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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.

CuriousCat
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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.

Charles Francis
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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.

Guy Inchbald
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