The solid will start to tip or flop over to its flat side once its centroid is vertically above the lowest point on the solid. This lowest point is the bottom of the semicircle at the foot of the solid.
The half-cylinder solid:
radius = r
Length = unknown = say, L long.
centroid is at L/2 and at, say, u from the flat face
Solving for u:
Consider a semicircle such that the flat "arc', or the diameter, is on the x-axis, and the center of the diameter is at (0,0).
so the equation of the circle of the semicircle is x^2 +y^2 = r^2
and so, x = sqrt[r^2 -y^2].
Taking moment of areas about the x-axis,
the dA element is horizontal, and dA is 2x wide by dy thick,
u*A = INT.(0 to r)[y*dA]
uA = INT.(0 to r)[y*2x dy]
uA = (2)INT.(0 to r)[y*sqrt(r^2 -y^2)]dy
uA = (2)INT.(0 to r)[sqrt(r^2 -y^2)](y dy)
uA = -[(2/3)(r^2 -y^2)^(3/2)] | (0 to r)
uA = (2/3)r^3
u = [(2/3)r^3] / [(1/2)pi*r^2]
u = 4r /(3pi) ------------------------------**
Now draw the figure on a new (x,y) set of axes.
It is a tilted r by L rectangular, leaning to the right, such that the bottomest point is at (0,0) and the centroid is on the y-axis..
From the centroid, draw a line segment perpendicular to the right L side of the rectangle....such that this line segment will bisect the L.
Call the angle made by the y-axis and the right L side of the rectangle as alpha.
tan(alpha) = (r -u) / (L/2)
tan(alpa) = [r -(4r / 3pi)] / (L/2) = [6pi*r -8r] / [3pi*L] = [(6pi -8)/(3pi)](r/L)
alpha = arctan[(r/L)(6pi -8)/(3pi)]
And if we call beta the tilting angle by which the half-cylinder solid begins to flop over its flat face, then,
beta = (pi/2) -alpha -------------------answer.