A modern aluminum sculpture consists of a hollow cylinder that is capped on one end by a solid hemisphere. The cylinder has an outer diameter of 100 cm and thickness of 1 cm, and the hemisphere has the same diameter as the outside of the cylinder.
If, on a level surface, the sculpture balances in stable equilibrium at any point on its hemispherical surface, how long is the cylinder, and what is the minimum ceiling height in the museum to permit the sculpture to assume any stable position?
I did it a similar way to Larry's. But, I started by looking-up the
integral part: a hemisphere's center of mass is at 3/8 R.
For unit density, the mass of the hemisphere is M = (4/6) pi 50^3
For cylinder of height h, the center of mass is on the centerline at
height h/2 and m= pi (r_out^2-r_in*2) h = pi (50^2-49^2) h
m= 99 pi h
So, to get the combined center of mass to be on the top of the cylinder, I weighted them:
[Mass] x [Distance] = [mass] x [distance]
(similar to Larry's torque calc, the moment arms are set to cancel)
[ (4/6) pi 50^3 ] [ (3/8) R ] = [99 pi h] [h/2]
h = 1250 sqrt(2/11) / 3 = 177.67 cm
So they should be sure the ceiling is higher than hemisphere + cylinder:
50 + 177.67 = 227.767cm, 2.277 m (or about 7' 5")
BUT
Looking at Larry's work I see my error. The tipped gizmo can be
taller than when straight up (like any good cylinder). the base adds
R=50 to the diagonal from the top rim to the new center of mass
at sqrt(177.67^2+50^2) +50 = 184.568 +50
= 234.568 cm (7' 8")
But this is unstable equilibrium, no? Will the display work?
Edited on January 23, 2025, 6:04 pm
soln
