The surface area of the hemispherical shell is 2\pi r^2 and its centroid is r/2 from its center. Except for one dimension, the free-body diagram for this problem looks just like the one for Problem 64.
Moment equilibrium about the point of contact with the ground gives
So
Where did I come up with the centroid of the hemispherical surface? In this case, I actually did the integration. To make life easier, I did the problem in spherical coordinates, (r, \theta, \phi), which are defined in this drawing that I pulled from the MathWorld page:
A differential patch on the surface has the area
and its z coordinate is r \cos\phi
The centroid of the upper hemispherical surface, then, is
Last modified: January 22, 2009 at 8:32 PM.