= Solution
Write $\omega$ for the imposed constant <angular velocity>. In inertial polar coordinates, the bead has $v^2=\dot r^2+\omega^2r^2$. Its constrained <Lagrangian> is
$$
L=\frac m2(\dot r^2+\omega^2r^2),\qquad
\ddot r=\omega^2r.
$$
The reduced <Lagrangian> has no explicit time dependence, so its <Lagrangian energy function> is conserved:
$$
\boxed{J=\dot r\frac{\partial L}{\partial\dot r}-L
=\frac m2(\dot r^2-\omega^2r^2).}
$$
This is the <Jacobi integral> $J=E-\omega L_z$, since the bead's actual inertial <energy> and axial <angular momentum> are
$$
E=\frac m2(\dot r^2+\omega^2r^2),\qquad L_z=m\omega r^2.
$$
The motor maintaining the stick's <angular velocity> supplies <torque> and does <work>. Indeed $\dot E=2m\omega^2r\dot r=\omega\dot L_z$, generally nonzero. A time-independent reduced <Lagrangian> therefore need not imply conservation of the bead's inertial <kinetic energy> when its imposed constraint is moving.
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