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I am wondering about this question since I asked myself: why do people feel more weightless in the rear car of a roller-coaster than in the front car?

To feel the effect of weightlessness, you must accelerate at the acceleration of the gravity (around 9.8m/s^2). Thus, you do not feel that effect in the front car but more likely in the rear car. But all the cars are connected together, and one individual car cannot accelerate faster or go faster because they will get pulled/pushed from the other cars.

I am stuck right now to get the answer. If all cars must go at the same acceleration or same speed at different points on the tracks, why does the rear-car feel more weightless? To have that feeling you must accelerate near the gravitational acceleration ... it doesn't make sense!

I have put the air friction, the frictional forces outside of this since I am guessing their force shouldn't be taken in consideration in that kind of situation.

IconDaemon
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Citizen602214085
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    You could say the absolute acceleration/speed is equal in each car, but the acceleration and speed vectors are different. – Paul Oct 10 '17 at 13:04

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The acceleration along the track is always equal for every car, but for each car that acceleration aligns with the hills/gravity in different ways. As the front car crests a hill, the coaster is decelerating; the front car is being pulled backward by the other cars. But as the rear car crests a hill, it's being pulled forward by the rest of the cars.

The front car is accelerated down hills. The rear car is accelerated over hills. This is why they feel different to ride.

Craig Gidney
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    So it appears that for a given rollercoaster track and a given rollercoaster that the coaster will be decelerating as the front car is going over a crest, and will continue to decelerate until the coaster is halfway over the crest. After the coaster is halfway over the crest, it will then start accelerating again. So the front car and the back car go over a crest at about the same speed and therefore experience about the same negative g's? Seems that the middle of the coaster is the place to sit if one wants to go over a crest at the smallest speed and experience less negative g forces. –  Oct 09 '17 at 18:22
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    And if the average speed over a full circuit of the rear car is different from the front car, well, there is a problem... – Jon Custer Oct 09 '17 at 19:23
  • @SamuelWeir I don't think so. You're accounting for acceleration due to following the track at speed, but not for acceleration along the track. This gives the correct negative Gs only at the very top of the hill. Just past the top, the front is being pulled back (against the fall) whereas the back is pulled forward (into the fall). I'm guessing that the follow-through after the crest is where the different sensation comes from. – Craig Gidney Oct 09 '17 at 20:25
  • @CraigGidney - Yes, I'm assuming that the acceleration along the direction of the track is small compared to the acceleration lateral to the track. Those lateral accelerations are the ones that I most remember after a rollercoaster ride, particularly when going over crests and experiencing negative or near-zero g's, so I'm assuming that along-the-track accelerations are small in comparison, at least near crests. I could be wrong, though. –  Oct 09 '17 at 21:03
  • @JonCuster Well, the rear car has become the front car :-| – user207421 Oct 10 '17 at 01:09
  • Very good answer. – Lambda Oct 10 '17 at 01:49
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    @JonCuster: Actually, if the instantaneous speed ever differs between the rear car and the front car, then is a problem! – dotancohen Oct 10 '17 at 09:42
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    @SamuelWeir, You are correct. The front car plunges into the "valleys" at a higher speed than any other car, and so a rider in the front experiences the highest"positive Gs." The rear car is whipped over the peaks faster than any other car, and the rider there experiences the highest "negative Gs." The longer the train, the greater the difference between the two positions. If you wish to avoid either extreme, then the middle car is the place to sit. – Solomon Slow Oct 10 '17 at 12:49
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    @dotancohen - there is some slack in the couplings, but not much. – Jon Custer Oct 10 '17 at 13:04
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    @jameslarge no, that actually does not make sense. By symmetry, the front car goes over the top at least as fast as the rear car (because in either case, potential energy is the same, and because the front passes the top earlier and the train isn't powered, it can only have at most that much kinetic energy by the time the rear car gets there). That's assuming the hill itself is symmetric – not a good assumption for most coasters, but if that's the crucial thing then the explanations so far are wrong. – leftaroundabout Oct 10 '17 at 22:55
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    @leftaroundabout you are correct. The middle car will experience the highest positive Gs in depressions and the lowest negative Gs over hills. You do not need to look at each individual car but can just look at the center of gravity for the whole train. When the center of gravity is at the highest point speed is lowest, when it is at the lowest point speed is highest. The center of gravity however should be roughly equal to the middle car. – Adwaenyth Oct 11 '17 at 10:34
  • @Adwaenyth yes, that would be my line of reasoning too. But how does any of the stuff said here make sense then? Riding in front should feel much the same as riding in the back, and both different only from the center. – leftaroundabout Oct 11 '17 at 10:39
  • @leftaroundabout The acceleration differs for the first and the last car. Gravity is constant for each car at every position, acceleration vector is not as it follows the track. The upward / downward forces are roughly equal, the forward /backward are not. – Adwaenyth Oct 11 '17 at 10:46
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    @Adwaenyth, Hmm... Y'know where it's not symmetric, is at the top of the very first hill. There, the train approaches the crest at a constant speed as it is pulled up by the drive chain, and the passenger in the first seat goes over much more slowly than the passenger in the last seat. Maybe that's the only hill that I am able to clearly remember. Maybe the ride scrambles my memory of all the other hills and valleys. – Solomon Slow Oct 11 '17 at 14:27
  • The front car accelerates after it reaches the top of the hill. If, by the time the center of mass reaches the top of the hill, the front car has reached the bottom of the hill, then it won't accelerate down; it will be be flat. It is the center of mass whose acceleration matches the hill. – Acccumulation Oct 16 '17 at 01:36
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If you draw a free body diagram with the center of gravity at the center of the train, but with the front car just over the crest, you will see that the net force is to decelerate the train (downwards and backwards as it is cresting the hill).

