Rolling Without Slipping of Annular Disk Under Impulse

To determine which statements are correct, we analyze the motion of the annular disk under the given impulse $J_0 \hat{x}$.

1. Calculation of $h_m$ for Pure Rolling Immediately After the Impulse

The moment of inertia $I$ of an annular disk of mass $M$, inner radius $a$, and outer radius $b$ about its central transverse axis through its center of mass is: $I = \frac{1}{2} M (a^2 + b^2)$

When an impulse $J_0 \hat{x}$ is applied at a height $h = h_m$ above the center of mass, the disk begins pure rolling immediately without any impulsive friction force from the ground.

Applying the linear impulse-momentum principle: $J_0 = M v_0 \implies v_0 = \frac{J_0}{M}$ where $v_0$ is the linear velocity of the center of mass immediately after the impulse.

Applying the angular impulse-momentum principle about the center of mass: $J_0 h_m = I \omega_0 \implies \omega_0 = \frac{J_0 h_m}{I}$ where $\omega_0$ is the initial angular velocity about the center of mass.

For pure rolling along the $x$-axis without slipping, the velocity of the lowest point of contact with the ground must be zero: $v_0 = \omega_0 b$

Substituting $v_0$ and $\omega_0$ into the rolling condition: $\frac{J_0}{M} = \left( \frac{J_0 h_m}{I} \right) b$

Solving for $h_m$: $h_m = \frac{I}{M b}$

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

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2. Evaluation of Statements

• Option A: For $\mu \neq 0$ and as $a \rightarrow 0$: $h_m = \lim_{a \rightarrow 0} \frac{a^2 + b^2}{2b} = \frac{b}{2}$ Therefore, Option A is correct.

• Option B: For $\mu \neq 0$ and as $a \rightarrow b$: $h_m = \lim_{a \rightarrow b} \frac{a^2 + b^2}{2b} = \frac{2b^2}{2b} = b$ Therefore, Option B is correct.

• Option C: When $h = h_m$, the initial angular velocity $\omega_0$ is: $\omega_0 = \frac{v_0}{b} = \frac{J_0}{M b}$ This expression depends only on $J_0$, $M$, and $b$, and is independent of the inner radius $a$. Therefore, Option C is correct.

• Option D: For $\mu = 0$ and $h = 0$:

  • Linear velocity imparted: $v_0 = \frac{J_0}{M} > 0$
  • Angular velocity imparted: $\omega_0 = \frac{J_0 \times 0}{I} = 0$ Since the surface is frictionless ($\mu = 0$), no horizontal force or torque acts on the disk after the impulse. Thus: $v(t) = v_0 \quad \text{and} \quad \omega(t) = 0 \quad \text{for all } t \ge 0$ Since $v(t) \neq \omega(t) b$, the bottom-most point of the disk always slips relative to the ground. Thus, the wheel always slides without rolling. Therefore, Option D is correct.

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Conclusion

Statements (A), (B), (C), and (D) are all correct.