Precession of a Gyroscope — Question 4

Question 4

A slow-precession approximation needs a dimensionless check

A gyroscope has $Mg\ell=.20\,\mathrm{N\,m}$ and $I_s=.0040\,\mathrm{kg\,m^2}$. Use the slow estimate $\Omega=Mg\ell/(I_s\omega_s)$. An experiment adopts $\Omega/\omega_s\le.010$ as a rate-separation criterion.

1. Find the minimum spin speed satisfying this criterion.

2. Does $\omega_s=20\,\mathrm{rad/s}$ satisfy it, and is rate separation alone a complete guarantee of steady precession?

Question 4 - Solution

A slow-precession approximation needs a dimensionless check

(a) Derivation and calculation

The rate ratio is $$\frac{\Omega}{\omega_s}=\frac{Mg\ell}{I_s\omega_s^2}=\frac{.20}{.0040\omega_s^2}=\frac{50}{\omega_s^2}.$$ The adopted criterion therefore requires $$50/\omega_s^2\le.010\Rightarrow\omega_s^2\ge5000,$$ $$\boxed{\omega_{s,\min}=\sqrt{5000}=70.71\ \mathrm{rad/s}}.$$

(b) Result and interpretation

At the proposed lower speed, $$\boxed{\Omega/\omega_s=50/(20)^2=.125>.010}.$$ It fails the criterion. Even a small rate ratio alone does not guarantee steady precession: the preparation and transverse-inertia contribution must also be consistent with spin dominance, and arbitrary initial conditions can produce nutation.

Physical check

Doubling spin speed reduces this rate ratio by a factor of four.