A 12 year-old skater, Tom Schaar, recently becamethe first person to ever land a 1080 on a skateboard. It's an impressive feat, which made some people wonder if it wasn't easier because Schaar is a kid. Our friends at Wired examine the question.
The 1080 (which is short for 1080° - just to be clear) was landed by a 12-year-old skater named Tom Schaar. Awesome indeed. But what about the original question? Would it be easier for a kid to do this? Time for a quick estimation.
What is important in a 1080? Well, you have to be in the air so you can spin. If I assume that both the child and the adult would be in the air for the same time, this means that angular velocity is important. Let's not get into the numbers; let's just say that you would need to rotate 1080° in a time interval Δt. Since the skater starts with zero angular speed, the skater would need to have some angular acceleration during the launch part.
Let me start with a simple model. Suppose the kid and the adult are both cylinders. (If they were cows, they would be spherical.) Here is a diagram.
I am just assuming the adult has the same proportions as the kid, but is bigger. One and half times bigger in this case. So, how do you get a cylinder to have an angular acceleration? You need torque. Suppose there is a force pushing on these cylinders during the jumping phase to increase the angular velocity. I can write:
Here, you can see that I am making some approximations. In this case, I am assuming the force is perpendicular to the axis and at the edge of the human-disk. I haven't said what Fa and Fk are (for the adult and kid forces respectively). I will get to that later. However, in rotational rigid body dynamics I can say the following:
These should all really be vectors, but I am trying to keep it simple. Here, the I is the moment of inertia about the axis of rotation (or as I like to call it, the rotational mass) and α is the angular acceleration. For a disk rotating about its center axis, I has a value:
A smaller I means that it will be easier to change the angular motion of the object. Putting this in and solving for the angular acceleration, I get:
Here you can already guess that the kid might have an advantage since he or she will have a smaller R. Of course, the kid will also not be able to push as hard (F) so, what does that say? It says to keep on working. Let me assume that both the child and the adult have the same density (ρ). This means that I can write the mass as:
Now my expression for the angular acceleration is:
How does that affect the child and the adult? Here are the two angular accelerations using the values from above.
In order for the adult to have the same angular acceleration as the child, the adult force would have to be about 5 times the value of the kid's force. Could this be? I don't think so. Why? If you are rolling on a skateboard, what would cause you to start spinning? Really, the force would come from the sideways frictional force between the skate wheels and the track. If this is the case, I would assume the following expression for the maximum frictional force (but not really, since they wouldn't be on a flat surface).
Like I said, this is wrong. The normal force would not be equal to the weight because during the jump, they are not on a flat surface. However, it doesn't matter. Why? Because there will be some coefficient in front of the mg term. This coefficient will be the same for both the child and the adult since it will mostly depend on the angle. The point is that the maximum frictional force will be mostly proportional to the mass. Let me say the frictional force is just proportional to the mass and put that into the expression. (I will use a proportionality constant of K:)
The adult still loses. The loss just doesn't look as bad. But wait. This is just an estimate. I didn't include things like the time of acceleration. Maybe the adult will have a longer "jumping time" so that even with a smaller angular acceleration, the adult could get to the same angular speed. Or maybe the adult can jump higher. Or maybe the kid could use a bigger skateboard and increase the R for just the part where the force is applied and have an even bigger advantage.
Overall, I would mark this down as one of those things were bigger doesn't mean the same. I fall in this trap too. It seem like if you build a small version of a train, it could be just like a big train. Unfortunately, things don't always scale the way you think they would.
Back of the Envelope
When this question was first asked on Twitter, I did a super quick version of this calculation on a scrap sheet of paper. Here is what I wrote.
Same idea, but I left off some stuff. Please note: I am NOT saying that in order to break 1440 you should train your 6-year-old brother or sister to skateboard. If they want to skateboard, that would be OK. As long as they are safe. Be safe, kids.
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