= Rigid rotation minimizes kinetic energy at fixed angular momentum
{title2=$K\geq L^2/(2I)$}
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> by
$$
K=\frac12\int v^2\,dm,\qquad L=\int rv\,dm,\qquad I=\int r^2\,dm>0.
$$
The <Cauchy-Schwarz inequality> gives $L^2\leq(\int v^2dm)(\int r^2dm)=2KI$. More precisely, direct expansion gives the nonnegative energy gap
$$
K-\frac{L^2}{2I}=\frac12\int\left(v-\frac LI r\right)^2dm.
$$
Thus 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> $V=\pi a^2h$, the <moment of inertia> is $I=Va^2/2$. Therefore $K\geq L^2/(Va^2)$, with equality for $\omega(r)=2L/(Va^2)$. Multiplying by a uniform nonunit <density> changes the mass measure and $I$ consistently.
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