= Solution
For particles in an <inertial frame>, total <momentum> and <angular momentum> about a fixed origin are
$$
\boxed{\mathbf P=\sum_i m_i\mathbf v_i,\qquad \mathbf L_O=\sum_i\mathbf r_i\times m_i\mathbf v_i.}
$$
The <momentum> derivative is the total external <force> when internal forces cancel pairwise. <Angular momentum> is conserved when the total external <torque> vanishes and the internal torques sum to zero, as for pairwise equal-and-opposite central forces. Equal-and-opposite internal forces alone need not cancel internal <torque> if their lines of action are different. An isolated system with these internal-<force> properties conserves both quantities; if fields carry <momentum>, the complete system must include them.
For a straight rocket, let its instantaneous remaining <mass> be $M$ and let $dM<0$ be ejected during a <velocity> increase $dv$. The exhaust's inertial <velocity> is $v-v_e$ to first order. <Conservation of momentum> gives
$$
Mv=(M+dM)(v+dv)+(-dM)(v-v_e),\qquad M\,dv=-v_e\,dM.
$$
Integrating from initial <mass> $m+m_f$ to dry <mass> $m$ yields the <rocket equation>
$$
\boxed{v=v_e\log\frac{m+m_f}{m}.}
$$
It is the rocket-plus-exhaust system that is isolated; the rocket alone is a variable-<mass> open system.
For the firework, the original diagram places one tangentially firing point rocket at the end of each radial rod. The <parallel axis theorem> changes each rod's central inertia to $ma^2/12+m(a/2)^2=ma^2/3$. Adding all rods and fueled rockets gives
$$
\boxed{I_i=3\left[\frac{ma^2}{3}+(m+m_f)a^2\right]=(4m+3m_f)a^2.}
$$
When fuel is exhausted, $I_f=4ma^2$.
Let $M_f$ now denote the total remaining fuel in all three rockets, so $I=4ma^2+a^2M_f$. Fuel at the rim co-rotates before ejection. Its backward tangential exhaust <velocity> is $a\omega-v_e$, and its <angular momentum> per unit <mass> about the axle is $a(a\omega-v_e)$. For an infinitesimal ejection,
$$
I\omega=(I+dI)(\omega+d\omega)+(-dM_f)a(a\omega-v_e).
$$
Since $dI=a^2dM_f$, the co-rotation terms cancel and leave $I\,d\omega=-av_e\,dM_f$. The <tangential-exhaust rotor with a fixed dry inertia> therefore obeys
$$
\omega_f=\frac{v_e}{a}\log\frac{I_i}{I_f},\qquad \boxed{\omega_f=\frac{v_e}{a}\log\beta,\quad \beta=\frac{4m+3m_f}{4m}=1+\frac{3m_f}{4m}.}
$$
Neglecting the exhaust's initial tangential motion would give an incorrect angular-<momentum> balance. The rod <mass> remains in the denominator throughout the burn.
After the burn, the opposing constant axle <torque> $G>0$ acts on the fixed inertia $I_f$. Its <work> over the remaining turn equals the initial <rotational kinetic energy>:
$$
G\Delta\theta=\frac12I_f\omega_f^2,\qquad \boxed{\Delta\theta=\frac{2ma^2\omega_f^2}{G}=\frac{2mv_e^2}{G}(\log\beta)^2.}
$$
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