Exercises in Classical Mechanics 1 Asymmetric Top Via Euler Angles 2

Exercises in Classical Mechanics CUNY GC, Prof. D. Garanin No.3 Solution —————————————————————————————————————————

1 Asymmetric top via Euler angles

Set up Lagrange equations in terms of Euler angles for an asymmetric top rotating around an axis close to one of its principal axes.

2 Nonholonomic systems: Sphere on a plane

Consider a body of spherical shape with radius R and mass M that is rolling on a plane without slipping. All moments of inertia Iα, α = 1, 2, 3, are diﬀerent and the center of mass is on the 3-axis at the distance a from the geometrical center. This sphere is not allowed to jump, too. a) Obtain the Lagrange equations in terms of the Euler angles for this sphere using Lagrange multipliers. (Do not work out the derivative d/dt explicitly, it results into too many terms). If you have access to mathematical software, solve these equations numerically with diﬀerent initial conditions. b) Do the same using the Newtonian dynamics and d’Alembert’s principle, considering the reaction force F from the plane on the sphere.

Solution: a) In general, a system with Nα non-holonomic constraints is described by conditions of the type

XN XN ∂Φ (q, q˙) Φ (q, q˙) ≡ c (q)q ˙ ≡ α q˙ = 0, α = 1, 2,....,N . (1) α αi i ∂q˙ i α i=1 i=1 i In the least-action principle, variations of q(t) are not independent but subject to the similar constraint as above: X ∂Φ (q, q˙) α δq = 0. (2) ∂q˙ i i i Using the method of Lagrange multipliers, one can add the constraint terms to the variation of the action: Z X X ∂Φ (q, q˙) δS0 = δS + dt λ α δq , (3) α ∂q˙ i α i i where the multipliers λα can depend on time and δS is the standard term Z Z µ ¶ X ∂L d ∂L δS = dt L(q, q˙) = dt − δq . (4) ∂q dt ∂q˙ i i i i Stationarity requirement δS0 = 0 leads to the Lagrange equations with constraints d ∂L ∂L X ∂Φ − − λ α = 0. (5) dt ∂q˙ ∂q α ∂q˙ i i α i

Together with Eqs. (1) these equations form a well-determined system of N + Nα equations. Considering constraint equations as one vector equation, one can rewrite Eq. (5) as d ∂L ∂L ∂Φ − − λ· = 0. (6) dt ∂q˙i ∂qi ∂q˙i

Let us apply this method to the sphere rolling over the plane. In this case the constraint condition enforces that the velocity of the contact point of the sphere with the plane v is zero, i.e.,

v = V0 + R [ω × n] = 0. (7)

1 Here V0 is the velocity of the geometrical center of the sphere, ω is the angular velocity of the sphere, and n is the unit vector directed downwards, n = −e(z). The velocity of the center of mass is given by h i (3) V = V0 + a ω × e , (8) that, combined with Eq. (7), yields the constraint equation for the velocity of the center of mass

V = [ω × RCM] , (9) where (z) (3) RCM = Re + ae (10) is the vector from the contact point to the center of mass. The constraint equation can be rewritten as

Φ ≡ V− [ω × RCM] = 0. (11)

Note that there are three components of the constraint vector Φ that correspond to the three components of the velocity of the center of mass V. The kinetic energy of the system is the sum of the translational kinetic energy of the center of mass

MV2 T = (12) tr 2 and the rotational kinetic energy 1 1 1 T = I ω2 + I ω2 + I ω2. (13) rot 2 1 1 2 2 2 2 3 3 The potential energy is due to the changing height Z of the center of mass,

U = MgZ. (14)

The Lagrange function is L =Ttr + Trot − U. (15) In terms of the Euler angles one has

ω1 =ϕ ˙ sin θ sin ψ + θ˙ cos ψ

ω2 =ϕ ˙ sin θ cos ψ − θ˙ sin ψ

ω3 =ϕ ˙ cos θ + ψ˙ (16) and

ωx = ψ˙ sin θ sin ϕ + θ˙ cos ϕ

ωy = −ψ˙ sin θ cos ϕ + θ˙ sin ϕ

ωz = ψ˙ cos θ +ϕ. ˙ (17)

