= Solution
During <quasi-static tether contraction>, <momentum conservation> keeps $\mathbf V_{\rm CM}$ constant and <conservation of angular momentum> keeps $L$ constant. To leading order in the slow radial motion, the <kinetic energy> is
$$
E=mV_{\rm CM}^2+\frac{L^2}{2I_{\rm CM}}=mV_{\rm CM}^2+\frac{L^2}{ml^2}.
$$
Its <derivative> at fixed $L$, not fixed $\omega$, is
$$
\boxed{\frac{dE}{dt}=-\frac{2L^2}{ml^3}\dot l>0\quad(\dot l<0).}
$$
To compute the <power> supplied by the <tension>, use the instantaneous unit vector $\mathbf n$ from the lower to the upper ship. Their radial <velocities> relative to the <centre of mass> are $\pm(\dot l/2)\mathbf n$, and the <tension> forces are $\mp T\mathbf n$. Consequently
$$
\dot W=(-T\mathbf n)\cdot\left(\frac{\dot l}{2}\mathbf n\right)+(T\mathbf n)\cdot\left(-\frac{\dot l}{2}\mathbf n\right)=-T\dot l=\boxed{-\frac{2L^2}{ml^3}\dot l=\frac{dE}{dt}.}
$$
The translational contributions cancel between the two ships and the tangential contributions are zero. The contraction mechanism supplies positive <work>, increasing the rotational <kinetic energy>.
The slow-contraction approximation matters: for a prescribed smooth $l(t)$ with a taut tether, the exact <kinetic energy> and <tension> include radial terms,
$$
E_{\rm exact}=mV_{\rm CM}^2+\frac{L^2}{ml^2}+\frac m4\dot l^2,\qquad T_{\rm exact}=\frac{2L^2}{ml^3}-\frac m2\ddot l.
$$
These still satisfy $dE_{\rm exact}/dt=-T_{\rm exact}\dot l$. Thus the paper's displayed formula is the leading result for slow, smooth contraction, or is exact when the radial <kinetic energy> is constant; it is not an identity for every possible shrinking schedule.
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