Do you go "faster" if you ride in the back car of a roller coaster?

I just got back from a week of Disneyfication in Orlando.  See, I like Disney World.  From the time I left until I got back, I didn't once think about getting ready for the resumption of classes on Wednesday; and, I got to spend time with the seven year old Nachoboy without him complaining that I didn't play his game right.  The magic kingdom is special because it contains entertainment that he and I (and even Burrito Girl) all enjoy. 

In particular, the Nachoboy enjoyed Big Thunder Mountain Railroad (pictured to the right).  As roller coasters go, it's rather tame; no upside-downs, not an enormous drop.  Nevertheless, neither I nor Burrito Girl can ride more than about twice without feeling icky.  On a typical day last week, we would get up early to arrive precisely at the park's 7:00 opening so that we could take turns escorting the Nachoboy on Big Thunder Mountain over and over and over and over again.

At 7:00 in the morning, with only 20 people trying to ride, the operators were willing to give us our choice of seat, and were even willing to let us re-ride without getting off.  My first thought, to Burrito Girl's consternation, was to do a bit of physics...

Burrito Girl overheard an operator telling a rider that "you go faster if you ride in the back car."  Well, that's patent nonsense -- as you can see in the picture, the train of about six cars is attached together.  If the back car did go faster, then it would end up going through the cars in front of it.  That doesn't happen.  Q.E.D. 

Burrito Girl was unimpressed and unconvinced by this response.  "Look, I rode in the front and the back this morning.  When I was in the back, I felt myself thrown out of my seat when we went over the first big hill.  When I was in front, though, the ride was pretty tame."

And her observation makes perfect sense.  The whole train is always going the same speed at any given time, starting approximately from rest when the middle of the cart crests the first hill.  But by the time the back car crests the hill, the train has sped up significantly.

Consider a free body diagram of a rider when he crests the hill.  There's his weight pulling down, and the normal force pushing up.  The NET force, equal to his weight minus the normal force, must equal mv²/r, where r is the radius of the hill's curve, m is the person's mass, and v is the train car's speed.  Some brief algebra shows that the normal force is equal to the weight minus mv²/r.

So what is this "normal force?"  It's the force of the seat pushing up on the rider.  But it's also the "apparent weight" of the rider -- what a bathroom scale would read were the rider sitting on one.  The person's mass doesn't change regardless of the train's speed.  What does change is the v term.  Thus, the apparent weight of the rider gets SMALLER as the train's speed at the hill's crest gets bigger.

That's what Burrito Girl observed -- in the front, she crested the hill with minimal speed, and so her apparent weight was equal to just her regular weight.  Riding in the back, though, her train car had sped up by the time she crested the hill, so her apparent weight was smaller.  So, she felt lighter, like she was being thrown out of her seat.

Now, the front of the car is going just as fast as the back of the car all the time.  The difference is, roller coaster riders generally want to experience that feeling of a small apparent weight.  Riding in the back is thus the way to go -- not because you go faster than the front, but because you go faster AT THE TOP OF THE HILL.

Possibly another Disney-related roller coaster post soon.  Tomorrow I'll be in Birmingham, Alabama giving an AP physics workshop -- stop by if you're in town!

GCJ