Answer
$m=.303\ kg$
Work Step by Step
We know that the moment of intertia of a rod about this point is $\frac{1}{3}ML^2$, and we know that the moment of inertia of a point mass a distance R away from the axis of rotation is $mR^2$. Thus, we find:
$\frac{T^2K}{4\pi^2}=I$
$\frac{T^2K}{4\pi^2}=\frac{1}{3}ML^2+ mR^2+mR^2$
$\frac{T^2K}{4\pi^2}=\frac{1}{3}ML^2+ 2mR^2$
$\frac{T^2K}{4\pi^2}-\frac{1}{3}ML^2= 2mR^2$
$m= \frac{ \frac{T^2K}{4\pi^2}-\frac{1}{3}ML^2}{2R^2}$
$m= \frac{ \frac{5.6^2(.63)}{4\pi^2}-\frac{1}{3}(.85)(.75)^2}{2(.75)^2}$
$m=.303\ kg$
