$2\mathrm{.}2$ If z$($t$)=0$ for all t$,$ what is the constant angle $$ that perfectly cancels the appearance of the gravity term on the right hand side? Rewrite your solution to $(1\mathrm{.}1)/(1\mathrm{.}2)$ in terms of the relative angle @ that satisfies $0=+0,$ where $$ is the static equilibrium rotation.