# 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?