= Radius change at constant angular momentum
{title2=$I\omega=L_z,\quad K_{\mathrm{rot}}=L_z^2/(2I)$}
Place two equal <point masses> of $1\,\mathrm{kg}$ each on opposite sides of a fixed origin, at equal <radius> $r$, and let them share a signed <angular velocity> $\omega=\dot\theta$ in a plane. Their total <mass> is $M=2\,\mathrm{kg}$ and their <moment of inertia> about the perpendicular axis is $I=Mr^2$. Radial pulling <forces> have zero <torque> about that origin, so
$$
L_z=I\omega=Mr^2\omega=\mathrm{constant},\qquad
K_{\mathrm{rot}}=\frac12I\omega^2=\frac{L_z^2}{2I}.
$$
Take an initial <radius> $r_i=1\,\mathrm m$ and <angular velocity> $\omega_i=1\,\mathrm{rad/s}$, and a final <radius> $r_f=0.5\,\mathrm m$. Then
$$
I_i=2\,\mathrm{kg\,m^2},\quad I_f=0.5\,\mathrm{kg\,m^2},\quad
L_z=2\,\mathrm{kg\,m^2/s},\quad \omega_f=4\,\mathrm{rad/s},
$$
and
$$
K_i=1\,\mathrm J,\qquad K_f=4\,\mathrm J,\qquad W=K_f-K_i=+3\,\mathrm J.
$$
These are endpoint states with $\dot r_i=\dot r_f=0$. During the pull, the full <kinetic energy> also contains radial motion: $K=M\dot r^2/2+K_{\mathrm{rot}}$. Assume the pulling apparatus supplies all net <work> to the two <point masses>; the <work-energy theorem> then fixes its net <work> at $+3\,\mathrm J$, including any radial acceleration and braking. Each <mass> spirals inward during the pull, allowing its inward <force> to do positive <work>. <Conservation of angular momentum> therefore predicts the final <angular velocity> without requiring <kinetic energy> conservation. For an apparatus included in a larger system, its stored <energy> supplies the increase consistently with <conservation of energy>. OpenStax uses the same mechanism to explain a skater pulling her arms inward. https://openstax.org/books/university-physics-volume-1/pages/11-3-conservation-of-angular-momentum[OpenStax, section 11.3: work during a change of moment of inertia]{ref}
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