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.

  • 4
    $\begingroup$ You could say the absolute acceleration/speed is equal in each car, but the acceleration and speed vectors are different. $\endgroup$
    – Paul
    Oct 10 '17 at 13:04

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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – user93237
    Oct 9 '17 at 18:22
  • 9
    $\begingroup$ And if the average speed over a full circuit of the rear car is different from the front car, well, there is a problem... $\endgroup$
    – Jon Custer
    Oct 9 '17 at 19:23
  • 27
    $\begingroup$ @JonCuster: Actually, if the instantaneous speed ever differs between the rear car and the front car, then is a problem! $\endgroup$
    – dotancohen
    Oct 10 '17 at 9:42
  • 4
    $\begingroup$ @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. $\endgroup$ Oct 10 '17 at 12:49
  • 3
    $\begingroup$ @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. $\endgroup$ Oct 11 '17 at 14:27

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.

  • $\begingroup$ 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. $\endgroup$ Oct 11 '17 at 10:44
  • $\begingroup$ @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. $\endgroup$
    – xiaomy
    Oct 12 '17 at 16:43
  • $\begingroup$ @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. $\endgroup$ Oct 12 '17 at 17:02
  • $\begingroup$ @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. $\endgroup$
    – xiaomy
    Oct 12 '17 at 20:54
  • $\begingroup$ @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. $\endgroup$ Oct 13 '17 at 8:13

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


Not the answer you're looking for? Browse other questions tagged or ask your own question.