 Tangential-exhaust rotor with a fixed dry inertia

Let fuel of remaining mass $M_f$ be carried at radius $a$ on a rotor whose dry moment of inertia is $I_d$. Then $I=I_d+a^2{M}_f$. Fuel leaves tangentially backwards with speed $v_e$ relative to the rim, so its inertial tangential velocity is $a\omega -v_e$. Conserving angular momentum of rotor plus newly emitted fuel during $d{M}_f<0$ givesSince $dI=a^{2}d{M}_f$, integration yields ${\omega }_f-{\omega }_i=(v_e/a)\log (I_i/I_f)$. Including the fuel's co-rotation is essential: the rotor alone is an open system and its angular momentum is not conserved. For three radial rods of mass $m$ and length $a$, each carrying a point rocket of dry mass $m$ and fuel mass $m_f$, $I_i=(4m+3m_f)a^2$ and $I_f=4m{a}^2$.