Homework 8 # 5.4, 5.6, 5.7, 5.26

Homework 8: # 5.4, 5.6, 5.7, 5.26 5.4 Derive Euler’s equations of motion, Eq. (5.39’), from the Lagrange equation of motion, in the form of Eq. (1.53), for the generalized coordinate ψ . Answer: Euler’s equations of motion for a rigid body are: I1 ω̇1 − ω2 ω3 (I2 − I3 ) = N1 I2 ω̇2 − ω3 ω1 (I3 − I1 ) = N2 I3 ω̇3 − ω1 ω2 (I1 − I2 ) = N3 The Lagrangian equation of motion is in the form d ∂T ∂T ( )− = Qj dt ∂ q̇j ∂qj The kinetic energy for rotational motion is 3 T = Xi 1 2 Ii ω i 2 The components of the angular velocity in terms of Euler angles for the body set of axes are ω1 = ˙φ sin θ sin ψ + θ˙cos ψ ω2 = ˙φ sin θ cos ψ −θ˙sin ψ ω3 = φ̇ cos θ + ψ̇ Solving for the equation of motion using the generalized coordinate ψ : ∂T d ∂T ( = Nψ )− ∂ψ dt ∂ ψ̇ 3 Xi 3 Ii ω i ∂ωi d ∂ωi Ii ω i − = Nψ dt Xi ∂ψ ∂ ψ̇ 1 Now is a good time to pause and calculate the partials of the angular velocities, ∂ω1 = − ˙ θ sin ψ + φ̇ sin θ cos ψ ∂ψ ∂ω2 = − ˙ θ cos ψ − φ̇ sin θ sin ψ ∂ψ ∂ω3 = 0 ∂ψ and ∂ω2 ∂ω1 = = 0 ∂ ψ̇ ∂ψ˙ ∂ω3 = 1 ∂ ψ̇ Now we have all the pieces of the puzzle, explicitly 3 Xi 3 Ii ω i ∂ωi d ∂ωi Ii ω i − = Nψ dt Xi ∂ψ ∂ ψ̇ I1 ω1 (− ˙θ sin ψ + φ̇ sin θ cos ψ) + I2 ω2 (−θ˙cos ψ − φ̇ sin θ sin ψ) − d I3 ω 3 = N ψ dt This is, pulling out the negative sign on the second term, I1 ω1 (ω2 ) − I2 ω2 (ω1 ) − I3 ω̇3 = Nψ I3 ω̇3 − ω1 ω2 (I1 − I2 ) = Nψ And through cyclic permutations I2 ω̇2 − ω3 ω1 (I3 − I1 ) = N2 I1 ω̇1 − ω2 ω3 (I2 − I3 ) = N1 we have the rest of Euler’s equations of motion for a rigid body. 2 5.6 • Show that the angular momentum of the torque-free symmetrical top rotates in the body coordinates about the symmetry axis with an angular frequency ω. Show also that the symmetry axis rotates in space about the fixed direction of the angular momentum with angular frequency I3 ω 3 φ˙ = I1 cos θ where φ is the Euler angle of the line of nodes with respect to the angular momentum as the space z axis. • Using the results of Exercise 15, Chapter 4, show that ω rotates in space about the angular momentum with the same frequency φ̇, but that the angle θ′ between ω and L is given by sin θ′ = Ω sin θ′′ φ˙ where θ′′ is the inclination of ω to the symmetry axis. Using the data given in Section 5.6, show therefore that Earth’s rotation axis and axis of angular momentum are never more than 1.5 cm apart on Earth’s surface. • Show from parts (a) and (b) that the motion of the force-free symmetrical top can be described in terms of the rotation of a cone fixed in the body whose axis is the symmetry axis, rolling on a fixed cone in space whose axis is along the angular momentum. The angular velocity vector is along the line of contact of the two cones. Show that the same description follows immediately from the Poinsot construction in terms of the inertia ellipsoid. Answer: Marion shows that the angular momentum of the torque-free symmetrical top rotates in the body coordinates about the symmetry axis with an angular frequency ω more explicitly than Goldstein. Beginning with Euler’s equation for force-free, symmetric, rigid body motion, we see that ω3 = constant. The other Euler equations are I3 − I ω3 )ω2 I I3 − I ω̇2 = −( ω3 )ω1 I Solving these, and by already making the substitution, because we are dealing with constants, ω̇1 = −( 3 Ω = I3 − I ω3 I we get ( ω̇1 + i ω̇2 ) − iΩ(ω1 + iω2 ) = 0 Let q = ω1 + iω2 Now q̇ − iΩq = 0 has solution q(t) = AeiΩt this is ω1 + iω2 = A cos Ωt + iAsin Ωt and we see ω1 (t) = A cos Ωt ω2 (t) = A sin Ωt The x3 axis is the symmetry axis of the body, so the angular velocity vector precesses about the body x3 axis with a constant angular frequency Ω = I3 − I ω3 I φ̇ = I3 ω 3 I1 cos θ φ̇ = I3 