One of the curators at the art museum is tilting a large cylinder backward. At what angle (theta) will the cylinder of height h and radius r will tumble?

My thinking is that the

length = height = h
2r = w
It seems that this cylinder will tumble when the angle is large enough to cause the center of gravity to go beyond the supporting edge.

I am not sure how to go about this...?

2 answers

Your assumption is correct. The CG is along the central axis midway between top and bottom. It will be above the edge (point of contact with the ground) when the tilt angle is arctan (2r/h)
A square of side 2L is removed from one corner of a square sandwich that has sides of length L. The center of mass of the remainder of the sandwich moves from C to C’. The displacement of the x coordinate of the center of mass (from C to C’) is?

In the picture, C' moves up and to the right. So positive x and positive y direction.