# Roller Slide Physics Explained

On Monday, I posted a short video and asked about the underlying physics. Here's the clip again, showing SteelyKid and then me going down a slide made up of a whole bunch of rollers at a local playground:

The notable thing about this is that SteelyKid takes a much, much longer time to get down the slide than I do. This is very different than an ordinary smooth slide, where elementary physics says we should go down the slide at the same rate, and empirically I tend to be a little slower than she is.

So what's the difference?

First of all, let's be a little more quantitative about this. Here's a graph of the distance moved by each of us as a function of time, obtained from the Tracker video analysis program (stealing Rhett's shtick, here):

The black points are the position of SteelyKid's feet along the slide relative to a fairly arbitrary starting position, which the red points show the position of my feet relative to my starting position. (I used feet because they're easier to mark than the center of mass, and neither of us really changes our orientation during the sliding.) The lines are fits to the data.

As you can see, I'm clearly accelerating as I go down the slide, while SteelyKid is accelerating only a tiny bit (the fits give accelerations of 1.32 +/- 0.03 m/s/s and 0.065 +/- 0.008 m/s/s, respectively). What causes this?

A lot of people pointed to the way SteelyKid's feet hit each of the rollers as a possible cause, but I don't think that's the dominant effect. It probably does play a role, but her acceleration would be smaller than mine even without that.

The issue here is best identified by SimonW in comment #10. The rollers making up the slide have some mass, and after we pass each roller, it continues spinning-- in fact, it takes several seconds for them to stop afterwards. This means that the rollers have acquired some kinetic energy, which has to have come from the potential energy SteelyKid and I had at the start.

On an ideal smooth slide, of the sort you might encounter on a physics exam, the potential energy we start with is completely converted to kinetic energy. Both gravitational potential energy and kinetic energy increase with the mass in the same way, though, so the final speed does not depend on the mass. SteelyKid and I each end up moving at exactly the same speed. This is true even if the slide has some friction, at least in the standard intro physics assumption that friction only depends on the force between the sliding surfaces, and not the size and shape of the objects, because the frictional force also increases with the mass in the same way. On a slide with friction, some of the initial potential energy gets converted to heat energy warming up the slide and the slider a tiny bit, but this does not have any effect on the final speed. The same fraction of the initial energy ends up as kinetic energy for both of us, which means we have the same speed.

(The assumption that friction only depends on the mass is not a great one, and as many commenters noted, my surface area is much greater than SteelyKid's, so friction has a bigger effect on me, which is why empirically, I tend to be slower than she is on a smooth slide.)

The roller slide, though, introduces another energy sink, one that does not depend on the slider's mass in any simple way. That means that the fraction of the initial energy that ends up as kinetic energy of the slider is different for the two of us. The rate of spin of the rollers will be tied to the speed of the slider (assuming we roll without slipping), making the exact numbers a little tricky to figure out, but in the end, I keep more of my initial energy as kinetic energy than she does.

(You could try to explain the whole thing with forces, but it's even messier, because you need to think in terms of the torque on the rollers. Energy is the cleanest way to think about it.)

Can we do anything quantitative with this? Sadly, not without more information than I have (the roller diameter and mass would be useful). I could probably get that information, but I don't really have the time for it at the moment.

(What I really ought to do is go back there with a camera on a tripod and roll a bunch of different masses down the slide on a sled of some kind (so the surface area is constant). Given that data, I bet we could come up with something quantitative. But again, I have finite time available. I bet this could be turned into a great class exercise, but I really don't want to explain this to the elementary school where the slide is located...)

Anyway, that's my explanation of the roller slide mystery. Further elaborations or arguments with my phrasing are welcome in the comments.

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Well it looks like SK is going to gt more days at the park to build up replicates for the experiment. Go SK! (I'm on Team SteelyKid, with some coaching, a few years of practice, She'll dominate the slide at the Olympics)

By Sideshow Bill (not verified) on 14 Jul 2011 #permalink

What about the fact that the "normal" slide involves kinetic friction, whereas the "roller" slide involves static friction (you are only touching one point at a time on the rollers since they move) and the coefficients for friction are different for both types of slides.

I wonder if this is the same reason why, in coasting downhill on our bikes, I always end up going faster than my (much smaller mass) girlfriend, even though our bikes are similar.

What about the fact that the "normal" slide involves kinetic friction, whereas the "roller" slide involves static friction (you are only touching one point at a time on the rollers since they move) and the coefficients for friction are different for both types of slides.

You need to be a little careful about this. For one thing, static friction is, if anything, a stronger force than kinetic friction-- if it weren't bigger, there wouldn't be any need for the concept.

It's also not the only kind of friction involved in the roller slide. There's still kinetic friction going on, but the location of it has shifted. You don't slide across the rollers, true, but the rollers experience some friction in the bearings-- it takes several seconds for them to stop spinning, but they do stop eventually, due to the influence of friction.

The friction involved in the roller bearings is really small, so it basically gets taken out of the picture. But the mass of the rollers means that they can take up energy more effectively than the kinetic friction of a smooth slide.

