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When a car is negotiating a bend, frictional force between tires and road provides the centripetal force. When an object attached to a string is rotating (circular motion), the stress in the string provides the centripetal force. Suppose a car is moving up hill along a circular path. In this case, what provides the centripetal force on the car along the path? Is it the difference between its own gravitational force and the normal force exerted by the hill?

p.s: as an answer to this question, some considers the difference between the gravitational force and the normal force equals centripetal force. But if the normal force is the counter force given by the road, why the sum of centripetal and gravitational force (both downward forces) not equals normal upward force?

Figure below shows the situation. enter image description here

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    $\begingroup$ Draw a free body diagram and all should then be clear to you. $\endgroup$ – Farcher Oct 12 '16 at 20:53
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/252500/2451 and links therein. $\endgroup$ – Qmechanic Oct 12 '16 at 22:21
  • $\begingroup$ @Qmechanic It sounds like this question may be about a vertical circle. A diagram would be nice. $\endgroup$ – bpedit Oct 12 '16 at 22:25
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I find that an easy, never-fail way to analyze the forces acting on an object in circular motion is:

(forces toward the center) - (forces away from the center) = mv²/r.

mv²/r is the amount of center-directed force necessary to make the object, which constantly wants to move in a straight line, move in a circle instead.   It's often called the centripetal force, but that's just a label we put on the combination of identifiable forces like gravity, normal force, tension, etc., that gives us mv²/r.

Here is a force diagram (a.k.a. free body diagram) for any car-on-a-(circular)-hill scenario:

enter image description here

Notice that there are no forces labeled as centripetial force, or mv²/r.  Our equation becomes $$mg(sin\theta) - F_N = mv²/r,$$ where $\theta$ is the angle that the position vector of the car makes with the horizontal, and FN is the normal force exerted on the car by the hill, or road.

At the top of the hill, where $\theta$ is 90°, the equation reduces to $$mg - F_N = mv²/r$$

So yes, in the case of a car going over a circular hill, we see that the centripetal force is the difference between the car's own gravitational force and the normal force exerted on it by the hill, as you suggested. 

If we rearrange the equation: $$F_N = mg(sin\theta) -mv²/r$$

We see that only when v = 0 is the normal force exerted by the road on the car equal to the gravitational force on the car.  In that situation they are equal and they oppose each other, so they look like a Newton's Third Law pair.   But they are not.  Note that they act on the same object.

There are two Newton's Third Law pairs in the situation.   One pair consists of the force mg(sin$\theta$), exerted on the car by the mass of the Earth, and an equal force exerted on the mass of the Earth by the car.   The other pair consists of FN exerted on the car by the road, and an equal normal force exerted on the road by the car.  That pair gets smaller as the velocity increases, as shown by the latest equation above.  When FN gets to zero, the car leaves the road tangentially.

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  • $\begingroup$ I don't know what you mean by Fn. However the answer doesn't make sense as it is. May be you want to think it over. $\endgroup$ – kamran Oct 17 '16 at 21:22
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    $\begingroup$ I have thought about it--for fifteen years of teaching physics. That doesn't mean that I don't make mistakes, but I don't think that I have here. I did edit the post for clarity, so thanks for the feedback. What doesn't make sense to you? $\endgroup$ – D. Ennis Oct 18 '16 at 1:17
  • $\begingroup$ I don't see the difference between the two Newton's Third Law pairs as you describe them. $\endgroup$ – freecharly Oct 18 '16 at 2:00
  • $\begingroup$ One pair have a magnitude of mg sin(theta), and the other pair (the normal forces) have a magnitude mg sin(theta) - mv²/r. Note that when I say the Earth, I mean the center of mass of the Earth, not the road, or surface of the hill. I made another edit to clear that up. $\endgroup$ – D. Ennis Oct 18 '16 at 2:11
  • $\begingroup$ But then, in the first pair the force directed towards the center of the earthshould be $mg$ not $mgsin(theta)$. What would then be the other force of this pair? $\endgroup$ – freecharly Oct 18 '16 at 2:16
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I will assume you are talking about a loop (a vertical circle).

Centripetal force is the name for the total force that pulls towards the circle's centre. There will only be two forces acting, as you also point out: Normal force $n$ from the road and weight $w$ of the car. And remember that normal force is always perpendicular to the road.

  • At the bottom of the vertical circle, $n$ pushes up and $w$ down. $n$ is larger, so it overcomes the weight and the result is a small but large enough push upwards towards the circle centre; this resulting force is the centripetal force. In this case the difference between $n$ and $w$ happens to be the centripetal force.

  • If it was a loop, then at the top of the vertical circle, $n$ pushes downwards and weight also downwards. There is no doubt here that their sum (not difference) is the centripetal force.

  • At the middle point between the bottom and the top (when the car is at the far-left side of the circle), only $n$ points inwards towards the centre. $w$ does not have any component towards the centre and has no influence on the centripetal force. The centripetal force is here equal to $n$.

So no, there is not really a golden rule saying that the centripetal force must be the difference. It purely depends on their directions. The normal force changes direction all the time here, and the weight gradually has less and less or more and more influence depending on the position in the circle.

