Behavior of a loose coin on a free fall ride

Question

At Six Flags over Texas, there used to be a free-fall drop ride called "the cliffhanger". Simply put, you got into a metal box, got elevated up to about 75 feet or so, and got dropped straight down.

One of the things riders would sometimes do is, while seated in the car, place a penny or coin on their thigh. During the drop, the penny would fly up and appear to levitate at or near eye level while the riders were seated while free falling. I actually saw numerous people do this during the many times I rode the ride, before it was torn down (i.e. the phenomenon is verified, at least by me).

I have a degree in physics and understand perfectly well that everything accelerates at $g = 9.8\, \text{m}/\text{s}^2$ near the Earth's surface notwithstanding its weight if we disregard air resistance. But I'm curious to know a little more about what everyone thinks about this - according to basic physics 101 equations, if the ride-car is dropped, it accelerates at g just like the penny, and the penny should stay put on the person's leg while it accelerates at exactly the same rate as the ride-car.

I'm curious if anyone out there knows a thing or two about constructing rides like "the cliffhanger" - is there some lever or something that pushes the ride-car down just a little faster or what?

I'm wondering what people have to say about this.

Answer 1

Suppose the penny is sitting on a thigh, and the thigh is sitting on a seat, and the seat is sitting on some shock absorber, and the shock absorber is sitting on the frame of the cart, and the exciting part of the ride is that the cart goes into free fall when some support is removed.

All of these pieces have some degree of elastic compressibility. The squashiest ones are the thigh and the shock absorber, but they are all a little compressible.

Prior to the release, each little vertical slice of this pile was the equilibrium with the pieces above and below it. No little piece of this pile will start to accelerate downwards until the force from below has changed. The information that the floor has gone is a low-pressure wave which travels up from the removed support at a speed which is related to the speed of sound (in the low-frequency limit). Here is a nice video showing this compression wave propagating downward from the top of a dropped Slinky; each part of the spring remains stationary until the part just above it has changed.

I can crudely measure this compression propagation time in my thigh by slapping my thigh with my hand and watching the jiggle propagate around to the other side. The time scale is a quarter-second.

The speed versus time for an object which is at rest prior to time $t_0$ and then accelerates with $a$ is

$$ v(t) = (t-t_0)a $$

If the penny starts to fall later than the thigh, because the thigh must decompress before the penny becomes unsupported, the penny’s speed is

$$ v_p(t)=(t-(t_0+t_\text{delay}))a $$

At a given instant, then, the penny is falling not so fast as the thigh:

$$ v-v_p=t_\text{delay}a $$

For a person attached to the thigh, they will see the penny drifting upwards, moving away from the contact point at (for a quarter-second delay) about 2.5 meters per second. The thigh will fall away from the penny until the brakes on the ride kick in, at which point the free-falling penny will start to catch up to the thigh again.

Note that no “twitch” is required here, and at no point does any part of the falling system actually accelerate upward. The issue is solely that different parts of the system begin to accelerate downwards at different times.

Answer 2

Firstly, you would not need to rely on some random small twitch.

The coin, in resting on the flesh of the thigh in the pre-launch phase, is causing some elastic compression of the skin. Upon launch, the skin will flatten out, giving the coin an upward velocity.

Perhaps more importantly, the force of gravity would have deformed the muscle of the upper leg; when gravity "disappears" the leg would round out, again launching the coin.

Now we have the coin moving in free fall through still air. Ordinary friction should slow the coin to a stop w. r. t. the air.

Answer 3

People are not comfortable above $2g$, so the ride probably does not subject you to more than that. So let us suppose you are in free fall (accelerating downward at $1g$) until you reach a maximum v, and then a braking force of $2g$ slows you to a stop.

The total force is the braking force plus gravity. So the net upward acceleration during braking is $1g$.

$$2a_1d_1 = v_{max}^2 = 2a_2d_2$$

$a_2 = a_1$, so

$$d_1 = d_2 = 37.5 ft \approx 12m$$

From $d = 1/2 a t^2$,

$$t_1 = t_2 = 1.5 \space sec$$

It doesn't sound like the coin really hovered. More likely it rose and sank. It is hard to keep track of what is motionless when your frame of reference is changing so quickly.

The coin rising can easily be accounted for by leg motion. It is hard to sit perfectly still when being dropped off a cliff.

Answer 4

It's not clear to me that the word "twitch" is the right word, but that has given me an idea that I think answers the question. I'm pretty sure that the passenger's leg would shutter just a little bit as the free fall began, and I'm sure this would be sufficient to explain the coin not staying put.

Thanks to everyone who contributed to this rather elementary post.