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By Newton's 3rd Law, every action has an opposite and equal reaction. When the object is at the point depicted in the circle, we know for sure that gravity is pointing straight down. Then, if a normal force existed at that point, it would be completely perpendicular to gravity. What force is then the opposite reaction to this normal force?

It can't be gravity as gravity is completely perpendicular to normal force at that point. Is the opposite force then centrifugal force? But I thought centrifugal force was a fictitious force?

I would greatly appreciate any help.

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    $\begingroup$ Ask yourself this, what force will keep the object moving in circular motion? That is which force is required to yield the appropriate centripetal force here? $\endgroup$
    – Triatticus
    Commented Mar 31, 2020 at 18:50
  • $\begingroup$ Here is something on the difference between centripetal and centrifugal forces. physics.stackexchange.com/q/93599/37364 $\endgroup$
    – mmesser314
    Commented Mar 31, 2020 at 19:13
  • $\begingroup$ You are right - gravity is pulling downward on that point. But that isn't the only force on that point. It is part of a rigid object, so other points apply forces that keep it from just falling to the floor. If the point is moving at uniform speed in a circle, the total force on the point must be a centripetal force, toward the center of the circle. $\endgroup$
    – mmesser314
    Commented Mar 31, 2020 at 19:16

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If a particle is constrained in circular motion by a circular wall then there is a normal force accelerating it inwards. This is called a centripetal force. The equal and opposite force is the force which it exerts on the wall.

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  • $\begingroup$ The title says that the object is travelling in a circle. $\endgroup$ Commented Mar 31, 2020 at 18:55
  • $\begingroup$ Oh, yes, that went off the screen when I read the rest of the question. Oops. $\endgroup$ Commented Mar 31, 2020 at 19:03
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    $\begingroup$ Hi, Thank you so much for your answer. Just to confirm, then the opposite and reaction force to the normal force is just the particle pushing against the walls of the circle? $\endgroup$
    – CuriousCat
    Commented Mar 31, 2020 at 20:24
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An object travelling in a circle of radius $r$ with angular velocity $\omega$ (expressed in radians per second) is under a constant acceleration $a$ towards the centre equal to $\omega^2 r$ (the acceleration can also be expressed as $v^2/r$ where $v$ is the speed of the rim). According to Newton's $F=ma$, this means there must be an inward force, which we call centripetal force. For an object of mass $m$, this force is obviously $m\omega^2 r$.

There is no force opposing centripetal force. If it were to stop, the object would simply stop accelerating and travel in a straight line instead. (This is often the hardest part for newcomers to grasp, but if $F=0$ then, according to Newton, $a=0$ too).

At the point in the diagram, the centripetal force is acting sideways. If it were to stop, the object would continue in its current direction, under the influence of gravity alone.

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  • $\begingroup$ You should indicate where the centripetal force is coming from, in this situation it is the normal force by the surface of course. Remember that centripetal force doesn't just happen, it is the name of the net inward directed force caused by other forces $\endgroup$
    – Triatticus
    Commented Mar 31, 2020 at 18:52
  • $\begingroup$ @Triatticus We are not told what is exerting the centripetal force. This does not matter, as the reason for the circular motion is irrelevant to the question. For all we know it could be a rocket engine firing outwards; I have seen such things on Fireworks Night! $\endgroup$ Commented Mar 31, 2020 at 18:57
  • $\begingroup$ We are told it is a normal force, which by definition means a force normal to a surface. The equal and opposite reaction force described by Newton's third law is exerted by the particle, it is not to be confused with opposing forces in equilibrium. $\endgroup$ Commented Mar 31, 2020 at 19:10
  • $\begingroup$ What I mean is you should be clear that the centripetal force is the net result of other forces, if it's a rocket firing outwards so be it, it doesn't matter. But OP mentions the normal force and as the above comment mentions that is due to a surface. $\endgroup$
    – Triatticus
    Commented Apr 1, 2020 at 2:05
  • $\begingroup$ @Triatticus which relevant comment mentions what surface? $\endgroup$ Commented Apr 2, 2020 at 12:10

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