If you draw the diagram, but instead now the rear car is at the top of the hill, then the net force is to accelerate the train (downwards and forwards) this will result in the "weightlessness" feeling of reduced g (maybe even negative depending on the coaster type).

So while the net acceleration and velocity of the train is uniform across the train, you are looking at the forces at different times, and thus different values.

epic ms paint free body diagram

James
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If I sit on the rim of a spinning wheel, I'm being continually accelerated towards the hub of the wheel, even though the wheel itself has constant motion. The same is true of the roller coaster.

The last car is pulled faster over the curve, and so experiences greater acceleration tangential to the track, even though its linear acceleration along the direction of the track is the same as the first car.

So yes and no. Acceleration along the track is the same for all cars. Acceleration towards or away from the track is greater for the last car when going over a hump, and greater for the first car when traversing a valley.

superluminary
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  • Why “the last car is pulled faster over the curve”? Energy considerations suggests it should be just as fast as the front car when that is at the same curve, unless the track is deliberately built with asymmetric hills. – leftaroundabout Oct 11 '17 at 10:44
  • @leftaroundabout Energy conservation doesn't work for individual cars here because they are not isolated. It works for the roller coaster train as a whole. – xiaomy Oct 12 '17 at 16:43
  • @xiaomy exactly, that's what I meant: when a) the first car is at the apex, the rest of the train is still lower and hence more energy in kinetic form. When b) the middle of the train is at the apex, most of the mass is high up and thus the speed lower. When c) the last car is at the apex, the kinetic energy is again higher, but it's only higher than in a) if the hill is asymmetric, i.e. steeper down than up. Not that this I consider this in any way unlikely, just if that's the important thing it should be properly discussed in the answers. – leftaroundabout Oct 12 '17 at 17:02
  • @leftaroundabout I see what you meant. Indeed at the instant a) and c) the kinetic energy is the same, assuming symmetry. However, at a) kinetic energy is being converted to potential energy whereas c) it's the other way, so the last car would be going over the apex speeding up. Subjectively it might feel different, but certainly both the first and last cars would be faster going over the apex than the ones in between. – xiaomy Oct 12 '17 at 20:54
  • @leftaroundabout - All the cars are travelling at the same speed at any instance in time, but the train is not a point, so all the cars are in different locations at that instance. Imagine - at one particular instance in time - that the train is falling down a hill. The first car is likely to be traversing a straight section of track, whereas the last train is likely to be being pulled at speed over a hump. Therefore the last car will be experiencing greater acceleration because its upward motion is being converted to downward motion. – superluminary Oct 13 '17 at 08:13
  • @leftaroundabout - This would be true even if the first car was not accelerating and had constant downward motion. For want of a better word, the last car would be experiencing centripetal force, which is just acceleration towards a hub counteracting inertia. – superluminary Oct 13 '17 at 08:15
  • @superluminary I'm fully aware of all that. Sure, the last car is experiencing centripetal force when it goes over the apex. My point is that at the time where the first car passes that apex, the train would be just as fast, and hence the first car should experience the same centripetal force as the last car does later. Only in between, when the middle car is at the apex, will the speed be lower because the center of gravity is higher and hence less energy in kinetic form. – leftaroundabout Oct 13 '17 at 10:05
  • @leftaroundabout Ah, I see. This would be correct assuming symmetrical hills. Rollercoasters are rarely symmetrical. Think of the first drop - a long slow drag, then a fall. The first car is barely moving; the last races over the curve. – superluminary Oct 16 '17 at 13:03