Kinetic Energy in Rotation

Consider a rigid body rotating with angular velocity $\omega$ about a fixed axis. Every particle of the body shares this same angular velocity $\omega$, but has its own distinct tangential (linear) velocity, depending on its own distance from the axis. Take a representative particle of mass $m_i$ at distance $r_i$ from the axis; its tangential velocity is $v_i=r_i\omega$, so its own kinetic energy is

Summing over every particle making up the whole rigid body (all sharing the same $\omega$):

Recognising $\sum_i m_ir_i^2$ as the moment of inertia $I$ of the whole body about the axis, this gives the compact result

analogous to the translational kinetic energy $KE=\frac12Mv^2$.

Relation between rotational kinetic energy and angular momentum. Since $L=I\omega$, multiplying both numerator and denominator of $KE=\frac12I\omega^2$ by $I$ gives

This alternative form,

$\boxed{KE=\frac{L^2}{2I}},$ …