Before setting up the Lagrange formalism with constraints for our problem, we have to note the following. One can eliminate V2 in Eq. (12) using Eq. (9). This leads to ³ ´ h i ³ ´ 2 2 2 2 V2 = R2 θ˙ + ψ˙ sin2 θ + 2Ra ω2 cos θ − (ϕ ˙ 2 cos θ + ψ˙ cos θ +ϕ ˙ ψ˙ (1 + cos2 θ) + a2 θ˙ +ϕ ˙ 2 sin2 θ ³ ´ h i ³ ´ 2 2 2 2 = R2 θ˙ + ψ˙ sin2 θ + 2Ra θ˙ cos θ − ϕ˙ ψ˙ (1 − cos θ)2 + a2 θ˙ +ϕ ˙ 2 sin2 θ ¡ ¢ 2 2 = R2 + 2Ra cos θ + a2 θ˙ + R2 sin2 θ ψ˙ − 2Ra (1 − cos θ)2 ϕ˙ ψ˙ + a2 sin2 θ ϕ˙ 2. (18)

Also one can eliminate Z in Eq. (14) by using Z = a cos θ, so that

U = Mga cos θ. (19)

Thus one obtains the Lagrange function solely in terms of the rotational variables, the Euler angles. Now apparently one can write Lagrange equations for the Euler angles without considering the constraint any- more. It turns out, however, that the Lagrange function and ensuing Lagrange equations obtained on this

2 way are wrong. In the simplest case of the sphere with the center of mass in the geometrical center, a = 0, and all moments of inertia equal to each other, the solution of these equations shows that the center of the sphere is making circles. This£ is a wrong¤ result since it is apparent that the sphere should move straight with a constant speed V =R ω × e(z) . An idea of why this method could be wrong can be obtained if one adds a potential energy depending on the X and Y components of the center of mass of the sphere. These components cannot be eliminated since the constraint is non-holonomic. If one eliminates V as above, one obtains a model in which there is a potential energy of X and Y but no corresponding kinetic energy! There will be additional wrong Lagrange equations ∂U/∂X = 0 and ∂U/∂Y = 0. On the other hand, this method seems to align well with the whole ideology of the Lagrange mechanics. Its failure throws some shadow on the Lagrange formalism in general. Let us now implement the Lagrange equations with constraints, Eq. (6). For the translational motion, using (β) V =Vβe one obtains ∂Φ (β) ∂Φ = e , λ· = λβ (20) ∂Vβ ∂Vβ so that with the notation Vβ = R˙ β (not to be confused with R) the ﬁrst three of Eqs. (6) become d ∂L ∂L − − λβ = 0 (21) dt ∂Vβ ∂Rβ where ∂L ∂L ∂L = = 0, = −Mg (22) ∂X ∂Y ∂Z or ∂L = Mg, g = − ge(z). (23) ∂R Finally one obtains MR¨β = −Mgδβz + λβ, β = x, y, z (24) or MR¨ =Mg+λ. (25) One can see that the constraint vector λ is just the reaction force acting from the plane on the sphere. There are three rotational Lagrange equations of the type d ∂L ∂L − = Kθ, (26) dt ∂θ˙ ∂θ where Kθ is the generalized torque, ∂Φ ³ ´ ∂Φ Kθ = λ· = M R¨ − g · (27) ∂θ˙ ∂θ˙

