{title2=$K=\frac12I\omega^2$}
{wiki}

<Rotational kinetic energy> is the <kinetic energy> of a <rigid body>'s rotation. For pure rotation about an axis fixed in an <inertial frame>,
$$
K_{\mathrm{rot}}=\frac12I\omega^2,
\qquad I=\int r_\perp^2\,dm,
\qquad \omega=|\boldsymbol\omega|.
$$
Here $r_\perp$ is the perpendicular distance of a <mass> element $dm$ from that axis, $I$ is the <moment of inertia> about that same axis, and $\omega$ is the common <angular speed>, measured in <radians> per second. The <rigid body> assumption fixes the relative positions of the <mass> elements; they share <angular velocity> but have different <speeds> $v=\omega r_\perp$. Summing their <kinetic energies> $dm\,v^2/2$ gives the formula. This is a nonrelativistic model; deformation or internal motion can contribute additional <kinetic energy>.<University Physics Volume 1 by William Moebs and others/10.4 Moment of Inertia and Rotational Kinetic Energy>[§10.4]{ref}

For a computed example, take a uniform disc and an ideal thin ring, each of <mass> $M=2\,\mathrm{kg}$ and radius $R=0.30\,\mathrm m$, rotating about its central perpendicular axis at <angular speed> $10\,\mathrm{rad/s}$. The <moment of inertia of a uniform disc> is $MR^2/2=0.090\,\mathrm{kg\,m^2}$, while the ring has <moment of inertia> $MR^2=0.180\,\mathrm{kg\,m^2}$. Their <rotational kinetic energies> are therefore $4.5\,\mathrm J$ and $9.0\,\mathrm J$. The ring stores twice the <energy> at the same <angular speed>, because more <mass> lies farther from the axis. At the same nonzero <rotational kinetic energy>, its <angular speed> would instead be $1/\sqrt{2}$ times the disc's. This makes radial placement useful for storing <energy> in a rotating wheel, although a real design must also withstand the stresses of rotation.

For fixed-axis rotation with constant <moment of inertia>, write the signed rotation rate as $\Omega=\dot\theta$, with <angle> $\theta$ in <radians>. The axial component of net external <torque> obeys $\tau_{\mathrm{axis}}=I\dot\Omega$, so the <work-energy theorem> becomes
$$
\frac{dK_{\mathrm{rot}}}{dt}=\tau_{\mathrm{axis}}\Omega,
\qquad \Delta K_{\mathrm{rot}}=\int\tau_{\mathrm{axis}}\,d\theta.
$$
An opposing <torque> removes <rotational kinetic energy>; bearing <friction> must be included in the net <torque>. The signed $\Omega$ matters when the direction reverses. Rotation about a fixed axis need not be about a <principal axis>: the scalar <energy> formula still holds, but transverse bearing <torques> can be needed to maintain that axis.

For general <rigid body> motion, use the <inertia tensor> $\mathsf I_G$ about the <centre of mass> $G$:
$$
K_{\mathrm{total}}=\frac12M|\mathbf V_G|^2+\frac12\boldsymbol\omega^{\mathsf T}\mathsf I_G\boldsymbol\omega,
\qquad \mathbf L_G=\mathsf I_G\boldsymbol\omega.
$$
This is the <kinetic-energy decomposition about the centre of mass>. In a <principal body frame>, the rotational term is $\tfrac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)$, using the <principal moments of inertia>. Away from a <principal axis>, <angular momentum> and <angular velocity> need not be parallel. A freely rotating asymmetric <rigid body> can change the direction of <angular velocity> relative to the body even while <rotational kinetic energy> remains constant; <Euler equations for a torque-free rigid body> govern that motion. One constant axial <moment of inertia> is then insufficient.<The Feynman Lectures on Physics, Volume I by Richard Feynman and others/Chapter 20: Rotation in space>{ref}


 Rotational kinetic energy (source code)