2. We use Poinsot’s construction to see how the angular velocity vector ω moves. This gives us no information on how the angular momentum vector L moves in the body system (Note: In space system, L is fixed in direction, since it is conserved!) To understand how the L vector moves in the principal axis system of body, use another geometric construction (the Binet construction). In the principal axis system, angular momentum: L = I  ω is: L 1 = I 1 ω 1, L 2 = I 2 ω 2, L 3 = I 3 ω 3 (1) KE: T = (½)ω  I  ω is: T = (½)I 1 (ω 1 ) 2 + (½)I 2 (ω 2 ) 2 + (½)I 3 (ω 3 ) 2 (2) Combining (1) & (2) gives: T = (½)[(L 1 ) 2 /I 1 ] + (½)[(L 2 ) 2 /I 2 ] + (½)[(L 3 ) 2 /I 3 ] (3) T = const, since it is conserved! Binet’s Construction

3. T = (½)[(L 1 ) 2 /I 1 ] + (½)[(L 2 ) 2 /I 2 ] + (½)[(L 3 ) 2 /I 3 ] (3) T = const, since it is conserved!  (3): Defines an ellipsoid  Binet Ellipsoid. Also known as the kinetic energy ellipsoid. Fixed in the body axes. NOT the same as the Inertia Ellipsoid! In what follows, assume: I 3  I 2  I 1. Also, change notation (again!): L 1  L x, L 2  L y, L 3  L z (Put (3) into the standard form for an ellipsoid):  (½)[(L x ) 2 /(T I 1 )] + (½)[(L y ) 2 /(T I 2 )] + (½)[(L z ) 2 /(T I 3 )] = 1 (3)

4. Binet (or KE) Ellipsoid: (½)[(L x ) 2 /(TI 1 )] + (½)[(L y ) 2 /(TI 2 )] + (½)[(L z ) 2 /(TI 3 )] = 1 (3) Or: [(L x ) 2 /a 2 ] + [(L y ) 2 /b 2 ]+ [(L z ) 2 /c 2 ] = 1 (3  ) See figure: Semimajor axes (Decreasing size): a = (2T I 1 ) ½, b = (2T I 2 ) ½ c = (2T I 3 ) ½ Also have Conservation of total angular momentum: L 2 =(L x ) 2 +(L y ) 2 +(L z ) 2 = const. Or: [(L x ) 2 +(L y ) 2 +(L z ) 2 ]/L 2 =1 (4) (4): A sphere in (L x,L y,L z ) space! (3) & (4) must be satisfied simultaneously!

5. Simultaneously we have the Binet (KE) Ellipsoid: (½)[(L x ) 2 /(TI 1 )] + (½)[(L y ) 2 /(TI 2 )] + (½)[(L z ) 2 /(TI 3 )] = 1 (3) & the angular momentum sphere: [(L x ) 2 +(L y ) 2 +(L z ) 2 ]/L 2 =1 (4)  The path of the L vector with time is the curve formed in L space by the intersection of the KE ellipsoid & angular momentum sphere. Equating (3) & (4): (½)[(L x ) 2 /(TI 1 )] + (½)[(L y ) 2 /(TI 2 )] + (½)[(L z ) 2 /(TI 3 )] = [(L x ) 2 +(L y ) 2 +(L z ) 2 ]/L 2 (5) Can show the ellipsoid (3) & the sphere (4) intersect so that (5) has a solution if: L is larger than the smallest semimajor axis & smaller than largest semimajor axis:  Solution if (2TI 3 ) ½ < L < (2TI 1 ) ½ (6)

6. Path of L: The curve from the intersection of the KE ellipsoid & the angular momentum sphere. Or, L must satisfy: (½)[(L x ) 2 /(TI 1 )] + (½)[(L y ) 2 /(TI 2 )] + (½)[(L z ) 2 /(TI 3 )] = [(L x ) 2 +(L y ) 2 +(L z ) 2 ]/L 2 (5) Solution if & only if (2TI 3 ) ½ < L < (2TI 1 ) ½ (6) The sphere is outside the ellipsoid on the L z axis & inside along the L x axis. Figure: Shows, for various L, curves on the ellipsoid where the sphere intersects it. Fig b  View as seen Fig a  Perspective view. along the L y axis Straight lines in Fig b: Correspond to L = (2TI 2 ) ½

7. Symmetrical Body Back to Euler’s Equations under force, torque free conditions: I 1 (dω 1 /dt) = ω 2 ω 3 (I 2 -I 3 ) (1) I 2 (dω 2 /dt) = ω 3 ω 1 (I 3 -I 1 ) (2) I 3 (dω 3 /dt) = ω 1 ω 2 (I 1 -I 2 ) (3) Special case: Steady rotation (constant angular velocity ω)  (dω i /dt) = 0 (i = 1,2,3)  ω 1 ω 2 (I 1 -I 1 ) = ω 2 ω 3 (I 2 -I 3 ) = ω 3 ω 1 (I 3 -I 1 ) = 0 (4)  All components of ω can be constant only if at least 2 of the ω i = 0. In other words: The vector ω can be constant only if it is along one of the principal axes! Take this along with geometric construction just discussed.

