# Hessian pendulum Simulated movement of a Hessian pendulum with center of gravity axis and angular momentum plane (black), main axes (blue), angular momentum (red) and angular velocity (green)

The Hess pendulum according to Wilhelm Hess is an asymmetrical top in the gyro theory , in which the center of gravity moves like a spherical pendulum , only the gravitational acceleration has to be replaced by a gyro-specific gravitational acceleration, see animation. The speed of a point on the main axis with the medium main moment of inertia always includes the same angle with the center of gravity axis from the base to the center of gravity, which is why the Hess pendulum is also called the loxodromic pendulum . The Hess pendulum is a direct generalization of the spherical pendulum.

The angular momentum is always perpendicular to the center of gravity. In addition, as with the Staude rotations, the center of gravity, the angular momentum and the angular velocity lie in one plane. The main moments of inertia A, B, C around the first, second and third main axis and the center of gravity coordinates s 1,2,3 must meet the conditions

${\ displaystyle s_ {2} = 0, \ quad {\ frac {s_ {1} ^ {2}} {s_ {3} ^ {2}}} = {\ frac {{\ frac {1} {B} } - {\ frac {1} {A}}} {{\ frac {1} {C}} - {\ frac {1} {B}}}} = {\ frac {C (AB)} {A ( BC)}}}$ adhere to. Here is wlog A> B> C is assumed.

Hess, a professor at the Lyceum in Bamberg, discovered this analytically describable movement in 1890. Russian mathematicians later deepened his study. The Hessian pendulum could also be transferred to the game top and Mlodzjejowsky found another generalization of the spherical pendulum.

## Conditions at the top and the initial conditions

So that the angular momentum always remains in a fixed plane e during the movement, it must be perpendicular to the center of gravity axis. The angular momentum only remains in plane e if the support point, center of gravity, angular momentum and angular velocity are coplanar. Then the plane e intersects the MacCullagh ellipsoid in a circle, which limits the possible mass distribution in the top.

### The solid plane containing the angular momentum

The external torque , formed from the cross product × the center of gravity axis with the weight force, is equal to the speed of the end point of the angular momentum according to the principle of twist . This speed is therefore always perpendicular to the center of gravity axis and must be contained in e. So the plane e is perpendicular to the centroid axis and defined by . ${\ displaystyle {\ vec {s}} \ times {\ vec {G}}}$ ${\ displaystyle {\ vec {s}}}$ ${\ displaystyle {\ vec {G}}}$ ${\ displaystyle {\ vec {L}}}$ ${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0}$ ### The angular velocity

From the above plane equation follows with the product rule

${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0 \ quad \ rightarrow \ quad {\ frac {\ mathrm {d}} {\ mathrm {d} t}} ({\ vec {L}} \ cdot {\ vec {s}}) = {\ dot {\ vec {L}}} \ cdot {\ vec {s}} + {\ vec {L}} \ cdot {\ dot {\ vec {s}}} = 0}$ in which, like the dot, symbolize the derivative of time . The first summand always disappears. In the second summand the velocity of the center of gravity forms with the angular velocity of the gyro: . So that the second summand is zero at all times, the center of gravity, support point, angular momentum and angular velocity must be coplanar: ${\ displaystyle {\ tfrac {\ mathrm {d}} {\ mathrm {d} t}}}$ ${\ displaystyle {\ dot {\ vec {L}}} \ cdot {\ vec {s}} = ({\ vec {s}} \ times {\ vec {G}}) \ cdot {\ vec {s} } = 0}$ ${\ displaystyle {\ vec {\ omega}}}$ ${\ displaystyle {\ dot {\ vec {s}}} = {\ vec {\ omega}} \ times {\ vec {s}}}$ ${\ displaystyle {\ vec {L}} \ cdot ({\ vec {\ omega}} \ times {\ vec {s}}) = 0}$ ### The intersection of the plane and the MacCullagh ellipsoid MacCullagh ellipsoid (yellow) with swirl ball (green), main axes (blue), center of gravity axis (light blue), base point O and center of gravity S.

The body-fixed plane e intersects the MacCullagh ellipsoid in a circle. Because the plane e perpendicular to the center of gravity through the support point intersects the MacCullagh ellipsoid in any case in a conic section (red in the picture). The angular momentum lies in the plane e and touches the ellipsoid at point P and let t be the tangent to the intersection curve in P. The tangent t is contained in e. With a second tangent t 'perpendicular to t, the tangential plane e' is spanned on the ellipsoid in P. This plane is perpendicular to the angular velocity. Because t is the line of intersection of the planes e and e 'which are perpendicular to the axis of the center of gravity or the angular velocity, t is also perpendicular to the plane that is generated by the axis of the center of gravity and the angular velocity. This plane also contains the angular momentum, which is why t is also perpendicular to it. The conic section therefore turns out to be a circle, because its tangents are always perpendicular to the radius vector.

