Exercises — Gyroscope — steady precession derivation

WHAT: plug straight into the fast-top formula. WHY: the wheel spins fast, so the simple formula applies. $\Omega =\frac{mg\ell }{I_3{\omega }_s}=\frac{2.0\cdot 9.8\cdot 0.20}{0.10\cdot 40}=\frac{3.92}{4.0}=0.98\text{rad/s}.$ The axis sweeps a slow horizontal circle at $0.98$ rad/s (prograde, since $\theta <90^{\circ }$).

WHAT: convert an angular rate to a period. WHY: one full circle is $2\pi$ radians, so time $=2\pi /\Omega$. $T=\frac{2\pi }{\Omega }=\frac{2\pi }{0.98}\approx 6.41\text{s}.$

False. The fast-top $\Omega =mg\ell /(I_3{\omega }_s)$ contains no $\theta$ at all — the $\sin \theta$ from the torque and the $\sin \theta$ from the rotating horizontal Angular Momentum cancel exactly.

WHAT: re-apply the formula with the smaller spin. WHY: $\Omega \propto 1/{\omega }_s$, so halving the spin doubles $\Omega$. $\Omega =\frac{3.92}{0.10\cdot 20}=\frac{3.92}{2.0}=1.96\text{rad/s}.$ As real tops spin down, they precess faster — exactly what you see just before a dying top wobbles and flops.

WHAT: invert the fast-top formula for ${\omega }_s$. WHY: $\Omega$ is measured, everything else known, so solve algebraically. ${\omega }_s=\frac{mg\ell }{I_3\Omega }=\frac{0.30\cdot 9.8\cdot 0.040}{1.2\times 10^{-4}\cdot 2.5}=\frac{0.11760}{3.0\times 10^{-4}}=392\text{rad/s}.$

WHAT: compare $\Omega \propto \ell$ (all else equal). WHY: a bigger lever arm $\ell$ means more gravitational torque, and torque drives precession. $\frac{{\Omega }_B}{{\Omega }_A}=\frac{{\ell }_B}{{\ell }_A}=2.$ Gyroscope B precesses twice as fast.

WHAT: demand the discriminant in the exact quadratic be non-negative. WHY: if the square root's argument goes negative, $\Omega$ is complex — no real steady precession, so the top must nutate or fall. Require $(I_3{\omega }_s{)}^2\ge 4I_{1}mg\ell \cos \theta$: ${\omega }_s\ge \frac{\sqrt{4I_{1}mg\ell \cos \theta }}{I_3}=\frac{\sqrt{4\cdot 0.012\cdot 0.5\cdot 9.8\cdot 0.05\cdot \cos 30^{\circ }}}{0.006}.$ Inside the root: $4\cdot 0.012\cdot 0.5\cdot 9.8\cdot 0.05\cdot 0.8660=0.010184$. So ${\omega }_s\ge \frac{\sqrt{0.010184}}{0.006}=\frac{0.10092}{0.006}\approx 16.8\text{rad/s}.$ Below $16.8$ rad/s, no steady precession is possible.

WHAT: at minimum spin the discriminant is zero, so both roots collapse to one. WHY: $\sqrt{0}=0$ leaves $\Omega =I_3{\omega }_s/(2I_1\cos \theta )$. With ${\omega }_s=16.82$ rad/s: $\Omega =\frac{I_3{\omega }_s}{2I_1\cos \theta }=\frac{0.006\cdot 16.82}{2\cdot 0.012\cdot 0.8660}=\frac{0.10092}{0.020785}\approx 4.86\text{rad/s}.$ The fast and slow precessions merge into a single "borderline" rate here.

WHAT: expand the square root for large ${\omega }_s$. WHY: we want to see the exact theory contain the simple one. With $a=I_3{\omega }_s$, $b=4I_{1}mg\ell \cos \theta$: $\sqrt{a^2-b}\approx a-\frac{b}{2a}$, so ${\Omega }_{-}\approx \frac{a-(a-b/2a)}{2I_1\cos \theta }=\frac{b/2a}{2I_1\cos \theta }=\frac{mg\ell }{I_3{\omega }_s}.$ Numeric check at ${\omega }_s=100$: fast-top gives $\Omega =(0.5\cdot 9.8\cdot 0.05)/(0.006\cdot 100)=0.2450/0.6=0.4083$ rad/s. Exact $-$ root: $a=0.6$, $b=0.010184$, $\sqrt{a^2-b}=\sqrt{0.36-0.010184}=\sqrt{0.349816}=0.59145$; ${\Omega }_{-}=(0.6-0.59145)/(2\cdot 0.012\cdot 0.8660)=0.008554/0.020785=0.4116$ rad/s. Agreement to about $0.8\%$. ✓

