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Centripetal Force and G Forces while travelling over a hill

  1. Apr 7, 2008 #1
    [SOLVED] Centripetal Force and G Forces while travelling over a hill

    1. The problem statement, all variables and given/known data

    Hi everyone - I've had some problems wrapping my head around the ideas behind centripetal force while travelling over a hill, and, at wits end after journeying for hours through sites which are either ridiculously complex or disagree with each other, I thought I would come for help here. I've got to write a piece of coursework about the physics of a rollercoaster, and rather than follow all my classmates' examples and write something about conservation of energy, I thought I'd be adventurous and try out centripetal force.
    We've been given data about a specific rollercoaster - a rather simple one. A large initial force is provided to the cart, it goes over a steep incline (~60m), then drops the same distance very soon after. We've even been given all the calculations relevant to centripetal force while travelling on this path (including the solutions). However, I simply don't understand where some of the assumptions used in calculations are coming from, and I don't want to copy+paste them into a piece of work without actually understanding what they mean. Thus the problem arises.

    2. Relevant equations

    These are the equations for the rollercoaster that have been given. The complexity of the ride has been simplified somewhat: the incline, small section at the peak of the hill and decline have been reduced to one equation for each, presumably representing the average radius and velocity at these points. The mass of the person travelling on the ride is assumed to be 100Kg (the mass of the cart itself is about 10 tonnes).

    Calculation 1: Incline of the "hill"
    Curve radius = 35m
    Centripetal force = (mv^2)/r
    Centripetal force on 100Kg rider= 100 x 38 x 38 / 35
    Centripetal force = 4125N
    Force of 1g on a 100Kg person = 1000N
    4125N = 4.1g
    Total force on rider = 4.1g + 1.0g = 5.1g

    Calculation 2: Peak of the "hill"
    Outside curve radius 8m
    Centripetal force = (mv^2)/r
    Centripetal force = 100 x 12 x 12 / 8
    Centripetal force = 1800N
    Force of 1g on a 100Kg person = 1000N
    Resultant force 800N = 0.8g

    Calculation 3: Decline of the "hill"
    Curve radius 40m
    Centripetal force = (mv^2)/r
    Centripetal force = 100 x 35 x 35 / 40
    Centripetal force = 3062N
    Force of 1g on a 1000Kg person = 1000N
    Resultant force on rider = 3.1g + 1.0g = 4.1g

    3. The attempt at a solution

    With calculations 1 and 3, I don't understand how the weight of the rider is being added to the centripetal force to calculate the G force involved. Gravity itself would possibly have some impact on G force involved, as at some times both forces will have a vertical component downwards - however, these would be added as vectors, so it wouldn't be a constant +1g (and it wouldn't reach that either). Another possible solution is that the normal is moved because of the combination of tangential and radial acceleration (the tangential acceleration coming primarily from gravity but also from friction and air resistance). However, it was my understanding that the normal is the force that stops the track from buckling under the inward force presented by the curve, and it is this force acting on the passenger that causes the actual sensation. It wouldn't make sense for it to move. So how is mg being added to the centripetal force in order to determine the total force pulling the passenger towards the centre?

    With the second calculation, mg is subtracted from the centripetal force to give the resultant force. Once again, I am confused as to why. If the cart became inverted whilst travelling over the hill, I could see how a force would act in a direction generally away from the centre of motion in order to prevent the cart from falling, but this is not the case - the cart remains upright while travelling over it. If anything, my understanding is that the centripetal force would now be more in-line with gravity than before, and so the two effects would be added. However, seemingly not. I would really appreciate if anyone could explain where I've gone wrong in intepreting these calculations.

    I do apologise if my questions have some simple answer that I've missed, or if that this is an issue too basic for discussion on this forum - pretty much all of the other topics I've seen here make very little sense to me. However, my other attempts at gathering information have been fruitless, and I would really like to talk about something more complex than energy conservation for this work. Thanks in advance for any help I do get.
  2. jcsd
  3. Apr 7, 2008 #2


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    Welcome to PF!

    Hi Zach! Welcome to PF! :smile:

    Yes, I agree … forces are added as vectors, and the weight is not generally parallel to the centripetal force. :frown:

    I suggest you take the numbers, but do your own vector addition. :smile:
    Yes … energy will only tell them how high and how fast … centripetal force will actually answer the safety question of whether the car will stay on the track! :smile:

    btw, now you've joined PF, do a forum search (from the menu bar at the top) for "centripetal" or "rollercoaster" … you'll find lots of helpful threads on similar cases! :smile:
  4. Apr 7, 2008 #3
    Okay, some searches here, more browsing of the tubes and an hour of relaxation has lead me to a crudely formed understanding of how my problem works. I'd appreciate if anyone could correct me where I've gone wrong and fill in the gaps.

