Symmetric top

A symmetric top is a rigid body with two equal principal moments of inertia, conventionally $I_1=I_2$. Its third principal axis is its symmetry axis.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

For a symmetric top with axial unit vector $\mathbf{e}$ and angular velocity $\mathbf{\omega }$, the classical axial spin is the scalar $n=\mathbf{\omega }\cdot \mathbf{e}$. The identity $\dot{\mathbf{e}}=\mathbf{\omega }\times \mathbf{e}$ implies $\mathbf{\omega }=\mathbf{e}\times \dot{\mathbf{e}}+n\mathbf{e}$. If the transverse and axial principal moments of inertia are $A$ and $C$, then the angular momentum about the fixed vertex is $\mathbf{L}=A\mathbf{e}\times \dot{\mathbf{e}}+Cn\mathbf{e}$. A torque perpendicular to $\mathbf{e}$ conserves $n$. During steady precession at rate $\Omega$ and inclination $\alpha$, the general angular velocity is $\Omega \mathbf{k}+(n-\Omega \cos \alpha )\mathbf{e}$; the precession rate by itself does not fix $n$.

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.

A rotor with two equal transverse moments and a smaller symmetry-axis moment, $I_a<I_b=I_c$. Its rotational constants obey $A>B=C$.

An oblate symmetric top has two equal smaller principal moments of inertia and a larger moment of inertia about its symmetry axis. A planar trigonal molecule, such as boron trifluoride, is an example. Its axial rotational constant is smaller than its transverse rotational constant.

For an axially symmetric rigid body with $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, no external torque means that angular momentum $\mathbf{L}$ is fixed in an inertial frame. The axial angular velocity ${\omega }_3$ is constant by Euler equations for a torque-free rigid body, and the symmetry axis obeysIts angle to $\mathbf{L}$ is constant and it precesses in space at $L/I_{\perp }$. The transverse angular-velocity components instead rotate in body coordinates at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. These are different frame descriptions and must not be substituted for one another.

For principal inertias $I_{\perp },I_{\perp },I_{\parallel }$, the transverse angular-velocity components rotate in body axes at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. The symmetry axis precesses about the inertially fixed angular momentum at rate $L/I_{\perp }$. These are different frame-dependent descriptions of the same motion.

A heavy symmetric top is a symmetric top with one point fixed in a uniform gravitational field. Its vertical and body-axis rotational symmetries give two cyclic Euler angles.

After fixing the two cyclic angular momenta of a heavy symmetric top, its inclination has an effective potential $U(u)$ with $u=\cos \theta$. Write the reduced energy equation in the form $P(u)=2A(E-U(u))(1-u^2)$. A steady nonvertical inclination has $E=U(u_0)$ and $U^{\prime }(u_0)=0$, the latter coming from the reduced second-order Euler-Lagrange equation. Therefore $P(u_0)=P^{\prime }(u_0)=0$.

This proves the double-root condition without dividing by a velocity that vanishes in the very solution under study. A simple turning point has $P(u_0)=0$ but nonzero derivative, and is not a steady inclination. Exactly vertical spinning corresponds to $u=\pm 1$, where the inclination coordinate and its reduced denominator are singular; these endpoint configurations require a separate treatment and do not supply an interior double root.

Writing the unit symmetry axis as $n$ and axial spin momentum as $s=I_3{\omega }_3$, angular momentum is $L=I_{1}n\times \dot{n}+sn$. Taking its derivative and applying gravitational torque gives $I_{1}n\times \ddot{n}+s\dot{n}=-mg\ell n\times \hat{z}$.

Variation in the inclination of a rotating body's axis, distinguished from its azimuthal precession. For a heavy symmetric top, the cyclic angular momenta reduce the motion to ${\dot{\mu }}^2=F(\mu )$ with $\mu =\cos \theta$ and $F$ cubic. Its admissible interval and turning points describe the nutation.

A heavy symmetric top precesses uniformly when its inclination is constant and its azimuthal speed is constant. For a nonvertical inclination, the condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$. The real-root condition depends on whether the body axis points above or below the horizontal.

When $s^2\gg 4mg\ell I_1\cos \theta$, the two nonvertical steady-precession roots are approximately $mg\ell /s$ and $s/(I_1\cos \theta )-mg\ell /s$. These asymptotic descriptions do not apply at a singular denominator without taking the original quadratic limit.