Toppling Curved Towers

Instead of rectangular prisms, now suppose the tower is part of an annulus. More specifically, it’s the region between two arcs of angle $\theta$ in circles of radius $1$ and $2,$ as shown below.

For small values of $\theta,$ the tower balances on one of its flat sides. But when $\theta$ exceeds some value, the tower no longer balances on a flat side. What is this critical value of $\theta$?

Again, we need to define the center of gravity $(\tilde{x}(\theta), \tilde{y}(\theta))$ and find for what value of $\theta$ do we have $\tilde{x}(\theta) \lt 1.$ Here we should use polar coordinates \begin{align*}x&=\rho \cos \phi \\ y&=\rho \sin \phi\end{align*} to integrate the annular region. We have \begin{align*}\tilde{x}(\theta) &= \frac{\int_0^\theta \int_1^2 \left( \rho \cos \phi \right) \rho \, d\rho \, d\phi}{\int_0^\theta \int_1^2 \rho \,d\rho \,d\phi} \\ &= \frac{ \int_0^\theta \left( \int_1^2 \rho^2 \,d\rho \right) \cos \phi \,d\phi }{ \int_0^\theta \left(\int_1^2 \rho \,d\rho \right) \, d\phi } \\ &= \frac{ \int_0^\theta \frac{2^3 - 1^3}{3} \cos \phi \,d\phi }{ \int_0^\theta \frac{2^2 - 1^2}{2} \,d\phi } \\ &= \frac{ \frac{7}{3} \sin \theta }{ \frac{3}{2} \theta } = \frac{14 \sin \theta}{9 \theta}.\end{align*}

We need to find the solution to the implicit equation $\tilde{x}(\theta) = \frac{14 \sin \theta}{9\theta} = 1.$ Therefore, we find that the point at which the curved tower will topple over is approximately $\theta^* \approx 1.55537048\dots$ radians.