8. Steady Rotation: ω 1 ω 2 (I 1 -I 1 ) = ω 2 ω 3 (I 2 -I 3 ) = ω 3 ω 1 (I 3 -I 1 ) = 0 (4)  ω can be constant only if it is along one of the principal axes! But, not all rotations with ω along a principal axis are stable! Must simultaneously satisfy (4) & the conditions just discussed that the angular momentum sphere intersects the KE ellipsoid. We can use this fact to determine whether a given motion is a stable or unstable rotation. Stable Rotation  If a small perturbation causes the rotation axis of the body to move only slightly away from the principal axis. Similar to the stable & unstable circular orbit criterion discussed in the central force chapter!

9. Example of a stable, steady rotation: Steady rotation about principal axis 3 (smallest principal moment I 3, z direction, L z ). The discussed relations  This happens if the radius of the angular momentum sphere is L  L s = (2TI 3 ) ½. Small deviations from this  Angular momentum sphere radius L is slightly < L s = (2TI 3 ) ½  The intersection of this sphere with KE ellipsoid is a small circle about the L z axis. (See figure):  The motion is stable, because L is never far from L z.   

10. Another example of a stable, steady rotation: Steady rotation about the principal axis 1 (the largest principal moment I 1, x direction, L x ). The discussed relations  This happens if the radius of the angular momentum sphere is L  L s = (2TI 1 ) ½. Small deviations from this  Angular momentum sphere radius L is slightly > L s = (2TI 1 ) ½  The intersection of this sphere with the KE ellipsoid is a small ellipse about the L x axis. (Figure):  The motion is stable, because L is never far from L x.   

11. An example of an unstable, steady rotation: Steady rotation about the principal axis 2 (the intermediate principal moment I 2, y direction, L y ). The discussed relations  Happens if the radius of the angular momentum sphere is L  L s = (2TI 2 ) ½ Small deviations from this  The angular momentum sphere radius L is slightly different than L s.  Intersection of the sphere with the KE ellipsoid has 2 possibilities. Figure. 2 curves: The “orbits” circle the ellipsoid & cross each other where the  L y axes pass through the ellipsoid. 2 curves with L slightly L s. Each has a long path on the surface  L can deviate significantly from L y  The motion is unstable. See paragraph & footnote, p. 205 about applications to stability of spinning spacecraft!  

12. Euler’s Eqtns under force, torque free conditions: I 1 (dω 1 /dt) = ω 2 ω 3 (I 2 -I 3 ) (1) I 2 (dω 2 /dt) = ω 3 ω 1 (I 3 -I 1 ) (2) I 3 (dω 3 /dt) = ω 1 ω 2 (I 1 -I 2 ) (3) Special Case: Symmetrical body. Now an analytical solution (which will be consistent with Poinsot’s Geometric construction, of course!). Take the rotation about the 3 or z axis. Symmetrical: I 1 = I 2. (1), (2), (3) become: I 1 (dω 1 /dt) = ω 2 ω 3 (I 2 -I 3 ) (1) I 1 (dω 2 /dt) = - ω 3 ω 1 (I 1 -I 3 ) (2) I 3 (dω 3 /dt) = 0 (3) NOTE: A minus sign typo in Goldstein 3rd Edition in Eq. (2)! ↖ | |  Analytic Solution for a Symmetrical Body

13. Symmetrical Body I 1 (dω 1 /dt) = ω 2 ω 3 (I 2 -I 3 ) (1) I 1 (dω 2 /dt) = - ω 3 ω 1 (I 1 -I 3 ) (2) I 3 (dω 3 /dt) = 0 (3) (3)  ω 3 = const. Treat as a known initial condition. Solve (1) & (2) simultaneously. –Define: Ω  [(I 1 -I 3 )/I 1 ]ω 3 (1)  (dω 1 /dt) = - Ωω 2 (a) (2)  (dω 2 /dt) = + Ωω 1 (b) –Take derivatives of (a) & (b). Substitute back & get: (d 2 ω 1 /dt 2 ) + Ω 2 ω 1 = 0 (a) (d 2 ω 2 /dt 2 ) + Ω 2 ω 2 = 0 (b) (a) & (b): Standard simple harmonic oscillator eqtns!

