Why does particle leave circular motion after string slacks?

Question

If a particle is attached to a string and made to move in a vertical circle with initial velocity of $\sqrt{4gl}$ $m/s$ where l is the length of string, at some angle (approx $131°$ with the initial position), the string slacks and the particle leaves the circular path and undergoes projectile motion. Why does this phenomenon occur even though the component of weight can provide centripetal acceleration?

If we throw the particle at $\sqrt{2gl}$ $m/s$ , the velocity and tension both become 0 when the string is in horizontal direction. How can we know whether the particle will now oscillate or leave the circular path?

Answer 1

Gravity governs projectile motion, so it is still accelerating the object downwards. The velocity at the time of exit is that it had at some point in the circle, which, instantaneously, was tangent to the circle at that point. That is the direction it moves off in, but gravity will accelerate it downwards.

In circular motion the velocity and acceleration are linked by a formula, derivable by twice differentiating the position vector to get acceleration. The centripetal acceleration is that component of acceleration perpendicular to the velocity, which (is the only one that) changes the velocity's direction. Gravity provides exclusively this kind of force at the top of the circle, when the velocity is horizontal, but as the velocity changes due to gravitational acceleration, or at some other point in the circle, the gravitational force has a smaller and smaller component perpendicular to the velocity.

When the particle attains a vertical velocity gravity has no component perpendicular to this and cannot accelerate it to turn any more; alternatively, the object might reach land first. The acceleration is thus sufficient to make a parabolic arc, but not circular motion.

So it is because gravity does not act perpendicular to velocity at every point, just downwards, that it cannot create circular motion.

Answer 2

Why does this phenomenon occur even though the component of weight can provide centripetal acceleration?

It occurs precisely because the weight can provide centripetal acceleration. More to the point, the weight provides too much centripetal acceleration.

The force from the string is naturally regulated by the fact that it comes from the string's tension. A tension force in a non-stretched object is automatically reduced to the weakest possible force that is still strong enough to prevent the object from stretching or breaking. If the force required to maintain circular motion decreases, the force exerted by the string decreases to match.

The acceleration from gravity, however, is constant regardless of motion. At the point in the circle that you noted, the particle's speed is greatly reduced from its initial speed because gravity has been slowing it down on its upward path. That reduction in speed reduces the required centripetal acceleration to maintain a circular path. The string decreases the tension force it exerts accordingly, but the force from the string cannot decrease below zero. At the point you noted, the string's force reaches zero, but the acceleration required for circular motion continues to decrease - and that requirement is for a precise match, not an upper or lower limit.

Beyond that point, the unchanging acceleration of gravity is too strong to match the greatly diminished amount needed for circular motion, and the string would have to instead be a rigid rod to exert the force that would compensate for the difference.

Answer 3

Following is not a full answer but a comment to help OP figure out the answer on their own.

Gravity does not provide centripetal acceleration to the particle. This is because centripetal acceleration always points towards a definite point while gravity always points towards a fixed direction.

Answer 4

Answering the second part first: While the force of gravity provides exactly the required centripetal force at the highest point, the force of gravity does not change direction as required for the centripetal force. The centripetal force has a horizontal component everywhere except for the highest and lowest points, and that horizontal component is required to force the weight into a horizontal oscillation. Because of the lacking horizontal force after the string has been released, the weight does not slow down horizontally, widening its trajectory into an inverted parabola.

Concerning the first part, note that the force of gravity is not in the direction of the string during the entire upswing. So, while gravity does supply increasing amounts of centripetal force (only after the point of vertical motion has been passed), it also supplies a braking force that slows the circular motion down. A the point where the string slacks, the circular motion becomes so slow that requires less centripetal force than gravity provides, and gravity is still slowing the weight down. As such, the radius of the actual curve grows every smaller as the weight becomes even slower. Until the weight moves horizontally, gravity switches from braking upward motion to accelerating downward motion, and the radius rises again so that the weight follows an inverted parabola trajectory.

Answer 5

Because tension and weight works in tandem to break the circle this time, and tension is not supported by any of centripetal forces, so after some time circle collapses :