Assume that the point mass, $m$ has two tiny thrusters, mounted so as to exert purely tangential force in the plane of the circular motion, one clockwise, and the other counter-clockwise.
The magnitude of the constant velocity of the mass is $v$, and the radius of the circle is $r$.
Measure the position of the point mass in the standard Cartesian coordinate way: angles are measured from the positive X-axis, counter-clockwise positive.
At the point where the mass is at a position angle $\theta$. the total radial force inward on the mass, $F_R$ is given by the centripetal force equation:$$F_R=\frac{mv^2}{r}$$
There are two forces that supply this radial force: the tension, $T$ in the string, and the inward radial component of the force of gravity:$$F_{G-R}=mg\sin(\theta)$$So:$$\frac{mv^2}{r}=T+mg\sin(\theta)$$and:$$T=\frac{mv^2}{r}-mg\sin(\theta)$$Note that this implies that:$$v >=\sqrt{rg}$$ or the string tension will become negative near the top of the circle, an impossibility.
The conditions of the question also require that at all times the net tangential force, $F_T$, be zero. The tangential component of the force of gravity, $F_{G-T}$ is given by:$$F_{G-T}=-mg\cos(\theta)$$where a positive force implies counter-clockwise force. The thrusters are needed to supply the exact opposite force to the mass at all times.
There are two senses to $v_{\mathrm{min}}$. Firstly it means that the speed of the particle must be greater or equal to $v_{\mathrm{min}}$ at $\theta = \pi$ or else it will not complete the cycle. Instead it will follow a ballistic trajectory until such time that tension comes back into the string. Secondly, the speed of the particle is not constant: it is fastest at the very bottom and slowest at the very top. Thus $v_{\mathrm{min}}$ it is the slowest speed of the particle at any point around the cycle for a particle that just completes the cycle.
EDIT:
The reason that there is no tension in the string is that the centripetal acceleration is perfectly balanced by the particle's weight when $v=v_{min}$ at the very top. If the particle were travelling faster here then the "excess" centripetal acceleration is taken up by tension in the string.
Best Answer
The tension in the rod does become zero, at exactly the same point as for the string. The difference is that the rod does not buckle and go limp like the string when the tension decreases even further and becomes -ve. It continues moving in a circle, keeping the bob exactly the same distance from the pivot. That is because the rod is rigid and the string is not. The string can only prevent the bob from moving further away from the pivot, it cannot stop it from moving closer.
When a pendulum reaches the end of its swing the only force pulling the bob away from the pivot point, and causing the tension in the string, is the component of the bob's weight. When the end of the pendulum swing reaches $90^{\circ}$ this component becomes zero, so the tension becomes zero.
If the pendulum swings any higher than $90^{\circ}$ the component of weight now acts toward the bob, causing compression of the string. Strings fold up when you try to compress them, so the pendulum bob leaves the circle it was swinging on, and moves like a projectile until the string becomes taut again.
However, this does not happen if the pendulum bob is moving fast enough and its speed $v$ never reaches zero when it goes up above $90^{\circ}$. Then the pendulum swings in a full circle and keeps going in the same direction into the next circle.
The difference with a rod is that, because it is rigid, it doesn't fold up or buckle when compressed by a force. So the bob continues to move in a circular arc after the tension in the rod has changed to compression.