I saw it from the book which talk about centripetal force in circular motion

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The main idea here is the balance in the term of relation between velocity and centripetal acceleration - centripetal force to maintain the uniform circular motion. However what I don't understand here is the normal force component which have the same direction with the gravitational force. The normal force exert on Diavolo from the top but according to Newton's 3rd law there must be a force exert on the loop to make the action-reaction pair. As I consider there is nothing from Diavolo exert on the loop so what make the normal force here

In another case, not on top of the loop but in the place of the red dot

enter image description here

In this case, there is nothing to nullify the effect of gravitational force and combining it with the centripetal force that will make the net force vector not perpendicular with the velocity vector so it will not the uniform circular motion. So how Diavolo can keep the velocity constant but not slowing down. Assumed that he reached the required velocity to no falling down from the loop at the first place so he don't need to use any force to accelerate his speed

Sorry for the ridiculous title, pls change it to the more suitable one

  • $\begingroup$ Your title is fine. $\endgroup$
    – BMS
    Oct 1, 2014 at 16:10
  • $\begingroup$ Are you following Flying Circus of physics by Jearl Walker or Resnick, Halliday's Principles of Physics?? $\endgroup$
    – user36790
    Dec 10, 2014 at 8:05

1 Answer 1


Diavolo does exert a force on the loop by Newton's 3rd law, just as you said. The reason this force isn't shown on the free body diagram is that only forces on the object of interest are shown, not forces by the object. Diavolo is the object of interest, so we don't include forces that he exerts on other things.

This is all justified because only forces acting on an object cause that object's motion to change. Forces exerted by objects don't directly affect their motion (except by 3rd law).

As for the side of the loop, I'm sure Diavolo did gain speed.

  • $\begingroup$ normaly, when Diavolo stand on the ground the gravitational force erxert on him and so on his weight exert the force on the ground but when he is on the loop how can his weight exert a force on the loop? As the side of the loop, I'm talking about the uniform circular motion which have only one force which is perpendicular to the velocity vector but in this case the force here is not perpendicular to it $\endgroup$
    – aukxn
    Oct 1, 2014 at 16:23
  • $\begingroup$ @aukxn: It's not his weight that exerts that force but the acceleration of his mass. $\endgroup$
    – CuriousOne
    Oct 1, 2014 at 16:27

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