Conservation of angular momentum during rolling

The question implies that when the disc is initially put down it is not rolling without slipping yet. In fact, at $t=0$ it's not yet rolling at all.

Simple conservation of momentum doesn't apply here because friction energy is not conserved.

The disc exerts a force $mg$ (its weight) on the surface, which in turn exerts an equal but opposite Normal force $F_N$, which prevents the ring from penetrating the surface:

$$F_N=mg$$

In the simple friction model a friction force $F_f$ is exerted in the opposite sense of motion, acc.:

$$F_f=\mu F_N=\mu mg,$$

where $\mu$ is the friction coefficient.

We can now set up two equations of motion:

1. Translation:

The force $F_f$ causes acceleration:

$$ma=\mu mg$$

So that:

$$v(t)=\mu gt$$

2. Rotation:

The force $F_f$ also causes a torque $\tau$:

$$\tau=I\alpha,$$

where $\tau=-R \mu mg$, $I=\frac{mR^2}{2}$ and $\alpha=\frac{d\omega}{dt}$, so:

$$-R \mu mg=\frac{mR^2}{2} \frac{d\omega}{dt}$$

$$-2\mu g=R\frac{d\omega}{dt}$$

$$\omega(t)=\omega-\frac{2\mu g}{R}t$$

---

Rolling without slipping occurs when $v(t)=\omega(t) R$, so with the expressions above:

$$\mu gt=\omega R- 2\mu gt$$

$$t=\frac{\omega R}{3\mu g}$$

Inserting this into $\omega(t)$ we get:

$$\large{\omega(t)=\frac{\omega}{3}}$$

The ring will lose half of its initial rotational momentum before rolling without slipping is achieved.

The initial kinetic energy was:

$$K_0=\frac{mR^2\omega^2}{4}$$

The final kinetic energy $K(t)$, including translational energy is:

$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$