 \,. \ end {align} $
Comparison of the components of the angular momentum in Euler angles (see #Consideration in the rotating reference system for a symmetrical top) yields in the moving system $ \ hat {g} _ {1,2,3} $:
$ \ begin {align} \ vec L = & L \ hat {e} _z = L \ begin {pmatrix} \ sin (\ beta) \ sin (\ gamma) \ \ sin (\ beta) \ cos (\ gamma ) \ \ cos (\ beta) \ end {pmatrix} _ {\ hat {g} _i} = \ begin {pmatrix} L_1 \ L_2 \ L_3 \ end {pmatrix} _ {\ hat {g} _i} = \ mathbf {I} \ cdot \ vec \ omega = \ begin {pmatrix} I_1 \ omega_1 \ I_2 \ omega_2 \ I_3 \ omega_3 \ end {pmatrix} _ {\ hat {g} _i} = \ begin {pmatrix } I_1A_1 \ operatorname {cn} (at; k) \ I_2A_2 \ operatorname {sn} (at; k) \ I_3A_3 \ operatorname {dn} (at; k) \ end {pmatrix} _ {\ hat {g} _i} \ \ rightarrow \ cos (\ beta) = & \ frac {L_3} {L} = \ frac {I_3A_3} {L} \ operatorname {dn} (at; k) = \ sqrt {\ frac {I_3} {I_3-I_1} \ frac {L ^ 2-2I_1E_ \ text {red}} {L ^ 2}} \ operatorname {dn} (at; k) \ \ tan (\ gamma) = & \ frac {L_1} {L_2} = \ frac {I_1A_1 \ operatorname {cn} (at; k)} {I_2A_2 \ operatorname {sn} (at; k)} = \ sqrt {\ frac {I_1 (I_3-I_2)} {I_2 (I_3 -I_1)}} \ frac {\ operatorname {cn} (at; k)} {\ operatorname {sn} (at; k)} \ end {align} $
The angle α is determined by $ L_1 ^ 2 + L_2 ^ 2 = I_1 ^ 2 \ omega_1 ^ 2 + I_2 ^ 2 \ omega_2 ^ 2 = L ^ 2 \ sin ^ 2 (\ beta) $ and cn² = 1-sn²

$ \ begin {align} \ vec \ omega = & \ begin {pmatrix} \ omega_1 \ \ omega_2 \ \ omega_3 \ end {pmatrix} _ {\ hat {g} _i} = \ begin {pmatrix} \ dot \ alpha \ sin (\ beta) \ sin (\ gamma) + \ dot \ beta \ cos (\ gamma) \ \ dot \ alpha \ sin (\ beta) \ cos (\ gamma) - \ dot \ beta \ sin ( \ gamma) \ \ dot \ alpha \ cos (\ beta) + \ dot \ gamma \ end {pmatrix} _ {\ hat {g} _i} \ \ rightarrow \ dot \ alpha = & \ frac {\ omega_1 \ sin (\ gamma) + \ omega_2 \ cos (\ gamma)} {\ sin (\ beta)} = L \ frac {\ omega_1 \ overbrace {L \ sin (\ beta) \ sin (\ gamma)} ^ {L_1 = I_1 \ omega_1} + \ omega_2 \ overbrace {L \ sin (\ beta) \ cos (\ gamma)} ^ {L_2 = I_2 \ omega_2}} {L ^ 2 \ sin ^ 2 (\ beta)} = L \ frac {I_1 \ omega_1 ^ 2 + I_2 \ omega_2 ^ 2} {I_1 ^ 2 \ omega_1 ^ 2 + I_2 ^ 2 \ omega_2 ^ 2} \ = & L \ frac {I_1A_1 ^ 2 \ operatorname {cn} ^ 2 ( at; k) + I_2A_2 ^ 2 \ operatorname {sn} ^ 2 (at; k)} {I_1 ^ 2A_1 ^ 2 \ operatorname {cn} ^ 2 (at; k) + I_2 ^ 2A_2 ^ 2 \ operatorname {sn} ^ 2 (at; k)} = L \ frac {I_3-I_2 + (I_2-I_1) \ operatorname {sn} ^ 2 (at; k)} {I_1 (I_3-I_2) + I_3 (I_2-I_1) \ operatorname {sn} ^ 2 (at; k)} \,. \ end {align} $

Movement on the Separatrix

Fig. 8: Path of a point on the 2-axis (red) around the angular momentum axis (vertical line) along a loxodrome

