1. Interpretation of the motion

The ring is twirled while the finger remains in contact with the inner rim. In the frame rotating with the finger, the point of contact is instantaneously at rest and the ring is in steady position relative to that frame.

The contact point moves in a horizontal circle of radius $r$, while the center of the ring moves in a circle of radius $R-r$ about the axis of the cone.

Hence the center of mass of the ring has centripetal acceleration

$a_c = \omega_0^2 (R-r).$

So the required horizontal centripetal force on the ring is

$F_c = M\omega_0^2 (R-r).$

2. Forces acting on the ring

At the contact between finger and ring, two contact forces act:

• Normal reaction $N$ along the radius of the ring through the contact point.
• Friction $f$ tangential to the ring at contact.

Also, weight $Mg$ acts downward.

Since the ring is not sliding down, friction must support the weight. At the limiting condition for minimum angular speed,

$f = Mg, \qquad f_{\max} = \mu N.$

Thus at the threshold,

Mg = \mu N. \tag{1}

3. Relation between normal force and centripetal force

The only horizontal force available to provide the centripetal force of the center of mass is the normal reaction $N$.

Therefore,

N = M\omega_0^2 (R-r). \tag{2}

4. Use limiting friction condition

Substitute (2) into (1):

$Mg = \mu \big(M\omega_0^2 (R-r)\big).$

Cancel $M$:

$g = \mu \omega_0^2 (R-r).$

So,

$\omega_0^2 = \frac{g}{\mu (R-r)}.$

Hence the minimum angular velocity is

$\boxed{\omega_{0,\min} = \sqrt{\frac{g}{\mu (R-r)}}}.$

5. Match with options

This corresponds to:

$\boxed{\text{Option C}}$

6. Comparison with stored correct answer

Stored correct answer: $C$

Our derived answer: $C$

They agree.