The equations for ϕ and ψ are similar. In Kθ one can eliminate R¨ = V˙ by diﬀerentiating Eq. (9) with the use of Eq. (11) one obtains µ ¶ d ∂ Kθ = M [ω × RCM] −g · [ω × RCM] , (28) dt ∂θ˙ where RCM is deﬁned by Eq. (10). Now one has to work out the scalar product using components of all vectors in the laboratory frame ½ ¾ d ∂ Kθ = M [ω × RCM] +gδαz [ω × RCM] dt α ∂θ˙ α ½ ³ ´ ¾ ³ ´ d (z) (3) ∂ (z) (3) = M R²αβγωβe + a²αβγωβe +gδαz R² 0 0 ω 0 e 0 + a² 0 0 ω 0 e 0 dt γ γ ∂θ˙ αβ γ β γ αβ γ β γ ³ ´ ³ ´ d (3) ∂ (3) ∂ (3) = M²αβγ² 0 0 Rωβδγz + aωβe Rω 0 δγ0z + aω 0 e 0 + Mga² 0 0 ω 0 e 0 . (29) αβ γ dt γ ∂θ˙ β β γ zβ γ ∂θ˙ β γ Here the components of e(3) are given by

(3) (3) (3) ex = sin θ sin ϕ, ey = − sin θ cos ϕ, ez = cos θ. (30)

3 Thus the equations of motion for our problem are completely deﬁned in terms of the Euler angles. It does not make sense to work out the lhs and rhs of Eq. (26) by means of manual analytics since it can be much more conveniently done by mathematical software. Remember that in the lhs of Eq. (26) we use components of ω in the body frame, Eq. (16), while in the rhs of this equation we use components of ω in the laboratory frame, Eq. (17). The latter is dictated by the gravity term. Without gravity one could also write Kθ in the body frame. Numerical solution of the above expression shows a rich behavior including dynamical chaos in the case a 6= 0 and I1 6= I2 6= I3. For a = 0 the geometrical center of the sphere moves straight on average but makes excursions around the straight line. For a 6= 0, with gravity or not, and two of the moments of inertia coinciding with each other, the motion is conﬁned to some region and the geometrical center makes quasiperiodic orbits. In the case of dynamical chaos the sphere wanders irregularly.

Computer-aided simpliﬁcation of Kθ yields

2 2 Kθ = MR Aθ + Ma Bθ + MRa Cθ + Mga Dθ, (31) where ¨ ˙ Aθ = θ +ϕ ˙ ψ sin θ,

Aϕ = 0, ¨ 2 ˙ ˙ ˙ Aψ = ψ sin θ − θϕ˙ sin θ + θψ sin θ cos θ ¨ 2 Bθ = θ − ϕ˙ sin θ cos θ, 2 Bϕ =ϕ ¨ sin θ + 2θ˙ϕ˙ sin θ cos θ, B = 0 ψ ³ ´ ¨ ˙ 2 2 ˙ Cθ = 2θ cos θ − θ +ϕ ˙ sin θ +ϕ ˙ ψ sin θ cos θ 2 Cϕ = −ψ¨ sin θ + θ˙ϕ˙ sin θ − θ˙ψ˙ sin θ cos θ 2 ˙ Cψ = −ϕ¨ sin θ − 2θϕ˙ sin θ cos θ

Dθ = − sin θ

Dϕ = 0

Dψ = 0. (32) A particular case of our model is a massless sphere with a point mass M attached at its surface, a = R. In this case one has I1 = I2 = I3 = 0 and thus Trot = 0, so that the equations simplify to

2 MR (Aθ + Bθ + Cθ) + MgRDθ = 0 2 MR (Aϕ + Bϕ + Cϕ) + MgR Dϕ = 0 2 MR (Aψ + Bψ + Cϕ) + MgR Dψ = 0 (33) or ³ ´ 2 g 2¨θ(1 + cos θ ) − θ˙ sin θ +ϕ ˙ −ϕ˙ + ψ˙ sin θ(1 + cos θ ) = sin θ ³ ´ R ϕ¨ − ψ¨ sin2 θ + θ˙ϕ˙ sin θ(1 + 2 cos θ ) − θ˙ψ˙ sin θ cos θ = 0 ³ ´ ϕ¨ − ψ¨ sin2 θ + θ˙ϕ˙ sin θ(1 + 2 cos θ ) − θ˙ψ˙ sin θ cos θ = 0. (34)

