- #1

Zach Smith

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**[SOLVED] Centripetal Force and G Forces while traveling over a hill**

## Homework Statement

Hi everyone - I've had some problems wrapping my head around the ideas behind centripetal force while traveling 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 traveling 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.

## Homework 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 traveling 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

## 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 traveling 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 traveling 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.