Rigid Body Motion - Dynamics and Relativity - page 229

Setting ω p = φ and using Eq. (10.103) gives us

ω p (L 3 sin θ − Iω p sin θ cos θ) = τ. (10.104)

Eq. (10.104) is a quadratic equation in ω p with the solution

L 3 −

L 3 ± 4 Iτ/ tan θ

2 I cos θ

ω p = . (10.105)

Eq. (10.96) gives us the magnitude of the torque

τ = mgR sin θ, (10.106)

which we can use together with Eq. (10.102) to rewrite Eq. (10.105) as

L 3 −

I 3 ω 3 ± 4 ImgR cos θ

2 I cos θ

ω p = . (10.107)

What is the meaning of the two solutions given by Eq. (10.107)? The higher value of
ω p corresponds to taking the plus sign in the numerator and will give us a frequency
ω p ∼

ω 3 , which is fast precession, given that ω t is large. The second solution
corresponds to taking the minus sign in the numerator. We shall soon show that
this 'slow' solution corresponds to the ω p we found previously, i.e. in Eq. (10.99).
That there are two solutions to the equation of motion is a feature of gyroscopic
motion that we missed with our simple analysis. To gain a deeper insight, let us
look at both solutions in the limit that the torque is very small, as would be the
case for a gyroscope of low mass. In this case

4 Iτ
tan θ

L 3 (10.108)

and the high frequency solution to Eq. (10.105) is

I 3 ω 3
I cos θ ,

ω p ≈ (10.109)

which is independent of the torque and represents the precession of a free symmetric
top (cf. Eq. (10.83)). The slow solution may be examined, in the limit of small
torque, by a binomial expansion to first order of the form:

1
2 x,

x) 1 / 2 ( 1 + 1 +

giving, (to first order in τ )

1 .

L 3 −

4 Iτ
2 L 3

4 Iτ/ tan θ ≈ L 3 − (10.110)

tan θ