# Spinning bicycle wheel in air and letting it go

Let's say that we have a bicycle wheel with radius $R = 25 cm$ and mass $m = 1 kg$. We spin the wheel in air so, that the tread is moving at the speed of $v_o = 50 km / h$.
Then, we let it roll on asphalt, given the coefficient of static friction: $f_s = 0.9$ and kinetic friction: $f_k = 0.7$. (I am making these numbers up, I hope they work)
I wonder what the final velocity of the wheel will be (after traction is "restored").

This is how I attempted to solve this problem:

Once this wheel is put on asphalt, it will start moving with the initial speed of $v_o$. Then, the force of kinetic friction $F = f_k mg$ will reduce its speed by $\Delta v$. By the momentum-impulse rule,
$$m \Delta v = F \Delta t$$

The torque generated by the force of kinetic friction will affect the angular velocity like this:
$$f_k mg \cdot R =I \frac{\Delta \omega}{\Delta t}$$

Now, using the fact that once the terminal velocity is reached, the wheel will no longer skid - this relation holds
$$v_{terminal} = \omega_{terminal} \cdot R$$

Now, this system of equations is actually solvable, but I am not sure whether my reasoning is correct. Any ideas?

## Answers and Replies

rcgldr
Homework Helper
Similar problems have been done here before, based on spheres, and possibly hollow cylinders, which would be the same as a bicycle wheel. The initial linear speed of the wheel will be zero. The wheel then accelerates linearly due to the kinetic friction which also decreases the wheel's angular velocity. Eventually the wheel will transition into pure rolling.