= Tangential-exhaust rotor with a fixed dry inertia
{title2=$\Delta\omega=(v_e/a)\log(I_i/I_f)$}
Let fuel of remaining mass $M_f$ be carried at radius $a$ on a rotor whose dry <moment of inertia> is $I_d$. Then $I=I_d+a^2M_f$. 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 $dM_f<0$ gives
$$
I\,d\omega+\omega\,dI-dM_f\,a(a\omega-v_e)=0,\qquad I\,d\omega=-av_e\,dM_f.
$$
Since $dI=a^2dM_f$, 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 $I_f=4ma^2$.
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