1. Use conservation of angular momentum

Since the rings are gently placed and there is no external torque about the vertical axis, angular momentum about the axis is conserved:

$I_i\omega_i = I_f\omega_f$

We need the initial and final moments of inertia about the original vertical axis.

2. Initial moment of inertia of the disc

For a uniform disc about its own central vertical axis,

$I_{\text{disc}} = \frac{1}{2}MR^2$

Given:

• $M = 50\,\text{kg}$
• $R = 0.4\,\text{m}$

So,

$I_{\text{disc}} = \frac{1}{2}(50)(0.4)^2 = 25 \times 0.16 = 4\,\text{kg m}^2$

Thus,

$I_i = 4\,\text{kg m}^2$

and

$\omega_i = 10\,\text{rad s}^{-1}$

Hence initial angular momentum:

$L_i = I_i\omega_i = 4 \times 10 = 40\,\text{kg m}^2\text{/s}$

3. Position of each ring on the disc

Each ring has radius $0.2\,\text{m}$. They touch each other along the axis of the disc, so the center of each ring is at a distance

$d = 0.2\,\text{m}$

from the disc axis.

4. Moment of inertia of one ring about the disc axis

For a thin circular ring of mass $m$ and radius $r$, about its own central axis perpendicular to its plane,

$I_{\text{cm}} = mr^2$

Here,

• $m = 6.25\,\text{kg}$
• $r = 0.2\,\text{m}$

So,

$I_{\text{cm}} = 6.25(0.2)^2 = 6.25 \times 0.04 = 0.25\,\text{kg m}^2$

Now use the parallel axis theorem to shift to the disc axis:

$I_{\text{one ring}} = I_{\text{cm}} + md^2$

Since $d = 0.2\,\text{m}$,

$I_{\text{one ring}} = 0.25 + 6.25(0.2)^2 = 0.25 + 6.25\times 0.04 = 0.25 + 0.25 = 0.5\,\text{kg m}^2$

For two rings,

$I_{\text{rings}} = 2 \times 0.5 = 1.0\,\text{kg m}^2$

5. Final moment of inertia

$I_f = I_{\text{disc}} + I_{\text{rings}} = 4 + 1 = 5\,\text{kg m}^2$

6. Find the new angular velocity

By conservation of angular momentum,

$I_i\omega_i = I_f\omega_f$

$4 \times 10 = 5\omega_f$

$\omega_f = \frac{40}{5} = 8\,\text{rad s}^{-1}$

7. Final answer

$\boxed{8}$

The derived answer matches the stored correct answer.