Precession of a Gyroscope — Question 10

Question 10

An exact steady-precession model reveals a second branch

For a symmetric top at constant nonvertical tilt $\theta$, an exact regular-precession condition is supplied: $I_\perp\Omega^2\cos\theta-L_s\Omega+Mg\ell=0$. Here $I_\perp$ is transverse inertia about the pivot and $L_s$ is the axial spin angular momentum. Let $I_\perp=.50\,\mathrm{kg\,m^2}$, $L_s=5\,\mathrm{kg\,m^2/s}$, $Mg\ell=1\,\mathrm{N\,m}$ and $\theta=60^\circ$.

1. Find both regular-precession rates and compare the smaller one with the slow approximation.

2. State the condition for real roots and explain why the quadratic alone does not say which motion follows an arbitrary release.

Question 10 - Solution

An exact steady-precession model reveals a second branch

(a) Derivation and calculation

Insert the supplied parameters: $$.50\cos60^\circ\,\Omega^2-5\Omega+1=0 \Rightarrow .25\Omega^2-5\Omega+1=0.$$ The quadratic formula gives $$\boxed{\Omega_\pm=\frac{5\pm\sqrt{25-1}}{.50}=10\pm4\sqrt6\ \mathrm{rad/s}}.$$ Thus $\Omega_-=.20204$ rad/s and $\Omega_+=19.79796$ rad/s. The slow estimate $Mg\ell/L_s=.200$ rad/s approximates only the smaller branch (about $1.01\%$ low).

(b) Result and interpretation

Real roots require nonnegative discriminant: $$\boxed{L_s^2-4I_\perp Mg\ell\cos\theta\ge0}.$$ Here the discriminant is $25-1=24>0$. The quadratic assumes constant tilt and compatible initial motion. It identifies candidate regular-precession states; it does not determine their stability or the transient motion after an arbitrary release.

Physical check

The root sum is $20$ rad/s and their product is $4$ rad$^2$/s$^2$, consistent with the quadratic coefficients.