Rigid rotation minimizes kinetic energy at fixed angular momentum ($K\ge L^2/(2I)$)

For rotational motion about a fixed axis, let $r$ be cylindrical radius, $v$ the signed tangential velocity, and $dm$ the mass measure. Define the kinetic energy, axial angular momentum, and moment of inertia byThe Cauchy-Schwarz inequality gives $L^2\le (\int v^{2}dm)(\int r^{2}dm)=2KI$. More precisely, direct expansion gives the nonnegative energy gapThus the minimum at fixed $L$ and fixed mass distribution occurs exactly when $v=(L/I)r$ almost everywhere with respect to $dm$. This means constant angular velocity $\Omega =L/I$, or solid-body rotation. If the profile is continuous, the equality holds at every point in the mass support.

For a uniform unit-density cylinder of radius $a$, height $h$, and volume $V=\pi a^{2}h$, the moment of inertia is $I=V{a}^2/2$. Therefore $K\ge L^2/(V{a}^2)$, with equality for $\omega (r)=2L/(V{a}^2)$. Multiplying by a uniform nonunit density changes the mass measure and $I$ consistently.