Binets Construction We use Poinsots construction to see
Binet’s Construction • 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 1ω1, L 2 = I 2ω2, L 3 = I 3 ω 3 (1) KE: T = (½)ω I ω is: T = (½)I 1(ω1)2 + (½)I 2(ω2)2 + (½)I 3(ω3)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] T = const, since it is conserved! (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 Lx, L 2 Ly, L 3 Lz (Put (3) into the standard form for an ellipsoid): (½)[(Lx)2/(T I 1)] + (½)[(Ly)2/(T I 2)] + (½)[(Lz)2/(T I 3)] = 1 (3 )
• Binet (or KE) Ellipsoid: (½)[(Lx)2/(TI 1)] + (½)[(Ly)2/(TI 2)] + (½)[(Lz)2/(TI 3)] = 1 Or: [(Lx)2/a 2] + [(Ly)2/b 2]+ [(Lz)2/c 2] = 1 See figure: Semimajor axes (Decreasing size): a = (2 T I 1)½ , b = (2 T I 2)½ c = (2 T I 3)½ Also have Conservation of total angular momentum: L 2 =(Lx)2+(Ly)2+(Lz)2 = const. Or: [(Lx)2+(Ly)2+(Lz)2]/L 2 =1 (4): A sphere in (Lx, Ly, Lz) space! (3 ) & (4) must be satisfied simultaneously! (3 ) (4)
• Simultaneously we have the Binet (KE) Ellipsoid: (½)[(Lx)2/(TI 1)] + (½)[(Ly)2/(TI 2)] + (½)[(Lz)2/(TI 3)] = 1 (3 ) & the angular momentum sphere: [(Lx)2+(Ly)2+(Lz)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)[(Lx)2/(TI 1)] + (½)[(Ly)2/(TI 2)] + (½)[(Lz)2/(TI 3)] = [(Lx)2+(Ly)2+(Lz)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 (2 TI 3)½ < L < (2 TI 1)½ (6)
• Path of L: The curve from the intersection of the KE ellipsoid & the angular momentum sphere. Or, L must satisfy: (½)[(Lx)2/(TI 1)] + (½)[(Ly)2/(TI 2)] + (½)[(Lz)2/(TI 3)] = [(Lx)2+(Ly)2+(Lz)2]/L 2 (5) Solution if & only if (2 TI 3)½ < L < (2 TI 1)½ (6) The sphere is outside the ellipsoid on the Lz axis & inside along the Lx 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 Ly axis Straight lines in Fig b: Correspond to L = (2 TI 2)½
Symmetrical Body • Back to Euler’s Equations under force, torque free conditions: I 1(dω1/dt) = ω2ω3(I 2 -I 3) (1) I 2(dω2/dt) = ω3ω1(I 3 -I 1) (2) I 3(dω3/dt) = ω1ω2(I 1 -I 2) (3) • Special case: Steady rotation (constant angular velocity ω) (dωi/dt) = 0 (i = 1, 2, 3) ω1ω2(I 1 -I 1) = ω2ω3(I 2 -I 3) = ω3ω1(I 3 -I 1) = 0 (4) All components of ω can be constant only if at least 2 of the ωi = 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.
• Steady Rotation: ω1ω2(I 1 -I 1) = ω2ω3(I 2 -I 3) = ω3ω1(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!
• Example of a stable, steady rotation: Steady rotation about principal axis 3 (smallest principal moment I 3, z direction, Lz). The discussed relations This happens if the radius of the angular momentum sphere is L Ls = (2 TI 3)½. Small deviations from this Angular momentum sphere radius L is slightly < Ls = (2 TI 3)½ The intersection of this sphere with KE ellipsoid is a small circle about the Lz axis. (See figure): The motion is stable, because L is never far from Lz.
• Another example of a stable, steady rotation: Steady rotation about the principal axis 1 (the largest principal moment I 1, x direction, Lx). The discussed relations This happens if the radius of the angular momentum sphere is L Ls = (2 TI 1)½. Small deviations from this Angular momentum sphere radius L is slightly > Ls = (2 TI 1)½ The intersection of this sphere with the KE ellipsoid is a small ellipse about the Lx axis. (Figure): The motion is stable, because L is never far from Lx.
