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Roller Coaster: Centripetal Forces

  1. Dec 10, 2006 #1
    Hey there. Say we have a roller coaster trail which is roughly circular.

    When the roller coaster reaches the top of the trail, then the only forces acting on it are the force of gravity, the centripetal force, and the normal force, which are all acting downwards.

    Thus, the net force acting on the roller coaster is also downwards, then why doesn't the roller coaster fall down?

    Can anyone please tell me what's going on?
     
  2. jcsd
  3. Dec 10, 2006 #2

    arildno

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    Sure, it falls down, but not quickly enough, so the roller coaster stays on track.
    Remember that the track as well curves downwards.
     
  4. Dec 10, 2006 #3
    OK. But if all the forces are acting downwards, how do we calculate the maximum speed which is needed for the roller coaster to prevent it from falling down?

    In linear motion, I used to set the net force to zero (for equilibrium) and then solve for whatever variable. But I donno what to do here since the net force is non-zero...
     
  5. Dec 10, 2006 #4
    This problem is probably much easier to do using the concept conservation of energy.

    The reason I said "probably" and not "definitely" is just in case you have to account for friction, in which case it becomes slightly more complicated.
     
  6. Dec 10, 2006 #5
    I know that it is easier to do by using the conservation of energy but I really wanna know what is happening in terms of forces.

    BTW, I am not really worried about friction.
     
  7. Dec 10, 2006 #6

    arildno

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    The forces must provide the CENTRIPETAL acceleration, which is given by:
    [tex]\frac{v^{2}}{R}[/tex]
    where v is the speed of the roller coaster, and R the track radius.
     
    Last edited: Dec 10, 2006
  8. Dec 10, 2006 #7

    rcgldr

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    I think the minimum speed at the top is the speed where there is zero centripetal force, and the roller coaster is in free fall for the instant it's at the top of the loop, but quickly expriences centripetal force from the curving track.

    In real life, modern roller coasters have wheels on the far side of the track to prevent them from falling off.
     
  9. Dec 10, 2006 #8

    Doc Al

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    Only two forces act: gravity and the normal force. Their combined force net force is the centripetal force. ("Centripetal" just means "toward the center".)
     
  10. Dec 10, 2006 #9

    Doc Al

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    The minimum speed at the top happens when the normal force (not the centripetal force) is zero.
     
  11. Dec 10, 2006 #10
    Thanks Doc Al. So if the net force is the centripetal force then

    [tex]F_N + F_g = m\frac{v^2}{R}[/tex]

    In this sense, it doesn't seem like it is a fundamental "force" like force of gravity or electromagnetic force. Is this right?

    Also, in this sense there would only be two kinds of "real" forces that is causing the actual motion: cetripetal force (one that only changes the direction) and a -- newly invented -- "linear" force (one that only changes the magnitude). Am I making sense?
     
  12. Dec 10, 2006 #11

    Doc Al

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    Right!

    Centripetal force is no more a type of force than is "vertical force". "Centripetal" refers to the direction that the force acts, it is not any new kind of force, much less a fundamental force. In others words: Something in circular motion is accelerating towards the center, thus is has a centripetal acceleration. A centripetal acceleration requires a centripetal force. As far as what real force provides that acceleration, that depends on the situation. (In this particular problem, gravity + normal force provides the centripetal force. In another problem, maybe tension in a string provides a centripetal force.)

    I think I understand what you are saying. Let me rephrase it. Again, we are only speaking of the direction of the force. The centripetal force acts towards the center--if that were the only force, then only the direction of motion would change, since the force is always perpendicular to the velocity. But if the speed is also changing at any point, there is also a tangential force acting in the direction of motion.

    Another example may help: Consider a race car driving on a circular track. Since he's going in a circle, he must have centripetal acceleration and a centripetal force; if he's also stepping on the gas (speeding up) he will also have a tangential acceleration and thus a tangential force. (What real force provides the centripetal and tangential forces in this case? Friction!)
     
