Step-by-Step Solution

The total kinetic energy of the ring is the sum of its translational kinetic energy (due to the motion of its center of mass) and its rotational kinetic energy (due to its rotation about the center of mass).

$KE_{total} = KE_{trans} + KE_{rot}$

1. Analyze the Geometry and Kinematics

• The problem states that the finger rotates with an angular velocity ω₀ and traces a circular path of radius r. This means the line from the center of revolution O to the contact point P rotates with angular velocity ω₀.
• Figure 2 provides a top-down view of the motion. It labels the radius of the path of the center of the ring, C, as $R_C = R-r$. We will proceed with this information, as it resolves ambiguities in the problem's text.
• The fact that the center C and the contact point P both revolve around O with the same angular velocity ω₀ implies that O, C, and P are collinear.
• With OC = R-r and OP = r, for these points to be collinear, O must be between C and P. Let's check the distance CP: CP = CO + OP = (R-r) + r = R. This is consistent with P being a point on the rim of the ring of radius R centered at C.

2. Translational Kinetic Energy ($KE_{trans}$)

• The center of mass (CM) of the ring is at its center, C.
• The CM moves in a circle of radius $R_C = R-r$ with a constant angular velocity ω₀.
• The linear velocity of the CM, $v_C$, is given by: $v_C = R_C \omega_0 = (R-r)\omega_0$
• The translational kinetic energy is: $KE_{trans} = {1 \over 2} M v_C^2 = {1 \over 2} M ((R-r)\omega_0)^2 = {1 \over 2} M \omega_0^2 (R-r)^2$

3. Rotational Kinetic Energy ($KE_{rot}$)

• The rotational kinetic energy about the center of mass C is given by: $KE_{rot} = {1 \over 2} I_C \omega^2$ where $I_C$ is the moment of inertia of the ring about its center, and ω is the absolute angular velocity of the ring.
• For a ring of mass M and radius R, the moment of inertia about an axis perpendicular to its plane passing through its center is $I_C = MR^2$.
• To find ω, we use the condition that the ring rolls without slipping at the contact point P. This means the velocity of the point P on the ring must be equal to the velocity of the finger.
• Let's use a coordinate system where the center of revolution O is the origin. At a particular instant, let C be at (R-r, 0) and P at (-r, 0). The revolution is counter-clockwise with angular velocity $\vec{\omega}_0 = \omega_0 \hat{k}$.
• Velocity of the finger at P: $\vec{v}_P = \vec{\omega}_0 \times \vec{r}_{OP} = (\omega_0 \hat{k}) \times (-r \hat{i}) = -r \omega_0 \hat{j}$
• Velocity of the center of mass C: $\vec{v}_C = \vec{\omega}_0 \times \vec{r}_{OC} = (\omega_0 \hat{k}) \times ((R-r) \hat{i}) = (R-r) \omega_0 \hat{j}$
• The velocity of point P on the ring is given by $\vec{v}_{P,ring} = \vec{v}_C + \vec{\omega} \times \vec{r}_{CP}$, where $\vec{\omega} = \omega \hat{k}$ is the absolute angular velocity of the ring.
• The position vector from C to P is $\vec{r}_{CP} = \vec{r}_{OP} - \vec{r}_{OC} = (-r \hat{i}) - ((R-r) \hat{i}) = -R \hat{i}$.
• So, $\vec{v}_{P,ring} = (R-r) \omega_0 \hat{j} + (\omega \hat{k}) \times (-R \hat{i}) = (R-r) \omega_0 \hat{j} - R \omega \hat{j}$.
• Applying the no-slip condition, $\vec{v}_{P,ring} = \vec{v}_P$: $((R-r) \omega_0 - R \omega) \hat{j} = -r \omega_0 \hat{j}$ $(R-r) \omega_0 - R \omega = -r \omega_0$ $R \omega_0 - r \omega_0 - R \omega = -r \omega_0$ $R \omega_0 = R \omega$ $\omega = \omega_0$
• The absolute angular velocity of the ring is ω₀.
• Now we can calculate the rotational kinetic energy: $KE_{rot} = {1 \over 2} I_C \omega^2 = {1 \over 2} (MR^2) \omega_0^2$

4. Total Kinetic Energy ($KE_{total}$)

• Summing the translational and rotational kinetic energies: $KE_{total} = KE_{trans} + KE_{rot}$ $KE_{total} = {1 \over 2} M \omega_0^2 (R-r)^2 + {1 \over 2} M R^2 \omega_0^2$ $KE_{total} = {1 \over 2} M \omega_0^2 [ (R-r)^2 + R^2 ]$

This matches option D.

Conclusion

The total kinetic energy is the sum of the translational kinetic energy of the center of mass and the rotational kinetic energy about the center of mass. By analyzing the kinematics based on the given diagram and applying the no-slip condition, we find the total kinetic energy to be ${1 \over 2}M\omega _0^2[{(R - r)^2} + {R^2}]$.