## Essential University Physics: Volume 1 (3rd Edition)

$x=\frac{1}{\sqrt2}$
We use the equation for the period to find: $T = \sqrt{\frac{I}{mgx}}$ $T = \sqrt{\frac{m(x^2+\frac{1}{2}R^2)}{mgx}}$ $T = \sqrt{\frac{(x^2+\frac{1}{2}R^2)}{gx}}$ We take the derivative of the inside of the radical and set it equal to zero to find: $T = \sqrt{\frac{(x+\frac{R^2}{2x})}{g}}$ $0=\frac{1}{g}(1+\frac{-2R^2}{4x^2})$ $0=(1+\frac{-2R^2}{4x^2})$ $-1=(\frac{-2R^2}{4x^2})$ $x =\sqrt{\frac{R^2}{2}}$ Using the first derivative test, we obtain that the minimum will occur at: $x=\frac{1}{\sqrt2}$