Physics - Rotation Question with Solution | TestHub

Let $I$ be the moment of inertia of one disk, so $I=\frac{1}{2}M{R}^2$. Initially, the angular momentum is $L_i=I\omega$. The initial kinetic energy is $K_i=\frac{1}{2}I{\omega }^2$.

When the second disk is dropped, the total moment of inertia becomes $I_{total}=I+I=2I$. By conservation of angular momentum, $L_i=L_f$, so $I\omega =(2I){\omega }_f$. This gives the final angular speed ${\omega }_f=\frac{\omega }{2}$. The final kinetic energy is $K_f=\frac{1}{2}(2I){\omega }_f^2=I{\left(\frac{\omega }{2}\right)}^2=I\frac{{\omega }^2}{4}=\frac{1}{4}I{\omega }^2$. The loss in kinetic energy is $$\Delta K=K_i-K_f=\frac{1}{2}I{\omega }^2-\frac{1}{4}I{\omega }^2=\frac{1}{4}I{\omega }^2$$.