Two discs of same moment of inertia rotating about their regular axis passing through centre and per

Question

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc has angular velocities $\omega_{1}$ and $\omega_{2}$. They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is\

Professor PC Thomas Classes · Answer

(a) : Initial angular momentum $=I \omega_{1}+I \omega_{2}$ Let $\omega$ be angular speed of the combined system.\ Final angular momentum $=2 I \omega$\ $	herefore \quad$ According to conservation of angular momentum $$ I \omega_{1}+I \omega_{2}=2 I \omega 	ext { or } \omega=\frac{\omega_{1}+\omega_{2}}{2} $$ Initial rotational kinetic energy, $$ E=\frac{1}{2} I\left(\omega_{1}^{2}+\omega_{2}^{2}ight) $$ Final rotational kinetic energy\ $E_{f}=\frac{1}{2}(2 I) \omega^{2}=\frac{1}{2}(2 I)\left(\frac{\omega_{1}+\omega_{2}}{2}ight)^{2}=\frac{1}{4} I\left(\omega_{1}+\omega_{2}ight)^{2}$\ $	herefore \quad$ Loss of energy $\Delta E=E_{i}-E_{f}$ $$ \begin{aligned} & =\frac{I}{2}\left(\omega_{1}^{2}+\omega_{2}^{2}ight)-\frac{I}{4}\left(\omega_{1}^{2}+\omega_{2}^{2}+2 \omega_{1} \omega_{2}ight) \ & =\frac{I}{4}\left[\omega_{1}^{2}+\omega_{2}^{2}-2 \omega_{1} \omega_{2}ight]=\frac{I}{4}\left(\omega_{1}-\omega_{2}ight)^{2} \end{aligned} $$