Why does less friction in a wheel result in more speed?

Question

Everyone knows that racing bikes have narrower wheels and that, by definition, friction reduces speed. What I don't understand is that since at no time is the wheel of the bike in horizontal motion relative to the road, surely any friction between the wheel and the road will have no negative effect and in fact aid the forward propulsion of the bike as a whole?

Answer 1

when a rubber tire's tread surface comes into contact with the ground as it rotates, the rubber deforms or "squishes" against the pavement surface and then "unsquishes" when that portion of the tire rotates out of contact with the pavement. This means that as any tire rotates, the rubber in it is getting dynamically flexed continuously, and since rubber is not perfectly elastic, part of that deformation work gets transformed into heat.

This in turn means that it takes work to rotate a rubber tire, to make up for the hysteresis losses in the rubber itself.

How to minimize these losses? First, if you pump more air into the tire, you reduce the rolling deformation and hence the hysteresis losses. Second, if you reduce the amount of rubber in contact with the road by narrowing down the tire, you can reduce those losses. This is why tires on racing bikes are very skinny and have 120PSI air inside them.

Third, if you design the tire so as to minimize the squish, you also minimize the losses. This is why radial-belted tires furnish better gas mileage than bias-ply tires. This is also why high-speed tires have large diameter and narrow width, compared to (for example) tractor or wagon tires which have small diameter and broad width (this also increases their load-carrying capacity- in an application where high-speed flexure is not a problem).

Answer 2

Suppose that $\vec{F}_1$ is the force produced by the biker and $\vec{F}_2$ is the friction due to the floor.

Let $m$ be the mass of the wheel, $I$ its moment of inertia, $R$ its radius, $a$ its acceleration along $x$ and $\alpha$ its angular acceleration.

We know that:

$$\begin{cases} ma =F_{1,x} - F_{2,x}\\ I\alpha = -RF_{2,x}\\ a + R\alpha = 0 \end{cases}.$$

The equation $a + R\alpha = 0$ means that the wheel does not slip on the floor.

We can find the three unknown of such system, namely $a$, $\alpha$ and $F_{2,x}$:

$$\begin{cases} a =\displaystyle\frac{F_{1,x}R^2}{mR^2 +I}\\ \alpha = -\displaystyle\frac{F_{1,x}R}{mR^2 +I}\\ F_{2,x} = \displaystyle\frac{F_{1,x}I}{mR^2 +I}\\ \end{cases}.$$

As expected, a "narrower" wheel, i.e., smaller $m$ and hence a smaller $I$, implies an higher acceleration $a$. Indeed, the denominator of $a$ depends on both $m$ and $I$.