How the calculate and/of measure the applied force between a motor and a flywheel? I am making a model for a flywheel. For this model, I found a way to calculate the torque and thus eventually the velocity of the flywheel. The flywheel itself will be floating due to magnetic bearings. Against the flywheel at the side the electrical motor will be placed. The flywheel itself will be horizontally placed.  For the torque, I still need the applied pressure of the motor on the flywheel. Can I calculate this if I know the output torque of the motor. If so how would one do that?
In the picture the thing at the side will be the casing for the motor and the black circle in the middle is the floating flywheel.

 A: Make a basic free body diagram. From the top down you have the following geometry and forces. I have separated the two bodies for clarity.

Point A is for actuator, point B for bearing and point C for contact. Equal and opposite forces act on the contact point. There is also motor torque $\tau_A$, bearing reactions at B and motor reactions at A.
Assume the mass moment of inertia of the motor is $I_A$ and the mass moment of inertia of the flywheel is $I_B$. The rotational speed of the motor is $\omega_A$ and the rotational speed of the flywheel $\omega_B$.

*

*The equations of motion for the motor are
$$ \begin{aligned}
A_x - C_x & = 0 \\
A_y - C_y & = 0 \\
\tau_A -r C_y & = I_A \dot{\omega}_A \\
\end{aligned}$$
The first two equations mean that $A_x=C_x$ and $A_y=C_y$.


*The equations of motion for the flywheel are
$$ \begin{aligned}
C_x - B_x & = 0 \\
C_y - B_y & = 0 \\
-r C_y & = -I_B \dot{\omega}_B \\
\end{aligned} $$
The first two equations mean that $B_x=C_x$ and $B_y=C_y$.


*Since we need a no slip contact, the contact normal force $C_x$ must large enough to provide sufficient traction. The contact conditions are
$$ \begin{aligned}
  r \omega_A & = R \omega_B \\
 C_x & \ge \mu C_y \end{aligned} $$
Solving in terms of the motor rotation $\dot{\omega}_B = \frac{r}{R} \dot{\omega}_A$ you have the following two performance equations
$$ \begin{aligned}
  \dot{\omega}_A & = \frac{R}{I_A R + I_B r} \tau_A \\
  C_y & = \frac{I_B}{R} \dot{\omega}_A  \\
  C_x & \ge \mu \frac{I_B}{R} \dot{\omega}_A \\
\end{aligned}$$
which you interpret the above as, motor torque $\tau_A$ results in motor shaft acceleration $\dot{\omega}_A$ (and corresponding flywheel acceleration $\dot{\omega}_B = \frac{r}{R} \dot{\omega}_A$). Also contact frictional force of $A_y = B_y  = C_y$ and the minimum required clamping force $A_x = B_x = C_x$ to keep traction.
