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?

I know Larry already introduced this method for solving it, but if memory serves there must have been an error somewhere because he didn't get the same answer as when he tried the PE method.  So the purpose of this comment is to show an FBD solution in full, and show you get the same answer.

Sidenote:  My biggest hurdle was figuring out a way to simplify cos(2t)/cos(t).  I had to use iteration to zero in on the answer (which is 23.213318 degrees or 0.40514883 radians).  Ok, here goes:

Ok, so I will call the straight edge of the bowl my horizontal and view it from the side (so all I see is a half-circle)

Call the angle the rod makes with the bowl, Theta (t). (For reference, I have the end of the rod sticking out the RIGHT side of the bowl)
Call the radius of the bowl, r.
So the length of the rod is 3r.
Call the mass of the rod m.

Let O be the center of the spherical bowl.
Let E be the point where the rod touches the edge of the bowl, and D be the point on the opposite side of the edge of the bowl (So ED = 2r)
Let B be the point where the rod touches the bowl’s inner surface.
Call the length EB, L.

Notice that triangle EOB is an isosceles triangle because OE = OB = r (definition of a sphere).
We said angle OEB = t, so angle OBE = t as well.
This means that angle EOB = 180 – 2t
Also, angle DOB = 180 – (18 – 2t) = 2t

Ok, let’s put some forces in.
We will have a force at point E (Fe) that is perpendicular to the rod.
A force at point B (Fb) that is perpendicular to the bowl’s surface.
And a force due to gravity at the midpoint of the rod (Fg) perpendicular to the horizontal.

Ok, so let’s determine all of their angles relative to the rod (CCW is positive, CW is negative, x is parallel to rod, y is perpendicular to rod).

I already said the angle between the rod and the horizontal was t, but here it is -t
Since Fe is perpendicular to the rod, the angle of Fe = +90
Since Fb is perpendicular to the bowl, that means it is pointing at O, so it has the same angle as OBE which is +t
Since Fg is perpendicular to the horizontal, its angle is –90–t.

Fex = Fe*cos(90) = 0
Fey = Fe*sin(90) = Fe*1 = Fe
Fbx = Fb*cos(t)
Fby = Fb*sin(t)
Fgx = mg*cos(-90-t) = mg*cos(90+t)
Fgy = mg*sin(-90-t) = -mg*sin(90+t)

Some helpful trig identities are:

sin(-x) = -sin(x)
cos(-x) = cos(x)
sin(x +/- y) = sin x cos y +/- cos x sin y
cos(x +/- y) = cos x cosy -/+ sin x sin y
cos(2x) = cos^2(x) - sin^2(x)

Fgx = mg[cos(90)*cos(t) – sin(90)*sin(t)] = mg[0*cos(t) – 1*sin(t)] = -mg*sin(t)
Fgy = -mg[sin(90)*cos(t) + cos(90)*sin(t)] = -mg[1*cos(t) + 0*sin(t)] = -mg*cos(t)

Ok, so the sum of forces along the rod are:
0 = Fex + Fbx + Fgx = 0 + Fb*cos(t) – mg*sin(t)
Fb = mg*sin(t)/cos(t) = mg*tan(t)

The sum of moments about point B are:
0 = Fey*L + Fgy*(3r)/2 = Fe*L – mg*cos(t)*3r/2
Fe = mg*cos(t)*3r/(2L)

Look at triangle EOB, remembering that the angle at O is 180-2t. The relationship between r, t, and L is the following:
sin((180-2t)/2) = (L/2)/r
sin(90-t) = L/(2r)
sin(90)*cos(t) – cos(90)*sin(t) = cos(t) = L/(2r)
So L = 2r*cos(t)
So Fe = mg*cos(t)*3r/(2*2r*cos(t)) = mg*3/4

The sum of forces perpendicular to the rod are:
0 = Fey + Fby + Fgy = Fe + Fb*sin(t) – mg*cos(t)
0 = Fe + mg*tan(t)*sin(t) – mg*cos(t)
Fe = mg[cos(t) – tan(t)*sin(t)] = mg*3/4

So
3/4 = cos(t) – tan(t)*sin(t) = [cos^2(t) – sin^2(t)]/cos(t) =cos(2t)/cos(t)

3/4 = cos(2t)/cos(t)

As I said above, I didn't know how to simplify that, so I used iteration to solve for t and got t = 23.213318 degrees or 0.40514883 radians

Posted by nikki on 2005-01-06 19:19:01