Minimum speed for stable motorbike ride along a straight line on flat ground

Question

What is the critical minimum angular speed of wheels or connected linear speed of motorbike along a straight line to avoid falling and remain stable on a motorbike in terms of moment of inertia $I$ of each of the two wheels of mass $m$, mass of rider plus bike frame $M$, height of their center of gravity $h?$

Is there a differential equation of motion for stability? Or do we just equate rotational (PE) and linear kinetic energies (KE) ?

EDIT

Hypothetically its handlebar is welded to the frame and cannot rotate, only open loop dynamics required for track holding. Thanks.

Answer 1

There is a misconception that gyroscopic motion is the main contributor to the vertical stability of a bicycle-like vehicle. In fact, the stability is largely a result of the geometry of the front wheel and steering mechanism, and the way that, as the bike tilts, the wheel will turn toward the direction of the tilt and "correct" it.

It is easily possible to design a bike (or motorbike) that is inherently unstable and can only be balanced by a gymnast.

Wikipedia:

If the steering of a bike is locked, it becomes virtually impossible to balance while riding. On the other hand, if the gyroscopic effect of rotating bike wheels is cancelled by adding counter-rotating wheels, it is still easy to balance while riding.[5][6] One other way that a bike can be balanced, with or without locked steering, is by applying appropriate torques between the bike and rider similar to the way a gymnast can swing up from hanging straight down on uneven parallel bars, a person can start swinging on a swing from rest by pumping their legs, or a double inverted pendulum can be controlled with an actuator only at the elbow.[33]

Answer 2

Here’s my concise answer, but see Wikipedia for a longer article with figures: https://en.wikipedia.org/wiki/Bicycle_and_motorcycle_dynamics

Let $\theta$ denote the lean of the bike, $\varphi$ the turn of the handlebars, $\alpha \approx 73{}^\circ$ the “head angle” formed by the axis of the steering column to the ground, $x\approx$ 107 cm the wheelbase, $M\approx$ 70+10 kg the mass of the rider plus bike, $h\approx$ 110 cm the altitude of the CM, $m\approx$ 2 kg the mass of the front wheel, and $r\approx$ 35 cm its radius. The relevant moments of inertia will be ${{I}_{\omega }}=m{{r}^{2}}$, ${{I}_{\theta }}=M{{h}^{2}}$, and ${{I}_{\varphi }}=???$ for the steering assembly plus wheel.

The key issue is whether centrifugal forces are sufficient to keep the bike from tipping over to the side, but this will depend on the curvature of the trajectory, which depends on the turn of the handlebars and the wheelbase. Gyroscopic torques are less important.

Gyroscopic effects aside, the tipping torques are $${{I}_{\theta }}\ddot{\theta }=HMg\sin (\theta )-HM\tfrac{{{v}^{2}}}{x}\cos (\theta )\sin (\alpha )\sin (\varphi )$$

The front struts are curved to make the wheel self-righting when in contact with the ground, with some natural frequency $\Omega $. Its equilibrium turn angle exceeds the bike’s lean angle: $$\tan ({{\varphi }_{0}})/\tan (\theta )=\sec (\alpha )$$
Linearizing, we get $$ \left[ \begin{matrix} {\ddot{\theta }} \\ {\ddot{\varphi }} \\ \end{matrix} \right]=\left[ \begin{matrix} +g/h & -{{v}^{2}}/hx \\ +{{\Omega }^{2}}\sec (\alpha ) & -{{\Omega }^{2}} \\ \end{matrix} \right]\ \left[ \begin{matrix} \theta \\ \varphi \\ \end{matrix} \right] $$

We can infer the signs of the eigenvalues from the determinant without knowing the value of ${{I}_{\varphi }}$ or $\Omega$, and we may conclude that the coupled system will be unstable unless ${{v}^{2}}>gx\cos (\alpha )$, so $v>$ 1.75 m/s or 6.3 km/h. It is also essential that ${{\Omega }^{2}}>g/h$, which depends on having sufficient “rake” of the forks, weight on the front wheel, and low ${{I}_{\varphi }}$.

Gyroscopic torques contribute small correction terms proportional to the rates of change of steering, lean, and direction: $$\begin{align} & {{I}_{\theta }}\ddot{\theta }=\ldots +{{I}_{\omega }}\tfrac{v}{r}[\dot{\varphi }\sin (\alpha )+\tfrac{v}{x}\varphi ] \\ & {{I}_{\varphi }}\ddot{\varphi }=\ldots -{{I}_{\omega }}\tfrac{v}{r}[\dot{\theta }\cos (\alpha )] \\ \end{align}$$