This system of equations is degenerate as the last two equations are the same. Thus the problem is not well defined if all moments of inertia are zero. Similar situation takes place for any a, although it is more difficult to see. Additional analysis is needed to clear up this situation. A practical way to numerically solve the problem is to solve the general problem with small but nonzero valies of I1,I2, and I3. This solution essentially depends on I1,I2, and I3 and shows a very complicated behavior. For all moments of inertia different from each other and sufficiently small the behavior is typically chaotic for all a 6= 0. If the moments of inertia are not very small, in the absence of gravity and a essentially smaller than R, the motion of the geometrical center is confined and quasiperiodic. If a becomes close to R, the motion becomes unconfined and chaotic. For a = R numerical solution diverges. Adding gravity makes the motion faster. Details can be found in the accompanying Mathematica program and the Mathematica archive file.

4 (b) The dynamics of the center of mass of the sphere is described by MV˙ =Mg + F, (35) where F is the reaction force acting on the sphere from the plane. The velocity of the center of mass obeys the constraint given by Eq. (9). This allows to calculate the torque acting on the sphere from the plane h ³ ´i K = − [RCM × F] = −M RCM × −g + V˙ · µ ¶¸ d = −M R × ge(z)+ [ω × R ] . (36) CM dt CM Substituting this torque in the Euler equations dω I 1 − (I − I )ω ω = K = K · e(1) 1 dt 2 3 2 3 1 dω I 2 − (I − I )ω ω = K 2 dt 3 1 3 1 2 dω I 3 − (I − I )ω ω = K , (37) 3 dt 1 2 1 2 3 and expressing everything in terms of the Euler angles, one obtains a closed system of equations. In particular, one uses Eqs. (16), (17), as well as the relation between the laboratory-frame and the body- frame vectors given by the general rotation matrix     Axx Axy Axz cos ψ cos ϕ − cos θ sin ϕ sin ψ cos ψ sin ϕ + cos θ cos ϕ sin ψ sin ψ sin θ A =  Ayx Ayy Ayz  =  − sin ψ cos ϕ − cos θ sin ϕ cos ψ − sin ψ sin ϕ + cos θ cos ϕ cos ψ cos ψ sin θ  Azx Azy Azz sin θ sin ϕ − sin θ cos ϕ cos θ (38) that rotates any vector x as x0=A · x. (39) Laboratory basis vectors are e(x,y,z) play the role of x (initial vectors) while the body-frame basis vectors e(1,2,3) play the role of x0 (rotated vectors), so that e(1) = e(x)0 = A · e(x) etc. One thus has (1) (x) (y) (z) e = Axxe + Axye + Axze (2) (x) (y) (z) e = Ayxe + Ayye + Ayze (3) (x) (y) (z) e = Azxe + Azye + Azze (40) and ³ ´ ³ ´ (1) (x) (2) (x) e · e = Axx, e · e = Ayx ³ ´ ³ ´ (1) (y) (2) (y) e · e = Axy, e · e = Ayy ³ ´ ³ ´ (1) (z) (2) (z) e · e = Axz, e · e = Ayz. (41) In the particular case of the point mass attached to a sphere all moments of inertia are equal to zero, so that the dynamic equation has the form K = 0. (42) It is convenient to project this equation onto the laboratory frame,

Kα = 0, α = x, y, z, (43) i.e., · µ ¶¸ µ ¶ (z) d (z) d 0 = RCM × ge + [ω × RCM] = ²αβγRCM,β geγ + [ω × RCM]γ dt α dt d = ² R g + ² R ² 0 0 ω 0 R 0 . (44) αβz CM,β αβγ CM,β dt α β γ α CM,β In the case a = R from Eq. (10) one obtains ³ ´ ³ ´ h³ ´ i (3) (3) d (3) 0 = g² δ + e + R² ² 0 0 δ + e δ 0 + e ω 0 , (45) αβz βz β αβγ α β γ βz β dt β z β0 α and simpliﬁcation of this equation leads to Eq. (34).