An annular disk of mass M, inner radius a, and outer radius b is placed on a horizontal surface with a coefficient of friction µ, as shown in the figure. At some time, an impulse $J_0\hat{x}$ is applied at a height h above the center of the disk. If h = h_m, then the disk rolls without slipping along the x-axis. Which of the following statement(s) is/are correct?

Correct Options:

1. For μ ≠ 0 and a → 0, $h_m=\frac{b}{2}$
2. For μ ≠ 0 and a → b, h_m = b
3. For h = h_m, the initial angular velocity does not depend on the inner radius a.
4. For μ = 0 and h = 0, the wheel always slides without rolling.

Detailed Step-by-Step Solution:

Let an annular disk of mass $M$, inner radius $a$, and outer radius $b$ be placed on a horizontal surface. An impulse $J_0$ is applied horizontally at a height $h$ above the center of the disk.

1. Moment of Inertia of an Annular Disk:

The moment of inertia of the annular disk about its central axis perpendicular to its plane is given by:

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

2. Linear and Angular Impulse Equations:

Let $v_0$ be the initial linear velocity of the center of mass right after the impulse, and ${\omega }_0$ be the initial angular velocity.

Linear impulse-momentum relation along the x-axis gives:

$J_0=M{v}_0\Rightarrow v_0=\frac{J_0}{M}$

Angular impulse-angular momentum relation about the center of mass gives:

$J_{0}h=I{\omega }_0\Rightarrow {\omega }_0=\frac{J_{0}h}{I}=\frac{2J_{0}h}{M(a^2+b^2)}$

3. Condition for Pure Rolling Immediately After Impulse:

For pure rolling without slipping along the surface immediately after the impulse, the bottom-most point in contact with the ground must have zero velocity. Thus:

$v_0={\omega }_{0}b$

Substituting $v_0$ and ${\omega }_0$:

$\frac{J_0}{M}=\left(\frac{2J_0{h}_m}{M(a^2+b^2)}\right)b$

Solving for $h_m$:

$h_m=\frac{a^2+b^2}{2b}$

4. Evaluating the Given Options:

Option 1: For $\mu \ne 0$ and solid disk limit ($a\to 0$):

$h_m=\frac{0^2+b^2}{2b}=\frac{b}{2}$

Hence, Statement 1 is correct.

Option 2: For $\mu \ne 0$ and thin ring limit ($a\to b$):

$h_m=\frac{b^2+b^2}{2b}=\frac{2b^2}{2b}=b$

Hence, Statement 2 is correct.

Option 3: When $h=h_m$, since pure rolling is established immediately, we have:

${\omega }_0=\frac{v_0}{b}=\frac{J_0}{Mb}$

As seen, ${\omega }_0$ depends only on $J_0$, $M$, and $b$, and does not depend on the inner radius $a$. Hence, Statement 3 is correct.

Option 4: For smooth surface ($\mu =0$) and impulse applied through the center of mass ($h=0$):

The torque about the center of mass is zero ($J_{0}h=0$), so ${\omega }_0=0$. Since there is no friction ($\mu =0$), no torque can ever be exerted on the disk. Thus, it only translates with velocity $v_0$ without rotating ($\omega =0$), meaning the wheel always slides without rolling. Hence, Statement 4 is correct.