Precession of a Gyroscope — Question 6

Question 6

A precession period can infer axial inertia

A gyroscope with $M=1.0\,\mathrm{kg}$ and $\ell=.10\,\mathrm m$ spins at $100\,\mathrm{rad/s}$. Its measured slow-precession period is $4\pi\,\mathrm s$. Use $g=10$ and neglect nutation.

1. Infer axial inertia.

2. Predict the period if the spin speed doubles with all other parameters fixed, and explain the trend.

Question 6 - Solution

A precession period can infer axial inertia

(a) Derivation and calculation

Convert the measured period to precession rate: $$\Omega=\frac{2\pi}{T_p}=\frac{2\pi}{4\pi}=.500\ \mathrm{rad/s}.$$ Rearrange the slow-precession law: $$\boxed{I_s=\frac{Mg\ell}{\Omega\omega_s} =\frac{1.0(10)(.10)}{.500(100)}=.0200\ \mathrm{kg\,m^2}}.$$

(b) Result and interpretation

The period is $$T_p=\frac{2\pi I_s\omega_s}{Mg\ell}.$$ At fixed geometry and inertia it is proportional to spin speed, so $$\boxed{T_{p,\mathrm{new}}=2T_p=8\pi=25.13\ \mathrm s}.$$ The doubled spin momentum requires twice as long for the same gravitational torque to turn it through a complete precession cycle.

Physical check

The new precession rate is .250 rad/s, half the original value.