# 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.

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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.

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I'm being very dim here, but do you guys mean parabolic motion while still being tied (via the string) to O? What I had visualised was that upon leaving circular motion, motion is downwards until the string becomes tight again, whereupon the particle re-enters circular motion, but now in the opposite direction. –  Ryan Apr 24 '12 at 14:45
Yep, you would have parabolic movement within the area of the circle. For $\theta_0$ large enough, particle would "land" back to circle in such a way that it would continue in the circulating same direction. For smaller $\theta_0$ it would change direction upon landing. Extremely complicated... In either case, particle would not just fall down, but do projectile movement with initial vertical component of speed directed upwards! Particle completely stops only for $\theta_0<90^\circ$. –  Pygmalion Apr 24 '12 at 14:49
I think I've finally understood. Thanks, Henry Higgins! However, why does the particle completely stop for $\ThetaO <90$? –  Ryan Apr 24 '12 at 15:15
This is also a tough question. I will add explanation in the answer, because it is too long. –  Pygmalion Apr 24 '12 at 15:26

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 $T\leq0$

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thanks for your concise answer. –  Ryan Apr 25 '12 at 0:27