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Homework Help: Circular Motion: What's the Source of Centripetal force in this?

  1. Sep 17, 2016 #1
    [Moderator's Note: Thread moved from forum General Physics hence no formatting template]

    I am trying to study Circular Motion for my exams and I'm kind of unsure about one question. The question asks what's keeping the truck in circular motion. It has to be gravity I know, but gravity being directed towards the center, shouldn't that just result in the truck falling? What keeps it INTACT to the track? I am giong to quote the actual question now.

    "Figure 18.17 shows part of the track of a roller-coaster ride in which a truck loops the loop. When the truck is at the position shown there is no reaction force between the wheels of the truck and the track. The diameter of the loop in the track is 8.0 m.

    a) Explain what provides the centripetal force to keep the truck moving in a circle.
    b) Given that the acceleration due to gravity g is 9.8 m s-2, calculate the speed of the truck.

    A-LEVEL4.png "
     
  2. jcsd
  3. Sep 17, 2016 #2

    andrewkirk

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    Why gravity?

    What keeps the truck moving in a circle when it's at the 3 o'clock or 9 o'clock position? What about at the 5 and 7 o'clock positions? What other force(s) are acting on the truck - in particular, pushing it towards the centre of the circle (which is what centripetal means)?
     
  4. Sep 17, 2016 #3

    PeroK

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    It is gravity and, as you conclude, at this point (for an instant) the truck is in free fall. To see what's happening, imagine the track was only the right half of the loop, so that after the highest point, there is no track to the left. What would happen to the truck?

    Hint: look at question b). Note that question b) refers to the speed of the truck at the highest point.
     
  5. Sep 17, 2016 #4

    CWatters

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    I think this is sufficiently close to homework that it should be moved to that section.
     
  6. Sep 17, 2016 #5
    I assume, the truck will go straight for a while because of inertia. So how exactly am I going to put this into words?
    "It is gravity that keeps the truck in it's circular motion and inertia that drives it away from the circle. The net sum of these two forces is what keeps the truck in circular motion"
     
  7. Sep 17, 2016 #6

    PeroK

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    Inertia is not a force. The truck won't go straight for a while, it will fall in a parabola. Try drawing parabolas for the truck:

    a) If it's still moving fast at the top of the loop.

    b) If it's moving "too slowly" at the top of the loop.

    We're still imagining there is no track on the left here, so the truck is falling through the air.
     
  8. Sep 17, 2016 #7

    CWatters

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    Centripetal force is the net force. What two forces sum to make it? Try drawing a free body diagram to the car.
     
  9. Sep 17, 2016 #8

    PeroK

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    The problem here is that the question deals with a special situation at the top of the loop where there is no normal force: this is an assumption in the question. There is only gravity acting on the truck at this point. The conundrum is why the truck doesn't fall off the track at this point.
     
  10. Sep 17, 2016 #9

    CWatters

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    Fair enough but I think it helps to understand the general principles first. It's useful to think of the centripetal force as the force required for an object to move in a circle and then consider what must happen to change the radius of motion (not to be confused with the radius of the track).

    Perhaps we should let the op reply to your question.
     
  11. Sep 17, 2016 #10
    At that point the truck is running parallel to the track. Its velocity is purely horizontal, as is the track(d(track)=0) ;there's no net force holding it there, nor one pushing it apart. It is neither rising, nor falling. Just like a tossed ball, at the apex, it's "weightless".
     
  12. Sep 17, 2016 #11
    The definitions of both centripetal and centrifugal tend to be a little vague, but yes, gravity is providing a positive centripetal force on the upside down bit, and a negative centripetal force on the rightside up bits.
     
  13. Sep 17, 2016 #12
    @PeroK If I imagine the two cases in both of them the horizontal motion will be unaffected by gravity (i am assuming there is no air resistance), the truck will vertically accelerate downwards at the same time until it hits the ground. In case a) where it is moving too fast the point x-intercept of the parabola will be far away from the track and in the latter case it will be closer to the tracks. I know that there is supposed to be a case where it will travel in a circle but i can't fathom how that would be the case and why it would not just get out of the orbit. We are not talking about geo-stationary satallites who are in orbit because of the curvature of earth falling under their path.


    @hmmm27: So the answer to question 'a' would be gravity, Right?
     