On the separatrix is ​​$ 2I_2 E _ {\ rm rot} = L ^ 2 $ and the movement is aperiodic because the angular velocity does not assume a state a second time. The movement of the top is characterized here by the fact that the plane spanned by the 2-axis and the angular momentum has a constant angular velocity L./I.2 circles around the angular momentum axis and the end point of the 2-axis is on a loxodrome with the direction angle

$ \ cos \ eta = \ sqrt {\ frac {(I_3-I_2) (I_2-I_1)} {I_2 (I_1 + I_3-I_2)}} $

approaches the axis defined by the angular momentum, see Fig. 8. The formulas of the previous section are valid here, but because the elliptical modulus is the extreme value

$ k = \ sqrt {\ frac {I_2-I_1} {I_3-I_2} \, \ frac {2I_3E_ \ text {red} -L ^ 2} {L ^ 2-2I_1E_ \ text {red}}} = \ sqrt {\ frac {I_2-I_1} {I_3-I_2} \, \ frac {2I_3E_ \ text {red} -2I_2 E _ {\ rm red}} {2I_2 E _ {\ rm red} -2I_1E_ \ text {red}}} = $ 1

assumes, the elliptic functions go into the aperiodic hyperbolic functions:

$ \ operatorname {cn} (at; 1) = \ operatorname {dn} (at; 1) = \ frac {1} {\ cosh (at)} $ and $ \ operatorname {sn} (at; 1) = \ tanh (at). $

The frequency and the angular velocities of the previous section specialize in:

$ \ begin {align} a = & \ sqrt {\ frac {I_3-I_2} {I_1I_2I_3} (L ^ 2-2I_1E_ \ text {rot})} = \ sqrt {\ frac {(I_3-I_2) (I_2- I_1)} {I_1I_2 ^ 2I_3}} L \ \ omega_1 = & \ sqrt {\ frac {2I_3E_ \ text {red} -L ^ 2} {I_1 (I_3-I_1)}} \ operatorname {cn} (at; k) = \ sqrt {\ frac {I_3-I_2} {I_1I_2 (I_3-I_1)}} \ frac {L} {\ cosh (at)} \ \ omega_2 = & \ sqrt {\ frac {2I_3E_ \ text { red} -L ^ 2} {I_2 (I_3-I_2)}} \ operatorname {sn} (at; k) = \ frac {L} {I_2} \ tanh (at) \ \ omega_3 = & \ sqrt {\ frac {L ^ 2-2I_1E_ \ text {rot}} {I_3 (I_3-I_1)}} \ operatorname {dn} (at; k) = \ sqrt {\ frac {I_2-I_1} {I_2I_3 (I_3-I_1) }} \ frac {L} {\ cosh (at)} \ end {align} $

Go as time goes on ω1 and ω3 towards zero and ω2 against L./I.2. The movement comes as close as desired to a rotation around the 2-axis without ever reaching this state. In reality, this form of movement will hardly occur, because with the smallest deviation from the ideal case $ 2I_2 E _ {\ rm rot} = L ^ 2 $ k≠ 1 and the angular velocities become the periodic ones of the previous section. A movement near the separatrix shows the Dschanibekow effect.

Fig. 9: Basic system used $ \ hat {h} _ {1,2,3} $ and Meridian $ \ hat m $

For the calculation of the movement, unlike in the previous section, the approach $ \ hat {h} _ {1,2,3} = \ hat {e} _ {Y, Z, X} $ is used for the local base system, see Fig. 9 and see Fig. 6.

The Euler angles - see #Motion function of the symmetrical top - result from an angular momentum in the z-direction and a start with ω2= 0 to

$ \ begin {align} \ dot \ alpha = & \ frac {L} {I_2} \ \ cos (\ beta) = & \ tanh (at) \ \ tan (\ gamma) = & \ sqrt {\ frac {I_3 (I_2-I_1)} {I_1 (I_3-I_2)}} \,. \ End {align} $

The plane spanned by the 2-axis and the angular momentum circles at a constant angular velocity L./I.2 around the angular momentum axis and the angle β goes to zero as time goes on.