• An example of an unstable, steady rotation: Steady rotation about the principal axis 2 (the intermediate principal moment I 2, y direction, Ly). The discussed relations Happens if the radius of the angular momentum sphere is L Ls = (2 TI 2)½ Small deviations from this The angular momentum sphere radius L is slightly different than Ls. Intersection of the sphere with the KE ellipsoid has 2 possibilities. Figure. 2 curves: The “orbits” circle the ellipsoid & cross each other where the Ly axes pass through the ellipsoid. 2 curves with L slightly < L & 2 with L slightly > L. Each s s has a long path on the surface L can deviate significantly from Ly The motion is unstable. See paragraph & footnote, p. 205 about applications to stability of spinning spacecraft!
Analytic Solution for a Symmetrical Body • Euler’s Eqtns under force, torque free conditions: I 1(dω1/dt) = ω2ω3(I 2 -I 3) I 2(dω2/dt) = ω3ω1(I 3 -I 1) I 3(dω3/dt) = ω1ω2(I 1 -I 2) (1) (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(dω1/dt) = ω2ω3(I 2 -I 3) (1 ) I 1(dω2/dt) = -↖| ω3ω1(I 1 -I 3) (2 ) I 3(dω3/dt) = 0 | (3 ) NOTE: A minus sign typo in Goldstein 3 rd Edition in Eq. (2 )!
Symmetrical Body I 1(dω1/dt) = ω2ω3(I 2 -I 3) I 1(dω2/dt) = - ω3ω1(I 1 -I 3) I 3(dω3/dt) = 0 (1 ) (2 ) (3 ) ω3 = const. Treat as a known initial condition. • Solve (1 ) & (2 ) simultaneously. – Define: Ω [(I 1 -I 3)/I 1]ω3 (1 ) (dω1/dt) = - Ωω2 (a) (2 ) (dω2/dt) = + Ωω1 (b) – Take derivatives of (a) & (b). Substitute back & get: (d 2ω1/dt 2) + Ω 2ω1 = 0 (d 2ω2/dt 2) + Ω 2ω2 = 0 (a ) (b ) (a ) & (b ): Standard simple harmonic oscillator eqtns!
(d 2ω1/dt 2) + Ω 2ω1 = 0 (d 2ω2/dt 2) + Ω 2ω2 = 0 Ω [(I 1 -I 3)/I 1] ω3 (a ) (b ) • Solutions: Sinusoidal functions: ω1 = A cos(Ωt) ω2 = A sin(Ωt) A depends on the initial conditions Interpretation: The angular velocity vector: ω = ω1 i + ω2 j + ω3 k Magnitude: ω = [(ω1)2+(ω2)2 + (ω3)2]½ = [A 2 + (ω3)2]½ = const Component of the angular velocity vector in the xy plane: ω = ω1 i + ω2 j Magnitude: ω = [(ω1)2+(ω2)2]½ = A = const
ω1 = A cos(Ωt), ω2 = A sin(Ωt) Interpretation: Angular velocity vector in xy plane: ω = ω1 i + ω2 j. Magnitude: ω = [(ω1)2 + (ω2)2]½ = A = const The vector ω rotates uniformly (fig. ) about the body z axis at frequency Ω Total angular velocity ω = ω1 i + ω2 j + ω3 k (also const in magnitude) precesses about z axis at precession frequency Ω. Ω depends As predicted by Poinsot’s on I 1, I 3, ω3 construction! Note: This precession is relative to the body axes. These are themselves, rotating with respect to the space axes at frequency ω!
ω1 = A cos(Ωt), ω2 = A sin(Ωt) Precession frequency Ω [(I 1 -I 3)/I 1]ω3 • 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]ω3. The Ch. 4 relation: (d/dt)s = (d/dt)b + ω (dω/dt) = ω Ω Gives the same (precessing ω) results!
ω1 = A cos(Ωt), ω2 = A sin(Ωt) Precession frequency Ω [(I 1 -I 3)/I 1]ω3 • 2 constants ω3 & 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(ω1)2 + (½)I 2(ω2)2 + (½)I 3(ω3)2 L 2 = (I 1ω1)2 + (I 2ω2)2 + (I 3ω3)2 Here: I 1 = I 2 , (ω1)2 + (ω2)2 = A 2. Or: T = (½)I 1 A 2 + (½)I 3(ω3)2 L 2 = (I 1 A)2 + (I 3ω3)2 (ω3)2 = [2 TI 1 - L 2]/[I 3(I 1 -I 3)] A 2 = [2 TI 3 - L 2]/[I 1(I 3 -I 1)]
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]ω3 – Numbers: (I 1 -I 3)/I 1 0. 00327 Ω 0. 00327 ω3 ω3/(306) – Also, ω3 ω = (2π)/(1 day) Precession period Tp = (2π)/ Ω 306 days Or (expected): Tp 10 months
• 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.
- Slides: 19