Normal forces on a rotating wheel in contact with two perpendicular surfaces Let's suppose that a wheel of mass M and radius R is rotating with an initial angular velocity $\omega_o$ in contact with the floor and a wall, as shown in the picture, with a kinetic coefficient of friction $\mu$ between both surfaces and the wheel.
If the point of contact with the floor is 1 and the point of contact with the wall is 2, $F_{r_{1}}$ would act pulling the wheel against the wall and thus increasing the normal force on 2, whereas  $F_{r_{2}}$ would act lifting the wheel and decreasing the normal force in 1.
Then, how could they be determined?

 A: 
*Problem Statement: A wheel (mass $m$, moment of inertia $I_{CM}$, and radius $r$) is kept in contact with both the ground and the wall, with an initial angular velocity of $\omega_0$ (anti-clockwise direction). The coefficient of kinetic friction is  $\mu$ for both the surfaces (assume $\mu<1$). What is the subsequent motion of the wheel?

The following outlines the process that details how the normal reactions from the ground and the wall change quickly in response to each other:
$$ \underline{\text{Iteration 1:}}\;\;N^{(1)}_{\text{ground}}=mg \Rightarrow f^{(1)}_{\text{ground}}=\mu N^{(1)}_{\text{ground}}=\mu mg$$
$$N^{(1)}_{\text{wall}}=f^{(1)}_{\text{ground}} \Rightarrow f^{(1)}_{\text{wall}}=\mu N^{(1)}_{\text{wall}}=\mu^2 mg$$
$$\underline{\text{Iteration 2:}}\;\; N^{(2)}_{\text{ground}}=mg-f^{(1)}_{\text{wall}} \Rightarrow f^{(2)}_{\text{ground}}=\mu mg(1-\mu^2)$$
$$N^{(2)}_{\text{wall}}=f^{(2)}_{\text{ground}} \Rightarrow f^{(2)}_{\text{wall}}=\mu^2mg(1-\mu^2)$$
$$\ldots$$
$$f_{\text{wall}}=\mu^2mg \;[(1-\mu^2(1-\mu^2(\;\ldots \;)))]=mg\frac{\mu^2}{1+\mu^2} \tag{1}$$
$$\text{Similarly, } f_{\text{ground}}=mg\frac{\mu}{1+\mu^2} \tag{2} $$
The back-and-forth modifications happen instantaneously. $(1)$ and $(2)$ are the values attained by the frictional forces due to both the surfaces. The iterations are performed using equations $(3)$ and $(4)$.
$$\text{Horizontal Direction: }N_{\text{wall}} = f_{\text{ground}}=\mu N_{\text{ground}} \tag{3}$$
$$\text{Vertical Direction: }f_{\text{wall}}=\mu N_{\text{wall}}=mg-N_{\text{ground}} \tag{4}$$ 
The net torque acting on the wheel (in clockwise direction) is given by $(5)$ and you can determine when the wheel will come to rest using it. 
$$\tau_{CM}=\frac{\mu+\mu^2}{1+\mu^2} mg r \tag{5}$$
Please do let me know in the comments if I've made a mistake somewhere in my reasoning.

$^*$ Do let me know if I've made a mistake in identifying the problem statement.
