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Intrinsic Wobble of Spinning Objects

It's known that the Earth's axis of spin is not perfectly aligned
with its polar axis of symmetry, so it "wobbles" a little bit.
This wobble is not to be confused with the precession of the
equinox, which is related to the fact that the Earth's axis of
spin is tilted quite a bit with respect to its orbital plane around
the Sun, and with respect to the Moon's orbital plane, which is a
forced precession with a period of about 25,800 years.  In contrast,
the wobble is just due to the free dynamics of the spinning Earth,
due to the tilt between its angular momentum and it axis of symmetry.

To analyze this problem, suppose we're given an axially symmetrical 
body with moment of inertia I1 about the axis of symmetry and I2 
about each of the principal perpendicular axes.  The body is floating 
freely in space, with no external forces, so we can take both the 
center of mass and the angular momentum vector H about that point 
to be fixed.

Let's define an orthogonal (and rotating) system of coordinates
xyz with their origin at the center of mass, and such that the 
z axis is always pointing in the direction of the body's axis of
symmetry, and the x axis is in the same plane as both the z axis 
and the angular momentum vector H.  The components of the angular
momentum in the x, y, and z directions are then simply

H_x = -|H|sin(a)     H_y = 0      H_z = |H|cos(a)

where "a" is the angle between the z axis and H.  Also, since x, 
y, and z are the principal axes of inertia, we have

H_x = I2 w_x       H_y = I2 w_y       H_z = I1 w_z

where w_x, w_y, and w_z are the components of the total angular
velocity of the body about the center of mass.  Combining these
relations gives

w_x = -|H|sin(a)/I2      w_y = 0      w_z = |H|cos(a)/I1

The component w_y denotes the rate of change of "a", and since 
this is zero, we see that the angle "a" remains constant, so the
configuration  can only precess about the total angular momentum
vector H.  Furthermore, since w_x/w_z equals the tangent of the angle
"b" between the z axis and the total angular velocity vector, it
follows that
I1 tan(a)  =  I2 tan(b)

This suffices to give us a complete qualitative as well as
quantitative picture of the motion.  Notice that the total angular
velocity vector w makes an angle "b" with the z axis, and it also
makes an angle a-b with the total H vector.  Consequently, the vector
w sweeps out a "cone" about the fixed vector H, and it sweeps out
another tangent cone about the rotating z axis (in accord with the
well-known kinematical fact that the most general kind of free motion
for a rigid body with a fixed point is generated by a "body cone"
rolling on a "space cone").  The axis of the latter is the H vector,
and the axis of the former is the z axis.  The line of tangency of
these cones is the direction of the total angular velocity w at each
instant.  

Also, notice that, depending on whether I1 is less than or greater
than I2, the outside or inside surface of the body cone rolls 
around the outside of the space cone, giving forward or retrograde
precession, respectively.  In either case, the relation between the
total angular velocity w, the relative spin rate s, and the free
precession rate p can be inferred from the parallelogram

_______________ w
p /        b  .  /
/         . a-b/
/       .      /
/a-b  .        /
/   .          /
/ .   b    pi-a/
/______________/
CG              s

The law of sines immediately gives

w           p          s
------  =  ------  =  --------
sin(a)     sin(b)     sin(a-b)

Thus we have the free precession rate

sin(b)       I1 cos(b)       I1  w_z
p  =  ------ w  =  -- ------ w  =  -- ------
sin(a)       I2 cos(a)       I2 cos(a)

and the relative spin rate

sin(a-b)         sin(a-b)
s  =  -------- w  =  ------------ w_z
sin(a)        sin(a)cos(b)

Using the trig identity sin(a-b) = sin(a)cos(b) - cos(a)sin(b) 
this reduces to

s  =  [ 1  -  tan(b)/tan(a) ] w_z

The relative spin rate s is the rate at which the angular velocity
vector w rotates, with respect to the body, about the body's axis 
of symmetry.  So, if we define this as the "rate of wobble", and if 
we recall that tan(b)/tan(a) = I1/I2, we get the well-known result
that this rate equals 1 - I1/I2 times the projection of the angular
velocity of the body onto the axis of symmetry (i.e., w_z).  For the
Earth we have I1/I2 = 1.0033, so the free precession is retrograde,
i.e., the angular velocity vector w rotates in a cone in direction
opposite the direction of w_z.  We have s = -0.0033 w_z, so the 
period is about 303 days (since w_z has a period of about 1 "day").

It turns out that the actual observed period of this precession is 
substantially longer, about 432.5 mean solar days.  This differs from
our computed value mainly because we assumed an ideal rigid body,
whereas the Earth is not perfectly rigid.
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