  13. Dec 11, 2006 #12
    Thanks Doc Al. Your explanations have helped me a lot.

    I have one more question though. If the track is circular, then is the normal force going to be zero when the coaster is anywhere in the top half of the circle and non-zero when it is anywhere on the bottom half?
     
  14. Dec 11, 2006 #13

    Doc Al

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    No, the normal force will be maximum at the very bottom, minimum at the very top. But not zero "anywhere in the top half". If the coaster is moving at the lowest possible speed at which it can still maintain contact at the top, that means the normal force is zero at the very top--but non-zero elsewhere.
     
  15. Dec 11, 2006 #14
    I'm sure it's given by a factor multiplied by [tex]1-\cos(\theta)[/tex]
     
  16. Dec 11, 2006 #15
    Really!? What provides that force (not the centrifugal force I presume because it doesn't exist)?

    Isn't the normal force exist because of the force of gravity (the force of gravity pushes the roller coaster which in turn pushes the track and the track then pushes back the roller coaster and hence producing a normal force)? And at the very top, there is no force pushing the coaster up and thus no foce pushing the track and hence the normal force is zero.

    What's wrong with my analysis?
     
    Last edited: Dec 11, 2006
  17. Dec 11, 2006 #16

    Doc Al

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    Absolutely. (For the minimum speed case.)

    The normal force is the force that the track exerts on the coaster, so it's the track that exerts the normal force. But I suspect what you really want to know is something like: What makes the track exert such a normal force?

    Think of it like this: We are forcing a high-speed coaster to move along a curved track. We know this requires a centripetal force. Unless something pushes it, forcing it to turn, the coaster will go straight (Newton's 1st law). The track will not let that happen--you try to blast straight through it, and it pushes you aside. That's the normal force. Sometimes gravity is able to provide the full centripetal force (for example, at the top of the loop in the minimum speed case), in which case the track does not have to strain to turn the coaster velocity in a circle.

    Imagine the coaster entering that loop with a much faster speed than the minimum. Now at the top of the loop the normal force is not zero, since gravity is not enough to provide the needed centripetal force. Get the coaster going faster and faster... at some point the structure of the track will not be strong enough to create enough normal force--the track falls apart and the coaster goes flying.

    The normal force is a "passive" force. It will be whatever it needs to be, up to the limits of the strength of the material (and support structure).

    Here's another example of a passive force providing the needed centripetal force: Imagine a rock tied to a string being twirled in a vertical circle (like the roller coaster track). The tension in the string now plays the role that the normal force did in the roller coaster example--it will be whatever it needs to be to keep from being pulled apart. And in the process of resisting being torn apart, it provides the needed centripetal force. But if you spin the rock too fast, the string will not have the strength to keep itself together--it will break and the rock will go flying. (Note that in this case there is again a minimum speed that will keep the string taut as it goes around the top of the loop. The very same force equations apply, replacing "normal force" with string tension.)

    Let me know if this helps.
     
  18. Dec 18, 2006 #17
    Thanks again Doc Al. And yes, your explanations were very helpful. I understand the concept of centripetal acceleration a lot better than before. :smile:

    However, I still don't exactly get the origin of the normal force exerted by the track. From you post, I infer that it is the inertia of the coaster (i.e. its tendency to go in a straight line) that *causes* the coaster to push on the track which then pushes the roller coaster back by exerting a normal force. But this whole idea of a normal force emerging from inertia sounds weird to me. Shouldn't there be a corresponding force that is equal and opposite to the normal force that the coaster is exerting on the track (by Newton's third law)?


    P.S. Sorry for the late response. :redface:
     
  19. Dec 18, 2006 #18

    Doc Al

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    Rest assured that Newton's 3rd law still holds. If the track exerts a force on the coaster, the coaster exerts an equal and opposite force on the track.
     
  20. Dec 20, 2006 #19
    Whew!...I am relieved that Newton's 3rd law still holds. Anyways, I am still a little worried about the whole inertia and normal force thing, but I guess it'll make more sense one I do a variety of different problems. :rolleyes:

    Thanks again for your hep Doc Al. :smile:
     
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