Lean angle of a turning bicycle

Question

I'm asked to derive a relationship for the leaning angle of a bicycle with the following specs:

• Center of gravity for bike and rider is a distance $L$ above the ground when vertical, and the total mass of bike and rider is $M$.
• The wheel radius is $r$ and the wheel mass is $m$
• The radius of curvature for the turn is $R$, the speed is $v$

I am to show that $$\tan \theta = \frac{v^2}{R} \frac{2mr + ML}{MLg}$$ but I am unsure. If I take $M$ to be the total mass, including the wheels, then I would guess that the condition is that centrifugal force and gravity act in such a way that the total torque is zero, i.e. $$M g L \sin \theta = M \frac{v^2}{R} \cos \theta L$. But then the mass of the wheels and their radius doesn't enter.

If, on the other hand, I exclude the wheels from $M$, wouldn't I still have to add them to both sides, i.e. $$(MgL + 2mgr) \sin \theta = \frac{v^2}{R} \cos \theta (ML + 2mr)$$

or where does the extra term 2mr come from?

Answer 1

Aha, I got it. :) Most resources on bicycle physics dismiss gyroscopic effects because those are too weak.

Here, we need them:

The wheels turn with angular velocity $\omega = v/r$ and hence have an angular momentum which we approximately can write as $L = mr^2 \omega$. In cylindrical coordinates, $L$ has a component in $z$-direction which stays constant in time, and a component in $r$ direction which changes with time. Let us look at the radial component:

$\vec{L}_r = mr^2 \omega \cos \theta \vec{e}_r$. The change of it over time involves a torque: $$\vec{T} = mr^2 \omega \Omega \cos \theta \vec{e}_r$$ where $\Omega = v/R$ is the angular frequency of the overall circular path. Using this, and $r\omega = v$, we obtain $$\vec{T} = mr \frac{v^2}{R} \cos \theta \vec{e}_r$$

Hence, for the total torque to be $0$, we need $$MgL \sin \theta = \frac{v^2}{R}\left( 2mr + ML\right) \cos \theta$$ which immediately yields the desired result.