  14. Sep 17, 2016 #13

    PeroK

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    This isn't helping.
    You're over-thinking this a little. If, at the top of the loop, the truck has sufficient speed, then it will fly off in a parabola outside where the circular track would be. So, when you add the track, the truck is forced into the circular path.

    If, however, the truck is moving too slowly, the parabola will lie inside the circle. In which case, when you add the track, the truck will leave the track until at some point it crashes back into it.

    That's one explanantion why there is a minimum speed required at the highest point. If the truck isn't moving fast enough, then it does fall off the track.
     
  15. Sep 17, 2016 #14
    Oh, I wouldn't say that if I was getting marked on it, since gravity at no time is actually responsible for keeping it in the curve. It's not the Moon being kept from flying off by the Earth's gravity.

    As far as visiualization is concerned, the (half) parabola's pretty good. Say you have a raised horizontal track that joins to a quarter circle track which plunges into the ground. If you push the cart fast enough that it flies off the end without actually touching the quarter circle, then that's a speed you want to be at to make it through the top of the loop in the diagram.
     
  16. Sep 17, 2016 #15

    PeroK

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    The more I look at this, the less I like question a).

    There is only one point on the track (at the top) where gravity is centripetal. Hence, gravity alone acts only instantaneously. There is no finite period of time where gravity alone acts, so you cannot really say that gravity causes anything to move in a circle (unless it's an orbit, which is a very different scenario). You can only describe motion as being in a circle over a finite period of time. All motion is instantaneously in a single direction.

    My example of the missing track proves this point. The motion is the same at the top of the track whether the track is missing on the left or not. If it's missing, the truck follows a parabola; and if the track is present, it moves in a circle. So, you cannot say what shape it's moving in at the top.

    Anyway, the important point is to understand the physics. It is possible, if you get the speed of the truck just right, for there to be no normal force (instantaneously) at the top. Any faster and there is a normal force; any slower and the truck will leave the track. That's the key point here.

    The expected answer to a) is gravity, but the wording of the question is perhaps dubious.
     
  17. Sep 17, 2016 #16
    I doubt the expected answer would be gravity, since gravity never actually helps keep the cart in the track.
     
  18. Sep 17, 2016 #17

    PeroK

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    Then you'd be wrong! Gravity is a centripetal force at the single point at the top of the loop. In this case, it is the only force at that point.
     
  19. Sep 17, 2016 #18
    I said I doubted it would be the expected answer. [edit: I am allowed to change my mind, right ? yes, I imagine it would be the expected answer : I misread the problem]

    If you're going to define "centripetal force" as simply something that works against inertia, then a vector component of gravity applies to the entire upside-down bit. This could be demonstrated as an increasingly flimsy track that would break apart if the full "centrifugal force" wasn't compensated for.
     
    Last edited: Sep 17, 2016
  20. Sep 17, 2016 #19

    PeroK

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    This is a homework thread, in which we are trying to help the OP.

    If you want to ask questions about what is and is not a centripetal force, you need to start your own thread. Leave this one for the OP's questions.
     
  21. Sep 17, 2016 #20
    right, sorry, got sidetracked - given the simplicity of the 'b' problem, I figured 'a' would be just a general question and didn't read it through properly; answer edited.

    Okay, forget parabolas for a minute.

    The truck sticks to the track because of the normal force; in space, the normal force would only be created by the truck running into the curvature and getting squished into the track with a centripetal acceleration ##a=v^2/r##. Give the slightest push to the truck and it will go 'round and 'round the track forever. But, it's not an orbit in the sense of gravitational pull, as you point out. (As an aside, don't dwell on it, the track could be elliptical, or it could be some massive 3D assembly of twisty inside curves)

    On Earth, however we have gravity. It's the most common source of the "normal force", eg. between the person and the chair he or she's sitting on. If you bring the track down to Earth then gravity will affect whatever's put on it, as it does everything, including the normal force of truck on track.

    In order to stick to the track, the normal force has to be > 0. The total force perpendicular to the tangent - rotational force plus gravitational component - must be positive. In terms of acceleration ##a_r + g_θ > 0## needs be achieved to keep it stuck to the track.

    For example, at the midpoints of the circle halfway up/down, gravity - a purely vertical force - has no effect on the normal force - which is purely horizontal at that point - so the rotational velocity only has to be greater than 0. At the bottom of the circle (indeed any point in the bottom semicircle), gravity will always contribute to the normal force to an extent, the most at the bottommost point.

