= Solution
At time $t$, the remaining water has <angular momentum>
$$
L=\frac12mR^2\omega.
$$
During $dt$, the ejected mass is $-dm$, and its signed tangential speed in the inertial frame is $R\omega+u$. Conservation of <angular momentum> gives, to first order,
$$
0=d\left(\frac12mR^2\omega\right)
+(-dm)R(R\omega+u).
$$
After simplification, the <angular rocket equation> is
$$
\boxed{m\dot\omega
=\dot m\left(\omega+\frac{2u}{R}\right)}.
$$
Here $u$ is signed relative to the direction of rotation; reversing the spray changes its sign.
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