Precession of a Gyroscope — Question 8

Question 8

Balancing the center at the pivot removes gravity-driven precession

A fast rotor has spin momentum $2.0\,\mathrm{kg\,m^2/s}$. Initially the total center of mass of the assembly lies at its pivot. A small $0.10\,\mathrm{kg}$ counterweight is then mounted $0.30\,\mathrm m$ along the horizontal rotor axis from the pivot. Use $g=10$; neglect counterweight contribution to angular momentum compared with rotor spin.

1. Find the gravity-driven slow-precession rate before and after adding the weight.

2. What happens to precession sense if the weight is placed the same distance on the opposite side of the pivot?

Question 8 - Solution

Balancing the center at the pivot removes gravity-driven precession

(a) Derivation and calculation

Initially the total center of mass is at the pivot: $$\vec r_C=0\Rightarrow\vec\tau_g=0.$$ There is no gravity-driven precession in the ideal prepared spin state. The added mass supplies moment $$|\tau_g|=mg\ell=.10(10)(.30)=.300\ \mathrm{N\,m},$$ $$\boxed{|\Omega|=|\tau_g|/L_s=.300/2.0=.150\ \mathrm{rad/s}}.$$

(b) Result and interpretation

Moving the weight to the opposite side reverses its position vector: $$(-\vec r)\times m\vec g=-\vec\tau_g.$$ For unchanged spin momentum, $$\boxed{\vec\Omega_{\mathrm{opposite}}=-\vec\Omega_{\mathrm{original}}}.$$ Balanced support removes gravitational torque; it does not exclude torque-free wobble arising from a different initial preparation.

Physical check

The calculation neglects the counterweight angular momentum as stipulated, retaining only its gravitational moment.