The Ring is retarded to 0 rad/s

Since the ring does not translate on the wall, let us consider translational equilibrium.

$\displaystyle \sum F_x = \sum F_y = 0$

Let the frictional force by the wall on the ring be $f$ , Normal reaction be $N$ and Tension is the string be $T$ .

$\displaystyle \sum F_x = N - T\sin \alpha = 0$

$\displaystyle \sum F_y = f + T\cos \alpha - mg = 0$

$f = \mu N$

Now the Rotational motion is to be analysed. The Friction would give angular retardation, say $\beta$ (since $\alpha$ is used) about the centre of the ring.

Writing Torque equation about centre of ring,

$f R = m R^2 \beta$

To find time taken to retard to $0 rad/s$ :

$0 - \omega_0 = -\beta t \Rightarrow t = \dfrac{\omega_0}{\beta}$

Solving these equations, you get

$f = \dfrac{\mu mg \sin \alpha}{\mu \sin \alpha + \cos \alpha}$

$\beta = \dfrac{\mu g \sin \alpha}{(\mu \sin \alpha + \cos \alpha) R}$

$t = \dfrac{ \omega_0 (\mu \sin \alpha + \cos \alpha) R}{\mu g \sin \alpha} = \boxed{2}$

draw Free body Diagram, resolve forces in x and y axis . ten apply dL/dt=rXF. which gives. Iw/t=r f = r (mu)(Tsin(alpha)) and T=mg/(cos(alpha)+(mu)sin(alpha)) using these two, we get, t=wr(1+(mu)tan(alpha))/((mu) tan(alpha) g).