    However at the top the full force of gravity is at play against the normal force, so ##a_r > g## is the only thing that will keep it on track.

    (As an aside, I can't help but note that if the truck had a motor and brakes, it could stop at the midpoints, before continuing. Alternatively It could even loop around at a constant angular velocity. But it's easiest just to let it zoom up to the top, slowing down all the way, then come back down, speeding up all the way.)
     
    Last edited: Sep 17, 2016
  22. Sep 18, 2016 #21
    The truck is not "falling" [edit: in the usual sense of a downward velocity] : the slope of its motion is 0: no vertical component. Furthermore, it has not and will not escape the track: even though the normal force is 0 (ie: it's not stuck to the track), since it isn't falling it isn't leaving, and the instant after that instant (pardon the semantics). the vertical component of the rotational force is greater than g, and it continues to not fall.

    Having addressed most of the OP's concerns (reposted to avoid up/down paging)

    we can see that, there is no explicitly stated connection between the diagram and problems a and b, concerning the position of the truck, however one can safely assume that b refers to the diagram.

    a on the other hand is problematic, since the wording states that there is something "providing a centripetal force to keep the truck moving in a circle" and at that point there is not - gravity is present on the centripetal vector, but it's doing squat as far as moving in a circle is concerned.
     
    Last edited: Sep 19, 2016
  23. Sep 18, 2016 #22

    haruspex

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    Depends what you mean by falling. In everyday use, it would probably be taken to imply moving downwards in free fall, but in a physics context it would be more usual just to require it to be in free fall. Throw a stone up at an angle. It is in free fall all the time until it hits the ground.
    Nonsense.
    The question states that there is no normal force at this particular point in the track, and yet the truck retains its circular trajectory. There must be a centripetal force to achieve that, and gravity is the only source.
     
  24. Sep 18, 2016 #23

    andrewkirk

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    Interestingly, if we focus solely on the question of the OP
    then I think it is correct to say that gravity plays no role in keeping the truck in circular motion. With a track like this, the only thing forcing the truck to travel in a circle is the track. The only thing gravity can do to the locus of the truck is to disrupt the circular motion, by making the truck fall downwards off the track if it's not going fast enough.

    The discussion seems to have moved from the original question to the related but different question of what is the source of the centripetal force at the top of the track. The answer to that question is 'only gravity', but that doesn't mean that without gravity the truck would not follow the circle. In fact, without gravity, the truck would be travelling faster at the top and so would have a greater normal force than ##9.8 ms^{-2}## exerted on it by the track.

    In the absence of gravity and friction, the truck would make a complete loop and then continue, without ever varying speed.

    In the absence of gravity but with friction present, the truck would continue around the loop, getting gradually slower, until it came to a stop because of friction. It would then simply remain wherever it had stopped. Whether it completed the loop before that happened would depend on the initial speed and the coefficient of friction.

    In the presence of gravity and friction, the truck starts slowing as soon as it enters the loop. If it comes to a stop before it becomes vertical, it will just roll backwards down the way it came. If it gets past there but its speed ever falls below that needed to keep it on the track, gravity will make it fall off the track. In the absence of friction, that would have to happen - if it happens at all - before it reaches the apex. With friction present, it could happen after the apex.

    Actually, as PeroK pointed out, the question doesn't actually make sense, because it is asking what is keeping the truck in circular motion at the single point that is the apex. The question is meaningless because in physics there is no such thing as an instantaneous effect, so the absence of either gravity or the track at such a dimensionless point would make no difference whatsoever to the locus of the truck.
     
  25. Sep 18, 2016 #24

    haruspex

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    As I wrote in post #22, the question is asking the student to explain how come it could remain on a circular trajectory even when there is no normal force. I.e., where is the centripetal force coming from? The answer is gravity. Sure, if you took away gravity it would still move in a circle, but then the normal force would be nonzero. The question is not asking whether gravity is necessary to produce the circular motion.
    The curve is presumed smooth, with a continuous second derivative. Therefore it has a centripetal acceleration at all points, including the apex. Your argument is similar to Zeno's arrow paradox.
     
  26. Sep 18, 2016 #25
    Nothing. The force of gravity will 'keep the truck moving' in a parabola.

    At the next instantaneous point over - where we can define the motion as being definitively circular - the normal force is reasserted.
     
    Last edited: Sep 19, 2016
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