**Foreword/disclaimer**: This is my derivation (as I mentioned I've never seen this procedure written ), so I would appreciate any constructive feedback. 

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I am assuming that the viscous "damping" coefficient for this system produces a torque which is proportional to the angular velocity. 

$$M(\omega(t) ) = c_t \omega(t) \tag{eq.1}$$ 

(this is an analogy to the viscous damping $F=c\cdot \dot{x}$.)

At any given point the rotational motion of the wheel is given by:

$$\sum M = I\cdot \alpha \tag{eq.2}$$

where:
- $\sum M = - M_t$ is the sum to torsional moments on the wheel (I am using the minus because the torque decelerates the wheel). 
- $I$ is mass moment of inertia of the wheel 
- $\alpha = \frac{d\omega}{dt}$ is the angular  acceleration of the wheel. 

From eq.1, eq.2 we obtain that:

$$-c_t \omega(t)  = I\cdot \frac{d\omega}{dt} $$
$$c_t dt  = -I\cdot \frac{d\omega}{\omega(t) } $$

by integrating both parts we obtain:

$$\int_{t_0}^{t_1} c_t dt  = -\int_{\omega_0}^{\omega_1}  I\cdot \frac{d\omega}{\omega(t) } $$


$$c_t (t_1 -t_0)  = -I\left[\ln \omega\right]_{\omega_0}^{\omega_1}   $$
$$c_t (t_1 -t_0)  = -I\ln \left(\frac{\omega_1}{\omega_0}\right)$$

Now, provided you know the mass moment of area A, you can make an estimation of the $c_t$.


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## numerical example

| revs | elapsed time | lap period | $ω$           | $-\ln\frac{ω_ι}{ω_0}$      | $c_t$         |
|:-----:|:-----:|:-----:|:-----:|:-----:|:-----:|
| 1   | 0.6  | 0.6    | 10.47| 0           |     NA        |
| 2   | 1.31 | 0.71   | 4.79 | 0.780 | 0.39 |
| 3   | 2.19 | 0.88   | 2.87 | 1.294 | 0.43 |