Minimum-period offset of a physical pendulum ($a_{\min }=\sqrt{I_{\mathrm{CM}}/M}$)

With a pivot at distance a from the centre of mass, the parallel axis theorem and small-angle restoring torque give $T^2=(4{\pi }^2/g)[a+I_{\mathrm{CM}}/(Ma)]$. Differentiation, or the arithmetic–geometric mean inequality, minimizes the period at the radius of gyration $\sqrt{I_{\mathrm{CM}}/M}$. 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.