An equilateral triangle $ABC$ formed from a uniform wire has two small identical beads initially located at $A$. The triangle is set rotating about the vertical axis $AO$. Then the beads are released from rest simultaneously and allowed to slide down. One along $AB$ and the other along $AC$ as shown. Neglecting frictional effects, the quantities that are conserved as beads slide down are
(a) angular velocity and total energy (kinetic and potential)
(b) total angular momentum and total energy
(c) angular velocity and moment of inertia about the axis of rotation
(d) total angular momentum and moment of inertia about the axis of rotation.
In this problem, the moment of inertia of system is increasing. Correct option given is (b) and it is argued that no external torque acts. But If moment of inertia is changing, there should be a torque by $$T= \omega \frac{\mathrm dI}{\mathrm dt}.$$ How can total angular momentum be conserved then? And in such a case total (mechanical) energy should also change.
What I am confused with, I can explain more clearly through another problem:
A thin uniform rod, pivoted at $O$, is rotating in the horizontal plane with constant angular speed $\omega$, as shown in the figure. At time $t = 0$, a small insect starts from $O$ and moves with constant speed $v$ with respect to the rod towards the other end. It reaches the end of the rod at $t = T$ and stops. The angular speed of the system remains $\omega$ throughout. The magnitude of the torque ($\lvert\vec\tau\rvert$) on the system about $O$, as a function of time is best represented by which plot?
Here too there is no external torque (considering the insect + rod as a system all forces exerted by the insect are internal). Yet there is a net torque on the system. So how are the two cases different?