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An object in a circular motion will fly out tangentially when released. As per Newton's Third Law of motion, there is a reaction in every action. The moment the object is released, where is the reaction force?

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  • $\begingroup$ I think all you're asking is this: consider the hammer throw (olympic sport). Yes, at the moment the person let's go, indeed, the person is "thrown" backwards a bit, because the force they are applying (inwards) disappears. $\endgroup$
    – Fattie
    Sep 22 at 15:29

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While an object is in uniform circular motion, the force that is acting upon it is directed to the center. This force is what prevents the object moving linearly if the force suddenly vanished. If this were the case, the object will move in a (straight) line tangent to the circle with constant velocity.

Hence it would be wrong to assume that there is a force that acts tangential. That is, there would be no reaction force in that direction since as stated, you can think of the object trying to move in a straight line with no force$^1$ (by virtue of it's own inertia), in which case Newton's third law is not applicable since this law applies to forces (paired forces).

$^1$ Of course not including the force toward the center which keeps it in uniform circular motion.

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    $\begingroup$ . . . . . the only force that is acting upon it is directed to the center (centripetal force). There could be many forces acting and it is a net force towards the centre which provides the centripetal acceleration $\endgroup$
    – Farcher
    Sep 21 at 7:11
  • $\begingroup$ A fun demonstration by Tom Scott: youtube.com/watch?v=bJ_seXo-Enc $\endgroup$
    – Nayuki
    Sep 21 at 15:29
  • $\begingroup$ I believe the OP is simply asking what happens when the "rope breaks" when something is being spun around from a central point. $\endgroup$
    – Fattie
    Sep 22 at 15:29
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This is part of why centrifugal forces are often called "fictitious." Newton's laws of motion only apply in an inertial reference frame. The moment you move into accelerating reference frames you introduce "fictitious" forces. Such "forces" are better described as fixes to your coordinate system than actual forces because they don't obey Newton's second law.

Point being, from the view of an inertial frame the object that is being "flung out" is simply moving in a straight line at constant velocity. If it had been moving in circular motion before that, then something was exerting a centripetal force on it to make it do so, and it was the cessation of that force that lead to the appearance of it being flung out.

Now, all reference frames are valid, but some principles, like conservation of energy and momentum, only apply in inertial frames. Think about it: if I'm moving along at 60 km/hr on the road, in my frame the whole world is rotating underneath me in a way that has a tremendous amount of kinetic energy. When I come to a stop, though, that energy is all gone. Where did it go? The answer is: it wasn't there in the first place. To identify the things that are "real" you need to find quantities that all observers agree on. That's where things like "proper time," "proper length," and "center of mass energy" come in.

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  • $\begingroup$ Such "forces" are better described as fixes to your coordinate system . . . . They are introduced to enable one to use Newton's second law. $\endgroup$
    – Farcher
    Sep 21 at 7:09
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    $\begingroup$ @Farcher They patch up laws 1 and 2, but fail on 3 because there is no body to exert the force and experience a reaction force. $\endgroup$ Sep 21 at 7:21
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If you're wondering how momentum is conserved in this situation, consider two objects bound together by a string/force and spinning, in a vacuum, with their shared center of mass at rest (zero momentum). Both bodies orbit this common barycenter, and when the string/force connecting them is severed, they will both proceed in straight, inertial trajectories in opposite directions with momenta that sum to zero in the original rest frame of their barycenter.

As others have explained, there's no "action-reaction pair" that forces them apart in the sense you're imagining, but that doesn't spoil the conservation of momentum, which is arguably the more important meaning of Newton's 3rd law.

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Very short answer: in order to stay in circular motion, there had to be a force to keep it in such a trajectory. If the object is let go, that force must cease to exist. It's not a matter of some force (which you are asking about) changing the state of motion of the object, but a change of forces.

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