What happens when the string slackens during vertical circular motion? 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.
 A: 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 $T\leq0$
A: As soon as particle leaves circular movement you have movement with constant gravitational acceleration - that is particle will continue with parabolic movement with the initial speed dependent on the moment when particle leaves circular movement.
P.S. Your question is not trivial at all.  For example, let's suppose that we measure angle $\theta$ from the bottom of the circle, and that $v(\theta=0) = v_0$.  And suppose that the angle at which particle either stops or leaves circular movement is $\theta_0$.
For certain speeds $v_0^2 \le 2 g l$ you will have simple (but not necessarily harmonic) oscillation and $\theta_0 \le 90^{\circ}$.  Particle never leaves circle and $v(\theta_0) = 0$.
For larger speeds you have movement which is by no means simple.  Particle leaves circle while still having non-zero velocity $v(\theta_0) > 0$ and continues with parabolic movement.  When particle returns to the circle it necessarily loses some kinetic energy, so $v_0$ in the next cycle shall be smaller.  $v_0$ keeps getting smaller each cycle not until $\frac{1}{2} v_0^2 \le g l$ and particle finish in the oscillation (which theoretically continues to infinity).
P.P.S. In order to stay on the circle, force of the string (centripetal force) must be non-zero.  In the lower half the circle, this is ensured by the fact that force of string opposes gravitational force, even if particle is at rest.  In upper half of the circle force of string cannot oppose gravitational force, so it can only exists, if it is necessary to create centripetal force needed for circulation.  And centripetal force exist only if velocity is non-zero.
