 Symmetric top under constant transverse body torque

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 $(\mathrm{Re}z,\mathrm{Im}z,c)$.