 Bottom-heavy spherical-cell orientation dynamics

For a spherical cell with its centre of mass displaced by $-{\ell }_g\mathbf{p}$ from its geometric centre, gravity produces torque $mg{\ell }_g\mathbf{p}\times \mathbf{k}$. Rotational resistance ${\zeta }_r$ gives angular velocity $\mathbf{\omega }/2+(mg{\ell }_g/{\zeta }_r)\mathbf{p}\times \mathbf{k}$. Taking its cross product with $\mathbf{p}$ gives the displayed orientation equation with $B={\zeta }_r/(mg{\ell }_g)$. Steady orientation therefore requires the two terms on the right to sum to zero; placing the vorticity term alone on the other side changes its sign.