## Bicycle Wheels, Inertia, and Energy

From: andrew cooke <andrew@...>

Date: Sun, 19 Apr 2015 17:44:45 -0300

[This post is in development. I will be editing it after it’s published to my site, because I want to view it formatted correctly]

Introduction

It’s common to hear that larger bicycle wheels accelerate more slowly because they have a larger moment of inertia. But a thread on Reddit raised interesting questions that pushed me to think more carefully about the physics involved.

Physical Model

First, we need a model for a wheel. I will ignore the hub, and consider just two components - spokes and rim/tyre (for simplicity I’ll sometimes refer to the latter as just the rim below, but the tyre is obviously important too).

(I’m excluding disks because I am not sure how they are made. I may return to them later).

I’m assuming that spokes are of length $$r$$, while the rim and tyre are of length $$2\pi r$$ which is an approximation - the spokes are shorter and the rim is slightly shorter than the tyre, and of finite depth. But it’s good enough.

Masses and Momentts of Inertia

To get masses I’ll use linear density for all the spokes (summed). In other words, $$M_S = r \rho_S$$. Similarly, for the rim, $$M_R = 2 \pi r \rho_R$$. So the total mass $$M = r (2 \pi \rho_R + \rho_S)$$

The moment of interia of the rim, $$I_R$$, about the axle is simple, just $$M_R r^2$$ or $$2 \pi r^3 \rho_R$$.

The moment of inertia of the spokes is a little more complex because they are not a disk (the effective density decreases with radius). So let’s go back to basics:

$I_S = \int_0^r x^2 \rho_S dx = r^3 \rho_S / 3$

So the total moment of inertia of the wheel is $$I = r^3 (2 \pi \rho_R + \rho_S / 3)$$.

Energy

To accelerate a bicycle to velocity $$v$$ we need to expend energy in both accelerating the centre of mass and in rotating the wheel. When the bike is moving with velocity $$v$$ the wheel rotates with angular velocity $$\dot\theta = v / r$$. The total energy required is

$E = \frac{1}{2} (M v^2 + I \dot{\theta}^2) = \frac{1}{2} r v^2 [(2 \pi \rho_R + \rho_S) + (2 \pi \rho_R + \rho_S / 3)]$

If we compare this to the equivalent energy for non-rotating mass, $$E_{NR} = \frac{1}{2} M v^2$$ then we can write $$E = X E_{NR}$$ where

$X = \frac{(2 \pi \rho_R + \rho_S) + (2 \pi \rho_R + \rho_S / 3)} {2 \pi \rho_R + \rho_S} = 1 + \frac{2 \pi \rho_R + \rho_S / 3}{2 \pi \rho_R + \rho_S}$

Discussion

First, pretty clearly, $$X>1$$. In other words, as is commonly known, rotating mass hurts. It’s close to twice as “bad” in terms of the total energy required to reach a particular speed.

Second, because $$M$$ increases with $$r$$, so does $$E$$ (because $$E = X M v^2$$). That means that larger wheels mean you need to do more work. But how much more?

There are two ways to look at this. One way is to note that because $$X > 1$$, a larger wheel is worse than you would expect from the weight alone. The “effective mass” is almost twice as bad. But the other way is to note that, perhaps counter-intuitively, that relationship is only linear (through $$M$$). You might expect increasing wheel size to hit you “quadratically” - more than you’d expect from the effective mass alone. But it doesn’t. And one way to explain this is given in the Reddit thread I linked to earlier - for a given speed over the ground, a larger wheel rotates more slowly.

Andrew

(I had to reconstruct this page from fragments after a mix-up with email. I hope it is still coherent).