14. (d 2 ω 1 /dt 2 ) + Ω 2 ω 1 = 0 (a) (d 2 ω 2 /dt 2 ) + Ω 2 ω 2 = 0 (b) Ω  [(I 1 -I 3 )/I 1 ] ω 3 Solutions: Sinusoidal functions: ω 1 = A cos(Ωt) ω 2 = A sin(Ωt) A depends on the initial conditions Interpretation:  The angular velocity vector: ω = ω 1 i + ω 2 j + ω 3 k Magnitude: ω = [(ω 1 ) 2 +(ω 2 ) 2 + (ω 3 ) 2 ] ½ = [A 2 + (ω 3 ) 2 ] ½ = const Component of the angular velocity vector in the xy plane: ω  = ω 1 i + ω 2 j Magnitude: ω  = [(ω 1 ) 2 +(ω 2 ) 2 ] ½ = A = const

15. ω 1 = A cos(Ωt), ω 2 = A sin(Ωt) Interpretation: Angular velocity vector in xy plane: ω  = ω 1 i + ω 2 j. Magnitude: ω  = [(ω 1 ) 2 + (ω 2 ) 2 ] ½ = A = const  The vector ω  rotates uniformly (fig.) about the body z axis at frequency Ω  Total angular velocity ω = ω 1 i + ω 2 j + ω 3 k (also const in magnitude) precesses about z axis at precession frequency Ω. As predicted by Poinsot’s construction! Note: This precession is relative to the body axes. These are themselves, rotating with respect to the space axes at frequency ω! Ω depends on I 1, I 3, ω 3

16. ω 1 = A cos(Ωt), ω 2 = A sin(Ωt) Precession frequency Ω  [(I 1 -I 3 )/I 1 ]ω 3 The closer I 1 is to I 3, the smaller the precession frequency Ω is compared to the total angular velocity ω. Demonstrate precession another way: Define a vector Ω in the z direction with magnitude Ω = [(I 1 -I 3 )/I 1 ]ω 3. The Ch. 4 relation: (d/dt) s = (d/dt) b + ω   (dω/dt) = ω  Ω Gives the same (precessing ω) results!

17. ω 1 = A cos(Ωt), ω 2 = A sin(Ωt) Precession frequency Ω  [(I 1 -I 3 )/I 1 ]ω 3 2 constants ω 3 & A: Obtained from the initial conditions. Can get them in terms of 2 constants of the motion: Total KE T & total angular momentum L 2 are conserved. In general: T = (½)I 1 (ω 1 ) 2 + (½)I 2 (ω 2 ) 2 + (½)I 3 (ω 3 ) 2 L 2 = (I 1 ω 1 ) 2 + (I 2 ω 2 ) 2 + (I 3 ω 3 ) 2 Here: I 1 = I 2, (ω 1 ) 2 + (ω 2 ) 2 = A 2.  T = (½)I 1 A 2 + (½)I 3 (ω 3 ) 2 L 2 = (I 1 A) 2 + (I 3 ω 3 ) 2 Or: (ω 3 ) 2 = [2TI 1 - L 2 ]/[I 3 (I 1 -I 3 )] A 2 = [2TI 3 - L 2 ]/[I 1 (I 3 -I 1 )]

18. Application to the Earth Expectation, (partially) confirmed by observation! The Earth’s axis of rotation should exhibit this type of precession. –External torques acting on the Earth are very weak. –The Earth is  A symmetrical rigid body (flattened at poles!)  I 1 < I 3, I 1  I 2  Earth’s rotational motion is  A torque & force free symmetrical rigid body. Precession frequency Ω  [(I 1 -I 3 )/I 1 ]ω 3 –Numbers: (I 1 -I 3 )/I 1  0.00327  Ω  0.00327 ω 3  ω 3 /(306) –Also, ω 3  ω = (2π)/(1 day)  Precession period T p = (2π)/ Ω  306 days Or (expected):T p  10 months

19. So, if there were some disturbance of the axis of rotation of Earth, that axis should precess around N pole once/10 mos. –Would be seen as periodic change in the apparent latitude of points on the Earth’s surface. –Something like this is seen. However, it is more complex ! Observations over ~ 100 years show: –Deviations between the N pole axis & the rotation axis are more like a wobble than a precession. –Also, the period is ~ 420 days, not 306 days. –Why the deviations? The Earth is not a symmetric rigid body! See discussion, p. 208. The wobble actually appears to be damped! Also, appears to be driven by some excitation! Don’t confuse this free precession (wobble) with the slow astronomical precession about the normal to the ecliptic (precession of equinoxes!). Period of ~ 26,000 years! Due to gravitational torques from Sun & Moon.