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:
$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$
A. Angular momentum “knows” nothing about the axis of rotation. Angular momentum is a physical quantity $J=\int (r-r_0)\times v dm$, where $r_0$ is the coordinates of the centre in respect to which you calculate it. It's easy to see that angular momentum changes when you just change the centre, you don't even need to change the velocity of the reference frame.
B. Now, let's speak about formula $J=I\omega$. In case of flat motion:
$$
J = \int (r-r_0)\times v dm=\int (r-r_0)\times \big(v_0+\omega\times(r-r_0)\big) dm =\\
\int (r-r_0)\times v_0 dm + \omega\int (r-r_0)^2 dm = M(r_{CM}-r_0)\times v_0+I\omega.
$$
We can see that $J=I\omega$ works only in two cases:
- When $r_{CM}=r_0$. In other words, we calculate angular momentum with respect to centre of mass.
- When $v_0=0$. In other words, we calculate the angular momentum with respect to fixed point.
When you change the frame of reference, the centre is no longer fixed and it wasn't CoM in the first place, so the formula is not valid.
C. Now let's speak about what is $\omega$? If you are given a particle moving with the speed $v$, what is its angular velocity? Of course, the answer will depend on the centre you choose (same arguments as with angular momentum applies). However, if you have many points coming from a rigid body like wheel, then you have an obvious choice of a centre: the point with zero velocity. Then the velocity of every point can be calculated as $v=\omega\times(r-r_0)$. Now if you change the frame of reference, the old centre will gain the speed and can no longer be considered a natural centre. However, in case of flat motion we can always find such $\Delta r$, so
$$
v'=v+v_{FR}=\omega\times(r-r_0) + v_{FR} = \omega\times(r-(r_0+\Delta r))
$$
so the velocity field can be considered rotational with the same angular velocity around some new centre $r_0+\Delta r$. I leave the proof of this fact to you.
Best Answer
My initial reasoning was wrong, sorry, I can't justify the use of angular momentum conservation in that non-inertial frame. One can write a balance equation for the angular momentum, but after all it should be the same as solution №2.
But ok, let's choose a single inertial frame. Let it be the one moving with the ultimate ball velocity $v_f$. This will lead the ball having some initial angular momentum $L_0$ and recalculated value for added momentum J. Then the conservation of angular momentum will be
$$(\beta+1) r (J + mvr) + L_0 = \frac{7}{5} m r^2 \omega$$
where
$$ \begin{multline} L_0 = - \int_0^{2r} \rho z v \, \pi \left[r^2 - (r-z)^2\right] dz = \\ - v \frac{m}{\frac{4}{3}\pi r^3} \pi \int_0^{2r} z \left[r^2 - (r-z)^2\right] \, dz = -mvr \end{multline} $$
Hence
$$(\beta+1) r J = \frac{7}{5} m r^2 \omega$$
Instead of direct integration one could use the expression for angular momentum in a moving frame, but I don't remember how to do it.