Physics - Rotation Question with Solution | TestHub

Initial moment of inertia is $I$ and initial angular velocity is $\omega$.

Initial kinetic energy is $K_i=\frac{1}{2}I{\omega }^2$.

When another identical ring is gently placed on it, the total moment of inertia becomes $I^{\prime }=I+I=2I$.

By conservation of angular momentum, $I\omega =I^{\prime }{\omega }^{\prime }$.

$I\omega =2I{\omega }^{\prime }\Rightarrow {\omega }^{\prime }=\frac{\omega }{2}$.

Final kinetic energy is $K_f=\frac{1}{2}{I}^{\prime }({\omega }^{\prime }{)}^2=\frac{1}{2}(2I){\left(\frac{\omega }{2}\right)}^2=\frac{1}{4}I{\omega }^2$.

Loss in kinetic energy is $\Delta K=K_i-K_f=\frac{1}{2}I{\omega }^2-\frac{1}{4}I{\omega }^2=\frac{1}{4}I{\omega }^2$.

Answer: $\frac{\mathrm{I}{\omega }^2}{4}$