I really ought to be doing other things, but this roller slide business kept nagging at me, and I eventually realized I could mock up a crude simulation of the results. This led to the production of this graph:

This looks pretty similar to the Tracker Video data from the previous post, which I’ll reproduce below the fold, along with an explanation of the math that went into the model:

The two graphs are qualitatively similar, other than, you know, the godawful Excel aesthetic of the simulation results graph. One line looks relatively straight, like motion at constant speed, while the others show signs of some acceleration. So, how did I do this?

The key was to think about the effect of passing over a single roller, in terms of the energy. At the start of the roller, the energy of the slider is:

E=_{i}Mgh+ 1/2Mv_{i}^{2}

where *M* is the mass of the person on the slide, and *v _{i}* is their initial speed. The first term is gravitational potential energy, the second kinetic energy due to their motion down the slide.

After passing over one roller, the energy becomes:

E= 1/2_{f}Mv_{f}^{2}+ 1/4mv_{f}^{2}

where *m* is the mass of a roller. This makes use of the fact that the speed of a point on the outside of the roller must be the same as the speed of the rider as they leave that roller, and assumes that the roller is a solid cylinder. In this equation, the potential energy from the first equation has vanished, as the rider has moved to a lower altitude during the sliding, so the first term corresponds to the kinetic energy of the rider moving down the slide, while the second term is the rotational kinetic energy of the spinning roller.

Because energy is conserved, we can set these two equations equal to one another, and solve for the final velocity. To get an exact number would require us to plug in numbers for the masses, which we don’t know, but we can re-arrange things to put everything in terms of the ratio of the mass of the rider to the mass of the roller. This gives:

v_{f}^{2}= (M/m)(2gh-v_{i}^{2})/((M/m)+(1/2))

(apologies for the dreadful calculator notation, but we don’t have LaTeX support on ScienceBlogs, and I don’t want to take the time to make nicer-looking equations as graphical objects). This lets you find the speed after passing over each roller. That, in turn, becomes the initial speed for the next roller, and you can repeat the process (this is why it’s done in Excel, because it’s less annoying to code that in there than in SigmaPlot). To get a graph of position vs. time, you just use those speeds to calculate the time needed to pass over the distance covered by each roller, and add them all up.

To get numbers in quasi-real units, I had to assume some change in elevation for each roller. I pulled a number out of the air, and said it was 2cm, which is probably an overesitmate. For the time conversion, I used the fact that the angle of the slide is around 20 degrees, so the distance covered along the slide during the passage over one roller would be around 6cm.

The simulated curves on the graph at the top of this post are for mass ratios (*M/m*) of 1, 5, and 50, going down a roller slide with 36 rollers. So, you see, a light rider moves at a fairly constant speed, while a heavier rider clearly accelerates toward the bottom.

This is, of course, a ridiculously simplistic model– it assumes a single roller in motion at any given time, neglects friction in the roller, assumes zero initial speed, etc.– but it does the right thing, qualitatively. And that’s enough to allow me to stop thinking about the damn thing and get back to the work that I should’ve been doing when I was thinking about this business all afternoon.

For those who saw seacretseason’s comment on the last post, the formulae here give a terminal velocity that is different from the one in that comment by a factor of the square root of two, which comes from different assumptions about the composition of the rollers– I said they were solid, secretseasons said they were hollow. I did it in terms of a height drop rather than the radius of the rollers, but the conversion between those is simple, and left as an exercise for the reader.

I’m really not sure what those roller-cylinders are like, maybe the solid moment of inertia is a better assumption… it seems to lead to a more reasonable number, anyway.

1. Your equation can be solved as an ODE i.e. http://www.wolframalpha.com/input/?i=d%5E2x%2Fdt%5E2+%2B+m%2F%284M%29+%28dx%2Fdt%29%5E2+-+g+%3D+0

where m is simply the mass of all of the rollers under your legs or Steelykid’s legs. Note, Steely kid will have less roller’s under her than you. This decreases the mass of the rollers and offsets, in part, your weight advantage.

2. I doubt there is enough mass in the roller to cause such a straight line curve right off the bat for Steelykid.

3. I think the real reason for the lack of speed/acceleration is work she is doing. The sit bones on her butt are sinking down between the rollers creating an impulse friction force. To a lesser extent, so are her feet banging on the front rollers.

You should note that your kid’s legs are not straight and are flopping a bit. Have you ever tried to slide down a set of stairs? First thing you do is make you legs board like. This allows you to sail down on the tops of the stairs. Try going down with legs that are a little bent and loose. You drop from one stair to the other.

Here, you had nice straight legs. Plus, you are bigger and even you rear end can spread over a couple of rollers. Your kid simply sinks between the rollers just enough to create a lot of resistance.

And what do I know of such things? Just yesterday my 9 year old challenge me to a race down a long steep steel side to side set of slides. I got a great start, and got the lead, but I was going too fast and grabbed the railing for safety. It instantaneously burnt a blister onto my palm. I’ve been dealing with slides and kids for a little longer than you. ðŸ˜‰

“God! Why aren’t you doing ‘real’ science!?”

“Because science is fun, and anyone who disagrees can fuck off.”