A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad $s^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad $s^{-1})$ of the system is

Step 1: Identify the conservation law.

The net external torque about the vertical axis of rotation is zero because all external forces act vertically.

$$\tau_{\text{ext}} = 0$$ $$L_i = L_f$$

Step 2: Calculate the moment of inertia of the disc.

The moment of inertia of the disc about its central vertical axis is determined using its mass and radius.

$$I_{\text{disc}} = \frac{1}{2}MR^2$$ $$I_{\text{disc}} = \frac{1}{2}(50)(0.4)^2$$ $$I_{\text{disc}} = 4\ \text{kg m}^2$$

Step 3: Calculate the initial angular momentum.

The initial angular momentum is the product of the disc's moment of inertia and its initial angular velocity.

$$L_i = I_{\text{disc}}\omega_i$$ $$L_i = 4 \times 10$$ $$L_i = 40\ \text{kg m}^2\text{s}^{-1}$$

Step 4: Calculate the moment of inertia of each ring.

Since the rings touch along the axis of the disc, the distance from the center of each ring to the axis of rotation is $d = r = 0.2\ \text{m}$.

$$I_{\text{ring}} = mr^2 + md^2$$ $$I_{\text{ring}} = 2mr^2$$ $$I_{\text{ring}} = 2(6.25)(0.2)^2$$ $$I_{\text{ring}} = 0.5\ \text{kg m}^2$$

Step 5: Calculate the final moment of inertia of the system.

The final moment of inertia is the sum of the moments of inertia of the disc and the two rings.

$$I_f = I_{\text{disc}} + 2I_{\text{ring}}$$ $$I_f = 4 + 2(0.5)$$ $$I_f = 5\ \text{kg m}^2$$

Step 6: Determine the final angular velocity.

Equate the initial and final angular momenta of the system.

$$I_{\text{disc}}\omega_i = I_f\omega_f$$ $$4 \times 10 = 5\omega_f$$ $$\omega_f = 8\ \text{rad s}^{-1}$$

Answer: $8\ \text{rad s}^{-1}$