Precession of a Gyroscope — Question 2

Question 2

The tilt factors cancel in the slow-precession rate

A rapidly spinning symmetric top has mass $M$, center offset $\ell$, spin momentum $L_s$ along its axis, and fixed tilt $0<\theta<\pi$ from upward vertical. Assume steady slow precession about vertical with spin-dominated angular momentum.

1. Derive the precession rate, retaining the tilt factors until the final step.

2. Explain why the exactly vertical case cannot be obtained by blindly cancelling $\sin\theta$.

Question 2 - Solution

The tilt factors cancel in the slow-precession rate

(a) Derivation and calculation

The gravitational torque magnitude is $$|\tau|=Mg\ell\sin\theta.$$ During precession, the tip of the spin vector traces a horizontal circle of radius $L_s\sin\theta$, so $$\left|\frac{d\vec L_s}{dt}\right|=|\Omega|L_s\sin\theta.$$ For the stipulated nonvertical tilt, cancel the nonzero sine factor: $$\boxed{\Omega=\frac{Mg\ell}{L_s}}$$ with the precession sense set by the signed spin direction.

(b) Result and interpretation

At exactly vertical alignment, $$\sin\theta=0\Rightarrow|\tau|=0,\qquad L_s\sin\theta=0.$$ The equation reduces to $0=0$, not a determination of $\Omega$. Changing azimuth does not change a vertical axis, so its azimuthal precession rate is geometrically undefined. The cancellation is valid only before taking the vertical limit.

Physical check

Tilt independence in the slow formula follows from equal sine factors in torque and vector-tip radius.