Q.A rigid body rotates with an angular momentum $L$. If its kinetic energy is halved, the angular momentum becomes,

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). …