Answering my own above question, there seems to be a slight dependence of the velocity on the mass of the rider, but the effect is largest when the riders' masses are on the order of the bike wheel mass -- not very realistic.

You can actually get something from the slope of SK's graph, if you also have the height of the slide and the number of rollers. Using SK as the reference frame, each of the rollers passing underneath undergo identical acceleration/deceleration (she's at constant speed). So each of n rollers receives the same total energy, which is (m.g.h-KE(SK))/n.

You can work out SK's final KE, and therefore the energy soaked up by each roller. Since the roller rotation rate must match SK's speed, you can determine moment of inertia (if we neglect bearing friction).

Anyway, you could further go on and use your accelerating graph (although we're straight into calculus territory because your speed is not constant, but you should come up with a second equation (ODE?) for MofI + friction, and that might allow separating MofI from friction.

This is fun!

By Lurker #753 (not verified) on 14 Jul 2011 #permalink

@Excited State, it might have more to do with grip strength, brake friction, and nerves :)

By Sideshow Bill (not verified) on 14 Jul 2011 #permalink

@Sideshow Bill
Nerves has something to do with it, as I don't ride the brakes at all. But I start pulling ahead pretty quickly (before she would tap the brakes), so I'd think there has to be something else going on.

You can actually get something from the slope of SK's graph, if you also have the height of the slide and the number of rollers. Using SK as the reference frame, each of the rollers passing underneath undergo identical acceleration/deceleration (she's at constant speed). So each of n rollers receives the same total energy, which is (m.g.h-KE(SK))/n.

It looks like about 36 rollers, give or take a little, and the total height is probably around a meter. SteelyKid's around 16kg (at a guess), and her speed is around 0.46m/s. If you put all that together, though, you get an implausibly large value for the mass of a roller-- something like 60 kg. That, or I'm making an arithmetic error somewhere. But I really, really need to do book revisions, and have to stop playing with this. Really.

There's a great intro physics lab hiding in this post.

By Chris Goedde (not verified) on 14 Jul 2011 #permalink

I commented on the other post but my comments never showed up.

I've noticed exactly this difference between me on the roller-slide vs my 2-yr-old son. I've thought a lot about it and think the terminal velocity analysis of gravitational potential energy being turned into rotational kinetic energy is correct.

Making some simplifying assumptions that the rollers' bearings are frictionless and the butt-roller interface rolls without slipping, the person reaches terminal velocity when the rollers acquire rotational kinetic energy (1/2)*I*omega^2 equal to a roller's-worth of fall Mperson*g*Hroller.

More assumptions: The rollers are hollow cylinders so I=Mroller*r^2 and the rollers are adjacent so the drop from roller-to-roller is 2*r*sin(theta) where theta is the angle the slide makes from the horizontal. [Incidentally, theta is 22 degrees for the Landscape Structures brand slide, on their website.]

Setting omega=v/r we obtain an expression for the terminal velocity:

(1/2)*I*omega^2 = Mperson*g*Hroller

(1/2)*Mroller*r^2*v^2/r^2 = Mperson*g*2*r*sin(theta)

which gives us an expression for v
v = 2 * SQRT ( r*g*(Mperson/Mroller)*sin(theta) )

Initially I had taken wild guesses at r (1-2cm?) and Mroller (2kg?) to get that my son reaches a terminal velocity of ~1-1.5 m/s and I reach a terminal velocity (or more likely, never reach it) of 2-3 times that.

But we could turn this around like in comment #9 and see if we can solve for a reasonable value of Mroller. From SK's 0.46 m/s terminal velocity I get ~9kg for Mroller. Is that reasonable? They're nearly meter-wide cylinders of steel, right?

By secretseasons (not verified) on 14 Jul 2011 #permalink

While the last comment was being written, I was doing my own mathematical model, which is in the next post. I made some slightly different assumptions, but I think the basic process is sound.

secretseaons: that's a very nice calculation of the terminal velocity. Thanks.

By Anonymous Coward (not verified) on 14 Jul 2011 #permalink

"I bet this could be turned into a great class exercise, but I really don't want to explain this to the elementary school where the slide is located..."

Leaving aside the possibility of simply laying your hands on a bed of rollers somewhere, perhaps something along the lines of: "I'm a professor at... would you terribly mind if I use your playground to do a 'physics in the real world' lab for my introductory physics course?" in a letter or phone call well before you do it.

By Rick Pikul (not verified) on 14 Jul 2011 #permalink

Comment 11 has it right in the detailed analysis, but it's worthwhile to take a step back and look at the bigger picture. You're basically colliding with each roller as you slide down, yes? That's actually very analogous to the simplified picture of air friction where the retarding force is proportional to v^2. So you can get effectively the same answer as post 11 without the detailed analysis, though the details are fun. :)

By BlackGriffen (not verified) on 15 Jul 2011 #permalink

Just a note, since I can only hazard rough guesses as to the physics myself, but these roller slides are a big thing in Japan. Here's a link to one in Okinawa City, just outside of Kadena Air Force base, that I've actually ridden on.

You really get moving on a roller slide that long!