forceshomework-and-exercisesnewtonian-mechanics
Consider a particle attached to one end of a string of length $l$ moving anti-clockwise in a vertical circle whose centre is $O$. What exactly happens physically when the string becomes slack and leaves circular motion?
I'm guessing that the particle falls downwards until its distance from $O$ is again $l$, whereupon it will re-enter circulation motion but now in a clockwise direction. But I think that this is either incorrect or incomplete.
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.
What is the configuration needed for a circular-motion?
The answer is, there must be an inward(towards the center) force perpendicular to the instantaneous velocity. The minimum velocity required for the bob initially to loop the whole loop is $\sqrt{5gR}$. In this situation, the tension at the topmost point becomes zero, but since there is velocity, which is $\sqrt{Rg}$ & there is inward force, which is the weight $mg$; this matches up the configuration of circular-motion & hence continues to complete the circle.
Now, if the initial velocity is less than $\sqrt{5gR}$, then the bob will have attained zero velocity before reaching the top. Now, using this velocity, we can find $h_1 \; , \; h_2$, the heights where the tension becomes zero & the height where the tangential instantaneous velocity of the bob becomes zero respectively. Using kinematics equation, we find $$h_1 = \frac{v_\text{initial}^2 + Rg}{3g} \\ h_2 = \frac{v_\text{initial}^2}{2g}$$.
Now, the case you mentioned can be described as $h_1 \lt h_2$ i.e. the tension becomes zero before velocity becomes & this implies certain velocity does exist when tension becomes zero.
The gravitational force never does point inwards i.e. towards the center of the circular trajectory. The component $mg\sin \theta$ opposes the tangential velocity & remains collinear with the instantaneous velocity vector but in the opposite direction; $mg\cos \theta$, on the other hand, points outward i.e. away from the center just opposite to the centripetal force i.e. tension. You can see in the pic$^1$ how the components of gravitational force are specified; no one is acting inwards i.e. towards the center of the circle.
When tension becomes zero, velocity still exists & $mg\cos\theta$ remains perpendicular to it & is pointing inward to some point; hence it matches the configuration of circular-motion. The bob thus breaks its old circular trajectory & undergoes circular motion around a new loop where the centripetal force is $mg\cos\theta$. But it can't undergo a pure circular motion as the bob is retarded by the opposing force $mg\sin\theta$ & thus it follows a parabolic trajectory.
So, in a nut-shell, it is gravity which lets the bob to move away from the trajectory when tension becomes zero.
$^1$ Courtsey: Understanding Physics vol.I by D.C.Pandey.
Best Answer
When the string becomes slack, the centripetal force disappears. So, the particle just undergoes normal parabolic projectile motion.
To get the exact motion, using the initial velocity and angle, find the parabola of its motion. Then see when it again cuts the circle of the original motion. At this point, it will re-enter circular motion keeping the tangential component of its velocity.
Note that string becomes slack when your equations say $T\leq0$