This problem involves applying the impulse-momentum theorem for both linear and angular motion to an annular disk. The key condition is that for a specific height of impulse application, $h = h_m$, the disk begins to roll without slipping immediately.

Step 1: Equations for Impulse and Momentum

Let the mass of the annular disk be $M$, its inner radius be $a$, and its outer radius be $b$. An impulse $J_0$ is applied horizontally at a height $h$ above the center of mass (CM).

1. Linear Impulse-Momentum Theorem: The impulse $J_0$ causes a change in the linear momentum of the center of mass. Let $v_{cm}$ be the velocity of the CM just after the impulse. $J_0 = M v_{cm} - 0 \implies v_{cm} = \frac{J_0}{M} \quad (1)$

2. Angular Impulse-Momentum Theorem (about the CM): The impulse $J_0$ applied at a distance $h$ from the CM creates an angular impulse. Let $\omega$ be the angular velocity just after the impulse. The moment of inertia of the annular disk about its center is $I_{cm}$. The angular impulse is $J_0 h$. This causes a change in angular momentum. $J_0 h = I_{cm} \omega - 0 \implies \omega = \frac{J_0 h}{I_{cm}} \quad (2)$

Step 2: Condition for Rolling Without Slipping

The condition for the disk to roll without slipping immediately after the impulse is that the velocity of the point of contact with the ground is zero. The velocity of the contact point is the sum of the translational velocity of the CM and the velocity due to rotation. $v_{contact} = v_{cm} - \omega b$ For no slipping, $v_{contact} = 0$, which gives: $v_{cm} = \omega b \quad (3)$ This condition is met when the impulse is applied at the special height $h = h_m$.

Step 3: Determine the height $h_m$

Substitute equations (1) and (2) into equation (3), with $h = h_m$: $\frac{J_0}{M} = \left( \frac{J_0 h_m}{I_{cm}} \right) b$ Simplifying for $h_m$: $h_m = \frac{I_{cm}}{Mb} \quad (4)$

Step 4: Moment of Inertia of an Annular Disk

The moment of inertia of an annular disk about its central axis is given by: $I_{cm} = \frac{1}{2} M (a^2 + b^2) \quad (5)$

Substituting (5) into (4): $h_m = \frac{\frac{1}{2} M (a^2 + b^2)}{Mb} = \frac{a^2 + b^2}{2b}$

Step 5: Evaluate the Given Statements

A: For $\mu \neq 0$ and $a \rightarrow 0, h_m = b / 2$. When $a \rightarrow 0$, the annular disk becomes a solid disk. Let's find the limit of our expression for $h_m$: $\lim_{a \to 0} h_m = \lim_{a \to 0} \frac{a^2 + b^2}{2b} = \frac{0 + b^2}{2b} = \frac{b}{2}$ For a solid disk, $I_{cm} = \frac{1}{2}Mb^2$, so $h_m = \frac{\frac{1}{2}Mb^2}{Mb} = \frac{b}{2}$. The result is consistent. The condition $\mu \neq 0$ ensures that rolling is a possible state of motion. Thus, statement (A) is correct.

B: For $\mu \neq 0$ and $a \rightarrow b, h_m = b$. When $a \rightarrow b$, the annular disk becomes a thin ring or hoop. Let's find the limit of our expression for $h_m$: $\lim_{a \to b} h_m = \lim_{a \to b} \frac{a^2 + b^2}{2b} = \frac{b^2 + b^2}{2b} = \frac{2b^2}{2b} = b$ For a thin ring, $I_{cm} = Mb^2$, so $h_m = \frac{Mb^2}{Mb} = b$. The result is consistent. Thus, statement (B) is correct.

C: For $h=h_m$, the initial angular velocity does not depend on the inner radius $a$. Let's find the expression for the angular velocity $\omega$ when $h=h_m$. From the no-slip condition (3) and the linear momentum equation (1): $v_{cm} = \omega b \implies \frac{J_0}{M} = \omega b$ $\omega = \frac{J_0}{Mb}$ This expression for $\omega$ depends on the impulse $J_0$, the mass $M$, and the outer radius $b$. It does not depend on the inner radius $a$. Thus, statement (C) is correct.

D: For $\mu=0$ and $h=0$, the wheel always slides without rolling. If $\mu=0$, the surface is frictionless. No frictional force can be exerted on the disk. If $h=0$, the impulse $J_0$ is applied at the center of mass.

• Linear motion: The impulse gives the CM a velocity $v_{cm} = J_0/M$.
• Angular motion: Since the impulse is applied at the center ($h=0$), the torque about the CM is zero ($\ au_{cm} = r \times F = 0$). Therefore, the angular impulse is zero, and there is no change in angular velocity. Since the disk starts from rest ($\omega_{initial}=0$), the final angular velocity is also zero ($\omega_{final}=0$).

The disk moves with a translational velocity $v_{cm}$ but does not rotate ($\omega=0$). This is the definition of pure sliding. Thus, statement (D) is correct.

All four statements are correct.