ω 3 I1 cos θ . To prove We may look at the two cone figure angular momentum components, where L is directed along the vertical space axis and θ is the angle between the space and body vertical axis. L1 = 0 L2 = L sin θ L3 = L cos θ If α is the angle between ω and the vertical body axis, then 4 ω1 = 0 ω2 = ω sin α ω3 = ω cos α The angular momentum components in terms of α may be found L1 = I 1 ω 1 = 0 L2 = I1 ω2 = I1 ω sin α L3 = I3 ω3 = I3 ω cos α Using the Euler angles in the body frame, we may find, (using the instant in time where x2 is in the plane of x3 , ω, and L, where ψ = 0) , ω2 = ˙φ sin θ cos ψ −θ˙sin ψ ω2 = φ̇ sin θ This is ω sin α L2 L ω2 L = = ω( ) φ˙ = = I1 sin θ sin θ I 1 ω L2 Plugging in L3 L L3 I3 ω 3 φ˙ = = = I1 I1 cos θ I1 cos θ A simple way to show sin θ′ = Ω sin θ′′ φ˙ may be constructed by using the cross product of ω × L and ω × x3 . 2 x+ωy q ′ 2 × L| = ωLsin θ = L ω |ω Using the angular velocity components in terms of Euler angles in the space fixed frame, this is equal to ωLsin θ′ = L ˙φ sin θ with θ fixed, and θ˙ = 0. For ω × x3 we have 2 x′ + ω y ′ q ′′ 2 × x | = ω sin θ = ω |ω 3 components in terms of Euler angles in the body Using the angular velocity fixed frame, this is equal to 5 ω sin θ′′ = φ˙ sin θ Using these two expressions, we may find their ratio Lφ˙ sin θ ωLsin θ′ = ω sin θ′′ φ̇ sin θ sin θ′ ψ˙ = sin θ′′ φ̇ Because ψ˙ = Ω sin θ′ = Ω sin θ′′ φ˙ To show that the Earth’s rotation axis and axis of angular momentum are never more than 1.5cm apart on the Earth’s surface, the following approximations may be made, sin θ′ ≈ θ′ , cos θ ≈ 1, sin θ′′ ≈ θ′′ , and I1 /I3 ≈ 1. Earth is considered an oblate spheroid, I3 >I1 and the data says there is 10m for amplitude of separation of pole from rotation axis. Using sin θ′ = φ̇ = Ω = Ω sin θ′′ φ˙ I3 ω 3 I1 cos θ I3 − I1 ω3 I1 we have sin θ′ = I1 cos θ I3 − I1 sin θ′′ ω3 I1 I3 ω 3 Applying the approximations θ′ = θ′ = I3 − I1 ′′ θ I1 I3 − I1 d = R I1 s R where R is the radius of the Earth, and s is the average distance of separation, which we will assume is half the amplitude, 5 m. d= I3 − I1 s = (.00327)(5) = 1.6 cm I1 Force free motion means the angular momentum vector L is constant in time and stationary, as well as the rotational kinetic energy. (because the center of mass of the body is fixed). So because T = ω1 · L is constant, ω precesses 2 6 around with a constant angle. This tracing is called the space cone, only if L is lined up with x3 space axis. Proving that L, x3 and ω all lie in the same plane will show that this space cone is traced out by ω. This results from I1 = I2 as shown below: L · (ω × e3 ) = 0 because ω × e3 = ω 2 e1 − ω 1 e2 L · (ω × e3 ) = I1 ω1 ω2 − I2 ω1 ω2 = 0 Because I1 = I2 . Now the symmetry axis of the body has the angular velocity ω precessing around it with a constant angular frequency Ω. Thus another cone is traced out, the body cone. So we have two cones, hugging each other with ω in the direction of the line of contact. 5.7 For the general asymmetrical rigid body, verify analytically the stability theorem shown geometrically above on p. 204 by examining the solution of Euler’s equations for small deviations from rotation about each of the principal axes. The direction of ω is assumed to differ so slightly from a principal axis that the component of ω along the axis can be taken as constant, while the product of components perpendicular to the axis can be neglected. Discuss the boundedness of the resultant motion for each of the three principal axes. Answer: Marion and Thornton give a clear analysis of the stability of a general rigid body. First lets define our object to have distinct principal moments of inertia. I1