Rotational velocity of the ring

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1. Aug 13, 2017

Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Wrt inertial frame with origin at the pivot,

The final angular velocity of the ring, the bug are $\vec ω_r$ and $\vec ω_b$ and the final velocity of the bug is $\vec v_b$.
Since, there is no net external torque and force is acting on the system , conservation of angular momentum and energy could be applied.
Conservation of angular momentum about the pivot gives,
0 = I_{piv} $\vec ω _r + 2 \vec R \times m \vec v_b$
assuming that $\vec v_b$ is perpendicular to $\vec R$,
This gives $v_b = \frac M m ω_r R$
Conservation of energy gives,
$\frac 1 2 m v^2 = \frac 1 2 I_{piv} {ω_r}^2 + \frac 1 2 m {v_b}^2$
Substituting the value of $v_b$ in the above eqn. gives
$ω_r = \frac {mv} {R\sqrt { M ( 2m + m) }}$
Is this correct so far?
Is
$\vec v_b = \vec ω_b \times \vec R$ ?

2. Aug 13, 2017

haruspex

Not sure what you mean by that. Do you mean the velocity of the bug in the lab frame when the bug is halfway round?
Yes, but only if you are taking moments about the pivot.
The bug does work. You cannot use conservation of energy.

3. Aug 14, 2017

Pushoam

I got it: The mechanical energy of the system due to bug's internal interactions gets changed. So, conservation of mechanical energy cannot be applied here.