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In this problem, a guy is moving along this circular track at constant speed. When he is at B, the centre of motion is 'downwards', isn't it? So there must be centripetal force on the cyclist that is pulling him downwards towards that centre 'O'. But, all the solutions of this problem say that the centripetal force on cyclist is 'upwards'.

I'm confused because, my current notion says centripetal force acts towards the centre and hence it should be acting downwards when cyclist is at B. Centrifugal force is the one that acts outwards, but it's from the rotating frame, and in this situation I'm sitting in inertial frame, observing this cyclist.

All in all, in which direction is $mv^2/r$ acting? Downwards (red arrow) or Upwards (green arrow)? If it is green arrow then why is this centripetal force acting 'away' from centre?

Here's the problem (I'm only concerned with part (a) of the question.)

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    $\begingroup$ What solution? The centripetal force is a center seeking source. There is no solution where this is not the case $\endgroup$
    – user310742
    Feb 20, 2023 at 4:51
  • $\begingroup$ @Obama2020 I have added link to the problem and its solution, please have a look. Thanks. $\endgroup$ Feb 20, 2023 at 6:40
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    $\begingroup$ Please don't use external links that could break in the future. Provide the gist of the problem here. $\endgroup$
    – Miyase
    Feb 20, 2023 at 7:47
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    $\begingroup$ Unfortunately some students seem to think that "Centripetal Force" is like "Centrifugal Force", but in the opposite direction (inward instead of outward). They have not been asked (or not often enough) "what real force is pushing or pulling towards the centre?" $\endgroup$
    – Peter
    Feb 20, 2023 at 11:14
  • $\begingroup$ I find this question interesting because I can't understand what the confusion is. Part (a) asks for the magnitude normal force. Also, I don't see where, in the solution, it is said that the centripetal force is pointing up, unless you are referring to point D. $\endgroup$
    – garyp
    Feb 20, 2023 at 12:26

5 Answers 5

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Suggest you first put to one side the terms 'centrifugal' and 'centripetal' and just concentrate on Newton's second law, which in this situation can be written $$ {\bf f} = m {\bf a}. $$ The total force on the bicycle is acting in the direction in which the bicycle is accelerating. If the bicycle is following a circular path at constant speed then its acceleration vector is pointing towards the centre of the circle. Therefore the net force on it is also pointing in that direction.

(The diagram is rather odd-looking. It suggests the situation is one of a bicycle going over a circular bump. If that is what it is showing then gravity is acting downwards, the normal reaction from the road is acting upwards, and during this part of the ride these two forces are not exactly balanced, resulting in the downwards acceleration of the bicycle.)

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The question asks you to compute the normal force of the road onto the cycle, which points upwards. The centripetal force, which points downward, therefore has a negative sign in that computation.

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  • $\begingroup$ True. But watch out: the way that this is phrased could be interpreted to say that the centripetal force is a real, mechanical force between two objects. $\endgroup$
    – garyp
    Feb 20, 2023 at 12:20
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Downwards. Centripetal force isn't some new force, here it's provided by one of the forces you have already studied: gravity. The equation on the website can be modified to:

$mg=mv^2/r+N$

As you can see, gravity both balances the normal reaction and provides the centripetal force.

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I think using the term centripetal force can be the cause of some confusion.
At position $B$ are two forces acting of the rider, the downward gravitational attractive force of the Earth, $mg$, and the upward normal contact force due to the ground.
The force $mg-N$ downwards and is the net (do not use the word centripetal) force on the cyclist and that causes a centripetal acceleration $v^2/R$ downwards.
Applying Newton's second law, $F=ma$ with forces on one side and accelerations on the other leads to the equation of motion $mg-N = m\cdot v^2/r$.

The cyclist's frame of reference is a non-inertial one (it is accelerating) and so without doing anything else Newton's laws of motion "do not work" / "cannot be used".
The reason for that is in his own frame of reference the cyclist is not acceleration and yet there is a force $mg-N$ downwards acting, thus $mg-N\ne m\cdot 0$.
In order to allow the use of Newton's laws in an accelerating frame an extra "fictitious" force is added which is often called the centrifugal force.
In this case the force is $m\cdot v^2/R$ upwards and with this fictitious force included Newton's second law "works" as $mg-N-mv^2/R = m\cdot 0$, which you will note, mathematically is the same equation as $mg-N = m\cdot v^2/R$.

The given solution "At B, $mg-mv^2/R =N$" is set out that way for the sake of brevity and comes from the rearrangement of the application of Newton's second law $mg-N = m\cdot v^2/R$

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By right the green arrow should be the correct answer, by right the centripetal force would be pointing inwards to the center (the red arrow), the centrifugal force would be pointing outwards (the green arrow), mv^2/r, which is the centrifugal force, should be pointing outwards, which means the answer is in some sense correct, except that mv^2/r is the centrifugal and not centripetal force.

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