Let $m$ be the mass of the stone, $r$ be the radius of the vertical circle, and $v$ be the speed of the stone. The forces acting on the stone are the tension from the string, $\vec{T}$, and gravity, $\vec{F}_g = m\vec{g}$. The net force provides the required centripetal force, $F_c = \frac{mv^2}{r}$, directed towards the center of the circle.

Let's analyze the forces in the radial direction at an arbitrary angle $\theta$ measured from the bottom of the circle. The tension $\vec{T}$ always points radially inward. The gravitational force has a radial component of $-mg\cos\theta$ (pointing outward).

Applying Newton's second law in the radial direction:

$$ \sum F_{radial} = T - mg\cos\theta = \frac{mv^2}{r} $$

Solving for the tension $T$:

$$ T = \frac{mv^2}{r} + mg\cos\theta $$

[Q1] Where is the stone on its path when the string breaks?

The string breaks when the tension $T$ reaches its maximum value, $T_{max}$. We need to find the position $\theta$ where $T$ is maximized. The expression for tension, $T = \frac{mv^2}{r} + mg\cos\theta$, depends on both speed $v$ and position $\theta$.

1. The term $mg\cos\theta$ is maximized when $\cos\theta$ is maximum. This occurs at the bottom of the circle, where $\theta = 0$ and $\cos(0) = 1$.
2. The speed $v$ is also maximum at the bottom of the circle due to the conversion of potential energy to kinetic energy. Since both terms in the expression are maximized at the same point, the tension $T$ is greatest at the bottom of the vertical circle. Therefore, the string will break when the stone is at the lowest point of its path.

[Q2] What is the speed of the stone as the string breaks?

The string breaks at the bottom of the circle ($\theta = 0$) when the tension equals the maximum withstandable tension, $T_{max}$. Let the speed at this point be $v$. Substituting $T = T_{max}$ and $\theta=0$ into our tension equation:

$$ T_{max} = \frac{mv^2}{r} + mg\cos(0) $$ $$ T_{max} = \frac{mv^2}{r} + mg $$

Now, we solve this equation for the speed $v$:

$$ T_{max} - mg = \frac{mv^2}{r} $$ $$ v^2 = \frac{r}{m}(T_{max} - mg) $$ $$ v = \sqrt{\frac{r}{m}(T_{max} - mg)} $$

This is the speed of the stone at the moment the string breaks.