= Minimum-period offset of a physical pendulum
{title2=$a_{\min}=\sqrt{I_{\rm CM}/M}$}
With a pivot at distance a from the centre of mass, the <parallel axis theorem> and small-angle restoring torque give $T^2=(4\pi^2/g)[a+I_{\rm CM}/(Ma)]$. Differentiation, or the arithmetic–geometric mean inequality, minimizes the period at the radius of gyration $\sqrt{I_{\rm CM}/M}$. The minimum is $T_{\min}=2\pi\sqrt{2a_{\min}/g}$. The allowed pivot locations must include this point; a constraint keeping the pivot outside the body can move the physical optimum to a boundary.
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