WHAT: read off which factors control $\Omega$ and scale one. WHY: $\Omega =mg\ell /(I_3{\omega }_s)$ is linear in $m$, $\ell$, and $1/I_3$. To double $\Omega$ at fixed ${\omega }_s$, any one of:

• Double $\ell$: $\ell =0.40$ m ⟹ $\Omega =3.92\cdot 2/4.0=1.96$ rad/s.
• Double $m$: $m=4.0$ kg ⟹ $\Omega =2\cdot 3.92/4.0=1.96$ rad/s.
• Halve $I_3$: $I_3=0.05{\text{kg⋅m}}^2$ ⟹ $\Omega =3.92/(0.05\cdot 40)=1.96$ rad/s. Any of the three works; each gives $\Omega =1.96$ rad/s.

WHAT: use the pre-cancellation torque balance to build $\Omega$, then check the $\sin \theta$ really cancels. WHY: this makes the "remarkable cancellation" concrete rather than symbolic. First extract $mg\ell$: since $mg\ell \sin \theta =3.0$ and $\sin 60^{\circ }=0.8660$, we get $mg\ell =3.0/0.8660=3.4641$ N·m. Balance: $\Omega L_s\sin \theta =mg\ell \sin \theta$, so $\Omega =\frac{mg\ell \sin \theta }{L_s\sin \theta }=\frac{mg\ell }{L_s}=\frac{3.4641}{5.0}=0.6928\text{rad/s}.$ Cross-check with the boxed formula: $mg\ell /(I_3{\omega }_s)=3.4641/5.0=0.6928$ rad/s. ✓ The two $\sin \theta$'s cancelled exactly.

WHAT: evaluate the full quadratic's two roots. WHY: the exact theory predicts a fast (nutation-like) branch and a slow (gravity-driven) branch — see Lagrangian Mechanics for the derivation. Ingredients: $a=I_3{\omega }_s=0.006\cdot 30=0.18$; $b=4I_{1}mg\ell \cos 45^{\circ }=4\cdot 0.012\cdot 0.5\cdot 9.8\cdot 0.05\cdot 0.7071=0.0083160$. Discriminant: $a^2-b=0.0324-0.0083160=0.0240840$; $\sqrt{}=0.155190$. Denominator: $2I_1\cos 45^{\circ }=2\cdot 0.012\cdot 0.7071=0.016971$. ${\Omega }_{+}=\frac{0.18+0.155190}{0.016971}=\frac{0.335190}{0.016971}=19.75\text{rad/s (fast)},$ ${\Omega }_{-}=\frac{0.18-0.155190}{0.016971}=\frac{0.024810}{0.016971}=1.462\text{rad/s (slow)}.$ The fast-top formula approximates the slow root ${\Omega }_{-}$: check $mg\ell /(I_3{\omega }_s)=(0.5\cdot 9.8\cdot 0.05)/0.18=0.245/0.18=1.361$ rad/s — close to $1.462$. Both roots are positive here (prograde) because $\cos 45^{\circ }>0$.

WHAT: apply Vieta's relations. WHY: for $A{\Omega }^2+B\Omega +C=0$, the roots satisfy ${\Omega }_{+}+{\Omega }_{-}=-B/A$ and ${\Omega }_{+}{\Omega }_{-}=C/A$ — a fast cross-check. Here $A=I_1\cos \theta$, $B=-I_3{\omega }_s$, $C=mg\ell$. So: ${\Omega }_{+}+{\Omega }_{-}=\frac{I_3{\omega }_s}{I_1\cos \theta },{\Omega }_{+}{\Omega }_{-}=\frac{mg\ell }{I_1\cos \theta }.$ Check L4.3 sum: $\frac{I_3{\omega }_s}{I_1\cos \theta }=\frac{0.18}{0.012\cdot 0.7071}=\frac{0.18}{0.0084853}=21.21$. Direct: $19.75+1.462=21.21$. ✓ Check L4.3 product: $\frac{mg\ell }{I_1\cos \theta }=\frac{0.245}{0.0084853}=28.87$. Direct: $19.75\cdot 1.462=28.88$. ✓

WHAT: set ${\omega }_s=0$ in sum and product. WHY: the sum ${\Omega }_{+}+{\Omega }_{-}=I_3{\omega }_s/(I_1\cos \theta )\to 0$, so ${\Omega }_{-}=-{\Omega }_{+}$. The product ${\Omega }_{+}{\Omega }_{-}=mg\ell /(I_1\cos \theta )>0$ for $\theta <90^{\circ }$ — but if ${\Omega }_{-}=-{\Omega }_{+}$ then the product is $-{\Omega }_{+}^2<0$. Contradiction ⟹ no real steady precession exists for a non-spinning top (discriminant $-4I_{1}mg\ell \cos \theta <0$). Interpretation: without spin there is no gyroscopic stiffness; a released rod simply falls (that is pure Nutation amplitude, not steady precession). Steady precession is fundamentally a spin phenomenon.