proof
Because in the basic system

$ \ begin {align} \ hat {h} _1 = & \ hat {e} _Y = \ hat {g} _2 = \ begin {pmatrix} - \ cos (\ alpha) \ sin (\ gamma) - \ sin ( \ alpha) \ cos (\ beta) \ cos (\ gamma) \ - \ sin (\ alpha) \ sin (\ gamma) + \ cos (\ alpha) \ cos (\ beta) \ cos (\ gamma) \ \ \ sin (\ beta) \ cos (\ gamma) \ end {pmatrix} \ \ hat {h} _2 = & \ hat {e} _Z = \ hat {g} _3 = \ begin {pmatrix} \ sin ( \ alpha) \ sin (\ beta) \ - \ cos (\ alpha) \ sin (\ beta) \ \ cos (\ beta) \ end {pmatrix} \ \ hat {h} _3 = & \ hat { e} _X = \ hat {g} _1 = \ begin {pmatrix} \ cos (\ alpha) \ cos (\ gamma) - \ sin (\ alpha) \ cos (\ beta) \ sin (\ gamma) \ \ sin (\ alpha) \ cos (\ gamma) + \ cos (\ alpha) \ cos (\ beta) \ sin (\ gamma) \ \ sin (\ beta) \ sin (\ gamma) \ end {pmatrix} \ end {align} $
surrendered:
$ \ begin {align} \ vec L = & L \ hat {e} _z = L \ begin {pmatrix} \ sin (\ beta) \ cos (\ gamma) \ \ cos (\ beta) \ \ sin ( \ beta) \ sin (\ gamma) \ end {pmatrix} _ {\ hat {h} _i} = \ begin {pmatrix} L_1 \ L_2 \ L_3 \ end {pmatrix} _ {\ hat {h} _i} = \ mathbf {I} \ cdot \ vec \ omega = \ begin {pmatrix} I_1 \ omega_1 \ I_2 \ omega_2 \ I_3 \ omega_3 \ end {pmatrix} _ {\ hat {h} _i} \ \ rightarrow \ cos (\ beta) = & \ frac {L_2} {L} = \ frac {I_2 \ omega_2} {L} = \ frac {I_2} {L} \ frac {L} {I_2} \ tanh (at) = \ tanh (at) \ \ tan (\ gamma) = & \ frac {L_3} {L_1} = \ frac {I_3} {I_1} \ frac {\ omega_3} {\ omega_1} = \ frac {I_3} {I_1} \ frac {\ sqrt {\ frac {I_2-I_1} {I_2I_3 (I_3-I_1)}} \ frac {L} {\ cosh (at)}} {\ sqrt {\ frac {I_3-I_2} {I_1I_2 (I_3 -I_1)}} \ frac {L} {\ cosh (at)}} = \ sqrt {\ frac {I_3 (I_2-I_1)} {I_1 (I_3-I_2)}} \ end {align} $
With the new basic system, the components of the angular velocity are:
$ \ begin {align} \ vec \ omega = & [\ dot \ alpha \ sin (\ beta) \ sin (\ gamma) + \ dot \ beta \ cos (\ gamma)] \ hat {g} _1 + [\ dot \ alpha \ sin (\ beta) \ cos (\ gamma) - \ dot \ beta \ sin (\ gamma)] \ hat {g} _2 + [\ dot \ alpha \ cos (\ beta) + \ dot \ gamma ] \ hat {g} _3 \ = & [\ underbrace {\ dot \ alpha \ sin (\ beta) \ sin (\ gamma) + \ dot \ beta \ cos (\ gamma)} _ {\ omega_3}] \ hat {h} _3 + [\ underbrace {\ dot \ alpha \ sin (\ beta) \ cos (\ gamma) - \ dot \ beta \ sin (\ gamma)} _ {\ omega_1}] \ hat {h} _1 + [\ underbrace {\ dot \ alpha \ cos (\ beta) + \ dot \ gamma} _ {\ omega_2}] \ hat {h} _2 \,. \ end {align} $
Now the angular velocity $ \ dot \ alpha $ can be calculated with $ \ dot \ gamma = 0 $ from $ \ omega_2 = \ dot \ alpha \ cos (\ beta) + \ dot \ gamma = \ dot \ alpha \ cos (\ beta) $ to be determined to
$ \ dot \ alpha = \ frac {\ omega_2} {\ cos (\ beta)} = \ frac {\ frac {L} {I_2} \ tanh (at)} {\ tanh (at)} = \ frac {L } {I_2} \ ,. $
The axis around which the angle β rotates is $ \ hat {u} _ \ beta = \ cos (\ gamma) \ hat {h} _3- \ sin (\ gamma) \ hat {h} _1 $ and the meridian thus has the direction
$ \ hat {m} = \ hat {h} _2 \ times \ hat {u} _ \ beta = \ cos (\ gamma) \ hat {h} _1 + \ sin (\ gamma) \ hat {h} _3 \, . $
The rate of the 2 axis is
$ \ dot {\ hat h} _2 = \ vec \ omega \ times \ hat {h} _2 = \ omega_1 \ hat {h} _3- \ omega_3 \ hat {h} _1 \ ,. $
With the above intermediate results and the trigonometric formulas, the direction angle between the meridian and the rate of the 2-vector to the constant is calculated