An annular disk of mass 𝑀, inner radius 𝑎 and outer radius 𝑏 is placed on a horizontal surface with coefficient of friction 𝜇, as shown in the figure. At some time, an impulse ℐ0𝑥̂ is applied at a height ℎ above the center of the disk. If ℎ = ℎ𝑚 then the disk rolls without slipping along the 𝑥-axis. Which of the following statement(s) is(are) correct? (A) For 𝜇 ≠ 0 and 𝑎 → 0, ℎ𝑚 = 𝑏/2. (B) For 𝜇 ≠ 0 and 𝑎 → 𝑏, ℎ𝑚 = 𝑏. (C) For ℎ = ℎ𝑚, the initial angular velocity does not depend on the inner radius 𝑎. (D) For 𝜇 = 0 and ℎ = 0, the wheel always slides without rolling.

 To analyze the problem, let's break it down step by step.

Given:

• Annular disk with mass $M$, inner radius $a$, and outer radius $b$.

• The disk is placed on a horizontal surface with a coefficient of friction $\mu$.

• An impulse ${\mathcal{I}}_0\widehat{x}$ is applied at a height $h$ above the center of the disk.

• If $h=h_m$, the disk rolls without slipping along the $x$-axis.

Key Concepts:

1. Rolling without slipping: This condition requires that the linear velocity $v$ of the center of mass and the angular velocity $\omega$ satisfy $v=\omega R$, where $R$ is the effective radius.

2. Impulse and angular momentum: The impulse ${\mathcal{I}}_0$ imparts both linear momentum ${\mathcal{I}}_0=Mv$ and angular momentum ${\mathcal{I}}_{0}h=I\omega$, where $I$ is the moment of inertia of the disk.

3. Moment of inertia for an annular disk: The moment of inertia about the center is $I=\frac{1}{2}M(a^2+b^2)$.

Analysis:

1. Condition for rolling without slipping:

   • Linear momentum: ${\mathcal{I}}_0=Mv$.

   • Angular momentum: ${\mathcal{I}}_{0}h=I\omega$.

   • Rolling condition: $v=\omega b$ (since the outer radius $b$ is the effective radius for rolling).

   Substituting $v=\omega b$ into the angular momentum equation:

   $${\mathcal{I}}_{0}h=I\omega =I\left(\frac{v}{b}\right)=I\left(\frac{{\mathcal{I}}_0}{Mb}\right)\mathrm{.}$$

   Solving for $h$:

   $$h=\frac{I}{Mb}\mathrm{.}$$

   Substituting $I=\frac{1}{2}M(a^2+b^2)$:

   $$h_m=\frac{\frac{1}{2}M(a^2+b^2)}{Mb}=\frac{a^2+b^2}{2b}\mathrm{.}$$

2. Evaluate the statements:

   • (A) For $\mu \mathrm{\ne }0$ and $a\to 0$, $h_m=\frac{b}{2}$:

     • If $a\to 0$, $h_m=\frac{0+b^2}{2b}=\frac{b}{2}$. This is correct.

   • (B) For $\mu \mathrm{\ne }0$ and $a\to b$, $h_m=b$:

     • If $a\to b$, $h_m=\frac{b^2+b^2}{2b}=b$. This is correct.

   • (C) For $h=h_m$, the initial angular velocity does not depend on the inner radius $a$:

     • The angular velocity $\omega =\frac{v}{b}=\frac{{\mathcal{I}}_0}{Mb}$ does not depend on $a$. This is correct.

   • (D) For $\mu =0$ and $h=0$, the wheel always slides without rolling:

     • If $\mu =0$, there is no friction to provide the torque needed for rolling. If $h=0$, the impulse is applied at the center, so no torque is generated. The disk will slide without rolling. This is correct.

Final Answer:

All statements (A), (B), (C), and (D) are correct.