    The centripetal force is not a "real" force itself in this case, but rather the resultant force of all of the forces acting on the rollercoaster cart. The most notable of these forces are gravity and the normal - a force acting on the cart caused by its motion. Other forces which will also have an impact are air resistance and friction. The normal is what affects the perceived "g" forces acting on a person within the ride - the greater the magnitude of the force, the stronger the feeling. The direction of the force is generally irrelevant in the sensation it causes.
    For these purposes, I'll assume that the effects of air resistance and friction are negligible, and that the object is travelling over a single curve (the logic is largely the same whether it's one, three, or hundreds, since in this scenario it's still retaining its hilly shape and personality).
    The centripetal force will always be acting towards the centre of the circle which the curve lies on. Its magnitude can change as the speed of the cart decreases (and the cart will decelerate as it is not being pushed forward in any way, only being pulled back), but the point it goes towards will remain constant.
    When the cart is reaching the base of the curve, it is likely slightly below the centre of the circle. At this point, the centripetal force will be strong, as the vehicle is travelling at high speeds. The normal will also be very large - if we represent the three forces as a vector diagram, we can see this easily, as it also has to overcome the effect of gravity to provide the centripetal force as the resultant vector.
    As the cart climbs the hill, the normal will decrease for two reasons. Firstly, the centripetal force is reducing because the speed of the cart is decreasing. Secondly, the angle between the centre of the circle and the object gets closer to horizontal, then begins going above horizontal. Because of this, the force of gravity has the same direction as the vertical component of the centripetal force, so the normal force no longer has to to "undo" the effects of gravity in order to get the correct resultant force. The increased angle above the horizontal means that more and more of the centripetal force consists of a vertical (downwards) force, so gravity reduces the normal more and more as you climb.
    As the very summit of the climb, gravity is pointing in the same direction as the centripetal force. This, combined with the effect of a reduced speed (and in my specific case, the smaller radius of the curve at the top) means that a small centripetal force is largely performed by gravity, meaning that the normal can be below 1g.
    As the object once again falls down the side of the hill, it is now gaining speed thanks to gravity. The normal acting on the cart will increase again, though not to a point as large as it was before - because the drop is the same height as the climb, and some of the initial energy given to the cart has been lost to waste.

    The overall effect (in my case) is for the person to be shunted into a very high force acting on them due to a steep, sudden incline. The force goes down slowly, then is heavily reduced as the cart reaches the much flatter peak of the ride (a large velocity in the direction is also rapidly negated, causing a moment of being flung up in one's seat). Soon after, a fairly rapid acceleration and increasing forces are felt as the cart moves down the side of the hill.

    It kinda makes sense to me now, I guess, assuming that I'm correct. However, there is one gaping question: Where is the normal force that is acting on the cart and passengers to pull them towards the centre of it actually coming from? I suppose the likeliest answer is that the path of the track forces the cart to begin down it, then the motion causes the normal force to take place. After which point the track itself provides another normal (opposed to this normal) equal and opposite to the centripetal force in order to stop the ride from buckling. Is that an answer that makes sense?

    Ah well, thanks for the help, tim, and for anyone who helps me out here. I'm happy to know I wasn't completely insane when I thought there was something dodgy about the calculations iI was given.
  5. Apr 8, 2008 #4


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    Hi Zach! :smile:

    Yes, that looks fine! :smile:

    Just a few comments:

    I don't think you've dealt with the difference between hills (where the curvature is downward) and valleys (where it's upward) … even if it's not technically part of the question, it's still worth a quick mention! :smile:

    And perhaps you should put in some calculations.
    You haven't grasped how the normal force is calculated … there are two forces on the cart, and they equal its mass times its acceleration (Newton's second law).

    So the radial acceleration (times mass) is the vector sum of the normal force and the radial component of the weight: mv^2/r = mgsinθ - N.

    Since the acceleration is fixed by the speed, and the weight is fixed, that means that the normal force simply adjusts itself to fit them.

    In particular, for a hill, the normal force must always be less than 1mg (in fact, it must be less than … what … ? :smile:), or the acceleration would be upward, and the cart would leave the track! :eek:

    Finally, don't forget that there is also a linear acceleration … which could produce a substantial g-force on the sharp drop! :smile:
  6. Apr 8, 2008 #5
    I believe I understand the majority of what's going on now, which is great. However, I do have a few issues left over:

    Firstly, why is it that the passenger can "feel" the normal force acting on them, not the centripetal force? It would make more sense to me that they feel the overall effect on them, not just one of the forces.

