If a vehicle is rolling down a hill, will its speed depend on the size of the wheel?

Question

If I am sitting on a skateboard and travel downhill, will the velocity depend at all on the size of the wheel?

The skateboard is only powered by gravity. There are a lot of variables to consider. Given the best conditions, where the surface of the road is smooth, and the tires are roughly equivalent.

I've heard answers that larger wheels are faster because they offer more springiness, and larger wheels are slower because they deform more and thus have higher rolling resistance. Larger wheels are faster because they roll over imperfections in the road better than smaller wheels.

I am curious from a theoretical standpoint, does the size of the wheel matter? That is perhaps, ignoring things like rolling resistance. And also from a practical standpoint.

Answer 1

Well, you can't directly compare speeds, because the vehicle has different speeds at different times (or equivalently, at different positions). But you can compare the accelerations. It turns out that if you look at the theoretically simplest effect, moment of inertia, the acceleration does not depend on the wheel size, but it does depend on the shape of the wheels.

Suppose a wheel has a moment of inertia $I$, which for a solid cylinder would be $\frac{1}{2}mr^2$ ($m$ is the wheel's mass and $r$ is its radius). The kinetic energy of the wheel is

$$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Now, suppose you have $N$ of these wheels mounted to a chassis of mass $M$. The kinetic energy of the entire ensemble is

$$K = \frac{1}{2}(M + Nm)v^2 + \frac{1}{2}NI\omega^2$$

If the wheels roll without slipping, then $v = r\omega$ and that becomes

$$K = \frac{1}{2}\left(M + Nm + \frac{NI}{r^2}\right)v^2$$

And if this whole thing is rolling down a slope of inclination $\theta$ (where $\theta = 0$ corresponds to horizontal), the total energy is

$$E = \frac{1}{2}\left(M + Nm + \frac{NI}{r^2}\right)v^2 - (M + Nm)gd\sin\theta = 0$$

where $d$ is the distance traveled along the slope from the point at which the vehicle started rolling. If you take the time derivative of this and do a little algebra, you get

$$a = \frac{(M + Nm)g\sin\theta}{M + Nm + \frac{NI}{r^2}}$$

At this point it's worth noting that moment of inertia for any kind of rotationally symmetric object (like a wheel) is given by a formula of the form

$$I = bmr^2$$

where $b$ is some number. For instance, $b = \frac{1}{2}$ for the solid cylinder I mentioned above. For a hollow wheel (no hubcaps or anything) you'd have $b = 1$, for a sphere $b = \frac{2}{5}$, and so on. So the above formula reduces to

$$a = \frac{(M + Nm)g\sin\theta}{M + Nm(1 + b)}$$

Notice that it doesn't depend on the size of the wheel ($r$) anymore. But it does depend on that constant $b$, which is determined by the wheel's shape. A wheel with a larger value of $b$ will undergo less acceleration.

Answer 2

From a purely theoretical standpoint, the radius of the wheel doesn't matter, but heavy wheels are slow.

In an idealized scenario, the skateboard conserves energy. This means that its total energy when it gets to the bottom of the hill is the same regardless of how it goes down.

As a wheel rolls down the hill, it picks up kinetic energy. Some of that goes into its translational motion, while some goes in its rotation. The energy in the rotation is essentially wasted from the point of view of going fast. A solid cylinder or disk, for example, will move $1/\sqrt{1.5} = 0.81$ times as fast at the bottom of a hill it has rolled down as it would go if it slid down without rolling (and without friction).

If you have heavy wheels compared to the weight of the skateboard and rider, then you suffer most of this slowdown. If you have light wheels, the energy of the wheels hardly matters and you can approach the ideal sliding speed.

Next we want to know if this matters. On Wikipedia I found that rolling resistances can be about $0.01$. When you roll down an incline of angle $\theta$, you lose about $0.01 \cot\theta$ of the energy you pick up to friction. For now set rolling resistance equal to $c$ instead of the number $0.01$. For cylindrical wheels of total mass $m$ and total wheels and rider of mass $M$, a fraction $m/3M$ is used in rotational rather than translational kinetic energy. Thus, the rotational energy stored in the wheels becomes important when, roughly speaking,

$$m/M > 3 c \cot\theta$$

The question we wanted to answer was not when rolling resistance becomes important, but what size wheels are faster. So imagine that $c$ is a function of $R$, the wheel radius, and that $m = \lambda R^2$, saying that we'll consider wheels of the same density and thickness, but different radius.

If we differentiate both sides of the previous expression with respect to $R$, we get a condition for the extra rotational energy stored in the wheel size to start being a bad trade off for any improvement in rolling resistance.

$$2\lambda R/M > - 3 c'(R) \cot\theta$$

or

$$ R > -\frac{3 M c'(R) \cot\theta}{2\lambda}$$

When this inequality is satisfied, making the wheels large will slow you down. Otherwise larger wheels are better. Note that if $c'(R)$ is zero or positive, larger wheels are always worse.

Unfortunately, I can't think of good ways to estimate $c'(R)$ without experimentation. One thing I can think of is this (it's highly speculative): A skateboard has pretty hard wheels that probably don't deform much under the weight of a rider, but skateboards are treated roughly and may get some grime in their bearings. So I would guess that friction in the bearing could be the most important factor in rolling resistance for a skateboard. I don't have much experience with them, but I think that if I take a skateboard wheel and spin it while holding it up in the air, it won't spin and spin for a minute or more like a bicycle wheel will. Larger wheels mean fewer rotations and less motion in the bearing, so this would give $c(R) = \alpha/R$ for some constant $\alpha$. This guess would give

$$ R > \frac{3 M \alpha \cot\theta}{2\lambda R^2}$$

or

$$R > \left(\frac{3 M \alpha \cot\theta}{2\lambda}\right)^{1/3}$$

When I let the density of the wheels be $1g/cm^3$, the mass of the rider by $75 kg$, the slope $10 ^\circ$, the width of the wheels $2 cm$, and the rolling resistance $0.01$ when the wheels are $5cm$ tall, I get that larger wheels slow you down when $R > 17 cm$. That would indicate that larger wheels are actually better in this case up to a pretty big size wheel (for a skateboard), but take it with a large grain of salt. There are lots of unjustified assumptions in there.

Answer 3

Let me try to make an explanation without formulas.

Suppose you assume no rolling friction, and you start at the top of a hill at height h and you and your wheels weigh w. Then your initial potential energy is wh (uh-oh, a formula).

Then you roll down the hill, and at the bottom your kinetic energy is wh, obviously.

But that consists of two parts, 1) you, traveling at a velocity v, plus 2) the wheels, which are spinning and have their own kinetic energy. Their rims are moving at velocity v with respect to their hubs.

So any energy in the spinning wheels will subtract from your bodily kinetic energy. In the extreme, if your wheels were connected to a big flywheel (like a "friction motor" in a toy) you would come down slowly and all the energy would be in the flywheel.

So to get down as fast as possible, you want to minimize the mass in the wheel rims. But if the wheel mass and distribution is held constant, the wheel radius should not matter. The reason the wheel radius does not matter is the energy in the wheels depends on v and the mass in the wheel rims, and nothing else.

So that would argue for smaller, lighter wheels. On the other hand, if you can't ignore rolling friction, I would suppose smaller wheels have more friction, depending on the type of surface (rough cement vs. something very smooth). If the surface were very smooth and firm, it would seem that tiny lightweight wheels (needle bearings?) would work best.