From my experience riding, at low speeds (between 0 and 10 mph) you mostly steer the bike with the handlebars. What I mean by this is if you want to turn left you rotate the handlebars counterclockwise, and the bike turns as desired. There is little to no lean caused by this. At medium speeds (between 10 and 20 mph) turning a bike seems to be a combination of steering with the handlebars and leaning. What I typically notice is that if I want to turn left I will push forward on the left handlebar ever-so-slightly, presumable turning the wheel clockwise just a bit, which will initiate a lean to the left, and I will then turn the handlebars counterclockwise to continue the turn. At speeds higher than 20 mph if I want to turn left I push the left handlebar slightly forward which initiates a lean to the left and I hold it like this to keep turning, never really turning the handlebars counterclockwise as in the previous two situations.
I don't really understand why turning a bike is different at different velocities. Why is lean more important in turning at higher speeds, whereas handlebar steering is more important at lower speeds. If I push the left handlebar forward at low speeds why don't I lean to the left like I do at higher speeds. At medium velocities, why does a combination of both seem to be the most efficient way to turn, and after initiating the lean left why doesn't the counterclockwise rotation of the handlebars put my bike back upright? In fact, since I rotate the handlebars counterclockwise to turn, how do I get back upright at all? Do I simply turn the handlebars counterclockwise more? If I was to try and steer the bike by simply shifting my weight, what would happen at these different speeds? For instance, if I am at a high speed and I shift my weight to the left to make the bike lean to the left will the handlebars automatically turn ever-so-slightly clockwise to initiate the lean even without any physical input to the handlebars?
Finally, how would one go about analyzing this quantitatively? If I am to set up the Lagrangian I presume I have nine degrees of freedom: the linear and angular coordinates of the bike and the angular coordinates of the front wheel. So does the make the Lagrangian
$$L=K_{bike} + K_{wheel} – mgh_{cm}?$$
It's not particularly clear to me what the constraints would be in a problem like this.
Best Answer
This is a classic example of torque as a vector cross product. When you push forward on the left handlebar, the torque $\textbf{r}\times\textbf{F}$ is down. The angular momentum vector of the wheel points to the left. A downward torque deflects the angular momentum vector down, which tilts the bike to the left, and causes you to start turning to the left.