Q.A disc of moment of inertia $I_a$ is rotating in a horizontal plane about its symmetry axis with a constant angular speed $\omega$. Another disc, initially at rest, of moment of inertia $I_b$, is dropped coaxially on to the rotating disc. Then both the discs rotate with the same constant angular speed. The loss of kinetic energy due to friction in this process is,

Deriving KE = ½Iω². Consider a rigid body rotating about a fixed axis with angular velocity $\omega$. A representative particle of mass $m_i$ at distance $r_i$ from the axis has tangential speed $v_i=r_i\omega$, and hence kinetic energy $KE_i=\frac12m_iv_i^2=\frac12m_ir_i^2\omega^2$. Summing over every particle making up the rigid body (all sharing the same $\omega$):

recognising $\sum m_ir_i^2=I$, the body's moment of inertia about that axis. This is the direct rotational analogue of the translational kinetic energy formula $KE=\frac12Mv^2$.

Relation to angular momentum. Since $L=I\omega\ \Rightarrow\ \omega=L/I$, substituting into $KE=\frac12I\omega^2$ and simplifying (multiplying numerator and denominator by $I$) gives the equally useful alternative form

This relation is especially convenient whenever it is the angular momentum $L$, rather than the angular velocity $\omega$, that is held fixed or known — e.g. if a system's kinetic energy is halved while its angular momentum $L$ stays the same (governed by no external torque), $L^2$ stays fixed while $I$ must have doubled, so a graph of $KE$ against $I$ for fixed $L$ is a rectangular hyperbola ($KE\propto1/I$): plotting two bodies of the same $L$ but different $I$ on this curve immediately shows which has the larger moment of inertia (the one with the lower kinetic energy). …