Balancing Hemispheres

Let $\theta$ be the angle formed with the axis of hemisphere $B$ and the center of the flat surface of hemisphere $A$ . Since all forces which do work in this scenario (gravity) are conservative, we will analyze this problem using conservation of energy, and determine for which a give a gravitational potential energy well about $\theta=0$ .

Let the $U(\theta)=0$ line be the plane of hemisphere $B$ 's flat surface, and let the center of mass position of each hemisphere with respect to its flat face be $a'$ and $b'$ respectively. For example, hemisphere $B$ is flat on the ground with its curved surface up, the center of mass position would be at $b'$ above the plane.

For hemisphere $A$ , we need some more variables - Let $\phi$ be the angle formed by the axis of hemisphere $A$ , and the line connecting the centers of both hemisphere's flat surfaces. Then $\phi + \theta$ gives the angle formed by the axis of hemisphere $A$ and the vertical. Here, it is important to note that due to the no-slip condition, the arcs traced out by $\theta$ and $\phi$ are the same, and so $a\phi = b\theta \Rightarrow \phi = \frac{b}{a}\theta$ . Then $\theta + \phi = (1+\frac{b}{a})\theta$ . Let $k = 1+\frac{b}{a} \rightarrow \theta + \phi = k\theta$ .

We can then find the gravitational potential energy of the system to be

$U(\theta) = \rho V_{b} g b' + \rho V_{a} g \left[(a+b)\cos(\theta) - a'\cos(\theta+\phi) \right] = \rho V_{b} g b' + \rho V_{a} g \left[(a+b)\cos(\theta) - a'\cos(k\theta) \right]$

Let $\rho V_{a} g = \bar{c}$ . Then

$U' (\theta) = \bar{c} \left[ a'k \sin(k\theta) - (a+b)\sin(\theta) \right] \Rightarrow U''(\theta) = \bar{c} \left[ a'k^{2} \cos(k\theta) - (a+b) \cos(\theta) \right]$

For stable equilibrium to hold, $U''>0$ . We then plug in our equilibrium theta and find our constraint on $a$

$U''(0) = \bar{c} \left[ ak^{2}- (a+b) \right] >0 \Rightarrow a+b<a'k^{2}$

We then use the fact that the center of mass position of a hemisphere lies on the axis a distance $\frac{3}{8}r$ from the flat face.

$a+b< \frac{3}{8}a \left( 1 + \frac{b}{a} \right)^{2} \Rightarrow a^{2}(a+b) < \frac{3}{8}a(a+b)^{2} \Rightarrow a<\frac{3}{8}a + \frac{3}{8}b \Rightarrow a<\frac{3}{5}b$

And so our answer is $\frac{3}{5} = \boxed{.600}$