The slippery soup bowl.

This is an ideal puzzle. It can be simply described and easily visualized. It is a simple device with only one moving part. It can be simply solved an has an answer that is simple, though not easily guessed. It has a small amount of input data that generates many mathematical relations, and many methods of solution. Many of these solutions are messy, encouraging blunders along the way.

Fig. 1. The slippery soup bowl.

A soup bowl has a hemispherical inner shape of radius R. A slender metal rod of length 2R and weight W has its center of mass at its center. (point C in Fig. 1). The rod is put into the bowl. The bowl-rod contact points are nearly frictionless, but eventually the rod comes to rest. At what angle (to the horizontal) is the rod at rest in stable equilibrium?

The bowl (green) is fixed and unmoving. Only the rod (red) can move.

Solution 1.

Fig. 2. Mortar and pestle.

This puzzle isn't as unrealistic or contrived as you might think. The mortar and pestle has been used since the time of the alchemists for grinding solid materials into powder. Cooks use it today for the same purpose. They don't always have hemispherical bowls, but some "wide" tea or coffee cups do.

Let L be the length of the portion of the rod measured between its two points of contact with the bowl.

There are three forces acting on the rod. (1) gravity, W, downward. (2) The force on the rod at the rim of the bowl, A, perpendicular to the rod. (3) The force at the contact between the end of the rod and the curved bowl. B, perpendicular to the bowl.

Fig. 3. The geometry of the soup bowl.
Schematic only. This is not the equilibrium angle.

The net (vector) force on a body in equilibrium is zero. The net torque on it is also zero.

When two bodies are in frictionless contact the force there is perpendicular to the tangent line there. So the force (A) at the rim is perpendicular to the rod, and the force (B) is at the right end of the rod is along the radius of the bowl.

Taking torques about the right end of the rod, A L = R W cos α.

The rod is stationary, so the horizontal components of A and B must be equal in size: A sin α = B cos β.

Eliminating L and cos α we get A = W/2.

The vertical components must satisfy A cos α + B sin β = W.

From Fig 3, there's an isosceles triangle with two sides of length R, and its base along the rod is L, so L = 2 R cos α.

Also from geometry, β = 2 α.

Eliminate W, A, and B from these equations. This might be messy, so lets first solve them by iteration using my old version of MathCad.

Clearly at equilibrium the value of α cannot depend on W or R. So we expect that they will drop out.

Here's a Mathcad spreadsheet of the interation model.

Fig. 4. The solution.

The last equation is the result of including dummy values for R and W. The answer will not depend on these, of course. This method allows us to check the intermediate results and confirm that these values of R and W do not affect the result. Using them contributes data truncation errors that only show up in the 6th decimal place.

Fig. 5. More geometry. This is accurately drawn
with the rod drawn at the equilibrium angle.

The result is α = 32.53333...°. This was a simple system, a machine with only two parts, only one of them movable, but the strategy necessary to calculate the answer was not simple. It had many opportunities for blunders. I know, I made them. One has the suspicion that there's an easier, shorter and more intuitive answer. One strategy might use the fact that if only three forces act on a body in static equilibrium, their lines of action must pass through a common point, since the net torque must be zero. In this problem, this point will lie directly above the center of mass of the rod, since W acts through the center of mass. Fig 4 gives you a start.

One may reduce the equations above to simpler forms:

cos α + [sin α][tan (2 α)] = 2

or, simplest of all,

cos(α) = 2 cos(2 α)

Solution 2.

Fig. 6. Center of mass at various angles.

Another puzzle, the sliding slot describes other methods for this sort of problem. One is to trace the curve of the system center of mass for various angles, then find the lowest point of the curve. Fig 6 shows that curve for this problem. It is not a semicircle, but is close. Point L is the lowest point on the curve. The red rod is at angle 30°, clearly not the equilibrium angle.

Experimental check.

Fig. 7. Steel construction set model. Fig. 8. Weight (red) hung at the srip's center of gravity.

This answer has been experimentally confirmed with the small model made with a metric steel construction set. Its parts have holes 1 cm apart. The white string is for visibility, a narrow string works better. The upper left bolt lets the slotted strip slide freely, and is not quite at the end of the slot.

This model can test the proposition that the weight of the slotted strip doesn't affect its equilibrium angle. Just hang a weight from the slotted strip's center of gravity. The strip's angle doesn't change. (Fig. 8).

I welcome other solutions. Send email to me at the address to the right.

  • Donald Simanek, August 10, 2018.

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