In the bottom half of the circle, the centripetal force purely consists of (a small part of) the normal force, where weight has reduced it a bit. And in the top half, the centripetal force consists of both the normal force and (a gradually larger part of) the weight. Since we are talking about a normal road, the upper half is never reached - not even close to the mid-point is reached. So in that case: centripetal force is always purely a part of the normal force only.

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    $\begingroup$ @ Steeven, the way you have understood the question is not what exactly my question is. My apology for not illustrating the situation. I will edit the question with a figure as soon as possible. $\endgroup$ – Kosala Oct 18 '16 at 16:42
  • $\begingroup$ I added a figure to elaborate the situation. $\endgroup$ – Kosala Oct 18 '16 at 18:02
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Without a picture of what exactly you mean by 'moving up hill along a circular path', it is hard to give a specific answer.

A general answer, however, starts in my opinion with discarding the term 'centripetal force' and replacing it by 'centripetal acceleration'. The latter term implicitly tells that it is not something on its own, but the result of several different forces.

Start from Newtons second law

$$\sum\vec{F} = m.\vec{a}$$

If you decompose this equation in a direction tangential (T) and perpendicular (P) to the velocity/direction of motion, you get

$$P: \sum F_{perp} = m.a_{perp}= m.a_{cp}=\pm m\frac{v^2}{R}\\ T: \sum F_{tang} = m.a_{tang}$$

This can be applied to any free body diagram:

enter image description here

Here the perpendicular part of Newtons second law (the part about the centripetal acceleration) becomes:

$$mg \sin\theta - F_N = m.a_{cp}$$

As D. Ennis states in one of his comments under his answer, $F_N$ is smaller than the perpendicular part of gravity: a part of the gravitational force is used to provide the necessary centripetal acceleration. The rest must be compensated by the normal force.

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  • $\begingroup$ Thanks Dries for your effort composing an answer. I have modified my question with a figure. However I have a problem of understanding why don't we equalize the sum of centripetal force (downward) and gravitational force (downward) to the normal force given by the road (upward force). E.g. Suppose a block that's on a table. When we apply a force to the block at its top, the normal counter force given by the table is the sum of gravitational force+ applied force. Why don't we apply the same idea here as well? where am I wrong? $\endgroup$ – Kosala Oct 18 '16 at 21:53
  • $\begingroup$ The major difference between your example and the situation on the hill, is that in the latter case there is an acceleration, while in the former there isn't. That's where the difference lies, you try to match a situation without acceleration with a situation with acceleration $\endgroup$ – Dries Oct 18 '16 at 22:11
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Reading the above answers and the pertaining comments I have the impression that there is still some confusion regarding the explanation of this problem. Here I also use an inertial frame of reference like the other answers and I use the notation of D. Ennis' answer. (In my prior answer below I used a co-rotational frame of reference.)

The centripetal force $F_{cp}=mv²/r$ in the direction of center of the circular motion is exactly the force that is required to keep the car that has the tangential speed $v$ on the circular trajectory. What provides this force? It is derived from the component of the gravitational force normal to the road (in the direction to the center of the rotation) $F_{gn}=mgsin{\theta}$. Only when the gravitational force component is larger(equal) than the required centripetal force $$F_{gn}≧F_{cp}$$ will the car follow the circular trajectory! What then causes the car to have exactly the required centripetal force $F_{cp}$? It is the counteracting (away from the rotation center) normal reaction force $F_n$ of the road on the car (preventing the penetration of the car into the road) that adjusts itself exactly to the value to produce together with $F_{gn}$ the required centripetal force: $$F_{gn}-F_n=mgsin{\theta}-F_n=mv²/r=F_{cp}$$ What happens when the inequality $F_{gn}≧F_{cp}$ is not fullfilled? Then the circular path cannot be followed, the car flies off with tangential speed $v$.

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The centripetal force is provided by the vector component of the gravitational force on the car normal to the hill surface which is in the direction of the center of the circular motion. This normal force minus the centrifugal force is equal to the normal force of the hill back on the car.

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    $\begingroup$ Thanks freecharly. But why we should talk about centrifugal force, which is factious? $\endgroup$ – Kosala Oct 13 '16 at 16:50
  • $\begingroup$ Because the car in its frame of reference experiences the centrifugal force, the centripetal gravitational force component and the opposite counterforce of the road surface. When the centrifugal force becomes larger than the gravitational force the car will not any longer follow the circular road profile. $\endgroup$ – freecharly Oct 13 '16 at 17:03
  • $\begingroup$ The downvoters seem not to have grasped that this answer is given assuming the standpoint of the rotational frame of the car. $\endgroup$ – freecharly Oct 20 '16 at 15:32
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    $\begingroup$ yeah. I understand your point. But still I am confused with answers other people provided with respect to an observer on the ground. $\endgroup$ – Kosala Oct 20 '16 at 16:50
  • $\begingroup$ Did you check my alternative answer for an ground based inertial system (observer on the ground) given above? $\endgroup$ – freecharly Oct 20 '16 at 18:34

protected by Qmechanic Oct 18 '16 at 10:48

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