Problem source

A firework consists of a uniform rod of mass $M$ and length $2a$, pivoted smoothly at one end so that it can rotate in a fixed horizontal plane, and a rocket attached to the other end. The rocket is a uniform rod of mass $m(t)$ and length $2l(t)$, with $m(t)=2\alpha l(t)$ and $\alpha$ constant. It is attached to the rod by its front end and it lies at right angles to the rod in the rod's plane of rotation.
The rocket burns fuel in such a way that $\mathrm{d}m/\mathrm{d}t=-\alpha\beta,$ with $\beta$ constant. The burnt fuel is ejected from the back of the rocket, with speed $u$ and directly backwards relative to the rocket. Show that, until the fuel is exhausted, the firework's angular velocity $\omega$ at time $t$ satisfies 
\[
\frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{3\alpha\beta au}{2[Ma^{2}+2\alpha l(3a^{2}+l^{2})]}.
\] 

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\r{2};
        \def\l{0.75};
        \coordinate (O) at (0,0);
        \coordinate (A) at (0, \r);
        \coordinate (B) at (\l, \r);
        \coordinate (G) at ($(A)!0.5!(B)$);


        \filldraw (O) circle (1pt) node[right] {$O$};
        \draw[dashed] (O) circle (\r);
        \draw (O) -- (A);
        \draw[ultra thick] (A) -- (B);
        \filldraw[red] (G) circle (1pt) node[above] {$G$};

        \draw[dashed] (O) -- (G);
        \node[<->] at ($(O)!0.5!(G)$) [right] {$d$};
        
    \end{tikzpicture}
\end{center}

The rocket principle states that the thrust generated by the rocket is $-\frac{\d m}{\d t}u = \alpha \beta u$

This force is acting at a distance $2a$ from $O$ and therefore is generating a torque of $2a \alpha \beta u$ on the system.

Let's also consider the moments of inertia about $O$.

The fixed rod will have moment of inertia $\frac13 M (2a)^2 = \frac43 M a^2$.

The rocket will have moment of inertia $I_{G} + md^2 = \frac1{12}m(t)(2l(t))^2  + m(t) ((2a)^2 + l(t)^2)= \frac43 ml^2+ 4ma^2$. Since our final equation doesn't involve $m$, lets replace all the $m$ with $2al$ to obtain a total $\displaystyle I = \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2$.

Since $\tau$ is constant, we can note that $I\omega = 2a \alpha \beta u t$ (by integrating) and so

\begin{align*}
&& \dot{\omega} &= \frac{\d }{\d t} \left ( \frac{2a \alpha \beta u t}{ \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2} \right) \\
&&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2Ma^2 +4\alpha l^3 + 4 \cdot 3 \cdot \alpha la^2} \right) \\
&&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2[Ma^2 +2\alpha l(l^2 + 3 a^2)]} \right) \\
\end{align*}

This is, close, but not quite what they are after since the denominator also has a dependency on $t$ we wont get exactly what they've asked for