Mechanics |15-10-2020|

We note that the hoop is most likely to bound when body $A$ is in the highest position. Let the velocity of the center of the hoop be $v$ when body $A$ is at the top position. Using the center of the hoop as reference, the translational velocity of body $A$ when it is at the bottom is $0$ , while that when it is at the top position is $2v$ . The moment of inertia of body $A$ is $I=mR^2$ and its angular velocity $\omega = \frac vR$ . By the conversation of energy on the two position, we have:

$\begin{aligned} \underbrace{0+\frac 12 mR^2\left(\frac {v_0}R\right)^2 - mgR}_{\text{body }A} + \underbrace{\frac 12 mv_0^2}_{\text{hoop}} & = \underbrace{\frac 12m(2v)^2+\frac 12 mR^2\left(\frac vR\right)^2 + mgR}_{\text{body }A} + \underbrace{\frac 12 mv^2}_{\text{hoop}} \\ v_0^2 - gR & = 3v^2 + gR \\ \implies v^2 & = \frac {v_0^2 - 2gR}3 \end{aligned}$

Consider the forces acting on the hoop and $A$ as a combined body, when $A$ is on top, we have $2mg = m\omega^2 R + N = \dfrac {mv^2}R + N$ , where $N$ is the normal reaction force. For the combined body not to bounce, $N \ge 0$ or

$\begin{aligned} 2mg & \ge \frac {mv^2}R \\ v^2 & \le 2gR \\ \frac {v_0^2 - 2gR}3 & \le 2gR \\ v_0^2 & \le 8gR \\ \implies v_0 & \le \sqrt{8gR} \end{aligned}$

Therefore $\alpha = \sqrt 8 \approx \boxed{2.83}$ .