Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Imagine you are on a theme park ride in which you sit in a car and are spun around in a circle, basically like a giant centrifuge.

An observer from the outside would say that there is no centrifugal force acting on the person in the car, only centripetal. However, for the person in the car, there is the force pulling them inwards (centripetal), and its reaction force pushing outwards against the side of the car (centrifugal). However, from this point of observation, the person in the car is stationary and everything else is moving around him.

Does this not mean that there is no circular motion in this frame of reference, and therefore the reaction force is not centrifugal, as there is no circle centre? Basically, I am confused as to why a fictitious centrifugal force will show up in a frame of reference where circular motion is not occurring.

share|cite|improve this question
The answer to that question refers to the forces as centripetal/centrifugal. What I am asking is why are the forces called this? 'Centrifugal' force implies a force acting away from the circle centre, yet in this frame of reference there is no circular motion and no circle centre, but it (the reaction force) is still called centrifugal force. Why is this so? – Dylan Cleaver Apr 22 '14 at 15:50
Your comment here appears to be entirely different from the question itself, then. – Kyle Kanos Apr 22 '14 at 15:52
The reason circular motion is not occurring in that frame of reference is because the fictitious centrifugal force was introduced, otherwise there would just be centripetal force and there would be circular motion. If it still doesn't make sense, that's good. This is meant to show that classical newtonian physics does not always apply to a frame undergoing constant acceleration – Jim Apr 22 '14 at 23:50

In the frame of the car it is not useful to talk about centripetal force. In the rotating frame, you have two forces: the centrifugal force, and the real force of the of the side of the car pushing against you as the centrifugal force accelerates you toward the outside. Note carefully that this force is not a reaction force to the centrifugal force.

You say that the centrifugal force is the reaction force to centripetal force. This is not true. In the stationary frame you have centripetal force, but no centrifugal. In the rotating frame you have centrifugal but no centripetal. In neither frame do the two forces exist at the same time, so they can't be reciprocal. Further, centrifugal force has no "agent". Forces occur as a result of an interaction between two systems. One we call the "object" or "system" or "system in question", the other is the agent. For gravity, I am the object, the earth is the agent. Of course Newton 3 allows us to say that the earth is the object and I am the agent, and the forces in the two cases are equal and opposite. Centrifugal force has no agent. With that observation, it's clear that it is not a real force. It does not arise from the interaction of two objects. There is no reaction force to centrifugal force.

share|cite|improve this answer
The real force pushing against you is centripetal as it is directed towards the center of rotation. I agree with you about the rest of your answer. – Valter Moretti Apr 22 '14 at 19:43
I see what you are saying. In the stationary frame, the force of the car on your body is a centripetal force, so you can stop there: it's a centripetal force. I would say two things: 1. In the frame of the car, the car is not rotating, so there can be no centripetal (center-pointing) force. The force of the car on me is the wall impeding the acceleration due to the centrifugal force. 2. I wouldn't classify "centripetal" as an intrinsic type of force. In either frame, the force is a normal force, whose origin is in the compression of chemical bonds. – garyp Apr 22 '14 at 21:14
Ok I see your point. – Valter Moretti Apr 23 '14 at 7:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.