Given a solid bowl (an idealized hemisphere, the opening facing straight up), let a slender rod rest in the bowl supported at two points: a point on the bowl's edge, and some point on the bowl's interior surface.

The bowl's edge exerts a force on the rod which is perpendicular to the rod's length.

The bowl's interior exerts, on the rod's end, a force which is perpendicular to the bowl's surface (at the point where they meet).

If the rod's length is three times the radius of the bowl, what is the angle between the rod, and the plane of the bowl's edge?

Here is a 4th equation, based on the fact that we can calculate the distance between the 2 points of contact as a function of radius r, and angle t.
t is the angle between rod and horizontal.
2t is the angle of the line from center of circle to where rod touches bowl
3r-L is the amount of rod inside the bowl
From Pythagoras:

(3r-L)^2 = [r sin(2t)]^2 + [r cos(2t) + r]^2
9r^2 - 6rL + L^2 = r^2 + 2r^2 cos(2t) + r^2
                           (since sin^2 + cos^2 = 1)

Eqn 4: 7r^2 - 6rL + L^2 = 2r^2 cos(2t)

(edited to correct math error)

Edited on January 1, 2005, 1:30 pm

Posted by Larry on 2005-01-01 12:35:26