Precession of a Gyroscope — Question 3

Question 3

Reversing spin reverses precession without reversing gravity torque

A horizontal gyroscope axis extends along $+x$ from its pivot; gravity is along $-z$. In trial A its spin momentum is $+L\hat x$; in trial B it is $-L\hat x$, with the same positive magnitude $L$. Assume the slow steady regime.

1. Find the signed precession vectors in both trials.

2. Explain why the gravity torque is the same even though precession reverses.

Question 3 - Solution

Reversing spin reverses precession without reversing gravity torque

(a) Derivation and calculation

In both trials the moment is $$\vec\tau=\ell\hat x\times(-Mg\hat z)=Mg\ell\hat y.$$ For trial A, $$\Omega_A\hat z\times L\hat x=Mg\ell\hat y \Rightarrow\boxed{\vec\Omega_A=(Mg\ell/L)\hat z}.$$ For trial B, $$\Omega_B\hat z\times(-L\hat x)=Mg\ell\hat y \Rightarrow\boxed{\vec\Omega_B=-(Mg\ell/L)\hat z}.$$

(b) Result and interpretation

The torque depends on $\vec r_C$ and gravity, not on spin. Reversing both factors of the precession cross product leaves it unchanged: $$(-\vec\Omega_A)\times(-\vec L_A)=\vec\Omega_A\times\vec L_A=\vec\tau.$$ Thus equal torque and opposite spin require opposite precession in the stated slow steady regime.

Physical check

Trial A precesses counterclockwise and trial B clockwise when viewed from above.