# The rod in the bowl.

Fig. 1. The slippery soup bowl. A fixed stationary hemispherical bowl (green) has radius R. A rod (red) of length L = (1.5)R has a weight (black) at its center of mass, C which is at distance R from its lower end. The total weight of rod and extra weight is W. What is the angle of the rod when it finds a static equilibrium position? Neglect friction.

Neglecting friction isn't as idealized as you might think. If you did this experiment with a real bowl and rod, you could find the equilibrium position by repeatedly tapping the bowl. The rod would settle to the same position as predicted by the frictionless analysis.

The physics. There are only three forces on the rod. The weight of the rod, acting at its center of mass, and the two forces of the bowl on the rod ends. These two forces are normal to the bowl surface at the point of contact and are therefore radial.

Since we want the equilibrium position, the rod is stationary and these three forces sum to zero. If we take the center of torques at the center of the arc of the bowl, where these two forces' lines of action cross, then the line of action of the weight also must pass through this point. All of the torques are zero around this point.

The geometry.

Fig. 2. The geometry. The rest is just mathematical drudgery. Fig. 2 shows the geometry with lines and angles labeled. R is the radius of the bowl. L is the length of the rod. X is the distance of the center of mass from the midpoint of the rod.

The black square at M is the position of the center of mass of the rod. This diagram shows the rod horizontal, and is certainly not a position of equilibrium. If the diagram is rotated until the dotted line is vertical, it will be the position of equilibrium.

Solution:

From the isosceles triangle with two sides equal to R:

 

sin(α) =L/(2R).

From Eq. 1:

 

α = arcsin(L/(2*R))
.

From the larger right triangle:

 

H = R cos(α).

From the smaller right triangle:

 

X = H tan(φ).

Eliminate H from Eqs. 3 and 4:

 

tan(φ) = X/(R cos(α).

Eliminate α from Eqs. 2 and 5. The result is: For the data given above α = 29.55...°.

Fig. 3. Solution, accurately drawn. Fig. 3 shows the solution, very accurately drawn using CAD software. The rod is shown rotated to 29.55...°. It should come as no surprise that this rotation of the rod automatically locates the center of mass directly below the center of curvature of the bowl, along the dotted vertical line, as we expected from physics. I welcome other solutions. Send email to me at the address to the right.

• Donald Simanek, Sept 17, 2018.