An annular disk of mass M,inner radius a and | vedclass

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(D) For an impulse $J_0$ applied at height $h$ from the center:$1$. Linear impulse equation: $J_0 = Mv \implies v = \frac{J_0}{M}$$2$. Angular impulse

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$A, B, C, D$

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(D) For an impulse $J_0$ applied at height $h$ from the center:$1$. Linear impulse equation: $J_0 = Mv \implies v = \frac{J_0}{M}$$2$. Angular impulse equation about the center of mass: $J_0 h = I_c \omega \implies \omega = \frac{J_0 h}{I_c}$For rolling without slipping,$v = \omega b$,where $b$ is the outer radius.Substituting $v$ and $\omega$: $\frac{J_0}{M} = \frac{J_0 h_m}{I_c} b \implies h_m = \frac{I_c}{Mb}$For an annular disk,$I_c = \frac{1}{2} M(a^2 + b^2)$.Thus,$h_m = \frac{\frac{1}{2} M(a^2 + b^2)}{Mb} = \frac{a^2 + b^2}{2b}$.$(A)$ If $a ightarrow 0$,$h_m = \frac{b^2}{2b} = \frac{b}{2}$. Statement $(A)$ is correct.$(B)$ If $a ightarrow b$,$h_m = \frac{b^2 + b^2}{2b} = \frac{2b^2}{2b} = b$. Statement $(B)$ is correct.$(C)$ $\omega = \frac{J_0 h_m}{I_c} = \frac{J_0}{I_c} \cdot \frac{I_c}{Mb} = \frac{J_0}{Mb}$. Since $M$ and $b$ are constants,$\omega$ does not depend on $a$. Statement $(C)$ is correct.$(D)$ If $\mu = 0$ and $h = 0$,the impulse passes through the center of mass,so no torque is produced $(	au = 0)$. Thus,$\omega = 0$ and the disk only translates. It slides without rolling. Statement $(D)$ is correct.

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