A steady precession of the heavy symmetric top checked on explicit data: the pivot balance, the two conserved momenta and the reduced energy

Numerical SI convention. Let $T$ denote physical elapsed time and fix $T_{\ast }=1\mathrm{s}$, $M_{\ast }=1\mathrm{k}\mathrm{g}$ and $D_{\ast }=1\mathrm{m}$. All formulas below use numerical SI values: $t:=T\mathrm{/}{T}_{\ast }$ is dimensionless, angles are in radians, positions and $d$ are divided by $D_{\ast }$, masses by $M_{\ast }$, inertia entries by $M_{\ast }{D}_{\ast }^2$, and $g$ by $D_{\ast }\mathrm{/}{T}_{\ast }^2$. Primes denote differentiation with respect to $t$. Restore physical angular velocity as ${\omega }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(T)=\omega (T\mathrm{/}{T}_{\ast })\mathrm{/}{T}_{\ast }$, angular momentum as $J_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(T)=(M_{\ast }{D}_{\ast }^2\mathrm{/}{T}_{\ast })J(T\mathrm{/}{T}_{\ast })$, energy as $(M_{\ast }{D}_{\ast }^2\mathrm{/}{T}_{\ast }^2)E$, and torque as $(M_{\ast }{D}_{\ast }^2\mathrm{/}{T}_{\ast }^2)\tau (T\mathrm{/}{T}_{\ast })$. Thus the displayed rates ${\phi }^{\mathrm{\prime }}=1$ and ${\psi }^{\mathrm{\prime }}=2$ mean physical rates $1{\mathrm{s}}^{-1}$ and $2{\mathrm{s}}^{-1}$, and every trigonometric argument is dimensionless. The physical placement is $x_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(T,{\rho }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}})=D_{\ast }x(T\mathrm{/}{T}_{\ast },{\rho }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}{D}_{\ast })$; its derivatives restore the pivot balance and heavy-top equations with the same SI data. These are unit conventions of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework, not measured inputs.

Adopt the heavy symmetric top branch Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot with declared parameters $A_c=2,C_c=1,m=1,d=1,g=1$ in the framework's units (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework), so that the body inertia matrix is $I=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(2,2,1)$ in the declared basis and the pivot transverse moment is $A=A_c+m{d}^2=2+1=3,C=C_c=1.$ Let ${\theta }_0:=\pi \mathrm{/}3$ and consider the angles $\phi (t)=t,\theta (t)={\theta }_0,\psi (t)=2t(t\in \mathbb{R}),$ the orientation path $R(t):=\mathrm{\Theta }(\phi (t),\theta (t),\psi (t))$ of The Euler-angle parametrization and the heavy symmetric top system, the centre of mass $c(t):=dR(t)e_3$, and the motion $x(t,\rho ):=c(t)+R(t)\rho$. Verify:

1. It is a motion of the branch satisfying the pivot balance. The pivot constraint $c(t)+R(t)(-d{e}_3)=0$ holds, and with $\omega (t)=(-\frac{\sqrt{3}}{2}\cos (2t),\frac{\sqrt{3}}{2}\sin (2t),\frac{5}{2})$ and $\gamma (t)=R(t{)}^{\mathsf{T}}{e}_3$, $\tilde{I}{\omega }^{\mathrm{\prime }}+\omega \times \tilde{I}\omega =-mgd(e_3\times \gamma )(t\in \mathbb{R}),$ where $\tilde{I}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(3,3,1)$; hence the motion satisfies the adopted balance of Adoption of the heavy symmetric top branch: axisymmetric rigid material, a fixed pivot material point, a uniform gravitational field and the balance of angular momentum about the pivot clause (v) and the conclusions of Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $\theta$ equation clauses 1-2 hold along it.
2. The conserved momenta and the reduced energy. With ${\phi }^{\mathrm{\prime }}=1$, ${\theta }^{\mathrm{\prime }}=0$, ${\psi }^{\mathrm{\prime }}=2$, $a=A{\phi }^{\mathrm{\prime }}{\sin }^2{\theta }_0+b\cos {\theta }_0=\frac{7}{2},b=C({\psi }^{\mathrm{\prime }}+{\phi }^{\mathrm{\prime }}\cos {\theta }_0)=\frac{5}{2},E=U({\theta }_0)=\frac{19}{4}\mathrm{.}$
3. The heavy-top system holds, and $\theta$ is constant. The three equations of The Euler-angle parametrization and the heavy symmetric top system hold for the triple, and $A{\theta }^{\mathrm{\prime }\mathrm{\prime }}=-U^{\mathrm{\prime }}({\theta }_0)=0$, so the motion is a steady precession: ${\phi }^{\mathrm{\prime }}=1$ and ${\psi }^{\mathrm{\prime }}=2$ are constant and the nutation angle is constantly $\pi \mathrm{/}3$.

• A steady precession, not a sleeping top. The nutation angle stays at $\pi \mathrm{/}3$ while the precession and spin rates are constant; the sleeping-top configuration $\theta =0$ is a pole of the chart and is not covered by Reduction of the heavy symmetric top branch: pivotal angular momentum, Euler-angle equations, the two conserved momenta and the one-dimensional $\theta$ equation, so this example makes no claim about it.

• What is declared. The parameters, angles and rates are declared model data in the framework's units; the verification is an exact computation inside the branch and asserts no measured mass, length, duration or field strength, and no claim that a concrete top realises this data.