Physics - Rotation Question with Solution | TestHub

Moment of inertia of ring A about its diameter is $I_A=\frac{1}{2}M{R}^2$.

Moment of inertia of ring B about a tangential axis in its own plane is obtained using the parallel axis theorem:
$I_B=I_A+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$.

Rotational kinetic energy is given by $K=\frac{1}{2}I{\omega }^2$.

Given that both rings have the same rotational kinetic energy, $K_A=K_B$:
$\frac{1}{2}{I}_A{\omega }_A^2=\frac{1}{2}{I}_B{\omega }_B^2$

$\frac{{\omega }_B^2}{{\omega }_A^2}=\frac{I_A}{I_B}$

$\frac{{\omega }_B}{{\omega }_A}=\sqrt{\frac{\frac{1}{2}M{R}^2}{\frac{3}{2}M{R}^2}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$

The ratio of the angular velocity of ring B to that of ring A is $1:\sqrt{3}$.

Answer: $1:\sqrt{3}$