Animated Solution for Physics - Rotational Motion: 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 other along $AC$ as shown. Neglecting frictional effects, the quantities that are conserved as beads slides down are

Imagine a beautifully symmetric setup: an equilateral triangle $ABC$ made of a uniform wire, spinning gracefully about its vertical axis $AO$. At the very peak, point $A$, sit two identical tiny beads. Suddenly, they are released, and they begin their descent, sliding down the arms $AB$ and $AC$.

This problem is a classic test of identifying which physical quantities remain invariant when a system evolves dynamically. To crack this, we need to put on our detective hats and hunt for torques and non-conservative forces.

The first major principle we must investigate is the conservation of angular momentum. We know that the angular momentum of a system about an axis is conserved if, and only if, the net external torque about that axis is zero.

Let's analyze the forces acting on our system (the triangle plus the two beads). The primary external force is gravity, which acts vertically downwards on the beads and the wire. However, our axis of rotation $AO$ is also perfectly vertical! Since the force of gravity is parallel to the axis of rotation, the perpendicular distance from the line of action of the force to the axis is zero. Consequently, gravity produces zero torque about the axis $AO$.

What about the normal forces between the beads and the wire? These are internal forces within our defined system. By Newton's Third Law, they cancel each other out and cannot produce a net external torque. Therefore, with ${\tau }_{\text{ext}}=0$, the total angular momentum ($L$) of the system is strictly conserved.

Next, we turn our attention to the mechanical energy of the system. The problem explicitly instructs us to "neglect frictional effects." This is a massive clue! Friction is the most common non-conservative force that dissipates mechanical energy into heat.

Without friction, the only forces doing work on the beads are conservative forces—specifically, gravity. The normal force does no net work on the system as a whole that would dissipate energy. Because only conservative forces are at play, the total mechanical energy (the sum of kinetic and potential energy) remains perfectly conserved as the beads slide down.

Now, let's address the remaining options involving angular velocity and moment of inertia. As the beads slide down the arms $AB$ and $AC$, they move further away from the central vertical axis $AO$.

The moment of inertia ($I$) of a particle is given by $m{r}^2$, where $r$ is the perpendicular distance from the axis. Since $r$ is increasing for both beads, the total moment of inertia of the system is increasing.

We already established that the total angular momentum $L=I\omega$ is conserved. If $I$ goes up, the angular velocity $\omega$ must correspondingly go down to keep the product constant. Therefore, neither the angular velocity nor the moment of inertia is conserved; they are in a continuous, inverse dance.

In conclusion, the only quantities that remain steadfast and conserved are the total angular momentum and the total energy.