= Solution
To match the printed order $df/dt=\{H,f\}$, define the <Poisson bracket> with the sign convention
$$
\boxed{\{F,G\}_{\rm paper}=\sum_i\left(F_{p_i}G_{q_i}-F_{q_i}G_{p_i}\right).}
$$
<Hamilton's equations> imply $\dot f=\sum_i(f_{q_i}H_{p_i}-f_{p_i}H_{q_i})=\{H,f\}_{\rm paper}$. With the other common <Poisson bracket> convention the correct statement is instead $\dot f=\{f,H\}_{\rm standard}$. For example $H=p^2/(2m)$, $f=q$ gives $\{H,q\}_{\rm standard}=-p/m$, contradicting the printed order unless its sign convention is adopted. Since $H_t=0$ and $\{H,H\}=0$, the <Hamiltonian> is conserved. Also $\dot p_\phi=-H_\phi=0$: this <canonical momentum> is the axial <angular momentum>.
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