• 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 = Iω is:

L1 = I1ω1, L2 = I2ω2, L3 = I3 ω 3 (1)

KE: T = (½)ωIω is:

T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2 (2)

• Combining (1) & (2) gives:

T = (½)[(L1)2/I1] + (½)[(L2)2/I2] + (½)[(L3)2/I3] (3)

T = const, since it is conserved!

Binet’s Construction

T = (½)[(L1)2/I1] + (½)[(L2)2/I2] + (½)[(L3)2/I3] (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: I3 I2 I1. Also, change notation (again!): L1 Lx, L2 Ly, L3 Lz

(Put (3) into the standard form for an ellipsoid):

(½)[(Lx)2/(T I1)] + (½)[(Ly)2/(T I2)] + (½)[(Lz)2/(T I3)] = 1 (3)

• Binet (or KE) Ellipsoid: (½)[(Lx)2/(TI1)] + (½)[(Ly)2/(TI2)] + (½)[(Lz)2/(TI3)] = 1 (3)

Or: [(Lx)2/a2] + [(Ly)2/b2]+ [(Lz)2/c2] = 1 (3)

See figure: Semimajor

axes (Decreasing size):

a = (2T I1)½ , b = (2T I2)½

c = (2T I3)½ Also have

Conservation of total

angular momentum:

L2 =(Lx)2+(Ly)2+(Lz)2

= const. Or: [(Lx)2+(Ly)2+(Lz)2]/L2 =1 (4)

(4): A sphere in (Lx,Ly,Lz) space!

(3) & (4) must be satisfied simultaneously!

• Simultaneously we have the Binet (KE) Ellipsoid:

(½)[(Lx)2/(TI1)] + (½)[(Ly)2/(TI2)] + (½)[(Lz)2/(TI3)] = 1 (3)

& the angular momentum sphere:

[(Lx)2+(Ly)2+(Lz)2]/L2 =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/(TI1)] + (½)[(Ly)2/(TI2)] + (½)[(Lz)2/(TI3)] = [(Lx)2+(Ly)2+(Lz)2]/L2 (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 (2TI3)½ < L < (2TI1)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/(TI1)] + (½)[(Ly)2/(TI2)] + (½)[(Lz)2/(TI3)] = [(Lx)2+(Ly)2+(Lz)2]/L2 (5)

Solution if & only if (2TI3)½ < L < (2TI1)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 = (2TI2)½

Symmetrical Body • Back to Euler’s Equations under force, torque free conditions:

I1(dω1/dt) = ω2ω3(I2 -I3) (1)

I2(dω2/dt) = ω3ω1(I3 -I1) (2)

I3(dω3/dt) = ω1ω2(I1 -I2) (3)

• Special case: Steady rotation (constant angular velocity ω)

(dωi/dt) = 0 (i = 1,2,3)

ω1ω2(I1 -I1) = ω2ω3(I2 -I3) = ω3ω1(I3 -I1) = 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(I1 -I1) = ω2ω3(I2 -I3) = ω3ω1(I3 -I1) = 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 I3, z direction, Lz). The discussed relations

This happens if the radius of the angular momentum sphere is L Ls = (2TI3)½ . Small deviations from this Angular momentum sphere radius L is slightly < Ls = (2TI3)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 I1, x direction, Lx). The discussed relations

This happens if the radius of the angular momentum sphere is L Ls = (2TI1)½. Small deviations from this

Angular momentum sphere radius L is slightly >

Ls = (2TI1)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 I2, y direction, Ly). The discussed relations Happens if the radius of the angular momentum sphere is L Ls = (2TI2)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

< Ls & 2 with L slightly > Ls. Each

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!

• Euler’s Eqtns under force, torque free conditions:

I1(dω1/dt) = ω2ω3(I2 -I3) (1)

I2(dω2/dt) = ω3ω1(I3 -I1) (2)

I3(dω3/dt) = ω1ω2(I1 -I2) (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: I1 = I2.

• (1), (2), (3) become:

I1(dω1/dt) = ω2ω3(I2 -I3) (1)

I1(dω2/dt) = - ω3ω1(I1 -I3) (2)

I3(dω3/dt) = 0 (3)

NOTE: A minus sign typo in Goldstein 3rd Edition in Eq. (2)!

↖| |

Analytic Solution for a Symmetrical Body

Symmetrical BodyI1(dω1/dt) = ω2ω3(I2 -I3) (1)

I1(dω2/dt) = - ω3ω1(I1 -I3) (2)

I3(dω3/dt) = 0 (3)

(3) ω3 = const. Treat as a known initial condition.

• Solve (1) & (2) simultaneously. – Define: Ω [(I1 -I3)/I1]ω3

(1) (dω1/dt) = - Ωω2 (a)

(2) (dω2/dt) = + Ωω1 (b)

– Take derivatives of (a) & (b). Substitute back & get:

(d2ω1/dt2) + Ω2ω1 = 0 (a)

(d2ω2/dt2) + Ω2ω2 = 0 (b)(a) & (b): Standard simple harmonic oscillator eqtns!

(d2ω1/dt2) + Ω2ω1 = 0 (a)

(d2ω2/dt2) + Ω2ω2 = 0 (b)

Ω [(I1 -I3)/I1] ω3

• Solutions: Sinusoidal functions:

ω1 = A cos(Ωt) ω2 = A sin(Ωt)

A depends on the initial conditions

Interpretation:

The angular velocity vector: ω = ω1i + ω2j + ω3k

Magnitude:

ω = [(ω1)2+(ω2)2 + (ω3)2]½ = [A2 + (ω3)2]½ = const

Component of the angular velocity vector in the xy plane:

ω = ω1i + ω2j

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 = ω1i + ω2j. 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 = ω1i + ω2j + ω3k

(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 I1, I3, ω3

ω1 = A cos(Ωt), ω2 = A sin(Ωt)

Precession frequency Ω [(I1 -I3)/I1]ω3

• The closer I1 is to I3, 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 = [(I1 -I3)/I1]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 [(I1 -I3)/I1]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 L2 are conserved. In general:

T = (½)I1(ω1)2 + (½)I2(ω2)2 + (½)I3(ω3)2

L2 = (I1ω1)2 + (I2ω2)2 + (I3ω3)2

Here: I1 = I2 , (ω1)2 + (ω2)2 = A2.

T = (½)I1A2 + (½)I3(ω3)2

L2 = (I1A)2 + (I3ω3)2

Or: (ω3)2 = [2TI1 - L2]/[I3(I1 -I3)]

A2 = [2TI3 - L2]/[I1(I3 -I1)]

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!)

I1 < I3, I1 I2

Earth’s rotational motion is A torque & force free symmetrical rigid body.

• Precession frequency Ω [(I1 -I3)/I1]ω3

– Numbers: (I1 -I3)/I1 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.