Part 1: Initial Acceleration

Let’s analyze the system immediately after bead B is released. We define the origin at the hinge O. Let the radius of the ring be $r$ and the mass of each bead be $m$. The ring itself is light (massless).

1. Kinematic Constraints

Let the ring have an instantaneous angular acceleration $\alpha$ in the counter-clockwise direction.

• Point P (Point on Ring at B): The point on the ring coincident with B accelerates due to rotation about O.
  Position relative to O: $\vec{r} = r\hat{i} – r\hat{j}$.
  Tangential acceleration $\vec{a}_P = \vec{\alpha} \times \vec{r}$. Since $\vec{\alpha} = \alpha \hat{k}$, the acceleration vector is perpendicular to the radius OB.
  Horizontal component of $\vec{a}_P$: Since B is at the horizontal diameter, the motion of the ring point is “up and right”. The horizontal component is $a_{Px} = \alpha r$.
• Bead B: Bead B is free to slide on the ring. The ring pushes it with a normal force $N_1$ which is horizontal at this instant. Friction is negligible.
  Constraint: B must stay on the ring. For standard circular motion constraints, the horizontal acceleration of B must match the horizontal acceleration of the ring at that point. $$a_{Bx} = \alpha r \quad \text{(Directed Right)}$$

2. Dynamics of Bead B

Applying Newton’s Second Law to Bead B:

• Vertical: Only gravity acts vertically. $$m a_{By} = mg \implies a_{By} = g \quad (\text{Down})$$
• Horizontal: The normal force $N_1$ provides the horizontal acceleration. $$N_1 = m a_{Bx} = m(\alpha r)$$

3. Dynamics of the Rigid System (Ring + Bead A)

We consider the Ring and fixed Bead A as a single rigid body rotating about the hinge O.
Torque Equation ($\tau_{net} = I_{sys}\alpha$):

• Moment of Inertia ($I_{sys}$): The ring is massless. Bead A (mass $m$) is at distance $\sqrt{r^2+r^2} = r\sqrt{2}$ from O. $$I_{sys} = m(r\sqrt{2})^2 = 2mr^2$$
• Torques about O (taking CCW as positive):
  • Torque due to A’s weight ($mg$): Lever arm is horizontal distance $r$. Force is down. Creates CCW torque. $\tau_A = +mgr$.
  • Torque due to Reaction Force on Ring from B: By Newton’s 3rd law, if Ring pushes B to the Right with $N_1$, B pushes Ring to the Left with $N_1$.
    Force $N_1$ acts Left at position $(r, -r)$.
    Torque $\vec{\tau} = \vec{r} \times \vec{F}$. The line of action is horizontal below O. This creates a Clockwise torque.
    $\tau_{reaction} = -N_1 r$.

Substituting into the torque equation:

$$mgr – N_1 r = (2mr^2)\alpha$$

Substitute $N_1 = m\alpha r$ from Bead B’s analysis:

$$mgr – (m\alpha r)r = 2mr^2 \alpha$$ $$mg – m\alpha r = 2m\alpha r$$ $$g = 3\alpha r \implies \alpha = \frac{g}{3r}$$

4. Calculating Accelerations

Acceleration of Bead A ($a_A$):

A is fixed to the ring. Its total acceleration is tangential (since $\omega=0$ initially).

$$a_A = \text{distance from O} \times \alpha = (r\sqrt{2}) \cdot \frac{g}{3r} = \frac{\sqrt{2}g}{3}$$

Acceleration of Bead B ($a_B$):

B has components $a_{Bx} = \alpha r = g/3$ and $a_{By} = g$.

$$a_B = \sqrt{(a_{Bx})^2 + (a_{By})^2} = \sqrt{\left(\frac{g}{3}\right)^2 + g^2} = \sqrt{\frac{g^2}{9} + g^2} = g\sqrt{\frac{10}{9}} = \frac{\sqrt{10}g}{3}$$

Part 2: Final Angle of Rotation

After a long time, the viscous friction dissipates the kinetic energy associated with the relative motion of bead B. The system reaches a stable static equilibrium. In this state:

• Bead B settles at the lowest possible point in space on the circular track provided by the ring. This point is vertically below the ring’s center.
• The center of mass (CM) of the entire system (A + B) must lie on the vertical line passing through the hinge O.

Let the ring rotate by an angle $\phi$ counter-clockwise.

Coordinates of Key Points (relative to O):

• Ring Center C: Originally at $(0, -r)$. After rotation $\phi$, it moves to: $$x_C = r \sin \phi$$
• Bead A (Fixed): Originally at $(-r, -r)$.
  Relative to center C, A is at $(-r, 0)$. In the rotated frame, vector $\vec{CA}$ makes an angle $\phi$ with the horizontal.
  Horizontal position of A: $$x_A = x_C – r \cos \phi = r \sin \phi – r \cos \phi$$
• Bead B (Free): B settles at the lowest point of the ring.
  Geometrically, this is directly below the center C. $$x_B = x_C = r \sin \phi$$

Equilibrium Condition:

For the system to be in equilibrium, the net torque about the hinge O due to gravity must be zero. Equivalently, the horizontal center of mass must be at $x=0$.

$$m x_A + m x_B = 0$$ $$x_A + x_B = 0$$ $$(r \sin \phi – r \cos \phi) + (r \sin \phi) = 0$$ $$2 \sin \phi – \cos \phi = 0$$ $$2 \sin \phi = \cos \phi$$ $$\tan \phi = \frac{1}{2}$$

Thus, the final angle of rotation is:

$$ \phi = \tan^{-1}\left(\frac{1}{2}\right) $$