Rotating pendulum

There are two forces acting on the bob, its weight $\vec{P}$ and the string tension $\vec{T}$ . When the bob follows a circular trajectory in a horizontal plane the vertical components of these forces have equal intensity and opposite direction. Let $z$ denote the vertical direction and $r$ the radial one. We have $\vec{P}=\vec{P_z}$ and $\vec{T}=\vec{T_z}+\vec{T_r}$ . As there is no vertical acceleration we have $\vec{T_z}=-\vec{P_z}=-\vec{P}$ . Along the radius we have $\vec{T_r}=m\vec{a}$ where $m$ is the bob's mass. Assuming the circular trajectory of the bob we know that $a=\omega^2 r$ where $\omega$ is the rotational speed and $r$ the radius of the bob's trajectory. We also know that $P=mg$ . Let $h$ be the altitude difference between the bob and the string attach (the height of the cone formed when the pendulum rotates). We have $\frac{h}{r}=\frac{T_z}{T_r}=\frac{mg}{m\omega^2 r}$ which simplifies to $h=\frac{g}{\omega^2}$ , so $\omega=\sqrt{\frac{g}{h}}$ which equals $2$ in our example.