    Secondly, while climbing and descending the hill (the climb and descent are actually a pair of "valleys", are they not?), the centre of motion will be directly above them for a significant portion of the trip. Does this not mean that the passenger is going to feel pulled upwards? Or is the sensation they feel being pushed back by the velocity?

    Isn't the cart taking off a bit of an exaggeration? For example, as the cart goes from being on the steep incline to being at the peak, it will lose a large amount of vertical velocity in a very short amount of time. It would have to be held on to the track to prevent it flying off, because there's also no track to apply a normal force to maintain the circular motion it'd like to keep. The track is applying the normal force, isn't it? Surely it has to come from somewhere? Even so, why would it do so if the normal was above 1g?

    Thanks for your help once again - I feel I understand it enough to be able to babble on about it now, which is good. I'm starting to feel a bit like a child screaming "why?" at every opportunity though. It'd just be nice to fully 'get' it. I hope I'm not being annoying >.<

    Edit: To answer my own questions - The reason passengers in the rollercoaster are pushed back in their seat is not because of the force, but rather the velocity - the cart is travelling it, and they're being held inside the cart as it moves, being dragged along with it and so pushed into the back of their seat. I think?
    And the reason we measure the normal is not because it is the force acting on them, but rather because when we talk about G-Forces, we don't include the effect of gravity in our calculations?
    Last edited: Apr 9, 2008
  7. Apr 9, 2008 #6


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    They do feel the overall effect.

    mv^2/r = mgsinθ - N: they feel the centripetal acceleration, which is mass times the combined forces.
    ah … couldn't work that out from you description.
    No … N = mgsinθ - mv^2/r … and in a valley, r is negative (compared with g), so N gets larger. On a hill, of course, r has the same sign as g, so N gets smaller.

    In other words: in a valley, the force felt gets larger, but on a hill it gets smaller … the passenger feels lighter on a hill (and may feel weightless, and leave the track, if v gets too large).
    Only if each hill is uniformly curved, and followed by a valley, or a hill of lesser curvature. If the curvature on the downward side of a hill increases, then N can become zero, and the cart will come off the track! :smile:

    Also if a valley "loops the loop" …

    (oh, and don't refer to a force of 1g … it's a force of 1mg, or an acceleration of 1g.

    Though for some reason, it's acceptable to refer to a g-force of 1g, or 2g, or … even though g-force is actually an acceleration!)

    Not at all! :smile:
    No no no … being pushed back is caused by the linear acceleration … which I warned you not to forget! :rolleyes:

    Newton's first law … velocity does not cause forces!! :smile:
    No … the only reason we measure the normal is to work out friction forces (if any), and to tell when the cart will leave the track.

    Otherwise, we're only really interested in the g-forces (including the 1g from gravity).
  8. Apr 12, 2008 #7
    Ah, sorry for my lack of a reply.

    Your response only confounds me further. The calculations that I have written down (and thus been given) end up calculating the value of the normal force. Does this mean that they were wrong in calculating this force? Or are they not calculating the normal force, but appear to be that way?

    I think this is a large stem of the confusion. The files I have been given suggest that the G-forces acting on someone are centripetal force +/- 1G. Why is this? Surely it would be just one of them?

    Thanks again for help; attached is a simplified diagram of the hill section of the ride, to prevent any further confusion when talking about the structure of the ride.

    Attached Files:

  9. Apr 14, 2008 #8


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    Hi Zach! Sorry for taking so long to reply. :redface:

    The known items are the gravitational force and the centripetal acceleration.

    The normal force is worked out from those known items: it is their difference, N = mgsinθ - mv²/r.

    So the "G-force" felt by the passengers is the ordinary 1g from gravity, plus or minus the centripetal acceleration.

    This will be the same as the normal force … so when the normal force is zero, the passengers will feel they are "floating".

    (btw, I do wish you wouldn't use "centripetal force". The v²/r is an acceleration, not a force … it is the "right-hand-side" of Newton's second law.

    The normal force is a force (obviously!), and it's centripetal, so "centripetal force" really ought to describe the normal force! :smile: )
    No, the normal usage is g ± centripetal aceleration.
    Are the "joins" vertical? If so, the car will leave the track at the uphill join … there's no reason for it do to anything other than go straight up, is there? :rolleyes:
  10. Apr 17, 2008 #9
    Alright, I've read some more, thought it over, and I think I finally get it now. Thanks for all of your help ;p. I feel a bit dense how I kept getting it wrong, but I'm confident with my idea now. Thanks again =) I have no idea how to set the thread title to solved, though. Or do I just edit it?

    EDIT: Ah, magic. Ta.
    Last edited: Apr 17, 2008
  11. Apr 17, 2008 #10

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