However, the rotational energy is not necessarily constant, which is why the MacCullagh ellipsoid then pulsates in its expansion. The angular momentum does not necessarily follow a circular path in the body-fixed system.

### The mass distribution in the top

The angular momentum lies at the momentary rotational energy E rot on the MacCullagh ellipsoid, that in angular momentum space with angular momentum components L 1,2,3 along the main axes by the equation

${\ displaystyle {\ frac {L_ {1} ^ {2}} {A}} + {\ frac {L_ {2} ^ {2}} {B}} + {\ frac {L_ {3} ^ {2 }} {C}} = 2E _ {\ text {red}}}$ is defined. This ellipsoid is intersected with a sphere in such a way that the cutting figure is flat . The sphere has the equation

${\ displaystyle {\ frac {1} {B}} (L_ {1} ^ {2} + L_ {2} ^ {2} + L_ {3} ^ {2}) = 2E _ {\ text {red}} }$ Subtraction gives:

${\ displaystyle {\ frac {1} {B}} (L_ {1} ^ {2} + L_ {2} ^ {2} + L_ {3} ^ {2}) - {\ frac {L_ {1} ^ {2}} {A}} - {\ frac {L_ {2} ^ {2}} {B}} - {\ frac {L_ {3} ^ {2}} {C}} = \ left ({ \ frac {1} {B}} - {\ frac {1} {A}} \ right) L_ {1} ^ {2} + \ left ({\ frac {1} {B}} - {\ frac { 1} {C}} \ right) L_ {3} ^ {2} = 0}$ In the 1-3 plane, this identity defines two straight lines through the origin which create two planes with the 2-axis. So that the centroid axis is perpendicular to one of these planes, must

${\ displaystyle s_ {2} = 0, \ quad \ left ({\ frac {1} {B}} - {\ frac {1} {A}} \ right) s_ {3} ^ {2} + \ left ({\ frac {1} {B}} - {\ frac {1} {C}} \ right) s_ {1} ^ {2} = 0}$ be.

### Inferences

The compiled conditions are two for the mass distribution (at s 2 and s 1 / s 3 ) and one for the initial conditions ( ). This degree of specialization is identical to that of the Euler gyro , the Lagrange gyro and the Kowalewskaja gyro , which each also place three conditions on the top, but only on its mass distribution. ${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0}$ It is possible to complete the integration of the Euler-Poisson equations in the Hess case. Joukowsky's geometrical theorems show that the center of gravity moves like a pendulum , only the acceleration due to gravity has to be replaced by a gyro- specific acceleration due to gravity.

## Joukowsky's geometrical theorems

N. Joukowsky formulated several sentences that illustrate the movement of Hess' pendulum. The sentences show that

1. the angular momentum lies in a plane fixed to the body, which was anticipated in the previous section,
2. the speed of a point on the 2-axis encloses a constant angle with the circular sections,
3. the kinetic energy of the top is equal to that of a mass point located in the center of gravity of the top, and that too
4. the angular momentum of the top is equal to the angular momentum of this mass point.

### The speed of a point on the 2-axis

Joukowsky's second theorem states that the speed of a point on the 2-axis encloses a constant angle with the circles, which represent successive positions of the circular section.

Because from the condition of the mass distribution in the form and ${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0}$ ${\ displaystyle \ left ({\ frac {1} {B}} - {\ frac {1} {A}} \ right) L_ {1} ^ {2} + \ left ({\ frac {1} {B }} - {\ frac {1} {C}} \ right) L_ {3} ^ {2} = 0}$ it follows that the ratio of the angular momentum L 3 to L 1 is constant. The ratio ω 3 to ω 1 of the angular velocities proportional to them is therefore also constant. The tangent vector to the circular section on the 2-axis has a constant amount because it is fixed to the body. The vector is because of ${\ displaystyle {\ vec {t}} = {\ vec {s}} \ times {\ hat {e}} _ {2}}$ ${\ displaystyle {\ dot {\ hat {e}}} _ {2} = {\ vec {\ omega}} \ times {\ hat {e}} _ {2}}$ ${\ displaystyle | {\ dot {\ hat {e}}} _ {2} | = | \ omega _ {1} {\ hat {e}} _ {3} - \ omega _ {3} {\ has { e}} _ {1} | = | \ omega _ {1} | {\ sqrt {1 + {\ frac {\ omega _ {3} ^ {2}} {\ omega _ {1} ^ {2}} }}}}$ proportional to the angular velocity ω 1 . The scalar product

${\ displaystyle {\ dot {\ hat {e}}} _ {2} \ cdot {\ vec {t}} = ({\ vec {\ omega}} \ times {\ hat {e}} _ {2} ) \ cdot ({\ vec {s}} \ times {\ hat {e}} _ {2}) = \ left (s_ {1} + s_ {3} {\ frac {\ omega _ {3}} { \ omega _ {1}}} \ right) \ omega _ {1}}$ is also proportional to ω 1 . So the cosine of the angle between and is constant because it is the ratio of the scalar product to the amounts of the vectors involved. Consequently, the angle between the path speed and the tangent or the complementary angle to the center of gravity is always the same. ${\ displaystyle {\ dot {\ hat {e}}} _ {2}}$ ${\ displaystyle {\ vec {t}}}$ ### The kinetic energy of the top

Joukowsky's third theorem says that the kinetic energy of the top is equal to that of a mass point located at the center of gravity of the top.

