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Bowl and Rod (Posted on 2004-12-31) Difficulty: 3 of 5
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?

See The Solution Submitted by Bractals    
Rating: 2.0000 (2 votes)

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Another method, Another answer | Comment 12 of 18 |
Change of tactics.  Looking at potential energy, let's assume that the center of mass of the rod will seek the lowest point.   Let y be the vertical distance from the plane of the edge of the bowl and the CM of the rod.

The vertical drop to point P is  - r sin(2t)
The vertical elevtion in the y direction from point P to the CM of the rod is:      + (1.5 r) sin(t)

y = 1.5 r sin(t) - r sin(2t)
y = 1.5 r sin(t) - 2 r sin(t) cos(t)
dy/dt = 1.5 r cos(t) + 2r sin^2(t) - 2r cos^2(t)
dy/dt = 2r(sin^2 + cos^2 - 2cos^2 + .75 cos)
         to minimize y, set dy/dt to zero
0 = 1 - 2 cos^2(t) + .75 cos(t)  a quadatic eqn in cos(t)
cos(t) = {.919044, -.54404}
angle t = {23.21 degrees, or 122.96 degrees}

So this method gives 23.21 degrees as the answer.
Is it valid?  Once again, not sure.  But it seems like a reasonable assumption.
  Posted by Larry on 2005-01-03 02:04:29
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