Consider a point mass on the end of a string of length $l$ near the surface of the Earth. Initially it is hanging straight down, but it has initial speed $v_0$, such that it will undergo a circular trajectory.
Initial mechanical energy is $$E_{m_0}=\frac{mv_0^2}{2}$$ Mechanical energy at some point along the trajectory is $$E_{m_1}=\frac{mv_1^2}{2}+mgl(1-\cos{\theta})$$
Where the second term is the potential energy and $l(1-\cos{\theta})$ is the height of the particle mass from the initial position. $$v_1=\sqrt{v_0^2-2gl(1-\cos{\theta})}$$
Applying Newton's 2nd law we can obtain an expression for the tension force $$mg\cos{\theta}-T=-m\frac{v_0^2-2gl(1-\cos{\theta})}{l}$$ $$\implies T=\frac{m}{l}(v_0^2+3gl\cos{\theta}-2gl)$$ Where I simply used the radial acceleration $a_R=\frac{v^2}{l}$.
At the top of the circle $\theta=\pi$ and thus $T(\pi)=\frac{m}{l}(v_0^2-5gl)$.
For the tension force to make sense physically, it must be greater than or equal to zero. $$T\geq0 \implies v_0 \geq \sqrt{5gl}$$
Now I ask the question: what must $v_0$ be for the speed at the top of the circular trajectory to be $0$?
$$v_1(\pi)=\sqrt{v_0^2-4gl}=0$$ $$\implies v_0=\sqrt{4gl}$$
But a $v_0$ of $\sqrt{4gl}$ means the tension force must be negative, which would mean that it would be a force pulling at the particle mass radially away from the center of the circle. This isn't possible in this problem.
Does this mean that no matter what, there is no trajectory that has zero speed at the top of the circle?