The kinetic energy of the top is equal to its rotational energy

${\ displaystyle E _ {\ text {kin}} = E _ {\ text {red}} = {\ frac {1} {2}} {\ vec {L}} \ cdot {\ vec {\ omega}}}$ In the case of a mass point with mass M and orbital velocity rotating at the same angular velocity , the kinetic energy is ${\ displaystyle {\ vec {v}} = {\ vec {\ omega}} \ times {\ vec {s}}}$ ${\ displaystyle E _ {\ text {kin}} = {\ frac {M} {2}} | {\ vec {v}} | ^ {2} = {\ frac {M} {2}} | {\ vec {\ omega}} \ times {\ vec {s}} | ^ {2}}$ Combination of the two equations leads to the position of the center of gravity and the orthogonality to the mass under the restrictions specified at the beginning${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0}$ ${\ displaystyle M = {\ frac {2E _ {\ text {kin}}} {| {\ vec {v}} | ^ {2}}} = {\ frac {{\ vec {L}} \ cdot {\ vec {\ omega}}} {| {\ vec {\ omega}} \ times {\ vec {s}} | ^ {2}}} = {\ frac {(AB) C} {(AC) s_ {1 } ^ {2}}} = {\ frac {B} {| {\ vec {s}} | ^ {2}}}}$ ### The angular momentum of the body

Joukowsky's fourth sentence says that the angular momentum of the body is equal to the angular momentum of the mass point from the third movement.

At the roundabout, the angular momentum results from the product of inertia tensor Θ with the angular velocity: . On the other hand, the angular momentum is at the mass point ${\ displaystyle {\ vec {L}}: = \ mathbf {\ Theta} \ cdot {\ vec {\ omega}}}$ ${\ displaystyle {\ vec {L}} _ {M}: = {\ vec {s}} \ times M {\ vec {v}} = {\ vec {s}} \ times M ({\ vec {\ omega}} \ times {\ vec {s}})}$ which with the mass of set 3 and the above-mentioned limitations for the location of the center of gravity and the orthogonality is identical to the angular momentum of the gyroscope: . ${\ displaystyle {\ vec {L}} \ cdot {\ vec {s}} = 0}$ ${\ displaystyle {\ vec {L}} = {\ vec {L}} _ {M}}$ ### Inferences

In the Hess pendulum, the rotational energy and the angular momentum are equal to that of a mass point with mass M in the center of gravity. This mass is not necessarily equal to the mass m of the top. A weight acts on this

${\ displaystyle mg = M \ left ({\ frac {m} {M}} g \ right) = M \ left ({\ frac {ms ^ {2}} {B}} g \ right) =: Mg ' }$ according to the gravitational acceleration g . The center of gravity of the gyro moves like a spherical pendulum with the mass M under the modified gravitational acceleration g '. In particular, the Hess pendulum adapts to the force-free Euler gyro when the center of gravity moves to the base and the modified gravitational acceleration thus approaches zero.

## Footnotes

1. Magnus (1971), p. 141 ff, Klein and Sommerfeld (2010), p. 197 ff.
2. a b c Magnus (1971), p. 142 f.
3. Klein and Sommerfeld (2010), p. 381.
4. Ulf Hashagen: Walther von Dyck: (1856-1934). Mathematics, technology and science organization at the TH Munich . Franz Steiner Verlag, Stuttgart 2003, ISBN 3-515-08359-6 , pp. 76 f . ( limited preview in Google Book search).
5. ^ Wilhelm Hess (1890), see literature.
6. see Klein and Sommerfeld (2010), p. 378. For the geometric interpretation, N. Joukowsky (1894) is quoted there (see literature) and for the analytical deepening
P. A. Nekrassoff : Recherches analytiques sur un cas de rotation d'un solid pesant author d'un point fixe . In: Mathematical Annals . tape 47 , 1896 (contains further references). ( eudml.org digitized
version )
7. Grammel (1920), p. 129.
8. a b Klein and Sommerfeld (2010), p. 379.
9. Klein and Sommerfeld (2010), p. 380.
10. Klein and Sommerfeld (2010), p. 381 and Magnus (1971), p. 141.
11. Klein and Sommerfeld (2010), p. 378.
12. a b N. Joukowski : Geometric interpretation of the Hessian case of the motion of a heavy rigid body around a fixed point . In: German Mathematicians Association (ed.): Annual report of the German Mathematicians Association . tape 3 . Reimer, 1894, ISSN  0012-0456 , p. 62–70 ( uni-goettingen.de ).