Symmetry-axis equation of a heavy symmetric top

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 $I_{1}n\times \ddot{n}+s\dot{n}=-mg\ell n\times \hat{z}$.