Final Angular Velocity of Rotating Disc with Two Added Rings

To find the new angular velocity of the system, we apply the principle of conservation of angular momentum about the vertical axis of rotation, as no external torque acts on the system.

1. Initial State of the System:

The system initially consists only of the rotating uniform circular disc.

• Mass of the disc, $M = 50\text{ kg}$
• Radius of the disc, $R = 0.4\text{ m}$
• Initial angular speed, $\omega_0 = 10\text{ rad s}^{-1}$

The initial moment of inertia of the disc about its vertical central axis is: $I_i = \frac{1}{2} M R^2$

Substituting the given values: $I_i = \frac{1}{2} \times 50 \times (0.4)^2 = 25 \times 0.16 = 4.0\text{ kg m}^2$

The initial angular momentum is: $L_i = I_i \omega_0 = 4.0 \times 10 = 40\text{ kg m}^2\text{s}^{-1}$

---

2. Final State of the System:

Two identical uniform circular rings are placed symmetrically on the disc:

• Mass of each ring, $m = 6.25\text{ kg}$
• Radius of each ring, $r = 0.2\text{ m}$

Since the rings are placed symmetrically such that they touch each other at the axis of rotation of the disc, the center of mass of each ring is located at a distance $d = r = 0.2\text{ m}$ from the axis of rotation.

The moment of inertia of a ring about an axis passing through its center of mass and perpendicular to its plane is: $I_{\text{cm}} = m r^2$

Using the parallel axis theorem, the moment of inertia of one ring about the axis of rotation of the disc is: $I_{\text{ring}} = I_{\text{cm}} + m d^2 = m r^2 + m r^2 = 2 m r^2$

Since there are two identical rings, the total moment of inertia contributed by both rings is: $I_{\text{rings}} = 2 \times I_{\text{ring}} = 4 m r^2$

Substituting the given values: $I_{\text{rings}} = 4 \times 6.25 \times (0.2)^2 = 25 \times 0.04 = 1.0\text{ kg m}^2$

The total moment of inertia of the composite system after the rings are placed is: $I_f = I_i + I_{\text{rings}} = 4.0 + 1.0 = 5.0\text{ kg m}^2$

---

3. Conservation of Angular Momentum:

Since no external torque acts on the system about the vertical axis: $L_i = L_f$ $I_i \omega_0 = I_f \omega$

Solving for the final angular velocity $\omega$: $\omega = \frac{I_i \omega_0}{I_f} = \frac{40}{5.0} = 8\text{ rad s}^{-1}$