WHAT: target $\Omega =2\pi /T$, then rearrange the quadratic to isolate ${\omega }_s$. WHY: the equation is quadratic in $\Omega$ but ${\omega }_s$ appears only in the single $-I_3{\omega }_s\Omega$ term, so it is linear in ${\omega }_s$ and solves cleanly. Step 1 — target rate: $\Omega =2\pi /T=2\pi /5.0=1.2566$ rad/s. Step 2 — isolate ${\omega }_s$: start from $I_1{\Omega }^2\cos \theta -I_3{\omega }_s\Omega +mg\ell =0$. Move the ${\omega }_s$ term alone to one side: $I_3{\omega }_s\Omega =I_1{\Omega }^2\cos \theta +mg\ell ⟹{\omega }_s=\frac{I_1{\Omega }^2\cos \theta +mg\ell }{I_3\Omega }.$ Step 3 — plug numbers. Numerator: $I_1{\Omega }^2\cos \theta =0.020\cdot (1.2566{)}^2\cdot 0.8660=0.020\cdot 1.5790\cdot 0.8660=0.027348$; add $mg\ell =1.0\cdot 9.8\cdot 0.15=1.470$; total $=1.497348$. Denominator: $I_3\Omega =0.010\cdot 1.2566=0.0125664$. $\boxed{{\omega }_s=\frac{1.497348}{0.0125664}=119.2\text{rad/s}}$ Step 4 — sanity vs fast-top: $mg\ell /(I_3\Omega )=1.470/0.0125664=116.98$ rad/s — the exact answer is slightly higher because the small centrifugal term $I_1{\Omega }^2\cos \theta$ adds to the numerator. ✓

WHAT: substitute $\cos 90^{\circ }=0$ and watch the exact formula's denominator vanish. WHY: $2I_1\cos \theta \to 0$ makes ${\Omega }_{\pm }=\frac{I_3{\omega }_s\pm \sqrt{(I_3{\omega }_s{)}^2-0}}{0}$ — a $\frac{\text{something}}{0}$, so the quadratic degenerates. Go back to the equation before dividing, at $\cos \theta =0$: $I_1{\Omega }^2\cdot 0-I_3{\omega }_s\Omega +mg\ell =0⟹-I_3{\omega }_s\Omega +mg\ell =0.$ The ${\Omega }^2$ term is gone: the quadratic collapses to a linear equation with a single finite root $\Omega =\frac{mg\ell }{I_3{\omega }_s},$ which is exactly the fast-top formula — no approximation needed at $\theta =90^{\circ }$. The "second root" ${\Omega }_{+}$ has run off to $\pm \infty$; physically the fast nutation-scale branch disappears because there is no vertical component of the spin axis for the transverse inertia $I_1$ to store precession angular momentum in. Interpretation: a horizontal gyroscope (axle level, CM offset sideways) precesses at precisely $mg\ell /(I_3{\omega }_s)$ with no correction. For $\theta$ just below $90^{\circ }$, ${\Omega }_{-}$ stays finite and near this value while ${\Omega }_{+}$ blows up; crossing to $\theta >90^{\circ }$ ($\cos \theta <0$) flips both roots' signs → retrograde precession, consistent with the sign convention above.

Fast-top precession rate ::: $\Omega =\frac{mg\ell }{I_3{\omega }_s}$, independent of tilt $\theta$. Which moment of inertia carries the spin angular momentum ::: $I_3$ (about the symmetry axis). Condition for real steady precession ::: $(I_3{\omega }_s{)}^2\ge 4I_{1}mg\ell \cos \theta$. Which exact root does the fast-top formula approximate ::: the slow root ${\Omega }_{-}$. What does a negative $\Omega$ mean ::: precession in the opposite (retrograde/clockwise-from-above) sense, not an error. What happens to the exact formula at $\theta =90^{\circ }$ ::: denominator $2I_1\cos \theta \to 0$; the quadratic degenerates to linear and $\Omega =mg\ell /(I_3{\omega }_s)$ exactly. What happens to precession as the spin dies ::: it speeds up ($\Omega \propto 1/{\omega }_s$), then nutates and falls.