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

$.494MR^2$
We add negative area, so we subtract the two moment of inertias. Thus, we obtain: $I=\frac{1}{2}MR^2-\frac{3}{2}M(\frac{1}{4}R)^2(\frac{1}{4})^2=.494MR^2$ (Note, we multiply by the $(1/4)^2$ due to the smaller mass of the missing hole.)